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On the Maximal Sum of Exponents of Runs in a String Maxime Crochemore, Marcin Kubica, Jakub Radoszewski, Wojciech Rytter, Tomasz Walen To cite this version: Maxime Crochemore, Marcin Kubica, Jakub Radoszewski, Wojciech Rytter, Tomasz Walen. On the Maximal Sum of Exponents of Runs in a String. Journal of Discrete Algorithms, 2012, 14 (-), pp.29-36.
hal-00742081 arXiv:1003.4866v1 [cs.DM] 25 Mar 2010 On the Maximal Sum of Exponents of Runs in a String Maxime Crochemore1,3, Marcin Kubica2, Jakub Radoszewski⋆2, Wojciech Rytter2,5, and Tomasz Wale´ n2 1 King’s College London, London WC2R 2LS, UK maxime.crochemore@kcl.ac.uk 2 Dept. of Mathematics, Computer Science and Mechanics, University of Warsaw, Warsaw, Poland [kubica,jrad,rytter,walen]@mimuw.edu.pl 3 Universit´ e Paris-Est, France 4 Dept. of Math. and Informatics, Copernicus University, Toru´ n, Poland Abstract. A run is an inclusion maximal occurrence in a string (as a subinterval) of a repetition v with a period p such that 2p ≤|v|. The exponent of a run is defined as |v|/p and is ≥2. We show new bounds on the maximal sum of exponents of runs in a string of length n. Our upper bound of 4.1 n is better than the best previously known proven bound of 5.6 n by Crochemore & Ilie (2008). The lower bound of 2.035 n, obtained using a family of binary words, contradicts the conjecture of Kolpakov & Kucherov (1999) that the maximal sum of exponents of runs in a string of length n is smaller than 2n.
1 Introduction Repetitions and periodicities in strings are one of the fundamental topics in combinatorics on words [1, 14]. They are also important in other areas: lossless compression, word representation, computational biology, etc. In this paper we consider bounds on the sum of exponents of repetitions that a string of a given length may contain. In general, repetitions are studied also from other points of view, like: the classification of words (both finite and infinite) not containing repetitions of a given exponent, efficient identification of factors being repetitions of different types and computing the bounds on the number of various types of repetitions occurring in a string. The known results in the topic and a deeper description of the motivation can be found in a survey by Crochemore et al. .
The concept of runs (also called maximal repetitions) has been introduced to represent all repetitions in a string in a succinct manner. The crucial prop-erty of runs is that their maximal number in a string of length n (denoted as ρ(n)) is O(n), see Kolpakov & Kucherov . This fact is the cornerstone of any ⋆Some parts of this paper were written during the author’s Erasmus exchange at King’s College London algorithm computing all repetitions in strings of length n in O(n) time. Due to the work of many people, much better bounds on ρ(n) have been obtained. The lower bound 0.927 n was first proved by Franek & Yang . Afterwards, it was improved by Kusano et al. to 0.944565 n employing computer experiments, and very recently by Simpson to 0.944575712 n. On the other hand, the first explicit upper bound 5 n was settled by Rytter , afterwards it was sys-tematically improved to 3.48 n by Puglisi et al. , 3.44 n by Rytter , 1.6 n by Crochemore & Ilie [2, 3] and 1.52 n by Giraud . The best known result ρ(n) ≤1.029 n is due to Crochemore et al. , but it is conjectured that ρ(n) < n. Some results are known also for repetitions of exponent higher than 2. For instance, the maximal number of cubic runs (maximal repetitions with exponent at least 3) in a string of length n (denoted ρcubic(n)) is known to be between 0.406 n and 0.5 n, see Crochemore et al. .
A stronger property of runs is that the maximal sum of their exponents in a string of length n (notation: σ(n)) is linear in terms of n, see Kolpakov & Kucherov . It has applications to the analysis of various algorithms, such as computing branching tandem repeats: the linearity of the sum of exponents solves a conjecture of concerning the linearity of the number of maximal tandem repeats and implies that all can be found in linear time. For other appli-cations, we refer to . The proof that σ(n) < cn in Kolpakov and Kucherov’s paper is very complex and does not provide any particular value for the con-stant c. A bound can be derived from the proof of Rytter but he mentioned only that the bound that he obtains is “unsatisfactory” (it seems to be 25 n).
The first explicit bound 5.6 n for σ(n) was provided by Crochemore and Ilie , who claim that it could be improved to 2.9 n employing computer experiments.
As for the lower bound on σ(n), no exact values were previously known and it was conjectured [11, 12] that σ(n) < 2n.
In this paper we provide an upper bound of 4.1 n on the maximal sum of exponents of runs in a string of length n and also a stronger upper bound of 2.5 n for the maximal sum of exponents of cubic runs in a string of length n. As for the lower bound, we bring down the conjecture σ(n) < 2n by providing an infinite family of binary strings for which the sum of exponents of runs is greater than 2.035 n.
2 Preliminaries We consider words (strings) u over a finite alphabet Σ, u ∈Σ∗; the empty word is denoted by ε; the positions in u are numbered from 1 to |u|. For u = u1u2 . . . um, let us denote by u[i . . j] a factor of u equal to ui . . . uj (in particular u[i] = u[i . . i]). Words u[1 . . i] are called prefixes of u, and words u[i . . |u|] suffixes of u.
We say that an integer p is the (shortest) period of a word u = u1 . . . um (notation: p = per(u)) if p is the smallest positive integer such that ui = ui+p holds for all 1 ≤i ≤m −p. We say that words u and v are cyclically equivalent (or that one of them is a cyclic rotation of the other) if u = xy and v = yx for some x, y ∈Σ∗.
A run (also called a maximal repetition) in a string u is an interval [i . . j] such that: – the period p of the associated factor u[i . . j] satisfies 2p ≤j −i + 1, – the interval cannot be extended to the right nor to the left, without violating the above property, that is, u[i −1] ̸= u[i + p −1] and u[j −p + 1] ̸= u[j + 1].
A cubic run is a run [i . . j] for which the shortest period p satisfies 3p ≤j −i+1.
For simplicity, in the rest of the text we sometimes refer to runs and cubic runs as to occurrences of the corresponding factors of u. The (fractional) exponent of a run is defined as (j −i + 1)/p.
For a given word u ∈Σ∗, we introduce the following notation: – ρ(u) and ρcubic(u) are the numbers of runs and cubic runs in u resp.
– σ(u) and σcubic(u) are the sums of exponents of runs and cubic runs in u resp.
For a non-negative integer n, we use the same notations ρ(n), ρcubic(n), σ(n) and σcubic(n) to denote the maximal value of the respective function for a word of length n.
3 Lower bound for σ(n) Tables 1 and 2 list the sums of exponents of runs for several words of two known families that contain very large number of runs: the words xi defined by Franek and Yang (giving the lower bound ρ(n) ≥0.927 n, conjectured for some time to be optimal) and the modified Padovan words yi defined by Simpson (giving the best known lower bound ρ(n) ≥0.944575712 n). These values have been computed experimentally. They suggest that for the families of words xi and yi the maximal sum of exponents could be less than 2n.
We show, however, a lower bound for σ(n) that is greater than 2n.
Theorem 1. There are infinitely many binary strings w such that σ(w) |w| > 2.035.
Proof. Let us define two morphisms φ : {a, b, c} 7→{a, b, c} and ψ : {a, b, c} 7→ {0, 1} as follows: φ(a) = baaba, φ(b) = ca, φ(c) = bca ψ(a) = 01011, ψ(b) = ψ(c) = 01001011 We define wi = ψ(φi(a)). Table 3 shows the sums of exponents of runs in words wi, computed experimentally.
Clearly, for any word w = (w8)k, k ≥1, we have σ(w) |w| > 2.035.
⊓ ⊔ i |xi| ρ(xi)/|xi| σ(xi) σ(xi)/|xi| 1 6 0.3333 4.00 0.6667 2 27 0.7037 39.18 1.4510 3 116 0.8534 209.70 1.8078 4 493 0.9047 954.27 1.9356 5 2090 0.9206 4130.66 1.9764 6 8855 0.9252 17608.48 1.9885 7 37512 0.9266 74723.85 1.9920 8 158905 0.9269 316690.85 1.9930 9 673134 0.9270 1341701.95 1.9932 Table 1. Number of runs and sum of exponents of runs in Franek & Yang’s words xi.
i |yi| ρ(yi)/|yi| σ(yi) σ(yi)/|yi| 4 37 0.7568 57.98 1.5671 8 125 0.8640 225.75 1.8060 12 380 0.9079 726.66 1.9123 16 1172 0.9309 2303.21 1.9652 20 3609 0.9396 7165.93 1.9856 24 11114 0.9427 22148.78 1.9929 28 34227 0.9439 68307.62 1.9957 32 105405 0.9443 210467.18 1.9967 36 324605 0.9445 648270.74 1.9971 40 999652 0.9445 1996544.30 1.9972 Table 2. Number of runs and sum of exponents of runs in Simpson’s modified Padovan words yi.
i |wi| σ(wi) σ(wi)/|wi| 1 31 47.10 1.5194 2 119 222.26 1.8677 3 461 911.68 1.9776 4 1751 3533.34 2.0179 5 6647 13498.20 2.0307 6 25205 51264.37 2.0339 7 95567 194470.30 2.0349 8 362327 737393.11 2.0352 9 1373693 2795792.39 2.0352 10 5208071 10599765.15 2.0353 Table 3. Sums of exponents of runs in words wi.
4 Upper bounds for σ(n) and σcubic(n) In this section we utilize the concept of handles of runs as defined in . The original definition refers only to cubic runs, but here we extend it also to ordinary runs.
Let u ∈Σ∗be a word of length n. Let us denote by P = {p1, p2, . . . , pn−1} the set of inter-positions in u that are located between pairs of consecutive letters of u. We define a function H assigning to each run v in u a set of some inter-positions within v (called later on handles) — H is a mapping from the set of runs occurring in u to the set 2P of subsets of P. Let v be a run with period p and let w be the prefix of v of length p. Let wmin and wmax be the minimal and maximal words (in lexicographical order) cyclically equivalent to w. H(v) is defined as follows: a) if wmin = wmax then H(v) contains all inter-positions within v, b) if wmin ̸= wmax then H(v) contains inter-positions between consecutive oc-currences of wmin in v and between consecutive occurrences of wmax in v.
Note that H(v) can be empty for a non-cubic-run v.
b a b a a b a b a a b a b a a 1 1 2 a a b b b b v 1 1 w w min1 max1 v1 2 Fig. 1. An example of a word with two highlighted runs v1 and v2. For v1 we have wmin1 ̸= wmax1 and for v2 the corresponding words are equal to b (a one-letter word).
The inter-positions belonging to the sets H(v1) and H(v2) are pointed by arrows Proofs of the following properties of handles of runs can be found in : 1. Case (a) in the definition of H(v) implies that |wmin| = 1.
2. H(v1) ∩H(v2) = ∅for any two distinct runs v1 and v2 in u.
To prove the upper bound for σ(n), we need to state an additional property of handles of runs. Let R(u) be the set of all runs in a word u, and let R1(u) and R≥2(u) be the sets of runs with period 1 and at least 2 respectively.
Lemma 1.
If v ∈R1(u) then σ(v) = |H(v)| + 1.
If v ∈R≥2(u) then ⌈σ(v)⌉≤|H(v)| 2 + 3.
Proof. For the case of v ∈R1(u), the proof is straightforward from the definition of handles. In the opposite case, it is sufficient to note that both words wk min and wk max for k = ⌊σ(v)⌋−1 are factors of v, and thus |H(v)| ≥2 · (⌊σ(v)⌋−2).
⊓ ⊔ Now we are ready to prove the upper bound for σ(n). In the proof we use the bound ρ(n) ≤1.029 n on the number of runs from .
Theorem 2. The sum of the exponents of runs in a string of length n is less than 4.1 n.
Proof. Let u be a word of length n. Using Lemma 1, we obtain: X v∈R(u) σ(v) = X v∈R1(u) σ(v) + X v∈R≥2(u) σ(v) ≤ X v∈R1(u) (|H(v)| + 1) + X v∈R≥2(u) |H(v)| 2 + 3 = X v∈R1(u) |H(v)| + |R1(u)| + X v∈R≥2(u) |H(v)| 2 + 3 · |R≥2(u)| ≤3 · |R(u)| + A + B/2, (1) where A = P v∈R1(u) |H(v)| and B = P v∈R≥2(u) |H(v)|. Due to the disjointness of handles of runs (the second property of handles), A + B < n, and thus, A + B/2 < n. Combining this with (1), we obtain: X v∈R(u) σ(v) < 3 · |R(u)| + n ≤3 · ρ(n) + n ≤3 · 1.029 n + n < 4.1 n.
⊓ ⊔ A similar approach for cubic runs, this time using the bound of 0.5 n for ρcubic(n) from , enables us to immediately provide a stronger upper bound for the function σcubic(n).
Theorem 3. The sum of the exponents of cubic runs in a string of length n is less than 2.5 n.
Proof. Let u be a word of length n. Using same inequalities as in the proof of Theorem 2, we obtain: X v∈Rcubic(u) σ(v) < 3·|Rcubic(u)|+n ≤3·ρcubic(n)+n ≤3·0.5 n+n = 2.5 n, where Rcubic(u) denotes the set of all cubic runs of u.
⊓ ⊔ References 1. J. Berstel and J. Karhumaki. Combinatorics on words: a tutorial. Bulletin of the EATCS, 79:178–228, 2003.
2. M. Crochemore and L. Ilie. Analysis of maximal repetitions in strings. In L. Kucera and A. Kucera, editors, MFCS, volume 4708 of Lecture Notes in Computer Science, pages 465–476. Springer, 2007.
3. M. Crochemore and L. Ilie. Maximal repetitions in strings. J. Comput. Syst. Sci., 74(5):796–807, 2008.
4. M. Crochemore, L. Ilie, and W. Rytter. Repetitions in strings: Algorithms and combinatorics. Theor. Comput. Sci., 410(50):5227–5235, 2009.
5. M. Crochemore, L. Ilie, and L. Tinta. Towards a solution to the ”runs” conjecture.
In P. Ferragina and G. M. Landau, editors, CPM, volume 5029 of Lecture Notes in Computer Science, pages 290–302. Springer, 2008.
6. M. Crochemore, C. Iliopoulos, M. Kubica, J. Radoszewski, W. Rytter, and T. Walen.
On the maximal number of cubic runs in a string.
In Proceedings of LATA, 2010 (to appear).
7. F. Franek and Q. Yang. An asymptotic lower bound for the maximal number of runs in a string. Int. J. Found. Comput. Sci., 19(1):195–203, 2008.
8. M. Giraud. Not so many runs in strings. In C. Mart´ ın-Vide, F. Otto, and H. Fernau, editors, LATA, volume 5196 of Lecture Notes in Computer Science, pages 232–239.
Springer, 2008.
9. D. Gusfield and J. Stoye. Simple and flexible detection of contiguous repeats using a suffix tree (preliminary version). In M. Farach-Colton, editor, CPM, volume 1448 of Lecture Notes in Computer Science, pages 140–152. Springer, 1998.
10. R. M. Kolpakov and G. Kucherov. Finding maximal repetitions in a word in linear time. In Proceedings of the 40th Symposium on Foundations of Computer Science, pages 596–604, 1999.
11. R. M. Kolpakov and G. Kucherov. On maximal repetitions in words. J. of Discr.
Alg., 1:159–186, 1999.
12. R. M. Kolpakov and G. Kucherov. On the sum of exponents of maximal repetitions in a word. Tech. Report 99-R-034, LORIA, 1999.
13. K. Kusano, W. Matsubara, A. Ishino, H. Bannai, and A. Shinohara. New lower bounds for the maximum number of runs in a string. CoRR, abs/0804.1214, 2008.
14. M. Lothaire. Combinatorics on Words. Addison-Wesley, Reading, MA., U.S.A., 1983.
15. S. J. Puglisi, J. Simpson, and W. F. Smyth. How many runs can a string contain?
Theor. Comput. Sci., 401(1-3):165–171, 2008.
16. W. Rytter. The number of runs in a string: Improved analysis of the linear upper bound. In B. Durand and W. Thomas, editors, STACS, volume 3884 of Lecture Notes in Computer Science, pages 184–195. Springer, 2006.
17. W. Rytter. The number of runs in a string. Inf. Comput., 205(9):1459–1469, 2007.
18. J. Simpson. Modified Padovan words and the maximum number of runs in a word.
Australasian J. of Comb., 46:129–145, 2010. |
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## Algorithms and Coding Interviews
·
Sharing methods to solve questions on leetcode, trying to systematize different types of questions
In-depth Backtracking with LeetCode Problems — Part 1
Introduction and Permutation
Li Yin
9 min readMar 21, 2018
Backtracking is a general algorithm for finding all (or some) solutions to some computational problems, that incrementally builds candidates to the solutions. As soon as it determines that a candidate cannot possibly lead to a valid complete solution, it abandons this partial candidate and “backtracks’’ (return to the upper level) and reset to the upper level’s state so that the search process can continue to explore the next branch. Backtracking is all about choices and consequences, this is why backtracking is the most common algorithm for solving constraint satisfaction problem (CSP, CSPs are mathematical questions defined as a set of objects whose state must satisfy a number of constraints or limitations, visit wiki for more information, such as Eight Queens puzzle, Map Coloring problem, Sudoku, Crosswords, and many other logic puzzles.
Properties and Applications
To generalize the characters of backtracking:
No Repetition and Completion: It is a systematic generating method that avoids repetitions and missing any possible right solution. This property makes it ideal for solving combinatorial problems such as combination and permutation which requires us to enumerate all possible solutions.
Search Pruning: Because the final solution is built incrementally, in the process of working with partial solutions, we can evaluate the partial solution and prune branches that would never lead to the acceptable complete solution: either it is invalid configuration, or it is worse than known possible complete solution.
In this blog, the organization is as follows:
Show Property 1: We will first show how backtrack construct the complete solution incrementally and how it backtrack to its previous state.
On implicit graph: We use permutation and combination as examples.
On Explicit Graph: Enumerating all pahts between the source and target vertex in a graph drawing
Show Property 2: We demonstrate the application of search pruning in backtracking through CSP problems such as sudoku.
Permutation
Before I throw you more theoretical talking, let us look at an example: Given a set of integers {1, 2, 3}, enumerate all possible permutations using all items from the set without repetition. A permutation describes an arrangement or ordering of items. It is trivial to figure out that we can have the following six permutations: [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], and [3, 2, 1].
This is a typical combinatorial problem, the process of generating all valid permutations is visualized in Fig. 1. To construct the final solution, we can start from an empty ordering shown at the first level, [ ]. Then we try to add one item where we have three choices :1, 2, and 3. We get three partial solutions , , at the second level. Next, for each of these partial solutions, we have two choices, for , we can either put 2 or 3 first. Similarly, for , we can do either 1 and 3, and so on. Given n distinct items, the number of possible permutations are n(n-1)…1 = n!.
Implicit Graph
In the graph, each node is either a partial or final solution. If we look it as a tree, the internal node is a partial solution and all leaves are final solutions. One edge represents generating the next solution based on the current solution. The vertices and edges are not given by an explicitly defined graph or trees, the vertices are generated on the fly and the edges are implicit relation between these nodes.
Backtracking and DFS
The implementation of the state transfer we can use either BFS or DFS on the implicit vertices. DFS is preferred because theoretically it took O(log n!) space used by stack, while if use BFS, the number of vertices saved in the queue can be close to n!. With recursive DFS, we can start from node [], and traverse to [1,2], then [1,2,3]. Then we backtrack to [1,2], backtrack to , and go to [1, 3], to [1, 3, 2]. To clear the relation between backtracking and DFS, we can say backtracking is a complete search technique and DFS is an ideal way to implement it.
We can generalize Permutation, Permutations refer to the permutation of n things taken kat a time without repetition, the math formula is A_{n}^{k} = n (n-1)(n-2)…k. In Fig.1, we can see from each level kshows all the solution of A_{n}^{k}. The generation of A_{n}^{k} is shown in the following Python Code:
Give the input of a=[1,2,3], we call the above function with the following code:
a = [1, 2, 3]n = len(a)ans = used = [False] len(a)ans = []A_n_k(a, n, n, 0, used, [], ans)print(ans)
The output is:
In the process, we add print before and after the recursive function call:
[1, 2][1, 2, 3]backtrack: [1, 2]backtrack: [1, 3][1, 3, 2]backtrack: [1, 3]backtrack: backtrack: [][2, 1][2, 1, 3]backtrack: [2, 1]backtrack: [2, 3][2, 3, 1]backtrack: [2, 3]backtrack: backtrack: [][3, 1][3, 1, 2]backtrack: [3, 1]backtrack: [3, 2][3, 2, 1]backtrack: [3, 2]backtrack: backtrack: []
Two Passes
Therefore, we can say backtrack visit these implicit vertices in two passes: First forward pass to build the solution incrementally, second backward pass to backtrack to previous state. We can see within these two passes, the curr list is used as all vertices, and it start with [] and end with []. This is the character of backtracking.
Time Complexity of Permutation
In the example of permutation, we can see that backtracking only visit each state once. The complexity of this is similar to the graph traversal of O(|V|+|E|), where |V| = \sum_{i=0}^{n}{A_{n}^{k}}, because it is a tree structure, |E| = |v|-1. This actually makes the permutation problem NP-hard.
Backtracking with LeetCode Problems — Part 2: Combination and all paths with backtracking
Backtracking with LeetCode Problems — Part 3: Constraint Satisfaction Problems with Search Pruning
LeetCode Examples
17. Letter Combinations of a Phone Number
Given a digit string, return all possible letter combinations that the number could represent.
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Join Medium for free to get updates from this writer.
A mapping of digit to letters (just like on the telephone buttons) is given below.
Input:Digit string "23"Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
Solution: this is not exactly backtracking problem, however, we recursively add the next digit to the previous combinations. Time complexity will be O(3^n), which came from O(3+3²+3³+…+3^n). The difference is we know it is possible solution, if we keep searching the graph, it works (no constraint)
def letterCombinations(self, digits): """ :type digits: str :rtype: List[str] """ mapping = {'2': 'abc', '3': 'def', '4': 'ghi', '5': 'jkl', '6': 'mno', '7': 'pqrs', '8': 'tuv', '9': 'wxyz'} def combine(rst, remain_digits): #end condition if len(remain_digits)==0: return rst if len(rst)==0: rst=[''] nxt_rst=[] digit = remain_digits.pop(0) for r in rst: for c in mapping[digit]: nxt_rst.append(r+c) return combine(nxt_rst,remain_digits) #nxt_rst = r+c return combine([],list(digits)) #first is current result
With backtrack:
78. Subsets
Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,3], a solution is:
[ , , , [1,2,3], [1,3], [2,3], [1,2], []]
Solution: because we dont care about the order, it is a combination (not a permutation). here we just use index+1 to pointer to the beignning of the possible paths. temp refers the curr: to record what we use, but when we return after the recursive call, we need to pop out. and keep adding the next element.
def subsets(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ #here we need a global wise list, each time we just append to the result rslt=[] def dfs(temp, idx): rslt.append(temp[:]) #pass temp[:] with shollow copy so that we wont change the result of rslt when temp is changed for i in range(idx, len(nums)): temp.append(nums[i]) #backtrack dfs(temp, i+1) temp.pop() dfs([],0) return rslt
46. Permutations
Given a collection of distinct numbers, return all possible permutations.
For example,
[1,2,3] have the following permutations:
[ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]]
Solution: The permutation is similar as the last power set, the difference is we use each element at least and only one time, and we dont care about the order. So for the remaining elements, it is different.
def permute(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ #here we need a global wise list, each time we just append to the result rslt=[] def dfs(temp, elements): #gather rslt if len(elements)==0: rslt.append(temp[:]) #still remember to use temp[:] for e in elements: temp.append(e) #backtrack next_elements=elements[:] next_elements.remove(e) elements.pop() dfs(temp, next_elements) temp.pop() dfs([],nums) #first is the current result return rslt
More summary
47. Permutations II
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,
[1,1,2] have the following unique permutations:
[ [1,1,2], [1,2,1], [2,1,1]]
Solution: The difference with the other permutation is, each time, we only append the unique element to temp.
def permuteUnique(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ #here we need a global wise list, each time we just append to the result rslt=[] def dfs(temp, elements): #gather rslt if len(elements)==0: rslt.append(temp[:]) #still remember to use temp[:] for e in list(set(elements)): #this is the only difference temp.append(e) #backtrack next_elements=elements[:] next_elements.remove(e) dfs(temp, next_elements) temp.pop() dfs([],nums) return rslt
301. Remove Invalid Parentheses
Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.
Note: The input string may contain letters other than the parentheses ( and ).
Examples:
"()())()" -> ["()()()", "(())()"]"(a)())()" -> ["(a)()()", "(a())()"]")(" -> [""]
Solution: at the beignning, check the number of left parathese and the right parentheses need to be removed. Then use DFS (try all possible ways) with back tracking to get all possible solutions (when l, r decrease to zero, check if it is valid). To be noted: we need to avoid duplicates
def removeInvalidParentheses(self, s): """ :type s: str :rtype: List[str] """ def isValid(s): left_paren_count = 0 for c in s: if c=='(': left_paren_count+=1 elif c==')': if left_paren_count==0: return False left_paren_count-=1 else: continue return left_paren_count==0 def getCount(s): rslt = True l, r = 0, 0 #extrac l or r parenthesis for c in s: l+=c=='(' if c==')': if l==0: r+=1 else: l-=1 return (l,r) rslt =[] def dfs(s, idx, l, r): if l==0 and r==0: if isValid(s): rslt.append(s) return #delete extra l or r, every time we only delete one for i in range(idx,len(s)): c=s[i] if i-1>=idx and c==s[i-1]: #to avoid duplication continue if c==')': new_s = s[:i]+s[i+1:] dfs(new_s, i,l,r-1) if c=='(': new_s = s[:i]+s[i+1:] dfs(new_s, i,l-1,r) l, r = getCount(s) dfs(s,0, l,r) return rslt
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5302 | https://fiveable.me/ap-calc/unit-6/integrating-using-integration-by-parts/study-guide/O4P3LchNoZnWElf8zETV | Integrating Using Integration by Parts - AP Calc Study Guide | Fiveable | Fiveable
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♾️AP Calculus AB/BC Unit 6 Review
6.11 Integrating Using Integration by Parts
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AP Calculus AB/BC
Unit 6 – Integration and Accumulation of Change
Topic: 6.11
♾️AP Calculus AB/BC Unit 6 Review
6.11 Integrating Using Integration by Parts
Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025
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APA
♾️AP Calculus AB/BC
Unit & Topic Study Guides
AP Calculus AB/BC Exams
Unit 1 – Limits and Continuity
Unit 2 – Fundamentals of Differentiation
Unit 3 – Composite, Implicit, and Inverse Functions
Unit 4 – Contextual Applications of Differentiation
Unit 5 – Analytical Applications of Differentiation
Unit 6 – Integration and Accumulation of Change
Unit 6 Overview: Integration and Accumulation of Change
6.1 Integration and Accumulation of Change
6.2 Approximating Areas with Riemann Sums
6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
6.4 The Fundamental Theorem of Calculus and Accumulation Functions
6.5 Interpreting the Behavior of Accumulation Functions Involving Area
6.6 Applying Properties of Definite Integrals
6.7 The Fundamental Theorem of Calculus and Definite Integrals
6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation
6.9 Integrating Using Substitution
6.10 Integrating Functions Using Long Division and Completing the Square
6.11 Integrating Using Integration by Parts
6.12 Integrating Using Linear Partial Fractions
6.13 Evaluating Improper Integrals
6.14 Selecting Techniques for Antidifferentiation (AB)
Unit 7 – Differential Equations
Unit 8 – Applications of Integration
Unit 9 – Parametric Equations, Polar Coordinates, and Vector–Valued Functions (BC Only)
Unit 10 – Infinite Sequences and Series (BC Only)
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6.11 Integrating Using Integration by Parts
Welcome back to AP Calculus with Fiveable! In this study guide, we'll delve into the technique of Integration by Parts. Integration by Parts is a powerful method used to integrate the product of two functions, and it often comes in handy when dealing with more complex integrals. We have a few techniques such as u-substitution and Riemann sums in our calculus toolbox, so let's keep building those integration skills! 🧱
🚧 This is an AP Calculus BC topic only! If you are taking Calculus AB, you can skip this material. If you’re taking AP Calculus BC, here you go! ⬇️
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🔄 Integration by Parts Basics
Take a look at the following integral:
∫x 2 sin(x)d x\int x^2 \sin(x)\,dx∫x 2 sin(x)d x
We can’t use any of our current integration tools to evaluate it: substitution fails, and we only know how to integrate x 2 x^2 x 2 and sin(x)\sin(x)sin(x) separately. Why don’t we try to use that to our advantage?
For the integral above, we will be using a method called Integration by Parts, which is based on the product rule for differentiation. It’s essentially the reverse process!
Here is the product rule, as u u u and v v v representing two different functions.
d d x u v=u⋅v′+v⋅u′\frac{d}{dx} uv = \textcolor{blue}{u}\cdot \textcolor{pink}{v'} +\textcolor{red}{ v}\cdot \textcolor{teal}{u'}d x duv=u⋅v′+v⋅u′
If we try to reverse the process, we take the integral of all of those terms. It will look like the following:
u v=∫u⋅v′+∫v⋅u′ uv = \int\textcolor{blue}{u}\cdot \textcolor{pink}{v'} +\int\textcolor{red}{ v}\cdot \textcolor{teal}{u'}uv=∫u⋅v′+∫v⋅u′
When we rearrange the terms, we get the following rule for Integration by Parts:
∫u d v=u v−∫v d u\int \textcolor{blue}{u} \, \textcolor{pink}{dv} = uv - \int \textcolor{red}{v} \, \textcolor{teal}{du} ∫u d v=uv−∫v d u
Where:
u\textcolor{blue}{u}u and d v\textcolor{pink}{dv}d v are selected parts of the integrand.
d u\textcolor{teal}{du}d u is the derivative of u\textcolor{blue}{u}u with respect to the variable of integration.
v\textcolor{red}{v}v is the antiderivative (integral) of d v\textcolor{pink}{dv}d v.
📚 Application and Strategy
To successfully apply Integration by Parts, follow these steps:
Select $u$ and $dv$ carefully. Choose the variables in a way that simplifies the integral or makes it easier to integrate! We typically use the LIATE mnemonic to quickly choose u u u because they tend to efficiently simplify the integral.
Differentiate and Integrate. Compute d u du d u and v v v by finding the derivative of u u u and the antiderivative of d v dv d v.
Apply the Formula. Plug the values of u u u, d v dv d v, v v v, and d u du d u into the Integration by Parts formula.
Evaluate the New Integral. The resulting integral on the right side, with v v v and d u du d u may still need simplification. Repeat the Integration by Parts process if necessary!
Solve for the Original Integral. If you end up with a similar integral on both sides of the equation, solve for the original integral with algebraic manipulation.
⚡ LIATE Mnemonic:
Logarithmic functions
Inverse trigonometric functions
Algebraic functions
Trigonometric functions
Exponential functions
🧮 Integration by Parts Practice Problems
Let's reinforce our understanding with some practice problems:
Integration by Parts Example 1
Evaluate the following integral using integration by parts.
∫x e x d x\int xe^x \, dx∫x e x d x
First, let's go through the LIATE acronym to make an educated guess on the best possible expression to use for u u u. Since Algebra comes before Exponential, we should start by choosing u=x u = x u=x, and then set d v=e x dv = e^x d v=e x.
Now let's differentiate and integrate to collect all the necessary information.
u=x u = x u=x
d u=d x du = dx d u=d x
v=e x v = e^x v=e x
d v=e x d x dv = e^x \,dx d v=e x d x
Now we can apply Integration by Parts:
∫x⋅e x d x=x⋅e x−∫e x d x\int x \cdot e^x \, dx = x \cdot e^x - \int e^x \, dx ∫x⋅e x d x=x⋅e x−∫e x d x
Now let's evaluate the new integral and simplify.
∫x⋅e x d x=x⋅e x−e x+C\int x \cdot e^x \, dx = x \cdot e^x - e^x + C ∫x⋅e x d x=x⋅e x−e x+C
Therefore,
∫x e x d x=x e x−e x+C\int xe^x \, dx = x e^x - e^x + C ∫x e x d x=x e x−e x+C
That was a great start. Let’s try another one!
Integration by Parts Example 2
Evaluate the following integral:
∫ln(x)\int \ln(x)∫ln(x)
When we look at this integral, it doesn't match the format that we’re used to seeing when working with integration by parts. Let’s do a little algebraic manipulation to make it fit!
If we multiply the integrand by 1, we get ∫1⋅ln(x)\int 1\cdot \ln(x)∫1⋅ln(x). Now, we have two distinct functions and can use LIATE to determine which to designate as u u u.
Since Logarithmic is more of a priority than Algebra, we can set u=ln(x)u = \ln(x)u=ln(x) and d v=1 dv = 1 d v=1. Then, let’s work on getting the information for v v v and d u du d u.
u=ln(x)u = \ln(x)u=ln(x)
d u=1 x d x du = \frac{1}{x} \,dx d u=x 1d x
v=x v = x v=x
d v=1 d x dv = 1 \,dx d v=1 d x
Now that we have all of the necessary information, we can work through integration by parts!
∫ln(x)=x ln(x)−∫x⋅1 x d x\int \ln(x) = x\ln(x)- \int x\cdot \frac{1}{x} dx ∫ln(x)=x ln(x)−∫x⋅x 1d x
We can recognize that the integral on the right simplifies to give us: x ln(x)−∫d x x\ln(x)- \int dx x ln(x)−∫d x.
Let's evaluate the integral.
∫d x=x\int dx = x∫d x=x
Therefore,
∫ln(x)=x ln(x)−x+C\int \ln(x) = x\ln(x)- x + C ∫ln(x)=x ln(x)−x+C
Amazing! Keep up the good work 🤩
Integration by Parts Example 3
Evaluate the Integral.
∫x 2 cos(x)d x\int x^2 \cos(x) \, dx ∫x 2 cos(x)d x
We’ve worked through a few of these problems by now, so we got the rhythm down. Let’s start by picking u u u and d v dv d v using LIATE, and then solving for d u du d u and v v v.
u=x 2 u = x^2 u=x 2
d u=2 x d x du = 2x dx d u=2 x d x
v=sinx v = \sin x v=sin x
d v=cosx dv = \cos x d v=cos x
Now we can use the Integration by Parts equation!
∫x 2 cos(x)d x=x 2⋅sin(x)−∫2 x⋅sin(x)d x\int x^2 \cos(x) \, dx = x^2 \cdot \sin(x) - \int 2x \cdot \sin(x) \, dx ∫x 2 cos(x)d x=x 2⋅sin(x)−∫2 x⋅sin(x)d x
We have to apply integration by parts again to the remaining integral: ∫2 x⋅sin(x)d x\int 2x \cdot \sin(x) \, dx∫2 x⋅sin(x)d x
Choose:
u=2 x u = 2x u=2 x
d u=2 d x du=2dx d u=2 d x
v=−cosx v =-\cos x v=−cos x
d v=sinx dv = \sin x d v=sin x
And then we can do integration by parts for the second integral:
∫2 x⋅sin(x)d x=−2 x⋅cos(x)−∫(−2)⋅cos(x)d x\int 2x \cdot \sin(x) \, dx = -2x \cdot \cos(x) - \int (-2) \cdot \cos(x) \, dx ∫2 x⋅sin(x)d x=−2 x⋅cos(x)−∫(−2)⋅cos(x)d x
Now, substitute this back into the original equation. We’re almost there!
∫x 2⋅cos(x)d x=x 2⋅sin(x)+2 x⋅cos(x)+2⋅∫cos(x)d x\int x^2 \cdot \cos(x) \, dx = x^2 \cdot \sin(x) + 2x \cdot \cos(x) + 2 \cdot \int \cos(x) \, dx∫x 2⋅cos(x)d x=x 2⋅sin(x)+2 x⋅cos(x)+2⋅∫cos(x)d x
Finally, integrate the remaining term:
∫cos(x)d x=sin(x)\int \cos(x) \, dx = \sin(x)∫cos(x)d x=sin(x)
Once we substitute this back into the equation, we get
∫x 2⋅cos(x)d x=x 2⋅sin(x)+2 x⋅cos(x)+2⋅sin(x)+C\int x^2 \cdot \cos(x) \, dx = x^2 \cdot \sin(x) + 2x \cdot \cos(x) + 2 \cdot \sin(x) + C∫x 2⋅cos(x)d x=x 2⋅sin(x)+2 x⋅cos(x)+2⋅sin(x)+C
Therefore,
∫x 2 cos(x)d x=x 2 sin(x)+2 x cos(x)+2 sin(x)+C\int x^2 \cos(x) \, dx = x^2 \sin(x) + 2x \cos(x) + 2 \sin(x) + C∫x 2 cos(x)d x=x 2 sin(x)+2 x cos(x)+2 sin(x)+C
You did amazing! 🙌
Integration by Parts Challenge Question
Evaluate the following integral:
∫e x cosx\int e^x \cos x∫e x cos x
Try working through this question and see if you can get the answer below.
Hint: Try using integration of parts twice!
∫e x cosx=e x sinx+e x cosx 2+C\int e^x \cos x = \frac{e^x\sin x + e^x\cos x}{2} + C∫e x cos x=2 e x sin x+e x cos x+C
If you do integration by parts twice, keeping e x e^x e x as the same variable, you will get an integral that matches the original. Use some algebraic manipulation to get both integrals on the same side, and then divide by two! If you can solve this one, you're an integration by parts expert!
🌟 Closing
Fantastic job! 🥳 Integration by Parts is a valuable tool in your calculus toolkit, allowing you to tackle a wide range of integrals. Remember to carefully choose u u u and d v dv d v to simplify the integration process. Keep practicing, and you'll become a master at integrating using Integration by Parts!
Frequently Asked Questions
How do I know when to use integration by parts instead of u-substitution?
Use u-sub when the integrand is essentially a composition f(g(x)) and you can set u = inner function so du matches the rest of the integrand (or a constant multiple). If that substitution turns the integral into a simple antiderivative, do it. Use integration by parts (IBP) when the integrand is a product of two types of functions where one part you want to differentiate (u) and the other you can integrate (dv). Common cases on the BC CED: logarithms or inverse trig functions (like ln x, arctan x), polynomials times exponentials or trig (x e^x, x sin x), or when you need repeated IBP or reduction formulas. A quick heuristic is LIATE: choose u in this order—Log, Inverse trig, Algebraic (polynomial), Trig, Exponential—but always check which choice makes v = ∫dv simple. Practical check: if u-sub doesn’t cleanly remove x-dependence (du doesn’t appear), try IBP. For definite integrals remember to evaluate the boundary term u·v (CED FUN-6.E)—it can be crucial. For AP BC practice and examples, see the Topic 6.11 study guide ( and the Unit 6 overview ( For more problems, try the AP practice bank (
What's the formula for integration by parts and how do I remember it?
Integration by parts comes from the product rule. Formula (indefinite): ∫ u dv = u·v − ∫ v du. For a definite integral on [a,b]: ∫_a^b u dv = [u·v]_a^b − ∫_a^b v du. How to remember it: start from d(uv) = u dv + v du. Integrate both sides → uv = ∫ u dv + ∫ v du, then solve for ∫ u dv to get ∫ u dv = uv − ∫ v du. A quick mnemonic: “∫ u dv → u·v minus ∫ v du” (swap dv→v and du stays with u, then subtract). Use the LIATE rule to pick u (Log, Inverse trig, Algebraic, Trig, Exponential) so du simplifies. For definite integrals don’t forget the boundary term [u·v]_a^b. Need more examples (like ∫ ln x dx or e^{x}·polynomial)? Check the AP BC study guide on integration by parts ( and practice problems (
I'm so confused about integration by parts - which part do I pick for u and which for dv?
Short answer: pick u so that differentiating it makes things simpler, and pick dv so you can integrate it easily. A quick rule is LIATE (Logarithmic, Inverse trig, Algebraic, Trig, Exponential)—choose u from the earliest category present. Always check: differentiate u (u′ should simplify) and integrate dv (v should be easy). Examples: ∫ x e^x dx → u = x, dv = e^x dx (because du = dx simplifies and v = e^x). ∫ ln x dx → u = ln x, dv = dx (ln x has no elementary antiderivative but differentiates to 1/x). For repeated polynomial×exponential you can use tabular integration. For definite integrals remember the uv|_a^b boundary term. This is BC-only material on the CED (FUN-6.E); practice selecting u/dv, using LIATE, and doing boundary terms. For a focused review, see the Topic 6.11 study guide ( and try problems at (
How do I solve integrals like ∫x·e^x dx using integration by parts?
Use integration by parts: pick u so its derivative simplifies, and dv so you can integrate it. LIATE suggests choosing u = x (algebra) and dv = e^x dx (exponential). Let u = x → du = dx. Let dv = e^x dx → v = e^x. Integration by parts formula: ∫ u dv = u v − ∫ v du. So ∫ x e^x dx = x·e^x − ∫ e^x dx = x e^x − e^x + C = e^x(x − 1) + C. That’s the standard exponential–polynomial case; sometimes you can use tabular integration for repeated products. This technique is exactly what Topic 6.11 (FUN-6.E) expects you to master on the BC exam. For a quick refresher and more examples, see the Fiveable study guide for this topic ( and try extra practice problems (
What does LIATE mean and how does it help me choose u in integration by parts?
LIATE is a quick heuristic for choosing u in integration by parts: Logarithmic, Inverse trig, Algebraic (polynomials like x), Trigonometric, Exponential. Pick u as the earliest (highest-priority) type in that list and let dv be the rest. The idea: choose u so its derivative is simpler and dv is something you can integrate easily. Why it helps: integration by parts uses ∫u dv = uv − ∫v du. If du is simpler, the new integral ∫v du is usually easier. Examples: ∫ln x dx → u = ln x, dv = dx (because ln is L); ∫x e^x dx → u = x (A), dv = e^x dx. For repeated polynomial × exponential/trig, use tabular integration (fast repeated parts). Remember this is an AP BC topic (FUN-6.E: antiderivatives and definite integrals by parts). For more examples and practice, see the Topic 6.11 study guide ( and grab extra problems at (
Can someone explain integration by parts step by step with an example?
Integration by parts comes from the product rule: d(uv) = u dv + v du, so ∫ u dv = uv − ∫ v du. Steps (use LIATE to pick u: Log, Inverse trig, Algebraic, Trig, Exponential): Example: ∫ x e^x dx (indefinite, BC-only topic) 1. Choose u and dv: let u = x (algebraic), dv = e^x dx (exponential). 2. Compute du and v: du = dx, v = ∫ e^x dx = e^x. 3. Apply formula: ∫ x e^x dx = u v − ∫ v du = x e^x − ∫ e^x dx. 4. Evaluate remaining integral: ∫ e^x dx = e^x, so result = x e^x − e^x + C = e^x(x − 1) + C. For a definite integral, use the same formula with limits: ∫_a^b u dv = [uv]_a^b − ∫_a^b v du (include boundary terms). For repeated polynomial·exponential cases, use tabular integration. Want more examples and AP-style practice? See the Topic 6.11 study guide ( and try practice problems (
When I do integration by parts, do I need to add the constant C at the end?
Yes—if you’re finding an antiderivative (an indefinite integral) you must include “+ C” at the end. Integration by parts just finds an antiderivative using the formula ∫ u dv = u·v − ∫ v du, so the result of that computation is still an antiderivative and needs the arbitrary constant C. If you do repeated parts, don’t add a new C at every step—one final “+ C” on the final antiderivative is enough. For definite integrals you don’t add C: use the definite-parts formula ∫_a^b u dv = [u·v]_a^b − ∫_a^b v du, so the boundary terms take care of constants (they cancel). This is exactly the kind of skill tested in Topic 6.11 (FUN-6.E)—see the integrating-by-parts study guide for examples ( For more practice problems, try Fiveable’s practice collection (
How do I do integration by parts with definite integrals - do the limits change?
Yes—for definite integrals you use the same integration-by-parts formula but apply the limits to the boundary term. In symbols (BC CED FUN-6.E): ∫_a^b u dv = [u·v]_a^b − ∫_a^b v du Steps: 1. Choose u and dv (LIATE helps). 2. Compute du and v (an antiderivative of dv). 3. Evaluate the boundary term u·v at b and a: u(b)v(b) − u(a)v(a). 4. Subtract the definite integral ∫_a^b v du (you may need to integrate by parts again). Quick example: ∫_1^e ln x dx. Let u = ln x, dv = dx ⇒ du = (1/x) dx, v = x. So ∫_1^e ln x dx = [x ln x]_1^e − ∫_1^e x·(1/x) dx = (e·1 − 1·0) − ∫_1^e 1 dx = e − 1 − (e − 1) = e − 1 − (e − 1) = 0 (actually here check: [x ln x]_1^e = e·1 − 1·0 = e; ∫_1^e 1 dx = e−1 ⇒ answer = e − (e−1)=1). Reminder: BC exam assesses definite integrals with boundary terms (FUN-6.E). For more examples and guided practice see the Topic 6.11 study guide ( and Unit 6 overview ( For lots of practice problems, use (
I don't understand why sometimes you have to use integration by parts twice in the same problem?
You sometimes apply integration by parts twice because the first application turns the integral into a new one that’s still not elementary—often one of the original kinds of pieces (like x, ln x, arctan x, or a polynomial times e^{x}) reappears. Doing parts again can either cancel terms or produce an equation you can solve for the original integral (common with ∫e^{x}cos x or ∫e^{x}sin x). Practically: pick u so differentiation simplifies (LIATE helps), let dv be the rest, do parts, and if the result brings back the original integral, move it to the left and solve algebraically. For repeated structure, tabular integration or reduction formulas speed this up. This is exactly the “repeated integration by parts” idea in the BC CED (FUN-6.E). For examples and step-by-step practice, check the Topic 6.11 study guide ( and the full Unit 6 page (
What's the difference between integration by parts and partial fractions?
Integration by parts and partial fractions are different tools for different integrands. - Integration by parts (IBP) comes from the product rule and is for integrals of products where one piece you can differentiate (u) and the other you can integrate (dv). Use LIATE/heuristics to pick u (log, inverse trig, algebraic, trig, exp). It’s the go-to for ∫x e^x, ∫ln x, ∫x arctan x, repeated/ tabular IBP, and definite integrals with boundary terms (CED FUN-6.E, keywords: product rule, u and dv selection, tabular integration). - Partial fractions decomposes a rational function (polynomial/ polynomial) into simpler rational pieces you can integrate (linear/quadratic factors). Use it when integrand is a rational function (degree numerator < degree denominator after division). It reduces ∫(polynomial)/(polynomial) into sums of logs and arctan terms. Quick check: ∫x e^x dx → IBP. ∫(x+1)/(x^2−1) dx → partial fractions. For BC Topic 6.11, practice IBP problems in the study guide ( and more practice at (
How do I integrate ∫ln(x) dx using integration by parts?
Use integration by parts. Choose u = ln(x) (because logs are good for u) and dv = dx. Compute: - u = ln(x) → du = (1/x) dx - dv = dx → v = x Then ∫ ln(x) dx = u·v − ∫ v du = x·ln(x) − ∫ x·(1/x) dx = x ln(x) − ∫ 1 dx = x ln(x) − x + C. So the antiderivative is x ln(x) − x + C. This is a standard example under FUN-6.E (integration by parts; antiderivative of ln x) in the AP BC curriculum. For a quick review of integration-by-parts choices and more examples, see the Topic 6.11 study guide ( For extra practice problems, try Fiveable’s practice page (
Why does my integration by parts answer look different from the answer key but still be correct?
Short answer: different-looking answers can both be correct because antiderivatives are only unique up to an added constant and because algebraic or trigonometric rearrangements can make two equivalent expressions look different. Why that happens (AP-BC relevant): - Indefinite integrals: if F(x) is an antiderivative, so is F(x)+C. Missing or different constants explain many “differences.” Always check by differentiating your result—if you get the integrand, it’s correct (FUN-6.E.1). - Algebraic/trig identities: integration by parts can produce expressions that simplify to the same function (e.g., combine ln(x) terms, use 1 = sin^2+cos^2, factor exponentials). Different choices of u and dv (LIATE) give different intermediate forms but same final antiderivative. - Definite integrals: boundary terms may cancel differently in two methods but yield the same numeric value—evaluate F(b)−F(a) to confirm. Quick checks: differentiate your answer; for definite integrals, compute both results numerically. For more on techniques and common pitfalls (including ln x and inverse-trig integrals), see the Topic 6.11 study guide ( and practice more at (
Do I need to memorize specific integration by parts formulas for the AP exam?
Short answer: no—don’t memorize a bunch of exotic “special” formulas, but you should know the integration-by-parts formula and common choices so you can apply it quickly. What to memorize/practice: - The formula ∫u dv = uv − ∫v du (know how it comes from the product rule). - Typical u choices (use the LIATE heuristic: Logarithmic, Inverse trig, Algebraic, Trig, Exponential) so you can pick u and dv fast. - Standard results that often come from parts (e.g., ∫ln x dx = x ln x − x + C, integrals of arctan, exponential×polynomial reductions)—not because they’re magic, but because they appear on BC problems. On the AP BC exam: you’ll be asked to find indefinite and definite integrals using parts, sometimes repeatedly or with reduction formulas (CED FUN-6.E). Practice the technique (including tabular integration) until picking u/dv is automatic. For a focused review and examples, see the Topic 6.11 study guide ( and lots of practice at (
How do I solve integrals with trig functions using integration by parts?
Use integration by parts when your integrand is a product and one piece is easier to differentiate (u) and the other is easy to integrate (dv). Pick u with the LIATE idea (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) so you usually choose algebraic or log first and leave trig as dv. Key tips from the CED: set u and dv, compute du and v, then ∫u dv = uv − ∫v du; use tabular integration for repeated differentiation (polynomial × trig or exponential × polynomial); keep boundary terms for definite integrals. Quick example: ∫ x cos x dx - u = x ⇒ du = dx; dv = cos x dx ⇒ v = sin x - ∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C For products of trig functions only (like sin^m x cos^n x) use trig identities or reduction formulas before parts. For inverse-trig or ln(x) factors, IBP is standard (u = ln x). For more practice and AP-aligned examples, see the Topic 6.11 study guide ( and Unit 6 overview ( For lots of practice problems, try Fiveable’s practice set (
I keep messing up the signs when doing integration by parts - any tips?
Short version: the minus sign comes from the product rule, so treat ∫u dv = uv − ∫v du like a “signed” bookkeeping rule—write every step and carry the minus explicitly. Practical tips: - Always write u, dv, then compute du and v on the next line (don’t mental-juggle signs). Example: let u = ln x, dv = dx → du = (1/x) dx, v = x. Then ∫ln x dx = x ln x − ∫x(1/x) dx. - Box the identity ∫u dv = uv − ∫v du and literally copy uv − ∫v du into your work before substituting; that prevents dropping the minus. - For definite integrals use uv |_a^b − ∫_a^b v du. Evaluate the boundary term separately first so you don’t lose a sign. - Use the tabular method for repeated IBP—signs alternate systematically there, so fewer sign errors. - Check by differentiating your antiderivative (product rule backward) or by a quick numeric check. For extra practice aligned with the AP CED (FUN-6.E), see the Topic 6.11 study guide ( and grab more practice problems (
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5303 | https://condor.depaul.edu/jmaresh/reference/Significant-Figure-Rules-for-Logs.pdf | 1 Significant Figure Rules for Logarithms • Things to remember: significant figures include all certain digits and the first uncertain digit. There is always some uncertainty in the last digit. • Regular sig fig rules are guidelines, and they don’t always predict the correct number of significant figures. We use the sig fig rules as a shortcut so that we don’t have to do a detailed error analysis on everything we calculate. • Logs are exponents. Playing with some numbers: log 0.00722 = -2.14146 log 0.00723 = -2.14086 log 0.00724 = -2.14026 The numbers we took the log of had three significant figures. The results of the calculations differed in the third decimal place. Here are some more. In this case, the numbers we are taking the log of have two significant figures. log 0.0056 = -2.2518 log 0.0057 = -2.2441 log 0.0058 = -2.2366 The results above start to differ in the second decimal place. Here is another set: log 0.00056 = -3.2518 log 0.00057 = -3.2441 log 0.00058 = -3.2366 Again, the numbers we took the log of have two significant figures, and the results differed in the second decimal place. Conclusion: When you take the log of a number with N significant figures, the result should have N decimal places. The number in front of the decimal place indicates only the order of magnitude. It is not a significant figure. The rule for natural logs (ln) is similar, but not quite as clear-cut. For simplicity, we will use the above rules for natural logs too. 2 Using natural logs: ln 0.0056 = -5.1850 ln 0.0057 = -5.1673 ln 0.0058 = -5.1499 Note that the numbers each had two significant figures, and the results started to differ in the second decimal place. Going the other way: The opposite of taking the log of a number is to raise 10 to the power of that number. This corresponds to the 10x button on your calculator. The sig fig rule for this function is the opposite of the sig fig rule for logs. Let’s try some numbers: 102.890 = 776.25 102.891 = 778.04 102.892 = 779.83 Notice that the original numbers had 3 digits behind the decimal place. The results differ in the third place. 101.2 = 15.8 101.3 = 19.9 101.4 = 25.1 In this case, the original numbers had 1 digit behind the decimal place. The results differ in the first place. Conclusion: When raising 10 to a power: If the power of 10 has N decimal places, the result should have N significant figures. The rule for raising e to a power is similar. For simplicity, we will use this rule for both situations. For example: e-2.55 = 0.078082 e-2.56 = 0.077305 e-2.57 = 0.076535 The numbers used in the power of e each had two decimal places. The results started to differ in the second place. The result should have 2 significant figures. |
5304 | https://mae.engr.ucdavis.edu/dsouza/Classes/ECS15-W13/counting.pdf | Counting in base 10, 2 and 16 1. Binary Numbers A super-important fact: (Nearly all) Computers store all information in the form of binary numbers. Numbers, characters, images, music files --- all of these are stored in a computer using binary numbers. But ... how? Well, before we get to that, let's study the binary number system. A binary number is a string of 0's and 1's. What is the meaning of a binary number? Let us look at the decimal system first. Consider the number 5281 • There is one "position" or "place" for each power of 10 • The "1" is in the 1's place • The "8" is in the 10's place • The "2" is in the 100's place • The "5" is in the 1000's place So we have: 5281 = 5 1000 + 2 100 + 8 10 + 1 1 Now let's move to the binary number system. Consider this number: 110101. There is a separate "place" for each power of 2. Starting at the right, we have: • a 1 in the 20=1's place • a 0 in the 21=2's place • a 1 in the 22=4's place • a 0 in the 23=8's place • a 1 in the 24=16's place • a 1 in the 25=32's place So we have: 110101=132 + 116 + 08 + 14 + 02 + 1 = 53 In other words: 110101 in binary is equivalent to 53 in our usual decimal system of numbers. A few more examples: • 1 1 1 1 = 18 + 14 +12 +11 = 15 • 0 1 0 1 = 08 + 14 + 02 + 11 = 5 • 1 0 1 1 1 = 116 + 08 + 14 + 12 + 11 = 23 • 1 1 = 12 + 1 = 3 • 1 1 1 0 = 18+ 14 + 12 = 14 Some Terminology: • Each digit of a binary number (each 1 or 0) is called a bit. • 1 byte = 8 bits. • 1 KB = 1 kilobyte = 2^10 bytes = 1024 bytes (approx 1 thousand bytes). • 1 MB = 1 Megabtye = 2^20 bytes = 1,048,580 bytes (approx 1 million bytes). • 1 GB = 1 Gigabyte = 2^30 bytes = 1,073,741,824 bytes (approx 1 billion bytes) Converting decimal to binary: To convert a decimal number to binary, keep dividing the number by 2 until you get down to 0. Keep track of the remainders. Example: Consider the decimal number 57. • 57 divided by 2 = 28 Remainder 1 • 28 divided by 2 = 14 Remainder 0 • 14 divided by 2 = 7 Remainder 0 • 7 divided by 2 = 3 Remainder 1 • 3 divided by 2 = 1 Remainder 1 • 1 divided by 2 = 0 Remainder 1 Now list the remainder values from bottom to top: 111001: this is the binary form of 57. Let us check our answer (111001) by converting it back to decimal: 111001 = 32 + 16 + 8 + 1 = 57, so we are correct. Maximum Value of a binary number: Consider a binary number with N bits (where N is a number). Its maximum possible value is 2N – 1 (2 to the power of N, minus 1) Example: let N = 3, for a 3-bit binary number, the maximum value is 111, i.e. 23-1=7 2. Counting Using Binary Numbers Consider how counting works in the decimal system. We start with 1 digit. We count using the numerals 0 through 9. After we reach 9, we've run out of numerals. So, we have to add a second digit. We start that digit at 1. Then we cycle the first digit through the numerals 0 through 9 again, to create the numbers 10-19. After we reach 19, we've run out of numerals in the "1's place" again, so we increment the second digit to 2. Eventually, we reach 99. We've run out of numerals in the "1's" place, so we want to increment the second digit again. But, now we've run out of numerals for the second digit as well. So, we have to introduce a 3rd digit, and we start it at 1. And so on. Counting using binary numbers works the same way, except that we only have 2 numerals (1 and 0) for each digit. So, we start with 1 digit. We count using the numerals 0 through 1: 0 1 We are already out of numerals. So, we have to add a second digit. We start that digit at 1, and then we can cycle the first digit through the numerals 0 through 1 again: 10 11 Next we add a 3rd digit, and start it at 1. Now we can cycle the 1st and 2nd digits as we did before: 100 101 110 111 And so on. Starting from the beginning again, the sequence of binary numbers looks like this: 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1111 10000 ... Let us rewrite this sequence, with the decimal value of each number listed to the right: binary decimal ------ ------- 0 0 1 1 10 2 11 3 100 4 101 5 110 6 111 7 1000 8 1001 9 1010 10 1011 11 1100 12 1101 13 1110 14 1111 15 10000 16 ... Here is a quick quiz. Consider this large binary number: 101010111. What is the next number? Answer: 101011000 3. Hexadecimal We've looked at the decimal number system (base 10) and the binary number system (base 2). The hexadecimal system has a base of 16, but it works the same way. One tricky point regarding the hexadecimal system is that it uses 16 different numerals: the regular numerals 0 through 9, and then the capital letters A through F. hexadecimal decimal equivalent ----------- ------------------ 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 A 10 B 11 C 12 D 13 E 14 F 15 Converting Hexadecimal to Decimal Consider this hexadecimal number: AC7. There is a separate position for each power of 16. Starting at the right, we have: o a 7 in the 1's place o a C in the 16's place o an A in the 162=256's place So we have: AC7 = A256 + C16 + 71 = 10256 + 1216 + 7 = 2759 In other words: AC7 in hexadecimal is equivalent to 2759 in our usual decimal system of numbers. Converting Binary to Hexadecimal (Hex) Consider the following table, which shows the hexadecimal equivalent of the 1st 16 binary numbers. Decimal Binary Hexadecimal 0 0000 0 1 0001 1 2 0010 2 3 0011 3 4 0100 4 5 0101 5 6 0110 6 7 0111 7 8 1000 8 9 1001 9 10 1010 A 11 1011 B 12 1100 C 13 1101 D 14 1110 E 15 1111 F Given this chart, it is easy to transform binary numbers into decimal form. Let us take an 8-bit example: 10001100 To convert to hex, we divide the number into groups of 4 bits (starting at the right-hand side): 1000 1100 Then, we convert each group of 4 bits to the hexadecimal equivalent, using the chart above: 8 C And that is all there is to it! Thus, the binary number 10001100 is equal to 8C in hex. This easy conversion from binary to hexadecimal makes hexadecimal notation very convenient. Instead of writing out a very long binary number, we can represent it compactly using hexadecimal. Example #2: Start with this binary number: 100101000110101 Divide the number into groups of 4 bits: 100 1010 0011 0101 Now add a 0 to the front of the 3-digit group on the left: 0100 1010 0011 0101 Now convert each 4-digit binary number to a hex number: 4 A 3 5 And this is our final answer: 4A35 Converting Hex to Binary Converting a hexadecimal number to binary is just as easy: just convert each hexadecimal digit to a 4-digit binary number, according to the same table! Example: D7A = D 7 A = 1101 0111 1010 = 110101111010 Counting in Hexadecimal Counting in hexadecimal is just like counting in the decimal system, except that we cycle each digit through the numerals 0 to F (instead of 0 to 9). Thus, counting in hexadecimal goes like this: hexadecimal decimal equivalent ----------- ------------------ 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 A 10 B 11 C 12 D 13 E 14 F 15 10 16 11 17 12 18 ...etc. 1F 31 20 32 21 33 ...etc. 2F 47 30 48 31 49 ...etc. FF 255 100 256 4. The ASCII Code (American Standard Code for Information Interchange) Computers store all information as numbers (binary numbers). So, how do we represent a keyboard character as a number? The answer is: we use the ASCII Code! This is in fact a very simple idea: The ASCII Code assigns a number (between 0 and 127 inclusive) to each keyboard character. Each ASCII code number is represented by a 7-bit binary number. However, 8 bits are used for each number, for convenience. Examples: (The numbers shown are decimal numbers.) • A to Z: 65 - 90 • 0 to 9: 48 - 57 • $ - 36 Application: D e a r ASCII code 68 101 97 114 Hexadecimal 44 65 61 72 Binary 01000100 01100101 01100001 01110010 5. Unicode The ASCI code is insufficient: 127 characters is not enough particularly for other languages than English. Unicode was developed to solve this problem. It works under the ame principle as ASCII code, with 1 unique number per character. However, the code supports many more characters (at least 65,536) |
5305 | https://math.stackexchange.com/questions/1907243/why-is-the-distance-between-two-circles-spheres-that-dont-intersect-minimised-a | Stack Exchange Network
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Why is the distance between two circles/spheres that don't intersect minimised at points that are in the line formed by their centers?
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From GRE 0568
From MathematicsGRE.Com:
I'm guessing the idea applies to circles also?
Is there a way to prove this besides the following non-elegant way?
Form a line between centers $C_1$ and $C_2$
Given a point on circle/sphere 1 $(x_1,y_1)$, minimise
$$f(x_2,y_2) = (x_1-x_2)^2 + (y_1-y_2)^2$$
to get $(x_2^,y_2^)$
Minimise
$$g(x_1,y_1) = (x_1-x_2^)^2 + (y_1-y_2^)^2$$
to get
$(x_1^,y_1^)$
Show that the $(x_2^,y_2^)$ and $(x_1^,y_1^)$ are on the line.
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edited Aug 29, 2016 at 10:43
BCLCBCLC
asked Aug 29, 2016 at 10:21
BCLCBCLC
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$\begingroup$ It is usually the case that the minimum is attained at those points. But the spheres (or circles) may intersect, when... $\endgroup$
Jyrki Lahtonen
– Jyrki Lahtonen
2016-08-29 10:26:27 +00:00
Commented Aug 29, 2016 at 10:26
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$\begingroup$ For a varying point $O_1$ from the first sphere to be as close as possible to a fixed point $O_2$ on the second it is necessary that the normal of the 1st sphere at $O_1$ points at $O_2$. Basically because this is what Lagrange multiplier formulation tells us. But IMO a more intuitive explanation is that otherwise we can bring $O_1$ closer to $O_2$ by a tangential motion in the appropriate direction. You can switch the roles of $O_1$ and $O_2$ in the argument - secure in the knowledge that a solution exists by compactness. $\endgroup$
Jyrki Lahtonen
– Jyrki Lahtonen
2016-08-29 10:34:20 +00:00
Commented Aug 29, 2016 at 10:34
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$\begingroup$ Purely on an intuitive level, this is a natural consequence of what it means to be a circle. It is easy to draw an arrangement of regular triangles where this will not be true. As you add more sides to a regular polygon, the variance from the center-line will be smaller and smaller. $\endgroup$
wberry
– wberry
2016-08-29 16:51:52 +00:00
Commented Aug 29, 2016 at 16:51
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The radical plane (3D version of radical line) of the spheres is $5x-y-z+6 = 0$. The distance of the centers of the spheres from this plane are $\dfrac{15}{\sqrt{27}}$ and $\dfrac{12}{\sqrt{27}}$. Hence the shortest distance between the spheres and the plane are $\dfrac{15}{\sqrt{27}}-1$ and $\dfrac{12}{\sqrt{27}}-2$. Since the spheres are on the opposite sides of this plane, the minimal distance is $\dfrac{15}{\sqrt{27}}+\dfrac{12}{\sqrt{27}} - 3= 3(\sqrt{3}-1) $
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edited Jan 5, 2018 at 23:03
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answered Aug 29, 2016 at 10:33
user348749user348749
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I have maybe misunderstood your question because the way you presented it was that you wanted an answer to GRE5168 question.
The answer is (E).
Proof: Let $O_1$ and $O_2$ be the centers of the spheres $S_1$ and $S_2$, with radii $R_1=1$ and $R_2=2$ resp.
The distance between the centers is
$d=O_1O_2=\sqrt{(2+3)^2+(1-2)^2+(3-4)^2}=\sqrt{27}\approx 5.2$ which is larger than $R_1+R_2=3$. Thus $S_1$ and $S_2$ do not intersect.
Consequently, if $A_1$ and $A_2$ are the closest points on $S_1$ and $S_2$ resp., they are on line segment $[O_1O_2]$ (see remark below) with:
$O_1A_1+A_1A_2+A_2O_2=1+\delta+2=d$. Consequently:
$$\delta=\sqrt{27}-3=3\sqrt{3}-3$$
is the (shortest) distance between $S_1$ and $S_2$.
Remark: if one among $A_1$ and $A_2$ wasn't on line segment $[O_1O_2]$, one could clearly find a closest pair, by reasoning on triangle's inequality.
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edited Aug 29, 2016 at 10:52
answered Aug 29, 2016 at 10:36
Jean MarieJean Marie
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I suppose you ask for a simple proof (geometric?) to show this?
Let $v=\vec{O_1O_2}$, $e=\frac{v}{|v|}$ and suppose $r_1+r_2<|v|$. One circle (sphere) is included in the halfspace of $P_1$'s for which $e\cdot \vec{O_1 P_1} \leq r_1$. The other in the half space $e\cdot \vec{O_2 P_2}\geq -r_2$. The distance between $P_1$ and $P_2$ is bounded from below by $$e\cdot \vec{P_1P_2} = e\cdot \left(\vec{O_1O_2} +\vec{O_2P_2} - \vec{O_1P_1}\right) \geq |v|-r_2-r_1$$ (making a drawing helps).
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edited Aug 29, 2016 at 10:57
answered Aug 29, 2016 at 10:35
H. H. RughH. H. Rugh
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These questions always involve the use of Pythagoras' equation 'twice', i.e., distance $d$ from centres is $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\tag{$\star$}$$ Now, if $r_1$, $r_2$ are the radii, check: if $dr_1+r_2$ they are apart from each other.
Best to draw a diagram.
$$S_1: (x-2)^2+(y-1)^2+(z-3)^2=1$$
$$S_2: (x+3)^2+(y-2)^2+(z-4)^2=4$$
So sphere $S_1$ has radius $r_1=1$ and centre $O_1=(2,1,3)$, while sphere $S_2$ has radius $r_2=2$ and centre $O_2=(-3,2,4)$. From this the distance between the spheres is found from $(\star)$ to be $d=3\sqrt{3}>r_1+r_2=3$, and so the spheres do not intersect. Thus the minimal distance between two points on the sphere is the difference between $d$ and $r_1+r_2$, which is $3\sqrt{3}-3$, and the answer is (E). Note: the reason why the distance is minimised at points that are in the line formed by their centres is made clear by the triangle inequality.
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edited Aug 29, 2016 at 11:45
answered Aug 29, 2016 at 11:35
Daniel BuckDaniel Buck
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Basically the question in your title is answered by the fact that a straight line is the shortest path between two points. If $P,Q$ are the points on your two spheres (or circles, if you are in a plane), and $C_i$ and $r_i$ are their respective centres and radii, for $i=1,2$, then $C_1-P-Q-C_2$ is a path from $C_1$ to $C_2$ of length $r_1+d(P,Q)+r_2$. Here only the middle term, the distance $d(P,Q)$ from $P$ to $Q$, depends on the choice of these points; the other two terms are constant. The shortest possible path from $C_1$ to $C_2$ is a straight line, and given that $d(C_1,C_2)\geq r_1+r_2$, this path can be obtained by choosing $P$ and $Q$ on the segment $[C_1,C_2]$. Every other choice of $P,Q$ gives a longer path, hence a larger value of $d(P,Q)$.
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edited Sep 3, 2016 at 15:42
answered Aug 29, 2016 at 16:47
Marc van LeeuwenMarc van Leeuwen
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Types of antibiotics
Antibiotics are crucial medications used to combat bacterial infections, categorized into various types based on their chemical structure and mechanism of action. The primary classes of antibiotics include beta-lactams, macrolides, tetracyclines, fluoroquinolones, glycopeptides, and aminoglycosides, each with distinct properties and applications. Beta-lactams, which include penicillins and cephalosporins, work by disrupting bacterial cell wall synthesis and are effective against a range of infections. Macrolides, like erythromycin, target protein synthesis and are particularly useful for respiratory infections. Tetracyclines are versatile and treat diverse infections, including skin conditions, but may have more side effects. Fluoroquinolones are synthetic antibiotics effective against both gram-positive and gram-negative bacteria, often used for urinary and respiratory infections. Glycopeptides, such as vancomycin, primarily target gram-positive bacteria and are critical for treating resistant strains like MRSA. Despite their effectiveness, the rise of antibiotic resistance poses a significant challenge to treatment, necessitating responsible use and ongoing research for new antibiotic development.
Published in: 2024
By: Hoey, Nicole M. Van, PharmD
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Types of antibiotics
Definition
Antibiotics are grouped by type or class to identify groups of similar antibiotics that act on specific bacteria types (such as gram-negative bacilli) and in the same manner (such as to kill cells or slow growth). The most common method of separating antibiotics by class is according to the type of chemical drug structure.
Beta-lactams
Penicillins and cephalosporins are two subclasses of beta-lactam antibiotics, as they share a five- or six-membered ring structure. All beta-lactams are bactericidal and work at the bacterial cell wall level. Beta-lactams irreversibly bind as a false substrate to an active site on the enzyme that is responsible for cell-wall peptide cross-linking; by preventing the cross-linking, beta-lactams prevent the completion of the bacterial cell wall.
Penicillin, the first beta-lactam, was identified as a mold spore, Penicillium notatum (now called P. chrysogenum), in 1928 by bacteriologist Alexander Fleming; the antibiotic itself was derived from P. chrysogenum in 1941 and was active against strains of the Staphylococcus bacterium. Although penicillin had only a narrow, gram-positive spectrum, later penicillin-related antibiotics, such as methicillin and ampicillin, provided expanded activity by avoiding bacterial resistance or by acting against select gram-negative organisms, respectively. Penicillins generally are used to treat skin, ear, respiratory, and urinary tract infections for which bacteria remain sensitive.
Cephalosporins provide much broader-spectrum coverage within the beta-lactam class compared with penicillins. Although their mechanism of action is like that of penicillin, they have varied spectrums of activity because of structural alterations. Cephalosporins are typically used to treat otitis media (ear), skin, and urinary tract infections but are also used in surgical prophylaxis and to treat bone infections and pneumonia.
The activity of cephalosporins can be defined by four subtypes, or generations, to provide wide bacterial coverage. First-generation drugs, such as cephalexin and cefazolin, provide primarily gram-positive activity; second-generation cephalosporins, such as cefuroxime and cefaclor, provide gram-negative and gram-positive activity but have a range of sensitivities. Third-generation examples include ceftriaxone, cefixime, and ceftibuten; these drugs provide wide gram-negative coverage but lose much of the class gram-positive coverage. Fourth-generation drugs cefepime and cefquinome have similar gram-positive activity as early cephalosporins but have better activity against beta-lactamase-resistant bacteria, and they cross the blood-brain barrier to treat meningitis and encephalitis.
All beta-lactams are well tolerated and are associated with the mild side effects of nausea and diarrhea. However, allergy to drugs in the beta-lactam class is not uncommon and may develop with both penicillin and cephalosporin use.
Macrolides
Unlike penicillins and cephalosporins, which act on the bacterial cell wall, macrolides interact with bacteria at protein synthesis, and they are typically bacteriostatic but may become bactericidal, depending on their concentrations and the bacteria types attacked. Macrolides such as erythromycin, clarithromycin, and azithromycin bind to the 50S section of the ribosome during bacterial protein development to change the ribosome and prevent peptide bonding. Erythromycin additionally may prevent the formation of the 50S subunit itself.
Macrolides are composed of a macrocyclic lactone and are derived from the bacterium Streptomyces. Erythromycin, the first-in-class macrolide, has similar activity to penicillin; conversely, the two newer macrolides have their best activity in lung diseases, and clarithromycin is particularly effective against Helicobacter pylori, which often causes stomach ulcers. Macrolides are used against Staphylococcus, Streptococcus, and Mycoplasma infections, and they are used to treat Legionnaires’ disease, which is caused by the Legionella bacterium. Side effects include mild nausea, diarrhea, and stomach upset.
Tetracyclines
Like macrolides, tetracyclines are derived from Streptomyces; they are made of four connected rings. Tetracyclines block the beginning of protein synthesis by binding the ribosome and preventing the addition of aminoacyl tRNA (transfer ribonucleic acid) building blocks. In addition, tetracyclines may change the ribosome itself to prevent successful protein synthesis. Tetracyclines provide bacteriostatic activity against a broader spectrum of bacteria than penicillins.
Tetracycline, minocycline, tigecycline, and doxycycline are common examples of drugs in this class. They have unique activity against Rickettsia and some amoebic parasites; they can treat sinus, middle ear, urinary tract, and intestinal infections. However, a common use of drugs in this class is to treat skin conditions such as rosacea or moderate acne. Tetracyclines have a greater risk of side effects, especially with prolonged use. Photosensitivity, cramps, diarrhea, and possible bone and tooth changes may occur with tetracycline use.
Fluoroquinolones
Fluoroquinolones, unlike beta-lactams, are synthetic rather than derived directly from a bacterial source. They are well absorbed, are distributed into bone, and can be given by mouth or intravenously. They consist of a dual ring and a fluor group that increases the antibiotic activity.
Fluoroquinolones are bactericidal against a broad spectrum of bacteria. Fluoroquinolones act by blocking deoxyribonucleic acid (DNA) building within the bacteria to prevent multiplication. Early examples, such as ciprofloxacin, are primarily active against gram-negative bacteria; newer agents, including levofloxacin, keep gram-negative activity and add activity against gram-positive bacteria such as pneumococcus (Streptococcus pneumoniae). They are often used to treat urinary tract and skin infections and respiratory infections such as bronchitis and bacterial pneumonia. Other fluoroquinolones, like ofloxacin and moxifloxacin, have additional activity against anaerobic bacteria.
Glycopeptides
Vancomycin, dalbavancin, oritavancin, telavancin, and teicoplanin are the most common glycopeptide antibiotics. Because their chemical makeup is so large and because these drugs cannot cross a cell membrane, they affect only gram-positive bacteria outside the cell. Each glycopeptide is made of two sugars and one aglycone moiety with a heptapeptide core that provides antibiotic action. Glycopeptides block the end of cell-wall peptidoglycan synthesis so that the cell wall cannot be completed, and the bacteria cannot survive. Vancomycin is useful in the treatment of methicillin-resistant Staphylococcus aureus (MRSA) in hospital settings; however, bacteria are also developing intermediate to full resistance to vancomycin.
Other Antibiotics
Aminoglycoside antibiotics, discovered in 1944, contain an amino and some sugar groups. They provide limited-spectrum coverage against gram-negative and gram-positive agents. Aminoglycosides insert themselves incorrectly into proteins during synthesis by binding to the ribosome. They are particularly active against Pseudomonas aeriginosa.
Lincosamides, such as clindamycin, have greater activity against anaerobes, such as those causing intestinal or gastric infections, and they are also used to treat gram-positive Staphylococcus skin infections, including moderate acne. Lincosamides are bacteriostatic and act by inhibiting protein synthesis by the bacterial ribosome.
Impact
With the development of bacterial resistance shortly after penicillin’s introduction in the 1940s, antibiotic drug development has greatly expanded within the beta-lactam class and beyond. However, bacterial resistance appears to be developing faster than new antibiotics are being discovered or developed in laboratories, so infections from common bacteria are once again complicated to treat. Antibiotic resistance has become a threat to global health, and although it can occur naturally, overuse of antibiotics has hastened the phenomenon.
Antibiotics are continually in development, but their efficacy against antibiotic-resistant bacteria varies. Humans must alter their behavior to prevent further antibiotic resistance. In the meantime, research continues to identify the best use of antibiotics within and among classes and to find the safest combination therapies against specific bacteria. For example, Cresolomycin is a synthetic compound in the lincosamide antibiotic family that attacks gram-positive and gram-negative bacteria. It is more effective in treating antibiotic-resistant infections than other antibiotics in its family.
Bibliography
"Antibiotic Resistance." World Health Organization (WHO), 21 Nov. 2023, www.who.int/news-room/fact-sheets/detail/antibiotic-resistance. Accessed 20 Oct. 2024.
Gilbert, David N., et al. The Sanford Guide to Antimicrobial Therapy. 53rd ed., pocket ed., Antimicrobial Therapy, Inc., 2023.
Mandell, Gerald L., et al., editors. Mandell, Douglas, and Bennett’s Principles and Practice of Infectious Diseases. 9th ed., Churchill Livingstone/Elsevier, 2020.
Murray, Patrick R. Murray’s Basic Medical Microbiology. 2nd ed., Elsevier, 2024.
Van Bambeke, Françoise, et al. “Antibiotics That Act on the Cell Wall.” Cohen and Powderly Infectious Diseases, edited by Jonathan Cohen, et al., 4th ed., Mosby/Elsevier, 2017.
Walsh, Christopher. Antibiotics: Actions, Origins, Resistance. ASM Press, 2003.
Related Topics
AntibioticsSide effectsTetracyclines (drug interactions)Fluoroquinolones (drug interactions)Glycopeptide antibioticsDrug resistance |
5307 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11060889/ | Neutrophils in physiology and pathology - PMC
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Annu Rev Pathol
. Author manuscript; available in PMC: 2024 May 1.
Published in final edited form as: Annu Rev Pathol. 2024 Jan 24;19:227–259. doi: 10.1146/annurev-pathmechdis-051222-015009
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Neutrophils in physiology and pathology
Alejandra Aroca-Crevillén
Alejandra Aroca-Crevillén
1 Program of Cardiovascular Regeneration, Centro Nacional de Investigaciones Cardiovasculares (CNIC), Madrid (Spain)
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1, Tommaso Vicanolo
Tommaso Vicanolo
1 Program of Cardiovascular Regeneration, Centro Nacional de Investigaciones Cardiovasculares (CNIC), Madrid (Spain)
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1, Samuel Ovadia
Samuel Ovadia
2 Program of Vascular Biology and Therapeutics and Department of Immunobiology, Yale University, New Haven (USA)
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2, Andrés Hidalgo
Andrés Hidalgo
1 Program of Cardiovascular Regeneration, Centro Nacional de Investigaciones Cardiovasculares (CNIC), Madrid (Spain)
2 Program of Vascular Biology and Therapeutics and Department of Immunobiology, Yale University, New Haven (USA)
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1 Program of Cardiovascular Regeneration, Centro Nacional de Investigaciones Cardiovasculares (CNIC), Madrid (Spain)
2 Program of Vascular Biology and Therapeutics and Department of Immunobiology, Yale University, New Haven (USA)
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Corresponding author: andres.hidalgo@yale.edu or ahidalgo@cnic.es
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PMCID: PMC11060889 NIHMSID: NIHMS1987090 PMID: 38265879
The publisher's version of this article is available at Annu Rev Pathol
Abstract
Infections, cardiovascular disease, and cancer are major causes of disease and death worldwide. Neutrophils are ineludibly associated with each of these health concerns, either by protecting from, instigating, or aggravating their impact to the host. However, each of these disorders have very different etiologies and understanding how neutrophils contribute to each of them implies understanding the intricacies of this immune cell type, including their immune and non-immune contributions to physiology and pathology. Here, we review some of these intricacies, from basic concepts in neutrophil biology such as production and acquisition of functional diversity, to the variety of mechanisms by which they contribute to preventing or aggravating infections, cardiovascular events, and cancer. We also review poorly explored aspects of how neutrophils promote health by favoring tissue repair and discuss how discoveries in their basic biology inform the development of new therapeutic strategies.
Keywords: neutrophil, inflammation, infection, cardiovascular disease, cancer
1. Origin and diversity of neutrophils, from subsets to states
The development of single cell technologies over the past decade has metamorphosed the way we categorize and understand the immune system. Immune cells have been classified in “subsets” when they could be identified by a defined set of markers and emerged from a distinct hematopoietic branch. This approach offered the possibility to isolate cells and build working simple models but fails at capturing the plasticity and functional dynamics that immune cells display across physiological and pathologic contexts or anatomical locations. Hence, efforts now focus on building landscapes of immune “states” by analysis of individual cells at high molecular and functional resolution. In this section, we review neutrophil diversity across maturation, anatomical and pathologic contexts, and discuss the possible sources of neutrophil diversity.
1.1. Granulopoiesis: origin of diversity and implication in pathology
Neutrophils emerge from a fine-tuned sequence of molecular and morphological transformations, or granulopoiesis, during which they progressively modify their transcriptomic, proteomic, structural, and functional properties. We will emphasize here the progressive acquisition of effector functions is, surprisingly, independent of the differentiation process. Indeed, neutrophil development can appear normal in terms of phenotype and abundance of precursors, and yet generate non-functional mature neutrophil (1).
Neutrophil commitment can be divided in two major steps, namely specification and determination. Pro-neutrophils (proNeu) 1 and 2 are already committed to the neutrophil lineage, but may replenish the monocytic compartment in the absence of two critical granulocytic transcription factors, C/EBPε (2) or GFI-1 (1) (Figure 1A). Conversely, the pre-neutrophil (preNeu) stage appears to be a determined stage (3). Contrasting with lineage specification, the acquisition of effector functions occurs only after determination and follows a precise series of molecular events (1–4) (Figure 1A). For instance, phagocytosis, chemotaxis and immune sensing and activation appear progressively during neutrophil maturation and peak at the so-called immature and mature stages of differentiation. Intriguingly, however, expression of genes encoding for the NADPH oxidase peak at the immature stage, and are drastically reduced as the cells advance to maturation in the marrow and in blood (1–4). Although these events take place after lineage determination, functional maturation is pre-established by chromatin remodeling at the stage of specification (proNeu), where GFI-1 plays a pivotal role (1). This duality in the role of GFI-1 (i.e. specification and functional maturation) is explained by the two mechanisms used by this protein to regulate gene expression; one dependent on the dose and one controlled by the Zn-Finger domain 6 (ZnF6) activity. Dose-dependent loci are enriched in genes involved in specification, whereas those dependent on ZnF6 are related to effector functions, including the NADPH oxidase, chemotaxis and defense response (1). This is exemplified in mouse models of Severe Congenital Neutropenia (SCN), a primary immune deficiency triggered by the presence of a single nucleotide polymorphism (SNP) in the Znf6 domain of Gfi1 that leads to profound defects in neutrophil specification due to reduced cytosolic accumulation of GFI-1. Rescuing the cytosolic accumulation of the mutated transcription factor is sufficient to rescue specification but generates non-functional mature neutrophils, demonstrating the importance of the ZnF6 domain for functional maturation (1). These elegant studies illustrate the complexity of implementing effector functions during granulopoiesis, and highlight the dichotomy between differentiation and functional maturation which may explain the remarkable adaptability of granulopoiesis to environmental changes, including during emergency granulopoiesis. This dichotomy is also seen in humans, in which G-CSF treatment rescues neutropenia in SCN patients but fails to generate functional mature neutrophils and to resolve the immune deficiency (5). Thus, chromatin remodeling independently controls specification and function and make granulopoiesis a major source of neutrophil heterogeneity.
Figure 1. Origin and Diversity of Neutrophils.
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A. Granulopoiesis is a stepwise process during which neutrophils acquire transcriptomic, proteomic, structural and functional properties. There are several features that allow categorizing neutrophil development, namely their precursors, proliferative capacity, transcriptional program driven by master transcription factors, nuclear morphology, granule formation, acquisition of effector functions, the cell surface proteome and type of granules. B. During normal granulopoiesis (1) precursors and mature neutrophils acquire diversity during development. Transcriptomic and phenotypic heterogeneity starts in the bone marrow but becomes prominent once in the periphery. It is likely that the exposure of neutrophils to inflammatory and microbial molecules and biochemical cues in the micro-circulation, might initiate this diversification. In tissues, neutrophils are reprogrammed and acquire tissue-tailored signatures. Neutrophils precursors may also egress into blood and migrate to the spleen to continue their differentiation (2). In some settings of infection or inflammation, neutrophil precursors are recruited and can differentiate in situ to generate a local pool of mature cells (3). Finally, neutrophil diversification may rely on the varying composition of the hematopoietic niches found in different bones. (4).
1.2. Emergency Granulopoiesis as a source of diversity and memory
Emergency Granulopoiesis (EG) is the process of de novo generation of neutrophils in response to systemic pathogen invasion or other major source of stress (such as trauma or cancer), resulting in significant increase in circulating neutrophils (6) (Figure 1B). Granulopoiesis relies on G-CSF, the main cytokine responsible for neutrophil maturation (7), and is accompanied by the serial action of different transcription factors, including PU.1, C/EBPs, GFI-1 (8) (Figure 1A).
EG does not only increases the number, but also shapes the identity of the precursors and mature cells, as demonstrated by the transcriptional reprogramming observed upon bacterial infection (4). For example, neutrophil precursors exhibit higher NADPH oxidase scores that suggest heightened capabilities of ROS production, whereas mature neutrophils feature greater phagocytic and chemotaxis capacities (4). Sterile inflammation also affects maturation, as seen in context of tumor (3) or myocardial infarction (9). Indeed, the presence of an early committed neutrophil progenitor (NeP) promotes tumoral growth in murine cancer models (10). Consistent with studies in mice, EG induced in humans after treatment with G-CSF alters the transcriptional landscape of precursors and mature neutrophils in blood and marrow: mature neutrophils disappear from the circulation and are replaced by low-density CD10− immature precursors (11). Functionally, G-CSF-elicited neutrophils exhibit reduced production of ROS and neutrophil extracellular traps (NET) but are better at producing cytokines than mature, normal-density neutrophils, implying that qualitative alterations may be as relevant as change in numbers during EG (11). The molecular mechanisms driving these qualitative changes remain to be elucidated, but appear to be associated with the different stages of maturation involved, as hinted by the transcriptional profile displayed by blood neutrophils depending on the type of perturbation (G-CSF treatment, bone marrow transplantation, or cancer) (11). Adaptation of granulopoiesis to disease is also evident in the context of coronavirus disease (COVID-19), in which symptomatology associates with mobilization of low-density immature neutrophils resembling proNeu and preNeu stages (12). These immature neutrophils are enriched in genes involved in NET formation and immunosuppression (12, 13), suggesting that granulopoiesis entrains new functional properties to the neutrophil pool. The bone marrow is also a site of early neutrophil re-education during cancer, as systemic factors present in tumor-bearing mice can distally modify the bone marrow stroma to promote the emergence of SiglecF high neutrophils that infiltrate the lung and favors tumor progression (14).
Normal granulopoiesis takes place in bone marrow niches, in dedicated locations recently shown to have oligoclonal composition (15). Variations in the composition of such niches and clones produced across different bones could conceivably result in the production of functionally different neutrophils. Indeed, meningeal neutrophils derived from the skull and vertebral marrows exhibit distinct transcriptional profiles compared with their blood counterparts, and infiltrate the central nervous system in the context of stroke (16) (Figure 1B). These findings suggest that other marrow compartments across different bones yield specialized populations of neutrophils that surveille distinct tissues. Likewise, granulopoietic precursors can be mobilized in the circulation and complete their differentiation outside of the bone marrow (Figure 1B). For example, during of S. aureus skin infection (17) neutrophil precursors terminate their differentiation in the infected tissue and in cancer a similar phenomenon of extramedullary granulopoiesis is observed in the spleen (18), where progenitors give rise to immunosuppressive neutrophils. Hence, extramedullary granulopoiesis adds another layer of diversification in neutrophil production.
1.3. Peripheral blood neutrophils
Neutrophils released into the circulation reach all perfused tissues and rapidly engage in immune responses when microbes or other threats are detected. Considered a homogenous population, peripheral blood neutrophils display a spectrum of transcriptional, phenotypic, and behavioral states.
Single cell transcriptomic analyses have reported at least three neutrophil populations in the murine circulation. One enriched in genes involved in migration and inflammatory responses, a second featuring interferon-induced signatures, and a third bearing traits of circadian aging and overall maturation (4). These populations have been identified across multiple studies in both mouse (1, 4, 19) and human (4, 11, 20). In mice, interferon signature gene (ISG)-related neutrophils are found in circulation but also in the spleen, where they exhibit a subcapsular localization that contrasts the uniform distribution of common S100a8+ neutrophils, demonstrating spatial compartmentalization (4). ISG+ neutrophils appear to be preferentially expanded and responsive during sterile and non-sterile inflammation, suggesting specialized roles in inflammatory pathophysiology. For instance expansion of ISG+ neutrophils correlates with disease severity in tuberculosis patients (21), whereas in COVID-19 patients the abundance of ISG+ neutrophils associates with mild symptomatology (13). Severe COVID-19 patients, in contrast, feature neutrophils with an immunosuppressive signature characterized by expression of the immune checkpoint receptor PD-L1 (11, 13), overall indicating that the transcriptional state of blood neutrophils associate with disease outcome.
In Systemic Lupus Erythematosus (SLE), a multifactorial autoimmune disease associated with vascular inflammation, an atypical population of granulocytes with altered buoyancy (low density neutrophils or LDN) appears in blood and exhibits distinct epigenetic, transcriptomic and functional properties when compared with autologous normal-density neutrophils (22). LDN are a mixture of mature and immature neutrophils, with mature cells featuring an ISG signature and enhanced NET forming capacity, and immature cells capable of degranulation, but reduced phagocytic capacity (22, 23). Current models propose that the distinct clinical manifestations in SLE patients are driven by the different neutrophil types. The origin and molecular drivers of these pathogenic neutrophils remain to be elucidated. Intriguingly, however, the epigenome of LDN from SLE patients suggests epigenetic reprograming already at the level of granulocytic precursors (23). ISG-related neutrophils are observed across species and disease with remarkable consistency and independent studies have found the ISG signature early in neutrophil development (4, 19), suggesting that a core program during maturation gives rise to this state. Formal proof, however, will require DNA barcoding-based tracing analyses.
While transcriptomic analyses dominate current efforts to classify immune cells, alternative approaches have emerged. Indeed, transcriptomic profiling fails to capture the full spectrum of neutrophil diversity, especially when defining immune function in dynamic environments, such as inflammation. In this context, transient properties (or “states”) may be more informative than fixed genetic programs (“subsets”). Recent studies have implemented imaging methods that measure the dynamic properties of leukocytes (24, 25). In particular, in vivo imaging of inflamed vessels in mice allowed extraction of tens of morphological and kinetic traits from individual cells that were computed to describe neutrophil “behaviors” (24, 26). These analyses identified three distinct behaviors, one of which described large neutrophils that remain sessile against the endothelium and correlated with the magnitude of inflammation. This pathogenic behavioral type was found to rely on the activity of the Fgr kinase, and deficiency or inhibition of this kinase was sufficient to prevent this behavior and to prevent vascular inflammation in the context of glomerulonephritis and myocardial ischemia (24). Importantly, neutrophils were found to transit through these behavioral archetypes although with low frequency, a finding that highlights the dynamism of neutrophils both in phenotype and function and raises the possibility that these different behavioral traits may be genetically encoded rather than driven by the environment (27).
1.4. Sources of diversity in blood
As discussed above, granulocytic progenitors or mature neutrophils can integrate cell-intrinsic and environmental cues that shape their phenotypic and functional identity. Among these, are signals that follow natural day/night oscillations and influence all aspects of organismal physiology, including the immune system (28). Indeed, once mobilized into blood, mature neutrophils undergo transcriptional and phenotypical alterations that align to circadian oscillations and are collectively referred to as “aging”. Circadian aging is driven by an internal circadian machinery that involves the transcription factor BMAL1, which controls expression of the chemokine CXCL2. This initiates signaling events by engaging the receptor CXCR2 in an autocrine fashion and, interestingly, is antagonized by signaling through another chemokine receptor, CXCR4, when the levels of its ligands CXCL12 naturally increase in plasma during the day (29, 30). The combination of these positive and negative signals modulates neutrophil function through the diurnal cycle and are critical to balance immune defense and protection of host cells. For example, aged neutrophils, which dominate at daytime in mice and humans, are better at controlling pathogen invasion. However, this comes at the cost of increased thrombo-inflammatory events, or other vascular-damaging processes, as seen in a model of myocardial infarction. In contrast, fresh neutrophils, which dominate at nighttime, may be less immune competent but are better at rapidly infiltrating inflamed tissues (30). The basis for this immune and migratory bias is, at least in part, dictated by regulation of the celĺs topology and granular content by the neutrophil circadian clock (30, 31).Circadian aging also requires the integration of signals from the microbiota. Indeed, microbiome-derived TLR4/TLR2 agonists regulate phenotypic properties of neutrophils (32), which in turn associates with more and more severe vaso-occlusive events in a mouse model of sickle cell disease (32).
Overall, the heterogeneity of mature neutrophils extends beyond granulopoiesis and into the circulation, in both mice and humans (19, 20), but the strategies used for diversification remain largely unknown (27).
1.5. Tissue neutrophils
Upon entry into naïve tissues from the circulation, neutrophils appear to localize in specific niches (33). For instance, in the pulmonary parenchyma neutrophils accumulate at the periphery of large vessels (33, 34), and bulk analyses of transcriptome and chromatin accessibility has revealed that distinct molecular landscapes of chromatin imprinted across different organs (34). However, this diversity in tissues does not seem to be caused by infiltration of neutrophils with pre-determined programs, at least as determined by the single cell transcriptional signatures of blood neutrophils (34), suggesting that microenvironmental cues co-opt the plasticity of neutrophils to mold their fates and functions (34).
A remarkable observation from these studies is that tissue neutrophils acquire non-immune properties. For instance, vascular regenerative programs, revealed by expression of genes encoding Apelin, Adamts, Vegfa, are associated with neutrophils from the lung and intestine, whereas factors that favor immunoglobulin production and maturation are produced by splenic neutrophils (34, 35). Interestingly, these signatures are not shared by other leukocyte subsets in the same organ (34), suggesting that the environment specifically enacts neutrophil programs. Thus, despite their reduced transcriptional activity, neutrophils share with macrophages the capacity to adapt to the environment to support tissue homeostasis.
A particularly interesting scenario is cancer. Indeed, the tumor micro-environment provides molecular, cellular and structural cues to the infiltrating cells that ultimately support tumor growth. Neutrophils that infiltrate tumors undergo reprograming that varies with tumor type and the different intratumoral microenvironments (36, 37). For example, in genetic mouse models and human patients with lung cancer, neutrophils acquire at least six different transcriptional programs, some of which can also be found in the healthy lung (38). The nature of such states and their potential role in the tumor remain unclear, although specific subsets with T cell-suppressive activity and angiogenic properties have been defined based on expression specific transcription factors, enzymes or growth factors (such as Bhlhe40, Ptgs2 and Vegfa, respectively) and have been functionally validated. Cancer provides a remarkable example of how neutrophils adapt to different environments as discussed in more detail below.
2. The reparative neutrophil
The recent discovery that neutrophils adapt to different tissues (34) highlights that their plasticity is broadly beneficial for the organism, and indeed neutrophils aggressively fight infections but also contribute to tissue repair. Here we focus on a typically neglected aspect of neutrophil biology related to interaction with the maintenance and repair of the tissue structure.
2.1. Neutrophils and the extracellular matrix
Neutrophils travel through the bloodstream and extravasate into most tissues under homeostatic conditions (33). This program is enhanced during infections or physical insults where neutrophils are massively recruited into the damaged tissues. Once neutrophils leave the blood stream, they encounter the ECM, a non-cellular component present in all tissues and organs that is an essential component of the interstitial matrix and the basement membrane. The interstitial matrix is a loose network of collagen fibers (mainly type I and III collagens, fibronectin, elastin, and proteoglycans) that creates a meshwork which connects structural cell types within tissues. The basement membrane is a protein-dense structure made up by type IV collagen, laminin and heparan sulfate proteoglycans. This ECM, which localizes at the base of epithelial or endothelial layers, separates cells from the surrounding matrix serving as a barrier for the transfer of materials (39).
During maturation in the bone marrow, neutrophils store a collection of proteases in their granules which degrade ECM proteins. Serine proteases, including elastase, proteinase 3, and cathepsin G, which are a substantial part of primary granules, can break down elastin, fibronectin, laminin, and collagen IV (40). Matrix metalloproteinases (MMPs), which degrade the matrix, are mostly found in secondary and tertiary granules, and can degrade all type of matrix proteins, including all types of collagens (41). Neutrophils dedicate such a significant portion of their granules to store matrix-degrading enzymes that enable their function and motility. For instance, migration through the vessel basement membrane is facilitated by neutrophil elastase and by finding ECM-poor gaps in the pericyte layer (42). Once out of blood vessels, neutrophils must migrate through the ECM mesh to penetrate and move inside tissues. They do so by employing their enzymatic repertoire to proteolytically degrade the matrix proteins. Reciprocally, the ECM regulates neutrophil behavior and recruitment: in the context of inflammation, neutrophils release MMP8 and MMP9 to breaks down collagen into bioactive ECM fragments endowed with chemotactic activity (43). Laminin in the basement membrane of ECM can send signals to upregulate surface expression of CD31 and the integrin α6β1 on neutrophils, which promotes transmembrane transport and infiltration into the inflamed tissue (44). The ECM also promotes antimicrobial functions in neutrophils, including ROS production, MPO release, and NET formation (45), and a recent study reported that fibrin deposits in the oral mucosa instruct neutrophil activation through direct engagement of Mac-1, which induces production of ROS and NETs for immune protection, but also cause considerable tissue damage (46).
2.2. Neutrophils as wound healers
The first immune cells to enter damaged tissues are neutrophils. They arrive in large numbers in response to DAMPs released from injured and necrotic cells, and orchestrate the precise and effective migration to the specific area of injury using a collective, feed forward mechanism called “swarming” (47). The contribution of neutrophils to wound healing is multifaceted: too few increase the risk of infection and delayed healing (48), whereas too many neutrophils remaining in the tissues can be toxic for host cells and delay healing. Impaired leukocyte trafficking is known to delay the healing of cutaneous wounds in mice (49), and experimental depletion of neutrophils in the context of intestinal injury associates with increased inflammation and impaired repair of the intestinal mucosal (50). Similarly, patients with neutropenia (or impaired neutrophil trafficking or function) display not only higher risk for developing infections but also impaired wound healing (51). These findings highlight the critical role of early immune responses and neutrophils in initiating wound repair.
A plethora of studies have unveiled links between neutrophils and ECM remodeling. During wound healing, degradation of the ECM facilitates neo-angiogenesis within the restored tissue, which is key to normal regeneration. MMPs exert a pro-angiogenic function as they can degrade ECM and release matrix-bound VEGF, which is then free to signal and promote angiogenic remodeling (52). Interestingly, neutrophils are the only cells that can release MMP9 free of its endogenous inhibitor, the tissue inhibitor of metalloproteinases (TIMPs), and can directly deliver MMP9 to sites of angiogenesis (52). Neutrophils also contribute to building new ECM in the wound by transporting preexisting matrix from the mesothelial layer surrounding visceral and parietal internal organs into the injured tissue, where they fuel tissue repair (53). This recently discovered phenomenon is mediated upregulating the collagen-binding integrin αMβ2, which binds and transports matrix proteins. In the absence of neutrophils, internal wounds fail to incorporate active fibroblasts and to mature into long-lasting scars. In this context, fibroblasts are part of a secondary response, but neutrophils have been estimated to account for about 80% of the matrix generated in the scar tissue (53). In addition to transporting preformed ECM, under specific pathological contexts neutrophils have been reported to actively produce ECM components. During myocardial infarction, neutrophils appear to contribute to ECM deposition and organization in the ischemic heart by producing fibronectin, fibrinogen and vitronectin (54). Similarly, neutrophils contribute to bone healing by synthesizing so-called emergency ECM before stromal cells infiltrate the fracture hematoma and synthesize bone matrix (55). Neutrophils within these early hematomas stain positive for cellular fibronectin, in contrast to neutrophils within coagulated peripheral blood (55). Thus, neutrophils are producers and transporters of ECM components, at least in the context of injury.
2.3. Remodeling the tumoral matrix
Alterations within the tumoral ECM can promote cancer by providing biomechanical and biochemical cues that drive cell growth, survival, invasion and metastasis, and by regulating angiogenesis and curbing immune function. Tumors often display enhanced stiffness, which is caused by altered deposition, cross-linking, and geometrical organization (e.g., linearization) of fibrous proteins, especially of collagen fibers that are often positioned perpendicular to the tumor boundary (56). Compared with benign lesions in which type I and type III collagen bundles are regularly organized, malignant human breast tumors show elevated expression of type I and type III collagen and newly synthesized collagens that are frequently arranged in abnormal bundles (56). ECM morphology and composition in the metastatic site is therefore important for the seeding and growing capacity of metastatic cells. For example, ECM stiffening facilitates colonization of cancer cells and infiltration of myeloid cells at the metastatic site (57). The contribution of neutrophils and other immune cells to the aberrant organization of the tumoral matrix is a field of emerging interest.
Neutrophils can also remodel the tumoral ECM and, in doing so, modulate cancer cell plasticity. Tumor-recruited neutrophils release MMP9 that remodels the ECM and induces angiogenesis or promotes metastasis (58). Indeed, MMP9-expressing neutrophils are predominantly found inside angiogenic islets in early stages of pancreatic carcinogenesis and mediate VEGF activation (59). Thus, neutrophil MMP9 is a potential therapeutic target in human cancers in which neutrophil infiltration is associated with enhanced tumor angiogenesis and poor prognosis. Neutrophils can also alter the ECM environment to facilitate tumor cell invasiveness; In a zebrafish xenograft model of in vivo tumorigenesis, neutrophil migration enhanced tumor cell invasion by providing collagen tracks that were exploited by cancer cells for migration (60) and, at least in vitro, neutrophils induce the formation of invadopodia and collagenous matrix degradation by oral squamous cell carcinoma cell lines (61). Interestingly, collagen-dense mammary tumors exhibit a cytokine milieu that promotes neutrophil recruitment and activation. In this setting, neutrophil ablation significantly reduces tumor burden and diminishes the formation of metastasis (62), suggesting that tumor progression in collagen-dense microenvironments is promoted by neutrophils.
Processing of the ECM can also participate at early stages of tumor formation. In the context of organismal stress (induced by LPS or tobacco inhalation) NETs produced by neutrophils induced the awakening of dormant cancer cells and promoted metastasis in mice. NETs concentrate neutrophil proteases (elastase and MMP9) which enable the sequential cleavage of laminin, and generates a neo-epitope for integrin a3b1 that triggers cancer cell activation (63). Collectively, these findings illustrate various mechanisms through which neutrophils modify the ECM to promote tumor progression and metastasis and may drive the design of new therapies against cancer.
3. Anti-microbial neutrophils: cost to the host
Neutrophils are fundamental for primary immune defense. After migrating to the site of inflammation, they kill, phagocytose and digest the invading microbes by generating reactive oxygen species (ROS) and by discharging the microbicidal content of their granules. They also produce cytokines, attract other immune cells; and release NETs, which can ensnare and kill pathogens (64, 65). The effective elimination of invading agents by neutrophils, however, comes at a cost as neutrophils actively promote inflammatory pathology (66), as prominently illustrated in the recent context of COVID19 (13, 67). Understanding the mechanisms of neutrophil-driven toxicity, interactions with other immune cells and how they drive or amplify inflammation is of major clinical relevance. Here, we focus on specific antimicrobial activities of neutrophils that may impact structural and functional integrity of the host.
3.1. Reactive oxygen species (ROS)
Among the array of antimicrobial products produced by neutrophils, reactive oxygen species (ROS), are essential for pathogen clearing after phagocytosis. The main mechanism for ROS generation involves activation of the NADPH oxidase complex (68), which comprises different NOX enzymes. NOX2, the main catalytic subunit of this complex, assembles in the membrane and generates the precursor superoxide anion (O 2−) by reduction of oxygen (69). This superoxide is released into the phagosome containing the microbe where it dismutates into hydrogen peroxide (H 2 O 2). Neutrophils additionally generate secondary oxidants such as hydroxyl radical (OH·), hypochlorous acid (HOCl), chloramines (R-NHCl) and hypothiocyanite (OSCN−) via the activity of myeloperoxidase (MPO) (69).
The production of ROS by neutrophils is variable. ROS production changes circadianally; a subset of phenotypically aged neutrophils has been shown to produce higher levels of ROS (32), although these differences are not apparent when comparing neutrophils collected at different times of the day (Zeitgeber or ZT5 and ZT13, corresponding to noon and 8pm, respectively) (30). During infections, ROS production associates with damage to multiple tissues, including lung, liver, heart and kidney injury in different sepsis models, as well as with endothelial dysfunction (70). In the context of COVID19, neutrophils from severe COVID19 patients and from non-COVID19 related sepsis exhibited a heightened capacity for ROS generation (71). This has been linked thrombosis and red blood cell dysfunction, which is likely to contribute to the higher mortality of these patients (72). Activation of neutrophils within vascularized tissues can also promote ROS production and cause vascular damage (see below; section 4). A particularly interesting example is that of reversed transmigrated neutrophils, which exhibit a distinct activation phenotype characterized by increased ROS production that disseminates systemic inflammation (73). In sum, the imbalance between ROS production and effective performance and pathogen removal is paramount for tissue protection against pathogens and to preserve homeostasis.
3.2. Degranulation and cytokine production
Neutrophil granules are formed during maturation in the bone marrow. The different types of granules with varying content and functions are distinguished based on the time of synthesis during granulopoiesis, or their major protein content. Azurophilic or primary granules are enriched in myeloperoxidase (MPO), specific or secondary granules contain lactoferrin and tertiary granules contain gelatinase and matrix metalloproteinases 9 (MMP9), among other proteins (74). During medullary maturation, expression of different transcription and growth factors determine the different stages of maturation that drive the acquisition of a particular granule type in neutrophils. High levels of the zinc finger transcription factor (GFI-1) and low PU.1 (75) guide the transition towards a myeloblast state and the production of primary granules, while low expression of ELF-1 and increased C/EBPε allow the transition into the myelocyte state and the formation of secondary granules. Finally, tertiary granules appear after the metamyelocyte state concomitant with loss of mitotic activity (76) (Figure 1A). This sequential production of granules has more recently been delineated by analyzing the transcriptional profile of neutrophils from the pre-neu to the immature state (3). The release of granule content into the environment is regulated and triggered by multiple stimuli but, interestingly, the process is very selective in that particular receptor-ligand coupling triggers the release of particular granule types. For example, fMLP only promotes the discharge of primary granules (77), whereas S100A9 discharges specific and gelatinase granules (78).
From a physiological perspective it is remarkable that basal degranulation occurs and is under circadian control. Basal release of granule content into circulation of healthy mice is driven by CXCR2 signaling, a component of the neutrophil clock, upon release from bone marrow (31), which implies that these granule proteins may serve purposes beyond immune defense. Although granule proteins can cause pathogenic inflammation, particularly by promoting NET formation and vascular inflammation (see more below), different studies have reported pathogenic roles in other scenarios. In a model of tuberculosis (TB), neutrophil-derived MMP8 released from secondary granules caused destruction of the collagen matrix both in cultured cells and in TB lung biopsies from patients (79). AMPK signaling in neutrophils was needed for MMP8 secretion and matrix destruction (79). In COVID19 patients, a neutrophil activation signature associated with severity and mortality reveals the presence of distinct granule proteins, including MMP8 and lipocalin-2 (80)have shown that, and studies in mice show that neutrophil elastase (NE) and cathepsin G (CG) may facilitate viral and bacterial infections (81).
Neutrophils promote both protective and pathological responses is by attracting other immune cells to the site of infection or injury. This is mediated by the production and release of mediators important for cell signaling, including cytokines, chemoattractants and alarmins (82). Cytokines, which are preformed within neutrophils, can be released under homeostatic and inflammatory conditions, though in most cases they are induced upon stimulation, following previous induction via mRNA accumulation and de novo synthesis (74). The array of cytokines produced by human neutrophils is vast; it comprises anti-inflammatory (IL-1ra, TGFβ1, TGFβ2) and pro-inflammatory cytokines (IL-1α, IL-1β, IL-6, IL-8, TNFα), chemokines (CXCL1, CXCL2, CXCL8, CXCL12) and immunoregulatory cytokines (IL-22, IL-23) (74). Uncontrolled release of pro-inflammatory cytokines can exacerbate inflammation and cause immunopathologies. This anomalous response, referred to as “cytokine storm”, can be initiated after non-infectious and infectious signals, and are common during bacterial and viral respiratory infections. Accordingly, the predictive mortality signature of COVID19 patients includes a neutrophil activation signature where the pro-inflammatory cytokine IL-8 is a predictor of severity (80), as is in patients with Avian influenza A (H5N1) virus infection with poor survival (83). In mice, neutrophil depletion during abdominal sepsis abolished the cytokine storm (IL-1β, IL-6 and TNF-α) fueled via IL-3 production by the so-called innate response activator B cells (84).
3.3. Neutrophil extracellular traps
NETs are structures formed by DNA, nuclear proteins, including citrullinated histones, and proteins originally contained in granules or in the cytoplasm of neutrophils, and were originally proposed to contain and kill microbes (65). The most common mechanism of NET formation is through a process that requires ROS production by the NADPH oxidase, chromatin decondensation, protein assembly and membrane permeabilization (85). The molecular mechanisms regulating NET formation have been extensively reviewed (86) and, in brief, involve MPO activation through ROS, leading to release of enzymes from azurophilic granules to the cytoplasm, and subsequently to the nucleus, where they promote chromatin decondensation. ROS also induce the enzyme peptidyl arginine deiminase 4 (PAD4), which induces histone citrullination, a hallmark of NETosis along with the presence of MPO and NE. As the nuclear membrane disassembles, the DNA mixes with granule and cytoplasmic content, and finally the plasma membrane disintegrates resulting in the choreographed release of DNA into the environment (86). While this phenomenon is well characterized in vitro, the mechanism, location and consequences of NET formation in living tissues remain somewhat mysterious, since tools to specifically detect NETs in vivo have been difficult to generate (87). NETs form in a variety of situations (see below) and strongly associate with damage to the host, particularly to the cardiovascular system (88). It is therefore not surprising that NET formation is highly regulated. For instance, the capacity of neutrophils to form NETs follows tight circadian oscillations both in human and mice, at least under homeostatic conditions. Indeed, neutrophils are more prone to produce NETs right after release from the marrow, and this susceptibility is progressively lost over time. This reduction is caused by the natural and progressive loss of granules and their content in the circulation, a “disarming” process that relies on the circadian factor Bmal1 and signaling through CXCR2 (31).
The lytic event of NETosis with consequent release of DNA and associated enzymes and nuclear proteins triggers an inflammatory response that irreversibly damage host tissues, and indeed NETs have been involved in many types of inflammatory disorders including psoriasis, gout, vasculitis, thrombosis, chronic lung diseases, cancer, atherosclerosis, lupus and diabetic wound healing (89). This is further illustrated in the context of experimental sepsis in mice, during which neutrophils migrate to the liver sinusoids (90, 91) or pulmonary capillaries (91) and produce NETs that may be important to trap bacteria and diminish systemic dissemination, but cause severe damage (90, 91). Accordingly, studies in mice and humans showed that preventing NET formation, through inhibition of the NET-inducer protein gasdermin D, reduces organ injury and death associated with sepsis (92). Similar results have been found in models of transfused related acute lung injury, in which NETs formation in lungs causes endothelial damage and organ dysfunction, both in human and mice (93, 94).
Although NETs are typically considered effective to combat and limit infections, they also contribute to pathologies associated with microbial infections. Indeed, while NETs kill large pathogens such as fungal hyphae (95), dysregulated control of NET formation can exacerbate organ damage if too much NETs are produced, as shown in Dectin-1 deficient mice in which excessive NET formation during fungal infection was harmful for the lung tissue (95). Other illustrative examples come from studies of genetic deficiencies associated with impaired NET formation. Patients with chronic granulomatous disease, which exhibit poor capacity to form NETs, are more susceptible to infections, and restoration of NET formation by gene therapy improves the response to Aspergillus nidulans (96). Interestingly, however, mice lacking PAD4 (which also controls NET formation) manifest lower fungal burden in lungs and associated injury during Aspergillus infection (97), suggesting that release of NETs can contribute to tissue damage and in some settings promote fungal outgrowth. The paradoxical susceptibility to infections induced by NETs also manifests in patients with cystic fibrosis, in which NET positively correlate with bacterial colonization, rather than elimination (98).
NETs are also produced during infections with a wide variety of viruses, including human immunodeficiency virus-1 (HIV), influenza A, encephalomyocarditis or respiratory syncital virus, among others, as well as against viral-derived molecules through mechanisms that typically involve neutrophil activation through Toll-like receptor signaling (99). Early studies demonstrated that, at least in vitro, NETs can physically capture viral particles to neutralize HIV infection (100) or directly inhibit influenza A replication (101). Nonetheless, there remains controversy since non-severe influenza A viral infection does not trigger NETosis in vivo (102). Further, LPS priming was shown to be needed for NET formation to capture myxoma virus in liver synusoids (103). This suggests that the production of antiviral NETs depends on the severity of infection and may be secondary to particular DAMP-PAMP interactions that are different from those seen during antibacterial and antifungal responses. As shown for bacterial and fungal infections, NETs produced after viral infections can also result in immunopathology. During severe influenza virus infection, excessive NET formation induces alveolar capillary damage, hemorrhagic lesions, and obstruction of the small airways in lungs from infected mice (104), that ultimately can favor bacterial outgrowth and be detrimental. Accordingly, studies have established a positive correlation between high levels of NETs and poor prognosis during severe influenza A infections (105).
The COVID19 pandemic has provided a particularly relevant context to understand how NETs propagate disease. The involvement of NETs in lung pathology was in fact the first evidence for the central role of neutrophils in coronavirus disease, and these structures were differentially detected in the serum of severe COVID-9 patients (106). NET aggregates and NET-rich thrombi were found in the lungs, hearts and kidneys of COVID19 patients, suggesting that NETs may promote systemic thrombosis and vascular occlusion (106). Endothelial activation by NETs can also compromise barrier function of different tissues such as the lung or the kidney and cause pulmonary edema, microthrombi and glomerular injury (67). These phenomena ultimately contribute to organ failure and may underlie the tight association between the abundance of NETs and clinical outcome (107).
In sum, while NETs are a fascinating and effective mechanism to contain pathogen spread, they can be extremely toxic and promote acute and chronic inflammation. Understanding the biological fundaments behind NETs, including not only their formation but also their clearance from tissues, will be paramount to effectively target the deleterious effects of neutrophils.
4. Neutrophils in sterile inflammation: cardiovascular and beyond
Neutrophils excel at antimicrobial defense because they combine active patrolling of the organism with a heavily cytotoxic armamentarium. Location in this context is critical, because timely and localized release of their cargo protects from microbial invasion. However, when stimulatory signals become systemic, or localize in damaged but otherwise uninfected organs, or the regulatory networks controlling neutrophil activity get faulty, the destructive activity of neutrophils can turn against host cells. Because vascular cells are often exposed to neutrophils, the cardiovascular system is a recurrent victim of neutrophil activation, as discussed in this section (Figure 2).
Figure 2. Pathogenic mediators of neutrophils.
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The effector functions of neutrophils may cause disease in different scenarios, from metabolic disorders and atherosclerosis to clonal hematopoiesis and organismal aging. For example, NETs have been shown to contribute to most, if not all, cardiovascular events and in chronic inflammation, whereas interactions with other cells have been reported mostly in the context of stroke, autoimmune disease and cancer. Please refer to the text for details about the multiple mechanisms by which neutrophils cause disease.
4.1. Cardiovascular disease
Atherosclerosis
Atherosclerosis arises from the accumulation of triglycerides, cholesterol, and other metabolites in the inner tunica intima of large vessels (108). It is the leading source of cardiovascular disease, which include ischemic events in the heart and brain, and therefore to much of the morbidity and mortality worldwide (108). Although the establishment of the atherosclerotic plaque is multifactorial (108), the number of neutrophils found in both human and mouse atherosclerotic lesions associates with poor prognosis (109, 110), thus providing evidence of their involvement in atherosclerosis.
During early stages of atherosclerosis, at the affected arterial wall, neutrophils adhere to the activated endothelium and infiltrate the intima (111). The patterns of neutrophil recruitment and adhesion differ in arteries and peripheral venules based on specific localization of chemoattractants, and has been shown to be under circadian control (recently reviewed in (111)). Neutrophil depletion in mice leads to reduced plaque size at early but not late stages of atherosclerosis (109), indicating early contribution to plaque formation. At these sites, neutrophils release granule proteins, cytokines or chemokines and ROS, all of which increase vascular permeability and favor secondary monocyte recruitment (111). For example, deposition of Cathepsin G on the arterial endothelium increased leukocyte adhesion, while its deficiency was sufficient to reduce arterial neutrophil adhesion and atherosclerotic plaque area (112). Similarly, interaction of neutrophils (and monocytes) with platelets also leads to delivery of platelet-derived chemokines, such as CCL5, on the atherosclerotic endothelium inside arteries, which further increases leukocyte adhesiveness and fuels disease progression (109). The presence of cholesterol crystals in early atherosclerotic lesions further promotes the release NETs, and their deposition on the affected endothelium has been reported to accelerate disease progression (113). Accordingly, preventing NET formation with a PAD4 inhibitor reduced atherosclerotic lesion size in mice (114). Mechanistically, NETs appear to function as platforms for the deposition of cytokines and granule proteins, which in turn activate endothelial cells, plasmacytoid dendritic cells (pDCs), platelets and macrophages (115). NETs also prime macrophages to release IL-1β, which amplifies immune cell recruitment in atherosclerotic plaques (113), and engages a feedback loop of disease progression. Neutrophils can also regulate hematopoiesis during atherosclerosis, boosting myeloid-biased hematopoiesis. For example, in conditions of sleep disruption, a population of pre-neutrophils in the bone marrow has been demonstrated to mediate myelopoiesis during atherosclerosis through CSF1 production in response to the neuropeptide hypocretin (116). This resulted in elevated leukocytes numbers in circulation and large atherosclerotic lesions (116).
In advanced stages of atherosclerosis, plaque destabilization leads to rupture of the fibrous cap, and neutrophils have been found to promote plaque instability and rupture (111). The abundance of neutrophils in the arterial intima correlate with plaque instability both in human and mice (110, 117), and this may be partly explained by the effects of NETs on structural cells of the arterial wall. Indeed, activated lesional smooth muscle cells attract neutrophils to the arterial lesion and instigate NET release. Extracellular histone 4 associated to NETs in turn binds and lyses smooth muscle cells thereby boosting plaque destabilization (117). Studies in endotoxemic and atherosclerotic mice similarly reported that leukotriene B4 (LTB4) production amplifies neutrophil recruitment increasing features of plaque destabilization (118).
In sum, atherosclerosis exemplifies the multiple stages at which neutrophils participate during localized sterile inflammation and highlight the relevance of neutrophil-driven death of host cells as a mechanism of disease.
Ischemic injury:
In addition to triggering thrombosis and ischemia, neutrophils contribute to the subsequent damage to the affected tissues, most notably the heart and brain. During myocardial infarction (MI), neutrophils are recruited to the ischemic tissue and initiate an aggressive inflammatory response. Infiltration of the infarcted myocardium follows circadian patterns with elevated infiltration at night in a model of permanent ischemia which correlates with more severe cardiac damage when ischemia is triggered at this time (119), suggesting coupling of the inflammatory response with the homeostatic dynamics of these cells in circulation. Indeed, CXCR2-dependent recruitment at night increased infarct size and cardiac dysfunction (119). Intriguingly, myocardial injury was more severe when infarction occurs in the morning in a model of ischemia-reperfusion (MI/R) (30), suggesting that the type of inflammation may in turn influence the immune dynamics. Humans also manifest circadian oscillations not only in onset but in infarct size, both of which are higher in the early morning (120).
How do neutrophils induce cardiac injury? ROS production and granule protein and cytokine release by neutrophils in the ischemic myocardium create a pro-inflammatory environment that directly affects myocyte viability and contractility, and impacts post-cardiac remodeling during MI (121). Release of the alarmins S100A8/A9 in the infarcted area and subsequent induction of IL-1β-mediated granulopoiesis has been shown to aggravate cardiac function after MI (9). NET release is another potential mechanism of injury; in MI patients, NET levels in coronary plasma correlate with the occurrence of adverse cardiac events and dysfunction (122) and, in mice, pharmacological or genetic targeting of NETosis reduced cardiac damage in a I/R surgical model, with improvement of contractile function (123).
Diversity in neutrophil populations has been also observed during MI. Recently, a marrow-originated late-stage neutrophil subset with high expression of SiglecF was found to appear at day 1 after MI and last up to day 4 in the permanent infarcted heart, probably due to its Myc-recovered signature that advantaged survival (124). Although the role of this population is still unclear, its late appearance could indicate a role for fibrosis and monocyte recruitment (124). In another study, a population of NLRP3-primed neutrophils were observed to be retained in the BM after MI (125). These reverse-migrating neutrophils that homed to the BM in a CXCR4-dependent manner were able to induce granulopoiesis by local production of IL-1β (125).
Beyond aggravating tissue damage, however, growing evidence suggests that neutrophils also promote cardiac repair by exerting pro-reparative and pro-angiogenic functions. Experimental neutrophil depletion demonstrated a beneficial role for neutrophils in the context of MI-induced healing (126). This was proposed to be mediated by neutrophil-driven polarization of macrophages towards a reparative phenotype characterized by increased expression of the phagocytic receptor MERTK, which enhanced clearance of apoptotic cells and improved cardiac remodeling (126). Neutrophils have also been shown to incite the production of vascular endothelial growth factor (VEGF)-A by macrophages through signaling that involved the receptor AnxA1, which was primarily derived from infiltrating neutrophils, and this positively impacted neovascularization and cardiac repair after MI (127).
The controversial data showing beneficial and detrimental roles for neutrophils during MI raise the possibility that these effects are mediated by different neutrophil subsets, which may be recruited at different times or by distinct inflammatory cues after ischemia. Understanding this potential dichotomy is of particular interest for future strategies aiming not only at inhibiting neutrophil function, but additionally potentiating the recruitment of beneficial subsets.
Within the myocardium, neutrophils can be also detrimental for the cardiac electrical system. Neutrophil counts positively correlate with cardiac arrhythmia in humans and mice (128). In a new model of spontaneous arrhythmia in mice, generated by combining hypokalemia with myocardial infarction, neutrophils increased ventricular tachycardia through lipocalin 2-mediated ROS production that caused cardiomyocyte damage. Interestingly, macrophages were protective in this model by removing dead cardiomyocyte and debris (128).
Stroke is another major clinical entity given its high prevalence, and shares mechanistic parallelism with myocardial ischemia, including its thrombotic etiology downstream of atherosclerotic plaque rupture (129). Studies addressing the recruitment dynamics of neutrophils to ischemic sites in a murine cerebral ischemia–reperfusion model revealed that neutrophil influx occurs within the first 24 hours and that neutrophils still represent the vast majority of immune cells in the ischemic hemisphere three days after reperfusion (130). Notably, depleting neutrophils or preventing their entry into the brain reduced cerebral injury and promoted functional recovery in hyperlipidemic mice (131), thereby revealing a prominent role for neutrophils during stroke. Neutrophils have been shown to aggravate cerebral damage via different mechanisms. Intravascular neutrophils recruited at early time points become primed upon interaction with activated platelets to boost entry and cerebral injury, and blocking these interactions by antibodies targeting the PSGL-1/P-selectin pair reduced infarct volumes in a permanent stroke model in mice (132). These interactions can be driven by different neutrophil-released proteins, such as cathelicidins (133), which enhance neutrophil and platelet activation, arterial thrombi formation and NET release (133) to worse stroke outcomes.
NETs also participate in stroke pathology. They are found throughout the brain of ischemic stroke patients, and elevated plasma NETs and high-mobility group box 1 (HMGB1) correlated with stroke outcomes (134). By studying mouse models, activated platelets were identified as a source of HMGB1, which promotes NET release during the acute phase. Consequently, targeting HMGB1, depletion of platelets or treatment with the neonatal NET-inhibitory factor (nNIF) improved stroke outcome (134). NET components, proteases and decondensed DNA all increase neuronal cell death in mice subjected to middle cerebral artery occlusion, a common model of stroke (135). NETs also impair cerebrovascular remodeling during stroke recovery by increasing blood-brain barrier breakdown and impairing neovascularization and vascular repair, ultimately hampering functional recovery (136).
But neutrophils can also protect the brain. At late stages after stroke a type of “alternative” neutrophil (typically referred to as N2, as they resemble alternatively polarized macrophages) appears in the infarcted brain (137). This type of polarization is mediated by activation of the peroxisome proliferator-activated receptor gamma (PPARγ) (137) or toll-like receptor 4 (TLR4) (138) signaling and associates with neuroprotection (137, 138). The mechanisms underlying this striking protective effect remain undefined but highlight the extreme functional heterogeneity of neutrophils even in the context of a single disease.
In sum, neutrophils are generally detrimental for cardiovascular health by promoting thrombosis or direct endothelial damage within vessels, and by dramatically compromising the integrity of tissues upon an infarction. The identification of multiple neutrophil subsets in the context of this family of diseases is proponent of the intriguing idea that beneficial neutrophils also exist that could be positively targeted to protect the host.
4.2. Chronic inflammation
When the initial response to infection or injury is not properly resolved, it can lead to persistent low-grade inflammation. This chronic inflammatory state can last for prolonged periods of time, from months to years, depending on the originating stimulus and the nature of the disrupted resolution program. Disorders associated with chronic inflammation affect a high percentage of the population, especially the elderly, and are a major cause of death worldwide (139). The contribution of neutrophils to chronic inflammation remains poorly defined, as discussed here for some of the most prevalent chronic disorders (Figure 2).
Cancer.
Many types of cancer are highly infiltrated by neutrophils, commonly referred to as tumor associated neutrophils (TANs). While anti-tumoral functions of neutrophils have been reported (reviewed in (36, 37)), particularly during early tumorigenesis, they are largely considered pro-tumoral cells in advanced cancer. In fact, their abundance in tumoral tissues has been shown to be the strongest predictor of adverse prognosis in a broad collection of human tumors (140).
Amplification of inflammation and tissue stress is a classical mechanism by which neutrophils are believed to incite carcinogenesis. In a model of experimental chemical carcinogenesis in the lungs, neutrophils contributed directly to neoplastic transformation by amplifying DNA damage in lung cells via ROS (141). In some RAS-driven cancers, neutrophils were found to release cytokines, proteins or factors that directly stimulated tumor cell proliferation, such as prostaglandin E 2 (PGE-2) (142) or NE (143). Similarly, the production of NETs also promotes tumorigenesis and participates in metastasis. For example, these structures act indirectly as safeguards and physically protect tumoral cells from defensive cells (144). Proteases bound to NETs also promote cleave of matrix proteins and appearance of binding sites that promote the “awakening” of dormant cancerous cells (63).
Another remarkable property associated with neutrophils in the context of cancer is the capacity to support vessel formation. Different mechanisms proposed include tumor infiltration of MMP9-expressing neutrophils that promoted VEGF activation or expression of the pro-angiogenic factor Bv8 that induced mobilization from bone marrow and local angiogenesis during murine pancreatic cancer (59). NETs can also increase vascularization of matrigel plugs in vivo (145) and show pro-angiogenic responses in classical in vitro matrigel tube formation assays after PMA (145). Furthermore, neutrophils can acquire potent immunosuppressive properties that hinder the anti-tumoral activity of other immune cells. Studies have associated these immunosuppressive properties with metabolic changes in neutrophils (146, 147). In mouse models of lung and colon carcinoma and pancreatic cancer, overexpression of the fatty acid transport protein 2 (FATP2) enables uptake of arachidonic acid and synthesis of PGE-2, resulting in the accumulation of immunosuppressive lipids (146). Consequently, blocking FATP2 reduced tumor progression and the capacity to suppress cytotoxic T cell responses (146). In mouse models of breast cancer (147), lung mesenchymal cells in the premetastatic niche repress lipase activity on neutrophils to favor the accumulation of lipids, which are delivered to tumoral cells to fuel their proliferative and metastatic capacities (147).
On the other hand, neutrophils can exert a variety of antitumoral functions, especially during the early stages of cancer (reviewed in depth in (37)). One clear example is the induction of DNA damage in cancer cells through neutrophil-released ROS (141). Elastase produced by mouse and human neutrophils displays remarkable and selective cytotoxic activity against cancer cells, and additionally elicits anti-tumoral activity of CD8 T cells (148). Neutrophils can also prevent tumor progression by orchestrating immune responses. In mouse models of lung cancer, neutrophils stimulated T cell proliferation and an IFNγ response (149). Similar findings were made in mouse and human sarcomas, where neutrophils promoted the activation of unconventional αβ T cells through IFN γ signaling (150) or suppressed the activation of pro-tumoral interleukin 17 (IL-17)–producing γδ T cells via induction of oxidative stress (151). This neutrophil activity is linked to the induction of antigen presentation and, in fact, a subset of neutrophils with antigen-presenting cell properties have been identified in early-stages of human lung cancer (152).
Overall, the function of neutrophils during cancer progression is a multi-layered process in which multiple factors contribute, including tumoral stage, cell-intrinsic features of the cancer cell, the particular immune landscapes of the TME, neutrophil reprogramming, location (circulating and tumor-associated) and species. Unravelling these relationships will allow the design of tailored strategies for cancer treatment.
Metabolic disease.
Obesity results in excessive accumulation of adipose tissue that meets with immunometabolic and functional transformation, and is therefore at the source of metabolic diseases, from type 2 diabetes mellitus (T2DM) to nonalcoholic fatty liver disease (NAFLD). In both cases, neutrophil infiltration and activation is a hallmark of disease (66).
T2DM patients feature elevated blood glucose levels and insulin resistance. Several potential mechanisms of neutrophil derangement contribute to adipose tissue inflammation (66), among which release of elastase has been directly associated with insulin resistance in diet-induced obese mice (153). Elastase degrades insulin receptor substrate 1 (Irs1) in hepatocytes after neutrophil infiltration of the mouse liver, leading to impaired insulin signaling, higher glucose production and insulin resistance. Genetic deletion of NE ceases obesity and improves glucose tolerance (153), and similar findings have been found in humans (66, 154). In addition to elastase, aggravation of vascular disease appears to be further propagated by NETs released in the circulation of obese mice (155).
Progression of NAFLDs from fatty liver to nonalcoholic steatohepatitis (NASH) to liver cirrhosis entails early accumulation of fat, followed by steatosis, inflammation and fibrosis, during most of which neutrophils have been causally associated. Studies performed in high fat diet (HFD)-fed mice showed that CXCL1-mediated infiltration of neutrophils in the liver promotes NASH through increase of neutrophil oxidative burst and activation of stress kinases (156). In line with these results, p38 MAP kinases are elevated in livers of obese NAFLD patients (157). Secreted granular proteins also contribute to acceleration of fatty liver damage, and neutrophil elastase promotes liver steatosis (154) while neutrophil-derived MPO accelerates hepatic fibrosis in response to HFD (158). The contribution of NETs is still controversial as these structures are found in the serum of NASH patients and mice, but inhibition of NET formation did not prevent disease progression despite reduced liver inflammation (159).
Autoimmune disease.
A hallmark of autoimmune disorders is the deregulated response of the immune system against self-antigens, which results in damage to host cells and tissues. A large body of studies has reported the participation of neutrophils in a broad range of autoimmune disorders, including rheumatoid arthritis (RA), SLE, type 1 diabetes mellitus (T1DM), primary Sjögren’s syndrome (pSS), multiple sclerosis (MS), crohn’s disease (CD), gout and inflammatory bowel disease (IBD). Tissue infiltration, release of effector molecules and pro-inflammatory cytokines, altered ROS production, autoantigen presentation and interaction with T or B cells are some of the multifaceted functions whereby neutrophils has been shown to contribute to the pathology of autoimmune disease (66). Particularly, during SLE, human neutrophils can extrude oxidized mitochondrial components (160), but also antimicrobial peptides and self-DNA (161) that act as interferogenic complexes, promoting chronic activation of plasmacytoid dendritic cells (pDCs) and triggering IFN-mediated autoimmune response (160, 161). A common mechanism in most scenarios, however, is the anomalous production of NETs. Enhanced NETosis has been observed in circulating and synovial neutrophils from RA patients (162). These NETs aggravate RA pathogenesis through externalization of citrullinated autoantigens that trigger the formation of autoantibodies against citrullinated protein antigens (ACPAs), in turn activating adaptive immune responses (162). More recently, studies have shown that fibroblast-like synoviocytes function as intermediaries that present citrullinated peptides from NETs to trigger adaptive responses (163). In lupus pathogenesis, a particular neutrophil subset (LDGs), was also reported to be endowed with the capacity to release NETs and to externalize different autoantigens (164), and neutrophils from lupus patients, were found to produce higher levels of NETs that could probably trigger a positive feedback loop to sustain inflammation (161).
Organismal aging.
With life expectancy increasing remarkably over the last decades, age has become the main risk factor for prevalent diseases in high-income countries. Aging remodels neutrophil biology such that chemotaxis, superoxide generation, phagocytosis and NET release are dysregulated (generally reduced) in elderly patients and in aged mice, which may explain higher susceptibility to infection and microbial dissemination (165). Vice versa, neutrophils from aged individuals can be involved in “inflammaging”. For example, during herpes viral infection excessive production of IL-17 by neutrophils in aged mice aggravates liver injury (166), and impaired neutrophil egress after resolution of inflammation can aggravate pulmonary injury, through altered expression of adhesion and chemotactic receptors in aged mice (167). Production of CXCL1 by senescent cells in aged mice also enhances neutrophil recruitment sustains inflammation and augments neutrophil reverse transmigration to induce remote organ damage (168). Neutrophils can also incite senescence in neighboring cells by damaging their telomeres via ROS (169), in turn affecting removal of apoptotic neutrophils and promoting a permanent state of low-grade inflammation. Thus, altered interactions of neutrophils with neighboring cells emerges as a common theme that promotes unresolved inflammation in aged environments.
Clonal hematopoiesis (CH).
Also associated with aging, accumulation of somatic mutations in hematopoietic progenitors drives CH, which precedes blood malignancies and promote cardiovascular disease (170). Although the role of neutrophils in CH is poorly understood, aberrant NET production has been observed associated with the JAK 2V617F mutation. Mice with conditional expression of JAK 2V617F have increased expression of PAD4 which predisposes for NET formation and associated thrombosis (171), a finding also validated in a large cohort of humans bearing the JAK2 V617F mutation (171). Other prevalent CH mutations, such as those affecting TET2 or DNMT3A, have not yet been shown to affect neutrophils, however their association with many forms of cardiovascular disease and autoimmunity suggests general alterations of the myeloid compartment, and possibility that merits further investigation.
5. Therapeutic neutrophils, a glimpse to the future
The substantial contribution of neutrophils to inflammatory disease has propelled a long-standing interest in targeting these cells for therapeutic benefit. However, this has been a frustrating path, which recognizes how poorly we still understand the biology of these cells. Indeed, inhibiting the pathogenic properties of neutrophils may cause unacceptable susceptibility to life-threatening infections. The emerging contribution of neutrophils to cancer has sparked renovated interest in targeting these cells through reprogramming rather than broad inhibition of their function. We now face the fascinating challenge of exploring the biology, plasticity and vulnerabilities of neutrophils to design new therapies, a topic that has been recently reviewed (172, 173). We close this review by discussing emerging therapeutic approaches that can harness the enormous potential of neutrophils in the clinic.
5.1. Modulating the neutrophil lifespan
Although neutrophils are already short-lived cells, strategies that promote neutrophil apoptosis have been amply exploited with promising results. The use of cyclin-dependent kinases (CDKs) inhibitors, specialized pro-resolving mediators (SPMs) or other effector molecules to interfere with transcription of the antiapoptotic protein Mcl-1 and expedite the resolution of inflammation has been long investigated in the context of different diseases (172) (Figure 3). During lung inflammation, the CDK9 inhibitor R-roscovitine decreased neutrophil-induced damage by reducing Mcl-1 levels and regulating transcription through phosphorylation of the RNA polymerase II in human neutrophils (174). In a mouse model of LPS-induced pleurisy, increase in Annexin A1 expression promoted resolution of neutrophilic inflammation via different cell apoptosis mechanisms, including inhibition of Mcl-1, ERK1/2, and NF-κB signaling (175). The PI3-kinase inhibitor LY294002 also increased the rate of apoptosis of human neutrophils and reduced the pro-survival effect of GM-CSF and TNF-α (176), which suggested potential translation to the clinic. The extrinsic apoptosis pathway is another potential target to blunt inflammation. Administration of TNF-related apoptosis-inducing ligand (TRAIL) induced neutrophil apoptosis in models of zymosan-induced peritonitis and LPS-induced lung injury (177). For discussion of other strategies to reduce the neutrophil lifespan the reader is directed to a focused review (172). Extending the lifetime or viability of neutrophils has also therapeutic advantages for immune-deficient individuals, such as those receiving blood transfusions (see below). For instance, combined targeting of several cell death pathways (caspases, lysosomal membrane permeabilization, oxidative environment and necroptosis) in combination with G-CSF (CLON-G) prolonged human and mouse neutrophil half-life in vitro for up to 5 days without evident loss of antimicrobial activity (178).
Figure 3. Strategies to target neutrophils.
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Many strategies have been used to interfere with the pathogenic function of neutrophils, from their production in and release from the bone marrow, to cytotoxic functions in tissues. Please refer to Section 5 for more details. Red and green boxes indicate strategies to block or to promote the associated process.
5.2. Targeting recruitment
The persistent recruitment of neutrophils at sites of inflammation can have deleterious consequences, as it can fuel chronic inflammation and irreversible tissue damage. However, full removal of apoptotic neutrophils can be counterproductive as they promote anti-inflammatory phenotypes in phagocytic macrophages (179). Therefore, blocking neutrophil recruitment to blunt inflammation should ideally target specific pathways or molecules that favor inflammation (Figure 3). While numerous mouse studies focused on blocking neutrophil adhesion, such as selectin and integrins, have shown benefits, these strategies have not translated successfully in the clinic (173). Human studies focused on the blockade of receptors involved in chemotaxis, such as CXCR1 and CXCR2, rendered more promising results. In fact, a wide variety of drugs targeting these molecules are been evaluated in clinical trials to improve the outcome of asthma, LPS-induced airway inflammation, ulcerative colitis, colon cancer, COPD or viral infections (180). CXCR2 antagonists were shown to be effective at diminishing neutrophil numbers and reduced inflammatory biomarkers in the sputum of patients with cystic fibrosis (181) or bronchiectasis (182). Importantly, these studies found no alterations in neutrophil mobilization from the marrow, phagocytosis or superoxide anion production against Escherichia coli (183). In patients on pump coronary artery bypass grafting, the use of reparixin, a CXCR1/2 antagonist, also reduced the proportion of circulating neutrophil after ischemia-reperfusion injury (184). Targeting other less intuitive targets, such as β1-adrenergic-receptor with metoprolol, was shown to decrease infarct size by impairing neutrophil-platelet interactions required for neutrophil recruitment (132, 185). Contrarily, when the number of neutrophils are limiting due to chemotherapy or recurrent infections, the use of G-CSF as a mobilization agent is prescribed to elevate production and mobilization of circulating neutrophils (173). Due to potential off-target effects of this cytokine in mobilizing immature and potential tumor-promoting neutrophils, however, other pathways such as the CXCR4 antagonist plerixafor, have becoming an attractive option (173).
5.3. Inhibiting NET formation
NETs have become a target of major interest for most conditions that involve neutrophils, and the reader is directed to dedicated reviews in the context of COVID19 infection and cancer (186, 187). Different strategies have been carried out to interrupt the NETs, such as preventing their formation by blocking the deiminase PAD4 or the pore forming protein Gasdermin D, or by promoting the elimination or cytotoxic function of its components, including elastase, DNA or histones (Figure 3). The use of inhibitors of peptidylarginine deiminases (PAD) to interfere with chromatin decondensation and prevent NET formation conferred in vivo protection in the context of systemic lupus (188) and atherosclerosis (114), and reduction of NET release in COVID19 patients (189). Despite the lack of specificity for NETs, deoxyribonucleases (DNase I) have been used to degrade the DNA backbone of NETs in the context of lupus (190) and transfusion-related acute lung injury (94). Histones associated to NETs are pro-thrombotic and highly cytotoxic (117) and therefore an appealing target within NETs. Blockade of NET-associated histone H4 prevented neutrophil-dependent lytic death of smooth muscle cells in the arterial wall and stabilized atherosclerotic lesions (117). Given the expanding list of NET-driven pathologies, targeting these structures will continue to be a major area of biomedical research, but we raise caution given the possible beneficial effects of NETs beyond microbial killing, for example in promoting resolution of inflammation (191).
5.4. Neutrophil transfer
In some instances, neutrophils can be used as a tool for therapeutic treatment. Neutrophil transfusions are considered in cases of neutropenia or defects of neutrophil activity against bacterial or fungal infections. However, this strategy has not gained sufficient attention in the clinic given the difficulty of obtaining large numbers of neutrophils, their short lifespan, the wide heterogeneity of these cells, and the challenges associated with ex vivo manipulation that typically cause cell activation and potential adverse effects. A possible alternative in these cases is ex vivo reprogramming of neutrophils towards a favoring phenotype (Figure 3). Ex vivo treatment of neutrophils with 4-phenylbutyrate (4-PBA), and inductor of peroxisome homeostasis, re-established neutrophil normal function and reduced atherosclerotic burden (192). Neutrophil-like particles have been also used as carriers of anti-inflammatory molecules; delivery of celastrol-loaded “neutrophil nanoparticles” (NNPs) to tumor-bearing mice showed specific accumulation at the tumor site, diminished off-target effects, and inhibited tumor growth and improved survival (193).
5.5. Reprogramming neutrophils
As discussed earlier, neutrophils are extremely plastic in terms of phenotype, transcription, and function, with properties that can even be antagonistic. Although still an incipient area of research, this plasticity has sparked enormous interest in the possibility to reprogram neutrophils for personalized medicine (Figure 3). For example, the demonstration that differential lipid handling could mediate the tumoral properties of neutrophils raised interest in inhibiting FATP2, a transporter that acts as intermediary of the protumoral transition in different cancer models. Accordingly, pharmacological inhibition of FATP2 improved mouse survival by specifically preventing the accumulation of pro-tumoral neutrophils (146). Using a different strategy, a study demonstrated that activation of tumoricidal activity in neutrophils could be achieved in living mice by delivery of a combination of drugs that promoted recruitment to the tumor (with TNF〈), cytotoxic activity (with an agonistic CD40 antibody) and specificity (with anti-tumoral antibodies). In mice, this treatment proved to be very efficient to blunt tumoral growth and metastatic seeding for a variety of cancer types (194). Care must be exerted, however, as improper reprogramming may aggravate disease, as illustrated in a model of low-dose endotoxin that induced proinflammatory polarization of neutrophils and exacerbated atherosclerosis through ROS-mediated damage (192).
Exciting new directions in this area involve targeting pre-existing mechanisms of natural reprogramming, such as those associated with circadian rhythms (195). The control of natural neutrophil activation by the circadian clock (30, 195) and the finding that neutrophils oscillate between states of high and low inflammatory potential, suggests that this may be a viable therapeutic approach (Figure 3). In macrophages, an agonist of the circadian repressor REV-ERB reduces proinflammatory phenotypes and attenuate cytokine production to improve endotoxin-induced cytokine responses, both in vivo and in vitro (196). Global examination of migratory factors controlled by the circadian clock further found leukocyte- and tissue-specific oscillations in pro-migratory factors both in mice and human. Genetic, myeloid-specific deletion of the core circadian gene Bmal1 affected the homeostatic circadian trafficking of neutrophils, which was further regulated by a defined set of receptors in the endothelium and neutrophils, such that blocking these receptors alleviated inflammation during systemic LPS challenge in mice (197). Extrinsic circadian cues also modulate neutrophil trafficking, and indeed, Bmal1 disruption in epithelial club cells limited the magnitude of pulmonary inflammation by modulating CXCL5 production in the lung (198). Strategies involving timed targeting of CCR2 demonstrated benefits in atherosclerosis when administered at night. These examples highlight the potential of exploiting the temporal physiology behind the functional and migratory reprograming of neutrophils for therapeutic benefit (199), including direct targeting of the recently reported “circadian timer” of neutrophils (30), a strategy that we are actively pursuing in our laboratory.
Acknowledgements
We thank all members of our laboratory and colleagues worldwide for discussion and contributions to the concepts summarized in this revision. Work in our lab is supported by grants R01AI165661 from NIH/NIAD, RTI2018-095497-B-I00 from Ministerio de Ciencia e Innovacion (MCIN), HR17_00527 from Fundación La Caixa and FET-OPEN (no. 861878) from the European Commission. A.A-C. and T.V. are supported by Fellowships from la Caixa Foundation (ID 100010434), with codes LCF/BQ/DR19/11740022 (A.A-C.) and LCF/BQ/DR21/11880022 (T.V.). The CNIC is supported by the MCIN and the Pro CNIC Foundation and is a Severo Ochoa Center of Excellence (CEX2020-001041-S).
Footnotes
Disclosure statement
The authors have no conflicts of interest to disclose.
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Abstract
1. Origin and diversity of neutrophils, from subsets to states
2. The reparative neutrophil
3. Anti-microbial neutrophils: cost to the host
4. Neutrophils in sterile inflammation: cardiovascular and beyond
5. Therapeutic neutrophils, a glimpse to the future
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5308 | https://www.bhu.ac.in/Content/Syllabus/Syllabus_3006312820200426110721.pdf | CHM 205: CHEMICAL BINDING TOPIC : Atomic Term Symbols Reference Books : (i) Pyshical Chemistry by: Donald A McQuarrie and J.D Simon (ii) Quantum Chemistry by: I.N. Levine (iii) Quantum Chemistry by: W.Kaufman ATOMIC TERM SYMBOLS AND RUSSEL-SAUNDERS COUPLING In an atom, an electron may have a spin angular momentum and an orbital angular momentum which may interact or couple to give a resultant angular momentum. The resultant angular momentum and energy of the system is expressed in term symbol. The term symbol for a particular atomic state is denoted as where S denotes the total spin quantum number and 2S+1 gives the multiplicity of the state, L is the total orbital angular momentum number and J gives the total angular quantum number. Like the symbols s, p, d, f, …….for l=0, 1, 2, 3, ….. the different values of L are designated as L=0(S), L=1(P), L=2(D), L=3(F) etc. Thus quantities are determined as follows:. a) When two electrons interact with each other, the total spin S will be S = , or , or , i.e. 1, 0 or -1 b) When two electrons with azimuthal quantum number l1, and l2 interact, L can have the values: L = l1 + l2, l1 + l2 –l, …. l1-l2 Where the modulus sign ‘II’ indicates t.he positive value, l1-l2 or l2-l1, is to be taken. c) The L and S values couple to give a series of J total angular quantum number values for all electrons. J = L+S, L+S-1, L+S-2, ……., L-S The interaction is called Russel-Saunders coupling. Coupling d) For a three-electron system, L is calculated by calculating L1 values for two electrons, which are then combined with the third value of l to get the resultant L vaues. S can be determineds as. S = , etc , i.e e) In general, multiplicity is 2S+1, except when L < S. In this case only one value of J is obtained. It may be noted that L-S coupling is applicable for atoms of light atomic mass. For atoms of heavy mass, J-J coupling occurs. In this case the li and si of each electron is combined to give the j values. Individual j values interact to give the total J for the atom. Problem 1: Write the term symbol for boron (1s22s22p1) and carbon (1s22s22p1) atoms. Solution: for closed and filled 1s and 2s shell, L=0 and S=0. The L and S values will be determined by 2p electrons only.` For this, l = 1, so that L = 1 S = 1/2 so that 2S+1 = 2 J = L + S, L + S – 1,….., L - S = , Term symbol for boron = 2P3/2, 2P1/2 For carbon: 1s22s22p2 For both p electrons, l = 1, so that L = 1 + 1, 1 + 1 – 1, 1 – 1 = 2, 1, 0 The corresponding symbols are D, P, S S = or , i.e 1, 0 Spin multiplicity = 3, 1 Now J = L + S, L + S – 1, ….. L - S For L = 2, S = 1, J = 2 + 1, 2 + 1 – 1, 2 – 1 = 3, 2, 1 For L = 2, S = 0, J = 2 For L = 1, S = 1 and J = 0 For L = 1, S = 0 and J = 1 For L = 0, S = 1 and J = 1 For L = 0, S = 0 and J = 0 J can have values 3, 2, 1 or 0 Thus term symbols are 3D3, 3D2, 3D1, 1D2, 3P0, 1P1, 3S1, 1S0, Problem 2: Determine the term symbol for , (i) L = 2, S = ½ (ii) L = 1, S = 1 (iii) L = 1, S = 3/2 Solution : (i) J = 5/2, 3/2 ; 2S + 1 = 2 ; terms 2D5/2, 2D3/2 (ii) J = 2, 1, 0 ; 2S + 1 = 2 ; terms 2P2, 3P1, 3P0 (iii) J = 5/2, 3/2, ½ ; 2S + 1 = 4 terms 4P5/2, 4P3/2, 4P1/2 Problem 3: determine the value of L, S, J and term symbols arising out of the coupling between an electron in p-orbital and othe in a d-orbital. ll = 1 and l2 = 2, therefore L = 3, 2, 1 s1 = s2 =1/2, therefore S = 1, 0 and 2S + 1 = 3, 1 Solution: For L = 3, S = 1, J = 4, 3, 2 ; term symbols are 3F4, 3F3, 3F2 For L = 3, S = 0, J = 3 ; term symbol is 3F3 For L = 2, S = 1, J = 3, 2, 1 ; term symbols are 3D3, 3D2, 3D1 For L = 2, S = 0, J = 2 ; term symbol is 1D2 For L = 1, S = 1, J = 2, 1, 0 ; term symbols are 3P2, 3P1, 3P0 For L = 1, S = 0, J = 1 ; term symbol is 1P1 Problem 4: determine the term symbol for the ground and first excited state of He L = l1 + l2 = 0 s1 = s2 =1/2 ; S = 1 (spin parallel), 0(spin paired) Solution: Ground state spins are paired i.e S = 0 and J = 0 ; term symbol : 1S0 Excited states (a) when spins are paired (S = 0), J = 0 ; term symbol : 1S0 (singlet) (b) when spins are parallel (S = 1), J = 1 ; term symbol : 3S1 (triplet) |
5309 | https://artofproblemsolving.com/wiki/index.php/Circumradius?srsltid=AfmBOoqU50W3NHywywTHYGKa1ZP3u2pUTGXNYS_hKo0JxQjuuMNAvOXD | Art of Problem Solving
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Circumradius
The circumradius of a cyclicpolygon is the radius of the circumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle.
Contents
[hide]
1 Formula for a Triangle
2 Proof
3 Formula for Circumradius
4 Circumradius, bisector and altitude
5 Euler's Theorem for a Triangle
6 Proof
7 Right triangles
7.1 Theorem
8 Equilateral triangles
9 If all three sides are known
10 If you know just one side and its opposite angle
11 See also
Formula for a Triangle
Let and denote the triangle's three sides and let denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply . This can be rewritten as .
Proof
We let , , , , and . We know that is a right angle because is the diameter. Also, because they both subtend arc . Therefore, by AA similarity, so we have or However, remember that . Substituting this in gives us and then simplifying to get and we are done.
Formula for Circumradius
Where is the circumradius, is the inradius, and , , and are the respective sides of the triangle and is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that .
But, if you don't know the inradius, you can find the area of the triangle by Heron’s Formula:
Circumradius, bisector and altitude
Circumradius and altitude are isogonals with respect bisector and vertex of triangle.
Euler's Theorem for a Triangle
Let have circumcenter and incenter .Then
Proof
See
Right triangles
The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle.
This results in a well-known theorem:
Theorem
The midpoint of the hypotenuse is equidistant from the vertices of the right triangle. The midpoint of the hypotenuse is the circumcenter of a right triangle.
Equilateral triangles
where is the length of a side of the triangle.
If all three sides are known
Which follows from the Heron's Formula and .
If you know just one side and its opposite angle
by the Law of Sines.
(Extended Law of Sines)
See also
Inradius
Semiperimeter
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5310 | https://www.hopkinsarthritis.org/patient-corner/drug-information/methotrexate/ | Skip to primary navigation
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Methotrexate Drug Information Sheet
What is Methotrexate?
Methotrexate is a drug used to treat rheumatoid arthritis (RA) and other inflammatory conditions. Nearly 60% of all rheumatoid arthritis patients are currently on or have been on methotrexate. Methotrexate is recommended as the first treatment for RA by the American College of Rheumatology.
How do I take it?
Methotrexate is taken ONCE A WEEK. You will choose a day of the week, for example Saturday. You will then take your methotrexate every Saturday. The pills are 2.5mg each and the usual dose is 3 to 10 pills. Follow your rheumatologist’s directions. Do not take more or less medicine than ordered. This medicine can be taken with or without food. You should also take 1mg of folate (folic acid) each day. Your rheumatologist will prescribe the folic acid. This will help prevent side effects.
What about side effects?
Methotrexate can cause mouth ulcers in a few patients, in the beginning of treatment. This should go away with time. Another possible side effect is nausea and vomiting. Methotrexate can also cause hair thinning or hair loss. In rare cases some people may develop lung problems.
Methotrexate can cause mild liver irritation. Please tell your doctor if you have a history of any alcohol abuse, hepatitis, yellow jaundice, or liver disease. While on methotrexate you must limit yourself to 2 alcoholic beverages per week. Blood work will be done every 4-12 weeks to check your liver function. This blood work will also include a complete blood count (CBC) since methotrexate can also cause a decrease in blood counts. This blood work is very important. It allows the rheumatologist to make timely changes to your dose of methotrexate if there is ever problem.
Methotrexate is known to cause birth defects in the children of both men and women taking this drug. If you are pregnant, considering having a child, or nursing, discuss this with your rheumatologist before beginning this medication. You must use an effective form of birth control while taking methotrexate and for at least 3 months after the methotrexate is stopped. Your rheumatologist or nurse can give you additional guidance.
What about other medications?
When you are taking methotrexate, it is very important that your doctors know if you are taking any other medicine. This includes prescription and non-prescription medicines as well as birth control pills, vitamins, and herbal supplements. Methotrexate can be taken with other medications. You should not take methotrexate while taking antibiotics containing trimethoprim-sulfa (Bactrim®, Sulfatrim®, etc…). If you are prescribed one of these medications for an infection, do not take your methotrexate that week.
What else should I know?
Always call your health care provider if you have any questions or concerns.
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5311 | http://clasesdeapoyonuevo.s3.amazonaws.com/soluciones_selectividad/4.3._campo_electrico._soluciones.pdf | 1 CAMPO ELÉCTRICO Modelo 2018. Pregunta 3A.- Considérese una carga puntual q = 5 nC situada en el centro de una esfera de radio R = 10 cm. Determine: a) El flujo del campo eléctrico a través de la superficie de la esfera. b) El trabajo que es necesario realizar para traer una carga de 2 nC desde el infinito hasta una distancia de 10 cm del centro de la esfera. Dato: Constante de Coulomb K=1/(4πεo) = 9·109 N m2 C ‒2. Solución. a. Según el teorema de Gauss, el flujo de líneas de campo que atraviesan una superpie es directamente proporcional al Alor absoluto carga que encierra dicha superficie e inversamente proporcional a la permitividad del vacío o constante dielectrica. o ε Q Φ = 2 1 2 12 9 o m N C 10 84 , 8 10 9 π 4 1 K π 4 1 ε − − − ⋅ = ⋅ ⋅ = = . m V 5 , 565 m C J 5 , 565 m N C 10 84 , 8 C 10 5 Φ 2 1 2 12 9 ⋅ = ⋅ = ⋅ ⋅ = − − − − b. El trabajo realizado para trasladar una carga dentro de un campo eléctrico viene dado por la expresión: V ∆ q W ⋅ − = Aplicando a las condiciones del problema: ( ) ( ) J 10 9 10 10 10 5 10 9 10 2 0 r q K q V V q V V q W 7 2 9 9 9 r cm 10 r o f − − − − ∞ = = ⋅ − = ⋅ ⋅ ⋅ ⋅ ⋅ − = − ⋅ ′ − = − ⋅ ′ − = − ⋅ ′ − = El signo negativo nos indica que el trabajo se esta realizando en contra el campo por medio de una fuerza exterior. Septiembre 2017. Pregunta 3A.- Dos cargas de +5 nC están separadas una distancia de 4 cm de acuerdo a la figura adjunta. Calcule: a) El campo eléctrico en el punto A y en el punto B creado por ambas cargas. b) El potencial eléctrico en el punto A y en el punto B, y el trabajo que hay que realizar sobre una carga de +3 nC para desplazarla desde el punto A al punto B. Dato: Constante de la Ley de Coulomb, K = 9·109 N m2 C‒2. Solución. a. Campo eléctrico en A. Aplicando el principio de superposición, el campo eléctrico en el punto A es la suma vectorial de los campos que generan cada una de las cargas. Como puede verse en la figura adjunta, y teniendo en cuenta que las cargas son iguales y están a igual distancia en direcciones simétricas, el campo eléctrico en el punto A es la suma de las componentes y E r de los campos eléctricos que crean cada una de las cargas ya que las componentes x E r se anulan entre ellas. En la figura inicial, puede observarse que las cargas y el punto A forman un triángulo equilátero, por lo que α = 60º. El módulo del campo eléctrico que crea cualquiera de las cargas en el punto A es: C N 125 28 04 , 0 10 5 10 9 r q K E E 2 9 9 2 = ⋅ ⋅ = = = − r En forma vectorial: j 60 sen 28125 i 60 cos 28125 j α sen E i α cos E E E E j 60 sen 28125 i 60 cos 28125 j α sen E i α cos E E E E 2 2 1 1 y x 2 y x 1 r r r r r r r r r r r r r r ⋅ + ⋅ − = ⋅ + ⋅ − = + = ⋅ + ⋅ = ⋅ + ⋅ = + = 2 El campo resultante es la suma vectorial de los campos que crea cada una de las cargas C N j 714 48 j 60 sen 28125 2 E E E 2 1 A r r r r r = ⋅ ⋅ = + = Campo eléctrico en B. Aplicando el principio de superposición, el campo en el punto B es la suma de los campos que crean cada una de las cargas. Por geometría, los campos se anulan entre si resultando que el campo eléctrico total en B es nulo. No es necesario, pero si se quiere se puede hacer el calculo numérico y de esa forma no tener que dar explicaciones. Módulo del campo eléctrico creado por una cualquiera de las cargas en el punto B: C N 500 112 02 , 0 10 5 10 9 r q K E E 2 9 9 2 = ⋅ ⋅ = = = − r Vectores: 0 E E E : i 500 112 i E E E i 500 112 i E E E 2 1 B x 2 x 1 2 1 = + = − = − = = = = = r r r r r r r r r r r b. Aplicando el principio de superposición y teniendo en cuenta que el potencial (V) es una magnitud escalar: v 2250 04 , 0 10 5 10 9 2 r q K 2 r r r q q q r q K r q K V V V 9 9 A A A 2 A 1 2 1 A 2 2 A 1 1 2 1 A = ⋅ ⋅ ⋅ = ⋅ = = = = = = ⋅ + ⋅ = + = − v 4500 02 , 0 10 5 10 9 2 r q K 2 r r r q q q r q K r q K V V V 9 9 B B B 2 B 1 2 1 B 2 2 B 1 1 2 1 B = ⋅ ⋅ ⋅ = ⋅ = = = = = = ⋅ + ⋅ = + = − El trabajo para desplazar una carga Q desde A a B es: ( ) ( ) J 10 75 , 6 2250 4500 10 3 V V Q V ∆ Q W 6 9 A B B A − − → ⋅ − = − ⋅ ⋅ − = − ⋅ − = ⋅ − = El signo negativo indica que el trabajo hay que hacerlo en contra del campo, lo cuál confirma que las cargas positivas tienden a desplazarse hacia regiones de menor potencial, por lo que si queremos desplazarla hacia una zona de mayor potencial deberemos realizar un trabajo en contra del campo Septiembre 2016. Pregunta 3B.- Dos esferas pequeñas tienen carga positiva. Cuando se encuentran separadas una distancia de 10 cm, existe una fuerza repulsiva entre ellas de 0,20 N. Calcule la carga de cada esfera y el campo eléctrico creado en el punto medio del segmento que las une si: a) Las cargas son iguales y positivas. b) Una esfera tiene cuatro veces más carga que la otra. Dato: Constante de la Ley de Coulomb, K = 9·109 N m2 C‒2. Solución. a. La fuerza entre cargas viene descrita por la Ley de Coulomb: u r Q Q K F 2 r r ′ ⋅ = El módulo de la fuerza es: { } 2 2 2 r Q K Q Q r Q Q K F = ′ = = ′ ⋅ = C 10 71 , 4 10 9 1 , 0 20 , 0 K r F Q 7 9 2 2 − × = × ⋅ = ⋅ = Campo eléctrico ( ) E r . Teniendo en cuenta que las cargas son iguales y el carácter vectorial de la magnitud, el campo eléctrico en el punto medio del segmento que une las cargas es nulo. 3 ( ) ( ) { } ( ) ( ) 0 i 2 r Q K i 2 r Q K Q Q i 2 r Q K i 2 r Q K E E E 2 2 2 2 = ⋅ − ⋅ = ′ = = ′ ⋅ − ⋅ = ′ + = r r r r r r r b. Aplicando la ley de Coulomb en módulo como en el apartado anterior: { } 2 2 2 r Q 4 K Q 4 Q r Q Q K F ′ = ′ = = ′ ⋅ = C 10 36 , 2 10 9 4 1 , 0 20 , 0 K 4 r F Q 7 9 2 2 − × = × ⋅ ⋅ = ⋅ = ′ ⇒ C 10 44 , 9 10 36 , 2 4 Q 7 7 − − × = × ⋅ = ( ) ( ) { } ( ) ( ) ( ) i 2 r Q 3 K i 2 r Q K i 2 r Q 4 K Q Q i 2 r Q K i 2 r Q K E E E 2 2 2 2 2 r r r r r r r r ′ ⋅ = ′ ⋅ − ′ ⋅ = ′ = = ′ ⋅ − ⋅ = ′ + = ( ) C N i 10 55 , 2 i 2 1 , 0 10 36 , 2 3 10 9 E 6 2 7 9 r r r × = × ⋅ ⋅ × = − Junio 2016. Pregunta 3A.- Dos cargas puntuales, q1 = 3 µC y q2 = 9 µC, se encuentran situadas en los puntos (0,0) cm y (8,0) cm. Determine: a) El potencial electrostático en el punto (8,6) cm. b) El punto del eje X, entre las dos cargas, en el que la intensidad del campo eléctrico es nula. Dato: Constante de la Ley de Coulomb, K = 9·109 N m2 C‒2. Solución. a. Según el principio de superposición, el potencial eléctrico creado por las cargas q1 y q2 en el punto C es: 2 2 1 1 i i r q K r q K r q K V ⋅ + ⋅ = ⋅ = ∑ v 10 62 , 1 06 , 0 10 9 1 , 0 10 3 10 9 r q r q K V 6 6 6 9 2 2 1 1 ⋅ = ⋅ + ⋅ ⋅ ⋅ = + ⋅ = − − b. Se busca el punto donde se cumpla: 0 E E 2 1 = + r r 2 1 E E r r − = 2 1 E E r r = Aplicando la ley de Coulomb: 2 2 2 2 1 1 r q K r q K ⋅ = ⋅ ( )2 2 2 1 x 8 q x q − = ( ) 1 2 2 2 q q x x 8 = − 3 10 3 10 9 q q x x 8 6 6 1 2 = ⋅ ⋅ = = − − − Despejando: x = 2,93 cm Modelo 2016. Pregunta 3A.- Una carga puntual, q = 3 µC, se encuentra situada en el origen de coordenadas, tal y como se muestra en la figura. Una segunda carga q1 = 1 µC se encuentra inicialmente en el punto P1(1,0) m y, recorriendo la espiral de la figura, llega al punto P2(0,2) m. Determine: a) La diferencia de potencial entre los puntos P1 y P2. b) El trabajo realizado para llevar la carga q1 del punto P1 al P2. Datos: Constante de la Ley de Coulomb; K = 9·109 N m2 C‒2 Solución. a. En este apartado se debe entender que se pide la diferencia de potencial entre los puntos P1 y P2 es debido a la presencia de la carga q. v 13500 2 1 1 1 10 3 10 9 r 1 r 1 Kq r q K r q K V V 6 9 2 1 2 1 P P 2 1 = − ⋅ × ⋅ × = − ⋅ = ⋅ − ⋅ = − − 4 b. El trabajo realizado por una carga al desplazarse en el seno de un campo magnético, solo depende de las posiciones iniciales y finales de la carga, y no del camino recorrido. ( ) 2 1 1 P P V V q W 2 1 − ⋅ = → 0 J 0135 , 0 500 13 10 1 W 6 P P 2 1 > = ⋅ × = − → El trabajo lo realiza el campo sobre la carga Septiembre 2015. Pregunta 3A.- Tres cargas iguales, cada una de 1µC, están situadas en los vértices de un triángulo equilátero de 10 cm de lado. Calcule: a) La energía potencial electrostática de cualquiera de las cargas. b) El potencial eléctrico en el punto medio de cualquier lado. Dato: Constante de la Ley de Coulomb, K = 9·109 N m2 C‒2. Solución. a. La energía potencial electrostática (U) de una de las cargas es la carga por el potencial que generan las otras dos en el punto donde esta. J 18 , 0 1 , 0 10 1 1 , 0 10 1 10 9 10 1 l q l q K q r q K r q K q V q U 6 6 9 6 3 2 1 l r 3 2 1 1 = ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ = + ⋅ = + ⋅ = ⋅ = − − − = b. Para calcular el potencial en el punto P, se calcula la distancia d mediante el teorema de Pitágoras 0075 , 0 05 , 0 1 , 0 d 2 2 = − = v 10 64 , 4 0075 , 0 10 05 , 0 10 05 , 0 10 10 9 r q K r q K r q K V V 5 6 6 6 9 3 3 2 2 1 1 i P ⋅ = + + ⋅ = + + = = − − − ∑ Junio 2015. Pregunta 3B.- Dos cargas de 2 nC se sitúan en los vértices de la base de un triángulo equilátero de lado 2 cm que se encuentra situada sobre el eje de abscisas. El punto medio de la base está en el origen de coordenadas y el vértice superior en el semieje positivo de ordenadas. Determine: a) El campo eléctrico y el potencial eléctrico creado por las cargas en el vértice libre. b) La fuerza que las cargas positivas ejercerían sobre una carga de -2 nC situada en el vértice libre del triangulo. Dato: Constante de la Ley de Coulomb, K = 9·109 N m2 C‒2. Solución. a. El campo eléctrico en el punto C se obtiene como suma vectorial de los campos que generan las cargas situadas en los puntos A y B. Por ser las cargas de igual intensidad y estar situadas a igual distancia, los módulos de los campos creados por ambas serán iguales. ( ) C N 45000 10 2 10 2 10 9 d q K E E E 2 2 9 9 2 B A = × × ⋅ × = ⋅ = = = − − r r La disposición de las cargas y el punto, determinan ángulos de 60º para los vectores campo (triángulo equilátero). C N j 3 , 77942 j 3 45000 E E E j 2 3 45000 i 2 1 45000 j º 60 sen E i º 60 cos E E j 2 3 45000 i 2 1 45000 j º 60 sen E i º 60 cos E E B A B A r r r r r r r r r r r r r r r = = + = ⋅ + ⋅ − = ⋅ + ⋅ − = ⋅ + ⋅ = ⋅ + ⋅ = El potencial que generan las dos cargas en el punto C es la suma escalar de los potenciales que crean cada una de las cargas en el punto. v 1800 10 2 10 2 10 9 10 2 10 2 10 9 d q K d q K V V V 2 9 9 2 9 9 B A B A C = × × ⋅ × + × × ⋅ × = + ⋅ = + = − − − − b. Si sobre el punto C se coloca una carga, esta se vera sometida a una fuerza que viene dada por la expresión: 5 N j 10 56 , 1 j 10 3 9 C N j 3 45000 C 10 2 E q F 4 5 9 r r r r r − − − × − = × − = ⋅ ⋅ − = ⋅ = Modelo 2015. Pregunta 3A.- Tres cargas puntuales, q1 = 3 µC, q2 = 1 µC y una tercera carga desconocida q3, se encuentran en el vacío colocadas en los puntos A (0,0), B(3,0) y C(0,4), respectivamente. El potencial que crean las tres cargas en el punto P(3,4) es V=10650 V. Calcule, teniendo en cuenta que las coordenadas vienen dadas en metros: a) El valor de la carga q3. b) La fuerza que experimentaría una carga de ‒7 µC colocada en el punto P, debido a la presencia de las otras tres. Datos: Constante de la Ley de Coulomb, K = 9×109 N m2 C‒2 Solución. a. Dado el carácter escalar del potencial: + + = + + = + + = 3 3 2 2 1 1 3 3 2 2 1 1 3 2 1 P r q r q r q K r q K r q K r q K V V V V 5 4 3 r 2 2 1 = + = + × + × × = − − 3 q 4 10 1 5 10 3 10 9 10650 3 6 6 9 ; C µ 1 C 10 1 q 6 3 = × = − b. Lo mas sencillo en estos casos es calcular el campo eléctrico que crean las tres cargas (q1, q2 y q3) en el punto P, y a continuación calcular la fuerza que experimentaría una carga situada en ese punto ( ) E q F r r ⋅ = . El modulo del campo eléctrico viene expresado por: 2 d q K E ⋅ = Mediante triángulos se pueden determinar las razones trigonométricas del ángulo α: 5 4 4 3 4 r r α sen 2 2 1 2 = + = = 5 3 r r α cos 1 3 = = 3 2 1 T E E E E r r r r + + = • C N j 864 i 648 j 5 4 5 10 3 10 9 i 5 3 5 10 3 10 9 j α sen E i α cos E E 2 6 9 2 6 9 1 1 1 r r r r r r r + = ⋅ × × + ⋅ × × = ⋅ + ⋅ = − − • C N j 5 , 562 j 4 10 1 10 9 j E E 2 6 9 2 2 r r r r = × × = = − • C N i 1000 i 3 10 1 10 9 i E E 2 6 9 3 3 r r r r = × × = = − j 6 , 1426 i 1648 i 1000 j 5 , 562 j 864 i 648 ET r r r r r r r + = + + + = La fuerza que experimenta la carga es: ( ) j 10 99 , 9 i 10 15 , 1 j 6 , 1426 i 1648 10 7 E q F 3 2 6 r r r r r r − − − × − × − = + ⋅ × − = ⋅ = Septiembre 2014. Pregunta 3B.- En el plano XY se sitúan tres cargas puntuales iguales de 2 μC en los puntos P1(1,-1) mm, P2(-1, -1) mm y P3(-1, 1) mm. Determine el valor que debe tener una carga situada en P4 (1, 1) mm para que: a) El campo eléctrico se anule en el punto (0,0) mm. En esas condiciones, ¿cuál será el potencial eléctrico en dicho punto? b) El potencial eléctrico se anule en el punto (0,0) mm. En esas condiciones, ¿cuál será el vector de campo eléctrico en dicho punto? Dato: Constante de Coulomb, K=9×109 N m2C‒2 6 Solución. a. El campo eléctrico resultante en un punto debido a una distribución de cargas puntuales, es la suma vectorial de los campos eléctricos que crean cada una de las cargas en el punto. 3 2 1 E E E E r r r r + + = Para calcular el campo eléctrico en un punto, se supone en dicho punto la unidad de carga positiva, lo cual permite establecer la dirección y sentido de los campos que crean las diferentes cargas de la distribución. La geometría de la distribución y el hecho de que las cargas son iguales, permite observar en el esquema adjunto, que los campo creados por las cargas situadas en los puntos P1 y P2 en el origen de ordenadas, son iguales y de signos contrarios por lo que se anulan entre si. Por lo tanto el campo resultante debido a las tres cargas en dicho punto es 2 E r . 2 3 2 1 d Q K E E E E ⋅ = = = = r r r ( ) ( ) 3 2 3 2 3 10 2 10 1 10 1 d − − − × = × + × = ( ) C N 10 9 10 2 10 2 10 9 E 9 2 3 6 9 × = × × ⋅ × = − − 2 1 α cos α sen = = j 2 1 10 9 i 2 1 10 9 E E E E j 2 1 10 9 i 2 1 10 9 j α sen E i α cos E E j 2 1 10 9 i 2 1 10 9 j α sen E i α cos E E j 2 1 10 9 i 2 1 10 9 j α sen E i α cos E E 9 9 3 2 1 T 9 9 3 9 9 2 9 9 1 r r r r r r r r r r r r r r r r r r r r r × + × = + + = × − × = ⋅ − ⋅ = × + × = ⋅ + ⋅ = × + × − = ⋅ + ⋅ − = Para que el campo sea nulo, 0 E E 4 T = + r r T 4 E E r r − = Para que el campo creado en P4 tenga sentido contrario al resultante de los otros tres, la carga en el punto 4 deberá ser positiva. Si T 4 E E r r − = ⇒ T 4 E E r r = ( ) 9 2 9 2 9 2 3 4 9 10 9 2 10 9 2 10 9 10 2 Q 10 9 × = × + × = × ⋅ × − ; C µ 2 C 10 2 10 9 10 2 10 9 Q 6 9 6 9 4 = × = × × ⋅ × = − − Este apartado, se puede explicar por geometría sin necesidad de realizar cálculos, llegando a la conclusión que por simetría, los campo eléctricos que generan las cargas situadas en P1 y P3 en el origen se anulan, por lo tanto deberá pasar lo mismo entre las cargas situadas en P2 y P4 por lo que la carga en P4 deberá ser igual a la de P2. El potencial (escalar) creado por la distribución de cargas en el origen de coordenadas es la suma escalar de los potenciales que crea cada carga en el origen de coordenadas: d Q K 4 d d d d d Q Q Q Q Q d Q K d Q K d Q K d Q K d Q K V V 4 3 2 1 4 3 2 1 4 4 3 3 2 2 1 1 i i i ⋅ = = = = = = = = = = ⋅ + ⋅ + ⋅ + ⋅ = ⋅ = = ∑ ∑ v 10 09 , 5 10 2 10 2 10 9 4 V 7 3 6 9 × = × × ⋅ × ⋅ = − − b. 0 d Q K d Q K 3 d d d d d Q Q Q Q d Q K d Q K d Q K d Q K V 4 4 3 2 1 3 2 1 4 4 3 3 2 2 1 1 = ⋅ + ⋅ = = = = = = = = = ⋅ + ⋅ + ⋅ + ⋅ = d Q K 3 d Q K 4 ⋅ ⋅ − = ⋅ C µ 6 C 10 6 10 2 3 Q 3 Q 6 6 4 − = × − = × ⋅ − = − = − − 7 Campo eléctrico: 4 3 2 1 E E E E E r r r r r + + + = ( ) C N 10 27 10 2 10 6 10 9 d Q K E E 9 2 3 6 9 2 4 4 4 × = × × ⋅ × = ⋅ = = − − r Suponiendo la unidad de carga positiva en el centro, se puede establecer la dirección y sentido del campo que crean cada carga. j 10 2 18 i 10 2 18 j 2 10 36 i 2 10 36 E E E E E j 2 1 10 27 i 2 1 10 27 j α sen E i α cos E E j 2 1 10 9 i 2 1 10 9 j α sen E i α cos E E j 2 1 10 9 i 2 1 10 9 j α sen E i α cos E E j 2 1 10 9 i 2 1 10 9 j α sen E i α cos E E 9 9 9 9 4 3 2 1 9 9 4 4 4 9 9 3 9 9 2 9 9 1 r r r r r r r r r r r r r r r r r r r r r r r r r r r r r × + × = × + × = + + + = × + × = ⋅ + ⋅ = × − × = ⋅ − ⋅ = × + × = ⋅ + ⋅ = × + × − = ⋅ + ⋅ − = Junio 2014. Pregunta 3B.- Un electrón se propaga en el plano XY con velocidad vo constante de 100 m s‒1 en el sentido negativo del eje X. Cuando el electrón cruza el plano x = 0 se adentra en una región del espacio donde existe un campo eléctrico uniforme de 8×10‒9 N C‒1 en el sentido negativo del eje X, tal y como se indica en la figura. a) Describa el tipo de movimiento que seguirá el electrón una vez se haya introducido en esa región del espacio. Discuta cual será la velocidad final del electrón. b) Calcule la fuerza ejercida sobre el electrón así como la aceleración que éste experimenta. Datos: Masa del electrón, me = 9,1×10‒31 kg ; Valor absoluto de la carga del electrón, e = 1,60×10‒19 C Solución. a. Al entrar el electrón en la región donde existe el campo eléctrico, se vera sometido a una fuerza en sentido opuesto al campo que proporcionará al electrón una aceleración en el sentido de i r + (negativa) hasta que lo pare, una vez parado, la fuerza seguirá actuando haciendo que el electrón se acelere en el sentido positivo de i r hasta salir de la región donde existe el campo eléctrico con la misma velocidad que entro pero con sentido opuesto, debido al carácter conservativo del campo eléctrico. El electrón describe un movimiento rectilíneo uniformemente acelerado, siendo su velocidad final 1 s m i 100 v − = r r . b. E q F r r ⋅ = ( ) N i 10 28 , 1 C N i 10 8 C 10 6 , 1 F 27 1 9 19 r r r − − − − × = × − ⋅ × − = a m F r r ⋅ = 2 31 27 s m i 6 , 1406 kg 10 9,1 N i 10 28 , 1 m F a − − − = × × = = r r r r Modelo 2014. Pregunta 3A. El campo electrostático creado por una carga puntual q, situada en el origen de coordenadas, viene dado por la expresión: 1 r 2 C N u r 9 E − = r r , donde r se expresa en m y r u r es un vector unitario dirigido en la dirección radial. Si el trabajo realizado para llevar una carga q´ desde un punto A a otro B, que distan del origen 5 y 10 m, respectivamente, es de − 9×10‒6 J, determine: a) El valor de la carga puntual q que está situada en el origen de coordenadas. b) El valor de la carga q´ que se ha transportado desde A hasta B. Dato: Constante de la Ley de Coulomb, K = 9×109 N m2 C‒2 8 Solución. a. Según la ley de Coulomb, el campo eléctrico viene dado por la expresión: r 2 u r q K E r r ⋅ = Si se identifica con la expresión que se da en el enunciado r 2 r 2 u r 9 u r q K E r r r = ⋅ = Se puede obtener el valor de la carga que genera el campo eléctrico. 9 q K = ⋅ nC 1 C 10 10 9 9 K 9 q 9 9 = = × = = − b. El trabajo realizado para trasladar una carga ( ) q′ dentro de un campo eléctrico cuya intensidad varia con el radio, viene dado por la expresión: 10 q 9 5 1 10 1 q 9 r 1 q 9 dr r 1 q 9 dr r 9 q r d E q W 10 5 10 5 2 10 5 2 r r B A B A ′ = − − − ⋅ ′ = − ⋅ ′ = ⋅ ′ = ⋅ ′ = ⋅ ′ = ∫ ∫ ∫ → r o r Igualando al valor del trabajo del enunciado, se despeja el valor de la carga que se traslada por el campo eléctrico. 6 10 9 10 q 9 − × − = ′ C µ 10 C 10 10 q 6 − = × − = ′ − Septiembre 2013. Pregunta 5A.- Se tiene un plano infinito con una densidad de carga superficial positiva σ. a) Deduzca, utilizando el teorema de Gauss, el vector campo eléctrico generado por la distribución. b) Calcule la diferencia de potencial eléctrico entre dos puntos, en el mismo semiespacio, separados una distancia d en la dirección perpendicular al plano cargado. Justifique si cambiaría su respuesta si la dirección fuera paralela al plano cargado. Solución. a. Según el teorema de Gauss, el flujo neto a través de una superficie cerrada cualquiera es igual a la suma algebraica de las cargas eléctricas encerradas en su interior dividida entre la constante dieléctrica del vacío. o ε Q ∑ = Φ Para un plano infinito, se toma como superficie gaussiana un paralelepípedo recto como el que muestra la figura. Sólo hay flujo a través de las caras S y S’ paralelas al plano. Las líneas de campo siempre salen de las cargas positivas, por lo que el campo creado por el plano será uniforme. El flujo a través de las superficies laterales es nulo (ninguna línea de campo las atraviesan). Aplicando el teorema de Gauss: o S S ε Q S E 2 0 cos S E 0 cos S E S d E S d E = ⋅ = ⋅ ′ ⋅ + ⋅ ⋅ = + = Φ ∫ ∫ ′ r o r r o r Teniendo en cuenta la densidad superficial de carga ( ) S σ Q ⋅ = o ε S σ S E 2 ⋅ = ⋅ o ε 2 σ E = b. La diferencia de potencial entre dos puntos viene dado por la expresión: ( ) d ε 2 σ r r ε 2 σ dr ε 2 σ º 0 cos dr E r d E V V o A B o r r o r r r r A B B A B A B A ⋅ − = − − = − = ⋅ ⋅ − = ⋅ − = − ∫ ∫ ∫ r r Si la línea que une los puntos fuese paralela al plano, 0 r r A B = − , y la diferencia de potencial entre ellos seria cero 9 Junio 2013. Pregunta 1B.- Dos cargas puntuales q1 y q2 están situadas en el eje X separadas por una distancia de 20 cm y se repelen con una fuerza de 2 N. Si las suma de la dos cargas es igual a 6 µC, calcule: a) El valor de las cargas q1 y q2. b) El vector campo eléctrico en el punto medio de la recta que une las cargas. Dato: Constante de la ley de Coulomb, K = 9×109 N m2 C‒2. Solución. a. Por repelerse y sumar 6 µC, las cargas deben tener igual signo, y ser positivas. Aplicando la Ley de Coulomb: t 2 2 1 u d q q K F r r ⋅ = En módulo 2 2 1 d q q K F ⋅ = ( ) 2 2 2 1 9 10 20 q q 10 9 2 − × ⋅ × = 11 2 1 10 9 8 q q − = ⋅ ( ) 11 1 6 1 11 2 1 6 2 1 10 9 8 q 10 6 q : 10 9 8 q q 10 6 q q − − − − × = − × ⋅ × = ⋅ × = + × = × = = × + × − − − − − 6 1 6 1 11 1 6 2 1 10 3 8 q 10 3 10 q : 0 10 9 8 q 10 6 q Si se toma como C 10 3 8 q C 10 3 10 q 6 2 6 1 − − × = ⇒ × = b. El campo eléctrico en el punto medio del segmento que une las cargas es la suma vectorial de los campos generan cada una de las cargas. ( ) i q q d K i d q K i d q K E E E 2 1 2 2 2 2 1 2 1 r r r r r r − = − = + = ( ) C N i 10 6 i 10 3 8 10 3 10 10 10 10 9 E 5 6 6 2 2 9 r r r × = × − × × × = − − − Modelo 2013. Pregunta 3B.- Una esfera maciza no conductora, de radio R = 20 cm, está cargada uniformemente con una carga de Q = +1×10‒6 C. a) Utilice el teorema de Gauss para calcular el campo eléctrico en el punto r = 2R y determine el potencial eléctrico en dicha posición. b) Si se envía una partícula de masa m = 3×10‒12 kg, con la misma carga +Q y velocidad inicial vo = 1×105 m s‒1, dirigida al centro de la esfera, desde una posición muy lejana, determine la distancia del centro de la esfera a la que se parará dicha partícula. Datos: K = 9×109 N m2 C‒2 Solución. a. Teorema de Gauss. “El flujo neto que atraviesa una superficie cerrada cualquiera es igual a la suma algebraica de las cargas eléctricas encerradas en su interior dividida entre la constante eléctrica del vacío = Φ o ε Q ” o GAUSS .
T 2 S S S ε Q r π 4 E ds E ds E S d E = ⋅ = = ⋅ = = Φ ∫ ∫ ∫ r o r ( ) m V 56250 10 20 2 10 10 9 r Q K r Q πε 4 1 E 2 2 6 9 2 2 K o = × ⋅ ⋅ × = = ⋅ = − − 3 2 1 V 22500 10 20 2 10 10 9 r Q K V 2 6 9 = × ⋅ ⋅ × = ⋅ = − − El campo eléctrico se puede expresar en V/m o en N/m. 10 b. La energía cinética que tiene la carga en un punto alejado (infinito) se transforma en trabajo que realiza para aproximarse a otra carga de igual signo. ( ) ( ) 2 o 2 o 2 o 2 c c mv 2 1 V Q : V Q V V Q V Q W mv 2 1 v v m 2 1 E E W − = ⋅ − ⋅ − = − ⋅ − = ∆ ⋅ − = − = − = ∆ ∆ = ∞ 2 mv 2 1 r Q K Q = ⋅ ( ) ( ) cm 60 m 6 , 0 10 10 3 10 10 9 2 mv Q K 2 r 2 5 12 2 6 9 2 2 = = ⋅ × ⋅ × ⋅ = ⋅ = − − Septiembre 2012. Pregunta 3A.- Dos cargas puntuales q1 = 2 mC y q2 = ‒4 mC están colocadas en el plano XY en las posiciones (‒1,0) m y (3,0) m, respectivamente: a) Determine en que punto de la línea que une las cargas el potencial eléctrico es cero. b) Es nulo el campo eléctrico creado por las cargas en ese punto? Determine su valor si procede. Dato: Constante de la ley de Coulomb, K = 9×109 N m2 C‒2 Solución a. Se pide calcular la posición de punto A como indica la figura, de manera que el potencial creado por las dos cargas en el punto sea nulo. Teniendo en cuenta la definición de potencial en un punto = d q K V y su carácter escalar: ( ) ∑ = 0 A Vi ; 0 V V 2 1 = + ; 0 3 x q K 1 x q K 2 1 = − + + ; 0 3 x q 1 x q 2 1 = − − = + 3 x 10 4 1 x 10 2 3 3 − × − − = + × − − 2 1 x 3 x ± = + − Resolviendo una vez con cada signo, se obtienen dos posibles posiciones. 2 1 x 3 x = + − ( ) 0 , 5 A 5 x − ⇒ − = 2 1 x 3 x − = + − ′ ⇒ = 0 , 3 1 A 3 1 x b. Dado el carácter vectorial del campo eléctrico, y los sentidos de los campos creados por cada carga en el punto A, el campo eléctrico en él no es nulo. i d q d q K i d q K i d q K i E i E E E E 2 2 2 2 1 1 2 2 2 2 1 1 2 1 2 1 A r r r r r r r r + − = + − = + − = + = 1 5 2 3 2 3 9 A Nm i 10 63 , 5 i 3 5 10 4 1 5 10 2 10 9 E − − − × − = − − × + + − × − × = r r r 1 6 2 3 2 3 9 A Nm i 10 178 , 15 i 3 3 1 10 4 1 3 1 10 2 10 9 E − − − × = − × + + × × = r r r 11 Junio 2012. Pregunta 3A.-. Un electrón que se mueve .con una velocidad 1 6 ms i 10 2 v − ⋅ × = penetra en una región en la que existe un campo eléctrico uniforme. Debido a la acción del campo, la velocidad del electrón se anula cuando éste ha recorrido 90 cm. Calcule, despreciando los efectos de la fuerza gravitatoria. a) El modulo, la dirección y el sentido del campo eléctrico existente en dicha región b) El trabajo realizado por el campo eléctrico en el proceso de frenado del electrón. Datos: Masa del electrón, kg 10 11 , 9 m 31 e − × = ; Valor absoluto de la carga del electrón, C 10 60 , 1 e 19 − × = Solución. a. Cuando una carga eléctrica entre en una región donde existe un campo eléctrico, se ve sometida a una fuerza que es proporcional a la intensidad del campo y al valor de su carga. La dirección de la fuerza será paralela al campo eléctrico y el sentido será el mismo que el del campo si la carga es positiva y opuesto si es negativa. E q F r r ⋅ = La fuerza a la que se ve sometida la carga se puede calcular teniendo en cuenta el segundo principio de la dinámica( ) a m F r r ⋅ = y el tipo de movimiento que realiza la carga (M.R.U.A). + + = + = 2 o o o at 2 1 t v s s at v v : MRUA : ( ) o 2 o 2 s s a 2 v v − = − Teniendo en cuenta que so = v = 0: s a 2 v2 o = − s 2 v a 2 o − = i s 2 v a 2 o r r − = i s 2 v m F 2 o r r − = Sustituyendo en la expresión del campo eléctrico: i s 2 mv E q 2 o r r − = ⋅ ( ) ( ) C N i 65 , 12 j 10 60 , 1 9 , 0 2 10 2 10 11 , 9 i sq 2 mv E 19 2 6 31 2 o r r r r = × − ⋅ ⋅ × ⋅ × − = − = − − b. El trabajo realizado por el desplazamiento de una carga en el seno de un campo eléctrico viene dado por: ( ) B A B A V V q V q W − ⋅ = ∆ ⋅ − = → El incremento de potencial (∆V) se puede obtener de la relación existente entre campo y potencial eléctrico. dx dV E − = : dx E dV − = Teniendo en cuenta que el campo es uniforme: x · E V ∆ − = ∆ ( ) x · E · q x · E q W : x · E V V q W B A B A ∆ = ∆ − ⋅ − = ∆ − = ∆ ∆ ⋅ − = → → J 10 -1,82 m 9 , 0 C N 65 , 12 C 10 6 , 1 W 18 19 B A − − → × = ⋅ ⋅ × − = También se puede resolver teniendo en cuenta que la carga se esta desplazando por un campo conservativo, y que por tanto U W ∆ − = , siendo en este caso ∆U el incremento de energía potencial. Por ser conservativo, la energía mecánica permanece constante. c P m E E E + = 0 E E E c P m = ∆ + ∆ = ∆ P c E E ∆ − = ∆ A medida que aumenta la energía cinética, disminuye la energía potencial ( ) ( ) J 10 82 , 1 10 2 0 10 11 , 9 2 1 v v m 2 1 E W : E E E W 18 2 6 31 2 o 2 c c P P − − × − = × − ⋅ × = − ⋅ = ∆ = ∆ = ∆ − ∆ − = 12 Modelo 2012. Pregunta 5A.- Se disponen tres cargas eléctricas puntuales en los vértices de un triángulo rectángulo cuyos catetos tienen una longitud L como indica la figura (L = 1,2 m, q1= q2 = 5 nC, q3= −5 nC). a) Calcule la fuerza total, F r , ejercida por las cargas q1 y q2 sobre la carga q3 , y dibuje el diagrama de fuerzas de la carga q3. b) ¿Cuál sería el trabajo necesario para llevar la carga q3 desde su posición actual al punto P de coordenadas x = 1,2 m, y = 1,2 m? Dato: Constante de la ley de Coulomb K = 9×109 N m2 C‒2. Solución. a. El módulo de la fuerza entre dos cargas se calcula mediante la Ley de Coulomb: 2 d q q K F ′ ⋅ = La dirección, la línea que une las cargas, y el sentido, por ser de signos contrarios, de atracción. N 10 56 , 1 2 , 1 10 5 10 5 10 9 L q q K F 7 2 9 9 9 2 3 1 1 − − − × = × ⋅ × × = ⋅ = N j 10 56 , 1 F 7 1 r r − × − = ( ) N 10 81 , 7 2 , 1 2 10 5 10 5 10 9 L q q K F 8 2 9 9 9 2 3 2 2 − − − × = × ⋅ × × = ⋅ = Por ser un triángulo isósceles rectángulo, el ángulo en valor absoluto es de 45º N j 45º sen 10 81 , 7 i 45º sen 10 81 , 7 F 8 8 2 r r r − − × − × = La fuerza ( ) F r a la que se ve sometida la carga q3 es la suma vectorial de 1 F r y 2 F r . ( ) j 45º sen 10 81 , 7 i 45º sen 10 81 , 7 j 10 56 , 1 F F F 8 8 7 2 1 r r r r r r − − − × − × + × − = + = ( ) j 45º sen 10 81 , 7 10 56 , 1 i 45º sen 10 81 , 7 F 8 7 8 r r r − − − × − × − + × = N j 10 11 , 2 i 10 56 , 5 F 7 8 r r r − − × − × = ( ) ( ) N 10 18 , 2 10 11 , 2 10 56 , 5 F 7 2 7 2 8 − − − × = × − + × = º 75 10 56 , 5 10 11 , 2 arctg F F arctg α 8 7 x y − ≈ × × − = = − − b. Por ser un campo conservativo: ( ) 3 4 3 P V V q V ∆ q E ∆ W − ⋅ − = ⋅ − = − = Siendo V4 el potencial creado por q1 y q2 en el punto P y V3 el potencial creado por q1 y q2 en la posición inicial de q3. d q K V = L 2 q K L q K V 2 1 3 + = ; L q K L 2 q K V 2 1 4 + = Teniendo en cuenta que q1 = q2 ⇒ V3 = V4, por lo tanto ∆V = 0 y el trabajo es nulo. 13 0 0 q V ∆ q W 3 = ⋅ − = ⋅ − = Septiembre 2011. Problema 2B.- En el punto de coordenadas (0, 3) se encuentra situada una carga, q1 = 7,11×l0−9 C y en el punto de coordenadas (4, 0) se encuentra situada otra carga, q2 = 3,0×10−9 C. Las coordenadas están expresadas en metros. a) Calcule la expresión vectorial de la intensidad del campo eléctrico en el punto (4, 3). b) Calcule el valor del potencial eléctrico en el punto (4, 3). c) Indique el valor y el signo de la carga q3 que hay que situar en el origen para que el potencial eléctrico en el punto (4, 3) se anule. d) Indique el valor y el signo de la carga q4 que hay que situar en el origen de coordenadas para que la intensidad del campo en el punto de coordenadas (4, 3) sea 0. Dato: Constante de la ley de Coulomb K = 9×109 N m2 C−2 Aclaración: No es necesario, pero si se desea que en el punto (4, 3) el campo eléctrico en el apartado d) sea un cero exacto, hay que considerar el valor de q1 como un número periódico, q1= (64/9)×10−9 C. Solución. a. El campo eléctrico en el punto (4, 3) es la suma vectorial de los campos eléctricos que generan cada una de las cargas en ese punto. 2 1 T E E E r r r + = i 4 i 4 10 9 64 10 9 i r q K i E E 2 9 9 2 1 1 1 1 r r r r r = × ⋅ × = ⋅ = = − i 3 j 3 10 3 10 9 j r q K i E E 2 9 9 2 2 2 2 2 r r r r r = × ⋅ × = ⋅ = = − j 3 i 4 E E E 2 1 r r r r r + = + = b. El potencial eléctrico en el punto (4, 3) es la suma algebraica de los potenciales que generan cada una de las cargas en el punto 2 1 T V V V + = V 25 3 10 3 4 10 9 64 10 9 r q r q K r q K r q K V V 9 9 9 2 2 1 1 2 2 1 1 i T = × + × ⋅ × = + ⋅ = ⋅ + ⋅ = = − − ∑ c. 0 r q K r q K r q K V V 3 3 2 2 1 1 i T = ⋅ + ⋅ + ⋅ = =∑ Simplificando la constante. 0 r q r q r q 3 3 2 2 1 1 = + + ; 2 2 1 1 3 3 r q r q r q − − = ; + − = 2 2 1 1 3 3 r q r q r q C 10 9 125 3 10 3 4 10 9 64 5 q 9 9 9 3 − − − × − = × + × ⋅ − = d. Para que el campo eléctrico en el punto (4, 3) sea nulo, la carga 3 situada sobre el punto (0, 0) deberá generar un campo eléctrico de igual modulo y dirección que el generado por las cargas 1 y 2 pero de sentido opuesto. Para que el campo generado por la carga 2 este dirigido hacia el punto (0, 0), la carga ha de se negativa. 0 E E E E 3 2 1 T = + + = r r r r ( ) j 3 i 4 E E E E E 2 1 2 1 3 r r r r r r r − − = + − = − − = 14 ( ) ( ) 5 r q K : r q K E 5 3 4 E 2 3 3 2 3 3 3 2 2 3 = ⋅ ⋅ = = − + − = r r ; K r 5 q 2 3 3 − = 5 3 4 r 2 2 3 = + = C 10 9 125 10 9 5 5 q 9 9 2 3 − × − = × ⋅ − = Junio 2011. Problema 2B.- Considérese un conductor esférico de radio R = 10 cm, cargado con una carga q = 5 nC. a) Calcule el campo electrostático creado en los puntos situados a una distancia del centro de la esfera de 5 y 15 cm. b) ¿A qué potencial se encuentran los puntos situados a 10 cm del centro de la esfera? c) ¿Y los situados a 15 cm del centro de la esfera? d) ¿Qué trabajo es necesario realizar para traer una carga de 2 nC desde el infinito a una distancia de 10 cm del centro de la esfera? Dato: Constante de Coulomb K = 1/(4π εo) = 9×109 N m2 C2. Solución. a. El campo eléctrico creado por un conductor esférico se calcula mediante el teorema de Gauss. Teniendo en cuenta que la dirección del campo eléctrico es radial y que la dirección del campo es perpendicular a la superficie esférica y su módulo es constante en toda la superficie, el flujo a través de la superficie es: 2 S R π 4 E dS E dS E S d E ⋅ = = = • = Φ ∫ ∫ ∫ r r Según el teorema de Gauss el flujo es igual a la carga encerrada dividida por o ε . o ε q = Φ Igualando 2 o R π 4 E ε q ⋅ = ; 2 o R q ε π 4 1 E ⋅ = En un conductor esférico, la carga se encuentra en la superficie y por tanto la carga en el interior de una esfera de radio menor al de la esfera conductora es cero, por lo que el campo será nulo. • Para r = 5 cm: q = 0; E = 0 • Para r = 15 cm: q= 5×10‒9 C; C N 2000 15 , 0 10 5 10 9 E 2 9 9 = × ⋅ × = − b. Para r = 10 cm: V 450 1 , 0 10 5 10 9 R q K V 9 9 = × ⋅ × = ⋅ = − c. Para r = 15 cm: V 300 15 , 0 10 5 10 9 R q K V 9 9 = × ⋅ × = ⋅ = − d. Por ser un campo conservativo: ( ) ( ) ( ) ( ) J 10 9 0 450 10 2 r V 1 , 0 r V q V q E W 7 9 p − × − = − ⋅ × − = ∞ = − = ⋅ − = ∆ ⋅ − = ∆ − = Se deberá hacer un trabajo de 9×10‒7 J para trasladar la carga. Modelo 2011. Problema 2A.- Se disponen dos cargas eléctricas sobre el eje X: una de valor Q1 en la posición (1,0) y otra de valor Q2 en (‒1,0). Sabiendo que todas las coordenadas están expresadas en metros, determine en los casos siguientes: 15 a) Los valores de las cargas Q1 y Q2 para que el campo eléctrico en el punto (0,1) sea C / N j 10 2 E 5 r r × = siendo j r el vector unitario en el sentido positivo del eje Y. b) La relación entre las cargas Q1 y Q2 para que el potencial eléctrico en el punto (2,0) sea 0. Dato: Constante de la ley de Coulomb K = 9×109 N m2 C‒2. Solución a. El campo eléctrico generado por ambas cargas en el punto (1, 0) es la suma vectorial de los campos que generan cada una de las cargas en el punto. Las dos cargas han de ser iguales para que se anulen las componentes x de los campo que crean cada una de las cargas, además, deben ser positivas para que el campo resultante tenga la dirección y sentido de j r + . Teniendo en cuenta que las dos cargas tienen el mismo valor y están a igual distancia del punto, los módulos de los campos creados por ambas son iguales. ( ) Q 10 5 , 4 2 Q 10 9 E : 2 1 1 r r Q K E E E 9 2 9 2 2 2 2 1 × = ⋅ × = = + = ⋅ = = = j θ sen E 2 j θ sen E i θ cos E j θ sen E i θ cos E E E E E E E E y 2 x 2 y 1 x 1 2 1 T r r r r r r r r r r r ⋅ = ⋅ + ⋅ + ⋅ + ⋅ − = + + + = + = j 2 1 Q 10 5 , 4 2 j 10 2 9 5 r r ⋅ × ⋅ = × C 10 14 , 3 10 5 , 4 2 2 10 2 Q 5 9 5 − ⋅ = × ⋅ ⋅ × = b. El potencial en un punto debido a una distribución de cargas puntuales, es la suma algebraica de los potenciales que crean cada una de las cargas en el punto. 0 r Q K r Q K V V V V 2 2 1 1 2 1 i T = ⋅ + ⋅ = + = =∑ 0 r Q r Q 2 2 1 1 = + ; 3 1 r r Q Q 2 1 2 1 − = − = Septiembre 2010 F.G. Cuestión 2A.- Dos cargas puntuales iguales, de valor 2×10‒6 C, están situadas respectivamente en los puntos (0, 8) y (6,0). Si las coordenadas están expresadas en metros, determine: a) La intensidad del campo eléctrico en el origen de coordenadas (0, 0). b) El trabajo que es necesario realizar, para llevar una carga q = 3×10−6 C desde el punto P (3, 4), punto medio del segmento que une ambas cargas, hasta el origen de coordenadas. Dato: Constante de la ley de Coulomb K = 9×109 N m2 C2 Solución. C 10 2 q q 6 2 1 − × = = a. j E i E j E i E E E E 2 1 2 1 2 1 T r r r r r r r − − = − − = + = j 5 , 281 i 500 j 8 10 2 10 9 i 6 10 2 10 9 j r q K i r q K E 2 6 9 2 6 9 2 1 1 2 2 2 T r r r r r r r − − = × ⋅ × − × ⋅ × − = ⋅ − ⋅ − = − − ( ) ( ) C N 67 , 573 5 , 281 500 E E 2 2 T T = − + − = = r b. El trabajo necesario para desplazar una carga entre dos puntos de un campo eléctrico es el producto de la carga por la diferencia de potencial entre los dos puntos. 16 ( ) A B B A V V q W − ⋅ − = − = = = = = ⋅ + ⋅ = + = A .
2 A .
1 2 1 A .
2 2 A .
1 1 2 .
A 1 .
A A r r q q q r q K r q K V V V = × ⋅ × ⋅ = ⋅ = − C J V 7200 5 10 2 10 9 2 r q K 2 6 9 A .
1 1 { } = + ⋅ = = = = ⋅ + ⋅ = + = B .
2 B .
2 2 1 B .
2 2 B .
1 1 2 .
B 1 .
B B r 1 r 1 q K q q q r q K r q K V V V = + × ⋅ × = − C J V 5250 6 1 8 1 10 2 10 9 6 9 ( ) J 10 85 , 5 7200 5250 10 3 W 3 6 B A − − − × = − ⋅ × − = Junio 2010 F.M. Cuestión 2B.- a) Enuncie y exprese matemáticamente el teorema de Gauss. b) Deduzca la expresión del módulo del campo eléctrico creado por una lámina plana, infinita, uniformemente cargada con una densidad superficial de carga σ. Solución. a. Teorema de Gauss. El flujo neto que atraviesa una superficie cerrada cualquiera es igual a la suma algebraica de las cargas eléctricas encerradas en su interior dividida entre la constante dieléctrica del vacío. o S Q S d E ε = = φ ∫ r o r b. En un plano infinito de carga constante la superficie gaussiana elegida tiene forma de un paralelepípedo como el que muestra la figura, y por lo tanto habrá flujo a través de las superficies S y S’(S = S’) paralelas al plano cargado. Aplicando el teorema de Gauss y teniendo en cuenta que el campo es constante y paralelo al vector de superficie: o ' S S S S Q S E 2 ' S E S E dS E dS E dS E S d E ε = = + = + = = = φ ∫ ∫ ∫ ∫ r o r o o 2 S Q S 2 Q E ε σ = σ = = ε = Expresión de la que se deduce que el campo e un punto del plano cargado es independiente de la distancia. Junio 2010 F.G. Problema 2B.- Tres cargas puntuales q1 = +3 nC, q2 = −5 nC y q3 = +4 nC están situadas, respectivamente, en los puntos de coordenadas (0, 3), (4, 3) y (4, 0) del plano XY. Si las coordenadas están expresadas en metros, determine: a) La intensidad de campo eléctrico resultante en el origen de coordenadas. b) El potencial eléctrico en el origen de coordenadas. c) La fuerza ejercida sobre una carga q = 1 nC que se sitúa en el origen de coordenadas. d) La energía potencial electrostática del sistema formado por las tres cargas q1, q2 y q3. Dato. Constante de la ley de Colulomb K = 9 × 109 N m2 C−2 Solución. En una distribución de cargas puntuales, como la que se pide, lo primero que hay que hacer es suponer la carga unidad positiva en el punto donde se pide calcular el campo eléctrico y establecer los vectores de campo eléctrico que genera cada una de las cargas en ese punto. 17 a. La intensidad de campo eléctrico generado por la distribución de cargas en el origen de ordenadas es el módulo del campo eléctrico generado por ellas. ( ) ( ) ( ) ( ) ( )j E E i E E i E j E i E j E E E E E E 1 y 2 3 x 2 3 y 2 x 2 1 3 2 1 i r r r r r r r r r r r − + − = − + + + − = + + = = ∑ El módulo del campo eléctrico generado por una carga q a una distancia r viene dado por la expresión: 2 r q K E = Aplicando a la distribución propuesta: C N 3 3 10 3 10 9 r q K E 2 9 9 2 1 1 1 = × × = = − = ⋅ = α = = ⋅ = α = = × × = = − 25 27 5 3 5 9 sen E E 25 36 5 4 5 9 cos E E : C N 5 9 5 10 5 10 9 r q K E 2 y 2 2 x 2 2 9 9 2 2 2 2 C N 4 9 4 10 4 10 9 r q K E 2 9 9 2 3 3 3 = × × = = − Sustituyendo en la expresión del campo eléctrico: ( ) ( ) j 25 48 i 100 81 j 3 25 27 i 4 9 25 36 j E E i E E E 1 y 2 3 x 2 r r r r r r r − − = − + − = − + − = C N 1 , 2 20 193 3 400 1737 25 48 100 81 E 2 2 ≈ = = − + − = b. El potencial en un punto debido a una distribución de cargas es la suma (escalar) de los potenciales de cada carga en el punto. ( ) v 9 1 1 1 9 4 10 4 5 10 5 3 10 3 10 9 r q r q r q K r q K V 9 9 9 9 3 3 2 2 1 1 i i = + − ⋅ = × + × − + × ⋅ × = + + ⋅ = ⋅ = − − − ∑ c. j 10 92 , 1 i 10 1 , 8 j 25 48 i 100 81 10 1 E q F 9 10 9 r r r r r r − − − × − × − = − − × = ⋅ = ( ) ( ) N 10 1 , 2 10 92 , 1 10 1 , 8 F 9 2 9 2 10 − − − × ≈ × − + × − = d. ( ) ( ) J 10 2 , 7 4 4 5 5 4 3 4 5 3 10 1 10 9 r q q r q q r q q K U 8 18 9 3 .
2 3 2 3 .
1 3 1 2 .
1 2 1 − − × − = ⋅ − + ⋅ + − ⋅ × ⋅ × = ⋅ + ⋅ + ⋅ ⋅ = Modelo 2010. Problema 2A.- Se disponen dos cargas eléctricas sobre el eje X: una de valor Q1 en la posición (1,0), y otra de valor Q2 en (−1,0). Sabiendo que todas las coordenadas están expresadas en metros, determine en los dos casos siguientes: a) Los valores de las cargas Q1 y Q2 para que el campo eléctrico en el punto (0,1) sea el vector 5 10 2 E × = r j r N/C, siendo j r el vector unitario en el sentido positivo del eje Y. 18 b) La relación entre las cargas Q1 y Q2 para que el potencial eléctrico en el punto (2,0) sea cero. Datos: Constante de la ley de Coulomb k = 9 × 109 N m2 C−2 Solución. a. POR SIMETRÍA: Para que el campo resultante sea en la dirección del eje OY, las dos cargas han de ser de igual signo y módulo, por ser en la dirección positiva del eje, las cargas serán positivas. y 2 y 1 x 2 1x 2 1 T E E E E simetria Por E E E r r r r r r r + = − = = + = Por definición: = = = 45º sen 2 Q k E 45º sen 2 Q k E : r Q k E 2 2 y 2 2 1 y 1 2 { } j 45º sen 2 Q k 2 Q Q j 45º sen 2 Q k j 45º sen 2 Q k E E j E E 2 2 1 2 2 2 1 y 2 y 1 T T r r r r r r r = = = + = + = = 45 sen Q k 45º sen 2 Q k 2 E 2 T = = C 10 1 ' 3 2 2 10 9 10 2 45º sen k E Q 5 9 5 T − × = × × = = b. El potencial eléctrico en el punto P, es la suma escalar de los potenciales que genera cada una de las cargas en dicho punto. 0 V V V 2 1 P = + = Por definición: r Q k V = 2 2 1 1 2 1 P r Q k r Q k V V V + = + = 1 Q k 3 Q k 0 2 1 + = 1 Q k 3 Q k 2 1 − = 2 1 Q 3 Q − = Tienen que ser de distinto signo, y Q2 de triple valor que Q1. Septiembre 2009. Cuestión 4.- Una superficie esférica de radio R tiene una carga eléctrica Q distribuida uniformemente en ella. a) Deduzca la expresión del módulo del vector campo eléctrico en un punto situado en el exterior a dicha superficie haciendo uso del teorema de Gauss. b) ¿Cuál es la razón entre los módulos de los vectores campo eléctrico en dos puntos situados a las distancias del centro de la esfera r1 = 2 R y r2 = 3 R? Solución. a. Según la ley de Columb, el campo eléctrico creado por un carga Q en un punto de una superficie esférica vale 2 r Q K E = , y es perpendicular a la superficie. El flujo a través de esta superficie será: o 2 2 S 2 S ε Q r π 4 r Q K dS r Q K S d E = = = = Φ ∫ ∫ r o r Si aplicamos el teorema de Gauss a dicha superficie esférica, el flujo a través de la superficie será: 2 S S S r π 4 E dS E dS E dS E = = = = Φ ∫ ∫ ∫ o r 19 Teniendo en cuenta que el flujo según el teorema de Gauss es: o ε Q = Φ 2 o r π 4 E ε Q = : 2 2 o r Q K r Q ε π 4 1 E = = Resultado que es idéntico al encontrado por la Ley de Coulomb para una carga puntual. b. Según la expresión deducida en el apartado a: ( ) 2 2 1 1 R 4 Q K R 2 Q K E : R 2 r = = = : ( ) 2 2 2 2 R 9 Q K R 3 Q K E : R 3 r = = = Comparando los módulos de los campos eléctricos: 4 9 R 9 Q K R 4 Q K E E 2 2 2 1 = = : 4 9 E E 2 1 = Junio 2009. Problema 2A.- Dos cargas puntuales de −3 µC y +3 µC se encuentran situadas en el plano XY, en los puntos (−1,0) y (1,0) respectivamente. Determine el vector campo eléctrico: a) En el punto de coordenadas (10,0). b) En el punto de coordenadas (0,10). Nota: Todas las coordenadas están expresadas en metros. Dato: Constante de la ley de Coulomb K=9×109 N m2 C‒2 Solución. Según el principio de superposición, el campo eléctrico creado por una distribución de cargas puntuales en un punto de espacio, es la suma vectorial de los campos eléctricos creados por cada carga de la distribución en ese punto. ∑ = i T E E El campo eléctrico (E) creado por una carga en un punto viene dado por la expresión: r 2 u r Q K E = r Donde K es la constante eléctrica, Q es la carga (C), r es la distancia (m) y r u es un vector unitario en la dirección de la recta que une la carga al punto, y sentido hacia la carga si es negativa, y en sentido opuesto si es positiva. a. En este apartado la distribución de cargas y los campos creados por ambas en el punto O(10, 0), ofrece una geometría unidimensional, tal como muestra la figura. C N i 1 , 223 i 11 10 3 10 9 i r q K E 2 6 9 2 1 1 1 − = ⋅ − × = = − C N i 3 , 333 i 9 10 3 10 9 i r q K E 2 6 9 2 2 2 2 = ⋅ × = = − C N i 110,2 i 333,3 i 1 , 223 E E E E 2 1 i T = + − = + = = ∑ b. La figura representa la distribución de cargas y los campos creados por ambas en el punto O(0, 10). Para que la representación quede más clara se ha tomado distinta escala en los ejes. Como el valor absoluto de las cargas y las distancias que las separan al 20 punto O son iguales, el módulo del campo creado por ambas cargas en O también lo es. ( ) C N 7 , 267 101 10 3 10 9 E : r q K E 2 6 9 2 = × × = = − Como en el apartado a, teniendo en cuenta el principio de superposición: 2 1 i T E E E E + = = ∑ Teniendo en cuenta las componentes trigonométricas de α y el cuadrante de cada ángulo: ( ) ( ) j sen E i cos E j sen E i cos E j E i E E y 1 x 1 1 α − α − = α − ⋅ + α − ⋅ = + = ( ) j sen E i cos E j sen E i cos E j E i E E y 2 x 2 2 α + α − = α ⋅ + α − ⋅ = + = Sumando los vectores se obtiene el campo en O. C N i 3 , 53 i 101 1 7 , 266 2 i cos E 2 E E E 2 1 T − = ⋅ − = α − = + = Nota: Por simetría se podría haber determinado que las componentes “y” de los campos se anulaban entre si. Modelo 2009. Problema 1B.- En el plano x = 0 existe una distribución superficial infinita de carga cuya densidad superficial de carga es 2 6 1 m / C 10− + = σ a) Empleando el teorema de Gauss determine el campo eléctrico generado por esta distribución de carga en los puntos del espacio de coordenadas (1, 0, 0) y (−1, 0, 0). Una segunda distribución superficial infinita de carga de densidad superficial 2 σ se sitúa en el plano x = 3. b) Empleando el teorema de Gauss determine el valor de 2 σ para que el campo eléctrico resultante de ambas distribuciones superficiales de carga en el punto (−2, 0, 0) sea i 10 E 4 r r + = N/C Nota: Todas las coordenadas están expresadas en unidades del SI Dato: Permisividad eléctrica del vacío 2 1 2 12 0 m N C 10 85 .
8 ε − − − × = Solución. a. Teorema de Gauss. El flujo neto que atraviesa una superficie cerrada cualquiera es igual a la suma algebraica de las cargas eléctricas encerradas en su interior dividida entre la constante dieléctrica del vacío. o S Q S d E ε = = φ ∫ r o r En un plano infinito de carga constante la superficie gaussiana elegida tiene forma de un paralelepípedo como el que muestra la figura, y por lo tanto habrá flujo a través de las superficies S y S’(S = S’) paralelas al plano cargado. Aplicando el teorema de Gauss y teniendo en cuenta que el campo es constante y paralelo al vector de superficie: o ' S S S S Q S E 2 ' S E S E dS E dS E dS E S d E ε = = + = + = = = φ ∫ ∫ ∫ ∫ r o r o o 2 S Q S 2 Q E ε σ = σ = = ε = Expresión de la que se deduce que el campo e un punto del plano cargado es independiente de la distancia.. Aplicando al caso que se propone: C N 10 65 .
5 10 85 .
8 2 10 ε 2 σ E 4 12 6 o × = ⋅ ⋅ = = − − b. Según el principio de superposición, el campo en un punto es la suma vectorial de los campos generados por cada una de las distribuciones 21 2 1 E E E r r r + = ( ) C N i 10 65 , 6 i 10 64 .
5 i 10 E E E 4 4 4 1 2 r r r r r r × = × − − = − = Aplicando la expresión obtenida en el apartado a: ( ) 2 6 12 4 o 2 2 o 2 2 m C 10 18 , 1 10 85 .
8 10 65 , 6 2 ε E 2 σ ε 2 σ E − − × − = × ⋅ × ⋅ − = − = ⇒ = Septiembre 2008. Cuestión 3. Se disponen tres cargas de 10 nC en tres de los vértices de un cuadrado de 1 m de lado. Determine en el centro del cuadrado: a) El módulo, la dirección y el sentido del vector campo eléctrico. b) El potencial eléctrico. Dato: Constante de la ley de Coulomb K = 9 × 109 N m2 C−2 Solución. a. La representación del sistema puede hacerse en la orientación que más nos interese, en este caso como aparece en la figura adjunta. La distancia entre el centro y cualquier vértice (R) es: 2 2 2 d R R = + : 2 2 d R 2 = : 2 d R = El módulo del campo eléctrico creado por una carga a una distancia r es: 2 r q k E ⋅ = En el centro del cuadrado, por ser las tres cargas iguales y las distancias a los vértices iguales, los módulos de los tres campos son iguales: ( ) C N 180 2 1 10 10 10 9 R q k E E E 2 9 9 2 3 2 1 = × ⋅ × = ⋅ = = = − El vector intensidad del campo eléctrico total es la suma vectorial de los vectores intensidad de campo creados por cada una de las cargas. C N j 180 j 180 i 180 i 180 E E E E 3 2 1 T r r r r r r r r + = + + − = + + = b. El potencial eléctrico es la suma escalar de los potenciales creados por cada una de las cargas, que son también iguales: Voltios 3 , 127 2 1 10 10 10 9 r q k V V V 9 9 3 2 1 = × ⋅ × = ⋅ = = = − Voltios 9 , 381 3 , 127 3 V V V V 3 2 1 T = ⋅ = + + = Septiembre 2008. Problema 1B.- Una carga de +10 nC se distribuye homogéneamente en la región que delimitan dos esferas concéntricas de radios rl = 2 cm y r2 = 4 cm. Utilizando el teorema de Gauss, calcule: a) El módulo del campo eléctrico en un punto situado a 6 cm del centro de las esferas. b) El módulo del campo eléctrico en un punto situado a 1 cm del centro de las esferas. Dato: Permitividad eléctrica del vacío εo = 8,85×10−12 N−1 m−2 C2 . Solución. Teorema de Gauss: El flujo neto que atraviesa una superficie cerrada cualquiera es igual a la suma algebraica de las cargas eléctricas encerradas en su interior dividida entre la constante dieléctrica del vacío. Dada la simetría del sistema, en todos los puntos que equidistan del centro el campo eléctrico será radial y del mismo valor, por lo que interesa coger como superficie de integración una esfera centrada en el origen. 22 o 2 2 S 2 2 S S Q Q K 4 r 4 r Q K dS r Q K r Q K E Coulomb de Ley dS E 1 0 cos S d paralelo E S d E ε = π = π ⋅ = = = = = = = = Φ ∫ ∫ ∫ r r r o r a. El flujo a través de una superficie esférica es: 2 S S S r 4 E dS E dS E S d E π ⋅ = = = = Φ ∫ ∫ ∫ r o r Según el teorema de Gauss, el flujo tiene un valor: o Q ε = Φ De ambas expresiones se deduce: o 2 Q r 4 E ε = π ⋅ : 2 o r Q 4 1 E ε π = Utilizando los datos del enunciado: r = 0,06 m. C 10 10 q 9 encerrada − × = C N 24977 06 , 0 10 10 10 ,85 8 4 1 E 2 9 12 = × × π = − − b. El procedimiento es idéntico al del apartado anterior pero ahora la superficie gaussiana es una superficie esférica de 1 cm de radio, esta superficie no encierra ninguna carga y por tanto el flujo es nulo y el campo eléctrico también es nulo. r = 0,02 m 0 qencerrada = C N 0 06 , 0 0 10 ,85 8 4 1 E 2 12 = × π = − Junio 2008. Problema 1A.- Dos cargas fijas Q1 = +12,5 nC y Q2 = -2,7 nC se encuentran situadas en los puntos del plano XY de coordenadas (2,0) y (-2,0) respectivamente. Si todas las coordenadas están expresadas en metros, calcule: a) El potencial eléctrico que crean estas cargas en el punto A (-2,3). b) El campo eléctrico creado por Q1 y Q2 en el punto A. c) El trabajo necesario para trasladar un ión de carga negativa igual a -2e del punto A al punto B, siendo B (2,3), indicando si es a favor o en contra del campo. d) La aceleración que experimenta el ión cuando se encuentra en el punto A. Datos: Valor absoluto de la carga del electrón e = 1,6×10-19 C Constante de la ley de Coulomb K = 9×109 N m2 C -2 Masa del ión M = 3,15×10-26 kg Solución. a. Según el principio de superposición, el potencial en un punto del campo creado por varias cargas puntuales es la suma algebraica de los potenciales debidos a cada una de las cargas puntuales. Para calcular el potencial en un punto hay que tener en cuenta que es un escalar, depende de la carga que crea el campo, de la distancia del punto a la carga y el signo será el de la carga. 2 1 A V V V + = V 5 , 22 m 5 C 10 5 , 12 C Nm 10 9 r Q K V 9 2 2 9 1 1 1 = × × = = − − V 1 , 8 m 3 C 10 7 , 2 C Nm 10 9 r Q K V 9 2 2 9 2 2 2 − = × − × = = − − ( ) V 4 , 14 1 , 8 5 , 22 VA = − + = 23 b. El campo eléctrico creado por varias cargas puntuales en un punto, es la suma vectorial de los campos que creados por cada una de las cargas en ese punto. El módulo el campo eléctrico se puede obtener del potencial. r V E r E V : r Q K E r Q K V 2 = ⇒ ⋅ = = = r C N 5 , 4 m 5 V 5 , 22 r V E 1 1 1 = = = C N 7 , 2 m 3 V 1 , 8 r V E 2 2 2 = = = Las componentes vectoriales se obtienen de las razones trigonométricas de los ángulos que forman los vectores. ( ) j 7 , 2 i 6 , 3 j 5 3 i 5 4 5 , 4 j sen i cos E E 1 1 + − = + − ⋅ = α + α ⋅ = ( ) ( ) ( ) j 7 , 2 j i 0 7 , 2 j 1 i 0 E E 2 2 − = − ⋅ = − + ⋅ = ( ) ( ) i 6 , 3 j 7 , 2 j 7 , 2 i 6 , 3 E E E 2 1 T − = − + + − = + = C N 6 , 3 ET = c. ( ) A B B A V V q W − ⋅ − = → Potencial en B: B .
2 B .
1 B V V V + = V 5 , 37 m 3 C 10 5 , 12 C Nm 10 9 r Q K V 9 2 2 9 B .
1 1 B .
1 = × × = = − − V 86 , 4 m 5 C 10 7 , 2 C Nm 10 9 r Q K V 9 2 2 9 B .
2 2 B .
2 − = × − × = = − − ( ) V 6 , 32 86 , 4 5 , 37 VA = − + = Sustituyendo en la expresión del trabajo: ( ) ( ) J 10 82 , 5 4 , 14 6 , 32 10 6 , 1 2 W 19 19 B A − − → × + = − ⋅ × ⋅ − − = Por ser positivo el trabajo es a favor del campo. d. Aplicando el segundo principio de la dinámica: F = m a La fuerza a la que se ve sometido el ión en un punto del campo es: E q F ión r r ⋅ = Igualando y se despeja la aceleración: a m E q ión ión r r = ⋅ m E q a ión r r = 2 7 26 19 s m i 10 6 6 , 3 10 15 , 3 i 6 , 3 10 6 , 1 2 a r r × = × ⋅ × ⋅ = − Modelo 2008. Cuestión 4.- a) Enuncie el teorema de Gauss y escriba su expresión matemática. b) Utilice dicho teorema para deducir la expresión matemática del campo eléctrico en un punto del espacio debido a una carga puntual. Solución. a. El flujo neto que atraviesa una superficie cerrada cualquiera es igual a la suma algebraica de las cargas eléctricas encerradas en su interior dividida entre la constante dieléctrica del vacío. 24 ∫ ∑ ⋅ = ε = φ c o S d E Q r r b. Para facilitar el cálculo del flujo, suponemos una superficie que encierre a la carga puntual con una simetría sencilla y adecuada, en este caso una esfera como la de la figura, de modo que el campo ( ) E en cualquier punto de la superficie es un vector con dirección radial y de módulo constante = = esfera la de radio R : R Q E 2 , y S d r es el vector representativo de la superficie diferencial a estudio, perpendicular a ella. El flujo a través de todo la esfera vendar dado por la expresión: ∫ ∫ ⋅ ⋅ = = c c α cos dS E S d E φ r o r α es el ángulo que forman E r y S d r , que es 0º por ser vectores paralelos. o 2 2 o c 2 c c Q R 4 R Q 4 1 S d R Q k S d E º 0 cos S d E ε = π ⋅ ε π = ⋅ = ⋅ = ⋅ ⋅ = φ ∫ ∫ ∫ r r r r r o Q ε = φ Septiembre 2007. Problema 2B.- Se disponen dos cargas eléctricas sobre el eje X: una de valor Q1 en la posición (1, 0), y otra de valor Q2 en (−1, 0). Sabiendo que todas las distancias están expresadas en metros, determine en los dos casos siguientes: a) Los valores de las cargas Q1 y Q2 para que el campo eléctrico en el punto (1, 0) sea el vector j 10 2 E 5 r r × = N/C, siendo j r el vector unitario en el sentido positivo del eje Y. b) La relación entre las cargas Q1 y Q2 para que el potencial eléctrico en el punto (2, 0) sea cero. Datos: Constante de la ley de Coulomb k = 9 × 109 N m2 C−2 Solución. a. POR SIMETRÍA: Para que el campo resultante sea en la dirección del eje OY, las dos cargas han de ser de igual signo y módulo, por ser en la dirección positiva del eje, las cargas serán positivas. y 2 y 1 x 2 1x 2 1 T E E E E simetria Por E E E r r r r r r r + = − = = + = Por definición: = = = 45º sen 2 Q k E 45º sen 2 Q k E : r Q k E 2 2 y 2 2 1 y 1 2 { } j 45º sen 2 Q k 2 Q Q j 45º sen 2 Q k j 45º sen 2 Q k E E j E E 2 2 1 2 2 2 1 y 2 y 1 T T r r r r r r r = = = + = + = = 45 sen Q k 45º sen 2 Q k 2 E 2 T = = C 10 1 ' 3 2 2 10 9 10 2 45º sen k E Q 5 9 5 T − × = × × = = 25 b. El potencial eléctrico en el punto P, es la suma escalar de los potenciales que generan cada una de las cargas en dicho punto. 0 V V V 2 1 P = + = Por definición: r Q k V = 2 2 1 1 2 1 P r Q k r Q k V V V + = + = 3 Q k 1 Q k 0 2 1 + = 3 Q k 1 Q k 2 1 − = 3 1 Q Q 2 1 − = Tienen que ser de distinto signo, y Q2 de triple valor que Q1. Junio 2007. Problema 2B.- Dos partículas con cargas de + 1 µC y de −1 µC están situadas en los puntos del plano XY de coordenadas (−1,0) y (1,0) respectivamente. Sabiendo que las coordenadas están expresadas en metros, calcule: a) El campo eléctrico en el punto (0,3). b) El potencial eléctrico en los puntos del eje Y. c) El campo eléctrico en el punto (3,0). d) El potencial eléctrico en el punto (3,0). Dato: Constante de la ley de Coulomb K = 9×109 N m2 C−2 Solución. a. Por el principio de superposición: ( ) ( ) ( ) 3 , 0 E 3 , 0 E 3 , 0 E 2 1 r r r + = Teniendo en cuenta el signo de las cargas: j α sen E i α cos E j α sen E i α cos E E E E E E 2 2 1 1 y 2 x 2 y 1 x 1 r r r r r r r r r − + + = + + + = ( ) ( )j α sen E α sen E i α cos E α cos E E 2 1 2 1 r r r − + + = 10 3 1 r r 2 2 2 1 = + = = 10 3 α sen = 10 1 α cos = ( ) C N 10 9 10 10 1 10 9 r q K E 2 2 6 9 2 1 1 1 × = × ⋅ × = ⋅ = − ( ) C N 10 9 10 10 1 10 9 r q K E 2 2 6 9 2 2 2 2 × = × ⋅ × = ⋅ = − j 10 3 10 9 10 3 10 9 i 10 1 10 9 10 1 10 9 E 2 2 2 2 r r r ⋅ × − ⋅ × + ⋅ × + ⋅ × = ( ) C N i 2 , 569 3 , 0 E r r = b. El potencial en un punto es la suma escalar de los potenciales que crean cada una de las cargas en ese punto. ( ) ( ) ( ) 2 6 9 1 6 9 2 2 1 1 2 1 i r 10 1 10 9 r 10 1 10 9 r q K r q K y , 0 V y , 0 V V y , 0 V − − × − ⋅ × + × ⋅ × = ⋅ + ⋅ = + = =∑ Teniendo en cuenta que en cualquier punto del eje de ordenadas r1 = r2 = r ( ) 0 r 10 9 r 10 9 y , 0 V 3 3 = × − × = c. Al igual que en el apartado a, el campo eléctrico en el punto (3, 0) se calcula como suma vectorial de los campos eléctricos que generan cada una de las cargas. Dada la geometría del problema, solo existe campo eléctrico en la componente i r . 26 ( ) ( ) ( ) 0 , 3 E 0 , 3 E 0 , 3 E 2 1 r r r + = Teniendo en cuenta el signo de las cargas: ( ) C N i 5 , 1687 i 2 10 1 10 9 i 4 10 1 10 9 j r q K i r q K 0 , 3 E 2 6 9 2 6 9 2 2 2 2 1 1 r r r r r r − = × ⋅ × − × ⋅ × = ⋅ − ⋅ = − − d. ( ) ( ) ( ) + = + ⋅ = + = 2 2 1 1 2 2 1 1 2 1 r q r q K r q K r q K 0 , 3 V 0 , 3 V 0 , 3 V ( ) V 2250 2 10 1 4 10 1 10 9 0 , 3 V 6 6 9 − = × − + × ⋅ × = − − Modelo 2007. Problema 1B.- Una carga positiva de 2 µ C se encuentra situada inmóvil en el origen de coordenadas. Un protón moviéndose por el semieje positivo de las X se dirige hacia el origen de coordenadas. Cuando el protón se encuentra en el punto A, a una distancia del origen de x = 10 m lleva una velocidad de 1000 m/s. Calcule: a) El campo eléctrico que crea la carga situada en el origen de coordenadas en el punto A. b) El potencial y la energía potencial del protón en el punto A. c) La energía cinética del protón en el punto A d) El cambio de momento lineal experimentado por el protón desde que parte de A y por efecto de la repulsión vuelve al mismo punto A. Datos: Constante de la ley de Coulomb 2 2 9 C m N 10 9 K − × = Masa del protón ; kg 10 67 , 1 m 27 p − × = Carga del protón C 10 6 , 1 q 19 p − × = Solución. a) Campo eléctrico es la región del espacio que se ve afectada por la presencia de una carga eléctrica. Según la Ley de Coulomb: i C N 180 m 10 C 10 2 C Nm 10 9 i r Q K E 2 2 6 2 2 9 2 A A r r r = × ⋅ × = ⋅ = − − . b) El potencial en un punto de un campo eléctrico, es el trabajo necesario para desplazar la carga unidad positiva desde ese punto al infinito. ( ) C J V 1800 m 10 C 10 2 C Nm 10 9 r Q K dr F q 1 V 6 2 2 9 r = × ⋅ × = ⋅ = ⋅ = − − ∞ ∫ r El potencial en un punto se puede expresar en función de la energía potencial en ese punto. J 10 88 ' 2 C 10 6 ' 1 C J 1800 q V E q E V 16 19 p p p − − × = × ⋅ = ⋅ = ⇒ = + c) ( ) J 10 35 ' 8 s m 10 kg 10 67 ' 1 2 1 v m 2 1 E 22 2 3 27 2 c − − × = ⋅ × ⋅ = ⋅ = d) ( ) o o A A p A p A p i f v v m v m v m p p p r r r r r r r − = − = − = ∆ + + + ( ) ( ) ( ) i s m 1000 i s m 1000 kg 10 67 ' 1 v v m p 27 A A p o r r r r r − − ⋅ × = − = ∆ − + 1 24 s m kg 10 34 ' 3 p − − × = ∆r 27 Septiembre 2006. Problema 2B.- Dos cargas eléctricas positivas e iguales de valor 6 10 3 − × C están situadas en los puntos A (0,2) y B (0,-2) del plano XY. Otras dos cargas iguales Q están localizadas en los puntos C (4,2) Y D (4,-2). Sabiendo que el campo eléctrico en, el origen de coordenadas es C N i 10 4 E 3r r × = , siendo i r el vector unitario en el sentido positivo del eje X, y que todas las coordenadas están expresadas en metros, determine: a) El valor numérico y el signo de las cargas Q. b) El potencial eléctrico en el origen de coordenadas debido a esta configuración de cargas. Datos: Constante de la ley de Coulomb 2 2 9 C m N 10 9 K − × = Solución. a. Por el principio de superposición, el campo eléctrico ( ) E r es la suma vectorial de los campos que generan todas las cargas en ese punto. Los campos eléctricos 1 E r y 2 E r creados por las cargas q1 y q2 respectivamente en el origen se anulan por simetría. Teniendo en cuenta que el T E r en el origen tiene dirección y sentido i r + , las cargas Q3 y Q4 han de ser iguales en módulo y signo (negativas), para que e esta forma, las componentes j r de ambos vectores se anulen por simetría y las componentes i r se sumen, tal y como se observa en la figura. El campo eléctrico creado por una carga según la ley de Coulomb es: u r q K E 2 r r = Siendo u r un vector unitario en la dirección que une el punto a la carga que crea el campo. ( ) ( ) ( ) ( ) i 5 2 20 Q 10 9 2 i α cos r Q K 2 20 r r r Q Q Q Q E Q E 0 , 0 E 2 9 2 4 3 4 3 4 x 3 x r r r r r × ⋅ = ⋅ = = = = = = = + = ( ) i 10 4 i Q 10 05 , 8 0 , 0 E 3 8 r r r × = × = C µ 97 , 4 C 10 97 , 4 Q 6 = × = − C µ 97 , 4 Q − = b. El potencial en un punto debido a una distribución de cargas es la suma escalar del potencial que genera cada una de las cargas en ese punto. 4 3 2 1 i T V V V V V V + + + = =∑ + + + = + + + = 4 4 3 3 2 2 1 1 4 4 3 3 2 2 1 1 T r Q r Q r q r q K r Q K r Q K r q K r q K V V 6996 20 10 97 , 4 20 10 97 , 4 2 10 3 2 10 3 10 9 V 6 6 6 6 9 T = × − + × − + × + × ⋅ × = − − − − 28 Junio 2006. Cuestión 3.- Una carga puntual de valor Q ocupa la posición (0,0) del plano XY en el vacío. En un punto A del eje X el potencial es V = ‒120 V y el campo eléctrico es i 80 E r r − = N/C siendo i r el vector unitario en el sentido positivo del eje X. Si las coordenadas están dadas en metros, calcule: a) La posición del punto A y el valor de Q. b) El trabajo necesario para llevar un electrón desde el punto B (2,2) hasta el punto A. Datos: Valor absoluto de la carga del electrón C 10 6 , 1 e 19 − × = Constante de la ley de Coulomb en el vacío 2 2 9 C m N 10 9 K − × = Solución. a. El campo creado por una carga Q a una distancia d sobre el eje X es i d q k E 2 = y el potencial en ese punto es d Q k V = Por tanto aplicando a cada uno de los datos tenemos: a. E: Q k d 80 i d Q k i 80 2 2 ⋅ = ⋅ − ⇒ = − b. V: kQ d 120 d Q k 120 = − ⇒ = − Igualando las expresiones: d 120 d 80 2 − = − Simplificado y ordenando se calculan las posibles soluciones ( ) = = − = = − ⋅ m 5 ' 1 d : 0 120 d 8 0 d : 0 120 d 80 d La solución 0 d = no tiene sentido porque en ese caso, ∞ = ∞ = V y E r , por lo tanto, solo queda la solución m 5 ' 1 d = Sabiendo que 2 2 9 C Nm 10 9 k × = y sustituyendo en la ecuación del potencial se despeja la carga. C 10 2 k d 120 Q 8 − × − = − = b. El trabajo para llevar una carga q desde B hasta A es igual al producto de la carga q por la diferencia de potencial entre los puntos A y B ( ) A B p V V q E W − = ∆ − = El potencial en B con la expresión: B q B d Q k V = Expresión de la que lo único que desconocemos es la distancia al punto B que se calcula por el teorema de Pitágoras Conocida la distancia B se calcula el potencial. V 6 , 63 m 2 2 C 10 2 C Nm 10 9 V 8 2 2 9 B − = × − × × = − y con el potencial en A y en B y el valor de la carga que se desplaza se calcula el trabajo. ( ) ( ) J 10 02 , 9 v 120 v 6 ' 63 C 10 6 , 1 W 18 19 − − × − = − − − ⋅ × − = 29 Junio 2005. Problema 2A.- Tres partículas cargadas Q1 = +2 µC, Q2 = +2 µC y Q3 de valor desconocido están situadas en el plano XY. Las coordenadas de los puntos en los que se encuentran las cargas son Q1: (1, 0), Q2: (−1, 0) y Q3: (0, 2). Si todas las coordenadas están expresadas en metros: a) ¿Qué valor debe tener la carga Q3 para que una carga situada en el punto (0,1) no experimente ninguna fuerza neta? b) En el caso anterior, ¿cuánto vale el potencial eléctrico resu1tante en el punto (0,1) debido a las cargas Q1, Q2 y Q3 ? Dato: Constante de la ley de Coulomb K = 9×109 N m2 C−2 Solución. a. Para que una carga situada en el punto (0, 1) no experimente fuerza neta, el campo E creado por las tres cargas en (0, 1) debe ser nulo. En el punto (0, 1) el campo creado por las cargas Q1 y Q2 es: 2 1 T E E E r r r + = ( ) ( ) ( ) ( ) + = + = + − = + − = j 2 2 i 2 2 r Q k j 45 sen i 45 cos r Q k 1 , 0 E j 2 2 i 2 2 r Q k j 45 sen i 45 cos r Q k 1 , 0 E 2 2 2 2 2 2 1 2 1 1 r r r r r r r r El campo total es la suma vectorial de los campos creados por ambas cargas. ( ) ( ) j 2 r Q k j 2 2 i 2 2 r Q k j 2 2 i 2 2 r Q k 1 , 0 E 1 , 0 E E 2 1 2 2 2 1 2 1 T r r r r r r r r = + + + − = + = ( ) m v j 10 2 9 j 2 2 10 2 10 9 j 2 r Q k E 3 2 6 9 2 1 2 .
1 T r r r r × = ⋅ × = = − El campo generado Q3 2n 2l punto (0, 1) será: ( ) r r Q k 1 , 0 E 2 3 3 ) r = ≡ r ) Vector unitario en la dirección del campo ( ) ( ) ( ) j r : j 1 , 0 2 0, 1 0, r r r r v ) v r r r ) − = − = − = − = = r = 1 m sustituyendo en la expresión del campo ( ) j Q 10 9 j Q k 1 , 0 E 3 9 3 3 v v r × − = ⋅ − = Para que el campo creado por las tres cardas en (0, 1) sea nulo se debe cumplir: 0 E E 3 2 .
1 T = + r r ( ) 0 j Q 10 9 j 10 2 9 3 9 3 = × − + × v r expresión de la que se puede despejar la carga Q3. 30 C 2 C 10 2 Q 6 3 µ = × = − b. El potencial = r q k V creado por las tres cargas en el punto (0, 1) es la suma de los potenciales creados por cada carga. ( ) v 10 2 18 2 10 2 10 9 r Q k V 3 6 9 1 1 × = × × = = − ( ) v 10 2 18 2 10 2 10 9 r Q k V 3 6 9 2 2 × = × × = = − ( ) v 10 2 9 1 10 2 10 9 r Q k V 3 6 9 3 3 × = × × = = − ( ) v 10 2 54 10 2 9 2 18 2 18 V V V V 3 3 3 2 1 T × = × + + = + + = Modelo 2005. Cuestión 3.- Dos cargas puntuales de C 6 -y C 6 µ µ + están situadas en el eje X, en dos puntos A y B distantes entre sí 12 cm. Determine: a) El vector campo eléctrico en el punto P de la línea AB, si AP = 4 cm. y PB = 8 cm. b) El potencial eléctrico en el punto C perteneciente a la mediatriz del segmento AB y distante 8 cm. de dicho segmento. Datos: Constante de la ley de Coulomb K = 9×109 Nm2 C‒2 Solución. a. El campo eléctrico en P, es la suma vectorial de los campos que generan cada una de las cargas en dicho punto: ( ) i 04 ' 0 10 6 10 9 U r q · k E E E E 2 6 9 r 2 A ) A ( p ) B ( p ) A ( p p − × ⋅ × = = + = ( ) i c N 10 38 ' 3 E 7 ) A ( p × = r 2 B ) B ( p U r q k E ⋅ = sustituyendo por los valores: ( ) ( ) i C N 10 84 ' 0 i 08 ' 0 C 10 6 10 9 E 7 2 6 9 ) B ( p × = × ⋅ × = − El campo total en P, es la suma vectorial de dos vectores de la misma dirección y sentido: i C N 10 22 ' 4 10 84 ' 0 10 38 ' 3 E 7 7 7 p r r × = × + × = La dirección de los vectores unitarios r u r se deduce suponiendo en el punto P, la unidad de carga positiva. b. El potencial, a deferencia del campo eléctrico, se trata de una magnitud escalar, por tanto sumamos escalarmente los potenciales producidos por ambas cargas en ese punto C: 31 V 000 54 1 ' 0 10 · 6 · 10 · 9 r q · k V V V V 6 9 A ) A ( C ) B ( C ) A ( C C = = = + = − Aplicando la misma expresión para la carga B: ( ) V 000 54 1 ' 0 10 · 6 · 10 · 9 r q · k V 6 9 B cB − = − = = − Por tanto: ( ) 0 V 54000 54000 V c c = ⇒ − + = El potencial eléctrico en C es nulo. Septiembre 2004. Problema 2B. Dos cargas eléctricas en reposo de valores , C 2 q y C 2 q 2 1 µ − = µ = están situadas en los puntos (0, 2) y (0, −2) respectivamente, estando las distancias en metros. Determine: a) El campo eléctrico creado por esta distribución de cargas en el punto A de coordenadas (3 ,0) b) El potencial en el citado punto A y el trabajo necesario para llevar una carga de C 3µ desde dicho punto hasta el origen de coordenadas. Solución. a. El campo eléctrico creado por una carga puntual q en un punto a distancia R de la carga viene dado por el vector: r 2 U R q K E r r ⋅ = en donde K es constante de Culomb y r U es un vector unitario en la dirección de la recta que une la carga q con el punto en el que se quiere calcular el campo, dependiendo el sentido del signo de la carga, repulsivo si es positiva y atractivo si fuese negativo. El campo creado por una distribución de cargas, es la suma vectorial de los campos creados por cada carga. 2 1 E E E r r r + = descomponiendo los campos creados por las cargas en sus componentes cartesianas: + = + = j E i E E j E i E E y 2 x 2 2 y 1 x 1 1 r r r r r r sustituyendo ( ) ( )j E E i E E j E i E j E i E E y 2 y 1 x 2 x 1 y 2 x 2 y 1 x 1 r r r r r r r + + + = + + + = Teniendo en cuenta la geometría de la distribución = − = y 1 y 2 x 1 x 2 E E E E la expresión del campo se reduce a: j R q K 2 j sen E 2 j E 2 E 2 1 1 y 1 r r r r = α = ⋅ = El sen α se calcula por triángulos rectángulos ( ) 13 2 2 3 2 sen 2 2 − = − + − = α sustituyendo todos los datos en la expresión: 32 C N 1536 j 13 2 13 10 2 10 9 2 E 2 6 9 − = − ⋅ × × ⋅ = − r r b. El potencial eléctrico creado por una carga puntual q en un punto a distancia R de la carga viene dado por: R q K V = en donde K es la constante de Culomb. El potencial eléctrico es una magnitud escalar. Para un sistema de cargas el potencial eléctrico del sistema se calcula como la suma algebraica de los potenciales creados por cada una de las cargas en el punto. En nuestro caso, el potencial en el punto A será: R q K R q K V V V 2 1 2 1 T + = + = Teniendo en cuenta que 1 2 q q − = , el potencial en el punto A es nulo v 0 R q K R q K VT = − = El trabajo necesario para llevar la carga desde el punto A al origen de ordenadas viene expresado por: ( ) ( ) ( ) A V 0 , 0 V q V q E W p − ⋅ − = ∆ ⋅ − = ∆ − = El potencial en el origen de ordenadas es: ( ) ( ) ( ) { } 0 q q 2 q K 2 q K 0 , 0 V 0 , 0 V 0 , 0 V 1 2 2 1 2 1 T = − = = + = + = sustituyendo en el trabajo ( ) 0 0 0 q W = − ⋅ − = El trabajo necesario para desplazar una carga entre dos puntos con igual potencial es nulo Modelo 2004. Cuestión 3.- Se crea un campo eléctrico uniforme de intensidad 6x104 N/C entre dos láminas metálicas planas y paralelas que distan entre si 2’5 cm. Calcule: a) La aceleración a la que esta sometido un electrón situado en dicho campo. b) Si el electrón parte del reposo de la lamina negativa, ¿con que velocidad llegara a la lamina positiva? Nota: Se desprecia la fuerza gravitatoria. Datos: Valor absoluto de la carga del electrón e = 1’6×10−19C Masa del electrón m = 9’1×10−31kg Solución. Se crea un campo eléctrico( ) C N 10 6 E 4 × = r entre dos placas paralelas: a. La aceleración de un electrón situado entre las placas, sí despreciamos la fuerza gravitatoria, será la producida por la única fuerza que actúa sobre el e− que es la fuerza electromagnética: E q F ⋅ = que será de sentido contrario a la dirección del campo E, ya que la carga es negativa. Igualamos esta fuerza, a la expresión de la fuerza según la segunda ley de Newton: a m E q F ⋅ = ⋅ = Despejando la aceleración: m E · q a = y sustituyendo valores numéricos: 2 16 s / m 10 · 055 ' 1 a = b. Si parte del reposo desde la lámina negativa, la trayectoria será recta y con MRUA. Aplicando la ecuación de este movimiento: t 055 ' 1 v t a v v f 0 f ⋅ = ⋅ + = 33 Puesto que se conoce la distancia entre placas cm 5 ' 2 d = y utilizando la ecuación: s a 2 v v 2 o 2 f ⋅ ⋅ = − teniendo en cuenta que la velocidad inicial es nula, se obtiene, sustituyendo los valores numéricos: s m 10 3 ' 2 10 5 ' 2 10 055 ' 1 2 v s a 2 v 7 2 16 f 2 f × = × ⋅ × ⋅ = ⇒ ⋅ ⋅ = − El vector velocidad: j 10 3 ' 2 v 7 f × − = (ya que solo existe movimiento en este eje) Septiembre 2003. Cuestión 1. a) Defina las superficies equipotenciales en un campo de fuerza conservativo. b) ¿Cómo son las superficies equipotenciales del campo eléctrico creado por una carga puntual? c) ¿Qué relación geométrica existe entre las líneas de fuerza de un campo conservativo y las superficies equipotenciales? d) Indique un ejemplo de campo de fuerzas no conservativo. Solución. a. Son aquellas superficies cuya característica principal es el valor constante del potencial creado por un campo conservativo en cualquier punto de dicha superficie. b. En el caso del campo eléctrico creado por una carga puntual, el potencial tiene la forma: r 1 V ∝ . Es decir, su valor solo depende del radio r (distancia de la carga al punto del campo), por tanto las superficies equipotenciales son superficies esféricas de radio r, con la carga en el centro de todas ellas. c. Las líneas de fuerza de un campo conservativo, son las trayectorias que seguiría una partícula (con carga, si es un campo eléctrico), abandonada en un punto del campo conservativo. Dicha línea de fuerza, siguen la dirección del gradiente del potencial, de modo que deben ser perpendiculares a las superficies equipotenciales, ya que esa dirección es la de máxima variación del potencial. Constituye un campo radial, de la forma: d. Un ejemplo seria el campo gravitatorio terrestre, si tuviéramos en cuenta el rozamiento con el aire, como no despreciable. En este caso, la energía de un cuerpo en este campo, ya no dependería sólo de la posición (campo conservativo), sino de la trayectoria seguida en un movimiento por dicho campo. Junio 2002. Problema 2B. Se tiene tres cargas situadas en los vértices de un triangulo equilátero cuyas coordenadas (expresadas en cm) son: ( ) ( ) ( ) 1 , 3 C , 1 , 3 B , 2 , 0 A − − − Sabiendo que las cargas situadas en los puntos B y C son idénticas e iguales a 2µC y que el campo eléctrico en el origen de coordenadas (centro del triángulo) es nulo, determine: a) El valor y el signo de la carga situada en el punto A. b) El potencial en el origen de coordenadas. Datos: Constante de la ley de Coulomb K = 9x109N m2/C2 Solución. a. Para hallar el valor de Q, se tendrá en cuenta que E (0,0)=0. el campo creado en el centro del triángulo por la carga EA viene expresado por: 2 A 2 · k E α = 34 El campo creado en el mismo punto por la carga B(EB) es: { } C 3 2 6 9 2 B E C N 10 · 2 9 2 10 2 10 9 2 1 3 r r Q k E = = × ⋅ × = = + = = ⋅ = − Si sumamos vectorialmente estos dos campos, solo nos queda componente “y”, ya que las componentes “x” se anulan por simétrica. La componente “y” es la suma de y B E y y C E , que son del mismo modulo, y de valor: α = = sen · E E E B C B y y Sumamos las dos componentes “y”: c N 10 · 2 9 º 30 sen 10 · 2 9 º 30 sen · 10 · 2 9 E E 3 3 3 C B y y = + = + La carga en el punto A, tiene que generar un campo en (0,0) de modo que el campo resultante sea nulo. Por tanto, el campo producido por la carga A, debe ser: j C N 10 · 2 9 E 3 A − = A partir del valor del campo, hallamos la carga: 2 3 A r Q · k 10 · 5 ' 4 E = = C 2·10 Q 10 · 9 4 · 10 · 5 ' 4 Q 6 9 3 − = = el signo tiene que ser positivo. Es decir, una carga igual a la que tenemos en B y C. b. Se calcula el potencial creado por cada carga en (0,0) y se suma escalarmente. v 10 · 27 v 10 · 9 v 10 · 9 v 10 · 9 V V V V : V 10 · 9 2 10 · 2 · 10 · 9 V V 10 · 9 2 10 · 2 · 10 · 9 V V 10 · 9 2 10 · 2 · 10 · 9 V : r Q · k V 3 3 3 3 C B A T 3 6 9 c 3 6 9 B 3 6 9 A i i = + + = + + = = = = = = = = − − − Septiembre 2001. Problema 2B.- Se tienen dos cargas puntuales sobre el eje X, q1 = −0’2 µC está situada a la derecha del origen y dista de él 1 m; q2 = +0,4 µC está a la izquierda del origen y dista de él 2 m. 35 a. ¿En qué puntos del eje X el potencial creado por las cargas es nulo? b. Si se coloca en el origen una carga q = +0,4 µC determine la fuerza ejercida sobre ella por las cargas q1 y q2. Datos: Constante de la ley de Coulomb en el vacio K = 9×109 N m2C−2 Solución. a. Para que el potencial en un punto del eje x se anule tiene que ocurrir que: ( ) ( ) 0 V V p p 2 1 = + Situando el punto P a la derecha de q2: 1 2 2 1 1 1 x 3 q k V x q · k V + = = sustituyendo en la condición de potencial nulo 0 x 3 C 10 4 ' 0 k x C 10 '2 0 k 0 x 3 q · k x q · k 1 6 1 6 1 2 1 1 = + × ⋅ − = × − ⋅ = + + − − Operando y simplificando la expresión anterior: ( ) ( ) ( ) 6 1 6 1 10 · 4 ' 0 x 10 · 2 ' 0 x 3 − − ⋅ = ⋅ + resolviendo la ecuación de 1º grado x1 = 3 m Si suponemos que el punto P se encuentra entre las cargas: 0 x 3 10 4 ' 0 · k x 10 2 ' 0 k 1 6 1 6 = − × − = × − ⋅ − − resolviendo x = 1 m coincide con el origen de coordenadas, P (0, 0) Repetimos el mismo procedimiento imponiendo un punto a la izquierda de la carga positiva: 1 6 1 6 x 3 C 10 2 ' 0 k x C 10 4 ' 0 k + × − ⋅ − = × ⋅ − − resolviendo la ecuación 36 x1 = −6 m No valido ya que x1 es una distancia y no puede ser negativa. No hay ningún punto a la izquierda de la q2 donde VT = 0. b. Si colocamos una carga q = +0’4µc en (0, 0) calcular la fuerza sobre ella. Calculamos el campo total E en el origen de coordenadas: C N 10 · 8 ' 1 E 10 2 ' 0 10 9 E 1 10 2 ' 0 · k E r q · k E 3 1 6 9 1 2 6 1 2 1 = × ⋅ × = × = = − − La dirección y sentido es i + (suponemos colocada en el origen una unidad de carga positiva) i C N 1800 E1 = Calculamos el campo creado por la carga 2: i C N 900 E 10 9 ' 0 E 2 10 4 ' 0 10 9 E r q · k E 2 3 2 2 6 9 2 2 2 = × = × ⋅ × = = − Puesto que ambos vectores tienen la misma dirección y sentido numeramos sus módulos: i C N 2700 E E E 2 1 T = + = La fuerza sobre la : c 10 4 ' 0 q 6 − × + = i 2700 10 4 ' 0 F E q F 6 ⋅ × = ⋅ = − i N 10 1 , 1 F 3 − × = Junio 2001. Problema 2B. Tres cargas positivas e iguales de valor q = 2 ηC cada una se encuentran situadas en tres de los vértices de un cuadrado de lado 10 cm. Determine: a) El campo eléctrico en el centro del cuadrado, efectuando un esquema gráfico en su explicación. b) Los potenciales en los puntos medios de los lados del cuadrado que unen las cargas y el trabajo realizado al desplazarse la unidad de carga entre dichos puntos. Datos: Constante de la ley de Coulomb en el vacío K = 9×109 Nm2C2 Solución. a. Como se observa en la figura, el campo resultante en el centro del cuadrado es el generado por la carga A( ) A E ya que por simetría los campos generados por las carga B y C se anulan entre si. La distancia de cualquier vértice al centro del cuadrado se calcula por teorema de Pitágoras m 2 05 ' 0 05 ' 0 05 ' 0 r 2 2 = + = Los módulos de los campos creados por B y por C son iguales ya que están creados por la misma carga y están a igual distancia del centro. ( ) C 6 2 6 9 2 B E C N 10 6 ' 3 2 05 ' 0 10 2 · 10 9 r Q · k E = × = ⋅ × × = = − Los vectores de campo generados por las cargas B y C ( ) C B E , E tienen igual módulo, dirección y sentido contrario, por lo que se anulan. El campo resultante es por tanto el generado por la carga A. Módulo: ( ) C N 10 6 ' 3 2 05 ' 0 10 2 · 10 9 r Q · k E 6 2 6 9 2 A × = ⋅ × × = = − El ángulo que forma el con la horizontal es: 37 45º 1 tg 05 ' 0 05 ' 0 tg = α = α = α Proyecciones del vector campo sobre los ejes coordenados: α = α = sen E E ·cos E E A A A A y x el campo resultante es: ( ) j 10 2 8 ' 1 i 10 2 8 ' 1 05 ' 0 , 05 ' 0 E 6 6 T × + × = b. El potencial en los puntos medios: V 10 2 ' 7 r Q k 2 r Q k r Q · k V V 10 2 ' 7 r Q k 2 r Q k r Q · k V 5 AB 5 CA × = = + = × = = + = los potenciales son los mismos yq que están creados por la mismas carga y a igual distancia. Por tanto : ( ) 0 V V · q W f i = − = ya que ambos puntos tienen el mismo potencial Septiembre 2000. Problema 2A.- Los puntos A, B y C son los vértices de un triángulo equilátero de 2 m de lado. Dos cargas iguales positivas de 2 mC están en A y B. a) ¿Cuál es el campo eléctrico en el punto C? b) ¿Cuál es el potencial en el punto C? c) ¿Cuánto trabajo se necesita para llevar una carga positiva de 5 mC desde el infinito hasta el punto C si se mantienen fijas las otras cargas? d) Responder al apartado anterior c) si la carga situada en B se sustituye por una carga de −2 mC. Datos: Permitividad del vacío εo = 8,85×10−12 N−1m−2C2 Solución. a. Se calcula el módulo del campo creado por cada carga en C, y se suman vectorialmente: C N 10 5 ' 4 2 10 2 10 85 ' 8 π 4 1 R q ε π 4 1 R q · k E 2 6 12 2 A o 2 A AC × = × ⋅ × ⋅ = ⋅ ⋅ = = − − r C N 10 5 ' 4 2 10 2 10 85 ' 8 π 4 1 R q ε π 4 1 R q · k E 2 6 12 2 A o 2 A BC × = × ⋅ × ⋅ = ⋅ ⋅ = = − − r Por simétrica, y puesto que los módulos de ambos campos son iguales, las componentes x se anulan, quedando únicamente las componente y: y C y B A A E º 60 sen E E r r r = ⋅ = C N j 10 9 E j º 60 sen E 2 E 3 T A T c C c r r r r × = ⋅ = b. Se calcula el potencial creado por cada carga en C y se suman escalarmente ( ) v C J 10 8 ' 1 2 10 2 · 10 85 ' 8 4 1 2 R q k 2 R q k R q k V 4 6 12 A B A T × = × × ⋅ π = ⋅ = ⋅ + ⋅ = − − c. El W para llevar una carga desde el infinito hasta el punto C se calcula como: ( ) C V V q W − = ∞ Sí q = 5×10−6, V∞ = 0 , VC = 1’8×104, sustituyendo en la ecuación anterior ( ) J 10 9 10 8 ' 1 0 10 5 W 2 4 6 − − × − = × − ⋅ × = Trabajo que se realiza contra el campo. d. Si la carga en B fuera C 10 · 2 q 6 B − − = , el potencial en C será: ( ) 0 2 10 · 2 10 85 ' 8 4 1 2 c 10 · 2 10 85 ' 8 4 1 2 q k 2 q k V 6 12 6 12 B A c = − × ⋅ π ⋅ + ⋅ × ⋅ π = + = − − − − Por tanto, el W necesario para traer del infinito la carga en este caso, será nulo: 38 ( ) 0 W V V · q W c = − = ∞ Junio 2000. Cuestión 3. Dos cargas puntuales e iguales de valor 2 mC cada una, se encuentran situadas en el plano XY en los puntos (0, 5) y (0, −5), respectivamente, estando las distancias expresadas en metros. a) ¿En qué punto del plano el campo eléctrico es nulo? b) ¿Cuál es el trabajo necesario para llevar una carga unidad desde el punto (l, 0) al punto (‒1, 0)? Solución. a. El campo es función de la carga que lo genera y de la distancia entre la carga que lo genera y el punto donde se calcula. Entre dos cargas iguales, el campo es nulo en el punto medio del segmento que une ambas cargas. El campo E es nulo en el origen, ya que 2 1 E y E son dos vectores en la misma dirección y sentidos contrarios, y de módulos iguales, de valor: 25 c 10 · 2 · 10 · 9 r q · k E E 3 9 2 2 1 − = = = ( ) 0 E C N 720000 E E 0 , 0 T 2 1 = = = b. ( ) ( ) B A A B V V q V V q V q W − ⋅ = − ⋅ − = ∆ ⋅ − = Hay que calcular el potencial ⋅ = r q K V en los puntos (1,0) y (-1,0) ( ) { } 26 q k 2 q q q 26 q · k 26 q · k V 2 1 2 1 0 , 1 ⋅ = = = = + = ( ) C J 26 10 · 36 26 10 2 10 9 2 V 6 3 9 0 , 1 = × × ⋅ = − repitiendo el cálculo para el punto (−1, 0) ( ) C J 26 10 · 36 26 q k 2 26 q k 26 q k V 6 2 1 0 , 1 = ⋅ ⋅ = ⋅ + ⋅ = − para la unidad de carga(q =1): { } 0 W V V V V W B A B A = = − = Lo que implica que se puede mover la carga q del punto (1, 0) al punto (−1, 0) sin realizar ningún tipo de trabajo, ya que ambos puntos tienen el mismo potencial. |
5312 | https://www.youtube.com/watch?v=1pND890hnAs | How to Calculate the Radius of a Circle Given Its Circumference
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hello this is Eva shova and today I will teach you how to calculate the radius of a circle given its circumference in this example the circumference given to you is 24.12 and you have to find the radius you will start with the formula for circumference c = 2 pi r you will start substituting what you have what you have here is your circumference so your circumference here is 2412 equals 2 pi you don't have R you're looking for R now you only have one unknown and that's R what you want to do is you want to get rid of 2 pi because you're multiplying R by 2 pi here you want to undo that multiplication by dividing both sides sides of the equation by 2 pi so ided 2 Pi / 2 pi 2 Pi / 2 pi will be one so you will be left with 1 R equals now on your calculator you will type 24.12 divided by in parentheses 2 pi close parentheses once again you equ will to 24.12 divided open parentheses 2 pi close parentheses enter on your calculator and your answer is 3838 and so on you have to run your answer to the nearest 10th that means one digit after the decimal point and that digit here is eight that means your radius is 3.8 in because your circumference was given in inches |
5313 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7090115/ | Local Management of Anogenital Warts in Non-Immunocompromised Adults: A Network Meta-Analysis of Randomized Controlled Trials - PMC
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Dermatol Ther (Heidelb)
. 2020 Feb 6;10(2):249–262. doi: 10.1007/s13555-020-00357-z
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Local Management of Anogenital Warts in Non-Immunocompromised Adults: A Network Meta-Analysis of Randomized Controlled Trials
Antoine Bertolotti
Antoine Bertolotti
1 Department of Infectious Disease, Saint-Pierre Hospital, Reunion Island, France
2 EA 4537, Antilles-Guyane University, Martinique, France
3 INSERM CICEC 1410, Reunion Island, France
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1,2,3,✉, Cyril Ferdynus
Cyril Ferdynus
3 INSERM CICEC 1410, Reunion Island, France
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3, Brigitte Milpied
Brigitte Milpied
4 Department of Dermatology and Pediatric Dermatology, National Center for Rare Skin Disorders, Saint-André and Pellegrin Hospitals, Bordeaux, France
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4, Nicolas Dupin
Nicolas Dupin
5 Department of Dermatology, Cochin Hospital, Paris Descartes University, Paris, France
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5, Laetitia Huiart
Laetitia Huiart
6 Department of Population Health, Luxembourg Institute of Health, Luxembourg, Luxembourg
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6, Christian Derancourt
Christian Derancourt
2 EA 4537, Antilles-Guyane University, Martinique, France
7 Department of Dermatology, Hôpital de Briançon, Briançon, France
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2,7
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1 Department of Infectious Disease, Saint-Pierre Hospital, Reunion Island, France
2 EA 4537, Antilles-Guyane University, Martinique, France
3 INSERM CICEC 1410, Reunion Island, France
4 Department of Dermatology and Pediatric Dermatology, National Center for Rare Skin Disorders, Saint-André and Pellegrin Hospitals, Bordeaux, France
5 Department of Dermatology, Cochin Hospital, Paris Descartes University, Paris, France
6 Department of Population Health, Luxembourg Institute of Health, Luxembourg, Luxembourg
7 Department of Dermatology, Hôpital de Briançon, Briançon, France
✉
Corresponding author.
Received 2019 Dec 16; Collection date 2020 Apr.
© The Author(s) 2020
Open Access This article is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, which permits any non-commercial use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit
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PMCID: PMC7090115 PMID: 32030564
Abstract
Introduction
No hierarchy of first-line treatments for anogenital warts (AGWs) is provided in international guidelines. This study aimed to determine the efficacy of topical treatments and ablative procedures for the management of AGWs.
Methods
Twelve electronic databases were systematically searched from inception to August 2018. All randomized controlled trials (RCTs) comparing immunocompetent adults with AGWs who received at least 1 provider-administered or patient-administered treatment in at least 1 parallel group were included. Risk of bias assessment followed the Cochrane Handbook. The study endpoint was complete lesion response after clearance and recurrence assessment. A network meta-analysis was performed.
Results
A network geometry was constructed based on 49 of the 70 RCTs included in our systematic review. All but 4 RCTs had a high risk of bias. The most efficacious treatments compared to placebo were surgery (RR 10.54; CI 95% 4.53–24.52), ablative therapy + imiquimod (RR 7.52; CI 95% 4.53–24.52), and electrosurgery (RR 7.10; CI 95% 3.47–14.53). SUCRA values confirmed the superiority of surgery (90.9%), ablative therapy + imiquimod (79.8%), and electrosurgery (77.1%). The most efficacious patient-administered treatments were podophyllotoxin 0.5% solution (63.5%) and podophyllotoxin 0.5% cream (62.2%).
Conclusions
With low-level evidence of most included RCTs, surgery and electrosurgery were superior to other treatments after clearance and recurrence assessment. Podophyllotoxin 0.5%was the most efficacious patient-administered treatment. Combined therapies should be evaluated in future RCTs in view of their identified effectiveness. The results of future RCTs should systematically include clinical type, number and location of AGWs, and sex of the patient, to refine therapeutic indications.
Protocol Registration
PROSPERO-CRD42015025827
Electronic supplementary material
The online version of this article (10.1007/s13555-020-00357-z) contains supplementary material, which is available to authorized users.
Keywords: Anogenital warts, Condyloma, Frequentist approach, Network meta-analysis, Sexually transmitted disease, Systematic review
Key Summary Points
Why carry out this study?
Anogenital warts (AGWs) are one of the most common sexually transmitted diseases, with an overall prevalence rate of around 1–5%.
No clinically meaningful hierarchy of first-line treatments for anogenital warts is provided in international guidelines.
What was learned from the study?
Based on a low level of evidence, surgery and electrosurgery achieved the best complete lesion response after clearance and recurrence assessment.
Podophyllotoxin 0.5% was the most efficacious patient-administered treatment.
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Introduction
Anogenital warts (AGWs) are benign epithelial skin lesions that are predominantly caused by the human papillomavirus (HPV types 6 and 11), but are sometimes associated with other types of oncogenic HPV . With an overall prevalence rate of around 1–5%, they are one of the most common sexually transmitted infections . AGWs are usually asymptomatic, but they can be painful or pruritic and can cause significant psychosocial distress depending on size and location [3, 4]. Numerous HPV vaccination campaigns have been conducted, but few studies have demonstrated their efficacy in reducing the number of AGWs . Moreover, in most countries, vaccination coverage is partial and has yet to be extended to men [6, 7].
Many treatments are available to treat AGWs. These can be divided into provider-administered treatments (ProTs) [trichloroacetic acid (TCA), podophyllin resin, CO 2 laser surgery, cryotherapy, surgical excision, electrosurgery, intralesional therapy, etc.] and patient-administered treatments (PaTs) [podophyllotoxin, imiquimod, sinecatechins, 5-fluorouracil (5-FU) cream, etc.]. The latest guidelines [8–11] recommend that treatment of AGWs be adapted to: size, number, and anatomic site of AGWs; patient preference; convenience; adverse effects; cost of treatment; and physician experience. These recommendations, however, are based on head-to-head randomized trials or on expert advice. Furthermore, RCTs comparing several major treatments for AGWs (cryotherapy vs podophyllotoxin cream or gel, imiquimod vs TCA, CO 2 laser vs surgery or electrosurgery, etc.) are lacking [10, 13–15] and may never be performed (because they are costly, time-consuming, less attractive than new treatments, etc.). Reliable evidence on the comparative efficacy of these treatments is nevertheless needed to make informed clinical decisions.
In this context, network meta-analyses (NMAs) can help compare the relative benefits associated with different types of intervention for the same disease [16, 17]. The only NMA on AGWs, which was conducted by Barton et al. based on a systematic review up to March 2018, concluded that ablative techniques were superior; it also found podophyllotoxin 0.5% gel to be the most cost-effective topical treatment. However, this NMA did not examine sinecatechins, 5-FU cream, and several RCTs on new treatments [citric acid, intralesional bleomycin, potassium hydroxide (KOH), photodynamic therapy (PDT), etc.].
Our NMA aims to establish a clinically meaningful hierarchy of PaTs and ProTs for the management of AGWs.
Methods
The study protocol is registered with PROSPERO (No. CRD42015025827). The systematic review, which was published earlier , adheres to the PRISMA Statement . The present study adheres to the PRISMA extension for NMA . This article is based on previously conducted studies and does not contain any studies with human participants or animals performed by any of the authors.
Systematic Review
Twelve electronic databases were systematically searched from inception to August 2018 by 2 independent reviewers (A.B. and C.D.). Search terms included 2 synonym groups, AGW and RCT, with adjustments for each database (Appendix S1). The reference lists of all published studies and all recent reviews and meta-analyses were also searched [8–10, 13, 20–22]. No language restriction was imposed. To be included in the NMA, RCTs had to: (1) have at least 1 treatment group composed of immunocompetent adults clinically diagnosed with AGWs and treated with a ProT (TCA, podophyllin, CO 2 laser, cryotherapy, surgical excision, electrosurgery, all intralesional treatments, KOH, PDT, citric acid) or a PaT [podophyllotoxin, imiquimod, sinecatechins, 5-FU, cidofovir, interferon (INF) cream]; and (2) provide original estimates with risk ratios and confidence intervals (CIs) or present sufficient data to allow calculation of these estimates. Complete lesion response (CLR) at the end of follow-up was assessed based on two outcome measures: clearance at 3 months and recurrence 3 months later.
An extraction grid was developed after collegial discussion. For all selected studies, variables of interest were extracted independently by 2 independent reviewers (A.B. and C.D.). These reviewers assessed the risk of bias in the selected RCTs using the Cochrane Collaboration Risk of Bias tool . When different RCTs involved the same patient cohort, the RCT with the longest follow-up period was considered.
Data Synthesis
An NMA was performed that combined the results of all selected comparisons of AGW treatments. This statistical technique is used to account for direct comparisons performed in single trials and to make indirect comparisons across trials based on a common comparator intervention . In our NMA, placebo and podophyllin served as comparators for indirect comparisons even though they are not used in clinical practice. For RCTs comparing treatments at lower or higher dosages than recommended in published guidelines, only recommended dosages were considered. All analyses were performed with a frequentist approach using a random effects model, with an equal heterogeneity variance assumed for all comparisons.
The network geometry was assessed by graphically examining the connections between interventions . Each node represented an intervention. The thickness of nodes was proportional to the number of allocated patients. The thickness of connecting lines was inversely proportional to the variance between 2 interventions.
Netmeta R package version 8.0 (available at: was used to perform head-to-head comparisons of different treatments to a placebo . Specifically, 2 forest plots using random effects models were generated by calculating point estimates of relative risk (RR) with a CI 95%. A heat mapping function (which is a type of matrix visualization) was created with the Netmeta R package to evaluate heterogeneity and inconsistency . Warmer or cooler colors indicated significant inconsistency.
The patient was the unit of analysis for all RCTs. The endpoint—CLR after clearance and recurrence assessment—was evaluated using per protocol analysis (cured patients/follow-up patients). Sensitivity analyses of 2 scenarios were performed: (1) a worst-case (intention to treat) scenario, in which patients lost to follow-up were considered to be failing treatment (cured patients/all included patients); and (2) a best-case scenario, in which patients lost to follow-up were considered cured [(cured patients + lost to follow-up patients)/all included patients].
The probability that each intervention achieved CLR was estimated based on the relative effect sizes estimated with the NMA. A hierarchy of compared interventions was performed using the Surface Under the Cumulative Ranking curve (SUCRA). SUCRA values are expressed as percentages and show the relative probability of an intervention being among the best options.
Results
Characteristics of Selected Trials
Seventy RCTs involving 9931 patients with a mean of 142 participants per study fulfilled the inclusion criteria (Appendix S2–S3). The overwhelming majority of included RCTs (66/70) were found to be of poor quality (Appendix S4) . Twenty-one RCTs were excluded from the NMA: 6 because they compared dosages that were lower than recommended [27–32], 14 because they did not evaluate recurrence [33–46], and 1 because it was disconnected from the network (intralesional bleomycin vs podophyllin + cryotherapy) .
Nine studies comparing a recommended dosage with a lower dosage were included, but without the treatment arm that received the lower dosage [48–56]. Ultimately, 29 treatments or combination therapies were included. One RCT compared 4 arms , 5 RCTs compared 3 arms [58–62], and 43 compared 2 arms [48–56, 63–96]. Following these inclusion criteria, only two of 4 low risk of bias RCTs were included [27, 45]. The median follow-up for the 6006 covered patients was 6 months (3–12 month range).
Network Geometry
The complex network generated from the 49 included RCTs is shown in Fig.1. Compared treatments were connected either directly or indirectly through 1 or more “comparators.” The level of evidence informing each comparison was evaluated. Treatment comparisons involving the largest number of patients were polyphenon vs placebo (3 trials; 767 patients receiving treatment) and podophyllin vs podophyllotoxin gel (6 trials; 1005 patients receiving treatment). Only 12 RCTs [50, 57, 58, 62, 77, 78, 81, 83, 84, 87, 90, 96] directly compared a ProT to a PaT; of these, 9 examined treatments that are not used in clinical practice (6 on podophyllin and 3 on intralesional therapies). The most commonly studied agents were placebo (18 trials; 939 patients receiving treatment) and podophyllin (13 trials; 716 patients receiving treatment).
Fig. 1.
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Evidence network of eligible comparisons for complete lesion response in network meta-analysis. The thickness of connecting lines represents the cumulative number of trials for each comparison, and the thickness of nodes is proportional to the number of enrolled participants. Cryo cryotherapy, ablative ablative treatment (surgery or electrosurgery or CO 2 laser or cryotherapy), imi imiquimod 5%, 5-FU 5 fluorouracil, 5-FU intra intralesional 5 fluorouracil, TCA trichloroacetic acid, podo podophyllin 20–25%, citric ac citric acid 9%, polyph polyphenon 15%, podotox cr podophyllotoxin 0.5% cream, podotox cr/gel podophyllotoxin 0.5% gel + cream, podotox gel podophyllotoxin 0.5% gel, PDT photodynamic therapy, mycobac intra intralesional Mycobacterium, KOH potassium hydroxide, electro electrosurgery, INF-1a intra intralesional interferon-1α, INF-2b intra intralesional interferon-2β
Complete Lesion Response
Figure 2 presents the CLR of all treatments and placebos compared using a random effects model. Most CIs were wide, but rarely included value 1. Cidofovir, citric acid 9%, intralesional INF, intralesional placebo, and polyphenon 15% achieved a CLR not significantly different from placebo. Surgery (RR 10.54; CI 95% 4.53–24.52), ablative therapy + imiquimod (RR 7.52; CI 95% 3.63–15.57), and electrosurgery (RR 7.10; CI 95% 3.47–14.53) achieved the best CLR compared to placebo. Other comparisons to placebo had RRs that ranged from 3.84 to 6.75
Fig. 2.
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Forest plot of the estimates of relative risk between each treatment and the reference placebo for complete lesion response. Data presented as RR (95% CI). Cryo cryotherapy, ablative ablative treatment (surgery or electrosurgery or CO 2 laser or cryotherapy), imi imiquimod 5%, 5-FU 5 fluorouracil, 5-FU intra intralesional 5 fluorouracil, TCA trichloroacetic acid, podo podophyllin 20–25%, citric ac citric acid 9%, polyph polyphenon 15%, podotox cr podophyllotoxin 0.5% cream, podotox cr/gel podophyllotoxin 0.5% gel + cream, podotox gel podophyllotoxin 0.5% gel, PDT photodynamic therapy, mycobac intra intralesional Mycobacterium, KOH potassium hydroxide, electro electrosurgery, INF-1a intra intralesional interferon-1α, INF-2b intra intralesional interferon-2β
Head-to-head comparisons using NMA are shown in the online supplement (Appendix S5). Surgery was more efficacious than imiquimod (RR 2.22; CI 95% 1.04–4.76), TCA (RR 2.28; CI 95% 1.09–4.75), KOH (RR 2.48; CI 95% 1.02–6.01), cryotherapy (RR 2.43; CI 95% 1.17–5.03), 5-FU (RR 2.44; CI 95% 1.07–5.56), and polyphenon (RR 7.07, CI 95% 2.82–17.72). No significant differences were found between surgery and other ablative therapies (electrosurgery, CO 2 laser), or between surgery and podophyllotoxin 0.5% solution or 0.5% cream. As regards direct comparisons (except for those involving a placebo or podophyllin), the only significant result was the superiority of CO 2 laser over cryotherapy (RR 2.40; CI 95% 1.29–4.46).
Examined RCTs presented both heterogeneity and inconsistency. The Netmeta R package provided an I 2 value of 60% from a Q statistic for the overall network of 70.7, which had a chi-square distribution with 28 degrees of freedom and yielded a p-value of 0.0001 . The Q statistic was further decomposed into heterogeneity and inconsistency components, valued at 14.7 and 56.0, respectively.
As shown in the net heat plot in Fig.3, a high inconsistency among mapping functions was found for RCTs comparing the following treatments: cryotherapy vs podophyllin 20–25% vs electrosurgery; 5-FU vs podophyllin; 5-FU vs CO 2 laser vs 5-FU + CO 2 laser; and CO 2 laser vs cryotherapy. Treatments examined in a single study were not evaluated.
Fig. 3.
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Net heat plot. Assessment of consistency between direct and indirect evidence. Horizontal: detached comparisons; vertical: comparisons observed in the network; warm color in the net heat plot indicates that significant inconsistency may arise from a specific comparison and this trend is illustrated by the intensity of the color; gray color: contribution of each direct comparison to the network estimates
Table 1 presents the SUCRA results that emerged from these data. These results confirm that ablative therapy, surgery (90.9%), and electrosurgery (77.1%) are the most efficacious treatments for AGWs. The SUCRA value of combination therapies was also good (PDT + CO 2 laser: 68.0%; CO 2 laser + 5-FU: 67.4%; Cryotherapy + podophyllotoxin 0.5% cream: 59.5%). Podophyllotoxin 0.5% solution (63.5%) and podophyllotoxin 0.5% cream (62.2%) had the highest SUCRA values of all PaTs. The SUCRA values of imiquimod, TCA, KOH, cryotherapy, and 5-FU ranged from 40 to 50%. The SUCRA value for polyphenon 15% was low at 13.1%.
Table 1.
Probabilities of treatment ranking
| | SUCRA | | SUCRA |
:--- :--- |
| Surgery | 0.909 | Imiquimod | 0.494 |
| Ablative therapy + Imiquimod | 0.798 | TCA | 0.480 |
| Electrosurgery | 0.771 | Podophyllin + TCA | 0.432 |
| Cidofovir | 0.710 | KOH | 0.430 |
| PDT | 0.707 | 5-FU | 0.426 |
| Podophyllotoxin 0.5% gel | 0.686 | Cryotherapy | 0.421 |
| PDT + CO 2 laser | 0.680 | Podophyllin 20–25% | 0.334 |
| CO 2 Laser + 5-FU | 0.674 | Intralesional IFN-2b | 0.334 |
| Intralesional 5-FU | 0.642 | Citric Acid 9% | 0.232 |
| Podophyllotoxin 0.5% solution | 0.635 | Intralesional placebo | 0.163 |
| Intralesional Mycobacterium | 0.632 | Podophyllotoxin 0.5% gel + cream | 0.132 |
| CO 2 laser | 0.626 | Polyphenon 15% | 0.131 |
| Podophyllotoxin 0.5% cream | 0.622 | Intralesional IFN-1a | 0.117 |
| Ablative therapy | 0.621 | Placebo | 0.066 |
| Cryotherapy + Podophyllotoxin 0.5% cream | 0.595 | | |
Open in a new tab
SUCRA surface under the cumulative ranking curve; ablative treatment: surgery or electrosurgery or CO 2 laser or cryotherapy, 5-FU 5 fluorouracil, TCA trichloroacetic acid, PDT photodynamic therapy, KOH potassium hydroxide, INF-1a interferon-1α, INF-2b interferon-2β
Sensitivity Analyses
Only polyphenon and podophyllin + TCA had a CI that included value 1 (Appendix S6).
Worst-case (intention to treat) scenario sensitivity analyses showed a superiority of surgery over podophyllotoxin 0.5% solution (RR 1.94; CI 95% 1.00–3.76), CO 2 laser (RR 2.20; CI 95% 1.05–4.60), electrosurgery (RR 2.28; CI 95% 1.10–4.74), and cryotherapy + podophyllotoxin 0.5% cream (RR 2.68; CI 95% 1.18–6.07). Ablative therapy + imiquimod was superior to imiquimod alone (RR 1.57; CI 95% 1.01–2.44) and to cryotherapy (RR 1.74; CI 95% 1.05–2.89). A superiority of podophyllotoxin 0.5% cream and podophyllotoxin 0.5% solution over cryotherapy was also found (RR 1.66; CI 95% 1.04–2.66 and RR 1.52; CI 95% 1.06–2.18, respectively) (Appendix S7).
Sensitivity analyses of SUCRA values confirmed the superiority of surgery and combination therapies. Worst-case scenario sensitivity analyses showed an increase in the efficacy of podophyllotoxin 0.5% cream and 0.5% solution (72.2% and 77.7%, respectively), as well as a decrease in the efficacy of electrosurgery due to the high number of patients lost to follow-up in the study by Stone et al. (Appendix S8).
Discussion
In our NMA, ProTs — mainly surgery and electrosurgery — achieved the best CLR, with a median follow-up of 6 months. These results differ from our pooled analysis, which found higher clearance for ProTs but lower recurrence at 12 months for PaTs . Few RCTs have used CLR as a study endpoint. This is unfortunate given that CLR, which assesses clearance until no recurrence, is more meaningful for patients undergoing treatment for AGWs. Combined with the more robust statistical methods of NMA, this endpoint yields more accurate results than pooled analyses. Cidofovir was ranked 4th in our SUCRA analysis. Yet, it is difficult to conclude on the efficacy of this treatment, as the only RCT on the topic found no significant difference with placebo use.
Our results are in line with the NMA of Barton et al. , which concluded that ablative techniques were superior. However, unlike us, Barton et al. recommended CO 2 laser as first-line treatment. This difference may be explained by the fact that their NMA was restricted to 39 RCTs, included immunocompromised patients and only one RCT using CO 2 laser , whereas our NMA compared 49 RCTs, focused on non-immunocompromised adults and compared 5 RCTs using CO 2 laser [59, 64, 69, 70, 92]. Moreover, Barton et al. found that podophyllotoxin 0.5% solution was the most cost-effective therapeutic solution, followed by CO 2 laser. In our NMA, podophyllotoxin 0.5% solution achieved the best CLR among all PaTs.
Unlike systematic reviews on AGW management [10, 13, 97], our NMA examined the efficacy of combination therapies, including ablative therapy + imiquimod, cryotherapy + podophyllotoxin 0.5% cream, and CO 2 laser + 5-FU. However, many combination therapies are missing from our NMA, including those most commonly recommended and used in practice: cryotherapy + imiquimod and cryotherapy + podophyllotoxin 0.5% solution. Combination therapies should be given greater consideration and should be adapted as best as possible to individual patients.
Our search was limited by restrictions on access to Chinese databases, especially regarding treatments like PDT. While our NMA results suggest that this treatment is highly efficacious, they are based on only 1 RCT (note that numerous non-randomized studies on MEDLINE have yielded the same finding [98, 99]). Other RCTs on PDT have likely been performed, but they remain inaccessible to the scientific community.
The management of AGWs is heterogeneous in terms of: the type of treatment used; the level of physician experience (for ProTs); the level of patient compliance (for PaTs); the clinical type of AGWs (papillary, flat, or pedunculated); the location of AGWs; the number of AGWs; and the sex of the patient [100, 101]. Such heterogeneity renders more difficult the establishment of a clinically meaningful hierarchy of treatments. In our systematic review, more than 90% of RCTs were found to have a high risk of bias , thus casting doubt on the validity of published recommendations. NMAs do not increase the level of evidence of risks of bias, as they remain dependent on the methodology of each RCT. But they do increase statistical power because they encompass all patients included in examined RCTs. Moreover, NMAs can be used to compare treatments that have never been compared before, to identify gaps in knowledge, and to help develop clinically meaningful hierarchies of treatments .
Only 2 RCTs in our NMA compared a recommended ProT to a recommended PaT (imiquimod and cryotherapy in both cases) [58, 90]. Future RCTs should compare recommended ProTs and PaTs—for instance, cryotherapy vs podophyllotoxin cream or solution; surgery vs imiquimod; surgery vs podophyllotoxin cream or solution; CO 2 laser vs imiquimod; and CO 2 laser vs podophyllotoxin cream or solution. Moreover, combination therapies should be more thoroughly assessed to help increase the efficacy of AGW management, and to make it better adapted to the number, type, and location of AGWs. New treatments (KOH, PDT alone or as an adjuvant) should also be evaluated further. Although 5-FU was not mentioned in guidelines until 2019 , it could be proposed as a second-line treatment in the future.
Our systematic review and our NMA should be updated regularly. Side effects should be assessed to help physicians personalize treatments for their individual patients. Lastly, study endpoints and ProT use practices (e.g., standardization of freezing or surgical procedures) should be homogenized to allow better comparison of RCTs.
Conclusions
To conclude, in our NMA, surgery and electrosurgery achieved the best CLR, and podophyllotoxin 0.5% was the most efficacious patient-administered treatment. Combined therapies should be evaluated more in future RCTs in view of their identified effectiveness. The results of future RCTs should systematically include clinical type, number and location of AGWs, and sex of the patient, to refine therapeutic indications.
Electronic supplementary material
Below is the link to the electronic supplementary material.
Supplementary file1 (PDF 2103 kb) (2.1MB, pdf)
Acknowledgements
We would like to thank Prof. Rodolphe Thiebaut for his continuous support, our research librarian Evelyne Mouillet, the dermatologists of the French Group of Dermato-Infectiology and Sexually Transmitted Diseases of the Société Française de Dermatologie and the Association des Dermatologues des Alpes du Sud, and our copy editor Arianne Dorval.
Funding
This study was supported by grants from the “Allocation jeunes chercheurs hospitaliers” 2015 and the “Programme Hospitalier de Recherche Clinique Interrégional” (No. 13 069). The journal’s Rapid Service Fee was funded by the Department of Research and Innovation of the Centre Hospitalier Universitaire de La Réunion.
Authorship
All named authors meet the International Committee of Medical Journal Editors (ICMJE) criteria for authorship for this article, take responsibility for the integrity of the work as a whole, and have given their approval for this version to be published.
Authorship Contributions
Christian Derancourt, Brigitte Milpied, and Nicolas Dupin conceptualized and designed the study. Antoine Bertolotti, Christian Derancourt, and Cyril Ferdynus participated in the acquisition, analysis, and interpretation of data. Antoine Bertolotti and Christian Derancourt drafted the initial manuscript. Laetitia Huiart, Nicolas Dupin, and Brigitte Milpied critically reviewed the manuscript. All authors read and approved the final manuscript.
Disclosures
Antoine Bertolotti, Cyril Ferdynus, Brigitte Milpied, Nicolas Dupin, Laetitia Huiart and Christian Derancourt have nothing to disclose.
Compliance with Ethics Guidelines
This article is based on previously conducted studies and does not contain any studies with human participants or animals performed by any of the authors.
Data Availability
The datasets analyzed during the current study are available from the corresponding author on reasonable request.
Footnotes
Enhanced Digital Features
To view enhanced digital features for this article go to 10.6084/m9.figshare.11604663.
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The datasets analyzed during the current study are available from the corresponding author on reasonable request.
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5314 | https://math.stackexchange.com/questions/468475/trigonometric-identities-like-a-sinx-b-cosy-cdots | Stack Exchange Network
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Trigonometric Identities Like $A \sin(x) + B \cos(y) = \cdots$
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$\begingroup$
Are there any identities for trigonometric equations of the form:
$$A\sin(x) + B\sin(y) = \cdots$$ $$A\sin(x) + B\cos(y) = \cdots$$ $$A\cos(x) + B\cos(y) = \cdots$$
I can't find any mention of them anywhere, maybe there is a good reason why there aren't identities for these? Thanks!
trigonometry
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edited Aug 15, 2013 at 18:33
Git Gud
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asked Aug 15, 2013 at 18:26
gandymeedgandymeed
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$\endgroup$
5
1
$\begingroup$ I have heard of asin(x) + bcos(x) = Acos(x+alpha) where A is in terms of a and b .Without a relation between x and y we cannot combine because of the different coefficients a and b . had they been 1 it is possible $\endgroup$
Harish Kayarohanam
– Harish Kayarohanam
2013-08-15 18:34:57 +00:00
Commented Aug 15, 2013 at 18:34
$\begingroup$ Perhaps you could be a little bit more specific in what kind of identity you are expecting. Personally I find your initial form simple enough but perhaps you are bothered by the $+$ which makes you unable to do a certain type of calculation? $\endgroup$
user88595
– user88595
2013-08-15 18:43:10 +00:00
Commented Aug 15, 2013 at 18:43
$\begingroup$ One simple special case is the identities saying $\sin x+\sin y$ $=\text{a single term, which is a product of half-angle functions}$, and similarly for $\cos x+\cos y$ and $\sin x+\cos y$. $\endgroup$
Michael Hardy
– Michael Hardy
2013-08-15 18:45:44 +00:00
Commented Aug 15, 2013 at 18:45
$\begingroup$ This seems related: Identity for a weighted sum of sines / sines with different amplitudes $\endgroup$
Martin Sleziak
– Martin Sleziak
2021-10-09 15:39:30 +00:00
Commented Oct 9, 2021 at 15:39
$\begingroup$ Also related: math.stackexchange.com/questions/4664525/… $\endgroup$
EthanAlvaree
– EthanAlvaree
2023-03-23 00:40:23 +00:00
Commented Mar 23, 2023 at 0:40
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4 Answers 4
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$\begingroup$
Since
$$ A \cos(a+b) = A \cos(a) \cos(b) - A \sin(a) \sin(b) \ \ \ \ \ \ (1) \ B \cos(a-b) = B \cos(a) \cos(b) + B \sin(a) \sin(b) \ \ \ \ \ \ (2) $$
(1) + (2) gives
$$ A \cos(x) + B \cos(y) = (A+B) \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + (B-A) \sin(\frac{x+y}{2}) \sin(\frac{x-y}{2}) $$
where
$$ x = a + b \ y = a - b $$
substitute
$$ Q = (A+B) \cos(\frac{x-y}{2}) \ R = (B-A) \sin(\frac{x-y}{2}) \ P = \frac{x+y}{2} $$
then
$$ A \cos(x) + B \cos(y) = Q \cos P + R \sin P = \sqrt{Q^2+R^2} \cos(P-\phi) $$
where
$$ \sin \phi = \frac{R}{\sqrt{Q^2+R^2}} \ \cos \phi = \frac{Q}{\sqrt{Q^2+R^2}} $$
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answered Oct 9, 2021 at 14:45
EfeEfe
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1
$\begingroup$ Thanks for providing a formula for $A \cos (x) + B \cos (y)$. Can you also share a formula for $A \sin (x) + B \sin (y)$? $\endgroup$
EthanAlvaree
– EthanAlvaree
2023-03-22 23:42:02 +00:00
Commented Mar 22, 2023 at 23:42
Add a comment |
1
$\begingroup$
there are no general formula for these expressions.but may exist when $A$ and $B$ are interrelated .
For example consider triangle $ABC$ where $a,b,\text{ and }c $ are the sides of the triangle and $A,B,\text{ and }C$ are the respective angles opposite to $a,b,\text{ and }c $ then $$c = a\cos B + b\cos A $$ here this is because $a,b ,A\text{ and }B$ are interrelated by laws of triangle.
therefore random values of the angles and the coefficients will not satisfy to form general formula.
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edited Aug 15, 2013 at 19:28
Michael Hardy
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answered Aug 15, 2013 at 18:53
Suraj M SSuraj M S
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$\begingroup$
Since the proof was asked for Asin(x)+Bsin(y), based on the prior answer, I expand this time with a Python example for some verification:
$$ A \sin(a+b) = A \sin(a) \cos(b) + A \cos(a) \sin(b) \ \ \ \ \ \ (1) \ B \sin(a-b) = B \sin(a) \cos(b) - B \cos(a) \sin(b) \ \ \ \ \ \ (2) $$
(1) + (2) gives
$$ A \sin(x) + B \sin(y) = (A+B) \sin(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + (A-B) \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2}) $$
where
$$ x = a + b \ y = a - b $$
substitute
$$ Q = (A+B) \sin(\frac{x+y}{2}) \ R = (A-B) \cos(\frac{x+y}{2}) \ P = \frac{x-y}{2} $$
then
$$ A \sin(x) + B \sin(y) = Q \cos P + R \sin P = \sqrt{Q^2+R^2} \sin(P+\phi) $$
where
$$ \tan \phi = \frac{Q}{R} \ $$
Proof in Python:
import numpy as np A, B, x, y = 3, 4, 5, 6 Q, R, P = (A+B)np.sin((x+y)/2), (A-B)np.cos((x+y)/2), (x-y)/2 print(Anp.sin(x)+Bnp.sin(y), np.sqrt(QQ+RR)np.sin(P+np.arctan2(Q, R)))
Result:
-3.9944348167851187 -3.9944348167851187
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answered May 24, 2023 at 21:13
Gregory MorseGregory Morse
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Add a comment |
-1
$\begingroup$
$A \, \cos(x) + B \, \cos(y)= C \, \cos(z)$, where, $$ C = \sqrt{(A \, \cos(x) + B \, \cos(y))^2 + (A \, \sin(x) + B \, \sin(y))^2}, $$ and $$ z = \tan^{-1}\left(\frac{A \, \sin(x) + B \, \sin(y)}{A \, \cos(x) + B \, \cos(y)}\right). $$
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edited Oct 27, 2022 at 0:44
Leucippus
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answered Oct 26, 2022 at 20:09
Walid HubbiWalid Hubbi
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$\begingroup$ Please format your math with MathJax. For some basic information about writing mathematics at this site see, e.g., here, here, here and here. $\endgroup$
Christian E. Ramirez
– Christian E. Ramirez
2022-10-26 20:25:46 +00:00
Commented Oct 26, 2022 at 20:25
$\begingroup$ Was this answer downvoted because it is incorrect, or because it was poorly formatted? Sources (or proofs) would be nice. I stumbled across this question and am also looking for formulas for $A \cos (x) + B \cos (y)$ and $A \sin (x) + B \sin (y)$. $\endgroup$
EthanAlvaree
– EthanAlvaree
2023-03-22 23:43:19 +00:00
Commented Mar 22, 2023 at 23:43
$\begingroup$ I assume it was downvoted because it is super artificial. We start with the formula $$u = \sqrt{u^2+v^2} \cos \varphi,$$ where $\varphi = \tan^{-1} \frac{v}{u}$, and randomly plug in $u = A \cos(x) + B \cos(y)$ and $v = A \sin(x) + B \sin(y)$. It obviously says nothing about the expression $A \cos(x) + B \cos(y)$ because it doesn't use its particular form. $\endgroup$
Adayah
– Adayah
2025-02-03 15:54:40 +00:00
Commented Feb 3 at 15:54
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5315 | https://en.wikibooks.org/wiki/en:Algebra/Chapter_12/Logarithms | Algebra/Chapter 12/Logarithms - Wikibooks, open books for an open world
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The Wikibooks community is developing a policy on the use of generative AI. Please review the draft policy and provide feedback on its talk page.
Contents
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Beginning
1 Logarithms
2 Properties of LogarithmsToggle Properties of Logarithms subsection
2.1 Basic properties
2.2 Proof
2.3 Examples
2.4 More properties
2.5 Ambiguity
3 Properties of Logarithms
4 Change of Base Formula
5 Swap of Base and Exponent Formula
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From Wikibooks, open books for an open world
<Algebra | Chapter 12
Logarithms (commonly called "logs") are a specific instance of a function being used for everyday use. Logarithms are used commonly to measure earthquakes, distances of stars, economics, and throughout the scientific world. It basically answers the question: what power do I have to raise this base to, to get this result.
Logarithms
[edit | edit source]
In order to understand logs, we need to review exponential equations. Answer the following problems:
1
What is 4 to the power of 3?
2
What is 3 to the power of 4?
3
2 5={\displaystyle 2^{5}=}
4
5 2={\displaystyle 5^{2}=}
Just like there is a way to say and write "4 to the power of 3" or "4 3{\displaystyle 4^{3}!}, there is a specific way to say and write logarithms.
For example, "4 to the power of 3 equals 64" can be written as: 4 3=64{\displaystyle 4^{3}=64!}
However, it can also be written as:
log 4(64)=3{\displaystyle \log _{4}(64)=3\ }
Once, you remember that the base of the exponent is the number being raised to a power and that the base of the logarithm is the subscript after the log, the rest falls into place. I like to draw an arrow (either mentally or physically) from the base, to the exponent, to the product when changing from logarithmic form to exponential form. So visually or mentally I would go from 2 to 5 to 32 in the logarithmic example which (once I add the conventions) gives us: 2 5=32{\displaystyle 2^{5}=32!}
So, when you are given a logarithm to solve, just remember how to convert it to an exponential equation. Here are some practice problems, the answers are at the bottom.
Properties of Logarithms
[edit | edit source]
The following properties derive from the definition of logarithm.
Basic properties
[edit | edit source]
For all real numbers a,b,c,d,y>0{\displaystyle a,b,c,d,y>0} with b≠1,d≠1{\displaystyle b\neq 1,d\neq 1}, we have
log b(y a)=a log b(y){\displaystyle \log {b}(y^{a})=a\log {b}(y)}
log b(b a)=a{\displaystyle \log _{b}(b^{a})=a}
log b(a c)=log b(a)+log b(c){\displaystyle \log {b}(ac)=\log {b}(a)+\log _{b}(c)}
log b(a/c)=log b(a)−log b(c){\displaystyle \log {b}(a/c)=\log {b}(a)-\log _{b}(c)}
log b(a)=log d(a)log d(b){\displaystyle \log {b}(a)={\frac {\log {d}(a)}{\log _{d}(b)}}\quad } (change of base rule).
Proof
[edit | edit source]
Let us take the log to base d of both sides of the equation b c=a{\displaystyle b^{c}=a}:
log d(b c)=log d(a){\displaystyle \log {d}(b^{c})=\log {d}(a)}.
Next, notice that the left side of this equation is the same as that in property number 1 above. Let us apply this property:
c log d(b)=log d(a){\displaystyle c\log {d}(b)=\log {d}(a)}
Isolating c on the left side gives
c=log d(a)log d(b){\displaystyle c={\frac {\log {d}(a)}{\log {d}(b)}}}
Finally, since c=log b(a){\displaystyle c=\log _{b}(a)}
log b(a)=log d(a)log d(b){\displaystyle \log {b}(a)={\frac {\log {d}(a)}{\log _{d}(b)}}}
Examples
[edit | edit source]
This rule allows us to evaluate logs to a base other than e or 10 on a calculator. For example,
log 3(12)=log 10(12)log 10(3)=2.262{\displaystyle \log {3}(12)={\frac {\log {10}(12)}{\log _{10}(3)}}=2.262}
Solve these logarithms
1
log 3(81)={\displaystyle \log _{3}(81)=}
2
log 6(216)={\displaystyle \log _{6}(216)=}
3
log 4(64)={\displaystyle \log _{4}(64)=}
4
Evaluate with a calculator (to 5dp)
log 4(6)={\displaystyle \log _{4}(6)=}
Find the y value of these logarithms
5
log 3(y)=3{\displaystyle \log _{3}(y)=3}
y=
6
log 5(y)=4{\displaystyle \log _{5}(y)=4}
y=
7
log 9(y)=4{\displaystyle \log _{9}(y)=4}
y=
More properties
[edit | edit source]
Logarithms are the reverse of exponential functions, just as division is the reverse of multiplication. For example, just as we have
5×6=30{\displaystyle 5\times 6=30}
and
30/6=5{\displaystyle 30/6=5}
we also have
7 3=343{\displaystyle 7^{3}=343}
and
log 7343=3{\displaystyle \log {7}343=3}
More generally, if a b=x{\displaystyle a^{b}=x}, then log ax=b{\displaystyle \log {a}x=b}. Also, if f(x)=a x{\displaystyle f(x)=a^{x}}, then f−1(x)=log ax{\displaystyle f^{-1}(x)=\log {a}x}, so if the two equations are graphed, each one is the reflection of the other over the line y=x{\displaystyle y=x}. (In both equations, _a is called the base.)
As a result, a log ab=b{\displaystyle a^{\log {a}b}=b} and log aa b=b{\displaystyle \log {a}a^{b}=b}.
Common bases for logarithms are the base of 10 (log 10x{\displaystyle \log {10}x} is known as the _common logarithm) and the base e (lnx{\displaystyle \ln x} is known as the natural logarithm), where e = 2.71828182846...
Natural logs are usually written as lnx{\displaystyle \ln x} or ln(x){\displaystyle \ln(x)} (ln is short for natural logarithm in Latin), and sometimes as log ex{\displaystyle \log {e}x} or log e(x){\displaystyle \log {e}(x)}. Parenthesized forms are recommended when x is a mathematical expression (e.g., ln(6 x+1){\displaystyle \ln(6x+1)}).
Logarithms are commonly abbreviated as logs.
Ambiguity
[edit | edit source]
The notation logx{\displaystyle \log x} may refer to either lnx{\displaystyle \ln x} or log 10x{\displaystyle \log {10}x}, depending on the country and the context. For example, in English-speaking schools, logx{\displaystyle \log x} usually means lnx{\displaystyle \ln x}, whereas it means log 10x{\displaystyle \log {10}x} in Italian- and French-speaking schools or to English-speaking number theorists. Consequently, this notation should only be used when the context is clear.
Properties of Logarithms
[edit | edit source]
log ax+log ay=log ax∗y{\displaystyle \log {a}x+\log {a}y=\log _{a}xy}
log ax−log ay=log ax y{\displaystyle \log {a}x-\log {a}y=\log _{a}{\frac {x}{y}}}
log ax b=b×log ax{\displaystyle \log {a}x^{b}=b\times \log {a}x}
Proof:
log ax+log ay=log ax∗y{\displaystyle \log {a}x+\log {a}y=\log _{a}xy}
log ax+log ay{\displaystyle \log {a}x+\log {a}y}
log ax=b{\displaystyle \log {a}x=b} and log ay=c{\displaystyle \log {a}y=c}
a b=x{\displaystyle \ a^{b}=x} and a c=y{\displaystyle \ a^{c}=y}
x y=a b a c{\displaystyle \ xy=a^{b}a^{c}}
x y=a(b+c){\displaystyle \ xy=a^{(b+c)}}
log ax y=b+c{\displaystyle \log _{a}xy=b+c}
and replace b and c (as above)
log ax y=log ax+log ay{\displaystyle \log {a}xy=\log {a}x+\log _{a}y}
Change of Base Formula
[edit | edit source]
log yx=log ax log ay{\displaystyle \log {y}x={\frac {\log {a}x}{\log {a}y}}} where _a is any positive number, distinct from 1. Generally, a is either 10 (for common logs) or e (for natural logs).
Proof:
log yx=b{\displaystyle \log _{y}x=b}
y b=x{\displaystyle \ y^{b}=x}
Put both sides to log a{\displaystyle \log _{a}}
log ay b=log ax{\displaystyle \log {a}y^{b}=\log {a}x}
b log ay=log ax{\displaystyle \ b\log {a}y=\log {a}x}
b=log ax log ay{\displaystyle \ b={\frac {\log {a}x}{\log {a}y}}}
Replace b{\displaystyle \ b} from first line
log yx=log ax log ay{\displaystyle \log {y}x={\frac {\log {a}x}{\log _{a}y}}}
Swap of Base and Exponent Formula
[edit | edit source]
a log bc=c log ba{\displaystyle a^{\log {b}c}=c^{\log {b}a}} where a or c must not be equal to 1.
Proof:
l o g a b=1 l o g b a{\displaystyle log_{a}b={\frac {1}{log_{b}a}}} by the change of base formula above.
Note that a=c l o g c a{\displaystyle a=c^{log_{c}a}}. Then
a l o g b c{\displaystyle a^{log_{b}c}} can be rewritten as
(c l o g c a)l o g b c{\displaystyle ({c^{log_{c}a}})^{log_{b}c}} or by the exponential rule as
c l o g c a∗l o g b c{\displaystyle c^{{log_{c}a}{log_{b}c}}}
using the inverse rule noted above, this is equal to
c l o g c a∗1 l o g c b{\displaystyle c^{{log_{c}a}{\frac {1}{log_{c}b}}}}
and by the change of base formula
c l o g b a{\displaystyle c^{log_{b}a}}
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5316 | https://topdrawer.aamt.edu.au/Patterns/Misunderstandings/Rectangular-grids/Rectangular-arrays | Geometric reasoning
Mental computation
Tartans and nursery rhymes
Mathematical patterns are regular
Find the regularity
Repeating patterns
One-dimensional repeating patterns
Two-dimensional repeating patterns
Repeating patterns and multiplication
Growing patterns
Making a staircase
Patterns in shapes
The language of shapes
Misunderstandings
Identical units of repeat
It's harder than you think
How do you teach the unit of repeat?
Making border patterns
Rectangular grids
Structure of rectangular grids
Drawing rectangular grids
Rectangular arrays
Number lines
Drawing and reading number lines
Number lines and repeating patterns
Drawing an analogue clock face
Number sequences
Introducing number sequences
Let's have a party!
Find a rule
Tessellations
Triangles tessellate
Making tessellations
Growing patterns can tessellate
Good teaching
Familiar situations
A pattern hunt
Patterns in school activities
Repeating patterns in school activities
Concrete to abstract
Making number sequences real
Abstraction and generalisation
Abstraction of repeating patterns
Generalisation in growing patterns
Using technology
Examples of ICT resources
Assessment
Approaches to assessment
Observation
Analysis of student work
Individual interviews
Group assessment
Activities
A hundred square
Bee-Bots
Drawing an analogue clock face
Folding patterns
Growing fractions
Growing patterns can tessellate
Growing polygons
Let's have a party!
Making an AAB pattern
Making border patterns
Monster choir patterns
Odds and evens
Picture books
Rectangular arrays
Repeating patterns in the environment
Sports day
Symmetrical patterns in the environment
When are two patterns the same?
Downloads
Acknowledgements
Home
Patterns
Misunderstandings
Rectangular grids
Rectangular arrays
Rectangular arrays
A rectangular array consists of a set of objects arranged in a similar way to a rectangular grid. It is one way of packing objects (e.g. eggs) efficiently.
This is an array of dots.
A 3 × 4 array.
Activities with arrays can help students understand the rectangular grid pattern. Here are some suggestions.
Supply students with drawings of various rectangular arrays of black dots and coloured counters.
Ask them to create their own colour patterns by placing the counters on the dots.
Ask students to draw arrays of various sizes from memory.
Supply students with drawings of various rectangular arrays of dots and ask them to count the number of dots in each array.
Then ask them to extend each array by one row and one column, and count the number of dots again.
Through such exercises, students will learn that the dots are aligned and equally spaced.
More advanced learners can be challenged to copy and extend a triangular grid pattern of dots.
A triangular array. |
5317 | https://www.drugs.com/dosage/pyrimethamine.html | Skip to main content
Pyrimethamine Dosage
Medically reviewed by Drugs.com. Last updated on Jul 31, 2025.
Applies to the following strengths: 25 mg
Usual Adult Dose for:
Toxoplasmosis
Toxoplasmosis - Prophylaxis
Pneumocystis Pneumonia Prophylaxis
Protozoan Infection
Usual Pediatric Dose for:
Toxoplasmosis
Toxoplasmosis - Prophylaxis
Pneumocystis Pneumonia Prophylaxis
Protozoan Infection
Additional dosage information:
Renal Dose Adjustments
Liver Dose Adjustments
Precautions
Dialysis
Other Comments
Usual Adult Dose for Toxoplasmosis
Starting dose: 50 to 75 mg orally once a day (with 1 to 4 g/day of a sulfapyrimidine-type sulfonamide [e.g., sulfadoxine])
Comments:
This starting dose is generally continued for 1 to 3 weeks, depending on patient response and tolerance to therapy.
After using the starting dose for 1 to 3 weeks, the dosage for each drug may be reduced to about one-half and continued for an additional 4 to 5 weeks.
The dose must be carefully adjusted to provide maximum therapeutic effect and minimum of side effects; young patients may tolerate higher doses than older patients.
Use: With a sulfonamide, for the treatment of toxoplasmosis
US CDC Recommendations:
Loading dose: 100 mg orally on the first day
Maintenance dose: 25 to 50 mg orally per day
Duration of therapy: 4 to 6 weeks
US CDC, National Institutes of Health (NIH), and HIV Medicine Association of the Infectious Diseases Society of America (HIVMA/IDSA) Recommendations for HIV-Infected Adults:
Loading dose: 200 mg orally once
Maintenance dose:
Weight less than 60 kg: 50 mg orally once a day
Weight at least 60 kg: 75 mg orally once a day
Duration of Therapy: At least 6 weeks
Comments:
US CDC: With leucovorin and sulfadiazine, recommended for ocular toxoplasmosis (the classic therapy)
US CDC: Treatment for ocular diseases should be based on a complete ophthalmologic evaluation; therapy should be followed by reevaluation of the patient's condition.
HIV-Infected Adults: With leucovorin and sulfadiazine, recommended as part of the preferred regimen for treating Toxoplasma gondii encephalitis; if this drug is unavailable or obtaining it is delayed, sulfamethoxazole-trimethoprim should be used instead of this drug and sulfadiazine.
HIV-Infected Adults: With leucovorin and (clindamycin or atovaquone), recommended as part of alternative regimens for treating T gondii encephalitis
HIV-Infected Adults: Duration of therapy may be longer if clinical/radiologic disease is extensive or incomplete response at 6 weeks.
HIV-Infected Adults: After completion of acute therapy, all patients should be continued on chronic maintenance therapy.
Current guidelines should be consulted for additional information.
Usual Adult Dose for Toxoplasmosis - Prophylaxis
US CDC, NIH, and HIVMA/IDSA Recommendations for HIV-Infected Adults:
Primary Prophylaxis: 50 or 75 mg orally once a week OR 25 mg orally once a day
Chronic Maintenance Therapy:
Preferred regimen and alternative regimen with clindamycin: 25 to 50 mg orally once a day
Alternative regimen with atovaquone: 25 mg orally once a day
Comments:
With (dapsone or atovaquone) and leucovorin, recommended as part of an alternative regimen for preventing the first episode of T gondii encephalitis (primary prophylaxis)
With leucovorin and sulfadiazine, recommended as part of the preferred regimen (and with leucovorin and [clindamycin or atovaquone], recommended as part of alternative regimens) for chronic maintenance therapy (secondary prophylaxis) for T gondii encephalitis
Current guidelines should be consulted for additional information.
Usual Adult Dose for Pneumocystis Pneumonia Prophylaxis
US CDC, NIH, and HIVMA/IDSA Recommendations for HIV-Infected Adults:
Primary and Secondary Prophylaxis: 50 or 75 mg orally once a week OR 25 mg orally once a day with food
Comments:
With (dapsone or atovaquone) and leucovorin, recommended as part of an alternative regimen for preventing the first episode of Pneumocystis pneumonia (PCP) (primary prophylaxis) and preventing subsequent episodes of PCP (secondary prophylaxis)
Current guidelines should be consulted for additional information.
Usual Adult Dose for Protozoan Infection
US CDC, NIH, and HIVMA/IDSA Recommendations for HIV-Infected Adults:
Acute Infection: 50 to 75 mg orally once a day
Chronic Maintenance Therapy: 25 mg orally once a day
Comments:
With leucovorin, recommended as alternative therapy for treating cystoisosporiasis (formerly isosporiasis) due to Cystoisospora belli (formerly Isospora belli); recommended for acute infection and for chronic maintenance therapy (secondary prophylaxis) in patients with CD4 count less than 200 cells/mm3
Current guidelines should be consulted for additional information.
Usual Pediatric Dose for Toxoplasmosis
1 mg/kg/day orally divided into 2 equal daily doses; after 2 to 4 days, may reduce to one-half and continue for about 1 month
Comments:
The dose must be carefully adjusted to provide maximum therapeutic effect and minimum of side effects; young patients may tolerate higher doses than older patients.
The usual pediatric sulfonamide dosage should be used with this drug.
Use: With a sulfonamide, for the treatment of toxoplasmosis
US CDC Recommendations for Pediatrics:
Congenitally-infected neonates and infants:
Loading dose: 2 mg/kg/day orally in 2 divided doses for the first 2 days
Maintenance dose (starting day 3): 1 mg/kg orally per day for 2 months (or 6 months if symptomatic), then 1 mg/kg orally 3 times a week
Duration of therapy: 12 months
Ocular toxoplasmosis:
Loading dose: 2 mg/kg orally on the first day
Maintenance dose: 1 mg/kg orally per day
Duration of therapy: 4 to 6 weeks
US CDC, NIH, HIVMA/IDSA, and Pediatric Infectious Disease Society (PIDS) Recommendations for HIV-Exposed and HIV-Infected Children:
Congenital Toxoplasmosis:
Loading dose: 2 mg/kg orally once a day for 2 days
Maintenance dose: 1 mg/kg orally once a day for 2 to 6 months, then 1 mg/kg orally 3 times a week
Duration of therapy: 12 months
Acquired Toxoplasmosis (Acute Induction Therapy):
Loading dose: 2 mg/kg orally once a day for 3 days
Maximum dose: 50 mg/dose
Maintenance dose: 1 mg/kg orally once a day
Maximum dose: 25 mg/dose
Duration of therapy: At least 6 weeks
US CDC, NIH, and HIVMA/IDSA Recommendations for HIV-Infected Adolescents:
Loading dose: 200 mg orally once
Maintenance dose:
Weight less than 60 kg: 50 mg orally once a day
Weight at least 60 kg: 75 mg orally once a day
Duration of Therapy: At least 6 weeks
Comments:
US CDC: With leucovorin and sulfadiazine, recommended for ocular toxoplasmosis (the classic therapy) and congenital toxoplasmosis
US CDC: Treatment for ocular diseases should be based on a complete ophthalmologic evaluation; therapy should be followed by reevaluation of the patient's condition.
HIV-Infected Children: With leucovorin and sulfadiazine, recommended as part of the first-choice regimen for congenital toxoplasmosis and acquired toxoplasmosis
HIV-Infected Adolescents: With leucovorin and sulfadiazine, recommended as part of the preferred regimen for treating T gondii encephalitis; if this drug is unavailable or obtaining it is delayed, sulfamethoxazole-trimethoprim should be used instead of this drug and sulfadiazine.
HIV-Infected Adolescents: With leucovorin and (clindamycin or atovaquone), recommended as part of alternative regimens for treating T gondii encephalitis
HIV-Infected Children (acquired toxoplasmosis) and Adolescents: Duration of therapy may be longer if clinical/radiologic disease is extensive or incomplete response at 6 weeks.
HIV-Infected Children (acquired toxoplasmosis) and Adolescents: After completion of acute induction therapy/acute therapy, all patients should be continued on chronic suppressive/maintenance therapy.
Current guidelines should be consulted for additional information.
Usual Pediatric Dose for Toxoplasmosis - Prophylaxis
US CDC, NIH, HIVMA/IDSA, and PIDS Recommendations for HIV-Exposed and HIV-Infected Children:
Primary and Secondary Prophylaxis: 1 mg/kg or 15 mg/m2 orally once a day
Maximum dose: 25 mg/dose
US CDC, NIH, and HIVMA/IDSA Recommendations for HIV-Infected Adolescents:
Primary Prophylaxis: 50 or 75 mg orally once a week OR 25 mg orally once a day
Chronic Maintenance Therapy:
Preferred regimen and alternative regimen with clindamycin: 25 to 50 mg orally once a day
Alternative regimen with atovaquone: 25 mg orally once a day
Comments:
Children: With leucovorin and (dapsone [aged 1 month or older] or atovaquone [aged 4 to 24 months]), recommended as part of an alternative regimen for primary prophylaxis of toxoplasmosis
Children: With leucovorin and sulfadiazine, recommended as part of the first-choice regimen for secondary prophylaxis (suppressive therapy) of toxoplasmosis; with leucovorin and (atovaquone [aged 4 to 24 months] or clindamycin), recommended as part of alternative regimens for secondary prophylaxis of toxoplasmosis
Children: Atovaquone plus leucovorin may be used with or without this drug for primary and secondary prophylaxis.
Adolescents: With (dapsone or atovaquone) and leucovorin, recommended as part of an alternative regimen for preventing the first episode of T gondii encephalitis (primary prophylaxis)
Adolescents: With leucovorin and sulfadiazine, recommended as part of the preferred regimen (and with leucovorin and [clindamycin or atovaquone], recommended as part of alternative regimens) for chronic maintenance therapy (secondary prophylaxis) for T gondii encephalitis
Current guidelines should be consulted for additional information.
Usual Pediatric Dose for Pneumocystis Pneumonia Prophylaxis
US CDC, NIH, and HIVMA/IDSA Recommendations for HIV-Infected Adolescents:
Primary and Secondary Prophylaxis: 50 or 75 mg orally once a week OR 25 mg orally once a day with food
Comments:
With (dapsone or atovaquone) and leucovorin, recommended as part of an alternative regimen for preventing the first episode of PCP (primary prophylaxis) and preventing subsequent episodes of PCP (secondary prophylaxis)
Current guidelines should be consulted for additional information.
Usual Pediatric Dose for Protozoan Infection
US CDC, NIH, HIVMA/IDSA, and PIDS Recommendations for HIV-Exposed and HIV-Infected Children:
Treatment: 1 mg/kg orally once a day for 14 days
Secondary Prophylaxis: 1 mg/kg orally once a day
Maximum dose: 25 mg/dose
US CDC, NIH, and HIVMA/IDSA Recommendations for HIV-Infected Adolescents:
Acute Infection: 50 to 75 mg orally once a day
Chronic Maintenance Therapy: 25 mg orally once a day
Comments:
Children: With leucovorin (folinic acid), recommended as an alternative regimen for the treatment of cystoisosporiasis and for secondary prophylaxis
Children (treatment): The optimum duration of therapy has not been established.
Adolescents: With leucovorin, recommended as alternative therapy for treating cystoisosporiasis due to C belli; recommended for acute infection in patients with sulfa intolerance and for chronic maintenance therapy (secondary prophylaxis) in patients with CD4 count less than 200 cells/mm3
Current guidelines should be consulted for additional information.
Renal Dose Adjustments
Renal dysfunction: Caution recommended.
Liver Dose Adjustments
Liver dysfunction: Caution recommended.
Precautions
CONTRAINDICATIONS:
Known hypersensitivity to the active component or any of the ingredients; documented megaloblastic anemia due to folate deficiency
Consult WARNINGS section for additional precautions.
Dialysis
Data not available
Other Comments
Administration advice:
All patients being treated for toxoplasmosis: Coadminister leucovorin (strongly recommended).
May administer with meals
General:
This drug should be used with a sulfonamide since synergism exists with this combination.
The manufacturer product information for the relevant sulfonamide (synergistic agent) should be consulted.
Monitoring:
Hematologic: Blood counts, including platelet counts, in patients receiving high dosage (semiweekly)
Patient advice:
Stop this drug and seek medical attention immediately at first sign of skin rash.
Stop this drug and seek medical treatment if sore throat, pallor, purpura, or glossitis develops; may be early indications of serious disorders
Females of childbearing potential: Avoid becoming pregnant during therapy.
Do not exceed recommended doses.
May minimize anorexia and vomiting by taking the drug with meals
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Toxoplasmosis
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5318 | https://www.youtube.com/watch?v=plcIjtaTNVs | OpenStax: Algebra and Trigonometry - Chapter 5, Section 1 | Quadratic Functions
Scalar Learning
117000 subscribers
30 likes
Description
2127 views
Posted: 25 Jun 2023
Welcome to Huzefa’s explanation video of OpenStax Algebra and Trigonometry textbook. This is a full walkthrough of Chapter 5, Polynomial and Rational Functions, Section 1, Quadratic Functions. Watch Huzefa as he reviews exercises 1-75 odd.
Credit:
Link to Exercises:
To skip to a particular question, use the chapters below:
00:00 Introduction
00:15 Exercise 1
01:04 Exercise 3
01:41 Exercise 5
02:21 Exercise 7
03:43 Exercise 9
05:05 Exercise 11
06:13 Exercise 13
08:04 Exercise 15
10:01 Exercise 17
11:35 Exercise 19
12:49 Exercise 21
14:14 Exercise 23
15:35 Exercise 25
16:44 Exercise 27
18:16 Exercise 29
19:19 Exercise 31
20:57 Exercise 33
21:51 Exercise 35
24:24 Exercise 37
26:46 Exercise 39
30:10 Exercise 41
31:09 Exercise 43
32:49 Exercise 45
34:23 Exercise 47
35:46 Exercise 49
37:31 Exercise 51
38:25 Exercise 53
39:14 Exercise 55
40:11 Exercise 57
40:46 Exercise 59
41:29 Exercise 61
43:05 Exercise 63
44:16 Exercise 65
45:53 Exercise 67
49:10 Exercise 69
50:32 Exercise 71
51:50 Exercise 73
52:33 Exercise 75
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2 comments
Transcript:
Introduction foreign what's up everybody and welcome back to openstack's algebra in trigonometry chapter 5 Section 1 quadratic functions let's do it question number one explain the advantage of writing a quadratic function in standard form so I just want Exercise 1 to put this out to start is that some textbooks consider standard form to be ax squared plus BX plus C but in this textbook standard form is what other textbooks call vertex form okay so this is going to be our standard form right now which is a times x minus H squared plus K and the great thing about standard form right in in this context is that you can easily identify the vertex as H comma K right if this was like minus two and plus five then the vertex would be at two comma five the other good thing is that you can also see which way this Parabola is going if a is positive it's going up if a is negative it's going down so that's that boom done explain why the condition of a equal does not equal zeros imposed on Exercise 3 the definition of a quadratic well the reason why is if we think about standard form and we've got a times x minus H squared plus K right if a is zero in this format this whole piece goes away so you have no x squared really it becomes a a constant linear function alternatively even in this other format ax squared plus BX plus C same thing if a is zero it kills the x squared component which means it would be a linear function x to the first degree so that's why we have the condition of a cannot equal zero what two algebraic methods can be used to find the horizontal intercepts or x-intercepts of Exercise 5 a quadratic function so the first one that I usually try is factoring and factoring can be done if and only if the quadratic is indeed factorable right and then you usually get nice whole number integers sorry not a whole number actually integer values for the for the zeros or for the x-intercepts when factoring doesn't work another method is using the quadratic formula okay and the quadratic formula is great because it work it always works whether it's factorable whether it's not it's gonna work to help you find those values so those are the those are the two methods boom done so for seven we're gonna Exercise 7 rewrite this in standard form again some other textbooks we'll call this vertex form so let's go through that process so the first thing I'm going to do is I'm going to kind of separate out the x squared and the X components and what I'm going to do is I'm going to do something called completing the square where I'm going to take this B value that middle value and I'm going to take half of it meaning divided by 2 and then I'm going to square that amount so 2 over 2 is 1 squared is 1. so I'm going to add 1 inside here while I'm adding one I'm also going to subtract one that's because I don't want it to actually change the equation that's going to balance it out if I'm adding and subtracting 1 it's like doing nothing but we're going to see how this helps us get this in standard form so then these combine to be negative 4 right negative 3 minus 1. and then this becomes a perfect square and that's the idea when this is a perfect square we can fact after it we've we've done that by Design and it becomes X Plus 1 squared right or another way to write it would be X Plus 1 times X plus one so then it becomes that nice perfect square so now this is the function G of X in vertex form and now we can identify the vertex which is right this is my H value but again it's x minus H so it's really a vertex of negative one for the x value and then for the Y value it's exactly what we have here which is negative four so there's the vertex boom done to put Exercise 9 this one into standard form we're going to reformat it as follows so I'm just kind of spacing that c term out here and then we're going to complete the square with these two components by taking half of that middle term which is five over two and then we're squaring it B over 2 squared which becomes 25 fourths but I can't just add 25 ports I must also subtract 25 4 out here to keep it all balanced because we don't want to actually change the function we're just reformatting it negative 2 is the same as uh eight over four and now this gives us common denominators so we can combine these guys negative eight over four minus 25 over 4 is negative 33 over four and then this is a nice perfect square trinomial which means we can Factor it perfectly to be X plus five over two so you notice when you are completing the square and reformatting it the that what you're going to put inside here is X plus whatever that b term are 2 was so here it's five over two and then if you foiled this out you'd see you'd come back to this so here this is in that nice standard form last but not least let's get that vertex it's the opposite of this value for the x coordinate of the vertex so it's negative 5 over 2 comma exactly this negative 33 over four there's the vertex boom done for this Exercise 11 one my first step to get it into standard form is to factor that leading coefficient that a value out from just these two so we have x squared minus 2x okay and then we still have that minus 9 out here now what we're going to do is we're going to do our typical completing the square we're going to take half of this value which is one or negative one and then Square it which is positive one so we're going to add one out here but notice this is not just adding a 1 because if you distribute that 3 and you're really adding a 3. so to counteract that we need to subtract 3 on the outside so that it's fully balanced out now what I'm going to do is I'm I'm going to combine negative 9 minus 3 which is negative 12 and then this is a perfect square trinomial and that factors as we should expect to be x minus 1 squared again just as a reminder this comes from B over 2. okay as a little shortcut so now we have this in this great standard form format and now we can identify the vertex which is positive one the opposite of this and negative 12 which is exactly what we have here there's the vertex boom done so for this one we're going to start by factoring out that leading term that a Exercise 13 value so it's x squared minus 5 over 3 right because we're dividing that by three five over three x and we're going to leave a little room and then out here is the minus one now we're going to complete the square so we're going to take half of this which is time like times one half so it's five over six negative but it doesn't matter so much because we're squaring it anyways and then that becomes 25 over 36 so that's what we add here but then we also have to subtract whatever we're adding in here out here so we keep it balanced but again it's not just a 25 over 36 it's 3 times that which is 75 over 36. so we have to minus 75 over 36. now I'm going to change that 1 to 36 over 36 it's of course negative so now I can combine negative 36 with negative 75 which is negative 111 over 36 on the outside and then up here we're going to factor this and again it's going to be x minus this value before we squared it 5 over 6. if you were to foil this out you'd see that we would come back to this so now we have this in this really great standard form and now let's get the vertex so the vertex would be the opposite of this which is positive 5 over 6 and then this value for for the for the Y value of the vertex but again actually this simplifies right both of these can be divided by three so let's do that 1 11 divided by 3 is 37 and 36 divided by 3 is 12. so we got a little simplified improper fraction now and then we get negative 37 over 12. that's the vertex boom done for this one we're meant to Exercise 15 figure out whether this has a minimum or a maximum then we're meant to find that minimum or maximum and then also identify the axis of symmetry so we can use a nice little trick when this is in this format to get the vertex which will be either the point of the minimum or the maximum but what is it well since a is positive that means this Parabola goes up like this which means the vertex is going to be at the minimum so this will have a minimum at x equals negative B over 2A right so negative B would be negative negative 10 which would be positive 10 over 2 times a so that becomes 10 over 4 or 5 over 2. so the vertex is occurring at 5 over 2 and then that that corresponding minimum function value we get by plugging 5 over 2 in to this equation so when we plug it in we get 2 times 5 over 2 squared which is 25 over 4 minus 10 times 5 over 2 plus 4 that simplifies to 25 over 2 right the two and the 4 cancel out minus 50 over 2 right 10 times 5 is 50 plus and then 4 is really eight over two so now we all got common denominators 25 minus 50 is negative 25 over 2 plus 8 over 2 is negative 17 over 2. so this is the minimum function value right and this is also here the coordinate of the minimum aka the vertex last but not least what is the axis of symmetry the axis of symmetry is x equals okay it's a vertical line x equals whatever the x value is of the vertex it goes right down the center like that of the parabola so the axis of symmetry is x equals five over two so there's the AOS here is the minimum boom done so for Exercise 17 this one first determining whether or not we have a Max or minimum I see that the a term here is positive which means this is a parabola opening upwards and that vertex therefore is a minimum so we know we're finding a minimum here now to find that minimum it is the vertex so first how do we find the vertex our best Avenue to doing that is knowing that the x value of the vertex is at negative B over 2A right so that's a b and c b is of course that invisible one in front of the X so it's negative 1 over 2 times 4 which is 8. so the vertex is occurring at negative 1 8 that's actually the x value the minimum and then the actual minimum value aka the Y value or the function value at negative 1 8 can be found by plugging this in as follows so when we have negative 1 8 squared that's 1 over 64 times 4 4 over 64 is 1 16 minus 1 8 which is the same as 2 6 teams right if we multiply top and bottom by two and then minus 1 which is 16 over 16 just so we have common denominators makes it a little easier one minus two is negative 1 minus 16 is negative 17 so we get negative 17 16. so here is the the minimum the minimum value of the function is negative 17 over 6 and the axis of symmetry is x equals whatever that x value is of the vertex it's that vertical line boom that goes down through the vertex so x equals negative 1 8 is the AOS there's your answer dot for this quadratic we see that it has a Exercise 19 positive a value in this format which means it's a parabola facing upwards which means we're going to have a minimum not a maximum right that vertex is going to be a minimum value now to find where that minimum occurs we're going to use the formula x equals negative B over 2A that's a B and C are those coefficients or those constants in the case of C so it's negative B which is three negative 3 over 2 times a which is one half that becomes two times one half is one negative three over that is negative three so the minimum occurs at x equals negative three and what is that minimum value well now we got to plug it in to the equation so then we get negative three squared is nine nine times one-half is 4.5 and then that's negative nine right three times negative three and then plus one so it's negative four point five plus one is negative 3.5 just as a decimal or we could do negative seven halves but there is your minimum value your minimum value is negative 3.5 it occurs at x equals negative three and the axis of symmetry is x equals whatever that x value of the vertex is which is negative three there's your answer done so for this one Exercise 21 we're meant to determine the domain and range of the quadratic function and it's in this nice standard form again some people call it vertex form so we can identify the vertex we can also determine the direction of the parab meaning whether it opens up or down so let's just look at this in in this format my vertex is the opposite of this value my H value which is three and then whatever this value is as is which is 2. so my vertex is at three comma two and also since there is really just a positive one out here the fact that's positive means it's going up like this so now we have a nice visual representation where that vertex is occurring at three comma two now we're talking about domain domain for these quadratics like this is going to be all real numbers there's no limitations on the X's you can plug in so we can say domain is all real numbers or in interval notation everything from negative Infinity to positive Infinity now what about the range by the way these are curve parentheses not squares because we're not actually including infinity and negative Infinity now we talk about the range well look at this this is the bottom in terms of the function values the Y values if that that's what the vertex is it's the minimum the bottom and the minimum value is two so the so the range starts at two but then goes off up to infinity and again we have a curved parenthesis because we don't include Infinity so there's the domain and range boom done for this function again the a value up in front of the x squared is positive so Exercise 23 we know it goes up like this and then when we're trying to find the range in particular we need to find the place of that vertex AKA what is that minimum value to help us know where the where the range is of all the Y values or the function values so let's find that vertex again we can find it when it's in this format as negative B over 2A or this is a that's one B is 6 and C is 4. so negative 6 over 2 times a which is two times one that becomes negative three and that's the x value enter of the vertex so that's going to help us find the corresponding y value of the vertex we now have to plug it in so negative 3 squared is 9 plus negative negative 3 times 6 is negative 18 or minus 18 n plus four nine minus 18 is negative 9 plus 4 is negative five so there's my vertex now I can do my domain and range domain once again is all real numbers like every quadratic right so it's just going to be negative Infinity to positive Infinity there's no limitations on the x value as it goes forever in that direction and that direction but the range there is a limitation because the minimum value is at negative five and then it goes up from there to Infinity going out forever and notice we have this square bracket to include the negative five so there's Exercise 25 your answer done so for this one once again it's got a positive a value which means this Parabola is opening upwards which is great and to find the domain and range really the range we just need to get that vertex so the vertex is occurring at negative B over 2A this is a b c so B is negative negative 6 which is positive 6 over 2A 2 a is two times three that's six over six which is one so the vertex occurs at one comma something and to figure out what that one is we have to plug and chug so plug in one here it's three times one squared which is three minus six times one which is six minus nine three minus six is negative three minus nine is negative twelve now we have enough information to get the domain and range domain again is all real numbers as are with all these quadratics right you can plug in any x value left right as much as you want so there's no limitations for the range we have a minimum value at negative 12 we're starting at negative 12 including it hence we have the square bracket all the way up to Infinity because then there's no upper bound it goes up and up and up forever and ever right so there's your domain and range boom done so for this one we have the vertex and then we Exercise 27 have another point so we're going to start in our standard form AKA vertex form so again we don't know what a is we need to figure that out but we do know what that H is and again in vertex form or standard standard form as they're calling it in this textbook x minus the x value of the vertex which is negative 2. so minus negative 2 is the same as plus 2 and then that y value of the vertex comes out here and then we're trying to find the a value so we're going to plug 3 in for y because it's a y value we're going to plug 4 negative 4 in for X now we're going to say negative 4 plus 2 is negative 2 negative 2 squared is 4 so we have 4A minus 1 equals three add one to both sides 4A equals 4 which means a equals 1. so then we come back to this and then our equation is subbing in one for a Y equals X plus 2 squared minus 1. so now to get the general form which is a x squared plus BX plus C we're just going to simplify by foiling this first so we have x squared plus 2X plus 2X right because we have X plus 2 times X Plus 2. so imagine it's foiling 2X and 2x which is 4X and then plus 4 minus 1 which gives us a final answer of and instead of Y we're going to say f of x as a function right f of x equals x squared 4X and then 4 minus 1 is 3. there is the general form boom done so for 29 we're going to start in our standard form Exercise 29 where we don't know what a is but we do know the vertex is 2 comma 3 which means it would be x minus 2 in here and then a plus 3 out here now to solve for a we're going to plug 12 in for y and 5 in for X like so and then we're going to solve for a 5 minus 2 is 3 3 squared is nine then we're going to subtract 3 from both sides we've got 9 equals 9A divide by 9 we got a equals 1. so so let me bring it back here we've got y equals 1 times x minus 2 squared plus three and then if we foil this we have x squared again that's x minus 2 x minus two so we have x squared then we have negative 2x negative 2x is negative 4X and then we have negative 2 times negative 2 which is positive four and we solve that 3 on at the end then we combine like terms and I'm going to write this again as a function of x f of x equals x squared minus 4X plus 7 there's the general form boom done so for this one we're going to Exercise 31 start with our standard form where we don't know what a is we know the vertex is three so we have x minus 3 and then the Y value is 2 so we have a plus two and then we're going to plug in 10 here and 1 here to solve for a 10 minus 3 is 7 7 squared is 49 a plus two equals one then we're going to minus 2 minus two and then we get negative 1 equals 4 49 a divided by 49 a equals negative 1 over 49. so now we're going to plug that in here then we're going to foil this guy which is really x minus 3 times x minus three so that becomes x squared minus 3x minus 3x which is minus 6X plus 9 and then we still have that 2 on the outside so now we're going to distribute and we get y equals negative 1 over 49 x squared negative 1 over 49 times negative 6 is positive 6 over 49 x and then that times 9 is negative 9 over 49 and then that 2 I'm going to make that 98 over 49 just so we can now combine like terms with common denominators so last but not least we're going to change that y to f of x okay and then we're going to combine like terms we got negative 1 over 49 x squared that stays as is the 6 over 49 stays as is with the X and then we got negative 9 plus 98 which is uh 89 positive positive 89 over 49 so there is the general form of the function quadratic function boom done so Exercise 33 for this one let's start again in our standard form like so where we have x minus 1 because that is the x value of the vertex Plus 0 that's the Y value of the vertex I don't have to write anything and then we're going to plug in 1 for y 0 for x and solve for a so 0 minus 1 is just negative 1 squared is one so a equals one a or a equals one and then we plug it back in here we get y equals 1 times x minus 1 squared and then we're just going to foil this out right so it's y equals and again x minus 1 squared just so you can see it is really this x minus 1 times x minus 1. so we get x squared then we get Negative X Negative X which is negative 2X and then negative 1 times negative 1 is positive one so we'd have f of x in general form is equal to x squared minus two X plus one boom done so for this one Exercise 35 we're going to start by finding the vertex and we're going to use x equals negative B over 2A because it's in this in this general form and a is one it's that coefficient B is negative six and C is negative one so x e equals negative negative six which is just six over two times one and six over two is just three so we know the vertex occurs at three comma something and to find that something we're going to plug three in here so we get 3 squared which is nine minus six times three which is eighteen minus one nine minus 18 is negative nine minus one is negative ten because that's the vertex we know the axis of symmetry is x equals the x value so it's a line kind of at x equals three a vertical line and then to get the X intercepts I'm going to use the quadratic formula that's going to be really nice the way it's laid out so it's negative B which is negative negative six plus or minus the square root of B squared negative six squared which is 36 minus 4 times a which is one times C which is negative one all over to a two times one which is just two then as we simplify we get 36 plus 4 which is 40 over two and then I can simplify 40 because 40 is real as a radical is really 4 4 times 10 right we're always trying to find those perfect squares 4 is a perfect square and take the square root of 4 which is 2 so now I have 6 plus or minus 2 that comes outside with the 10 stuck inside divided by 2 then I can simplify by dividing both of these pieces by two we'll go up here and it becomes 3 plus or minus radical 10. so my two x intercepts are three plus red ten comma zero three minus red ten comma zero and it doesn't ask for the Y intercept but it's really easy to find so I'm going to find it as well by plugging in 0 for x and that becomes zero that becomes zero so we get zero comma negative one for my y-intercept all right let's plot these so vertex is at three comma negative 10 Y intercept is at negative one and the x-intercepts come out to be just over six and just under zero and let's connect the dots so now as we plot the points we got 3 negative 10 for the vertex we got the x-intercept three plus red 10 which is just over six three minus red ten just under zero and then our Y intercept of negative one and then we can connect the dots so that's not the best drawing but you get the idea and the axis of symmetry is not really part of the graph but if you want to draw it in just to see what it looks like it would be kind of a dotted line going up and down there's your answer done for this one we're going to start off by finding the vertex and we're Exercise 37 going to use x equals negative B over 2A where we got a B and C A is one B is negative seven C is three so we got a equals negative negative seven over two times a which is seven over two and then to find the corresponding function value we're going to plug in seven halves and see what we get to get the Y value of the vertex so we get 7 over 2 squared which is 49 over 4 minus seven times seven over two 2 which is 49 over 2 and then plus 3. to get common denominators we have 49 over 4 minus 98 over 4 right doubling the top and the bottom plus 12 over 4 and when you add all this together you get Negative 49 plus 12 which is negative 37 over 4. that also gives us an axis of symmetry at x equals the x value of the vertex which is x equals seven over two and then to get the X intercepts we're going to use the quadratic formula so we get negative B which is 7 plus or minus the square root of B squared which is 49 minus 4 a is one C is three I'll just not write the one doesn't matter all over 2 times a which is 2 2 times 1 sorry which is 2. 49 minus 12 is 37 so 7 plus or minus square root of 37 over 2. okay so it's not the prettiest but that's our final answer so we can say our two x intercepts seven plus red 37 over 2 comma 0 and then 7 minus rad 37 over 2 comma zero and these two approximate to the top one is like about six and a half so let's say somewhere there if you you plug it into a calculator the bottom one is about one half plug it in there and then let's plug in that vertex so we got seven halves which is 3.5 comma negative 37 over four is just over negative nine it's negative nine point two five so maybe like somewhere here and then if we want the y-intercept just for fun you plug in 0 for x and everything zeros out and you get positive three so there would be my y-intercept and now let's connect the dots and here's our axis of symmetry at seven over two AKA 3.5 just lining it up with that vertex and there's your final answer boom done so once again to find the vertex we're Exercise 39 going to start with x equals negative B over two a where a is four B is negative 12 and C is negative three so we have negative of negative 12 which is positive 12 over 2 times 4 which is 8 so that simplifies down to 3 over 2. so to get the corresponding function value we're going to plug and chug so we got 4 times 3 over 2 squared minus 12 times 3 over 2 minus 3. 3 over 2 squared is nine over four nine over four times four is nine and let's see if we can make some room here so we got 9 minus 12 times 3 over 2 is divided by 2 is 6 times 3 is 18 minus three it's a nice whole number negative 9 or a nice integer negative 9 minus 3 is negative 12 so there's our vertex this also means our axis of symmetry is x equals three over two that x value of the vertex and then now we can find the X intercepts by again using the quadratic formula so negative B which is negative negative 12 which is 12 plus or minus the square root of B squared 12 squared is 1 or negative 12 squared is 144 minus four times a times C which is negative three all over two times four two times a which is eight so that becomes 144 plus 48 which is 192. so now I can rewrite 192 as 12 plus or minus square root of 3 times 64 over 8. the reason why this is awesome is because I can take the square root of 64 is 8. so now I get 12 plus or minus 8 square root 3 over 2 and sorry over 8 and then I can divide everything by four right this this and this we have to divide everything by four and then we get our two x intercepts which is which are excuse me um three plus or minus two rad three over two and in coordinate format would be three plus two red three over two comma zero three three minus two red three over two comma zero and last but not least just to find the y-intercept you can plug in zero for the x's and you get zero comma negative three so now we have a bunch of good points let's plot them so zero negative three is here this has an x-intercept approximates to 6.5 ish so just do roughly there this one would be like negative 0.2 so let's say roughly here and then this x-intercept approximates to like 3.23 so somewhere like around here maybe and then this one approximates to negative 0.23 so maybe something like here and then we got a Vertex at 1.5 comma negative 12. so one and a half comma negative 12 like down here then we'll connect the dots and then we're going to draw that axis of symmetry down at 1.5 down the middle like so there's our answer done so for this one we're going to start in J in the standard form and then we're going to convert it to general form so because it's got this nice obvious vertex at negative one comma two this is a great way to start so we have y equals a times x minus that x value of the vertex minus negative 1 is plus one squared and then plus that y value which Exercise 41 is 2. and then now we want to find a so if we can find one other point which this one looks great the y-intercept we can figure this out so I know that when x is 0 Y is 3. so now we can solve for a we got 3 equals zero plus one is one squared is just one so that's a plus two minus 2 from both sides a equals 1. so now we know the equation is Y equals X Plus 1 squared times one so we don't need to write it plus two then we're going to foil this out y equals x squared plus 2X plus 1 and then still we got that plus two so why is this this just as a reminder X Plus 1 squared is X Plus 1 times X plus one right x squared 1X 1X and then one and then last but not least we'll combine like terms then we'll write it in function notation so y can be replaced with f of x a function of X is x squared plus two X plus three boom done so here we're going to start Exercise 43 with this vertex of negative one comma two and we're going to start to write this in standard form AKA vertex form and then we'll we will convert it to general form so here we'd have x minus that vertex value x minus negative one which is the same as plus one and then we got that 2 out here and then we need to figure out what a is well we need another point which we have here zero comma negative one so let's plug and chug so negative 1 in for y and zero in for x and then we got zero plus one is one squared is one so we got negative one equals one a or just a plus two minus two from both sides we get negative three so now we plug it back in here y equals negative three times X plus one squared plus two again if we foil this out it's really X plus one times X plus one so that gives us negative three still keeping it all in parentheses of x squared X Plus X which is 2x who've made that two bit long and then plus 1 times 1 which is one plus two then we're going to distribute the negative three we get negative 3 x squared minus six x minus 3 plus 2. combine those like terms and we get our function f of x equals negative 3x squared and again just I'm just replacing the Y with f of x okay that's fine it's just that as a function versus a nice equation with y's this is they're they're equivalent negative 3x squared minus six x negative 3 plus 2 is negative one so there is your final answer boom done for this one we're Exercise 45 going to start out in standard form because we have a nice identifiable vertex at negative two positive three we also have a nice y-intercept which we'll use as well zero comma two so let's start with that form it's y equals a times x minus the x value of the vertex minus negative 2 which is Plus 2 squared and then plus that K value of the vertex which is 3. now we need to find a so we're going to take this second point and we're going to plug in 0 for X we're going to plug in 2 for y like so 0 plus 2 is just 2 2 squared is 4 then we're going to subtract 3 from both sides we get negative 1 equals 4A divided by 4 a equals negative 1 over 4. so now we can plug it in and simplify so X plus 2 squared again that's X plus 2 times X plus two so that's going to become and we'll keep that negative 1 4 on the outside that's going to become x squared x times x is x squared 2x 2x which is 4X and then 2 times 2 which is 4 and then we got the plus 3 out here and then we are going to distribute the negative 1 4 boom boom boom and we get negative 1 4 x squared negative 1 4 times 4 is negative 1 or just Negative X and then negative 1 4 times 4 is negative 1. and then we got the three at the end so last but not least f of x equals in general form negative 1 4 x squared minus X and then negative 1 plus 3 is 2 for the win boom Exercise 47 done so for this question we're meant to find the axis of symmetry in the vertex and then use that to create the equation of the line so what I'm going to use to find the axis of symmetries I'm going to see if there is symmetric values occurring at any point well I see the Y value equaling 1 here and here which means the center is exactly in between these two points which is right here so look we got our vertex it's going to happen at negative one comma zero we also know the axis of symmetry which is x equals whatever the x value is of the vertex now we're going to start with standard form AKA vertex form where we have x minus the x value the vertex x minus negative 1 is X Plus 1 squared plus that y value which is zero so that's good to go and then we just need to figure out what a is so I'm going to take any other point let's take this one this one looks pretty easy to plug in so when y equals 1 x equals zero okay so then we have one equals and then zero plus one is one squared is one so one equals a so check it out we just plug one in there so we have y equals just X plus one squared and then as we foil that out it becomes and again that's X Plus 1 times X Plus 1 so that becomes x squared plus X Plus X which is 2X and then 1 times 1 which is 1 and we'll change that back to f of x just for function notation there's our answer Exercise 49 so for this one again we're trying to figure out where the vertex is by seeing where the Symmetry happens you know where are we seeing any identical values of Y and we are right here and right here so it looks like it's going to 1 1 as my vertex that's that's kind of what we're bouncing around because then you go one to the left we get 0 1 to the right we get zero so that's our indication that our vertex is here and our axis of symmetry is x equals one again you know it's always x equals the x value of the vertex so now we can plug this into standard form which is a times x minus the x value of the vertex squared plus the Y value of the vertex one now to solve for a we can plug in any coordinate that we see here on the function y naught zero zero that's the easiest one so plug in 0 for y 0 for x and solve for a so then we get 0 equals a times negative 1 squared which is just one so that's one a plus one and then a equals by subtracting 1 from both sides negative 1. so then we have y equals negative 1 times x minus 1 squared plus 1. again x minus 1 squared is x minus 1 times x minus 1. so that becomes as we foil it out and don't forget that negative 1 stays there x squared and then we have negative x negative X that's negative 2x negative 1 times negative 1 is positive one and then we have that straggling one out there when we distribute that negative one inside and I'll change this over to f of x simultaneously we got negative x squared negative 1 times negative 2x is positive 2x negative 1 times 1 is negative 1 plus 1 is just zero which means this is the final answer boom Exercise 51 done so for this one we're going to graph all three in Desmos and we're going to talk about the changes that seem to be impacted by a different leading coefficient so as we see the regular x squared right here this is a typical Parabola now when we have a 2 in front look what happens it becomes skinnier so that a value is a vertical stretch it's making it skinnier okay comparatively if we look at the one-third x squared it's making it fatter so what we notice is that when that a value is larger than one two three four whatever it's going to make it skinnier when it's between 0 and 1 like a fraction like one-third right it becomes wider so that's how the the leading coefficient changes that coefficient either is a vertical stretch when it's greater than one like two three four and it's a vertical compression when it's between zero and one like one-third one-half Etc boom done so for this one we're going to Exercise 53 graph all three in Desmos and then we're gonna or all four I should say in Desmos and then we're going to compare them so here's x squared as our original function so what's happening when we do x minus 2 squared so when we look at that on the graph you'll notice it's the exact same Parabola but it's shifted two spots to the right right the vertex 0 0 here it's two zero then when we have minus three it's again shifted to the right right that might be counterintuitive you might be like minus three we should be going to the left but no it's reversed for those X for those horizontal shifts right applied to the X so that's three to the right of vertex of three zero and then when we have plus 4 as you now might imagine instead of going right we're going to the left by four okay so that's a horizontal shift to the left so whatever is in here that's the amount of that horizontal shift and that's how you do it so here we have a suspension bridge with the Exercise 55 height model by this function with this domain and absolute value of x is the number of feet from the center right so if you have a negative 1500 that'd be 1500 away from the center Etc so it says use the a trace feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet what I'm going to use instead is Desmos so it's not going to have a trace but I'm going to show you how to do it in Desmos pretty cool so now we have the graph here of 0.001.0001 x squared and it's kind of within the confines from negative 2000 to to 2000 and we want to know when is this going to have a height aka the Y value of 100. so I'm going to plot that line in and look this makes it really easy for us to figure it out in Desmos it's those intersection points so how far does it have to be it has to be a thousand feet to the right or to the left it doesn't matter so that is our answer it has to be a thousand feet from Exercise 57 the center boom done so here we want to imagine the vertex and the fact that it opens down to sort of visualize this Parabola so we're going like this and like this and the vertex is negative one comma two now as per usual with our domain for a quadratic it's going to be all real numbers AKA everything from negative Infinity to Infinity and then for that range we've got opening down meaning this is a Max Point okay which means it's everything from negative Infinity all the bottom values going up and including two which is why we'd have a square bracket including Exercise 59 that two boom done here we have a vertex of negative 100 100 and it opens up like so so we're going to write that out and then we're going to find the domain and the range so again it's a parabola okay it's a quadratic domain there's no restrictions on X values so the domain as per usual is everything all real numbers AKA negative Infinity to positive Infinity what about the range well since it opens up this is a minimum Point that's the bottom and the bottom is that function value so it goes as low as a hundred oh sorry but that should be including 100 so we don't want a curved parenthesis we want a nice square bracket up to in or up all the way to Infinity right it goes up and up and up forever so there's our domain and range boom done so so this one is interesting Exercise 61 it says it contains negative 1 4 and has the shape of 2x squared which I presume to mean it it has that same a value of two okay because that affects the vertical basically like vertical uh stretch or compression so it's a vertical stretch here and then we have a Vertex on the y-axis what that means is my vertex is zero comma something and it also contains negative one four so we can put this into our standard form and see if we can figure it out so we got y equals an a value of 2 times x minus the x value of the vertex squared plus our K value so it's the K value that we don't know and I think we can figure it out using this point so we're going to plug and chug plug 4 in because that's a y value so plug it in for y negative 1 in for X because that's the x coordinate and then we got 4 equals 2 times negative 1 minus zero squared plus K negative 1 minus 0 is negative one negative 1 squared is one so that's two plus K subtract 2 from both sides and we get k equal equals 2. so now I can plug this in here and we got a vertex of zero comma two now we can create this equation a little bit better because we can take that 2 and plug it in here we got everything we need we got y equals an a value of 2 times x minus 0 which is just x squared okay we'll move the parentheses in a second plus two well this is actually already in our nice general form which is what we want here and then I'll just change that back to f of x in our function notation f of x equals 2x squared plus 2 boom done so Exercise 63 this one says it contains this point and has this shape which means our a value we can already presume is going to be the same a value here which is negative one it says vertex is on the y-axis which means the vertex has zero for the x value we don't know what the k value is yet so we'll we'll leave that alone so now in our nice standard form we get y equals negative one for a times x minus the x value of the vertex which is zero so we can just say x you know x minus 0 is x squared and then plus that K value which we don't know and now since we have this nice coordinate we can plug and chug and solve for K so negative 1 times 1 squared is just negative 1 plus K and then add 1 to both sides to solve for k k equals negative two so now we can plug that in here so we have a vertex of negative two now we can write this out as y equals my a value negative 1 times x minus 0 squared which is just x squared plus negative 2 that y value of the vertex and then to simplify this out and also convert it back into function notation f of x equals negative 1 times x squared which is negative x squared minus 2 there's the winning answer boom Dot here it says Exercise 65 it contains this point with a shape like this meaning the a value is going to be the same as that which is three it says the vertex has an x coordinate of negative one so my vertex is negative one comma something we don't know what that is but we got enough information to figure this out so in standard form we got y equals my a value of 3 times x minus negative 1 which is plus one uh squared plus our K value which we don't know but now we can solve for that we're going to take this coordinate 1 negative 6. we're going to plug one in for x and we're going to plug negative 6 in for y because it's an x y coordinate so we got negative 6 equals 3 times 1 plus 1 which is 2 okay squared plus K and then we got negative 6 equals 2 squared is 4 times 3 is 12 plus K subtract 12 from both sides k equals negative 18. so that's the Y value of the vertex so now we have y equals 3 times again X Plus 1 squared minus 18. and now we're just going to put this into our general form so we got X Plus 1 squared which is X Plus 1 times X Plus 1 just a reminder that's y equals 3 times x squared and then 1X Plus 1X is 2X and then 1 times 1 is just one minus 18. distribute distribute distribute we get 3x squared plus six x and then 3 times 1 is 3 minus 18 is negative 15 and then put it back into function notation f of x equals this that's your final answer Exercise 67 boom done find the dimensions of the rectangular dog park split into two pens of the same size producing the greatest possible and closed area given 300 feet of fencing so right off the bat we're going to make a drawing and we've got this split down the middle into two pens don't have to be a data in line though we can just make it solid and all of this is going to add up to 300 right because we have 300 feet of fencing and then the other thing I can do which is pretty cool is I can call all these sides X because they're all equal and then I can call these two sides here Y and Y moreover I've got something cool happening which I know that 3x is plus two of the Y's gives me that total length of 300. so now I got some nice stuff here I got X and Y and a relationship between the two of them so let's do our thing so the first I'm going to get everything in terms of x's so I got 2y equals 300 minus 3x divide everything by 2 y equals 300 minus 3x over 2. so I can replace or say this equals 300 minus 3x over 2. and that's special because now I have everything in terms of one variable and I can represent this thing now as an area in terms of X so watch this my area I'll say a of X or you know area function is equal to the length x times the width which is y but we'll Now call it 300 minus 3x over 2. and when you multiply this out you'll notice you're going to get x to the second power that's a quadratic and then that's going to be cool for us to find the max value because guess what it's going to occur at the vertex okay so let's multiply this out so a of x equals 300 x and I'm going to split it up over 2 minus 3x squared right x times negative 3x is 3x squared over 2. I'm just splitting it up into two pieces just so we have a better understanding and then the area function therefore is I'm going to flip it around so we have this term in front negative 3x squared over 2 and then plus 150x so now to find that max value of area we need to find the vertex and again the vertex occurs at negative B over 2 to a well my negative B is negative 150 over 2A 2 times negative 3 halves is negative 3. so this simplifies to 50. so that means it's going to have a width of 50 or X equaling 50 feet and then to figure out the the length the corresponding y value and you can say with length interchangeable I think I called y the width earlier so either way but those are the two dimensions and then y would equal plug in what was it 50 for X right so plug in 50 that's 300 minus 3 times 50 is negative one there's 150. so 3 minus 150 is 150 divided by 2 is 75 feet so that would be the the dimensions and then the to get the greatest possible area we just have to multiply 70 by 50 which would give a maxed out area of 3750 feet squared that's just for fun that's you don't even need to write that for the for the answer but here are the dimensions 50 by 70 five boom done among Exercise 69 all the pairs of numbers whose sum is six find the pair with the largest product so if we're trying to find a product and I believe what they're referring to is whole numbers right so if we're trying to find those numbers we want to have a representation of the two numbers so the first one could be X the question is what's the second one well if the two are going to add together to make six it would make sense that this would be 6 minus X so for example X would be one the second number would be five x could be two the second number would be four so on and so forth so we're trying to find the largest product so we could say p of x equals x times six minus X and then we get this nice quadratic when we distribute and I'm just going to rearrange it x times negative X is negative x squared and then x times 6 is 6X so for try and we know that this is already going down because that's negative so if we're trying to find the maximum AKA largest product we're trying to find the vertex so again we're going to use negative B over 2A and this is a is negative 1 B is 6 so we got negative 6 over 2 times negative 1 which is negative two so that is three so the value where the product is maximized is when X is three and then the other number would be six minus three which would be three as well so the the two numbers would be three and three and that product would be three times three which is nine boom done it says suppose that the price per unit Exercise 71 dollars for of a cell phone production is modeled by this where x's and thousands of phones produce and the revenue represented by thousands of dollars is x times this find the production level that will maximize Revenue so to get Revenue it's saying we just take this function which is p 45 minus point zero one two five x and multiply it by X so as we distribute this we get and I'm going to rearrange it to put the the leading the highest degree term first negative point zero one two five x squared and then x times 45 is 45x so there's a quadratic and we want to maximize the revenue well again this is going to be a parabola going down like this so the max is going to occur at the vertex vertex occurs at negative B over 2A where this is a and this is B so it's negative 45 over 2 times negative 0.0125 so when you simplify this out x equals 1800 so that's the amount that's going to maximize the revenue is going to be x equals 1800 which is thousands of phones so we want eighteen hundred thousand AKA one million 800 000 phones will maximize Exercise 73 the revenue boom done a ball is thrown in the air from the top of a building its height is given by this function and says how long does it take to reach a maximum height AKA we're looking for the x value in this case the T value at the max meaning we're trying to find the the x value or in this case the T value of the vertex that's when the max is occurring so again we're going to use our nice formula T equals negative B over 2A where this is a b and c so that's negative 24 over 2 times negative 4.9 and when you plug that into your calculator you get approximately 2.449 and that's in seconds there's your answer boom Dot a farmer finds that if Exercise 75 she plants 75 trees per acre each tree will yield 20 bushels of fruit she estimates that for each additional tree planted it the the yield of each tree will decrease by three bushels how many trees should we plant in order to maximize or harvest so these questions are really interesting and they're pretty difficult but I'm gonna show you how to break it down so first let's think about this 20 bushels so we know that for every you know it currently we produce 20 bushels but that decreases by three bushels for every additional tree so if trees are the or X right every time I plant one tree I'm dropping by three so just imagine what we've done here so right now we have if we if we planted one tree we'd have a we'd have 17 bushels we planted two trees it'd be 14 bushels so this accurately represents our output and then what we're doing is we're multiplying this by the number of trees so right now we have 75 but it depends on how many extra trees we plant right so if if X is zero we don't plant any additional it's just 75 times 20 bushels that makes sense if we plant one additional tree now we get 76 times 20 minus 3 which is 17 so 76 times 17 and so on and so forth so now when we foil this out we'll get a nice general form which will be 1500 minus 225 x plus 20x minus 3x squared and then put it in a nice general form we're going to put the negative 3x squared first negative 225 plus 20 is negative 205 x and then plus 1500. so we want to know what when we're going to maximize this this amount right so we're going to again find the x value of the vertex that's going to be the number of trees per acre which is negative B which is positive 2 over 205 over 2 times a which is 2 times negative 3. and when we simplify this we get Negative 34.16 6 repeating okay so what does that mean it means that we need 34 less than this current 75 and again we're going to approximate it's going to give you a decimal but we're going to approximate and get to the closest Value that is going to is going to maximize our output so we'll just take it as negative 34 right 75 minus 34 would be 41 trees so we want to have and this is kind of making an assumption it says that like you know the yield will decrease by three bushels every time we increase so we have to assume that as we decrease the amount of trees the bushels increase right so this is the optimal point if we plant 41 trees per acre we get the max yield boom done I hope you guys enjoyed this video and if you did please click that like button and if you want to see more from the scalar Learning Channel make sure to click subscribe thank you guys so much for joining and I'll see you in the next video take it easy |
5319 | https://www.journalmeddbu.com/full-text/224 | Diagnosis of rheumatoid arthritis based on clinical and serological findings
Alican Yürük, Atakan Özdoğan, Rafet Bilgiç, Harun Akar
Department of Internal Medicine, University of Health Sciences, Tepecik Training and Research Hospital, Izmir, Turkey
Keywords: Inflammatory arthritis, interstitial lung disease, rheumatoid arthritis, symmetrical polyarticular
Abstract
In this report, we report a case who was hospitalized in the internal medicine clinic with pneumonia. Based on clinical and serological findings, the definitive diagnosis of rheumatoid arthritis was made and the inflammatory process was suppressed with steroid therapy.
Introduction
Although rheumatoid arthritis (RA) is a chronic systemic inflammatory disease with severe joint involvement, extra-articular involvement has been reported with a rate of 17.8 to 50% in various studies.[1,2] Pulmonary involvement is one of the common findings of RA and can cause serious mortality and morbidity. Rheumatoid lung disease may present with different clinical manifestations such as pleuritis, pleural effusion, pneumonia, interstitial lung disease, pulmonary nodules, and bronchiectasis. In this article, we report a case of RA presenting with pneumonia.
Case Report
A 75-year-old female patient who was evaluated in the emergency department with complaints of weakness, generalized body pain, and cough was initially hospitalized in the internal medicine clinic to investigate the etiology of acute renal failure and pneumonia. On further investigation, the patient complained of not being able to walk for two months. It was learned from the patient’s history that she was followed for 15 years for coronary artery disease and for 12 years due to hypertension. On physical examination, suprapubic tenderness, coarseness in bilateral lung sounds, and bilateral crepitant rales were detected. There was pain and tenderness in the right and left shoulder joints, right and left elbow joints, and all right and left metacarpophalangeal joints. Laboratory examinations were as follows: white blood count (WBC): 27.5x10 uL, neutrophil: 23.9x10 uL, hemoglobin: 9.9 g/dL, mean corpuscular volume: 79 fL, platelet count: 525x10 uL, urea: 163 mg/dL, creatinine: 2.6 mg/dL, sodium: 132 mmol/L, potassium: 5.89 mmol/L, chloride: 106 mmol/L, calcium: 8.7 mg/dL, phosphate: 4.5 mg/dL, magnesium: 2 mg/dL, procalcitonin: 1.53 µg/L, C-reactive protein (CRP): 345 mg/L, erythrocyte sedimentation rate (ESR): 129 mm/h, and ferritin: >450 µg/L. Arterial blood gas analysis results were as follows: pH: 7.28, partial pressure of carbon dioxide: 31.8 mmHg, bicarbonate: 15.5 mmol/L, lactate: 0.8 mmol/L, and potassium: 5.8 mmol/L. Thoracic computed tomography (CT) showed a 3-cm-thick loculated effusion in the right pleural space, and pleuroparenchymal, locally fibrotic changes in the upper, middle and lower zones in both hemithoraces, and signs of reticulation in the subpleural area, compatible with new-onset interstitial lung disease. The pleural fluid sampling revealed no growth in the pleural fluid culture, and pleural fluid biochemical examinations were as follows: lactate dehydrogenase: 952 U/L, glucose: 48 mg/dL, sodium: 142 mmol/dL, albumin: 1.7 g/dL, WBC: 0.6 103/uL, and neutrophil count: 0.3 103/uL. It was initially thought that the significant increase in inflammatory markers at the time of hospitalization might be due to urinary tract infection and pneumonia. Urinalysis and urine culture were performed. Piperacillin-tazobactam 2.25 g q.i.d. and 2 mL/kg/h intravenous hydration were started. Since leukocyte and nitrite were negative in the urinalysis and no growth was detected in urine culture, the preliminary diagnosis of urinary infection was ruled out. Control CRP values at 48 and 96 h of antibiotherapy were 274 mg/L and 259 mg/L, respectively. The control ESR was 118 mm/h. Since the desired acute phase reactant response could not be obtained despite antibiotherapy and the complaints of joint pain continued, rheumatological markers were analyzed and the results were as follows: rheumatoid factor (RF): 1150 U/mL, myeloperoxidase-antineutrophil cytoplasmic antibodies (ANCA): negative, C3: 1.53 g/L, C4: 0.48 g/L, anti-cyclic citrullinated peptide (anti-CCP): >200 U/mL, c-ANCA (-), p-ANCA (-), anti-double-stranded deoxyribonucleic acid: 5.5 IU/mL, and antinuclear antibody: negative. With possible RA diagnosis according to the rheumatology recommendation, 4 mg methylprednisolone was started. Since antihepatitis B core protein antibody immunoglobulin G was found to be positive, entecavir 0.5 mg was started for hepatitis B prophylaxis. After 4 mg methylprednisolone, control CRP and procalcitonin values decreased. The methylprednisolone dose was increased to 8 mg. The decline in the CRP values continued (CRP: 259.8 mg L-118.1 mg/L-67 mg/L-46.6 mg/L). Renal function tests and metabolic acidosis of the patient regressed after hydration and antibiotherapy. Antibiotherapy was discontinued on the seventh day, due to the decrease in joint pain and regression of inflammation markers under 8 mg methylprednisolone treatment. She was discharged with the recommendation of rheumatology outpatient clinic control. At one month after discharge, ESR was 66 mm/h and CRP was 25 mg/L.
A written informed consent was obtained from the patient for all diagnostic and therapeutic procedures.
Discussion
The experience gained from basic and clinical research over the last two decades has been enlightening for the diagnosis and management of RA. In an appropriate clinical picture, when RA is suspected, requesting anti-CCP and RF as prognostic biomarkers provides a great advantage in the diagnosis. It has been shown that positivity in biomarkers such as RF and anti-CCP may precede the clinical diagnosis of RA.
Initially, mostly symmetrical involvement of the small joints of the hands and feet, and normal physical activity following joint stiffness within 1 h of waking up in the morning are supportive for the diagnosis of RA. In our case, in addition to polyarticular symmetrical small joint involvement, anti-CCP and RF positivity, which may indicate an active RA disease, also supported the diagnosis.
Extraarticular manifestations are important in the clinical course of RA. Interstitial lung disease associated with RA is considered one of the major causes of mortality, although it is a rare extra-articular manifestation of RA. Uncontrolled RA disease is thought to be one of the main risk factors for the development of interstitial lung disease associated with RA, as in our patient.
The symptoms of interstitial lung disease associated with RA exhibit a wide spectrum ranging from asymptomatic to fatal acute interstitial pneumonia, while the most common symptoms are exertional dyspnea and dry cough. Pulmonary hypertension and subsequent hypoxic respiratory failure may develop in the advanced stages of the disease.
In our case, the fact that acute phase reactants did not regress with antibiotherapy, and no growth was found in the urine and pleural fluid cultures, rheumatological diseases were considered in the differential diagnosis. The regression in acute phase reactants after methylprednisolone treatment and the decrease in joint pain of the patient day by day were explained by RA in the light of clinical and serological findings. On the other and, the lung CT findings were also found to be compatible with the initial stage of interstitial lung disease.
In conclusion, the presence of symmetrical polyarticular inflammatory arthritis, the marked increase of clinical, specific serological, and inflammatory markers, and clinical and laboratory response after steroid therapy seem to be supportive of the diagnosis of RA.
The authors declared no conflicts of interest with respect to the authorship and/or publication of this article.
The authors received no financial support for the research and/or authorship of this article.
References
Issue: 2021, Volume 7 - Issue 1
Page: 045-047
DOI: 10.5606/fng.btd.2021.25046
Category: Case Report
Demiroglu Science University © 2025
Yazılım Parkı - Scientific Journal Publishing and Management System
This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. |
5320 | https://www.vedantu.com/maths/conversion-of-minutes-to-seconds | Courses for Kids
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Why Learn to Convert Minutes to Seconds?
Today we will learn about time. As we all know that time is very precious in our life, We should manage time very smartly in our daily life. Time is the invention of sundials. Time is used to measure our daily events and activities. Time has different units and moves in one direction.
Time
In a year there are 365 days and in a day there are 24 hours. In 1 hour there are always 60 minutes and in 1 minute there are always 60 seconds. In today’s topic we will learn how many seconds are there in 1 minute and how to convert minutes into seconds.
What is Time in Minutes?
There are 12 big divisions in a clock and 1 big division is equal to 5 small divisions and each small division is called Minute. In a clock there are three hands: Hour hand, Minute hand and Second hand. The bigger hand is called the minute hand.
How Many Seconds are There in 1 Minute?
We can understand it by an example: A teacher in a class distributes 60 toffees costing Rs. 60 to all the students and there was a brilliant student who always scored high marks in examinations. The teacher gave him a chocolate which cost Rs. 60.
So we can assume that a brilliant student is minute and the other students are seconds.
The cost of one chocolate is equal to 60 toffees.
So there are 60 seconds in 1 minute
Conversion of Time
How to Convert 1 Minute into Seconds?
1 minute is equal to 60 seconds
1 minute = 60 seconds
So for converting minutes into seconds we will multiply given minutes by 60.
For Example:
Q1. Convert 5 minutes into seconds?
Ans: 1 minute = 60 seconds
Then, 5 minutes = (5 × 60) seconds = 300 seconds
Q2. Convert 42 minutes into seconds?
Ans: 1 minute = 60 seconds
Then, 42 minutes = (42 × 60) seconds = 2520 seconds
Q3. Convert 58 minutes into seconds?
Ans: 1 minute = 60 seconds
Then, 58 minutes = (58 × 60) seconds = 3480 seconds
Q4. Convert 6 minutes 55 seconds into seconds?
Ans: 1 minute = 60 seconds
Then, 6 minutes 55 seconds = (6 × 60) seconds + 55 seconds
= 360 seconds + 55 seconds
= 415 seconds
Q5. Convert 36 minutes 18 seconds into seconds?
Ans: 1 minute = 60 seconds
Then, 36 minutes 18 seconds = (36 × 60) seconds + 18 seconds
= 2160 seconds + 18 seconds
= 2178 seconds
Q6. Rohit swims 8 m length of a swimming pool. His target is to swim the length in under 7 minutes. It takes him 440 seconds. Explain why Rohit did not achieve his target ?
Ans: The length of swimming pool is 8m
His target to swim is 7 minutes = (7 × 60) second = 420 seconds
But he takes 440 seconds = 7 minutes and 20 seconds
He did not complete his target because 440 seconds is more than 420 seconds.
Practice on Your Own:
Q1. Convert 48 minutes into seconds?
Ans: 2880 seconds
Q2. Convert 25 minutes 54 seconds into seconds?
Ans: 1554 seconds
Q3. Reena Swings 18m length of a swimming pool. His target is to swim the length in under 20 minutes. It takes him 1245 seconds. Explain why Reena did not achieve his target ?
Ans: 1245 seconds is more than 1200 seconds
Summary
There are 365 days in a year and 24 hours in a day. There are always 60 minutes in an hour and 60 seconds in a minute.
A clock has 12 main divisions, each of which is divided into 5 smaller divisions, each of which is referred to as a minute. The hour hand, minute hand, and second hand are the three hands on a clock. The minute hand refers to the larger hand.
One hour has 60 minutes, and one minute has 60 seconds.
FAQs on Convert Minutes to Seconds Easily: Step-by-Step Guide
What is the basic formula to convert minutes to seconds?
The fundamental relationship between minutes and seconds is that 1 minute = 60 seconds. To convert any number of minutes to seconds, you use the formula: Seconds = Minutes × 60. You simply multiply the number of minutes by 60 to find the equivalent value in seconds.
How do you convert 5 minutes into seconds using a step-by-step example?
To convert 5 minutes into seconds, you can follow these simple steps:
Step 1: Recall the conversion factor: 1 minute is equal to 60 seconds.
Step 2: Take the number of minutes you want to convert, which is 5.
Step 3: Multiply the number of minutes by 60. So, 5 minutes × 60 = 300 seconds.
Therefore, 5 minutes is equal to 300 seconds.
How do you convert a time given in both minutes and seconds (e.g., 3 minutes and 45 seconds) into total seconds?
To convert a time with both minutes and seconds into total seconds, you first convert the minutes part and then add the remaining seconds. For 3 minutes and 45 seconds:
First, convert the minutes to seconds: 3 minutes × 60 = 180 seconds.
Next, add the extra seconds to this result: 180 seconds + 45 seconds = 225 seconds.
So, 3 minutes and 45 seconds is equal to a total of 225 seconds.
How do you handle converting decimal or fractional minutes to seconds, for example, 2.5 minutes?
The conversion process remains the same even for decimal or fractional minutes. You still multiply by 60. For example, to convert 2.5 minutes:
Multiply the decimal value by 60: 2.5 × 60 = 150.
This means that 2.5 minutes is equal to 150 seconds. This works because 2.5 minutes is the same as 2 full minutes (120 seconds) and half a minute (30 seconds), which totals 150 seconds.
Why are there 60 seconds in a minute and 60 minutes in an hour?
The system of using 60 for time divisions (a sexagesimal system) comes from the ancient Babylonians. They used a base-60 numbering system for mathematics and astronomy, which was later adopted by other cultures. This system is highly divisible by many numbers (1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60), making calculations with fractions easier. This historical convention is why we have 60 seconds in a minute and 60 minutes in an hour today.
What is the difference between converting minutes to seconds and seconds to minutes?
The main difference lies in the mathematical operation you use. These are inverse operations:
Minutes to Seconds: Since a minute is a larger unit, you multiply the number of minutes by 60 to get the equivalent number of seconds.
Seconds to Minutes: Since a second is a smaller unit, you divide the number of seconds by 60 to get the equivalent number of minutes.
For example, 2 minutes × 60 = 120 seconds, while 120 seconds ÷ 60 = 2 minutes.
In what real-life situations is converting minutes to seconds important?
Converting minutes to seconds is a crucial skill in many real-world applications. Some key examples include:
Sports: Timing races, calculating lap times, or measuring performance in events where every fraction of a second counts.
Cooking and Baking: Following recipes that specify cooking times in seconds for precision, like for microwaves or specific chemical reactions in baking.
Science Experiments: Recording data for experiments where reactions happen quickly and need to be timed in smaller units.
Music and Video Editing: Calculating the length of tracks or video clips, where timelines are often measured in minutes and seconds.
How would you convert a larger time unit, like 2 hours, entirely into seconds?
To convert hours to seconds, you must perform a two-step conversion. First from hours to minutes, and then from minutes to seconds.
Step 1 (Hours to Minutes): Convert the hours to minutes. Since 1 hour = 60 minutes, you have 2 hours × 60 = 120 minutes.
Step 2 (Minutes to Seconds): Now, convert these minutes to seconds. Since 1 minute = 60 seconds, you have 120 minutes × 60 = 7200 seconds.
Therefore, 2 hours is equal to 7200 seconds.
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5321 | https://www.vedantu.com/maths/sum-of-even-numbers | Maths
Sum of Even Numbers – Formula, Examples & Quick Tricks
Sum of Even Numbers – Formula, Examples & Quick Tricks
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How to Calculate the Sum of Even Numbers with Formula and Stepwise Examples
The concept of sum of even numbers plays a key role in mathematics and is widely applicable to both real-life situations and exam scenarios. Whether you are calculating series, solving quick arithmetic questions, or preparing for competitive exams, knowing how to efficiently find the sum of even numbers can save you time and increase your problem-solving confidence.
What Is Sum of Even Numbers?
A sum of even numbers is the total you get when you add all the even integers in a sequence—either up to a certain count or within a specified range. Even numbers are integers divisible by 2 with no remainder. You’ll find this concept applied in areas such as arithmetic progression (AP) in sequences, mental maths tricks, and quick exam calculations.
Key Formula for Sum of Even Numbers
Here’s the standard formula: , where n is the number of consecutive even terms starting from 2.
For example, the sum of first 5 even numbers (2, 4, 6, 8, 10):
Cross-Disciplinary Usage
The sum of even numbers is not only useful in Maths but also plays an important role in Physics (for motion and patterns), Computer Science (loops and iterations), and everyday logical reasoning. Students preparing for exams such as JEE or Olympiads will see its relevance in multiple types of series and pattern-based questions.
Step-by-Step Illustration
List the even numbers you want to add, e.g., from 2 to 10:
2, 4, 6, 8, 10
Count how many even numbers there are:
There are 5 even numbers (n = 5)
Apply the formula:
S5 = 5 × (5 + 1) = 30
You can check by adding directly:
2 + 4 + 6 + 8 + 10 = 30
Table of Sums for Popular Ranges
| n (Number of Even Terms) | Sum S = n(n+1) | Expanded Series |
---
| 1 | 2 | 2 |
| 2 | 6 | 2 + 4 = 6 |
| 3 | 12 | 2 + 4 + 6 = 12 |
| 5 | 30 | 2 + 4 + 6 + 8 + 10 = 30 |
| 10 | 110 | 2 + 4 + ... + 20 = 110 |
| 20 | 420 | 2 + 4 + ... + 40 = 420 |
| 50 | 2550 | 2 + 4 + ... + 100 = 2550 |
Speed Trick or Vedic Shortcut
Here’s a quick shortcut for adding consecutive even numbers:
Count how many even numbers you need to sum (n).
Directly apply the formula S = n(n+1), instead of manually adding all terms.
This saves time especially for large n, like when finding the sum from 2 to 100: n=50, so S = 50 × 51 = 2550.
Tricks like this are practical for exams including Olympiads and JEE arithmetic sections. Vedantu’s live sessions give even more such quick techniques for speed and accuracy.
Try These Yourself
Write the first ten even numbers and find their sum.
Find the sum of even numbers from 1 to 50.
Is 48 an even number? Show why.
Using the formula, what is the sum of the first 15 even numbers?
Add all even numbers between 12 and 20.
Frequent Errors and Misunderstandings
Forgetting that the series starts at 2, not 0.
Counting the number of terms incorrectly—always check if n is the quantity of terms, not the largest even number itself.
Mixing up “sum of even numbers” with “sum of odd numbers” or total sum of all natural numbers.
Plugging the last even number as n (instead, use n = last even number ÷ 2).
Relation to Other Concepts
The idea of sum of even numbers connects closely with sum of odd numbers and arithmetic progression. Mastering this will also help you solve things like series, patterns and prepare for sums on sequences found in sequence and series topics.
Classroom Tip
A helpful way to remember the sum of even numbers formula is—count the number of terms, multiply it by the next number, and that’s your answer! For example: For 7 terms, 7 × 8 = 56. Vedantu’s teachers often use visual tables or practice quizzes in live classes to reinforce this logic.
We explored sum of even numbers—from definition, formula, examples, common mistakes, and connections to other important ideas. Keep practicing with Vedantu’s resources and topic quizzes to master even more maths concepts with confidence!
Related topics to boost your learning:
Sum of Odd Numbers Formula |
Sequence and Series |
Arithmetic Progression |
Even and Odd Numbers |
Maths Formulas for Class 8
FAQs on Sum of Even Numbers – Formula, Examples & Quick Tricks
What is the sum of even numbers formula?
The sum of the first n even numbers is given by the formula: S = n(n + 1). This formula works because even numbers form an arithmetic progression with a common difference of 2. The formula efficiently calculates the sum without needing to individually add each even number.
How do I calculate the sum of even numbers from 1 to 100?
First, determine how many even numbers are between 1 and 100 (there are 50). Then, substitute n = 50 into the formula S = n(n + 1): S = 50(50 + 1) = 2550. Therefore, the sum of even numbers from 1 to 100 is 2550.
What is the sum of the first 20 even numbers?
Using the formula S = n(n + 1), where n = 20 (the number of even numbers), we get: S = 20(20 + 1) = 420. The sum of the first 20 even numbers is 420.
Is there a formula for the sum of even numbers between any two given numbers?
Yes, you can use the formula for the sum of an arithmetic series: S = n/2 (a + l), where n is the number of terms, a is the first term, and l is the last term. First, identify the first and last even numbers in your range and calculate the number of terms (n).
How can I quickly calculate the sum of even numbers?
The quickest method is using the formula S = n(n + 1). However, for larger numbers, a calculator can help. Remember to first determine the value of n (the number of even numbers).
What are some common mistakes students make when calculating the sum of even numbers?
Common mistakes include: Incorrectly identifying the number of even terms (n); Misapplying the formula; Forgetting that the formula is for consecutive even numbers. Carefully counting the terms and correctly substituting into the formula will prevent these errors.
How does the sum of even numbers relate to arithmetic progressions (AP)?
Even numbers form an arithmetic progression (AP) with the first term a = 2 and a common difference of d = 2. The sum of even numbers formula is a specific case of the general formula for the sum of an AP.
Where are sums of even numbers used in real-world applications or advanced mathematics?
Sums of even numbers appear in various areas, such as: Calculating totals involving even quantities; Analyzing patterns and sequences in number theory; Solving problems related to arithmetic progressions; Applications in computer science and programming (e.g., loop iterations).
What is the difference between the sum of even numbers and the sum of odd numbers?
The formula for the sum of the first n even numbers is n(n + 1). The formula for the sum of the first n odd numbers is n². Notice the difference in the structure of these formulas. The sum of odd numbers grows more slowly than the sum of even numbers.
Can I use the sum of even numbers formula for negative even numbers?
Technically, yes, but you need to adjust the formula and carefully consider the sequence. The formula S = n(n+1) is typically applied to positive, consecutive even numbers starting from 2. For negative even numbers, you'll need to adapt the approach depending on the specific sequence.
How can I find the sum of even numbers in a given range using a programming language like Python?
Python offers efficient ways to calculate this. One method is using list comprehension and the sum() function: sum([x for x in range(start, end + 1, 2)]), where start and end define your range. Remember to adjust the start value if you don't want to include 0.
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5322 | https://www.esuppa.it/Articoli/Suppa_Inequalities%20from%20the%20word%201995-2005.pdf | Inequalities From Around the World 1995-2005 Solutions to ’Inequalities through problems’ by Hojoo Lee Autors: Mathlink Members Editor: Ercole Suppa Teramo, 22 October 2021 - Version 1.1 I Introduction The aim of this work is to provide solutions to problems on inequalities proposed in various countries of the world in the years 1990-2005.
In the summer of 2006, after reading Hoojoo Lee’s nice book, Topics in Inequalities - Theorem and Techniques, I developed the idea of demonstrating all the inequalities proposed in chapter 5, subsequently reprinted in the article Inequalities Through Problems by the same author. After a hard and tiresome work lasting over two months, thanks also to the help I mustered from specialised literature and from the website, I finally managed to bring this ambitious project to an end.
To many inequalities I have offered more than one solution and I have always provided the source and the name of the author. In the contents I have also marked with an asterisk all the solutions which have been devised by myself.
Furthermore I corrected the text of the problems 5, 11, 32, 79, 125, 140, 159 which seems to contain some typos (I think !).
I would greatly appreciate hearing comments and corrections from my read-ers. You may email me at Ercole Suppa ercsuppa@gmail.com To Readers This book is addressed to challenging high schools students who take part in mathematical competitions and to all those who are interested in inequalities and would like improve their skills in nonroutine problems. I heartily encourage readers to send me their own alternative solutions of the proposed inequalities: these will be published in the definitive version of this book. Enjoy!
Acknowledgement I’m indebted to Hojoo Lee, Vasile Cˆ ırtoaje, Massimo Gobbino, Darij Grinberg and many other contributors of Mathlinks Forum for their nice solu-tions. Without their valuable help this work would not have been possible.
II Contents 1 Years 2001 ∼2005 1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 Problem 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 Problem 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 Problem 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 Problem 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 Problem 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 Problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 Problem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 Problem 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 Problem 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 Problem 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 Problem 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24 Problem 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24 Problem 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 Problem 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26 Problem 28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 Problem 29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29 Problem 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29 Problem 31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 Problem 32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 Problem 33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 Problem 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34 Problem 35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37 Problem 36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 Problem 37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 Problem 38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 Problem 39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40 Problem 40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41 Problem 41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42 Problem 42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42 Problem 43 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43 III Problem 44 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44 Problem 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44 Problem 46 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 Problem 47 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46 Problem 48 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46 Problem 49 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48 Problem 50 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48 Problem 51 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49 Problem 52 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50 Problem 53 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52 Problem 54 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52 Problem 55 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53 2 Years 1996 ∼2000 55 Problem 56 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55 Problem 57 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55 Problem 58 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
56 Problem 59 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57 Problem 60 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58 Problem 61 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58 Problem 62 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62 Problem 63 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63 Problem 64 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63 Problem 65 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64 Problem 66 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65 Problem 67 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66 Problem 68 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67 Problem 69 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68 Problem 70 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 Problem 71 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 Problem 72 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 Problem 73 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
72 Problem 74 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73 Problem 75 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73 Problem 76 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74 Problem 77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75 Problem 78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
76 Problem 79 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78 Problem 80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80 Problem 81 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80 Problem 82 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81 Problem 83 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82 Problem 84 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84 Problem 85 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86 Problem 86 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86 Problem 87 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87 IV Problem 88 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87 Problem 89 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88 Problem 90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88 Problem 91 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89 Problem 92 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89 Problem 93 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
90 Problem 94 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91 Problem 95 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91 Problem 96 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93 Problem 97 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93 Problem 98 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98 3 Years 1990 ∼1995 100 Problem 99 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Problem 100 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Problem 101 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Problem 102 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Problem 103 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Problem 104 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 Problem 105 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 Problem 106 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Problem 107 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Problem 108 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 Problem 109 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 Problem 110 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Problem 111 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Problem 112 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Problem 113 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Problem 114 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Problem 115 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 4 Supplementary Problems 116 Problem 116 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 Problem 117 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 Problem 118 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Problem 119 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Problem 120 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Problem 121 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Problem 122 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Problem 123 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Problem 124 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Problem 125 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 Problem 126 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 Problem 127 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 Problem 128 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Problem 129 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 V Problem 130 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 Problem 131 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 Problem 132 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 Problem 133 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Problem 134 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 Problem 135 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 Problem 136 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Problem 137 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Problem 138 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 Problem 139 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 Problem 140 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 Problem 141 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Problem 142 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Problem 143 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Problem 144 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 Problem 145 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Problem 146 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 Problem 147 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Problem 148 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 Problem 149 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 Problem 150 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Problem 151 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Problem 152 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 Problem 153 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 Problem 154 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 Problem 155 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Problem 156 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Problem 157 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Problem 158 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 Problem 159 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 Problem 160 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 Problem 161 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 A Classical Inequalities 150 B Bibliography and Web Resources 155 VI Inequalities From Around the World 1995-2005 Solutions to ’Inequalities through problems’ by Hojoo Lee Mathlink Members 27 March 2011 1 Years 2001 ∼2005 1. (BMO 2005, Proposed by Duˇ san Djuki´ c, Serbia and Montenegro) (a, b, c > 0) a2 b + b2 c + c2 a ≥a + b + c + 4(a −b)2 a + b + c First Solution. (Ercole Suppa) Using the Cauchy-Schwartz inequality we have: a2 b + b2 c + c2 a −a −b −c = a2 b + b −2a + b2 c + c −2b + c2 a + a −2c = (a −b)2 b + (c −b)2 c + (a −c)2 a ≥(a −b + c −b + a −c)2 a + b + c = 4(a −b)2 a + b + c Equality holds if and only if a = b = c.
□ Second Solution. (Ciprian - ML Forum) With Lagrange theorem (for 3 num-bers) we have a2 b + b2 c + c2 a −(a + b + c) = 1 a + b + c · " ac −b22 bc + bc −a22 ab + ab −c22 ac # 1 So we have to prove that ac −b22 bc + bc −a22 ab + ab −c22 ac ≥4 (a −b)2 But (ab−c2) 2 ac ≥0 and ac −b22 bc + bc −a22 ab ≥ ac −b2 −bc + a22 b (a + c) = (a −b)2 (a + b + c)2 b (a + c) By AM-GM we have b (a + c) ≤(a + b + c)2 4 = ⇒ (a −b)2 (a + b + c)2 b (a + c) ≥4 (a −b)2 Then we get a2 b + b2 c + c2 a ≥a + b + c + 4(a −b)2 a + b + c □ Remark. The Binet-Cauchy identity n X i=1 aici ! n X i=1 bidi !
− n X i=1 aidi ! n X i=1 bici !
= X 1≤i 0) b + c a2 + c + a b2 + a + b c2 ≥1 a + 1 b + 1 c 2 Solution. (Ercole Suppa) By using the Cauchy-Schwarz inequality we have 1 a + 1 b + 1 c 2 = X cyc √ b + c a 1 √ b + c !2 ≤ ≤ X cyc b + c a2 ! 1 b + c + 1 a + c + 1 a + b ≤ (Cauchy-Schwarz) ≤ X cyc b + c a2 ! 1 a + 1 b + 1 c Therefore X cyc b + c a2 !
≥ 1 a + 1 b + 1 c □ 3. (Romania 2005, Traian Tamaian) (a, b, c > 0) a b + 2c + d + b c + 2d + a + c d + 2a + b + d a + 2b + c ≥1 First Solution. (Ercole Suppa) From the Cauchy-Schwartz inequality we have (a + b + c + d)2 ≤ X cyc a b + 2c + d X cyc a (b + 2c + d) Thus in order to prove the requested inequality is enough to show that (a + b + c + d)2 P cyc a (b + 2c + d) ≥1 The last inequality is equivalent to (a + b + c + d)2 − X cyc a (b + 2c + d) ≥0 ⇐ ⇒ a2 + b2 + c2 + d2 −2ac −2bd ≥0 ⇐ ⇒ (a −c)2 + (b −d)2 ≥0 which is true.
□ 3 Second Solution. (Ramanujan - ML Forum) We set S = a + b + c + d. It is a b + 2c + d + b c + 2d + a + c d + 2a + b + d a + 2b + c = = a S −(a −c) + b S −(b −d) + c S + (a −c) + d S + (b −d) But a S −(a −c) + c S + (a −c) = (a + c)S + (a −c)2 S2 −(a −c)2 ≥a + c S (1) and b S −(b −d) + d S + (b −d) ≥b + d S (2) Now from (1) and (2) we get the result.
□ 4. (Romania 2005, Cezar Lupu) a + b + c ≥1 a + 1 b + 1 c, a, b, c > 0 a + b + c ≥ 3 abc Solution. (Ercole Suppa) From the well-known inequality (x + y + x)2 ≥3(xy + yz + zx) it follows that (a + b + c)2 ≥ 1 a + 1 b + 1 c 2 ≥ ≥3 1 ab + 1 bc + 1 ca = = 3(a + b + c) abc Dividing by a + b + c we have the desidered inequality.
□ 5. (Romania 2005, Cezar Lupu) (1 = (a + b)(b + c)(c + a), a, b, c > 0) ab + bc + ca ≤3 4 4 Solution. (Ercole Suppa) From the identity (a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) −abc we have ab + bc + ca = 1 + abc a + b + c (1) From AM-GM inequality we have 1 = (a + b) (b + c) (c + a) ≥2 √ ab · 2 √ bc · 2√ca = 8abc = ⇒ abc ≤1 8 (2) and 2a + b + c 3 = (a + b) + (b + c) + (c + a) 3 ≥ 3 p (a + b) (b + c) (c + a) = 1 = ⇒ a + b + c ≥3 2 (3) From (1),(2),(3) it follows that ab + bc + ca = 1 + abc a + b + c ≤1 + 1 8 3 2 = 3 4 □ 6. (Romania 2005, Robert Szasz - Romanian JBMO TST) (a + b + c = 3, a, b, c > 0) a2b2c2 ≥(3 −2a)(3 −2b)(3 −2c) First Solution. (Thazn1 - ML Forum) The inequality is equivalent to (a + b + c)3(−a + b + c)(a −b + c)(a + b −c) ≤27a2b2c2 Let x = −a + b + c, y = a −b + c, z = a + b −c and note that at most one of x, y, z can be negative (since the sum of any two is positive). Assume x, y, z ≥0 if not the inequality will be obvious. Denote x + y + z = a + b + c, x + y = 2c, etc. so our inequality becomes 64xyz(x + y + z)3 ≤27(x + y)2(y + z)2(z + x)2 Note that 9(x + y)(y + z)(z + x) ≥8(x + y + z)(xy + yz + zx) and (xy + yz + zx)2 ≥3xyz(x + y + z) Combining these completes our proof!
□ 5 Second Solution. (Fuzzylogic - Mathlink Forum) As noted in the first solution, we may assume a, b, c are the sides of a triangle. Multiplying LHS by a + b + c and RHS by 3, the inequality becomes 16∆2 ≤3a2b2c2 where ∆is the area of the triangle. That is equivalent to R2 ≥1 3 since ∆= abc 4R , where R is the circumradius. But this is true since R = a + b + c 2(sin A + sin B + sin C) and sin A + sin B + sin C ≤3 √ 3 2 by Jensen.
□ Third Solution. (Harazi - Mathlink Forum) Obviously, we may assume that a, b, c are sides of a triangle. Write Schur inequality in the form 9abc a + b + c ≥a(b + c −a) + b(c + a −b) + c(a + b −c) and apply AM-GM for the RHS. The conclusion follows.
□ 7. (Romania 2005) (abc ≥1, a, b, c > 0) 1 1 + a + b + 1 1 + b + c + 1 1 + c + a ≤1 First Solution. (Virgil Nicula - ML Forum) The inequality is equivalent with the relation X a2(b + c) + 2abc ≥2(a + b + c) + 2 (1) But 2abc ≥2 (2) 6 and 3 + X a2(b + c) = X (a2b + a2c + 1) ≥ ≥3 X 3 √ a2b · a2c · 1 ≥ ≥3 X a · 3 √ abc ≥ ≥3 X a = = 2 X a + X a ≥ ≥2 X a + 3 3 √ abc ≥ ≥2 X a + 3 Thus we have X a2(b + c) ≥2 X a (3) From the sum of the relations (2) and (3) we obtain (1).
□ Second Solution. (Gibbenergy - ML Forum) Clear the denominator, the inequality is equivalent to: a2(b + c) + b2(c + a) + c2(a + b) + 2abc ≥2 + 2(a + b + c) Since abc ≥1 so a + b + c ≥3 and 2abc ≥2. It remains to prove that a2(b + c) + b2(a + c) + c2(a + b) ≥2(a + b + c) It isn’t hard since X (a2b + a2c + 1) −3 ≥ X 3 3 √ a4bc −3 ≥ ≥ X 3 3 √ a3 −3 = = 3 X a −3 ≥ ≥2(a + b + c) + (a + b + c −3) ≥ ≥2(a + b + c) □ Third Solution. (Sung-yoon Kim - ML Forum) Let be abc = k3 with k ≥1. Now put a = kx3, b = ky3, c = kz3, and we get xyz = 1. So 7 X 1 1 + a + b = X 1 1 + k(x3 + y3) ≤ ≤ X 1 1 + x3 + y3 ≤ ≤ X 1 xyz + x2y + xy2 = = X 1 xy · 1 x + y + z = = 1 xy + 1 yz + 1 zx 1 x + y + z = = 1 xyz = 1 and we are done.
□ Remark 1. The problem was proposed in Romania at IMAR Test 2005, Juniors Problem 1. The same inequality with abc = 1, was proposed in Tournament of the Town 1997 and can be proved in the following way: Solution. (See , pag. 161) By AM-GM inequality a + b + c ≥3 3 √ abc ≥3 and ab + bc + ca ≥3 3 √ ab · bc · ca ≥3 Hence (a + b + c) (ab + bc + ca −2) ≥3 which implies 2 (a + b + c) ≤ab (a + b) + bc (b + c) + ca (c + a) Therefore 1 1 + a + b + 1 1 + b + c + 1 1 + c + a −1 = = 2 + 2a + 2b + 2c −(a + b) (b + c) (c + a) (1 + a + b) (1 + b + c) (1 + c + a) = = 2a + 2b + 2c −ab (a + b) −bc (b + c) −ca (c + a) (1 + a + b) (1 + b + c) (1 + c + a) ≤0 □ Remark 2. A similar problem was proposed in USAMO 1997 (problem 5) Prove that, for all positive real numbers a, b, c we have 1 a3 + b3 + abc + 1 b3 + c3 + abc + 1 c3 + a3 + abc ≤ 1 abc 8 The inequality can be proved with the same technique employed in the third solution (see problem 87).
8. (Romania 2005, Unused) (abc = 1, a, b, c > 0) a b2(c + 1) + b c2(a + 1) + c a2(b + 1) ≥3 2 First Solution. (Arqady - ML Forum) Let a = x z , b = y x and c = z y, where x > 0, y > 0 and z > 0. Hence, using the Cauchy-Schwarz inequality in the Engel form, we have X cyc a b2(c + 1) = X cyc x3 yz(y + z) = = X cyc x4 xyz(y + z) ≥ ≥(x2 + y2 + z2)2 2xyz(x + y + z) Id est, remain to prove that (x2 + y2 + z2)2 2xyz(x + y + z) ≥3 2 which follows easily from Muirhead theorem. In fact x2 + y2 + z22 2xyz (x + y + z) ≥3 2 ⇐ ⇒ 2 X cyc x4 + 4 X cyc x2y2 ≥6 X cyc x2yz ⇐ ⇒ X sym x4 + 2 X sym x2y2 ≥3 X sym x2yz and X sym x4 ≥ X sym x2yz , X sym x2y2 ≥ X sym x2yz □ 9 Second Solution. (Travinhphuctk14 - ML Forum) Let a = x z , b = y x and c = z y, where x > 0, y > 0 and z > 0. We need prove X cyc a b2(c + 1) = X cyc x3 yz(y + z) ≥3 2 We have x3 + y3 ≥xy(x + y), . . . , etc. Thus the desidered inequality follows from Nesbit inequality: x3 yz(y + z) + y3 xz(x + z) + z3 xy(x + y) ≥ x3 y3 + z3 + y3 z3 + x3 + z3 x3 + y3 ≥3 2 □ 9. (Romania 2005, Unused) (a + b + c ≥a b + b c + c a, a, b, c > 0) a3c b(c + a) + b3a c(a + b) + c3b a(b + c) ≥3 2 Solution. (Zhaobin - ML Forum) First use Holder or the generalized Cauchy inequality. We have: a3c b(a + c) + b3a c(a + b) + c3b a(b + c) (2a + 2b + 2c)(a b + b c + c a) ≥(a + b + c)3 so: a3c b(a + c) + b3a c(a + b) + c3b a(b + c) ≥a + b + c 2 but we also have: a + b + c ≥a b + b c + c a ≥3 so the proof is over.
□ 10. (Romania 2005, Unused) (a + b + c = 1, a, b, c > 0) a √ b + c + b √c + a + c √ a + b ≥ r 3 2 10 First Solution. (Ercole Suppa) By Cauchy-Schwarz inequality we have 1 = (a + b + c)2 = X cyc √a 4 √ b + c √a 4 √ b + c !2 ≤ ≤ X cyc a √ b + c ! a √ b + c + b √ a + c + c √ a + b Therefore X cyc a √ b + c !
≥ 1 a √ b + c + b√a + c + c √ a + b (1) Since a + b + c = 1 we have a √ b + c + b √ a + c + c √ a + b = X cyc √a p a (b + c) !
≤ ≤ √ a + b + c √ 2ab + 2bc + 2ca = (Cauchy-Schwarz) = r 2 3 √ 3ab + 3bc + 3ac ≤ ≤ r 2 3 q (a + b + c)2 = r 2 3 and from (1) we get the result.
□ Second Solution. (Ercole Suppa) Since a + b + c = 1 we must prove that: a √1 −a + b √ 1 −b + c √1 −c ≥ r 3 2 The function f(x) = x √1−x is convex on interval [0, 1] because f ′′ (x) = 1 4 (1 −x)−5 2 (4 −x) ≥0 , ∀x ∈[0, 1] Thus, from Jensen inequality, follows that X cyc a √ b + c = f (a) + f (b) + f (c) ≥3f a + b + c 3 = 3f 1 3 = r 3 2 □ 11 11. (Romania 2005, Unused) (ab + bc + ca + 2abc = 1, a, b, c > 0) √ ab + √ bc + √ca ≤3 2 Solution. (See , pag. 10, problem 19) Set x = √ ab, y = √ bc, z = √ca, s = x + y + z. The given relation become x2 + y2 + z2 + 2xyz = 1 and, by AM-GM inequality, we have s2 −2s + 1 = (x + y + z)2 −2 (x + y + z) + 1 = = 1 −2xyz + 2 (xy + xz + yz) −2 (x + y + z) + 1 = = 2 (xy + xz + yz −xyz −x −y −z + 1) = = 2 (1 −x) (1 −y) (1 −z) ≤ (AM-GM) ≤2 1 −x + 1 −y + 1 −z 3 3 = 2 3 −s 3 3 Therefore 2s3 + 9s2 −27 ≤0 ⇔ (2s −3) (s + 3)2 ≤0 ⇔ s ≤3 2 and we are done.
□ 12. (Chzech and Slovak 2005) (abc = 1, a, b, c > 0) a (a + 1)(b + 1) + b (b + 1)(c + 1) + c (c + 1)(a + 1) ≥3 4 Solution. (Ercole Suppa) The given inequality is equivalent to 4(ab + bc + ca) + 4(a + b + c) ≥3(abc + ab + bc + ca + a + b + c + 1) that is, since abc = 1, to ab + bc + ca + a + b + c ≥6 The latter inequality is obtained summing the inequalities a + b + c ≥3 3 √ abc = 3 ab + bc + ca ≥3 3 √ a2b2c2 = 3 which are true by AM-GM inequality.
□ 12 13. (Japan 2005) (a + b + c = 1, a, b, c > 0) a (1 + b −c) 1 3 + b (1 + c −a) 1 3 + c (1 + a −b) 1 3 ≤1 First Solution. (Darij Grinberg - ML Forum) The numbers 1+b−c, 1+c−a and 1 + a −b are positive, since a + b + c = 1 yields a < 1, b < 1 and c < 1.
Now use the weighted Jensen inequality for the function f (x) = 3 √x, which is concave on the positive halfaxis, and for the numbers 1 + b −c, 1 + c −a and 1 + a −b with the respective weights a, b and c to get a 3 √ 1 + b −c + b 3 √1 + c −a + c 3 √ 1 + a −b a + b + c ≤ ≤ 3 r a (1 + b −c) + b (1 + c −a) + c (1 + a −b) a + b + c Since a + b + c = 1, this simplifies to a 3 √ 1 + b −c + b 3 √ 1 + c −a + c 3 √ 1 + a −b ≤ ≤ 3 p a (1 + b −c) + b (1 + c −a) + c (1 + a −b) But X cyc a (1 + b −c) = (a + ab −ca) + (b + bc −ab) + (c + ca −bc) = a + b + c = 1 and thus a 3 √ 1 + b −c + b 3 √ 1 + c −a + c 3 √ 1 + a −b ≤ 3 √ 1 = 1 and the inequality is proven.
□ Second Solution. (Kunny - ML Forum) Using A.M-G.M.
3 √ 1 + b −c = 3 p 1 · 1 · (1 + b −c) ≦1 + 1 + (1 + b −c) 3 = 1 + b −c 3 Therefore by a + b + c = 1 we have a 3 √ 1 + b −c + b 3 √ 1 + c −a + c 3 √ 1 + a −b ≤ ≦a 1 + b −c 3 + b 1 + c −a 3 + c 1 + a −b 3 = 1 □ 13 Third Solution. (Soarer - ML Forum) By Holder with p = 3 2 and q = 3 we have X cyc a (1 + b −c) 1 3 = X cyc a 2 3 [a (1 + b −c)] 1 3 ≤ ≤ X cyc a ! 2 3 "X cyc a (1 + b −c) # 1 3 = = X cyc a ! 2 3 X cyc a ! 1 3 = X cyc a = 1 □ 14. (Germany 2005) (a + b + c = 1, a, b, c > 0) 2 a b + b c + c a ≥1 + a 1 −a + 1 + b 1 −b + 1 + c 1 −c First Solution. (Arqady - ML Forum) 1 + a 1 −a + 1 + b 1 −b + 1 + c 1 −c ≤2 a b + b c + c a ⇔ X 2a b − 2a b + c −1 ≥0 ⇔ X cyc (2a4c2 −2a2b2c2) + X cyc (a3b3 −a3b2c −a3c2b + a3c3) ≥0 which is obviously true.
Second Solution. (Darij Grinberg - ML Forum) The inequality 1 + a 1 −a + 1 + b 1 −b + 1 + c 1 −c ≤2( b a + c b + a c ) can be transformed to 3 2 + a b + c + b c + a + c a + b ≤a c + c b + b a or equivalently to 14 ab c(b + c) + bc a(c + a) + ca b(a + b) ≥3 2 We will prove the last inequality by rearrangement. Since the number arrays ab c ; bc a ; ca b and 1 a + b; 1 b + c; 1 c + a are oppositely sorted (in fact, e. g., if c ≥a ≥b, we have ab c ≤bc a ≤ca b and 1 a+b ≥ 1 b+c ≥ 1 c+a), we have ab c · 1 b + c + bc a · 1 c + a + ca b · 1 a + b ≥ab c · 1 a + b + bc a · 1 b + c + ca b · 1 c + a i.e.
ab c (b + c) + bc a (c + a) + ca b (a + b) ≥ ab c (a + b) + bc a (b + c) + ca b (c + a) Hence, in order to prove the inequality ab c (b + c) + bc a (c + a) + ca b (a + b) ≥3 2 it will be enough to show that ab c (a + b) + bc a (b + c) + ca b (c + a) ≥3 2 But this inequality can be rewritten as ab ca + bc + bc ab + ca + ca bc + ab ≥3 2 which follows from Nesbitt.
□ Third Solution. (Hardsoul and Darij Grinberg - ML Forum) The inequality 1 + a 1 −a + 1 + b 1 −b + 1 + c 1 −c ≤2( b a + c b + a c ) can be transformed to 3 2 + a b + c + b c + a + c a + b ≤a c + c b + b a 15 or equivalently to ab c(b + c) + bc a(c + a) + ca b(a + b) ≥3 2 Now by Cauchy to s ab c(b + c), s bc a(c + a), r ca b(a + b) !
and √ b + c, √ a + c, √ a + b we have LHS · (2a + 2b + 2c) ≥ r ab c + r bc a + rca b !2 To establish the inequality LHS ≥3 2 it will be enough to show that r ab c + r bc a + rca b !2 ≥3 (a + b + c) Defining q bc a = x, p ca b = y, q ab c = z, we have yz = rca b r ab c = r ca b · ab c = √ a2 = a and similarly zx = b and xy = c, so that the inequality in question, r bc a + rca b + r ab c !2 ≥3 (a + b + c) takes the form (x + y + z)2 ≥3 (yz + zx + xy) what is trivial because (x + y + z)2 −3 (yz + zx + xy) = 1 2 · (y −z)2 + (z −x)2 + (x −y)2 □ Fourth Solution. (Behzad - ML Forum) With computation we get that the inequality is equivalent to: 2( X a3b3 + X a4b2) ≥6a2b2c2 + X a3b2c + a3bc2 which is obvious with Muirhead and AM-GM.
□ 16 15. (Vietnam 2005) (a, b, c > 0) a a + b 3 + b b + c 3 + c c + a 3 ≥3 8 Solution. (Ercole Suppa) In order to prove the inequality we begin with the following Lemma Lemma. Given three real numbers x, y, z ≥0 such that xyz = 1 we have 1 (1 + x)2 + 1 (1 + y)2 + 1 (1 + z)2 ≥3 4 Proof. WLOG we can assume that xy ≥1, z ≤1. The problems 17 yields 1 (1 + x)2 + 1 (1 + y)2 ≥ 1 1 + xy = z z + 1 Thus it is easy to show that 1 (1 + x)2 + 1 (1 + y)2 + 1 (1 + z)2 ≥ z z + 1 + 1 (1 + z)2 ≥3 4 and the lemma is proved.
□ The power mean inequality implies 3 r a3 + b3 + c3 3 ≥ r a2 + b2 + c2 3 ⇔ a3 + b3 + c3 ≥ 1 √ 3 a2 + b2 + c23/2 (1) Thus setting x = b a, y = c b, z = a c , using (1) and the Lemma we have: LHS = a a + b 3 + b b + c 3 + c c + a 3 ≥ ≥ 1 √ 3 " a a + b 2 + b b + c 2 + c c + a 2#3/2 ≥ ≥ 1 √ 3 " 1 (1 + x)2 + 1 (1 + y)2 + 1 (1 + z)2 #3/2 ≥ ≥ 1 √ 3 3 4 3/2 = 3 8 □ 17 Remark. The lemma can be proved also by means of problem 17 with a = x, b = y, c = z, d = 1.
16. (China 2005) (a + b + c = 1, a, b, c > 0) 10(a3 + b3 + c3) −9(a5 + b5 + c5) ≥1 Solution. (Ercole Suppa) We must show that for all a, b, c > 0 with a+b+c = 1 results: X cyc 10a3 −9a5 ≥1 The function f(x) = 10x3 −9x5 is convex on [0, 1] because f ′′ (x) = 30x 2 −3x2 ≥0, ∀x ∈[0, 1] Therefore, since f 1 3 = 1 3, the Jensen inequality implies f (a) + f (b) + f (c) ≥3f a + b + c 3 = 3 · f 1 3 = 1 □ 17. (China 2005) (abcd = 1, a, b, c, d > 0) 1 (1 + a)2 + 1 (1 + b)2 + 1 (1 + c)2 + 1 (1 + d)2 ≥1 First Solution. (Lagrangia - ML Forum) The source is Old and New Inequal-ities . The one that made this inequality is Vasile Cartoaje.
I will post a solution from there: The inequality obviously follows from: 1 (1 + a)2 + 1 (1 + b)2 ≥ 1 1 + ab and 1 (1 + c)2 + 1 (1 + d)2 ≥ 1 (1 + cd Only the first inequality we are going to prove as the other one is done in the same manner: it’s same as 1 + ab(a2 + b2) ≥a2b2 + 2ab which is true as 1 + ab(a2 + b2) −a2b2 −2ab ≥1 + 2a2b2 −a2b2 −2ab = (ab −1)2 18 This is another explanation: 1 (1 + a)2 + 1 (1 + b)2 − 1 1 + ab = ab(a −b)2 + (ab −1)2 (1 + a)2(1 + b)2ab) ≥0 Then, the given expression is greater than 1/(1 + ab) + 1/(1 + cd) = 1 with equality if a = b = c = d = 1.
□ Second Solution. (Iandrei - ML Forum) I’ve found a solution based on an idea from the hardest inequality I’ve ever seen (it really is impossible, in my opinon!). First, I’ll post the original inequality by Vasc, from which I have taken the idea.
Let a, b, c, d > 0 be real numbers for which a2 + b2 + c2 + d2 = 1. Prove that the following inequality holds: (1 −a)(1 −b)(1 −c)(1 −d) ≥abcd I’ll leave its proof to the readers. A little historical note on this problem: it was proposed in some Gazeta Matematica Contest in the last years and while I was still in high-school and training for mathematical olympiads, I tried to solve it on a very large number of occasions, but failed. So I think I will always remember its difficult and smart solution, which I’ll leave to the readers.
Now, let us get back to our original problem: Let a, b, c, d > 0 with abcd = 1. Prove that: 1 (1 + a)2 + 1 (1 + b)2 + 1 (1 + c)2 + 1 (1 + d)2 ≥1 Although this inequality also belongs to Vasc (he published it in the Gazeta Matematica), it surprisingly made the China 2005 TSTs, thus confirming (in my opinion) its beauty and difficulty. Now, on to the solution: Let x = 1 1 + a, y = 1 1 + b, z = 1 1 + c, t = 1 1 + d Then abcd = 1 ⇒1 −x x · 1 −y y · 1 −z z · 1 −t t = 1 ⇒(1−x)(1−y)(1−z)(1−t) = xyzt We have to prove that x2 +y2 +z2 +t2 ≥1. We will prove this by contradiction.
Assume that x2 + y2 + z2 + t2 < 1.
19 Keeping in mind that x2+y2+z2+t2 < 1, let us assume that (1−x)(1−y) ≤zt and prove that it is not true (I’m talking about the last inequality here, which we assumed to be true). Upon multiplication with 2 and expanding, this gives: 1 −2(x + y) + 1 + 2xy ≤2zt This implies that 2zt > x2 + y2 + z2 + t2 −2(x + y) + 1 + 2xy So, 2zt > (x+y)2−2(x+y)+1+z2+t2, which implies (z−t)2+(x+y−1)2 < 0, a contradiction. Therefore, our original assumption implies (1 −x)(1 −y) > zt.
In a similar manner, it is easy to prove that (1 −z)(1 −t) > xy. Multiplying the two, we get that (1−x)(1−y)(1−z)(1−t) > xyzt, which is a contradiction with the original condition abcd = 1 rewritten in terms of x, y, z, t.
Therefore, our original assumption was false and we indeed have x2 + y2 + z2 + t2 ≥1 □ 18. (China 2005) (ab + bc + ca = 1 3, a, b, c ≥0) 1 a2 −bc + 1 + 1 b2 −ca + 1 + 1 c2 −ab + 1 ≤3 First Solution. (Cuong - ML Forum) Our inequality is equivalent to: X a 3a(a + b + c) + 2 ≥ 1 3 a + b + c Since ab + bc + ca = 1 3, by Cauchy we have: LHS ≥ (a + b + c)2 3 (a + b + c) (a2 + b2 + c2) + 2 (a + b + c) = = a + b + c 3 (a2 + b2 + c2) + 2 (a + b + c) = = a + b + c 3 (a2 + b2 + c2 + 2ab + 2ac + 2bc) = = a + b + c 3 (a + b + c)2 = 1 3 a + b + c □ 20 Second Solution. (Billzhao - ML Forum) Homogenizing, the inequality is equivalent to X cyc 1 a(a + b + c) + 2(ab + bc + ca) ≤ 1 ab + bc + ca Multiply both sides by 2(ab + bc + ca) we have X cyc 2(ab + bc + ca) a(a + b + c) + 2(ab + bc + ca) ≤2 Subtracting from 3, the above inequality is equivalent to X cyc a(a + b + c) a(a + b + c) + 2(ab + bc + ca) ≥1 Now by Cauchy we have: LHS = (a + b + c) X cyc a2 a2 (a + b + c) + 2a (ab + bc + ca) ≥ ≥ (a + b + c)3 P cyc [a2 (a + b + c) + 2a (ab + bc + ca)] = 1 □ 19. (Poland 2005) (0 ≤a, b, c ≤1) a bc + 1 + b ca + 1 + c ab + 1 ≤2 First Solution. (See pag. 204 problem 95) WLOG we can assume that 0 ≤a ≤b ≤c ≤1. Since 0 ≤(1 −a) (1 −b) we have a + b ≤1 + ab ≤1 + 2ab ⇒ a + b + c ≤a + b + 1 ≤2 (1 + ab) Therefore a 1 + bc + b 1 + ac + c 1 + ab ≤ a 1 + ab + b 1 + ab + c 1 + ab ≤a + b + c 1 + ab ≤2 □ 21 Second Solution. (Ercole Suppa) We denote the LHS with f (a, b, c). The function f is defined and continuous on the cube C = [0, 1] × [0, 1] × [0, 1] so, by Wierstrass theorem, f assumes its maximum and minimum on C. Since f is convex with respect to all variables we obtain that f take maximum value in one of vertices of the cube. Since f is symmetric in a, b, c it is enough compute the values f(0, 0, 0), f(0, 0, 1), f(0, 1, 1), f(1, 1, 1). It’s easy verify that f take maximum value in (0, 1, 1) and f (0, 1, 1) = 2. The convexity of f with respect to variable a follows from the fact that f (x, b, c) = x bc + 1 + b cx + 1 + c bx + 1 is the sum of three convex functions. Similarly we can prove the convexity with respect to b and c.
□ 20. (Poland 2005) (ab + bc + ca = 3, a, b, c > 0) a3 + b3 + c3 + 6abc ≥9 Solution. (Ercole Suppa) Since ab + bc + ca = 1 by Mac Laurin inequality we have: a + b + c 3 ≥ r ab + bc + ca 3 = 1 (1) By Schur inequality we have X cyc a (a −b) (a −c) ≥0 = ⇒ a3 + b3 + c3 + 3abc ≥ X sym a2b and, from (1), it follows that a3 + b3 + c3 + 6abc ≥a2b + a2c + abc + b2a + b2c + abc + c2a + c2b + abc = = (a + b + c) (ab + bc + ca) = = 3 (a + b + c) ≥9 □ 22 21. (Baltic Way 2005) (abc = 1, a, b, c > 0) a a2 + 2 + b b2 + 2 + c c2 + 2 ≥1 Solution. (Sailor - ML Forum) We have X a a2 + 2 ≤ X a 2a + 1 We shall prove that P a 2a+1 ≤1 or X 2a 2a + 1 ≤2 ⇐ ⇒ X 1 2a + 1 ≥1 Clearing the denominator we have to prove that: 2 X a ≥6 wich is true by AM-GM.
□ 22. (Serbia and Montenegro 2005) (a, b, c > 0) a √ b + c + b √c + a + c √ a + b ≥ r 3 2(a + b + c) Solution. (Ercole Suppa) Putting x = a a + b + c, y = b a + b + c, z = c a + b + c the proposed inequality is exctly that of problem 10.
□ 23. (Serbia and Montenegro 2005) (a + b + c = 3, a, b, c > 0) √a + √ b + √c ≥ab + bc + ca Solution. (Suat Namly - ML Forum) From AM-GM inequality we have a2 + √a + √a ≥3a By the same reasoning we obtain 23 b2 + √ b + √ b ≥3b c2 + √c + √c ≥3c Adding these three inequalities, we obtain a2 + b2 + c2 + 2 √a + √ b + √c ≥3 (a + b + c) = = (a + b + c)2 = = a2 + b2 + c2 + 2 (ab + bc + ca) from which we get the required result.
□ 24. (Bosnia and Hercegovina 2005) (a + b + c = 1, a, b, c > 0) a √ b + b√c + c√a ≤ 1 √ 3 Solution. (Ercole Suppa) From the Cauchy-Schwarz inequality we have a √ b + b√c + c√a 2 = X cyc √a √ ab !2 ≤ ≤ X cyc a ! X cyc ab !
= = (ab + bc + ca) ≤ ≤1 3 (a + b + c)2 = 1 3 Extracting the square root yields the required inequality.
□ 25. (Iran 2005) (a, b, c > 0) a b + b c + c a 2 ≥(a + b + c) 1 a + 1 b + 1 c 24 First Solution. (Ercole Suppa) After setting x = a b , y = b c, z = c a the inequality become (x + y + z)2 ≥+x + y + z + xy + xz + yz ⇐ ⇒ x2 + y2 + z2 + xy + xz + yz ≥3 + x + y + z (1) Since xyz = 1, the AM-GM inequality yields xy + yz + zx ≥3 3 p x2y2z2 = 3 (2) so 3 x2 + y2 + z2 ≥(x + y + z)2 ≥3(x + y + z) ⇒ x2 + y2 + z2 ≥x + y + z (3) By summing (2) and (3) we get (1).
□ Second Solution. (Ercole Suppa) The inequality rewrites in the following form a2 b2 + b2 c2 + c2 a2 + b a + c b + a c ≥a b + b c + c a + 3 (1) Observe that for every positive real number x we have (x + 1)(x −1)2 x ≥0 ⇔ x2 + 1 x ≥x + 1 (2) By applying (2) to the numbers a b , b c, c a we get a2 b2 + b a ≥a b + 1 , b2 c2 + c b ≥b c + 1 , c2 a2 + a c ≥c a + 1 Summing these inequalities we obtain (1).
26. (Austria 2005) (a, b, c, d > 0) 1 a3 + 1 b3 + 1 c3 + 1 d3 ≥a + b + c + d abcd 25 Solution. (Ercole Suppa) WLOG we can assume that a ≥b ≥c ≥d so 1 a ≥1 b ≥1 c ≥1 d Since the RHS can be written as a + b + c + d abcd = 1 bcd + 1 acd + 1 abd + 1 abc from the rearrangement (applied two times) we obtain 1 a3 + 1 b3 + 1 c3 + 1 d3 ≥ 1 bcd + 1 acd + 1 abd + 1 abc = a + b + c + d abcd □ 27. (Moldova 2005) (a4 + b4 + c4 = 3, a, b, c > 0) 1 4 −ab + 1 4 −bc + 1 4 −ca ≤1 First Solution. (Anto - ML Forum) It is easy to prove that : X 1 4 −ab ≤ X 1 4 −a2 ≤ X a4 + 5 18 The first inequality follows from : 2 4 −ab ≤ 1 4 −a2 + 1 4 −b2 The second : 1 4 −a2 ≤a4 + 5 18 is equivalent to : 0 ≤4a4 + 2 −a6 −5a2 ⇐ ⇒0 ≤(a2 −1)2(2 −a2) which is true since a4 ≤3 and as a result a2 ≤2.
Thus X cyc 1 4 −ab ≤ X cyc a4 + 5 18 = a4 + b4 + c4 + 15 18 = 3 + 15 18 = 1 □ 26 Second Solution. (Treegoner - ML Forum) By applying AM-GM inequality, we obtain LHS ≤ X 1 4 − q a4+b4 2 = X 1 4 − q 3−c4 2 Denote u = 3 −c4 2 , v = 3 −c4 2 , w = 3 −c4 2 Then 0 < u, v, w < 3 2 and u + v + w = 3. Let f(u) = 1 4−√u. Then f ′(u) = 1 2√u(4 −√u)2 f ′′(u) = −u −3 4 (1 −3 4u 1 2 ) (4u 1 4 −u 3 4 )3 Hence f ′′(u) < 0 for every 0 < u < 3 2. By apply Karamata ’s inequality for the function that is concave down, we obtain the result.
□ 28. (APMO 2005) (abc = 8, a, b, c > 0) a2 p (1 + a3)(1 + b3) + b2 p (1 + b3)(1 + c3) + c2 p (1 + c3)(1 + a3) ≥4 3 First Solution. (Valiowk, Billzhao - ML Forum) Note that a2 + 2 2 = (a2 −a + 1) + (a + 1) 2 ≥ p (a2 −a + 1)(a + 1) = p a3 + 1 with equality when a = 2. Hence it suffices to prove a2 (a2 + 2)(b2 + 2) + b2 (b2 + 2)(c2 + 2) + c2 (c2 + 2)(a2 + 2) ≥1 3 and this is easily verified. In fact clearing the denominator, we have 3 X cyc a2(c2 + 2) ≥(a2 + 2)(b2 + 2)(c2 + 2) Expanding, we have 6a2 + 6b3 + 6c3 + 3a2b2 + 3b2c2 + 3c2a2 ≥ ≥a2b2c2 + 2a2b2 + 2b2c2 + 2c2a2 + 4a2 + 4b2 + 4c2 + 8 27 Recalling that abc = 8, the above is equivalent to 2a2 + 2b2 + 2c2 + a2b2 + b2c2 + c2a2 ≥72 But 2a2 + 2b2 + 2c2 ≥24 and a2b2 + b2c2 + c2a2 ≥48 through AM-GM. Adding gives the result.
□ Second Solution. (Official solution) Observe that 1 √ 1 + x3 ≥ 2 2 + x2 In fact, this is equivalent to (2 + x2)2 ≥4(1 + x3), or x2(x −2)2 ≥0. Notice that equality holds if and only if if x = 2. Then a2 p (1 + a3)(1 + b3) + b2 p (1 + b3)(1 + c3) + c2 p (1 + c3)(1 + a3) ≥ ≥ 4a2 (2 + a2)(2 + b2) + 4b2 (2 + b2)(2 + c2) + 4c2 (2 + c2)(2 + c2) ≥ ≥ 2 · S(a, b, c) 36 + S(a, b, c) = = 2 1 + 36 S(a,b,c) where S(a, b, c) = 2 a2 + b2 + c2 + (ab)2 + (bc)2 + (ca)2 By AM-GM inequality, we have a2 + b2 + c2 ≥3 3 q (abc)2 = 12 (ab)2 + (bc)2 + (ca)2 ≥3 3 q (abc)4 = 48 The above inequalities yield S(a, b, c) = 2 a2 + b2 + c2 + (ab)2 + (bc)2 + (ca)2 ≥72 Therefore 2 1 + 36 S(a,b,c) ≥ 2 1 + 36 72 = 4 3 which is the required inequality. Note that the equalitiy hold if and only if a = b = c = 2.
□ 28 29. (IMO 2005) (xyz ≥1, x, y, z > 0) x5 −x2 x5 + y2 + z2 + y5 −y2 y5 + z2 + x2 + z5 −z2 z5 + x2 + y2 ≥0 First Solution. (See , pag. 26) It’s equivalent to the following inequality x2 −x5 x5 + y2 + z2 + 1 + y2 −y5 y5 + z2 + x2 + 1 + z2 −z5 z5 + x2 + y2 + 1 ≤3 or x2 + y2 + z2 x5 + y2 + z2 + x2 + y2 + z2 y5 + z2 + x2 + x2 + y2 + z2 z5 + x2 + y2 ≤3.
With the Cauchy-Schwarz inequality and the fact that xyz ≥1, we have (x5 + y2 + z2)(yz + y2 + z2) ≥(x2 + y2 + z2)2 or x2 + y2 + z2 x5 + y2 + z2 ≤yz + y2 + z2 x2 + y2 + z2 .
Taking the cyclic sum and x2 + y2 + z2 ≥xy + yz + zx give us x2 + y2 + z2 x5 + y2 + z2 + x2 + y2 + z2 y5 + z2 + x2 + x2 + y2 + z2 z5 + x2 + y2 ≤2 + xy + yz + zx x2 + y2 + z2 ≤3.
□ Second Solution. (by an IMO 2005 contestant Iurie Boreico from Moldova, see pag. 28). We establish that x5 −x2 x5 + y2 + z2 ≥ x5 −x2 x3(x2 + y2 + z2).
It follows immediately from the identity x5 −x2 x5 + y2 + z2 − x5 −x2 x3(x2 + y2 + z2) = (x3 −1)2x2(y2 + z2) x3(x2 + y2 + z2)(x5 + y2 + z2).
Taking the cyclic sum and using xyz ≥1, we have X cyc x5 −x2 x5 + y2 + z2 ≥ 1 x5 + y2 + z2 X cyc x2 −1 x ≥ 1 x5 + y2 + z2 X cyc x2 −yz ≥0.
□ 30. (Poland 2004) (a + b + c = 0, a, b, c ∈R) b2c2 + c2a2 + a2b2 + 3 ≥6abc 29 First Solution. (Ercole Suppa) Since a + b + c = 0 from the identity (ab + bc + ca)2 = a2b2 + b2c2 + c2a2 + 2abc(a + b + c) follows that a2b2 + b2c2 + c2a2 = (ab + bc + ca)2 Then, from the AM-GM inequality we have a2b2 + b2c2 + c2a2 ≥3 · 3 √ a2b2c2 2 By putting abc = P we have a2b2 + b2c2 + c2a2 + 3 −6abc ≥ ≥9 (abc) 4 3 + 3 −6abc = = 9P 4 −6P 3 + 3 = = 3 " P 2 −1 2 + 2P 2 P −1 2 2 + 3 2P 2 # ≥0 □ Second Solution. (Darij Grinberg - ML Forum) For any three real numbers a, b, c, we have b2c2 + c2a2 + a2b2 + 3 −6abc = = (b + 1)2 (c + 1)2 + (c + 1)2 (a + 1)2 + (a + 1)2 (b + 1)2 −2 (a + b + c) (a + b + c + bc + ca + ab + 2) so that, in the particular case when a + b + c = 0, we have b2c2 + c2a2 + a2b2 + 3 −6abc = = (b + 1)2 (c + 1)2 + (c + 1)2 (a + 1)2 + (a + 1)2 (b + 1)2 and thus b2c2 + c2a2 + a2b2 + 3 ≥6abc □ 30 Third Solution. (Nguyenquockhanh, Ercole Suppa - ML Forum) WLOG we can assume that a > 0 e b > 0. Thus, since c = −a −b, we have a2b2 + b2c2 + c2a2 + 3 −6abc = = a2b2 + a2 + b2 (a + b)2 + 3 + 6ab (a + b) ≥ ≥ a2 + b2 a2 + ab + b2 + ab a2 + b2 + a2b2 + 3 + 6ab (a + b) ≥ ≥3ab a2 + b2 + 3 a2b2 + 1 + 6ab (a + b) = ≥3ab a2 + b2 + 6ab + 6ab (a + b) = = 3ab a2 + b2 + 2 + 2a + 2b = = 3ab h (a + 1)2 + (b + 1)2i ≥0 □ 31. (Baltic Way 2004) (abc = 1, a, b, c > 0, n ∈N) 1 an + bn + 1 + 1 bn + cn + 1 + 1 cn + an + 1 ≤1 Solution. (Ercole Suppa) By setting an = x, bn = y e cn = z, we must prove that 1 1 + x + y + 1 1 + y + z + 1 1 + z + x ≤1 where xyz = 1. The above inequality is proven in the problem 7.
□ 32. (Junior Balkan 2004) ((x, y) ∈R2 −{(0, 0)}) 2 √ 2 p x2 + y2 ≥ x + y x2 −xy + y2 First Solution. (Ercole Suppa) By using the two inequalities x + y ≤ p 2(x2 + y2) , x2 + y2 ≤2(x2 −xy + y2) we have: (x + y) p x2 + y2 x2 −xy + y2 ≤ p 2(x2 + y2) p x2 + y2 x2 −xy + y2 ≤ ≤ √ 2(x2 + y2) x2 −xy + y2 ≤ ≤2(x2 −xy + y2) √ 2 (x2 −xy + y2) = 2 √ 2 □ 31 Second Solution. (Darij Grinberg - ML Forum) You can also prove the in-equality by squaring it (in fact, the right hand side of the inequality is obviously ≥0; if the left hand side is ≤0, then the inequality is trivial, so it is enough to consider the case when it is ≥0 as well, and then we can square the inequality); this leads to (x + y)2 (x2 −xy + y2)2 ≤ 8 x2 + y2 This is obviously equivalent to (x + y)2(x2 + y2) ≤8(x2 −xy + y2)2 But actually, an easy calculation shows that 8(x2 −xy + y2)2 −(x + y)2(x2 + y2) = (x −y)2 2(x −y)2 + 5x2 + 5y2 ≥0 so everything is proven.
□ 33. (IMO Short List 2004) (ab + bc + ca = 1, a, b, c > 0) 3 r 1 a + 6b + 3 r 1 b + 6c + 3 r 1 c + 6a ≤ 1 abc First Solution. (Ercole Suppa) The function f(x) = 3 √x is concave on (0, +∞).
Thus from Jensen inequality we have: X cyc f 1 a + 6b ≤3 · f 1 a + 1 b + 1 c + 6a + 6b + 6c 3 (1) From the well-know inequality 3(xy + yz + zx) ≤(x + y + z)2 we have 3abc(a + b + c) = 3(ab · ac + ab · bc + ac · bc) ≤(ab + bc + ca)2 = 1 = ⇒ 6(a + b + c) ≤ 2 abc (2) The AM-GM inequality and (2) yields 1 a + 1 b + 1 c + 6a + 6b + 6c ≤ab + bc + ca abc + 2 abc = 3 abc (3) Since f(x) is increasing from (1) and (3) we get f 1 a + 6b + f 1 b + 6c + f 1 c + 6a ≤3 · f 1 abc = 3 3 √ abc ≤ 1 abc 32 where in the last step we used the AM-GM inequality 3 3 √ abc = 3 3 p (abc)2 abc = 3 3 √ ab · bc · ca abc ≤3 · ab+bc+ca 3 abc = 1 abc □ Second Solution. (Official solution) By the power mean inequality 1 3(u + v + w) ≤ 3 r 1 3(u3 + v3 + w3) the left-hand side does not exceed 3 3 √ 3 3 r 1 a + 6b + 1 b + 6c + 1 c + 6a = 3 3 √ 3 3 r ab + bc + ca abc + 6(a + b + c) (⋆) The condition ab + bc + ca = 1 enables us to write a + b = 1 −ab c = ab −(ab)2 abc , b + c = bc −(bc)2 abc , c + a = ca −(ca)2 abc Hence ab + bc + ca abc + 6(a + b + c) = 1 abc + 3[(a + b) + (b + c) + (c + a)] = = 4 −3 (ab)2 + (bc)2 + (ca)2 abc Now, we have 3 (ab)2 + (bc)2 + (ca)2 ≥(ab + bc + ca)2 = 1 by the well-known inequality 3(u2 + v2 + w2) ≥(u + v + w)2.
Hence an up-per bound for the right-hand side of (⋆) is 3/ 3 √ abc.
So it suffices to check 3/ 3 √ abc ≤1/(abc), which is equivalent to (abc)2 ≤1/27. This follows from the AM-GM inequality, in view of ab + bc + ca = 1 again: (abc)2 = (ab)(bc)(ca) ≤ ab + bc + ca 3 3 = 1 3 3 = 1 27.
Clearly, equality occurs if and only if a = b = c = 1/ √ 3.
□ Third Solution. (Official solution) Given the conditions a, b, c > 0 and ab + bc + ca = 1, the following more general result holds true for all t1, t2, t3 > 0: 3abc(t1 + t2 + t3) ≤2 3 + at3 1 + bt3 2 + ct3 3.
(1) 33 The original inequality follows from (2) by setting t1 = 1 3 3 r 1 a + 6b, t2 = 1 3 3 r 1 b + 6c, t3 = 1 3 3 r 1 c + 6a.
In turn, (1) is obtained by adding up the three inequalities 3abct1 ≤1 9 + 1 3bc + at3 1, 3abct2 ≤1 9 + 1 3ca + bt3 2, 3abct3 ≤1 9 + 1 3ab + ct3 3.
By symmetry, it suffices to prove the first one of them. Since 1 −bc = a(b + c), the AM-GM inequality gives (1 −bc) + at3 1 bc = a b + c + t3 1 bc ≥3a 3 r bc · t3 1 bc = 3at1.
Hence 3abct1 ≤bc(1 −bc) + at3 1, and one more application of the AM-GM in-equality completes the proof: 3abct1 ≤bc(1 −bc) + at3 1 = bc 2 3 −bc + 1 3bc + at3 1 ≤ bc + ( 2 3 −bc) 2 2 + 1 3bc + at3 1 = 1 9 + 1 3bc + at3 1.
□ 34. (APMO 2004) (a, b, c > 0) (a2 + 2)(b2 + 2)(c2 + 2) ≥9(ab + bc + ca) First Solution. (See , pag. 14) Choose A, B, C ∈ 0, π 2 with a = √ 2 tan A, b = √ 2 tan B, and c = √ 2 tan C. Using the well-known trigonometric identity 1 + tan2 θ = 1 cos2θ, one may rewrite it as 4 9 ≥cos A cos B cos C (cos A sin B sin C + sin A cos B sin C + sin A sin B cos C) .
One may easily check the following trigonometric identity cos(A + B + C) = = cos A cos B cos C −cos A sin B sin C −sin A cos B sin C −sin A sin B cos C.
Then, the above trigonometric inequality takes the form 4 9 ≥cos A cos B cos C (cos A cos B cos C −cos(A + B + C)) .
34 Let θ = A+B+C 3 . Applying the AM-GM inequality and Jensen’s inequality, we have cos A cos B cos C ≤ cos A + cos B + cos C 3 3 ≤cos3 θ.
We now need to show that 4 9 ≥cos3 θ(cos3 θ −cos 3θ).
Using the trigonometric identity cos 3θ = 4 cos3 θ −3 cos θ or cos 3θ −cos 3θ = 3 cos θ −3 cos3 θ, it becomes 4 27 ≥cos4 θ 1 −cos2 θ , which follows from the AM-GM inequality cos2 θ 2 · cos2 θ 2 · 1 −cos2 θ 1 3 ≤1 3 cos2 θ 2 + cos2 θ 2 + 1 −cos2 θ = 1 3.
One find that the equality holds if and only if tan A = tan B = tan C = 1 √ 2 if and only if a = b = c = 1.
□ Second Solution. (See , pag. 34) After expanding, it becomes 8 + (abc)2 + 2 X cyc a2b2 + 4 X cyc a2 ≥9 X cyc ab.
From the inequality (ab −1)2 + (bc −1)2 + (ca −1)2 ≥0, we obtain 6 + 2 X cyc a2b2 ≥4 X cyc ab.
Hence, it will be enough to show that 2 + (abc)2 + 4 X cyc a2 ≥5 X cyc ab.
Since 3(a2 + b2 + c2) ≥3(ab + bc + ca), it will be enough to show that 2 + (abc)2 + X cyc a2 ≥2 X cyc ab, which is proved in , pag.33.
□ 35 Third Solution. (Darij Grinberg - ML Forum) First we prove the auxiliary inequality 1 + 2abc + a2 + b2 + c2 ≥2bc + 2ca + 2ab According to the pigeonhole principle, among the three numbers a −1, b −1, c −1 at least two have the same sign; WLOG, say that the numbers b −1 and c −1 have the same sign so that (b −1)(c −1) ≥0. Then according to the inequality x2 + y2 ≥2xy for any two reals x and y, we have (b −1)2 + (c −1)2 ≥2(b −1)(c −1) ≥−2(a −1)(b −1)(c −1) Thus (1 + 2abc + a2 + b2 + c2) −(2bc + 2ca + 2ab) = = (a −1)2 + (b −1)2 + (c −1)2 + 2(a −1)(b −1)(c −1) ≥ ≥(a −1)2 ≥0 and the lemma is proved.
Now, the given inequality can be proved in the following way: (a2 + 2)(b2 + 2)(c2 + 2) −9(ab + bc + ca) = = 3 2 (b −c)2 + (c −a)2 + (a −b)2 + 2 (bc −1)2 + (ca −1)2 + (ab −1)2 + + (abc −1)2 + (1 + 2abc + a2 + b2 + c2) −(2bc + 2ca + 2ab) ≥0 □ Fourth Solution. (Official solution.) Let x = a + b + c, y = ab + bc + ca, z = abc. Then a2 + b2 + c2 = x2 −2y a2b2 + b2c2 + c2a2 = y2 −2xz a2b2c2 = z2 so the inequality to be proved becomes z2 + 2 y2 −2xz + 4 x2 −2y + 8 ≥9y or z2 + 2y2 −4xz + 4x2 −17y + 8 ≥0 Now from a2 + b2 + c2 ≥ab + bc + ca = y, we get x2 = a2 + b2 + c2 + 2y ≥3y 36 Also a2b2 + b2c2 + a2c2 = (ab)2 + (bc)2 + (ca)2 ≥ ≥ab · ac + bc · ab + ac · bc = = (a + b + c)abc = xz and thus y2a2b2 + b2c2 + a2c2 + 2xz ≥3xz Hence z2 + 2y2 −4xz + 4x2 −17y + 8 = = z −x 3 2 + 8 9(y −3)2 + 10 9 y2 −3xz + 35 9 x2 −3y ≥0 as required.
□ 35. (USA 2004) (a, b, c > 0) (a5 −a2 + 3)(b5 −b2 + 3)(c5 −c2 + 3) ≥(a + b + c)3 Solution. (See pag. 19) For any positive number x, the quantities x2 −1 and x3 −1 have the same sign. Thus, we have 0 ≤(x3 −1)(x2 −1) = x5 −x3 −x2 + 1 = ⇒ x5 −x2 + 3 ≥x3 + 2 It follows that (a5 −a2 + 3)(b5 −b2 + 3)(c5 −c2 + 3) ≥(a3 + 2)(b3 + 2)(c3 + 2) It suffices to show that (a3 + 2)(b3 + 2)(c3 + 2) ≥(a + b + c)3 (⋆) Expanding both sides of inequality (⋆) and cancelling like terms gives a3b3c3 + 3(a3 + b3 + c3) + 2(a3b3 + b3c3 + c3a3) + 8 ≥ ≥3(a2b + b2a + b2c + c2b + c2a + a2c) + 6abc By AM-GM inequality, we have a3+a3b3+1 ≥3a2b. Combining similar results, the desidered inequality reduces to a3b3c3 + a3 + b3 + c3 + 1 + 1 ≥6abc which is evident by AM-GM inequality.
□ 37 36. (Junior BMO 2003) (x, y, z > −1) 1 + x2 1 + y + z2 + 1 + y2 1 + z + x2 + 1 + z2 1 + x + y2 ≥2 Solution. (Arne - ML Forum) As x ≤1+x2 2 we have X 1 + x2 1 + y + z2 ≥ X 2(1 + x2) (1 + y2) + 2(1 + z2).
Denoting 1 + x2 = a and so on we have to prove that X a b + 2c ≥1 but Cauchy tells us X a b + 2c X a(2b + c) ≥ X a 2 and as X a 2 ≥3(ab + bc + ca) = X a(2b + c) we have the result.
□ 37. (USA 2003) (a, b, c > 0) (2a + b + c)2 2a2 + (b + c)2 + (2b + c + a)2 2b2 + (c + a)2 + (2c + a + b)2 2c2 + (a + b)2 ≤8 First Solution. (See pag. 21) By multipliyng a, b and c by a suitable factor, we reduce the problem to the case when a + b + c = 3. The desidered inequality read (a + 3)2 2a2 + (3 −a)2 + (b + 3)2 2b2 + (3 −b)2 + (c + 3)2 2c2 + (3 −c)2 ≤8 Set f(x) = (x + 3)2 2x2 + (3 −x)2 38 It suffices to prove that f(a) + f(b) + f(c) ≤8. Note that f(x) = x2 + 6x + 9 3 (x2 −2x + 3) = = 1 3 1 + 8x + 6 x2 −2x + 3 = = 1 3 1 + 8x + 6 (x −1)2 + 2 ≤ ≤1 3 1 + 8x + 6 2 = 1 3 (4x + 4) Hence f(a) + f(b) + f(c) ≤1 3 (4a + 4 + 4b + 4 + 4c + 4) = 8 as desidered, with equality if and only if a = b = c.
□ Second Solution. (See ) We can assume, WLOG, a + b + c = 1. Then the first term of LHS is equal to f(a) = (a + 1)2 2a2 + (1 −a)2 = a2 + 2a + 1 3a2 −2a + 1 (When a = b = c = 1 3, there is equality. A simple sketch of f(x) on [0, 1] shows the curve is below the tangent line at x = 1 3, which has the equation y = 12x+4 3 ).
So we claim that a2 + 2a + 1 3a2 −2a + 1 ≤12a + 4 3 for a < 0 < 1. This inequality is equivalent to 36a3 −15a2 −2a + 1 = (3a −1)2(4a + 1) ≥0 , 0 < a < 1 hence is true. Adding the similar inequalities for b and c we get the desidered inequality.
□ 38. (Russia 2002) (x + y + z = 3, x, y, z > 0) √x + √y + √z ≥xy + yz + zx Solution. (Ercole Suppa) See Problem 23.
□ 39 39. (Latvia 2002) 1 1+a4 + 1 1+b4 + 1 1+c4 + 1 1+d4 = 1, a, b, c, d > 0 abcd ≥3 First Solution. (Ercole Suppa) We first prove a lemma: Lemma. For any real positive numbers x, y with xy ≥1 we have 1 x2 + 1 + 1 y2 + 1 ≥ 2 xy + 1 Proof. The required inequality follows from the identity 1 x2 + 1 + 1 y2 + 1 − 2 xy + 1 = (x −y)2 (xy −1) (x2 + 1) (y2 + 1) (xy + 1) the proof of which is immediate.
□ In order to prove the required inequality we observe at first that 1 1 + a4 + 1 1 + b4 ≤1 = ⇒ a4b4 ≥1 = ⇒ a2b2 ≥1 Thus by previous lemma we have 1 1 + a4 + 1 1 + b4 ≥ 2 a2b2 + 1 (1) and similarly 1 1 + c4 + 1 1 + d4 ≥ 2 c2d2 + 1 (2) Since ab ≥1 e cd ≥1 we can add (1) and (2) and we can apply again the lemma: 1 = 1 1 + a4 + 1 1 + b4 + 1 1 + c4 + 1 1 + d4 ≥ ≥2 1 a2b2 + 1 + 1 c2d2 + 1 ≥ ≥ 4 abdc + 1 Thus abcd + 1 ≥4 so abcd ≥3.
□ 40 Second Solution. (See , pag. 14) We can write a2 = tan A, b2 = tan B, c2 = tan C, d2 = tan D, where A, B, C, D ∈ 0, π 2 . Then, the algebraic identity becomes the following trigonometric identity cos2 A + cos2 B + cos2 C + cos2 D = 1.
Applying the AM-GM inequality, we obtain sin2 A = 1 −cos2 A = cos2 B + cos2 C + cos2 D ≥3 (cos B cos C cos D) 2 3 .
Similarly, we obtain sin2 B ≥3 (cos C cos D cos A) 2 3 , sin2 C ≥3 (cos D cos A cos B) 2 3 and sin2 D ≥3 (cos A cos B cos C) 2 3 .
Multiplying these four inequalities, we get the result!
□ 40. (Albania 2002) (a, b, c > 0) 1 + √ 3 3 √ 3 (a2 + b2 + c2) 1 a + 1 b + 1 c ≥a + b + c + p a2 + b2 + c2 Solution. (Ercole Suppa) From AM-GM inequality we have 1 a + 1 b + 1 c = ab + bc + ca abc ≥3 3 √ a2b2c2 abc = 3 3 √ abc (1) From AM-QM inequality we have a + b + c ≤3 r a2 + b2 + c2 3 (2) From (1) and (2) we get a + b + c + √ a2 + b2 + c2 (a2 + b2 + c2) 1 a + 1 b + 1 c ≤ 3 √ 3 √ a2 + b2 + c2 + √ a2 + b2 + c2 (a2 + b2 + c2) 3 3 √ abc ≤ ≤3 + √ 3 √ 3 √ a2 + b2 + c2 3 √ abc 3 (a2 + b2 + c2) ≤ ≤3 + √ 3 3 √ 3 √ a2 + b2 + c2 q a2+b2+c2 3 a2 + b2 + c2 = = √ 3 + 1 3 √ 3 41 Therefore 1 + √ 3 3 √ 3 (a2 + b2 + c2) 1 a + 1 b + 1 c ≥a + b + c + p a2 + b2 + c2 □ 41. (Belarus 2002) (a, b, c, d > 0) p (a + c)2 + (b + d)2+ 2|ad −bc| p (a + c)2 + (b + d)2 ≥ p a2 + b2+ p c2 + d2 ≥ p (a + c)2 + (b + d)2 Solution. (Sung-Yoon Kim, BoesFX ) Let A(0, 0), B(a, b), C(−c, −d) and let D be the foot of perpendicular from A to BC. Since [ABC] = 1 2 det 0 0 1 a b 1 −c −d 1 = 1 2 |ad −bc| we have that AH = 2 [ABC] BC = |ad −bc| q (a + c)2 + (b + d)2 So the inequality becomes: BC + 2 · AH ≥AB + AC ≥BC ∠A is obtuse, since A(0, 0), B is in quadran I, and C is in the third quadrant.
Since ∠A is obtuse, BD + DC must be BC. By triangle inequality, AB + AC ≥BC, BD + AD ≥AB, DC + AD ≥AC So, AB + AC ≤BD + DC + 2AD = BC + 2AD and the inequality is proven.
□ 42. (Canada 2002) (a, b, c > 0) a3 bc + b3 ca + c3 ab ≥a + b + c 42 First Solution. (Massimo Gobbino - Winter Campus 2006) We can assume WLOG that a ≥b ≥c. Then from the rearrangement inequality we have a3 ≥b3 ≥c3 , 1 bc ≥1 ac ≥1 ab ⇒a2 b + b2 c + c2 a ≤a3 bc + b3 ac + c3 ab and a2 ≥b2 ≥c2 , 1 c ≥1 b ≥1 a ⇒a + b + c ≤a2 b + b2 c + c2 a Therefore a + b + c ≤a3 bc + b3 ac + c3 ab □ Second Solution. (Shobber - ML Forum) By AM-GM, we have a3 bc + b + c ≥3a Sum up and done.
□ Third Solution. (Pvthuan - ML Forum) The inequality is simple applications of x2 + y2 + z2 ≥xy + yz + zx for a2, b2, c2 and ab, bc, ca,. We have a4 + b4 + c4 ≥a2b2 + b2c2 + c2a2 ≥abc(a + b + c) □ Fourth Solution. (Davron - ML Forum) The inequality a4 + b4 + c4 ≥a2b2 + b2c2 + c2a2 ≥abc(a + b + c) can be proved by Muirheads Theorem.
□ 43. (Vietnam 2002, Dung Tran Nam) (a2 + b2 + c2 = 9, a, b, c ∈R) 2(a + b + c) −abc ≤10 43 First Solution. (Nttu - ML Forum) We can suppose, WLOG, that |a| ≤|b| ≤|c| ⇒ c2 ≥3 ⇒ 2ab ≤a2 + b2 ≤6 We have [2 (a + b + c) −abc]2 = [2 (a + b) + c (2 −ab)]2 ≤ (Cauchy-Schwarz) ≤ h (a + b)2 + c2i h 22 + (2 −ab)2i = = 100 + (ab + 2)2 (2ab −7) ≤100 Thus 2 (a + b + c) −abc ≤10 □ Second Solution. (See , pag. 88, problem 93) □ 44. (Bosnia and Hercegovina 2002) (a2 + b2 + c2 = 1, a, b, c ∈R) a2 1 + 2bc + b2 1 + 2ca + c2 1 + 2ab ≤3 5 Solution. (Arne - ML Forum) From Cauchy-Schwartz inequality we have 1 = a2 + b2 + c22 ≤ X cyc a2 1 + 2bc ! X cyc a2 (1 + 2bc) !
(1) From GM-AM-QM inequality we have: X cyc a2 (1 + 2bc) !
= a2 + b2 + c2 + 2abc (a + b + c) ≤ ≤1 + 2 sa2 + b2 + c2 3 3 · 3 r a2 + b2 + c2 3 = (2) = 1 + 2 3 a2 + b2 + c22 = 5 3 The required inequality follows from (1) and (2).
□ 45. (Junior BMO 2002) (a, b, c > 0) 1 b(a + b) + 1 c(b + c) + 1 a(c + a) ≥ 27 2(a + b + c)2 44 Solution. (Silouan, Michael Lipnowski - ML Forum) From AM-GM inequality we have 1 b(a + b) + 1 c(b + c) + 1 a(c + a) ≥ 3 XY where X = 3 √ abc and Y = 3 p (a + b)(b + c)(c + a). By AM-GM again we have that X ≤a + b + c 3 and Y ≤2a + 2b + 2c 3 So 3 XY ≥ 27 2(a + b + c)2 and the result follows.
□ 46. (Greece 2002) (a2 + b2 + c2 = 1, a, b, c > 0) a b2 + 1 + b c2 + 1 + c a2 + 1 ≥3 4 a√a + b √ b + c√c 2 Solution. (Massimo Gobbino - Winter Campus 2006) From Cauchy-Schwarz inequality, a2 + b2 + c2 = 1 and the well-knon inequality a2b2 + b2c2 + c2a2 ≤1 3 a2 + b2 + c22 we have a√a + b √ b + c√c 2 ≤ X cyc √a √ b2 + 1 !2 ≤ ≤ X cyc a b2 + 1 ! X cyc a2b2 + a2 !
= = X cyc a b2 + 1 !
1 + a2b2 + b2c2 + c2a2 ≤ ≤ X cyc a b2 + 1 ! 1 + 1 3 a2 + b2 + c2 = = 4 3 X cyc a b2 + 1 !
45 Hence X cyc a b2 + 1 !
≥3 4 a√a + b √ b + c√c 2 □ 47. (Greece 2002) (bc ̸= 0, 1−c2 bc ≥0, a, b, c ∈R) 10(a2 + b2 + c2 −bc3) ≥2ab + 5ac Solution. (Ercole Suppa) At first we observe that 1−c2 bc ≥0 if and only if bc 1 −c2 ≥0. Thus: 10 a2 + b2 + c2 −bc3 −2ab −5ac = = 5(b −c)2 + 5 2(a −c)2 + (a −b)2 + 10bc 1 −c2 + 4b2 + 5 2c2 + 13 2 a2 ≥0 □ 48. (Taiwan 2002) a, b, c, d ∈ 0, 1 2 abcd (1 −a)(1 −b)(1 −c)(1 −d) ≤ a4 + b4 + c4 + d4 (1 −a)4 + (1 −b)4 + (1 −c)4 + (1 −d)4 Solution. ( Liu Janxin - ML Forum) We first prove two auxiliary inequalities: Lemma 1. If a, b ∈ 0, 1 2 we have a2 + b2 ab ≥(1 −a)2 + (1 −b)2 (1 −a)(1 −b) Proof. Since 1 −a −b ≥0 (bacause 0 ≤a, b ≤1 2) we get a2 + b2 ab −(1 −a)2 + (1 −b)2 (1 −a)(1 −b) = (1 −a −b)(a −b)2 ab(1 −a)(1 −b) ≥0 Lemma 2. If a, b, c, d ∈ 0, 1 2 we have a2 −b22 abcd ≥ (1 −a)2 −(1 −b)2 (1 −a)(1 −b)(1 −c)(1 −d) 46 Proof. Since 0 ≤c, d ≤1 2 we get (1 −c)(1 −d) cd ≥1 (1) Since 0 ≤a, b ≤1 2 we get a2 −b22 ab − (1 −a)2 −(1 −b)22 (1 −a)(1 −b) = (a −b)4(1 −a −b) ab(1 −a)(1 −b) ≥0 Therefore a2 −b22 ab ≥ (1 −a)2 −(1 −b)22 (1 −a)(1 −b) (2) Multiplying (1) and (2) we have a2 −b22 abcd ≥ (1 −a)2 −(1 −b)2 (1 −a)(1 −b)(1 −c)(1 −d) and the Lemma 2 is proven.
□ Now we can prove the required inequality. By Lemma 2, we have a4 + b4 + c4 + d4 abcd − a2 + b2 b2 + c2 abcd = = a2 −c22 + a2 −d22 + b2 −c22 + b2 −d22 2abcd ≥ ≥ (1 −a)2 −(1 −c)22 + (1 −a)2 −(1 −d)22 + (1 −b)2 −(1 −c)22 + (1 −b)2 −(1 −d)22 2(1 −a)(1 −b)(1 −c)(1 −d) = = (1 −a)4 + (1 −b)4 + (1 −c)4 + (1 −d)4 (1 −a)(1 −b)(1 −c)(1 −d) − (1 −a)2 + (1 −b)2 (1 −c)2 + (1 −d)2 (1 −a)(1 −b)(1 −c)(1 −d) By Lemma 1, we have a2 + b2 b2 + c2 abcd ≥ (1 −a)2 + (1 −b)2 (1 −c)2 + (1 −d)2 (1 −a)(1 −b)(1 −c)(1 −d) Thus, addingthe last two inequalities, we get a4 + b4 + c4 + d4 abcd ≥(1 −a)4 + (1 −b)4 + (1 −c)4 + (1 −d)4 (1 −a)(1 −b)(1 −c)(1 −d) and the desidered inequality follows: abcd (1 −a)(1 −b)(1 −c)(1 −d) ≤ a4 + b4 + c4 + d4 (1 −a)4 + (1 −b)4 + (1 −c)4 + (1 −d)4 □ 47 49. (APMO 2002) ( 1 x + 1 y + 1 z = 1, x, y, z > 0) √x + yz + √y + zx + √z + xy ≥√xyz + √x + √y + √z Solution. (Suat Namly) Multiplying by √xyz, we have √xyz = rxy z + ryz x + rzx y So it is enough to prove that √z + xy ≥√z + rxy z By squaring, this is equivalent to z + xy ≥z + xy z + 2√xy ⇐ ⇒ z + xy ≥z + xy 1 −1 x −1 y + 2√xy ⇐ ⇒ x + y ≥2√xy ⇐ ⇒ √x −√y 2 ≥0 □ 50. (Ireland 2001) (x + y = 2, x, y ≥0) x2y2(x2 + y2) ≤2.
First Solution. (Ercole Suppa) Using AM-GM inequality we get 2 = x + y ≥2√xy ⇒ xy ≤1 Therefore applying the AM-GM inequality again we get x2y2(x2 + y2) = x2y2(4 −2xy) = 2(xy)(xy)(2 −xy) ≤2 xy + xy + 2 −xy 3 3 ≤2 1 + 2 3 3 = 2 □ 48 Second Solution. (Pierre Bornzstein - ML Forum) WLOG, we may assume that x ≤y so that x ∈[0, 1]. Now x2y2(x2 + y2) = x2(2 −x)2(x2 + (2 −x)2) = f(x) Straighforward computations leads to f ′(x) = 4x(1 −x)(2 −x)(2x2 −6x + 4) ≥0 Thus f is increasing on [0; 1]. Since f(1) = 2, the result follows. Note that equality occurs if and only if x = y = 1.
□ Third Solution. (Kunny - ML Forum) We can set x = 2 cos2 θ, y = 2 sin2 θ so we have x2y2(x2 + y2) = 2 −2 cos4 2θ ≦2 □ 51. (BMO 2001) (a + b + c ≥abc, a, b, c ≥0) a2 + b2 + c2 ≥ √ 3abc First Solution. (Fuzzylogic - ML Forum) From the well-know inequality (x + y + z)2 ≥3(xy + yz + zx) by putting x = bc, y = ca, z = ab we get ab + bc + ca ≥ p 3abc(a + b + c) Then a2 + b2 + c2 ≥ab + bc + ca ≥ p 3abc(a + b + c) ≥abc √ 3 □ 49 Second Solution. (Cezar Lupu - ML Forum) Let’s assume by contradiction that a2 + b2 + c2 < abc √ 3 By applying Cauchy-Schwarz inequality, 3(a2 + b2 + c2) ≥(a + b + c)2 and the hipothesys a + b + c ≥abc we get abc < 3 √ 3 On the other hand , by AM-GM we have abc √ 3 > a2 + b2 + c2 ≥3 3 √ a2b2c2 We get from here abc > 3 √ 3, a contradiction.
□ Third Solution. (Cezar Lupu - ML Forum) We have a + b + c ≥abc ⇔1 ab + 1 bc + 1 ca ≥1 We shall prove a stronger inequality ab + bc + ca ≥abc √ 3 ⇔1 a + 1 b + 1 c ≥ √ 3 Now, let us denote x = 1 a, y = 1 b, z = 1 c and the problems becomes: If x, y, z are three nonnegative real numbers such that xy + yz + zx ≥1, then the following holds: x + y + z ≥ √ 3 But, this last problem follows immediately from this inequality (x + y + z)2 ≥3(xy + yz + zx) □ 52. (USA 2001) (a2 + b2 + c2 + abc = 4, a, b, c ≥0) 0 ≤ab + bc + ca −abc ≤2 50 First Solution. (Richard Stong, see pag. 22) From the given condition, at least one of a, b, c does not exceed 1, say a ≤1. Then ab + bc + ca −abc = a(b + c) + bc(1 −a) ≥0 It is easy to prove that the equality holds if and only if (a, b, c) is one of the triples (2, 0, 0), (0, 2, 0) or (0, 0, 2).
To prove the upper bound we first note that some two of three numbers a, b, c are both greater than or equal to 1 or less than or equal to 1. WLOG assume that the numbers with this property are b and c. Then we have (1 −b)(1 −c) ≥0 (1) The given equality a2 + b2 + c2 + abc = 4 and the inequality b2 + c2 ≥2bc imply a2 + 2bc + abc ≤4 ⇐ ⇒ bc(2 + a) ≤4 −a2 Dividing both sides of the last inequality by 2 + a yelds bc ≤2 −a (2) Combining (1) and (2) gives ab + bc + ac −abc ≤ab + 2 −a + ac(1 −b) = = 2 −a(1 + bc −b −c) = = 2 −a(1 −b)(1 −c) ≤2 as desidered. The last equality holds if and only if b = c and a(1−b)(1−c) = 0.
Hence, equality for upper bound holds if and only if (a, b, c) is one of the triples (1, 1, 1), (0, √ 2, √ 2), ( √ 2, 0, √ 2) and ( √ 2, √ 2, 0).
□ Second Solution. (See ) Assume WLOG a ≥b ≥c.
If c > 1, then a2 + b2 + c2 + abc > 1 + 1 + 1 + 1 = 4, contradiction.
So c ≤1.
Hence ab + bc + ca ≥ab ≥abc.
Put a = u + v, b = u −v, so that u, v = 0. Then the equation given becomes (2 + c)u2 + (2 −c)v2 + c2 = 4 So if we keep c fixed and reduce v to nil, then we must increase u. But ab+bc+ ca−abc = (u2−v2)(1−c)+2cu, so decreasing v and increasing u has the effect of increasing ab+bc+ca−abc. Hence ab+bc+ca−abc takes its maximum value when a = b. But if a = b, then the equation gives a = b = √2 −c. So to establish that ab+bc+ca−abc ≤2 it is sufficient to show that 2−c+2c√2 −c = 2+c(2−c).
Evidently we have equality if c = 0.
If c is non-zero, then the relation is equivalent to 2√2 −c ≤3 −c or (c −1)2 ≥0. Hence the relation is true and we have equality only for c = 0 or c = 1.
□ 51 53. (Columbia 2001) (x, y ∈R) 3(x + y + 1)2 + 1 ≥3xy Solution. (Ercole Suppa) After setting x = y we have 3(2x + 1)2 + 1 −3x2 ≥0 ⇐ ⇒ (3x + 2)2 ≥0 (1) where the equality holds if x = −2 3. This suggest the following change of variable 3x + 2 = a , 3y + 2 = b Now for all x, y ∈R we have: 3 (x + y + 1)2 + 1 −3xy = 3 a + b −4 3 + 1 2 + 1 −3(a −2) (b −2) 9 = = (a + b −1)2 3 + 1 −ab −2a −2b + 4 3 = = a2 + b2 + ab 3 = = a2 + b2 + (a + b)2 6 = = (3x + 2)2 + (3y + 2)2 + [3 (x + y) + 4]2 6 ≥0 □ 54. (KMO Winter Program Test 2001) (a, b, c > 0) p (a2b + b2c + c2a) (ab2 + bc2 + ca2) ≥abc+ 3 p (a3 + abc) (b3 + abc) (c3 + abc) First Solution. (See , pag. 38) Dividing by abc, it becomes sa c + b a + c b c a + a b + b c ≥abc + 3 sa2 bc + 1 b2 ca + 1 c2 ab + 1 .
After the substitution x = a b , y = b c, z = c a, we obtain the constraint xyz = 1.
It takes the form p (x + y + z) (xy + yz + zx) ≥1 + 3 sx z + 1 y x + 1 z y + 1 .
52 From the constraint xyz = 1, we find two identities x z + 1 y x + 1 z y + 1 = x + z z y + x x z + y y = (z+x)(x+y)(y+z), (x + y + z) (xy + yz + zx) = (x+y)(y+z)(z+x)+xyz = (x+y)(y+z)(z+x)+1.
Letting p = 3 p (x + y)(y + z)(z + x), the inequality now becomes p p3 + 1 ≥ 1+p. Applying the AM-GM inequality, we have p ≥ 3 q 2√xy · 2√yz · 2√zx = 2.
It follows that (p3 + 1) −(1 + p)2 = p(p + 1)(p −2) ≥0.
□ Second Solution. (Based on work by an winter program participant, see pag. 43).
□ 55. (IMO 2001) (a, b, c > 0) a √ a2 + 8bc + b √ b2 + 8ca + c √ c2 + 8ab ≥1 Solution. (Massimo Gobbino - Winter Campus 2006) Let T is the left hand side of the inequality. We have (a + b + c)2 = X cyc √a 4 √ a2 + 8bc √a 4 p a2 + 8bc !2 ≤ (Cauchy-Schwarz) ≤T · X cyc a p a2 + 8bc !
= ≤T · X cyc √a√a p a2 + 8bc !
≤ (Cauchy-Schwarz) ≤T · (a + b + c) 1 2 X cyc a3 + 8abc ! 1 2 = = T · (a + b + c) 1 2 a3 + b3 + c3 + 24abc 1 2 Hence 53 T ≥ (a + b + c) 3 2 (a3 + b3 + c3 + 24abc) 1 2 ≥1 where in the last step we used the inequality (a + b + c)3 ≥a3 + b3 + c3 + 24abc which is true by BUNCHING, since (a + b + c)3 ≥a3 + b3 + c3 + 24abc ⇐ ⇒ 3 X sym a2b !
+ 6abc ≥24abc ⇐ ⇒ X sym a2b ≥6abc ⇐ ⇒ X sym a2b ≥ X sym abc □ 54 2 Years 1996 ∼2000 56. (IMO 2000, Titu Andreescu) (abc = 1, a, b, c > 0) a −1 + 1 b b −1 + 1 c c −1 + 1 a ≤1 Solution. (See , pag. 3) Since abc = 1, we make the substitution a = x y , b = y z , c = z x for x, y, z > 0. We rewrite the given inequality in the terms of x, y, z : x y −1 + z y y z −1 + x z z x −1 + y x ≤1 ⇔ xyz ≥(y + z −x)(z + x −y)(x + y −z) This is true by Schur inequality.
□ Remark. Alternative solutions are in , pag. 18, 19.
57. (Czech and Slovakia 2000) (a, b > 0) 3 s 2(a + b) 1 a + 1 b ≥3 ra b + 3 r b a First Solution. (Massimo Gobbino - Winter Campus 2006) After setting a = x3 a b = y3 the required inequality become x y + y x ≤ 3 s 2 (x3 + y3) 1 x3 + 1 y3 x2 + y2 xy ≤1 xy 3 q 2 (x3 + y3)2 x2 + y23 ≤2 x3 + y32 x2 + y2 1 2 ≤2 1 6 x3 + y3 1 3 r x2 + y2 2 ≤ 3 r x3 + y3 2 which is true by Power Mean inequality. The equality holds if x = y, i.e. if a = b.
□ 55 Second Solution. (Official solution.) Elevating to the third power both mem-bers of the given inequality we get the equivalent inequality a b + 3 3 ra b + 3 3 r b a + b a ≥4 + 2a b + 2 b a that is a b + b a + 4 ≥3 3 ra b + 3 r b a !
The AM-GM inequality applied to the numbers a b , 1, 1 implies a b + 1 + 1 ≥3 3 ra b Similarly we have b a + 1 + 1 ≥3 3 r b a Adding the two last inequalities we get the required result.
□ 58. (Hong Kong 2000) (abc = 1, a, b, c > 0) 1 + ab2 c3 + 1 + bc2 a3 + 1 + ca2 b3 ≥ 18 a3 + b3 + c3 First Solution. (Official solution) Apply Cauchy-Scwarz Inequality, we have 1 + ab2 c3 + 1 + bc2 a3 + 1 + ca2 b3 c3 + a3 + b3 ≥ X cyc p 1 + ab2 !2 It remain to prove X cyc p 1 + ab2 ≥ √ 18 The proof goes as follows p 1 + ab2 + p 1 + bc2 + p 1 + ca2 ≥ ≥ r (1 + 1 + 1)2 + √ ab2 + √ bc2 + √ ca2 2 ≥ (Minkowski Ineq) ≥ r 9 + 3 √ abc 2 = (AM-GM Ineq) = √ 18 □ 56 Second Solution. (Ercole Suppa) From AM-HM inequality we have 1 c3 + 1 a3 + 1 b3 ≥ 9 a3 + b3 + c3 (1) and ab2 c3 + bc2 a3 + ca2 b3 ≥3 3 r a3b3c3 a3b3c3 = 9 3 3 √ a3b3c3 ≥ 9 a3 + b3 + c3 (2) Adding (1) and (2) we get the required inequality.
□ 59. (Czech Republic 2000) (m, n ∈N, x ∈[0, 1]) (1 −xn)m + (1 −(1 −x)m)n ≥1 Solution. (See pag. 83) The given inequality follow from the following most general result: Let x1, . . . , xn and y1, . . . , yn be nonnegative real numbers such that xi + yi = 1 for each i = 1, 2, . . . , n. Prove that (1 −x1x2 · · · xn)m + (1 −ym 1 ) (1 −ym 2 ) · · · (1 −ym n ) ≥1 We use the following probabilistic model suggested by the circumstance that xi +yi = 1. Let n unfair coins. Let xi be the probability that a toss of the i−th coin is a head (i = 1, 2, . . . , n). Then the probability that a toss of this coin is a tail equals 1 −xi = yi.
The probability of n heads in tossing all the coins once is x1x2 · · · xn, because the events are independent. Hence 1 −x1x2 · · · xn is the probability of at least one tail. Consequently, the probability of at least one tail in each of m consecutive tosses af all the coins equals (1 −x1x2 · · · xn)m With probability ym i , each of m consecutive tosses of the i-th coin is a tail; with probability 1 −ym i , we have at least one head. Therefore the probability that after m tosses of all coins each coin has been a head at least once equals (1 −ym 1 ) (1 −ym 2 ) · · · (1 −ym n ) Denote the events given above in italics by A and B, respectively. It is easy to observe that at leat one of them must occur as a result of m tosses. Indeed, suppose A has not occurred. This means that the outcome of some toss has been n heads, which implies that B has occurred. Now we need a line more to 57 finish the proof. Since one of the events A and B occurs as a result of m tosses, the sum of their probabilities is greater than or equal to 1, that is (1 −x1x2 · · · xn)m + (1 −ym 1 ) (1 −ym 2 ) · · · (1 −ym n ) ≥1 □ Remark. Murray Klamkin - Problem 68-1 (SIAM Review 11(1969)402-406).
60. (Macedonia 2000) (x, y, z > 0) x2 + y2 + z2 ≥ √ 2 (xy + yz) Solution. (Ercole Suppa) By AM-GM inequality we have x2 + y2 + z2 = x2 + 1 2y2 + 1 2y2 + z2 ≥ ≥2x y √ 2 + 2 y √ 2z = = √ 2xy + √ 2yz = = √ 2 (xy + yz) □ 61. (Russia 1999) (a, b, c > 0) a2 + 2bc b2 + c2 + b2 + 2ca c2 + a2 + c2 + 2ab a2 + b2 > 3 First Solution. (Anh Cuong - ML Forum) First let f(a, b, c) = a2+2bc b2+c2 + b2+2ac a2+c2 + c2+2ab a2+b2 . We will prove that: f(a, b, c) ≥ 2bc b2 + c2 + b c + c b Suppose that: b ≥c ≥a. Since a2 + 2bc b2 + c2 ≥ 2bc b2 + c2 we just need to prove that: b2 + 2ac a2 + c2 + c2 + 2ab a2 + b2 ≥b c + c b 58 We have: b2 + 2ac a2 + c2 + c2 + 2ab a2 + b2 −b c −c b = = b3 + 2abc −c3 −ca2 b(c2 + a2) + c3 + 2abc −b3 −ba2 c(b2 + a2) ≥ ≥ b3 −c3 b(a2 + c2) + c3 −b3 c(a2 + b2) = = bc −a2 (b −c)2 b2 + bc + c2 bc (a2 + b2) (a2 + c2) ≥0 Hence: f(a, b, c) ≥ 2bc b2 + c2 + b c + c b But 2bc b2 + c2 + b c + c b ≥3 ⇔(b −c)2(b2 + c2 −bc) ≥0.
So we have done now.
□ Second Solution. (Charlie- ML Forum) Brute force proof: Denote T(x, y, z) = P sym axbycz. Expanding and simplifying yields 1 2 · T(6, 0, 0) + T(4, 1, 1) + 2 · T(3, 2, 1) + T(3, 3, 0) ≥2 · T(4, 2, 0) + 1 2 · T(2, 2, 2) which is true since 1 2 · T(6, 0, 0) + 1 2 · T(4, 1, 1) ≥T(5, 1, 0) by Schur’s inequality, and T(5, 1, 0) + T(3, 3, 0) ≥2 · T(4, 2, 0) by AM-GM (a5b + a3b3 ≥2a4b2), and 2 · T(3, 2, 1) ≥2 · T(2, 2, 2) ≥1 2 · T(2, 2, 2) by bunching.
□ Third Solution. (Darij Grinberg - ML Forum) Using the P cyc notation for cyclic sums, the inequality in question rewrites as X cyc a2 + 2bc b2 + c2 > 3 59 But X cyc a2 + 2bc b2 + c2 −3 = X cyc a2 + 2bc b2 + c2 −1 = = X cyc a2 b2 + c2 − X cyc (b −c)2 b2 + c2 Thus, we have to show that X cyc a2 b2 + c2 > X cyc (b −c)2 b2 + c2 Now, by the Cauchy-Schwarz inequality in the Engel form, we have X cyc a2 b2 + c2 = X cyc a22 a2b2 + c2a2 ≥ ≥ a2 + b2 + c22 (a2b2 + c2a2) + (b2c2 + a2b2) + (c2a2 + b2c2) = = a2 + b2 + c22 2 (b2c2 + c2a2 + a2b2) Hence, it remains to prove that a2 + b2 + c22 2 (b2c2 + c2a2 + a2b2) > X cyc (b −c)2 b2 + c2 i. e. that a2 + b2 + c22 > 2 b2c2 + c2a2 + a2b2 X cyc (b −c)2 b2 + c2 Now, 2 b2c2 + c2a2 + a2b2 X cyc (b −c)2 b2 + c2 = X cyc 2 b2c2 + c2a2 + a2b2 b2 + c2 (b −c)2 = = X cyc 2b2c2 b2 + c2 + 2a2 (b −c)2 The HM-GM inequality, applied to the numbers b2 and c2, yields 2b2c2 b2 + c2 ≤ √ b2c2 = bc 60 thus, 2 b2c2 + c2a2 + a2b2 X cyc (b −c)2 b2 + c2 = X cyc 2b2c2 b2 + c2 + 2a2 (b −c)2 ≤ ≤ X cyc bc + 2a2 (b −c)2 Hence, instead of proving a2 + b2 + c22 > 2 b2c2 + c2a2 + a2b2 X cyc (b −c)2 b2 + c2 it will be enough to show the stronger inequality a2 + b2 + c22 > X cyc bc + 2a2 (b −c)2 With a bit of calculation, this is straightforward; here is a longer way to show it without great algebra: X cyc bc + 2a2 (b −c)2 = = X cyc (a (a + b + c) −(c −a) (a −b)) (b −c)2 = = X cyc a (a + b + c) (b −c)2 − X cyc (c −a) (a −b) (b −c)2 = = (a + b + c) X cyc a (b −c)2 −(b −c) (c −a) (a −b) X cyc (b −c) | {z } =0 = = (a + b + c) X cyc a (b −c)2 = = (a + b + c) X cyc a ((b −c) (b −a) + (c −a) (c −b)) = = (a + b + c) X cyc a (b −c) (b −a) + X cyc a (c −a) (c −b) !
= = (a + b + c) X cyc c (a −b) (a −c) + X cyc b (a −b) (a −c) !
= = (a + b + c) X cyc (b + c) (a −b) (a −c) Thus, in order to prove that a2 + b2 + c22 > P cyc bc + 2a2 (b −c)2, we will show the equivalent inequality a2 + b2 + c22 > (a + b + c) X cyc (b + c) (a −b) (a −c) 61 In fact, we will even show the stronger inequality a2 + b2 + c22 > X cyc a2 (a −b) (a −c) + (a + b + c) X cyc (b + c) (a −b) (a −c) which is indeed stronger since P cyc a2 (a −b) (a −c) ≥0 by the Schur inequal-ity.
Now, this stronger inequality can be established as follows: X cyc a2 (a −b) (a −c) + (a + b + c) X cyc (b + c) (a −b) (a −c) = = X cyc a2 + (a + b + c) (b + c) (a −b) (a −c) = = X cyc a2 + b2 + c2 + (bc + ca + ab) + bc (a −b) (a −c) = = a2 + b2 + c2 + (bc + ca + ab) X cyc (a −b) (a −c) | {z } =(a2+b2+c2)−(bc+ca+ab) + X cyc bc (a −b) (a −c) | {z } =a2+bc−ca−ab<2a2+bc < < a2 + b2 + c2 + (bc + ca + ab) a2 + b2 + c2 −(bc + ca + ab) + X cyc bc 2a2 + bc = = a2 + b2 + c22 −(bc + ca + ab)2 + (bc + ca + ab)2 = a2 + b2 + c22 and the inequality is proven.
□ 62. (Belarus 1999) (a2 + b2 + c2 = 3, a, b, c > 0) 1 1 + ab + 1 1 + bc + 1 1 + ca ≥3 2 Solution. (Ercole Suppa) From Cauchy-Schwartz inequality we have 9 = a2 + b2 + c22 ≤ X cyc 1 1 + bc ! X cyc a2 (1 + bc) !
(1) From GM-AM-QM inequality we have: X cyc a2 (1 + bc) !
= a2 + b2 + c2 + abc (a + b + c) ≤ ≤3 + sa2 + b2 + c2 3 3 · 3 r a2 + b2 + c2 3 = (2) = 3 + 1 3 a2 + b2 + c22 = 3 + 3 = 6 The required inequality follows from (1) and (2).
□ 62 63. (Czech-Slovak Match 1999) (a, b, c > 0) a b + 2c + b c + 2a + c a + 2b ≥1 Solution. (Ercole Suppa) Using Cauchy-Schwartz inequality and the well-know (a + b + c)2 ≥3(ab + bc + ca) we have (a + b + c)2 ≤ X cyc a b + 2c · X cyc a(b + 2c) = (Cauchy-Schwarz) = X cyc a b + 2c · 3(ab + bc + ca) ≤ ≤ X cyc a b + 2c · (a + b + c)2 Dividing for (a + b + c)2 we get the result.
□ 64. (Moldova 1999) (a, b, c > 0) ab c(c + a) + bc a(a + b) + ca b(b + c) ≥ a c + a + b b + a + c c + b First Solution. (Ghang Hwan, Bodom - ML Forum) After the substitution x = c/a, y = a/b, z = b/c we get xyz = 1 and the inequality becomes z x + 1 + x y + 1 + y z + 1 ≥ 1 1 + x + 1 1 + y + 1 1 + z Taking into account that xyz = 1, this inequality can be rewritten as z −1 x + 1 + x −1 y + 1 + y −1 z + 1 ≥0 ⇐ ⇒ yz2 + zx2 + xy2 + x2 + y2 + z2 ≥x + y + z + 3 (∗) The inequality (∗) is obtained summing the well-know inequality x2 + y2 + z2 ≥x + y + z and yz2 + zx2 + xy2 ≥3 3 p x3y3z3 = 3xyz = 3 which follows from the AM-GM inequality.
□ 63 Second Solution. (Gibbenergy - ML Forum) We have L −R = abc h ab c2 + bc a2 + ac b2 −3 + b2 c2 + c2 a2 + a2 b2 − b c + c a + a b i (a + b)(b + c)(c + a) ≥0 because ab c2 + bc a2 + ac b2 −3 ≥0 by AM-GM inequality and b2 c2 + c2 a2 + a2 b2 − b c + c a + a b ≥0 by the well-know inequality x2 + y2 + z2 ≥x + y + z.
□ 65. (United Kingdom 1999) (p + q + r = 1, p, q, r > 0) 7(pq + qr + rp) ≤2 + 9pqr First Solution. (Ercole Suppa) From Schur inequality we have (p + q + r)3 + 9pqr ≥4(p + q + r)(pq + qr + rp) Therefore, since p + q + r = 1, we obtain 1 + 9pqr ≥4(pq + qr + rp) Hence 2 + 9pqr −7(pq + qr + rp) ≥2 + 4(pq + qr + rp) −1 −7(pq + qr + rp) = = 1 −3(pq + qr + rp) = = (p + q + r)2 −3(pq + qr + rp) = = 1 2 (p −q)2 + (q −r)2 + (r −p)2 ≥0 and the inequality is proven.
□ Second Solution. (See pag. 189) Because p + q + r = 1 the inequality is equivalent to 7(pq + qr + rp)(p + q + r) ≤2(p + q + r)3 + 9pqr ⇐ ⇒ 7 X cyc p2q + pq2 + pqr ≤9pqr + X cyc 2p3 + 6p2q + 6pq2 + 4pqr ⇐ ⇒ X cyc p2q + X cyc pq2 ≤ X cyc 2p3 = X cyc 2p3 + q3 3 + X cyc p3 + 2q3 3 This last inequality is true by weighted AM-GM inequality.
□ 64 66. (Canada 1999) (x + y + z = 1, x, y, z ≥0) x2y + y2z + z2x ≤4 27 First Solution. (See pag. 42) Assume WLOG that x = max(x, y, z). If x ≥y ≥z, then x2y + y2z + z2x ≤x2y + y2z + z2x + z [xy + (x −y) (y −z)] = = (x + y)2 y = 4 1 2 −1 2y 1 2 −1 2y y ≤4 27 where the last inequality follows from AM-GM inequality. Equality occurs if and only if z = 0 (from the first inequality) and y = 1 3, in which case (x, y, z) = 2 3, 1 3, 0 .
If If x ≥y ≥z, then x2y + y2z + z2x ≤x2z + z2y + y2x −(x −z) (z −y) (x −y) ≤ ≤x2z + z2y + y2x ≤4 27 where the second inequality is true from the result we proved for x ≥y ≥z (except with y and z reversed. Equality holds in the first inequality only when two of x, y, z are equal, and in the second inequality only when (x, z, y) = 2 3, 1 3, 0 . Because these conditions can’t both be true, the inequality is actually strict in this case.
Therefore the inequality is indeed true, and the equality olds when (x, y, z) equals 2 3, 1 3, 0 , 1 3, 0, 2 3 or 0, 2 3, 1 3 .
□ Second Solution. (CMO Committee - Crux Mathematicorum 1999, pag. 400) Let f(x, y, z) = x2y + y2z + z2x. We wish to determine where f is maximal.
Since f is cyclic WLOG we may assume that x = max(x, y, z). Since f (x, y, z) −f (x, z, y) = x2y + y2z + z2x −x2z −z2y −y2x = = (y −z) (x −y) (x −z) we may also assume y ≥z. Then f (x + z, y, 0) −f (x, y, z) = (x + z)2 y −x2y −y2z −z2x = = z2y + yz (x −y) + xz (y −z) ≥0 so we may now assume z = 0. The rest follows from AM-GM inequality f (x, y, 0) = 2x2y 2 ≤1 2 x + x + 2y 3 3 = 4 27 Equality occurs when x = 2y, hence when (x, y, z) equals 2 3, 1 3, 0 , 1 3, 0, 2 3 or 0, 2 3, 1 3 .
□ 65 Third Solution. (CMO Committee - Crux Mathematicorum 1999, pag. 400) With f as above, and x = max(x, y, z) we have f x + z 2, y + z 2, 0 −f (x, y, z) = yz (x −y) + xz 2 (x −z) + z2y 4 + z3 8 so we may assume that z = 0. The rest follows as for second solution.
□ Fourth Solution. (See pag. 46, problem 32) Assume WLOG that x = max(x, y, z). We have x2y + y2z + z2x ≤ x + z 2 2 y + z 2 (1) because xyz ≥y2z and x2z 2 ≥xz2 2 .
Then by AM-GM inequality and (1) we have 1 = x + z 2 2 + x + z 2 2 + y + z 2 ≥ ≥3 3 s y + z 2 x + z 2 2 4 ≥ ≥3 3 r x2y + y2z + z2x 4 from which follows the desidered inequality x2y + y2z + z2x ≤ 4 27.
□ 67. (Proposed for 1999 USAMO, [AB, pp.25]) (x, y, z > 1) xx2+2yzyy2+2zxzz2+2xy ≥(xyz)xy+yz+zx First Solution. (See pag. 67) The required inequality is equivalent to x2 + 2yz log x + y2 + 2xz log y + z2 + 2xy log z ≥ ≥(xy + yz + x + zx) (log x + log y + log z) that is (x −y)(x −z) log x + (y −z)(y −x) log y + (z −x)(z −y) log z ≥0 We observe that log x, log y, log z > 0 because x, y, z > 1. Furthermore, since the last inequality is symmetric, we can assume WLOG that x ≥y ≥z. Thus (z −x)(z −y) log z ≥0 (1) 66 and, since the function log x is increasing on x > 0, we get (x −y)(x −z) log x ≥(y −z)(x −y) log y (2) because each factor of LHS is greater or equal of a different factor of RHS. The required inequality follows from (1) and (2).
□ Second Solution. (Soarer - ML Forum) The required equality is equivalent to xx2+yz−xy−xzyy2+xz−xy−yzzz2+xy−xz−yz ≥1 x(x−y)(x−z)y(y−x)(y−z)z(z−x)(z−y) ≥1 x y x−y · y z y−z · x z x−z ≥1 (⋆) By symmetry we can assume WLOG tath x ≥y ≥z. Therefore (⋆) is verifyied.
□ 68. (Turkey, 1999) (c ≥b ≥a ≥0) (a + 3b)(b + 4c)(c + 2a) ≥60abc First Solution. (ML Forum) By AM-GM inequality we have (a + 3b) (b + 4c) (c + 2a) ≥4 4 √ ab3 · 5 5 √ bc4 · 3 3 √ ca2 = = 60 a 1 4 a 2 3 b 3 4 b 1 5 c 4 5 c 1 3 = = 60a 11 12 b 19 20 c 17 15 = = 60abc · a−1 12 b−1 20 c 2 15 ≥ ≥60abc · c−1 12 c−1 20 c 2 15 = = 60abc where the last inequality is true because c ≥b ≥a ≥0 and the function f(x) = xα (with α < 0) is decreasing.
□ Second Solution. (See pag. 176) By the AM-GM inequality we have a + b + b ≥3 3 √ ab2. Multiplying this inequality and the analogous inequalities for b + 2c and c + 2a yields (a + 2b)(b + 2c)(c + 2a) ≥27abc. Then (a + 2b)(b + 2c)(c + 2a) ≥ ≥ a + 1 3a + 8 3b b + 2 3b + 10 3 c (c + 2a) = = 20 9 (a + 2b)(b + 2c)(c + 2a) ≥60abc □ 67 69. (Macedonia 1999) (a2 + b2 + c2 = 1, a, b, c > 0) a + b + c + 1 abc ≥4 √ 3 First Solution. (Frengo, Leepakhin - ML Forum) By AM-GM, we have 1 = a2 + b2 + c2 ≥3 3 p (abc)2 ⇒(abc)2 ≤1 27 Thus, by AM-GM a + b + c + 1 abc = a + b + c + 1 9abc + 1 9abc + · · · + 1 9abc ≥ ≥12 12 s 1 99(abc)8 ≥ ≥12 12 s 1 99( 1 27)4 = 4 √ 3 Equality holds if and only if a = b = c = 1 9abc or a = b = c = 1 √ 3.
□ Second Solution. (Ercole Suppa) The required inequality is equivalent to abc a + b + c −4 √ 3 + 1 ≥0 From Schur inequality we have (a + b + c)3 + 9abc ≥4(a + b + c)(ab + bc + ca) Since ab + bc + ca = 1 2 (a + b + c)2 − a2 + b2 + c2 = 1 2 (a + b + c)2 −1 we get abc ≥1 9 (a + b + c)3 −2(a + b + c) After setting S = a + b + c from Cauchy-Schwarz inequality follows that S = a + b + c ≤ √ 1 + 1 + 1 p a2 + b2 + c2 = √ 3 and, consequently abc a + b + c −4 √ 3 + 1 ≥1 9 S3 −2S S −4 √ 3 + 1 = = 1 9 h S3 −2S S −4 √ 3 + 9 i = = 1 9 √ 3 −S 4 + 20S √ 3 −S ≥0 □ 68 Third Solution. (Ercole Suppa) After setting S = a+b+c, Q = ab+bc+ca, from the constraint a2 +b2 +c2 = 1 we have S2 = 1 + 2Q ≥1. Then S ≥1 and, by Cauchy-Schwarz inequality we get S = a + b + c ≤ √ 1 + 1 + 1 p a2 + b2 + c2 = √ 3 From the well-know inequality (ab + bc + ca)2 ≥3abc(a + b + c) follows that 1 abc ≥3S Q2 . Thus, to establish the required inequality is enough to show that S + 3S Q2 ≥4 √ 3 ⇐ ⇒ 4 S −4 √ 3Q2 Q2 + 12S ≥0 Sine 1 < S ≤ √ 3 we have 4 S −4 √ 3 Q2 + 12S = S −4 √ 3 S2 −1 2 + 12S = = √ 3 −S −S4 + 3 √ 3S3 + 11S2 + 3 √ 3S −4 ≥ ≥ √ 3 −S −S4 + 3S4 + 11S2 + 3S2 −4 ≥ ≥ √ 3 −S 2S4 + 14S2 −4 ≥12 √ 3 −S ≥0 □ Fourth Solution. (Ercole Suppa) Since a2 + b2 + c2 = 1, the inequality abc a + b + c −4 √ 3 + 1 ≥0 can be tranformed into a homogeneous one in the following way abc (a + b + c) −4 √ 3abc p a2 + b2 + c2 + a2 + b2 + c22 ≥0 Squaring and expanding the expression we get 1 2 X sym a8 + 4 X sym a6b2 + X sym a6bc + 2 X sym a5b2c + 3 X sym a4b4+ + X sym a4b3c + 13 2 X sym a4b2c2 ≥24 X sym a4b2c2 The last inequality can be obtained addind the following inequalities which are 69 true by Muirhead theorem: 1 2 X sym a8 ≥1 2 X sym a4b2c2 (1) 4 X sym a6b2 ≥4 X sym a4b2c2 (2) X sym a6bc ≥ X sym a4b2c2 (3) 2 X sym a5b2c ≥2 X sym a4b2c2 (4) 3 X sym a4b4 ≥3 X sym a4b2c2 (5) X sym a4b3c ≥4 X sym a4b2c2 (6) 13 2 X sym a4b2c2 ≥13 2 X sym a4b2c2 (7) □ Fifth Solution. (Tiks - ML Forum) a + b + c + 1 abc ≥4 √ 3 ⇐ ⇒a + b + c + (a2 + b2 + c2)2 abc ≥4 p 3(a2 + b2 + c2) ⇐ ⇒(a2 + b2 + c2)2 −3abc(a + b + c) abc ≥4( p 3(a2 + b2 + c2) −(a + b + c)) ⇐ ⇒(a2 + b2 + c2)2 −3abc(a + b + c) abc ≥43(a2 + b2 + c2) −(a + b + c)2 p 3(a2 + b2 + c2) + a + b + c ⇐ ⇒ X (a −b)2 (a + b)2 + 3c2 2abc ≥ X (a −b)2 4 p 3(a2 + b2 + c2) + a + b + c but we have that p 3(a2 + b2 + c2)+a+b+c ≥2(a+b+c) so we have to prove that X (a −b)2 (a + b)2 + 3c2 2abc ≥ X (a −b)2 2 a + b + c ⇐ ⇒ X (a −b)2 (a + b)2 + 3c2 2abc − 2 a + b + c ≥0 We have that (a + b + c) (a + b)2 + 3c2 ≥c(a + b)2 ≥4abc 70 hence (a + b)2 + 3c2 2abc − 2 a + b + c ≥0 So the inequality is done.
□ 70. (Poland 1999) (a + b + c = 1, a, b, c > 0) a2 + b2 + c2 + 2 √ 3abc ≤1 Solution. (Ercole Suppa) From the well-know inequality (x + y + z)2 ≥3(xy + yz + xz) by putting x = ab, y = bc e z = ca we have (ab + bc + ca)2 ≥3abc(a + b + c) = ⇒ ab + bc + ca ≥ √ 3abc (1) From the constraint a + b + c = 1 follows that 1 −a2 −b2 −c2 = (a + b + c)2 −a2 −b2 −c2 = 2ab + 2ac + 2ca (2) (1) and (2) implies 1 −a2 −b2 −c2 −2 √ 3abc = 2ab + 2bc + 2ca −2 √ 3abc ≥0 □ 71. (Canada 1999) (x + y + z = 1, x, y, z ≥0) x2y + y2z + z2x ≤4 27 Solution. (Ercole Suppa) See: problem n.66.
□ 72. (Iran 1998) 1 x + 1 y + 1 z = 2, x, y, z > 1 √x + y + z ≥ √ x −1 + p y −1 + √ z −1 71 Solution. (Massimo Gobbino - Winter Campus 2006) √ x −1 + p y −1 + √ z −1 = X cyc √x −1 √x √x !
≤ ≤ X cyc x −1 x ! 1 2 (x + y + z) 1 2 = (Cauchy-Schwarz) = 3 −1 x −1 y −1 z 1 2 √x + y + z = = √x + y + z □ 73. (Belarus 1998, I. Gorodnin) (a, b, c > 0) a b + b c + c a ≥a + b b + c + b + c a + b + 1 First Solution. (Ercole Suppa) The required inequality is equivalent to ab4 + a3c2 + a2b3 + b3c2 + b2c3 ≥a2b2c + 2ab3c + 2ab2c2 ⇔ ab4 + a3c2 + a2b3 + b3c2 + b2c3 + a2b2c ≥2a2b2c + 2ab3c + 2ab2c2 ⇔ ab4 + a3c2 −2a2b2c + a2b3 + b3c2 −2ab3c + b2c3 + a2b2c −2ab2c2 ≥0 ⇔ a b4 + a2c2 −2ab2c + b3 a2 + c2 −2ac + b2c c2 + a2 −2ac ≥0 ⇔ a b2 −ac 2 + b3(a −c)2 + b2c(a −c)2 ≥0 so it is verified, being a sum of non-negative terms.
The equality hold if and only if b2 = ac e a = c, i.e. a = b = c.
□ Second Solution. (Ercole Suppa) The required inequality is equivalent to a2b3 + ab4 + a3c2 + b3c2 + b2c3 ≥a2b2c + 2ab3c + 2ab2c2 (⋆) From AM-GM inequality we have a2b3 + b3c2 ≥2 √ a2b6c2 = 2ab3c (1) 1 2ab4 + 1 2a3c2 ≥2 r a4b4c2 4 = a2b2c (2) 1 2ab4 + 1 2a3c2 + b2c3 ≥a2b2c + b2c3 = 2 √ a2b4c4 = 2ab2c2 (3) The (⋆) is obtained adding (1),(2) and (3).
□ 72 74. (APMO 1998) (a, b, c > 0) 1 + a b 1 + b c 1 + c a ≥2 1 + a + b + c 3 √ abc Solution. (See pag. 174) We have 1 + a b 1 + b c 1 + c a = = 2 + a b + b c + c a + a c + b a + c b = = a b + a c + a a + b c + b a + b b + c a + c b + c c −1 ≥ ≥3 a 3 √ abc + b 3 √ abc + c 3 √ abc −1 = = 2 a + b + c 3 √ abc + a + b + c 3 √ abc −1 ≥ ≥2 a + b + c 3 √ abc + 3 −1 = = 2 1 + a + b + c 3 √ abc by two applications of AM-GM inequality.
□ 75. (Poland 1998) a + b + c + d + e + f = 1, ace + bd f ≥ 1 108 a, b, c, d, e, f > 0 abc + bcd + cde + def + efa + fab ≤1 36 Solution. (Manlio - ML Forum) Put A = ace + bd f and B = abc + bcd + cde + def + efa + fab.By AM-GM inequality we have A + B = (a + d)(b + e)(c + f) ≤(((a + d) + (b + e) + (c + f))/3)3 = 1/27 so B ≤1/27 −A ≤1/27 −1/108 = 1/36 □ Remark. (Arqady) This is a private case of Walther Janous’s inequality: If x1 + x2 + ... + xn = 1 where xi are non-negative real numbers and 2 ≤k < n, k ∈N, then x1x2...xk + x2x3...xk+1 + ... + xnx1...xk−1 ≤max{ 1 kk , 1 nk−1 } 73 76. (Korea 1998) (x + y + z = xyz, x, y, z > 0) 1 √ 1 + x2 + 1 p 1 + y2 + 1 √ 1 + z2 ≤3 2 First Solution. (See , pag. 14) We can write x = tan A, y = tan B, z = tan C, where A, B, C ∈ 0, π 2 . Using the fact that 1 + tan2 θ = 1 cos θ 2, we rewrite it in the terms of A, B, C : cos A + cos B + cos C ≤3 2 (⋆) It follows from tan(π −C) = −z = x+y 1−xy = tan(A+B) and from π −C, A+B ∈ (0, π) that π −C = A + B or A + B + C = π.
Since cos x is concave on 0, π 2 , (⋆) a direct consequence of Jensen’s inequality and we are done.
□ Second Solution. (See , pag. 17) The starting point is letting a = 1 x, b = 1 y, c = 1 z. We find that a + b + c = abc is equivalent to 1 = xy + yz + zx. The inequality becomes x √ x2 + 1 + y p y2 + 1 + z √ z2 + 1 ≤3 2 or x p x2 + xy + yz + zx + y p y2 + xy + yz + zx + z p z2 + xy + yz + zx ≤3 2 or x p (x + y)(x + z) + y p (y + z)(y + x) + z p (z + x)(z + y) ≤3 2.
By the AM-GM inequality, we have x p (x + y)(x + z) = x p (x + y)(x + z) (x + y)(x + z) ≤ ≤1 2 x[(x + y) + (x + z)] (x + y)(x + z) = = 1 2 x x + z + x x + z In a like manner, we obtain y p (y + z)(y + x) ≤1 2 y y + z + y y + x 74 and z p (z + x)(z + y) ≤1 2 z z + x + z z + y Adding these three yields the required result.
□ 77. (Hong Kong 1998) (a, b, c ≥1) √ a −1 + √ b −1 + √ c −1 ≤ p c(ab + 1) First Solution. (Ercole Suppa) After setting x = √a −1, y = √ b −1, z = √c −1, with x, y, z ≥0, by easy calculations the required inequality in trans-formed in x + y + z ≤ p (1 + z2) [(1 + x2) (1 + y2) + 1] ⇐ ⇒ (x + y + z)2 ≤ 1 + z2 x2y2 + x2 + y2 + 2 ⇐ ⇒ x2y2 + x2 + y2 + 1 z2 −2(x + y)z + x2y2 −2xy + 2 ≥0 (⋆) The (⋆) is true for all x, y, z ∈R because: ∆ 4 = (x + y)2 − x2y2 + x2 + y2 + 1 x2y2 −2xy + 2 = = (x + y)2 − x2y2 + (x + y)2 −2xy + 1 (xy −1)2 + 1 = = (x + y)2 − (xy −1)2 + (x + y)2 (xy −1)2 + 1 = = −(xy −1)4 −(xy −1)2 −(x + y)2(xy −1)2 = = −(xy −1)2 2 + x2 + y2 + x2y2 ≤0 □ Second Solution. (Sung-Yoon Kim - ML Forum) Use √ x −1 + p y −1 ≤√xy ⇐ ⇒ 2 p (x −1)(y −1) ≤(x −1)(y −1) + 1 Then √ a −1 + √ b −1 + √ c −1 ≤ √ ab + √ c −1 ≤ p c(ab + 1) □ Remark. The inequality used in the second solution can be generalized in the following way (see , pag. 183, n.51): given threee real positive numbers a, b, c con a > c, b > c we have p c(a −c) + p c(b −c) ≤ √ ab The inequality, squaring twice, is transformed in (ab −ac −bc)2 ≥0.
The equality holds if c = ab/(a + b).
75 78. (IMO Short List 1998) (xyz = 1, x, y, z > 0) x3 (1 + y)(1 + z) + y3 (1 + z)(1 + x) + z3 (1 + x)(1 + y) ≥3 4 First Solution. (IMO Short List Project Group - ML Forum) The inequality is equivalent to the following one: x4 + x3 + y4 + y3 + z4 + z3 ≥3 4(x + 1)(y + 1)(z + 1).
In fact, a stronger inequality holds true, namely x4 + x3 + y4 + y3 + z4 + z3 ≥1 4[(x + 1)3 + (y + 1)3 + (z + 1)3].
(It is indeed stronger, since u3 + v3 + w3 ≥3uvw for any positive numbers u, v and w.) To represent the difference between the left- and the right-hand sides, put f(t) = t4 + t3 −1 4(t + 1)3, g(t) = (t + 1)(4t2 + 3t + 1).
We have f(t) = 1 4(t −1)g(t). Also, g is a strictly increasing function on (0, ∞), taking on positive values for t > 0. Since x4 + x3 + y4 + y3 + z4 + z3 −1 4[(x + 1)3 + (y + 1)3 + (z + 1)3] =f(x) + f(y) + f(z) =1 4(x −1)g(x) + 1 4(y −1)g(y) + 1 4(z −1)g(z), it suffices to show that the last expression is nonnegative.
Assume that x ≥y ≥z; then g(x) ≥g(y) ≥g(z) > 0. Since xyz = 1, we have x ≥1 and z ≤1. Hence (x−1)g(x) ≥(x−1)g(y) and (z −1)g(y) ≤(z −1)g(z).
So, 1 4(x −1)g(x) + 1 4(y −1)g(y) + 1 4(z −1)g(z) ≥1 4[(x −1) + (y −1) + (z −1)]g(y) =1 4(x + y + z −3)g(y) ≥1 4(3 3 √xyz −3)g(y) = 0, because xyz = 1. This completes the proof. Clearly, the equality occurs if and only if x = y = z = 1.
□ 76 Second Solution. (IMO Short List Project Group - ML Forum) Assume x ≤y ≤z so that 1 (1 + y)(1 + z) ≤ 1 (1 + z)(1 + x) ≤ 1 (1 + x)(1 + y).
Then Chebyshev’s inequality gives that x3 (1 + y)(1 + z) + y3 (1 + z)(1 + x) + z3 (1 + x)(1 + y) ≥1 3(x3 + y3 + z3) 1 (1 + y)(1 + z) + 1 (1 + z)(1 + x) + 1 (1 + x)(1 + y) = 1 3(x3 + y3 + z3) 3 + (x + y + z) (1 + x)(1 + y)(1 + z).
Now, setting (x+y+z)/3 = a for convenience, we have by the AM-GM inequality 1 3(x3 + y3 + z3) ≥a3, x + y + z ≥3 3 √xyz = 3, (1 + x)(1 + y)(1 + z) ≤ (1 + x) + (1 + y) + (1 + z) 3 3 = (1 + a)3.
It follows that x3 (1 + y)(1 + z) + y3 (1 + z)(1 + x) + z3 (1 + x)(1 + y) ≥a3 · 3 + 3 (1 + a)3 .
So, it suffices to show that 6a3 (1 + a)3 ≥3 4; or, 8a3 ≥(1 + a)3. This is true, because a ≥1. Clearly, the equality occurs if and only if x = y = z = 1. The proof is complete.
□ Third Solution. (Grobber - ML Forum) Amplify the first, second and third fraction by x, y, z respectively. The LHS becomes X x4 x(1 + y)(1 + z) ≥ x2 + y2 + z22 x + y + z + 2(xy + yz + zx) + 3 ≥ x2 + y2 + z22 4 (x2 + y2 + z2) ≥3 4 I used the inequalities x2 + y2 + z2 ≥xy + yz + zx x2 + y2 + z2 ≥3 x2 + y2 + z2 ≥(x + y + z)2 3 ≥x + y + z □ 77 Fourth Solution. (MysticTerminator - ML Forum) First, note that X cyc x3 (1 + y)(1 + z) ≥ x2 + y2 + z22 x(1 + y)(1 + z) + (1 + x)y(1 + z) + (1 + x)(1 + y)z by Cauchy, so we need to prove: 4 x2 + y2 + z22 ≥3 (x(1 + y)(1 + z) + (1 + x)y(1 + z) + (1 + x)(1 + y)z) Well, let x = a3, y = b3, z = c3 (with abc = 1), and homogenize it to find that we have to prove: X cyc 4a12 + 8a6b6 ≥ X cyc 3a6b6c3 + 6a5b5c2 + 3a4b4c4 which is perfectly Muirhead.
□ Remark. None of the solutions 1 and 2 above actually uses the condition xyz = 1. They both work, provided that x + y + z ≥3. Moreover, the alternative solution also shows that the inequality still holds if the exponent 3 is replaced by any number greater than or equal to 3.
79. (Belarus 1997) (a, x, y, z > 0) a + y a + z x + a + z a + xy + a + x a + y z ≥x + y + z ≥a + z a + z x + a + x a + y y + a + y a + z z First Solution. (Soarer - ML Forum) First one X x a + z a + x = X a + z −a X a + z a + x = = x + y + z −a X a + z a + x −3 ≤ ≤x + y + z Second one X xa + y a + z ≥x + y + z ⇔ X y −z a + z x ≥0 ⇔ X xy a + z ≥ X xz a + z ⇔ X 1 z(a + z) ≥ X 1 y (a + z) which is rearrangement.
□ 78 Second Solution. (Darij Grinberg- ML Forum) Let’s start with the first inequality: a + z a + xx + a + x a + y y + a + y a + z z ≤x + y + z It is clearly equivalent to a + z a + xx + a + x a + y y + a + y a + z z −(x + y + z) ≤0 But a + z a + xx + a + x a + y y + a + y a + z z −(x + y + z) = = a + z a + x −1 x + a + x a + y −1 y + a + y a + z −1 z = = z −x a + xx + x −y a + y y + y −z a + z z = (z −x) x a + x + (x −y) y a + y + (y −z) z a + z = = z x a + x −x x a + x + x y a + y −y y a + y + y z a + z −z z a + z = = z x a + x + x y a + y + y z a + z − x x a + x + y y a + y + z z a + z thus, it is enough to prove the inequality z x a + x + x y a + y + y z a + z − x x a + x + y y a + y + z z a + z ≤0 This inequality is clearly equivalent to z x a + x + x y a + y + y z a + z ≤x x a + x + y y a + y + z z a + z And this follows from the rearrangement inequality, applied to the equally sorted number arrays (x; y; z) and x a+x; y a+y; z a+z (proving that these arrays are equally sorted is very easy: if, for instance, x ≤y, then a x ≥ a y, so that a+x x = a x + 1 ≥a y + 1 = a+y y , so that x a+x ≤ y a+y).
Now we will show the second inequality: x + y + z ≤a + y a + z x + a + z a + xy + a + x a + y z It is equivalent to 0 ≤ a + y a + z x + a + z a + xy + a + x a + y z −(x + y + z) 79 Since a + y a + z x + a + z a + xy + a + x a + y z −(x + y + z) = = a + y a + z −1 x + a + z a + x −1 y + a + x a + y −1 z = = y −z a + z x + z −x a + xy + x −y a + y z = (xy −zx) 1 a + z + (yz −xy) 1 a + x + (zx −yz) 1 a + y = = xy 1 a + z −zx 1 a + z + yz 1 a + x −xy 1 a + x + zx 1 a + y −yz 1 a + y = = xy 1 a + z + yz 1 a + x + zx 1 a + y − zx 1 a + z + xy 1 a + x + yz 1 a + y thus, it is enough to verify the inequality 0 ≤ xy 1 a + z + yz 1 a + x + zx 1 a + y − zx 1 a + z + xy 1 a + x + yz 1 a + y This inequality is equivalent to zx 1 a + z + xy 1 a + x + yz 1 a + y ≤xy 1 a + z + yz 1 a + x + zx 1 a + y But this follows from the rearrangement inequality, applied to the equally sorted number arrays (yz; zx; xy) and 1 a+x; 1 a+y; 1 a+z (proving that these arrays are equally sorted is almost trivial: if, for instance, x ≤y, then y ≥x and yz ≥zx, while on the other hand a + x ≤a + y and thus 1 a+x ≥ 1 a+y).
This completes the proof of your two inequalities.
□ 80. (Ireland 1997) (a + b + c ≥abc, a, b, c ≥0) a2 + b2 + c2 ≥abc Solution. (Ercole Suppa) See problem 51.
□ 81. (Iran 1997) (x1x2x3x4 = 1, x1, x2, x3, x4 > 0) x3 1 + x3 2 + x3 3 + x3 4 ≥max x1 + x2 + x3 + x4, 1 x1 + 1 x2 + 1 x3 + 1 x4 80 Solution. (See pag. 69) After setting A = 4 P i=1 x3 i , Ai = A −x3 i , from AM-GM inequality we have 1 3A1 ≥ 3 q x3 2x3 3x3 4 = x2x3x4 = 1 x1 Similarly can be proved that 1 3Ai ≥ 1 xi for all i = 2, 3, 4. Therefore A = 1 3 4 X i=1 Ai ≥ 4 X i=1 1 xi On the other hand by Power Mean inequality we have 1 4A = 1 4 4 X i=1 x3 i ≥ 1 4 4 X i=1 xi !3 = = 1 4 4 X i=1 xi ! 1 4 4 X i=1 xi !2 ≥ ≥ 1 4 4 X i=1 xi !
(in the last step we used AM-GM inequality: 4 P i=1 xi ≥ 4 √x1x2x3x4 = 1). Thus A ≥ 4 X i=1 xi and the inequality is proven.
□ 82. (Hong Kong 1997) (x, y, z > 0) 3 + √ 3 9 ≥xyz(x + y + z + p x2 + y2 + z2) (x2 + y2 + z2)(xy + yz + zx) Solution. (Ercole Suppa) From QM-AM-GM inequality we have x + y + z ≥ √ 3 p x2 + y2 + z2 (1) xy + yz + zx ≥3 3 p (xyz)2 (2) p x2 + y2 + z2 ≥ √ 3 3 √xyz (3) 81 Therefore xyz(x + y + z + p x2 + y2 + z2) (x2 + y2 + z2)(xy + yz + zx) ≤xyz p x2 + y2 + z2 √ 3 + 1 (x2 + y2 + z2)3 3 p (xyz)2 ≤ ≤ xyz √ 3 + 1 3 p x2 + y2 + z2 3 p (xyz)2 ≤ ≤ xyz √ 3 + 1 3 √ 3 3 √xyz 3 p (xyz)2 = = √ 3 + 1 3 √ 3 = 3 + √ 3 9 □ Remark. See: Crux Mathematicorum 1988, pag. 203, problem 1067.
83. (Belarus 1997) (a, b, c > 0) a b + b c + c a ≥a + b c + a + b + c a + b + c + a b + c First Solution. (Ghang Hwan - ML Forum, Siutz - ML Contest 1st Ed. 1R) The inequality is equivalent with 1 + b a 1 + c a + 1 + c b 1 + a b + 1 + a c 1 + b c ≤a b + b c + c a Let x = a/b, y = c/a, z = b/c and note that xyz = 1. After some boring calculation we see that the inequality become x2 + y2 + z2 −x −y −z + x2z + y2x + z2y −3 ≥0 This inequality is true. In fact the first and second term are not negative because x2 + y2 + z2 ≥(x + y + z)x + y + z 3 ≥x + y + z (by CS and AM-GM) and x2z + y2x + z2y ≥3 3 p x3y3z3 = 3 (by AM-GM) □ 82 Second Solution. (See , pag. 43, problem 29) Let us take x = a/b, y = c/a, z = b/c and note that xyz = 1. Observe that a + c b + c = 1 + xy 1 + y = x + 1 −x 1 + y Using similar relations, the problem reduces to proving that if xyz = 1, then x −1 y + 1 y −1 z + 1 + z −1 x + 1 ≥0 ⇐ ⇒ x2 −1 (z + 1) + y2 −1 (x + 1) + z2 −1 (y + 1) ≥0 ⇐ ⇒ X x2z + X x2 ≥ X x + 3 But this inequality is very easy. Indeed, using the AM-GM inequality we have P x2z ≥3 and so it remains to prove that P x2 ≥P x, which follows from the inequalities X x2 ≥(P x)2 3 X x □ Third Solution. (Darij Grinberg, ML Forum) We first prove a lemma: Lemma. Let a, b, c be three reals; let x, y, z, u, v, w be six nonnegative reals.
Assume that the number arrays (a; b; c) and (x; y; z) are equally sorted, and u (a −b) + v (b −c) + w (c −a) ≥0 Then, xu (a −b) + yv (b −c) + zw (c −a) ≥0 Proof Since the statement of Lemma is invariant under cyclic permutations (of course, when these are performed for the number arrays (a; b; c), (x; y; z) and (u; v; w) simultaneously), we can WLOG assume that b is the ”medium one” among the numbers a, b, c; in other words, we have either a ≥b ≥c, or a ≤b ≤c. Then, since the number arrays (a; b; c) and (x; y; z) are equally sorted, we get either x ≥y ≥z, or x ≤y ≤z, respectively. What is important is that (x −z) (a −b) ≥0 (since the numbers x−z and a−b have the same sign: either both ≥0, or both ≤0), and that (y −z) (b −c) ≥0 (since the numbers y −z and b −c have the same sign: either both ≥0, or both ≤0). Now, xu (a −b) + yv (b −c) + zw (c −a) = = u (x −z) (a −b) | {z } ≥0 +v (y −z) (b −c) | {z } ≥0 +z (u (a −b) + v (b −c) + w (c −a)) | {z } ≥0 ≥0 83 and the Lemma is proven. □ Proof of inequality. The inequality a b + b c + c a ≥a + b c + a + b + c a + b + c + a b + c can be written as X c + a c + b ≤ X a b ⇐ ⇒ X a b − X c + a c + b ≥0 But X a b − X c + a c + b = X a b −c + a c + b = X 1 b + c · c b · (a −b) So it remains to prove that X 1 b + c · c b · (a −b) ≥0 In fact, denote u = c/b; v = a/c; w = b/a. Then, X u (a −b) = X c b (a −b) = X ca b −c = X ca b − X c = = P c2a2 −P c2ab abc = 1 2 P (ca −ab)2 abc ≥0 Now, denote x = 1 b + c ; y = 1 c + a ; z = 1 a + b Then, the number arrays (a; b; c) and (x; y; z) are equally sorted (in fact, e. g., if a ≥b, then c + a ≥b + c, so that 1 b+c ≥ 1 c+a, or, equivalently, x ≥y); thus, according to the Lemma, the inequality X u (a −b) ≥0 implies X xu (a −b) ≥0 In other words, P 1 b+c · c b · (a −b) ≥0. And the problem is solved.
□ 84. (Bulgaria 1997) (abc = 1, a, b, c > 0) 1 1 + a + b + 1 1 + b + c + 1 1 + c + a ≤ 1 2 + a + 1 2 + b + 1 2 + c 84 Solution. (Official solution) Let x = a + b + c and y = ab + bc + ca. It follows from CS inequality that x ≥3 and y ≥3. Since both sides of the given inequality are symmetric functions of a, b and c, we transform the expression as a function of x, y. Taking into account that abc = 1, after simple calculations we get 3 + 4x + y + x2 2x + y + x2 + xy ≤12 + 4x + y 9 + 4x + 2y which is equivalent to 3x2y + xy2 + 6xy −5x2 −y2 −24x −3y −27 ≥0 Write the last inequality in the form 5 3x2y −5x2 + xy2 3 −y2 + xy2 3 −3y + 4 3x2y −12x + + xy3 3 −3x + (3xy −9x) + (3xy −27) ≥0 When x ≥3, y ≥3, all terms in the left hand side are nonnegative and the inequality is true. Equality holds when x = 3, y = 3, which implies a = b = c = 1.
□ Remark. 1 The inequality can be proved by the general result: if Q xi = 1 then X 1 n −1 + xi ≤1 Proof. f(x1, x2, ..., xn) = P 1 n−1+xi .As Q xi = 1 we may assume x1 ≥1, x2 ≤ 1. We shall prove that f(x1, x2, ..., xn) ≤f(1, x1x2, ..., xn). And this is true because after a little computation we obtain (1−x1)(x2−1)(x1x2+(n−1)2) ≥0 which is obviously true. So we have f(x1, x2, ..., xn) ≤f(1, x1x2, ..., xn) ≤... ≤ f(1, ..., 1) = 1.
□ Remark 2. (Darij Grinberg) I want to mention the appearance of the inequality with solution in two sources: 1. Titu Andreescu, Vasile Cˆ ırtoaje, Gabriel Dospinescu, Mircea Lascu, Old and New Inequalities, Zalau: GIL 2004, problem 99.
2. American Mathematics Competitions: Mathematical Olympiads 1997-1998: Olympiad Problems from Around the World, Bulgaria 21, p. 23.
Both solutions are almost the same: Brute force. The inequality doesn’t seem to have a better proof.
85 85. (Romania 1997) (xyz = 1, x, y, z > 0) x9 + y9 x6 + x3y3 + y6 + y9 + z9 y6 + y3z3 + z6 + z9 + x9 z6 + z3x3 + x6 ≥2 Solution. (Ercole Suppa) By setting a = x3, b = y3, c = z3 we have abc = 1.
From the know inequality a3 + b3 ≥ab(a + b) follows that a3 + b3 a2 + ab + b2 = a3 + b3 + 2 a3 + b3 3 (a2 + ab + b2) ≥ ≥a3 + b3 + 2ab(a + b) 3 (a2 + ab + b2) = = (a + b) a2 + ab + b2 3 (a2 + ab + b2) = = a + b 3 Similarly can be proved the following inequalities: b3 + c3 b2 + bc + c2 ≥b + c 3 ; c3 + a3 c2 + ca + a2 ≥c + a 3 Then, by AM-GM inequality we have: x9 + y9 x6 + x3y3 + y6 + y9 + z9 y6 + y3z3 + z6 + z9 + x9 z6 + z3x3 + x6 = = a3 + b3 a2 + ab + b2 + b3 + c3 b2 + bc + c2 + c3 + a3 c2 + ca + a2 = = a + b 3 + b + c 3 + c + a 3 = = 2(a + b + c) 3 ≥ ≥2 · 3 3 √ abc 3 = 2 □ 86. (Romania 1997) (a, b, c > 0) a2 a2 + 2bc + b2 b2 + 2ca + c2 c2 + 2ab ≥1 ≥ bc a2 + 2bc + ca b2 + 2ca + ab c2 + 2ab 86 Solution. (Pipi - ML Forum) Let I = a2 a2 + 2bc + b2 b2 + 2ca + c2 c2 + 2ab , J = bc a2 + 2bc + ca b2 + 2ca + ab c2 + 2ab We wish to show that I ≥1 ≥J. Since x2 + y2 ≥2xy we have a2 a2 + 2bc ≥ a2 a2 + b2 + c2 Similarly, b2 b2 + 2ca ≥ b2 a2 + b2 + c2 , c2 c2 + 2ab ≥ c2 a2 + b2 + c2 Then it is clear that I ≥1. Next, note that I +2J = 3 or I = 3−2J. By I ≥1, it is easy to see that J ≤1.
□ 87. (USA 1997) (a, b, c > 0) 1 a3 + b3 + abc + 1 b3 + c3 + abc + 1 c3 + a3 + abc ≤ 1 abc.
Solution. (ML Forum) By Muirhead (or by factoring) we have a3 + b3 ≥ab2 + a2b so we get that: X cyc abc a3 + b3 + abc ≤ X cyc abc ab(a + b + c) = X cyc c a + b + c = 1 □ 88. (Japan 1997) (a, b, c > 0) (b + c −a)2 (b + c)2 + a2 + (c + a −b)2 (c + a)2 + b2 + (a + b −c)2 (a + b)2 + c2 ≥3 5 Solution. (See ) WLOG we can assume that a+b+c = 1. Then the first term on the left become (1 −2a)2 (1 −a)2 + a2 = 2 − 2 1 + (1 −2a)2 87 Next, let x1 = 1 −2a, x2 = 1 −2b, x3 = 1 −2c, then x1 + x2 + x3 = 1, but −1 < x1, x2, x3 < 1. In terms of x1, x2, x3, the desidered inequality is 1 1 + x2 1 + 1 1 + x2 2 + 1 1 + x2 3 ≤27 10 We consider the equation of the tangent line to f(x) = 1 1+x2 at x = 1/3 which is y = 27 50(−x + 2). We have f(x) ≤27 50(−x + 2) for −1 < x < 1 because 27 50(−x + 2) − 1 1 + x2 = (3x −1)2(4 −3x) 50(x2 + 1) ≥0 Then f(x1) + f(x2) + f(x3) ≤27 10 and the desidered inequality follows.
□ 89. (Estonia 1997) (x, y ∈R) x2 + y2 + 1 > x p y2 + 1 + y p x2 + 1 Solution. (Ercole Suppa) We have: x − p y2 + 1 + y − p x2 + 1 ≥0 and, consequently, x2 + y2 + 1 ≥x p y2 + 1 + y p x2 + 1 The equality holds if and only if x = p y2 + 1 and y = √ x2 + 1, i.e.
x2 + y2 = x2 + y2 + 2 Since this last equality is impossible, the result is proven.
□ 90. (APMC 1996) (x + y + z + t = 0, x2 + y2 + z2 + t2 = 1, x, y, z, t ∈R) −1 ≤xy + yz + zt + tx ≤0 88 Solution. (Ercole Suppa) After setting A = xy + yz + zt + tx we have 0 = (x + y + z + t)2 = 1 + 2A + 2(xz + yt) = ⇒ A = −1 2 −xz −yt The required inequality is equivalent to −1 ≤−1 2 −xz −yt ≤0 ⇐ ⇒ −1 2 ≤xz + yt ≤1 2 ⇐ ⇒ |xz + yt| ≤1 2 and can be proved by means of Cauchy-Schwarz and AM-GM inequalities |xz + yt| ≤ p x2 + y2 · p t2 + z2 = (CS) = p (x2 + y2) (t2 + z2) ≤ (AM-GM) ≤x2 + y2 + t2 + z2 2 = 1 2 □ 91. (Spain 1996) (a, b, c > 0) a2 + b2 + c2 −ab −bc −ca ≥3(a −b)(b −c) Solution. (Ercole Suppa) We have: a2 + b2 + c2 −ab −bc −ca −3(a −b)(b −c) = a2 + 4b2 + c2 −4ab −4bc + 2ac = = (a −2b + c)2 ≥0 □ 92. (IMO Short List 1996) (abc = 1, a, b, c > 0) ab a5 + b5 + ab + bc b5 + c5 + bc + ca c5 + a5 + ca ≤1 Solution. (by IMO Shortlist Project Group - ML Forum) We have a5 + b5 = (a + b) a4 −a3b + a2b2 −ab3 + b4 = = (a + b) (a −b)2 a2 + ab + b2 + a2b2 ≥ ≥a2b2(a + b) 89 with equality if and only if a = b. Hence ab a5 + b5 + ab ≤ ab ab(a + b) + 1 = = 1 ab(a + b + c) = = c a + b + c Taking into account the other two analogous inequalities we have X ab a5 + b5 + ab ≤ c a + b + c + a a + b + c + b a + b + c = 1 and the required inequality is established. Equality holds if and only if a = b = c = 1.
□ 93. (Poland 1996) a + b + c = 1, a, b, c ≥−3 4 a a2 + 1 + b b2 + 1 + c c2 + 1 ≤9 10 Solution. (Ercole Suppa) The equality holds if a = b = c = 1/3. The line tangent to the graph of f(x) = x x2+1 in the point with abscissa x = 1/3 has equation y = 18 25x + 30 50 and the graph of f(x), per x > −3/4, if entirely below that line, i.e.
x x2 + 1 ≤18 25x + 30 50 , ∀x > −3 4 because 18 25x + 30 50 − x x2 + 1 = (3x −1)2(4x + 3) ≥0 , ∀x > −3 4 Therefore X cyc a a2 + 1 ≤f(a) + f(b) + f(c) = 9 10 □ Remark. It is possible to show that the inequality X cyc a a2 + 1 ≤9 10 is true for all a, b, c ∈R such that a + b + c = 1. See ML Forum.
90 94. (Hungary 1996) (a + b = 1, a, b > 0) a2 a + 1 + b2 b + 1 ≥1 3 Solution. (See , pag. 30) Using the condition a + b = 1, we can reduce the given inequality to homogeneous one, i. e., 1 3 ≤ a2 (a + b)(a + (a + b)) + b2 (a + b)(b + (a + b)) or a2b + ab2 ≤a3 + b3, which follows from (a3 + b3) −(a2b + ab2) = (a −b)2(a + b) ≥0. The equality holds if and only if a = b = 1 2.
□ 95. (Vietnam 1996) (a, b, c ∈R) (a + b)4 + (b + c)4 + (c + a)4 ≥4 7 a4 + b4 + c4 First Solution. (Namdung - ML Forum) Let f(a, b, c) = (a + b)4 + (b + c)4 + (c + a)4 −4 7 a4 + b4 + c4 .
We will show that f(a, b, c) ≥0 for all a, b, c. Among a, b, c, there exist at least one number which has the same sign as a + b + c, say a. By long, but easy computation, we have f(a, b, c)−f(a, b + c 2 , b + c 2 ) = 3a(a+b+c)(b−c)2+ 3 56 7b2 + 10bc + 7c2 (b−c)2 ≥0 So, it sufficient (and necessary) to show that f(a, t, t) >= 0 for all a, t.
Is equivalent to f(0, t, t) ≥0 and f(1, t, t) ≥0 (due homogeneousness). The first is trivial, the second because f(1, t, t) = 59t4 + 28t3 + 42t2 + 28t + 5 = = 6 59(20t + 7)2 + √ 59t2 + 14t √ 59 − 1 √ 59 2 > 0 □ 91 Remark. To find the identity f(a, b, c)−f(a, b + c 2 , b + c 2 ) = 3a(a+b+c)(b−c)2+ 3 56 7b2 + 10bc + 7c2 (b−c)2 ≥0 we can use the following well-known approach. Let h(a, b, c) = f(a, b, c) −f(a, b + c 2 , b + c 2 ) ≥0 The first thing we must have is h(0, b, c) ≥0. h(0, b, c) is symmetric homogenus polynomial of b, c and it’s easily to find that h(0, b, c) = 3 56 7b2 + 10bc + 7c2 (b −c)2 Now, take h(a, b, c) −h(0, b, c) and factor, we will get h(a, b, c) −h(0, b, c) = 3a(a + b + c)(b −c)2 Second Solution. (Iandrei - ML Forum) Let f(a, b, c) = (a + b)4 + (b + c)4 + (c + a)4 −4 7 a4 + b4 + c4 . It’s clear that f(0, 0, 0) = 0. We prove that f(a, b, c) ≥0. We have f(a, b, c) = 10 7 X a4 + 4 X ab a2 + b2 + 6 X a2b2 ≥0 ⇐ ⇒ 5 7 X a4 + 2 X ab a2 + b2 + 3 X a2b2 ≥0 ⇐ ⇒ 5 X a4 + 14 X ab a2 + b2 + 21 X a2b2 ≥0 We prove that 5 2 a4 + b4 + 14ab a2 + b2 + 21a2b2 ≥0 (⋆) Let x = ab , y = a2 + b2. Thus 5 2 y2 −2x2 + 14xy + 21x2 ≥0 ⇐ ⇒ 16x2 + 14xy + 5 2y2 ≥0 If x ̸= 0, we want prove that 32 + 28y x + 5 y x 2 ≥0.
If y/x = t with |t| > 2 , we must prove 32 + 28t + 5t2 ≥0. The latter second degree function has roots r1 = −28 + 12 10 = −1, 6 , r2 = −28 −12 10 = −4.
It’s clear that |t| > 2 implies 32 + 28t + 5t2 ≥0. If x = 0 then a = 0 or b = 0 and (⋆) is obviously verified.
□ 92 Remark. A different solution is given in , pag. 92, problem 98.
96. (Belarus 1996) (x + y + z = √xyz, x, y, z > 0) xy + yz + zx ≥9(x + y + z) First Solution. (Ercole Suppa) From the well-know inwquality (xy + yz + zx)3 ≥3xyz(x + y + z) and AM-GM inequality we have (xy + yz + zx)3 ≥3xyz(x + y + z) = = 3(x + y + z)3 ≥ (AM-GM) ≥3 (3 3 √xyz)3 = = 81xyz = 81(x + y + z)2 The required inequality follows extracting the square root.
□ Second Solution. (Cezar Lupu - ML Forum) We know that x + y + z = √xyz or (x + y + z)2 = xyz. The inequality is equivalent with this one: xyz( 1 x + 1 y + 1 z ) ≥9(x + y + z), or (x + y + z)2( 1 x + 1 y + 1 z ) ≥9(x + y + z).
Finally, our inequality is equivalent with this well-known one: (x + y + z)( 1 x + 1 y + 1 z ) ≥9.
□ 97. (Iran 1996) (a, b, c > 0) (ab + bc + ca) 1 (a + b)2 + 1 (b + c)2 + 1 (c + a)2 ≥9 4 93 First Solution. (Iurie Boreico, see , pag. 108, problem 114) With the sub-stitution y + z = a, z + x = b, x + y = c the inequality becomes after some easy computations X 2 ab −1 c2 (a −b)2 Assume LOG that a ≥b ≥c. If 2c2 ≥ab, each term in the above expression is positive and we are done. So, suppose 2c2 < ab. First, we prove that 2b2 ≥ac, 2a2 ≥bc. Suppose that 2b2 < ac. Then (b + c)2 ≤2(b2 + c2) < a(b + c) and so b + c < a, false. Clearly, we can write the inequality like that 2 ac −1 b2 (a −c)2 + 2 bc −1 a2 (b −c)2 ≥ 1 c2 −2 ab (a −b)2 We can immediately see that the inequality (a −c)2 ≥(a −b)2 + (b −c)2 holds and thus it suffices to prove that 2 ac + 2 bc −1 a2 −1 b2 (b −c)2 ≥ 1 b2 + 1 c2 −2 ab −2 ac (a −b)2 But is clear that 1 b2 + 1 c2 −2 ab < 1 b −1 c 2 and so the right hand side is at most (a −b)2(b −c)2 b2c2 Also, it is easy to see that 2 ac + 2 bc −1 a2 −1 b2 ≥1 ac + 1 bc > (a −b)2 b2c2 which show that the left hand side is at least (a −b)2(b −c)2 b2c2 and this ends the solution.
□ Second Solution. (Cezar Lupu - ML Forum) We take x = p−a, y = p−b, z = p −c so the inequality becomes: (p −a)(p −b) + (p −b)(p −c) + (p −c)(p −a) · ( 1 a2 + 1 b2 + 1 c2 ) ≥9 4 ⇐ ⇒ (p2 −16Rr + 5r2) (4R + r)(p2 −16Rr + 5r2) + 4r (3R(5R −r) + r(R −2r)) + +4R3(R −2r)2 ≥0 But using Gerrestein’s inequality p2 ≥16Rr−5r2 and Euler’s inequality R ≥2r we are done. Hope I did not make any stupid mistakes in my calculations.
□ 94 Remark. Gerrestein’s inequality.
In the triangle ABC we have p2 + 5r2 ≥16Rr.
Put a = x + y, b = y + z, c = z + x, x, y, z > 0. The inital inequality becomes (x + y + z)3 ≥4(x + y)(y + z)(z + x) −5xyz This one is homogenous so consider x + y + z = 1. So we only must prove that 1 ≥4(1 −x)(1 −y)(1 −z) −5xyz ⇔1 + 9xyz ≥4(xy + yz + zx) which is true by Shur . Anyway this one is weak , it also follows from s2 ≥2R2 + 8Rr + 3r2 which is little bit stronger.
Third Solution. (payman pm - ML Forum) (xy + yz + zx)( 1 (x + y)2 + 1 (y + z)2 + 1 (z + x)2 = = (xy + yz + xz)((x + y)2(y + z)2 + (y + z)2(z + x)2 + (z + x)2(x + y)2 (x + y)2(y + z)2(z + x)2 but we have (xy + yz + zx)((x + y)2(y + z)2 + (y + z)2(z + x)2 + (z + x)2(x + y)2) = X (x5y + 2x4y2 + 5 2x4yz + 13x3y2z + 4x2y2z2) and (x + y)2(y + z)2(z + x)2 = X (x4y2 + x4yz + x3y3 + 6x3y2z + 5 3x2y2z2) and by some algebra X (4x5y −x4y2 −3x3y3 + x4yz −2x3y2z + x2y2z2) ≥0 and by using Sschur inequality we have P(x3 −2x2y +xyz) ≥0 and if multiply this inequality to xyz : X (x4yz −2x3y2z + x2y2z2) ≥0 (1) and by usingAM −GM inequality we have X ((x5y −x4y2) + 3(x5y −x3y3)) ≥0 (2) and by using (1), (2) the problem is solved.
□ 95 Fourth Solution. (Darij Grinberg - ML Forum) I have just found another proof of the inequality which seems to be a bit less ugly than the familiar ones.
We first prove a lemma: Lemma 1. If a, b, c, x, y, z are six nonnegative reals such that a ≥b ≥c and x ≤y ≤z, then x (b −c)2 3bc + ca + ab −a2 + y (c −a)2 3ca + ab + bc −b2 + + z (a −b)2 3ab + bc + ca −c2 ≥0.
Proof of Lemma 1. Since a ≥b, we have ab ≥b2, and since b ≥c, we have bc ≥c2. Thus, the terms 3ca + ab + bc −b2 and 3ab + bc + ca −c2 must be nonnegative. The important question is whether the term 3bc + ca + ab −a2 is nonnegative or not. If it is, then we have nothing to prove, since the whole sum x (b −c)2 3bc + ca + ab −a2 + y (c −a)2 3ca + ab + bc −b2 + + z (a −b)2 3ab + bc + ca −c2 is trivially nonnegative, as a sum of nonnegative expressions. So we will only consider the case when it is not; i. e., we will consider the case when 3bc + ca + ab −a2 < 0. Then, since (b −c)2 ≥0, we get (b −c)2 3bc + ca + ab −a2 ≤0, and this, together with x ≤y, implies that x (b −c)2 3bc + ca + ab −a2 ≥y (b −c)2 3bc + ca + ab −a2 On the other hand, since 3ab + bc + ca −c2 ≥0 and (a −b)2 ≥0, we have (a −b)2 3ab + bc + ca −c2 ≥0, which combined with y ≤z, yields z (a −b)2 3ab + bc + ca −c2 ≥y (a −b)2 3ab + bc + ca −c2 Hence, x (b −c)2 3bc + ca + ab −a2 +y (c −a)2 3ca + ab + bc −b2 +z (a −b)2 3ab + bc + ca −c2 ≥y (b −c)2 3bc + ca + ab −a2 +y (c −a)2 3ca + ab + bc −b2 +y (a −b)2 3ab + bc + ca −c2 = y (b −c)2 3bc + ca + ab −a2 + (c −a)2 3ca + ab + bc −b2 + (a −b)2 3ab + bc + ca −c2 But (b −c)2 3bc + ca + ab −a2 + (c −a)2 3ca + ab + bc −b2 + (a −b)2 3ab + bc + ca −c2 ≥(b −c)2 −bc + ca + ab −a2 + (c −a)2 −ca + ab + bc −b2 + (a −b)2 −ab + bc + ca −c2 (since squares of real numbers are always nonnegative) = (b −c)2 (c −a) (a −b) + (c −a)2 (a −b) (b −c) + (a −b)2 (b −c) (c −a) = (b −c) (c −a) (a −b) (b −c) + (c −a) + (a −b) | {z } =0 = 0 96 thus, x (b −c)2 3bc + ca + ab −a2 + y (c −a)2 3ca + ab + bc −b2 + + z (a −b)2 3ab + bc + ca −c2 ≥0 and Lemma 1 is proven.
Now to the proof of the Iran 1996 inequality: We first rewrite the inequality using the P notation as follows: X 1 (b + c)2 ≥ 9 4 (bc + ca + ab) Upon multiplication with 4(bc + ca + ab), this becomes X 4 (bc + ca + ab) (b + c)2 ≥9 Subtracting 9, we get X 4 (bc + ca + ab) (b + c)2 −9 ≥0 which is equivalent to X 4 (bc + ca + ab) (b + c)2 −3 !
≥0 But 4 (bc + ca + ab) (b + c)2 −3 = (3b + c) (a −b) (b + c)2 −(3c + b) (c −a) (b + c)2 Hence, it remains to prove X (3b + c) (a −b) (b + c)2 −(3c + b) (c −a) (b + c)2 !
≥0 97 But X (3b + c) (a −b) (b + c)2 −(3c + b) (c −a) (b + c)2 !
= = X (3b + c) (a −b) (b + c)2 − X (3c + b) (c −a) (b + c)2 = = X (3b + c) (a −b) (b + c)2 − X (3a + c) (a −b) (c + a)2 = = X (3b + c) (a −b) (b + c)2 −(3a + c) (a −b) (c + a)2 !
= = X (a −b)2 3ab + bc + ca −c2 (b + c)2 (c + a)2 Thus, the inequality in question is equivalent to X (a −b)2 3ab + bc + ca −c2 (b + c)2 (c + a)2 ≥0 Upon multiplication with (b + c)2 (c + a)2 (a + b)2, this becomes X (a + b)2 (a −b)2 3ab + bc + ca −c2 ≥0 In other words, we have to prove the inequality (b + c)2 (b −c)2 3bc + ca + ab −a2 + (c + a)2 (c −a)2 3ca + ab + bc −b2 + + (a + b)2 (a −b)2 3ab + bc + ca −c2 ≥0 But now it’s clear how we prove this - we WLOG assume that a ≥b ≥c, and define x = (b + c)2, y = (c + a)2 and z = (a + b)2; then, the required inequality follows from Lemma 1 after showing that x ≤y ≤z (what is really easy: since a ≥b ≥c, we have (a + b + c) −a ≤(a + b + c) −b ≤(a + b + c) −c, what rewrites as b + c ≤c + a ≤a + b, and thus (b + c)2 ≤(c + a)2 ≤(a + b)2, or, in other words, x ≤y ≤z).
This completes the proof of the Iran 1996 inequality. Feel free to comment or to look for mistakes (you know, chances are not too low that applying a new method one can make a number of mistakes).
□ Remark. For different solutions proof see: , pag.306; , pag.163; Crux Mathematicorum [1994:108; 1995:205; 1996:321; 1997:170,367].
98. (Vietnam 1996) (2(ab + ac + ad + bc + bd + cd) + abc + bcd + cda + dab = 16, a, b, c, d ≥0) a + b + c + d ≥2 3(ab + ac + ad + bc + bd + cd) 98 Solution. (Mohammed Aassila - Crux Mathematicorum 2000, pag.332) We first prove a lemma: Lemma If x, y, z are real positive numbers such that x + y + z + xyz = 4, then x + y + z ≥xy + yz + zx Proof. Suppose that x + y + z < xy + yz + zx. From Schur inequality we have 9xyz ≥4(x + y + z)(xy + yz + zx) −(x + y + z)3 ≥ ≥4(x + y + z)2 −(x + y + z)3 = = (x + y + z)2 [4 −(x + y + z)] = = xyz(x + y + z)2 Thus (x + y + z)2 < 9 = ⇒ x + y + z < 3 and AM-GM implies xyz < x + y + z 3 3 = 1 Hence x+y +z +xyz < 4 and this is impossible. Therefore we have x+y +z ≥ xy + yz + zx and the lemma is proved.
□ Now, the given inequality can be proved in the following way. Put S1 = X a , S2 = X ab , S3 = X abc.
Let P(x) = (x −a)(x −b)(x −c)(x −d) = x4 −S1x3 + S2x2 −S3x + abcd.
Rolle’s theorem says that P ′(x) has 3 positive roots u, v, w. Thus P ′(x) = 4(x−u)(x−v)(x−w) = 4x3−4(u+v+w)x2+4(uv+vw+wu)x−4uvw Since P ′(x) = 4x3 −3S1x2 + 2S2x −S3 we have: S1 = 4 3(u + v + w) , S2 = 2(uv + vw + wu) , S3 = 4uvw (1) From (1) we have 2(ab + ac + ad + bc + bd + cd) + abc + bcd + cda + dab = 16 ⇐ ⇒ 2S2 + S3 = 16 ⇐ ⇒ 4(uv + vw + wu) + 4uvw = 16 ⇐ ⇒ uv + vw + wu + uvw = 4 (2) From the Lemma and (1) follows that u + v + w ≥uv + vw + wu ⇐ ⇒ 3 4S1 ≥1 2S2 ⇐ ⇒ S1 ≥2 3S2 i.e. a + b + c + d ≥2 3(ab + ac + ad + bc + bd + cd) and we are done.
□ Remark. A different solution is given in pag. 98.
99 3 Years 1990 ∼1995 99. (Baltic Way 1995) (a, b, c, d > 0) a + c a + b + b + d b + c + c + a c + d + d + b d + a ≥4 Solution. (Ercole Suppa) From HM-AM inequality we have 1 x + 1 y ≥ 4 x + y ; ∀x, y ∈R+ 0 .
Therefore: LHS = (a + c) 1 a + b + 1 c + d + (b + d) 1 a + b + 1 c + d = = (a + b + c + d)(a + c) 1 a + b + 1 c + d ≥ (HM-AM) ≥(a + b + c + d) 4 a + b + c + d = 4 □ 100. (Canada 1995) (a, b, c > 0) aabbcc ≥abc a+b+c 3 First Solution. (Ercole Suppa) From Weighted AM-GM inequality applied to the numbers 1 a, 1 b, 1 c with weights p1 = a a+b+c, p2 = b a+b+c, p3 = c a+b+c we have p1 · 1 a + p2 · 1 b + p3 · 1 c ≥ 1 a p1 · 1 b p2 · 1 c p3 ⇐ ⇒ 3 a + b + c ≥ 1 a+b+c √ aabbcc Thus, the AM-GM inequality yields: a+b+c √ aabbcc ≥a + b + c 3 ≥ 3 √ abc = ⇒ aabbcc ≥abc a+b+c 3 □ 100 Second Solution. (See pag. 15) We can assume WLOG that a ≤b ≤c. Then log a ≤log b ≤log c and, from Chebyshev inequality we get a + b + c 3 · log a + log b + log c 3 ≤a log a + b log b + c log c 3 Therefore a log a + b log b + c log c ≥a + b + c 3 (log a + log b + log c) = ⇒ aabbcc ≥abc a+b+c 3 □ Third Solution. (Official solution - Crux Mathematicorum 1995, pag. 224) We prove equivalently that a3ab3bc3c ≥(abc)a+b+c Due to complete symmetry in a, b and c, we may assume WLOG that a ≥b ≥ c > 0. Then a −b ≥0, b −c ≥0, a −c ≥0 and a/b ≥1, b/c ≥1, a/c ≥1.
Therefore a3ab3bc3c (abc)a+b+c = a b a−b b c b−c a c a−c ≥1 □ 101. (IMO 1995, Nazar Agakhanov) (abc = 1, a, b, c > 0) 1 a3(b + c) + 1 b3(c + a) + 1 c3(a + b) ≥3 2 First Solution. (See , pag. 17) After the substitution a = 1 x, b = 1 y, c = 1 z, we get xyz = 1. The inequality takes the form x2 y + z + y2 z + x + z2 x + y ≥3 2.
It follows from the Cauchy-Schwarz inequality that [(y + z) + (z + x) + (x + y)] x2 y + z + y2 z + x + z2 x + y ≥(x + y + z)2 so that, by the AM-GM inequality, x2 y + z + y2 z + x + z2 x + y ≥x + y + z 2 ≥3(xyz) 1 3 2 = 3 2.
□ 101 Second Solution. (See , pag. 36) It’s equivalent to 1 a3(b + c) + 1 b3(c + a) + 1 c3(a + b) ≥ 3 2(abc)4/3 .
Set a = x3, b = y3, c = z3 with x, y, z > 0. Then, it becomes X cyc 1 x9(y3 + z3) ≥ 3 2x4y4z4 .
Clearing denominators, this becomes X sym x12y12 + 2 X sym x12y9z3 + X sym x9y9z6 ≥3 X sym x11y8z5 + 6x8y8z8 or X sym x12y12 − X sym x11y8z5 !
+ 2 X sym x12y9z3 − X sym x11y8z5 !
+ + X sym x9y9z6 − X sym x8y8z8 !
≥0 and every term on the left hand side is nonnegative by Muirhead’s theorem. □ 102. (Russia 1995) (x, y > 0) 1 xy ≥ x x4 + y2 + y y4 + x2 Solution. (Ercole Suppa) From AM-GM inequality we have: x x4 + y2 + y y4 + x2 ≤ x 2 p x4y2 + y 2 p x2y4 = = 1 2xy + 1 2xy = 1 xy □ 103. (Macedonia 1995) (a, b, c > 0) r a b + c + r b c + a + r c a + b ≥2 102 Solution. (Manlio Marangelli - ML Forum) After setting A = 1, B = a b+c the HM-AM yields √ AB ≥ 2 1 A + 1 B so r a b + c ≥ 2 1 + b+c a = 2a a + b + c Similar inequalities are true also for the other two radicals. Therefore r a b + c + r b c + a + r c a + b ≥ 2a a + b + c + 2b a + b + c + 2c a + b + c = 2 □ 104. (APMC 1995) (m, n ∈N, x, y > 0) (n−1)(m−1)(xn+m+yn+m)+(n+m−1)(xnym+xmyn) ≥nm(xn+m−1y+xyn+m−1) Solution. (See pag. 147) We rewrite the given inequality in the form mn(x −y) xn+m−1 + yn+m−1 ≥(n + m −1) (xn −yn) (xm −ym) and divide both sides by (x −y)2 to get the equivalent form nm xn+m−2 + xn+m−3y + · · · + yn+m−2 ≥ ≥(n + m −1) xn−1 + · · · + yn−1 xm−1 + · · · + ym−2 We now will prove a more general result. Suppose P(x, y) = adxd + · · · + a−dyd is a homogeneous polynomial of degree d with the following properties (a) For i = 1, . . . , d, ai = a−i (equivalently P(x, y) = P(y, x)) (b) d P i=−d ad = 0, (equivalently P(x, x) = 0) (c) For i = 0, . . . , d −1, ad + · · · + ad−i ≥0.
103 Then P(x, y) ≥0 for all x, y > 0. (The properties are easily verified for P(x, y) equal to the difference of the two sides in our desidered inequality. The third property follows from the fact that in this case, ad ≥ad−1 ≥· · · ≥a0). We prove the general result by induction on d, as it is obvious for d = 0. Suppose P has the desidered properties, and let Q(x, y) = (ad + ad−1) xd−1 + ad−2xd−2y + · · · + a−d+2xyd−2 + (a−d + a−d+1) yd−1.
Then Q has smaller degree and satisfies the required properties, so by the in-duction hypotesis Q(x, y) ≥0. Moreover, P(x, y) −Q(x, y) = ad xd −xd−1y −xyd−1 + yd = = ad(x −y) xd −yd ≥0 since ad ≥0 and the sign of x −y is the same as the sign of xd −yd. Adding these two inequalities give P(x, y) ≥0, as desidered.
□ 105. (Hong Kong 1994) (xy + yz + zx = 1, x, y, z > 0) x(1 −y2)(1 −z2) + y(1 −z2)(1 −x2) + z(1 −x2)(1 −y2) ≤4 √ 3 9 First Solution. (Grobber - ML Forum) What we must prove is x + y + z + xyz(xy + yz + zx) ≤4 √ 3 9 + xy(x + y) + yz(y + z) + zx(z + x).
By adding 3xyz to both sides we get (we can eliminate xy + yz + zx since it’s equal to 1) x + y + z + 4xyz ≤4 √ 3 9 + x + y + z.
By subtracting x + y + z from both sides and dividing by 4 we are left with xyz ≤ √ 3 9 , which is true by AM-GM applied to xy, yz, zx.
□ Second Solution. (Murray Klamkin - Crux Mathematicorum 1998, pag.394) We first convert the inequality to the following equivalent homogeneous one: x T2 −y2 T2 −z2 + yx T2 −z2 T2 −x2 + z T2 −x2 T2 −y2 ≤ ≤4 √ 3 9 (T2) 5 2 104 where T2 = xy + yz + zx, and for subsequent use T1 = x + y + z, T3 = xyz.
Expanding out, we get T1T 2 2 −T2 X x y2 + z2 + T2T3 ≤4 √ 3 9 (T2) 5 2 or T1T 2 2 −T2 (T1T2 −3T3) + T2T3 = 4T2T3 ≤4 √ 3 9 (T2) 5 2 Squaring, we get one of the know Maclaurin inequalities for symmetric functions: 3 p T3 ≤ 3 r T2 3 There is equality if and only if x = y = z.
□ 106. (IMO Short List 1993) (a, b, c, d > 0) a b + 2c + 3d + b c + 2d + 3a + c d + 2a + 3b + d a + 2b + 3c ≥2 3 Solution. (Massimo Gobbino - Winter Campus 2006) From Cauchy-Schwarz inequality we have: (a + b + c + d)2 = X cyc √a √ b + 2c + 3d √a √ b + 2c + 3d !2 ≤ ≤ X cyc a b + 2c + 3d ! X cyc ab + 2ac + 3ad !
= = X cyc a b + 2c + 3d ! X sym ab !
≤ ≤ X cyc a b + 2c + 3d !
3 2 (a + b + c + d)2 (1) The inequality of the last step can be proved by BUNCHING principle (Muir-head Theorem) in the following way: 105 X sym ab ≤3 2 (a + b + c + d)2 X sym ab ≤3 2 X cyc a2 + 3 4 X sym ab 1 4 X sym ab ≤3 2 X cyc a2 X sym ab ≤6 X cyc a2 X sym ab ≤ X sym a2 From (1) follows that X cyc a b + 2c + 3d ≥2 3 □ 107. (APMC 1993) (a, b ≥0) √a + √ b 2 !2 ≤a + 3 √ a2b + 3 √ ab2 + b 4 ≤a + √ ab + b 3 ≤ v u u t 3 √ a2 + 3 √ b2 2 !3 Solution. (Tsaossoglou - Crux Mathematicorum 1997, pag. 73) Let A = 6 √a, B = 6 √ b. The first inequality √a + √ b 2 !2 ≤a + 3 √ a2b + 3 √ ab2 + b 4 is equivalent to √a + √ b 2 ≤ 3 √ a2 + 3 √ b2 3 √a + 3 √ b ⇐ ⇒ A3 + B32 ≤ A4 + B4 A2 + B2 which holds by the Cauchy inequality.
The second inequality a + 3 √ a2b + 3 √ ab2 + b 4 ≤a + √ ab + b 3 106 is equivalent to 3(a + b) + 3 3 √ ab 3 √a + 3 √ b ≤4 a + √ ab + b ⇐ ⇒ a + 3 3 √ a2b + 3 3 √ ab2 + b ≤2 a + √ ab + b ⇐ ⇒ 3 √a + 3 √ b 3 ≤2 √a + √ b 2 ⇐ ⇒ A2 + B2 2 3 ≤ A3 + B3 2 2 which holds by the power mean inequality.
The third inequality a + √ ab + b 3 ≤ v u u t 3 √ a2 + 3 √ b2 2 !3 is equivalent to A6 + A3B3 + B6 3 2 ≤ A4 + B4 2 3 For this it is enough to prove that A4 + B4 2 3 − A6 + A3B3 + B6 3 2 ≥0 or 9 A4 + B43 −8 A6 + A3B3 + B62 = =(A −B)4(A8 + 4A7B + 10A6B2 + 4A5B3 −2A4B4+ + 4A3B5 + 10A2B6 + 4AB7 + B8) ≥ ≥(A −B)4 A8 −2A4B4 + B8 = =(A −B)4 A4 −B42 ≥0 □ 108. (Poland 1993) (x, y, u, v > 0) xy + xv + uy + uv x + y + u + v ≥ xy x + y + uv u + v 107 Solution. (Ercole Suppa) Is enough to note that xy + xv + uy + uv x + y + u + v − xy x + y + uv u + v = =(x + u)(y + v) x + y + u + v − xy x + y + uv u + v = = (vx −uy)2 (x + y)(u + v)(x + y + u + v) ≥0 □ 109. (IMO Short List 1993) (a + b + c + d = 1, a, b, c, d > 0) abc + bcd + cda + dab ≤1 27 + 176 27 abcd Solution. (See pag. 580) Put f(a, b, c, d) = abc + bcd + cda + dab −176 27 abcd and note that f is symmetric with respect to the four variables a, b, c, d. We can write f(a, b, c, d) = ab(c + d) + cd(a + b −176 27 ab) If a + b −176 27 ab ≤0, then using AM-GM for a, b, c + d, we have f(a, b, c, d) ≤ab(c + d) = 1 27 If a + b −176 27 ab > 0 by AM-GM inequality applied to c, d we get f(a, b, c, d) ≤ab(c + d) + 1 4(c + d)2 a + b −176 27 ab = f a, b, c + d 2 , c + d 2 Consider now the following fourtplets: P0(a, b, c, d) , P1 a, b, c + d 2 , c + d 2 , P2 a + b 2 , a + b 2 , c + d 2 , c + d 2 P3 1 4, a + b 2 , c + d 2 , 1 4 , P4 1 4, 1 4, 1 4, 1 4 From the above considerations we deduce that for i = 0, 1, 2, 3 either f(Pi) = 1/27 or f(Pi) ≤f(Pi+1). Since f(P4) = 1/27, in every case we are led to f(a, b, c, d) = f(P0) = 1 27 Equality occurs only in the cases 0, 1 3, 1 3, 1 3 (with permutations) and 1 4, 1 4, 1 4, 1 4 .
□ 108 110. (Italy 1993) (0 ≤a, b, c ≤1) a2 + b2 + c2 ≤a2b + b2c + c2a + 1 First Solution. (Ercole Suppa) The given inequality is equivalent to a2 (1 −b) + b2 (1 −c) + c2 (1 −a) ≤1 The function f (a, b, c) = a2 (1 −b) + b2 (1 −c) + c2 (1 −a) after setting b, c is convex with respect to the variable a so take its maximum value in a = 0 or in a = 1. A similar reasoning is true if we fix a, c or a, b.
Thus is enough to compute f(a, b, c) when a, b, c ∈{0, 1}. Since f is symmetric (with respect to a, b, c) and: f (0, 0, 0) = 0, f (0, 0, 1) = 1, f (0, 1, 1) = 1, f (1, 1, 1) = 0 the result is proven.
□ Second Solution. (Ercole Suppa) We have a2 (1 −b) + b2 (1 −c) + c2 (1 −a) ≤a (1 −b) + b (1 −c) + c (1 −a) = = a + b + c −(ab + bc + ca) = = 1 −(1 −a) (1 −b) (1 −c) −abc ≤1 □ 111. (Poland 1992) (a, b, c ∈R) (a + b −c)2(b + c −a)2(c + a −b)2 ≥(a2 + b2 −c2)(b2 + c2 −a2)(c2 + a2 −b2) Solution. (Harazi - ML Forum) It can be proved observing that (a + b −c)2(c + a −b)2 ≥(a2 + b2 −c2)(c2 + a2 −b2) which is true because: (a + c −b)2(b + a −c)2 = (a2 −(b −c)2)2 = a4 −2a2(b −c)2 + (b −c)4 But (a2 + c2 −b2)(b2 + a2 −c2) = a4 −(b2 −c2)2. So, it is enough to prove that (b2−c2)2+(b−c)4 ≥2a2(b−c)2 ⇐ ⇒(b+c)2+(b−c)2 ≥2a2 ⇐ ⇒b2+c2−a2 ≥0 We can assume that b2 + c2 −a2 ≥0, c2 + a2 −b2 ≥0, a2 + b2 −c2 ≥0 (only one of them could be negative and then it’s trivial), so these inequalities hold.
Multiply them and the required inequality is proved.
□ 109 112. (Vietnam 1991) (x ≥y ≥z > 0) x2y z + y2z x + z2x y ≥x2 + y2 + z2 First Solution. (Gabriel - ML Forum) Since x ≥y ≥z ≥0 we have that, x2y z + y2z x + z2x y = x3y2 + y3z2 + z3x2 xyz ≥ ≥(x3 + y3 + z3)(x2 + y2 + z2) 3(xyz) ≥ ≥3xyz(x2 + y2 + z2) 3(xyz) = = x2 + y2 + z2 by Chebyshev’s inequality □ Second Solution. (Murray Klamkin - Crux Mathematicorum 1996, pag.111) Let z = a, y = a + b, x = a + b + c where a > 0 and b, c ≥0. Substituting back in the inequality, multiplying by the least common denominator, we get x2y z + y2z x + z2x y −x2 −y2 −z2 = = 1 a(a + b)(a + b + c)(a3b2 + 3a2b3 + 3ab4 + b5 + a3bc + 6a2b2c + 8ab3c+ + 3b4c + a3c2 + 3a2bc2 + 6ab2c2 + 3b3c2 + abc3 + b2c3 ≥0 and the inequality is proved.
□ Third Solution. (ductrung - ML Forum) First, note that X ab(a −b) c = (a −b)(a −c)(b −c)(ab + bc + ca) abc ≥0 Hence X a2b c ≥ X ab2 c and so 2 X a2b c ≥ X ab(a + b) c It remains to show that X ab(a + b) c ≥2(a2 + b2 + c2) 110 or equivalently X a2b2(a + b) ≥2abc(a2 + b2 + c2) But X a2b2(a + b) −2abc(a2 + b2 + c2) = X c3(a −b)2 ≥0 Remark. Different solutions are given in Crux Mathematicorum 1994, pag.43.
113. (Poland 1991) (x2 + y2 + z2 = 2, x, y, z ∈R) x + y + z ≤2 + xyz First Solution. (See , pag. 57, problem 50) Using the Cauchy-Schwarz inequality, we find that x + y + z −xyz = x(1 −yz) + (y + z) ≤ p [x2 + (y + z)2] [1 + (1 −yz)2] So, it is enough to prove that this last quantity is at most 2, which is equivalent to (2 + 2yz) 2 −2yz + (yz)2 ≤4 ⇐ ⇒ (2yz)3 ≤(2yz)2 which is clearly true because 2yz ≤y2 + z2 ≤2.
□ Second Solution. (Crux Mathematicorum 1989, pag. 106) Put S = x + y + z, P = xyz. It is enough to show that E = 4 −(S −P)2 ≥0 Now using x2 + y2 + z2 = 2 we have 4E = 23 −22 S2 −2 + 2(4SP) −4P 2 = = 23 −22(2xy + 2yz + 2zx) + 2 4x2yz + 4xy2z + 4xyz2 −8x2y2z2 + 4P 2 = = (2 −2xy)(2 −2yz)(2 −2zx) + 4P 2 Since 2 −2xy = z2 + (x −y)2 2 −2yz = x2 + (x −z)2 2 −2zx = y2 + (z −x)2 the above quantities are nonnegative. Thus, so also is E, completing the proof.
□ 111 Third Solution. (Crux Mathematicorum 1989, pag. 106) Lagrange multipliers provide a straighforward solution. Here the Lagrangian is L = x + y + z −xyz −λ x2 + y2 + z2 −2 Now setting the partial derivatives equal to zero we obtain 1 −yz = 2λx 1 −xz = 2λy 1 −xy = 2λz On subtraction, we get (x −y)(z −2λ) = 0 = (y −z)(x −2λ) Thus the critical points are x = y = z and x = y, z = 1 −x2 /x and any cyclic permutation. The maximum value corresponds to the critical point x = y, z = 1 −x2 /x. Since x2 + y2 + z2 = 2 this leads to 3x2 −1 x2 −1 = 0 Finally, the critical point (1, 1, 0) and permutations of it give the maximum value of x + y + z −xyz to be 2.
□ Fourth Solution. (See pag. 155) If one of x, y, z is nagative, for example z < 0 then 2 + xyz −x −y −z = (2 −x −y) −z(1 −xy) ≥0 since x + y ≤ p 2 (x2 + y2) ≤2 and 2xy ≤x2 + y2 ≤2. Thus, WLOG, we can suppose 0 < x ≤y ≤z. If z ≤1 then 2 + xyz −x −y −z = (1 −x)(1 −y) + (1 −z)(1 −xy) ≥0 If, on the contrary z > 1 then by Cauchy-Schwartz inequality we have x + y + z ≤ p 2 [(x + y)2 + z2] = 2 p xy + 1 ≤xy + 2 ≤xyz + 2 □ Remark 1. This inequality was proposted in IMO shortlist 1987 by United Kingdom.
Remark 2.
The inequality admit the following generalization: Given real numbers x, y, z such that x2 + y2 + z2 = k, k > 0, prove the inequality 2 k xyz − √ 2k ≤x + y + z ≤2 k xyz + √ 2k When k = 2, see problem 113.
112 114. (Mongolia 1991) (a2 + b2 + c2 = 2, a, b, c ∈R) |a3 + b3 + c3 −abc| ≤2 √ 2 Solution. (ThAzN1 - ML Forum) It suffices to prove (a3 + b3 + c3 −abc)2 ≤8 = (a2 + b2 + c2)3.
This is equivalent to (a2 + b2 + c2)3 −(a3 + b3 + c3)2 + 2abc(a3 + b3 + c3) −a2b2c2 ≥0 X (3a4b2 + 3a2b4 −2a3b3) + 2abc(a3 + b3 + c3) + 5a2b2c2 ≥0 X (a4b2 + a2b4 + a2b2(a −b)2 + a4(b + c)2) + 5a2b2c2 ≥0 □ 115. (IMO Short List 1990) (ab + bc + cd + da = 1, a, b, c, d > 0) a3 b + c + d + b3 c + d + a + c3 d + a + b + d3 a + b + c ≥1 3 First Solution. (See pag. 540) Let A, B, C, D denote b + c + d, a + c + d, a + b + d, a + b + c respectively.
Since ab + bc + cd + da = 1 the numbers A, B, C, D are all positive.
By Cauchy-Schwarz inequality we have a2 + b2 + c2 + d2 ≥ab + bc + cd + da = 1 We’ll prove the required inequality under a weaker condition that A, B, C, D are all positive and a2 + b2 + c2 + d2 ≥1. We may assume, WLOG, that a ≥b ≥c ≥d ≥0. Hence a3 ≥b3 ≥c3 ≥d3 ≥0 and 1 A ≥ 1 B ≥ 1 C ≥ 1 D ≥0.
Using Chebyshev inequality and Cauchy inequality we obtain a3 A + b3 B + c3 C + d3 D ≥1 4 a3 + b3 + c3 + d3 1 A + 1 B 1 C + 1 D ≥ ≥1 16 a2 + b2 + c2 + d2 (a + b + c + d) 1 A + 1 B 1 C + 1 D = = 1 48 a2 + b2 + c2 + d2 (A + B + C + D) 1 A + 1 B 1 C + 1 D ≥1 3 This complete the proof.
□ 113 Second Solution. (Demetres Christofides - J. Sholes WEB site) Put S = a + b + c + d A = a3 S −a + b3 S −b + c3 S −c + d3 S −d B = a2 + b2 + c2 + d2 C = a(S −a) + b(S −b) + c(S −c) + d(S −d) = 2 + 2ac + 2bd By Cauchy-Scwarz we have AC ≥B2 (1) We also have (a −b)2 + (b −c)2 + (c −d)2 + (d −a)2 ≥0 = ⇒ B ≥ab + bc + cd + da = 1 (2) and (a −c)2 + (b −d)2 ≥0 = ⇒ B ≥2ac + 2bd (3) If 2ac + 2bd ≤1 then C ≤3, so by (1) and (2) we have A ≥B2 C ≥1 C ≥1 3 If 2ac + 2bd > 1 then C > 3, so by (1), (2), and (3) we have A ≥B2 C ≥B C ≥2ac + 2bd C ≥C −2 C = 1 −2 C > 1 3 This complete the proof.
□ Third Solution. (Campos - ML Forum) By Holder we have that X a3 b + c + d X a(b + c + d) X 1 2 ≥ X a 4 = ⇒ X a3 b + c + d ≥ (a + b + c + d)4 16(P a(b + c + d)) but it’s easy to verify from the condition that (a + b + c + d)2 ≥4(a + c)(b + d) = 4 and 3(a + b + c + d)2 ≥4 X a(b + c + d) 114 because 3(a + b + c + d)2 −4 X a(b + c + d) = =3(a + b + c + d)2 −8(ab + ac + ad + bc + bd + cd) = =3 a2 + b2 + c2 + d2 −2(ab + ac + ad + bc + bd + cd) = =4 a2 + b2 + c2 + d2 −(a + b + c + d)2 ≥0 (by Cauchy-Schwarz) This complete the proof.
□ 115 4 Supplementary Problems 116. (Lithuania 1987) (x, y, z > 0) x3 x2 + xy + y2 + y3 y2 + yz + z2 + z3 z2 + zx + x2 ≥x + y + z 3 Solution. (Gibbenergy - ML Forum) Since 3xy ≤x2 + xy + y2 we have x3 x2 + xy + y2 = x − xy(x + y) x2 + xy + y2 ≥x −x + y 3 Then doing this for all other fractions and summing we obtain the inequality we want to prove.
□ Remark. This inequality was proposed in Tournament of the Towns 1998.
117. (Yugoslavia 1987) (a, b > 0) 1 2(a + b)2 + 1 4(a + b) ≥a √ b + b√a First Solution. (Ercole Suppa) From AM-GM inequality follows that a+b 2 ≥ √ ab. Therefore 1 2(a + b)2 + 1 4(a + b) −a √ b −b√a = = 1 2(a + b)2 + 1 4(a + b) − √ ab √a + √ b ≥ ≥1 2(a + b)2 + 1 4(a + b) −a + b 2 √a + √ b = = a + b 2 a + b + 1 2 −√a − √ b = = a + b 2 "√a −1 2 2 + √ b −1 2 2# ≥0 □ 116 Second Solution. (Arne- ML Forum) The left-hand side equals a2 2 + b2 2 + ab + a 4 + b 4 Now note that, by AM-GM inequality a2 2 + ab 2 + a 8 + b 8 ≥4 4 r a2 2 · ab 2 · a 8 · b 8 = a √ b Similarly b2 2 + ab 2 + a 8 + b 8 ≥4 4 r b2 2 · ab 2 · a 8 · b 8 = b√a Adding these inequalities gives the result.
□ 118. (Yugoslavia 1984) (a, b, c, d > 0) a b + c + b c + d + c d + a + d a + b ≥2 Solution. (See pag. 127) From Cauchy-Schwarz inequality we have (a + b + c + d)2 ≤ X cyc a(b + c) X cyc a b + c = = (ab + 2ac + bc + 2bd + cd + ad) · X cyc a b + c Then, to establish the required inequality it will be enough to show that (ab + 2ac + bc + 2bd + cd + ad) ≤1 2(a + b + c + d)2 This inequality it is true because 1 2(a + b + c + d)2 −(ab + 2ac + bc + 2bd + cd + ad) = = 1 2(a −c)2 + 1 2(b −d)2 ≥0 The equality holds if and only if a = c e b = d.
□ 119. (IMO 1984) (x + y + z = 1, x, y, z ≥0) 0 ≤xy + yz + zx −2xyz ≤7 27 117 First Solution. (See , pag. 23) Let f(x, y, z) = xy + yz + zx −2xyz.
We may assume that 0 ≤x ≤y ≤ z ≤1. Since x + y + z = 1, we find that x ≤1 3. It follows that f(x, y, z) = (1 −3x)yz + xyz + zx + xy ≥0. Applying the AM-GM inequality, we obtain yz ≤ y+z 2 2 = 1−x 2 2. Since 1 −2x ≥0, this implies that f(x, y, z) = x(y+z)+yz(1−2x) ≤x(1−x)+ 1 −x 2 2 (1−2x) = −2x3 + x2 + 1 4 .
Our job is now to maximize F(x) = 1 4(−2x3 + x2 + 1), where x ∈ 0, 1 3 . Since F ′(x) = 3 2x 1 3 −x ≥0 on 0, 1 3 , we conclude that F(x) ≤F( 1 3) = 7 27 for all x ∈ 0, 1 3 .
□ Second Solution. (See , pag. 31) Using the condition x+y+z = 1, we reduce the given inequality to homogeneous one, i. e., 0 ≤(xy + yz + zx)(x + y + z) −2xyz ≤7 27(x + y + z)3.
The left hand side inequality is trivial because it’s equivalent to 0 ≤xyz + X sym x2y.
The right hand side inequality simplifies to 7 X cyc x3 + 15xyz −6 X sym x2y ≥0.
In the view of 7 X cyc x3+15xyz−6 X sym x2y = 2 X cyc x3 − X sym x2y !
+5 3xyz + X cyc x3 − X sym x2y !
, it’s enough to show that 2 X cyc x3 ≥ X sym x2y and 3xyz + X cyc x3 ≥ X sym x2y.
We note that 2 X cyc x3− X sym x2y = X cyc (x3+y3)− X cyc (x2y+xy2) = X cyc (x3+y3−x2y−xy2) ≥0.
The second inequality can be rewritten as X cyc x(x −y)(x −z) ≥0, which is a particular case of Schur’s theorem.
□ 118 120. (USA 1980) (a, b, c ∈[0, 1]) a b + c + 1 + b c + a + 1 + c a + b + 1 + (1 −a)(1 −b)(1 −c) ≤1.
Solution. (See pag. 82) The function f(a, b, c) = a b + c + 1 + b c + a + 1 + c a + b + 1 + (1 −a)(1 −b)(1 −c) is convex in each of the three variables a, b, c, so f takes its maximum value in one of eight vertices of the cube 0 ≤a ≤1, 0 ≤b ≤1, 0 ≤c ≤1. Since f(a, b, c) takes value 1 in each of these points, the required inequality is proven.
□ 121. (USA 1979) (x + y + z = 1, x, y, z > 0) x3 + y3 + z3 + 6xyz ≥1 4.
Solution. (Ercole Suppa) The required inequality is equivalent to 4 x3 + y3 + z3 + 24xyz ≥(x + y + z)3 ⇐ ⇒ 3 x3 + y3 + z3 + 18xyz ≥3 X sym x2y ⇐ ⇒ X x3 + 3 X xyz ≥ X sym x2y which is true for all x, y, z > 0 by Schur inequality.
□ 122. (IMO 1974) (a, b, c, d > 0) 1 < a a + b + d + b b + c + a + c b + c + d + d a + c + d < 2 Solution. (Ercole Suppa) We have a a + b + d + b b + c + a + c b + c + d + d a + c + d < < a a + b + b b + a + c c + d + d c + d = 2 119 and a a + b + d + b b + c + a + c b + c + d + d a + c + d > > a a + b + c + d + b a + b + c + d + c a + b + c + d + d a + b + c + d = 1 □ Remark. In the problem 5 of IMO 1974 is requested to find all possible values of S = a a + b + d + b b + c + a + c b + c + d + d a + c + d for arbitrary positive reals a, b, c, d. A detailed solution is given in , pag.
203.
123. (IMO 1968) (x1, x2 > 0, y1, y2, z1, z2 ∈R, x1y1 > z12, x2y2 > z22) 1 x1y1 −z12 + 1 x2y2 −z22 ≥ 8 (x1 + x2)(y1 + y2) −(z1 + z2)2 Solution. (pag. 369) Define u1 = √x1y1+z1, u2 = √x2y2+z2, v1 = √x1y1−z1 and v2 = √x2y2−z2.
By expanding both sides we can easily verify (x1 + x2) (y1 + y2) −(z1 + z2)2 = (u1 + u2) (v1 + v2) + (√x1y2 −√x2y1)2 Thus (x1 + x2) (y1 + y2) −(z1 + z2)2 ≥(u1 + u2) (v1 + v2) Since xiyi −z2 i = uivi for i = 1, 2, it suffices to prove 8 (u1 + u2) (v1 + v2) ≤ 1 u1v1 + 1 u2v2 ⇐ ⇒ 8u1u2v1v2 ≤(u1 + u2) (v1 + v2) (u1v1 + u2v2) which trivially follows from the AM-GM inequalities 2√u1u2 ≤u1 + u2 , 2√v1v2 ≤v1 + v2 , 2√u1v1u2v2 ≤u1v1 + u2v2 Equality holds if and only if x1y2 = x2y1, u1 = u2 and v1 = v2, i.e. if and only if x1 = x2, y1 = y2 and z1 = z2.
□ 124. (Nesbitt’s inequality) (a, b, c > 0) a b + c + b c + a + c a + b ≥3 2 120 Solution. (See , pag. 18) After the substitution x = b + c, y = c + a, z = a + b, it becomes X cyc y + z −x 2x ≥3 2 or X cyc y + z x ≥6, which follows from the AM-GM inequality as following: X cyc y + z x = y x + z x + z y + x y + x z + y z ≥6 y x · z x · z y · x y · x z · y z 1 6 = 6.
Remark. In , are given many other proofs of this famous inequality.
125. (Polya’s inequality) (a ̸= b, a, b > 0) 1 3 2 √ ab + a + b 2 > a −b ln a −ln b Solution. (Kee-Wai Lau - Crux Mathematicorum 1999, pag.253) We can assume WLOG that a > b. The required inequality is equivalent to 1 3 2 √ ab + a + b 2 ≥a −b ln a b or, dividing both members by b : 1 3 2 ra b + a b + 1 2 ≥ a b −1 ln a b After setting x = p a b we must show that ln x −3 x2 −1 x2 + 4x + 1 ≥0 , ∀x ≥1 By putting f(x) = ln x −3 x2 −1 x2 + 4x + 1 a direct calculation show that f ′(x) = (x −1)4 x (x2 + 4x + 1)2 Thus f ′(x) > 0 for all x > 1 (i.e.
f(x) is increasing for all x > 1).
Since f(1) = 0, we have f(x) > 0 for all x > 1 and the the result is proven.
□ 121 126. (Klamkin’s inequality) (−1 < x, y, z < 1) 1 (1 −x)(1 −y)(1 −z) + 1 (1 + x)(1 + y)(1 + z) ≥2 Solution. (Ercole Suppa) From AM-GM inequality we have 1 (1 −x)(1 −y)(1 −z)+ 1 (1 + x)(1 + y)(1 + z) ≥ 2 p (1 −x2) (1 −y2) (1 −z2) ≥2 □ 127. (Carlson’s inequality) (a, b, c > 0) 3 r (a + b)(b + c)(c + a) 8 ≥ r ab + bc + ca 3 First Solution. (P. E. Tsaoussoglou - Crux Mathematicorum 1995, pag. 336) It is enough to prove that for all positive real numbers a, b, c the following inequality holds 64(ab + bc + ca)3 ≤27(a + b)2(b + c)2(c + a)2 or 64 · 3(ab + bc + ca)(ab + bc + ca)2 ≤81 [(a + b)(b + c)(c + a)]2 It is know that 3(ab + bc + ca) ≤(a + b + c)2. Thus, it is enough to prove one of the following equivalent inequalities 8(a + b + c)(ab + bc + ca) ≤9(a + b)(b + c)(c + a) 8(a + b)(b + c)(c + a) + 8abc ≤9(a + b)(b + c)(c + a) 8abc ≤(a + b)(b + c)(c + a) The last inequality is well-know and this complete the proof.
□ Second Solution. (See pag. 141) It is enough to prove that for all positive real numbers a, b, c the following inequality holds 64(ab + bc + ca)3 ≤27(a + b)2(b + c)2(c + a)2 122 Write s = a+b+c, u = ab+bc+ca, v = abc. Since a2+b2+c2 > ab+bc+ca = u we have s = p a2 + b2 + c2 + 2u ≥ √ 3u By AM-GM inequality s ≥3 3 √ abc , u ≥3 3 p (ab)(bc)(ca) = 3 3 √ v2 and hence su ≥9v. Consequently, (a + b)(b + c)(c + a) = (s −a)(s −b)(s −c) = s3 + su −v ≥ ≥su −1 9su = 8 9su ≥8 9u √ 3u = = 8 √ 3 9 p (ab + bc + ca)3 and raising both sides to the second power we obtain the asserted inequality.
Equality holds if and only if a = b = c.
□ Remark.
The problem was proposed in Austrian-Polish Competition 1992, problem 6.
128. (See , Vasile Cirtoaje) (a, b, c > 0) a + 1 b −1 b + 1 c −1 + b + 1 c −1 c + 1 a −1 + c + 1 a −1 a + 1 b −1 ≥3 Solution. (See , pag. 89, problem 94) Assume WLOG that x = max{x, y, z}.
Then x ≥1 3(x + y + z) = 1 3 a + 1 a + b + 1 b + c + 1 c −3 ≥1 3(2 + 2 + 2 −3) = 1 On the other hand, (x + 1)(y + 1)(z + 1) = abc + 1 abc + a + b + c + 1 a + 1 b + 1 c ≥ ≥2 + a + b + c + 1 a + 1 b + 1 c = 5 + x + y + z and hence xyz + xy + yz + zx ≥4 Since y + z = 1 a + b + (c −1)2 c > 0 two cases are possible.
123 (a) Case yz ≤0. We have xyz ≤0, and from xy + yz + zx ≥4 it follows that xy + yz + zx ≥4 > 3.
(b) Case y, z > 0. Let d = q xy+yz+zx 3 . We have to show that d ≥1. By AM-GM we have xyz ≤d3. Thus xyz + xy + yz + zx ≥4 implies d3 + 3d2 ≥4, (d −1)(d + 2)2 ≥0, d ≥1. Equality occurs for a = b = c = 1.
□ 129. ([ONI], Vasile Cirtoaje) (a, b, c, d > 0) a −b b + c + b −c c + d + c −d d + a + d −a a + b ≥0 Solution. (See , pag. 60, n. 54) By AM-HM inequality we have a −b b + c + b −c c + d + c −d d + a + d −a a + b = =a + c b + c + b + d c + d + c + a d + a + d + b a + b −4 = =(a + c) 1 b + c + 1 d + a + (b + d) 1 c + d + 1 a + b −4 ≥ ≥ 4(a + c) (b + c) + (d + a) + 4(b + d) (c + d) + (a + b) −4 = 0 □ 130. (Elemente der Mathematik, Problem 1207, ˜ Sefket Arslanagi´ c) (x, y, z > 0) x y + y z + z x ≥x + y + z 3 √xyz Solution. (Ercole Suppa) The required inequality is equivalent to x2z + y2x + z2y ≥(x + y + z) 3 p (xyz)2 The above inequality is obtained by adding the following 1 3x2z + 1 3x2z + 1 3xy2 ≥x 3 p (xyz)2 1 3xy2 + 1 3xy2 + 1 3yz2 ≥y 3 p (xyz)2 1 3yz2 + 1 3yz2 + 1 3x2z ≥z 3 p (xyz)2 which follows from AM-GM inequality.
□ 124 131. ( √ WURZEL, Walther Janous) (x + y + z = 1, x, y, z > 0) (1 + x)(1 + y)(1 + z) ≥(1 −x2)2 + (1 −y2)2 + (1 −z2)2 First Solution. (Ercole Suppa) By setting A = xy + yz + zx, B = xyz, since x + y + z = 1, we get x2 + y2 + z2 = (x + y + z)2 −2(xy + yz + zx) = 1 −2A and x4 + y4 + z4 = x2 + y2 + z22 −2(x2y2 + y2z2 + z2x2) = = (1 −2A)2 −2 (xy + yz + zx)2 −2 x2yz + xy2z + xyz2 = = (1 −2A)2 −2 A2 −2B(x + y + z) = = (1 −2A)2 −2A2 + 4B = = 2A2 −4A + 4B + 1 The required inequality is equivalent to 1 + x + y + z + xy + yz + zx + xyz ≥x4 + y4 + z4 −2 x2 + y2 + z2 + 3 ⇐ ⇒ 2 + A + B ≥2A2 −4A + 4B + 1 −2(1 −2A) + 3 ⇐ ⇒ A ≥2A2 + 3B ⇐ ⇒ xy + yz + zx ≥2(xy + yz + zx)2 + 3xyz ⇐ ⇒ (x + y + z)2(xy + yz + zx) ≥2(xy + yz + zx)2 + 3xyz(x + y + z) ⇐ ⇒ x2 + y2 + z2 (xy + yz + zx) ≥3xyz(x + y + z) (⋆) The inequality (⋆) follows from Muirhead theorem since x2 + y2 + z2 (xy + yz + zx) ≥3xyz(x + y + z) ⇐ ⇒ x3y + x3z + xy3 + y3z + xz3 + yz3 ≥2x2yz + 2xy2z + 2xyz2 ⇐ ⇒ X sym x3y ≥ X sym x2yz Alternatively is enough to observe that for all x, y, z ≥0 we get x2 + y2 + z2 (xy + yz + zx) −3xyz(x + y + z) ≥ ≥(xy + yz + zx)2 −3xyz(x + y + z) = =1 2 x2(y −z)2 + y2(z −x)2 + z2(x −y)2 ≥0 □ 125 Second Solution. (Yimin Ge - ML Forum) Homogenizing gives (x + y + z)(2x + y + z)(x + 2y + z)(x + y + 2z) ≥ X ((y + z)(2x + y + z))2 By using the Ravi-substitution, we obtain (a + b + c)(a + b)(b + c)(c + a) ≥2 X (a(b + c))2 which is equivalent to X sym a3b ≥ X sym a2b2 which is true.
□ Remark. This inequality was proposed in Austrian-Polish Competition 2000, problem 6.
132. ( √ WURZEL, Heinz-J¨ urgen Seiffert) (xy > 0, x, y ∈R) 2xy x + y + r x2 + y2 2 ≥√xy + x + y 2 Solution. (Campos - ML Forum) We have 2xy x + y + r x2 + y2 2 ≥√xy + x + y 2 ⇔ r x2 + y2 2 −√xy ≥x + y 2 −2xy x + y ⇔ (x −y)2 2 ≥(x −y)2 2(x + y) · r x2 + y2 2 + √xy !
⇔ x + y ≥ r x2 + y2 2 + √xy ⇔ (x + y)2 2 ≥2 r x2 + y2 2 √xy ⇔ x2 + y2 + 2xy 2 ≥ p (x2 + y2) · 2xy and this is AM-GM.
□ 126 133. ( √ WURZEL, ˇ Sefket Arslanagi´ c) (a, b, c > 0) a3 x + b3 y + c3 z ≥(a + b + c)3 3 (x + y + z) Solution. (Ercole Suppa) First we prove the following lemma: Lemma. If a1, · · · , an, b1, · · · , bn, c1, · · · , cn are real positive numbers, the fol-lowing inequality holds n X i=1 aibici !3 ≤ n X i=1 ai 3 ! n X i=1 bi 3 ! n X i=1 ci 3 !
Proof. By Holder and Cauchy-Schwarz inequalities we have X aibici ≤ X a3 i 1 3 X (bici) 3 2 2 3 ≤ ≤ X a3 i 1 3 X b3 i 1 2 X c3 i 1 2 2 3 = = X a3 i 1 3 · X b3 i 1 3 · X c3 i 1 3 □ In order to show the required inequality we put (a1, a2, a3) = a 3 √x, b 3 √y , c 3 √z (b1, b2, b3) = 3 √x, 3 √y, 3 √z (c1, c2, c3) = (1, 1, 1) and we use the Lemma: (a + b + c)3 = X aibici 3 ≤ ≤ X a3 i X b3 i X c3 i = = a3 x + b3 y + c3 z (x + y + z)(1 + 1 + 1) Finally, dividing by (x + y + z) we have a3 x + b3 y + c3 z ≥(a + b + c)3 3 (x + y + z) □ 127 134. ( √ WURZEL, ˇ Sefket Arslanagi´ c) (abc = 1, a, b, c > 0) 1 a2 (b + c) + 1 b2 (c + a) + 1 c2 (a + b) ≥3 2.
Solution. (Ercole Suppa) After setting a = 1 x, b = 1 y, c = 1 z we have xyz = 1 and the required inequality is equivalent to x y + z + y x + z + z x + y ≥3 2 which is the well-know Nesbitt inequality (see Problem 124).
□ 135. ( √ WURZEL, Peter Starek, Donauw¨ orth) (abc = 1, a, b, c > 0) 1 a3 + 1 b3 + 1 c3 ≥1 2 (a + b) (c + a) (b + c) −1.
Solution. (Ercole Suppa) After setting a = 1 x, b = 1 y, c = 1 z we have xyz = 1 and the required inequality is equivalent to x3 + y3 + z3 ≥1 2 · x + y xy · y + z yz · z + x zx −1 ⇐ ⇒ 2 x3 + y3 + z3 ≥x2y + x2z + xy2 + y2z + xz2 + yz2 ⇐ ⇒ X sym x3 ≥ X sym x2y The above inequality follows from Muirhead theorem or can be obtained adding the three inequalities x3 + y3 ≥x2y + xy2, y3 + z3 ≥y2z + yz2, z3 + x3 ≥z2x + zx2 □ 128 136. ( √ WURZEL, Peter Starek, Donauw¨ orth) (x+y+z = 3, x2+y2+z2 = 7, x, y, z > 0) 1 + 6 xyz ≥1 3 x z + y x + z y Solution. (Ercole Suppa) From the constraints x + y + z = 3, x2 + y2 + z2 = 7 follows that 9 = (x + y + z)2 = 7 + 2(xy + yz + zx) = ⇒ xy + yz + zx = 1 The required inequality is equivalent to 3xyz + 18 ≥x2y + y2z + z2x ⇐ ⇒ 3xyz + 6(x + y + x)(xy + yz + zx) ≥x2y + y2z + z2x ⇐ ⇒ 21xyz + 5 x2y + y2z + z2x + 6 x2z + xy2 + yz2 ≥0 which is true for all x, y, z > 0.
□ 137. ( √ WURZEL, ˇ Sefket Arslanagi´ c) (a, b, c > 0) a b + 1 + b c + 1 + c a + 1 ≥3 (a + b + c) a + b + c + 3.
Solution. (Ercole Suppa) Using Cauchy-Schwartz and the well known inequality (a+b+c)2 ≥3(ab+bc+ca) we have a b + 1 + b c + 1 + c a + 1 = a2 a(b + 1) + b2 b(c + 1) + c2 c(a + 1) ≥ (a + b + c)2 ab + bc + ca + a + b + c ≥ (a + b + c)2 1 3(a + b + c)2 + (a + b + c) = a + b + c 1 3(a + b + c) + 1 = 3(a + b + c) a + b + c + 3 Equality holds if and only if a = b = c.
□ 129 138. ([ONI], Gabriel Dospinescu, Mircea Lascu, Marian Tetiva) (a, b, c > 0) a2 + b2 + c2 + 2abc + 3 ≥(1 + a)(1 + b)(1 + c) Solution. (See , pag. 75, problem 74) Let f(a, b, c) = a2 + b2 + c2 + 2abc + 3 −(1 + a)(1 + b)(1 + c). We have to prove that all values of f are nonnegative. If a, b, c > 3, then we have 1 a + 1 b + 1 c < 1 = ⇒ ab + bc + ca < abc hence f(a, b, c) = a2 + b2 + c2 + abc + 2 −a −b −c −ab −bc −ca > > a2 + b2 + c2 + 2 −a −b −c > 0 So, we may assume that a ≤3 and let m = b+c 2 . Easy computations show that f(a, b, c) −f(a, m, m) = (3 −a)(b −c)2 4 ≥0 and so it remains to prove that f(a, m, m) ≥0 ⇐ ⇒ (a + 1)m2 −2(a + 1)m + a2 −a + 2 ≥0 This is cleary true, because the discriminant of the quadratic equation is ∆= −4(a + 1)(a −1)2 ≤0 □ 130 139. (Gazeta Matematic˜ a) (a, b, c > 0) p a4 + a2b2 + b4+ p b4 + b2c2 + c4+ p c4 + c2a2 + a4 ≥a p 2a2 + bc+b p 2b2 + ca+c p 2c2 + ab Solution. (See , pag. 43) We obtain the chain of equalities and inequalities X cyc p a4 + a2b2 + b4 = X cyc s a4 + a2b2 2 + b4 + a2b2 2 ≥ ≥ 1 √ 2 X cyc r a4 + a2b2 2 + r b4 + a2b2 2 !
= (Cauchy-Schwarz) = 1 √ 2 X cyc r a4 + a2b2 2 + r a4 + a2c2 2 !
≥ ≥ √ 2 X cyc 4 s a4 + a2b2 2 a4 + a2c2 2 ≥ (AM-GM) ≥ √ 2 X cyc r a4 + a2bc 2 = (Cauchy-Schwarz) = X cyc p 2a4 + a2bc □ 140. (C12362, Mohammed Aassila) (a, b, c > 0) 1 a(1 + b) + 1 b(1 + c) + 1 c(1 + a) ≥ 3 1 + abc Solution. (Crux Mathematicorum 1999, pag. 375, n.2362) We use the well-know inequality t + 1/t ≥2 for t > 0. Equality occurs if and only if t = 1. Note that 1 + abc a(1 + b) = 1 + a a(1 + b) + b(1 + c) 1 + b −1 1 + abc b(1 + c) = 1 + b b(1 + c) + c(1 + a) 1 + c −1 1 + abc c(1 + a) = 1 + c c(1 + a) + a(1 + b) 1 + a −1 1CRUX with MAYHEM 131 Then 1 + abc a(1 + b) + 1 + abc b(1 + c) + 1 + abc c(1 + a) ≥2 + 2 + 2 −3 = 3 by the above inequality. Equality holds when 1 + a a(1 + b) = 1 + b b(1 + c) = 1 + c c(1 + a) = 1 that is, when a = b = c = 1.
□ 141. (C2580) (a, b, c > 0) 1 a + 1 b + 1 c ≥ b + c a2 + bc + c + a b2 + ca + a + b c2 + ab Solution. (Crux Mathematicorum 2001, pag. 541, n.2580) Let D = abc a2 + bc b2 + ac c2 + ab . Clearly D > 0 and 1 a + 1 b + 1 c − b + c a2 + bc −c + a b2 + ac + a + b c2 + ab = =a4b4 + b4c4 + c4a4 −a4b2c2 −b4c2a2 −c4a2b2 D = = a2b2 −b2c22 + b2c2 −c2a22 + c2a2 −a2b22 2D ≥0 which shows that the given inequality is true.
Equality holds if and only if a = b = c.
□ 142. (C2581) (a, b, c > 0) a2 + bc b + c + b2 + ca c + a + c2 + ab a + b ≥a + b + c Solution. (Crux Mathematicorum 2001, pag. 541, n.2581) Let D = (a+b)(b+c)(c+a). Clearly D > 0. We show that the difference between 132 the left-hand side and the right-hand side of the inequality is nonnegative.
a2 + bc b + c −a + b2 + ca c + a −b + c2 + ab a + b −c = =a2 + bc −ab −ac b + c + b2 + ac + ab −bc a + c + c2 + ab −ac −bc a + b = =(a −b)(a −c) b + c + (b −a)(b −c) a + c + (c −a)(c −b) a + b = = a2 −b2 a2 −c2 + b2 −a2 b2 −c2 + c2 −a2 c2 −b2 D = =a4 + b4 + c4 −b2c2 −c2a2 −a2b2 D = = a2 −b22 + b2 −c22 + c2 −a22 2D ≥0 Equality holds if and only if a = b = c.
□ 143. (C2532) (a2 + b2 + c2 = 1, a, b, c > 0) 1 a2 + 1 b2 + 1 c2 ≥3 + 2(a3 + b3 + c3) abc Solution. (Crux Mathematicorum 2001, pag. 221, n.2532) We have 1 a2 + 1 b2 + 1 c2 −3 −2(a3 + b3 + c3) abc = =a2 + b2 + c2 a2 + a2 + b2 + c2 b2 + a2 + b2 + c2 c2 −3 −2 a2 bc + b2 ca + c2 ab = =a2 1 b2 + 1 c2 + b2 1 a2 + 1 c2 + c2 1 a2 + 1 b2 −2 a2 bc + b2 ca + c2 ab = =a2 1 b −1 c 2 + b2 1 c −1 a 2 + c2 1 a −1 b 2 ≥0 Equality holds if and only if a = b = c.
□ 133 144. (C3032, Vasile Cirtoaje) (a2 + b2 + c2 = 1, a, b, c > 0) 1 1 −ab + 1 1 −bc + 1 1 −ca ≤9 2 Solution. (Crux Mathematicorum 2006, pag. 190, problem 3032) Note first that the given inequality is equivalent to 3 −5(ab + bc + ca) + 7abc(a + b + c) −9a2b2c2 ≥0 3 −5(ab + bc + ca) + 6abc(a + b + c) + abc(a + b + c −9abc) ≥0 (1) By the AM-GM inequality we have a + b + c −9abc = (a + b + c) a2 + b2 + c2 −9abc ≥ ≥3 3 √ abc · 3 3 √ a2b2c2 −9abc = 0 (2) On the other hand, 3 −5(ab + bc + ca) + 6abc(a + b + c) = =3 a2 + b2 + c22 −5(ab + bc + ca) a2 + b2 + c2 + 6abc(a + b + c) = =3 a4 + b4 + c4 + 6 a2b2 + b2c2 + c2a2 + abc(a + b + c) −5 ab a2 + b2 + bc b2 + c2 + ca c2 + a2 = = h 2 X a4 + 6 X a2b2 −4 X ab a2 + b2i + + a4 + b4 + c4 + abc(a + b + c) −ab a2 + b2 −bc b2 + c2 −ca c2 + a2 = = (a −b)4 + (b −c)4 + (c −a)4 + + a2(a −b)(a −c) + b2(b −a)(b −c) + c2(c −a)(c −b) ≥0 (3) since (a −b)4 + (b −c)4 + (c −a)4 ≥0 and a2(a −b)(a −c) + b2(b −a)(b −c) + c2(c −a)(c −b) ≥0 is the well-knom Schur’s inequality. Now (1) follows from (2) and (3). The equality holds if and only if a = b = c = √ 3/3.
□ 134 145. (C2645) (a, b, c > 0) 2(a3 + b3 + c3) abc + 9(a + b + c)2 (a2 + b2 + c2) ≥33 First Solution. (Darij Grinberg - ML Forum) Equivalently transform our inequality: 2 a3 + b3 + c3 abc + 9 (a + b + c)2 a2 + b2 + c2 ≥33 ⇐ ⇒ 2 a3 + b3 + c3 abc −6 !
+ 9 (a + b + c)2 a2 + b2 + c2 −27 !
≥0 ⇐ ⇒ 2a3 + b3 + c3 −3abc abc + 9(a + b + c)2 −3 a2 + b2 + c2 a2 + b2 + c2 ≥0 Now, it is well-known that a3 + b3 + c3 −3abc = (a + b + c) a2 + b2 + c2 −bc −ca −ab and (a + b + c)2 −3 a2 + b2 + c2 = −2 a2 + b2 + c2 −bc −ca −ab , so the inequality above becomes 2(a + b + c) a2 + b2 + c2 −bc −ca −ab abc +9−2 a2 + b2 + c2 −bc −ca −ab a2 + b2 + c2 ≥0 Now, according to the well-known inequality a2 + b2 + c2 ≥bc + ca + ab, we have a2 + b2 + c2 −bc −ca −ab ≥0, so that we can divide this inequality by a2 + b2 + c2 −bc −ca −ab to obtain 2a + b + c abc + 9 −2 a2 + b2 + c2 ≥0 ⇐ ⇒ 2a + b + c abc − 2 · 9 a2 + b2 + c2 ≥0 ⇐ ⇒ a + b + c abc ≥ 9 a2 + b2 + c2 ⇐ ⇒ (a + b + c) a2 + b2 + c2 ≥9abc But this is evident, since AM-GM yields a + b + c ≥3 3 √ abc and a2 + b2 + c2 ≥ 3 3 √ a2b2c2, so that (a + b + c) a2 + b2 + c2 ≥3 3 √ abc · 3 3 √ a2b2c2 = 9abc.
Proof complete.
□ 135 Second Solution. (Crux Mathematicorum 2002, pag. 279, n.2645) On multiplying by the common denominator and performing the necessary cal-culations, we have that the given inequality is equivalent to 2 a3 + b3 + c3 a2 + b2 + c2 + 9abc(a + b + c)2 −33abc a2 + b2 + c2 ≥0 The left side of this is the product of a2 + b2 + c2 −ab −bc −ca (1) and 2 a3 + b3 + c3 + a2b + a2c + b2a + b2c + c2a + c2b −9abc (2) The product of (1) and (2) is nonnegative because a2 + b2 + c2 −ab −bc −ca = (a −b)2 + (b −c)2 + (c −a)2 2 ≥0 and (by AM-GM) 2 a3 + b3 + c3 + a2b + a2c + b2a + b2c + c2a + c2b −9abc ≥ ≥2 9 9 √ a9b9c9 −9abc = 0 Equality holds if and only if a = b = c.
□ Remark.
In order to prove that (2) is positive we can use also the S.O.S method (=sum of squares): 2 a3 + b3 + c3 + a2b + a2c + b2a + b2c + c2a + c2b −9abc = =(a −b)2(a + b + 3c) + (b −c)2(b + c + 3a) + (c −a)2(c + a + 3b) ≥0 □ 146. (x, y ∈R) −1 2 ≤(x + y)(1 −xy) (1 + x2)(1 + y2) ≤1 2 First Solution. (Ercole Suppa) The required inequality is equivalent to − 1 + x2 1 + y2 ≤2(x + y)(1 −xy) ≤ 1 + x2 1 + y2 ⇐ ⇒ −(x + y)2 −(1 −xy)2 ≤2(x + y)(1 −xy) ≤(x + y)2 + (1 −xy)2 which is true by the well-know inequalitie a2 + b2 ± 2ab ≥0.
□ 136 Second Solution. (See , pag. 185, n.79) Let ⃗ a = 2x 1 + x2 , 1 −x2 1 + x2 , ⃗ b = 1 −y2 1 + y2 , 2y 1 + y2 Then it is easy to verify that |⃗ a| = |⃗ b| = 1. The Cauchy-Schwarz |⃗ a ·⃗ b| ≤|⃗ a| · |⃗ b| inequality implies that |⃗ a ·⃗ b| = 2 · x 1 −y2 + y 1 −x2 (1 + x2) (1 + y2) = 2 · (x + y)(1 −xy) (1 + x2) (1 + y2) ≤1 Dividing by 2, we get the result.
□ 147. (0 < x, y < 1) xy + yx > 1 Solution. (See , pag. 198, n. 66) First we prove the following lemma: Lemma. If u, x are real numbers such that u > 0, 0 < x < 1, we have (1 + u)x < 1 + ux Proof. Let f(u) = 1 + xu −(1 + u)x. We have f(0) = 0 and f is increasing in the interval ]0, 1[ because f ′(u) = x −x(1 + u)x−1 = x 1 − 1 (1 + u)1−x > x > 0 Thus f(u) > 0 for all x ∈R and the lemma is proved.
□ Now, the given inequality can be proved in the following way: Let x = 1 1+u, y = 1 1+v, u > 0, v > 0. Then, by the Lemma, we have xy = 1 (1 + u)y > 1 1 + uy = 1 + v 1 + u + v yx > 1 (1 + v)x > 1 1 + vx = 1 + u 1 + u + v Thus xy + yx > 1 + v 1 + u + v + 1 + u 1 + u + v = 1 + 1 1 + u + v > 1 and the inequality is proven.
□ 137 148. (x, y, z > 0) 3 √xyz + |x −y| + |y −z| + |z −x| 3 ≥x + y + z 3 Solution. (Ercole Suppa) We can assume WLOG that x ≤y ≤z. Let a, b, c be three real numbers such that x = a, y = a + b, z = a + b + c con a > 0, b, c ≥0. The required inequality is equivalent to: 3 p a(a + b)(a + b + c) + b + c + b + c 3 ≥3a + 2b + c 3 ⇐ ⇒ 3 3 p a(a + b)(a + b + c) ≥3a −c ⇐ ⇒ 54a2b + 54a2c + 27ab2 + 27abc −9ac2 + c3 ≥0 ⇐ ⇒ 54a2b + 27ab2 + 27abc + 54a2c + c3 −9ac2 ≥0 The above inequality is satisfied for all a > 0, b, c ≥0 since AM-GM inequality yields 54a2c + c3 ≥2 √ 54a2c4 = 6 √ 6ac2 ≥9ac2 □ 149. (a, b, c, x, y, z > 0) 3 p (a + x)(b + y)(c + z) ≥ 3 √ abc + 3 √xyz Solution. (Massimo Gobbino - Winter Campus 2006) By generalized Holder inequality we have n X i=1 aibici ≤ n X i=1 ap i ! 1 p n X i=1 bq i ! 1 q n X i=1 cr i ! 1 r which is true for all p, q, r ∈R such that 1 p + 1 q + 1 r = 1. After setting a1 = a 1 3 , b1 = b 1 3 , c1 = c 1 3 e a2 = x 1 3 , b2 = y 1 3 , c2 = z 1 3 we get: 3 √ abc + 3 √xyz = X aibici ≤ ≤ X a3 i 1 3 X b3 i 1 3 X c3 i 1 3 = = 3 p (a + x) (b + y) (c + z) □ 138 150. (x, y, z > 0) x x + p (x + y)(x + z) + y y + p (y + z)(y + x) + z z + p (z + x)(z + y) ≤1 Solution. (Walther Janous, see , pag. 49, problem 37) We have (x + y)(x + z) = xy + x2 + yz + xz ≥xy + 2x√yz + xz = √xy + √xz 2 Hence X x x + p (x + y)(x + z) ≤ X x x + √xy + √xz = = X √x √x + √y + √z = 1 and the inequality is proved.
□ 151. (x + y + z = 1, x, y, z > 0) x √1 −x + y √1 −y + z √1 −z ≥ r 3 2 First Solution. (Ercole Suppa) The function f(t) = t √1−t is convex on ]0, 1[ because f ′′(t) = 4 −t 4(1 −t) 5 2 ≥0 Then by Jensen inequality f(x) + f(y) + f(z) ≥3f x + y + z 3 = 3f 1 3 ⇐ ⇒ x √1 −x + y √1 −y + z √1 −z ≥ r 3 2 □ 139 Second Solution. (Ercole Suppa) After setting a = √1 −x, b = √1 −y, c = √1 −z, we have 0 < a, b, c < 1, a2 + b2 + c2 = 2 and the required inequality is equivalent to: 1 −a2 a + 1 −b2 b + 1 −b2 c ≥ r 3 2 ⇐ ⇒ 1 a + 1 b + 1 c ≥ r 3 2 + a + b + c (⋆) From Cauchy-Schwarz inequality we have 2 = a2 + b2 + c2 ≥(a + b + c)2 3 = ⇒ a + b + c ≤2 r 3 2 (1) From AM-HM inequality we have 1 a + 1 b + 1 c ≥ 9 a + b + c ≥9 2 r 2 3 = 3 r 3 2 (2) By adding (1) and (2) we get (⋆) and the result is proven.
□ Third Solution. (Campos - ML Forum) Assume WLOG that x ≥y ≥z.
Then 1 √1 −x ≥ 1 √1 −y ≥ 1 √1 −z and, from Chebyshev and Cauchy-Schwarz inequalities, we have X x √1 −x ≥ P x · P 1 √1−x 3 = = 1 3 · X 1 √1 −x = = 1 3 · 9 P √1 −x ≥ ≥ 3 p 3 · P(1 −x) = = 3 √ 6 = r 3 2 □ Remark.
The inequality can be generalized in the following way (India MO 1995): 140 If x1, x2, ..., xn are n real positive numbers such that x1 + x2 + x3 + ... + xn = 1 the following inequality holds x1 √1 −x1 + x2 √1 −x2 + .... + xn √1 −xn ≥ r n n −1 152. (a, b, c ∈R) p a2 + (1 −b)2 + p b2 + (1 −c)2 + p c2 + (1 −a)2 ≥3 √ 2 2 Solution. (Ercole Suppa) After setting a + b + c = t, from the Minkowski inequality we have: p a2 + (1 −b)2 + p b2 + (1 −c)2 + p c2 + (1 −a)2 ≥ ≥ p (a + b + c)2 + (3 −a −b −c)2 = = p t2 + (3 −t)3 ≥3 √ 2 2 The last step is true since p t2 + (3 −t)3 ≥3 √ 2 2 ⇐ ⇒ t2 + (3 −t)2 ≥9 2 ⇐ ⇒ (2t −3)2 ≥0 □ 153. (a, b, c > 0) p a2 −ab + b2 + p b2 −bc + c2 ≥ p a2 + ac + c2 First Solution. (Ercole Suppa) We have: p a2 −ab + b2 + p b2 −bc + c2 ≥ p a2 + ac + c2 ⇐ ⇒ a2 −ab + b2 + b2 −bc + c2 + 2 p (a2 −ab + b2)(b2 −bc + c2) ≥a2 + ac + c2 ⇐ ⇒ 2 p (a2 −ab + b2)(b2 −bc + c2) ≥ab + bc + ac −2b2 ⇐ ⇒ 4(a2 −ab + b2)(b2 −bc + c2) ≥ ab + bc + ac −2b22 ⇐ ⇒ 3(ab −ac + bc)2 ≥0 and we are done.
□ 141 Second Solution. (Albanian Eagle - ML Forum) This inequality has a nice geometric interpretation: let O, A, B, C be four points such that ∠AOB = ∠BOC = 60◦and OA = a, OB = b, OC = c then our inequality is just the triangle inequality for △ABC.
Remark. The idea of second solution can be used to show the following in-equality (given in a Singapore TST competition): Let a, b, c be real positive numbers. Show that c p a2 −ab + b2 + a p b2 −bc + c2 ≥b p a2 + ac + c2 Proof. By using the same notations of second solution, the required inequality is exactly the Tolomeo inequality applied to the quadrilateral OABC.
□ Third Solution. (Lovasz - ML Forum) We have p a2 −ab + b2+ p b2 −bc + c2 = v u u ta 2 −b 2 + a √ 3 2 !2 + v u u t b −c 2 2 + c √ 3 2 !2 In Cartesian Coordinate, let the two vectors a 2 −b, b √ 3 2 and b −c 2, c √ 3 2 .
Then ⃗ a +⃗ b = a −c 2 , (a + c) √ 3 2 !
.
Now use ∥a∥+ ∥b∥≥∥a + b∥, we get: p a2 −ab + b2 + p b2 −bc + c2 ≥ r (a −c)2 4 + 3(a + c)2 4 = = p a2 + ac + c2 □ 154. (xy + yz + zx = 1, x, y, z > 0) x 1 + x2 + y 1 + y2 + z 1 + z2 ≥2x(1 −x2) (1 + x2)2 + 2y(1 −y2) (1 + y2)2 + 2z(1 −z2) (1 + z2)2 142 Solution. (See , pag.185, n.89) After setting x = tan α/2, y = tan β/2, z = tan γ/2, by constraint xy+yz+zx = 1 follows that tan γ 2 = 1 −xy x + y = 1 −tan α 2 tan β 2 tan α 2 tan β 2 = 1 tan α+β 2 = = cot α + β 2 = tan π 2 −α + β 2 Thus α + β + γ = π, so we can assume that α, β, γ are the angles of a triangle.
The required inequality is equivalent to cos α sen α + cos β sen β + cos γ sen γ ≤sen α + sen β + sen γ 2 sen 2α + sen 2β + sen 2γ ≤sen α + sen β + sen γ (1) By sine law, using the common notations, we have sen α + sen β + sen γ = a + b + c 2R = 2s 2R = sr Rr = S rR (2) If x, y, z are the distances of circumcenter O fromi BC, CA, AB we have sen 2α + sen 2β + sen 2γ = 2 (sen α cos α + sen β cos β + sen γ cos γ) = = a cos α + b cos β + c cos γ R = = a · x R + b · y R + c · z R R = 2S R2 (3) From (2), (3) and Euler inequality R ≥2r we get sen α + sen β + sen γ sen 2α + sen 2β + sen 2γ = R 2r ≥1 and (1) is proven.
□ 155. (x, y, z ≥0) xyz ≥(y + z −x)(z + x −y)(x + y −z) Solution. (See , pag. 2) The inequality follows from Schur’s inequality because xyz −(y + z −x)(z + x −y)(x + y −z) = =x(x −y)(x −z) + y(y −z)(y −x) + z(z −x)(z −y) ≥0 The equality hols if and only if x = y = z or x = y and z = 0 and cyclic permutations.
□ 143 156. (a, b, c > 0) p ab(a + b) + p bc(b + c) + p ca(c + a) ≥ p 4abc + (a + b)(b + c)(c + a) Solution. (Ercole Suppa) Squaring both members with easy computations we get that the required inequality is equivalent to: a p bc(a + b)(a + c) + b p ac(b + a)(b + c) + c p ab(c + a)(c + b) ≥3abc which is true by AM-GM inequality: a p bc(a + b)(a + c) + b p ac(b + a)(b + c) + c p ab(c + a)(c + b) ≥ ≥3 3 p (abc)2(a + b)(b + c)(c + a) ≥ ≥3 3 p 8(abc)3 = 6abc ≥3abc □ 157. (Darij Grinberg) (x, y, z ≥0) p x (y + z) + p y (z + x) + p z (x + y) ·√x + y + z ≥2 p (y + z) (z + x) (x + y) First Solution. (Darij Grinberg - ML Forum) Consider the triangle with sides a = y+z, b = z+x, c = x+y and semiperimeter s = a+b+c 2 = x + y + z. Then, our inequality becomes p (s −a) a + p (s −b) b + p (s −c) c · √s ≥2 √ abc or r s (s −a) bc + r s (s −b) ca + r s (s −c) ab ≥2 If we call A, B, C the angles of our triangle, then this simplifies to cos A 2 + cos B 2 + cos C 2 ≥2 i. e.
sin 90◦−A 2 + sin 90◦−B 2 + sin 90◦−C 2 ≥2 But 90◦−A 2 , 90◦−B 2 and 90◦−C 2 are the angles of an acute triangle (as one can easily see); hence, we must show that if A, B, C are the angles of an acute triangle, then sin A + sin B + sin C ≥2 144 (Actually, for any non-degenerate triangle, sinA+sinB +sinC > 2, but I don’t want to exclude degenerate cases.) Here is an elegant proof of this inequality by Arthur Engel: Since triangle ABC is acute, we have A −B ≤C, and cos A−B 2 ≥cos C 2 , so that sin A + sin B = 2 sin A + B 2 cos A −B 2 = = 2 cos C 2 cos A −B 2 ≥ ≥2 cos2 C 2 = 1 + cos C and sin A + sin B + sin C ≥1 + cos C + sin C ≥2 Hereby, we have used the very simple inequality cos C + sin C ≥1 for any acute angle C.
(I admit that I did not find the proof while trying to solve the problem, but I rather constructed the problem while searching for a reasonable application of the sin A + sin B + sin C ≥2 inequality, but this doesn’t matter afterwards...) □ Second Solution. (Harazi - ML Forum) Take x + y + z = 1. Square the inequality X p x(1 −x) ≥2 · p (1 −x)(1 −y)(1 −z) and reduce it to X xy −2xyz ≤ X p xy(y + z)(z + x) But X xy −2xyz ≤ X xy and X p xy(x + z)(y + z) ≥ X xy + X z · √xy □ Third Solution. (Zhaobin, Darij Grinberg - ML Forum) We have s x (y + z) (x + y + z) (y + z) (z + x) (x + y) ≥x (y + z) (x + y + z) (y + z) (z + x) (x + y) 145 then we get: X s x (y + z) (x + y + z) (y + z) (z + x) (x + y) ≥ X x (y + z) (x + y + z) (y + z) (z + x) (x + y) = = 2(y + z) (z + x) (x + y) + xyz (y + z) (z + x) (x + y) ≥2 □ 158. (Darij Grinberg) (x, y, z > 0) √y + z x + √z + x y + √x + y z ≥ 4 (x + y + z) p (y + z) (z + x) (x + y) .
Solution. (See , pag. 18) By Cauchy, we have p (a + b)(a + c) ≥a + √ bc. Now, X √ b + c a ≥ 4(a + b + c) p (a + b)(b + c)(c + a) ⇐ ⇒ X b + c a p (a + b)(a + c) ≥4(a + b + c) Substituting our result from Cauchy, it would suffice to show X (b + c) √ bc a ≥2(a + b + c) Assume WLOG a ≥b ≥c, implying b+c ≤c+a ≤a+b and √ bc a ≤ √ca b ≤ √ ab c .
Hence, by Chebyshev and AM-GM, X (b + c) √ bc a ≥ (2(a + b + c)) √ bc a + √ca b + √ ab c ≥2(a + b + c) as desidered.
□ 159. (Darij Grinberg) (a, b, c > 0) a2 (b + c) (b2 + c2) (2a + b + c) + b2 (c + a) (c2 + a2) (2b + c + a) + c2 (a + b) (a2 + b2) (2c + a + b) > 2 3.
146 Solution. (Zhaobin - ML Forum) Just notice (b + c)(a2 + bc) = ba2 + ca2 + b2c + bc2 = b(a2 + c2) + c(a2 + b2) then let x = a b2 + c2 , y = b a2+c2 , z = c a2+b2 . The given inequality is equiva-lent to the well-know Nesbitt inequality.
x y + z + y x + z + z x + y ≥3 2 □ 160. (Darij Grinberg) (a, b, c > 0) a2 2a2 + (b + c)2 + b2 2b2 + (c + a)2 + c2 2c2 + (a + b)2 < 2 3.
Solution. (Darij Grinberg - ML Forum) The inequality in question, X a2 2a2 + (b + c)2 < 2 3 rewrites as 2 3 − X a2 2a2 + (b + c)2 > 0 But 2 3 − X a2 2a2 + (b + c)2 = =2 3 · X a a + b + c − X a2 2a2 + (b + c)2 = = X 2 3 · a a + b + c − a2 2a2 + (b + c)2 !
= = X a (b + c −a)2 + a (b + c) (b + c −a) 3 (a + b + c) 2a2 + (b + c)2 = = X a (b + c −a)2 3 (a + b + c) 2a2 + (b + c)2 + X a (b + c) (b + c −a) 3 (a + b + c) 2a2 + (b + c)2 Now, it is obvious that X a (b + c −a)2 3 (a + b + c) 2a2 + (b + c)2 ≥0 147 What remains to be proven is the inequality X a (b + c) (b + c −a) 3 (a + b + c) 2a2 + (b + c)2 > 0 which simplifies to X a (b + c) (b + c −a) 2a2 + (b + c)2 > 0 Now, X a (b + c) (b + c −a) 2a2 + (b + c)2 = X ab (b + c −a) + ca (b + c −a) 2a2 + (b + c)2 = = X ab (b + c −a) 2a2 + (b + c)2 + X ca (b + c −a) 2a2 + (b + c)2 = = X bc (c + a −b) 2b2 + (c + a)2 + X bc (a + b −c) 2c2 + (a + b)2 = = X bc c + a −b 2b2 + (c + a)2 + a + b −c 2c2 + (a + b)2 !
= = X bc a (a + b + c)2 + a2 + 2bc + (b + c) (b −c)2 2b2 + (c + a)2 2c2 + (a + b)2 > 0 □ 161. (Vasile Cirtoaje) (a, b, c ∈R) (a2 + b2 + c2)2 ≥3(a3b + b3c + c3a) Solution. (Darij Grinberg - ML Forum) Vasile Cartoaje established his inequality a2 + b2 + c22 ≥3 a3b + b3c + c3a using the identity 4 a2 + b2 + c2 −(bc + ca + ab) a2 + b2 + c22 −3 a3b + b3c + c3a = = a3 + b3 + c3 −5 a2b + b2c + c2a + 4 b2a + c2b + a2c 2 + + 3 a3 + b3 + c3 − a2b + b2c + c2a −2 b2a + c2b + a2c + 6abc 2 148 Actually, this may look a miracle, but there is a very natural way to find this identity. In fact, we consider the function g (a, b, c) = a2 + b2 + c22 −3 a3b + b3c + c3a over all triples (a, b, c) ∈R3.
We want to show that this function satisfies g (a, b, c) ≥0 for any three reals a, b, c. Well, fix a triple (a, b, c) and translate it by some real number d; in other words, consider the triple (a+d, b+d, c+d). For which d ∈R will the value g (a + d, b + d, c + d) be minimal? Well, minimizing g (a + d, b + d, c + d) is equivalent to minimizing g (a + d, b + d, c + d)−g (a, b, c) (since (a, b, c) is fixed), but g (a + d, b + d, c + d) −g (a, b, c) = =d2 a2 + b2 + c2 −(bc + ca + ab) + + d a3 + b3 + c3 −5 a2b + b2c + c2a + 4 b2a + c2b + a2c so that we have to minimize a quadratic function, what is canonical, and it comes out that the minimum is achieved for d = − a3 + b3 + c3 −5 a2b + b2c + c2a + 4 b2a + c2b + a2c 2 ((a2 + b2 + c2) −(bc + ca + ab)) So this is the value of d such that g (a + d, b + d, c + d) is minimal. Hence, for this value of d, we have g (a, b, c) ≥g (a + d, b + d, c + d). Thus, in order to prove that g (a, b, c) ≥0, it will be enough to show that g (a + d, b + d, c + d) ≥0. But, armed with the formula d = − a3 + b3 + c3 −5 a2b + b2c + c2a + 4 b2a + c2b + a2c 2 ((a2 + b2 + c2) −(bc + ca + ab)) and with a computer algebra system or a sufficient patience, we find that g (a + d, b + d, c + d) = =3 a3 + b3 + c3 − a2b + b2c + c2a −2 b2a + c2b + a2c + 6abc 2 4 ((a2 + b2 + c2) −(bc + ca + ab)) what is incontestably ≥0. So we have proven the inequality. Now, writing g (a, b, c) = g (a + d, b + d, c + d) −(g (a + d, b + d, c + d) −g (a, b, c)) and performing the necessary calculations, we arrive at Vasc’s mystic identity.
□ 149 A Classical Inequalities Theorem 1. (AM-GM inequality) Let a1, · · · , an be positive real numbers. Then, we have a1 + · · · + an n ≥ n √a1 · · · an.
Theorem 2. (Weighted AM-GM inequality) Let λ1, · · · , λn real positive numbers with λ1+· · ·+λn = 1. For all x1, · · · , xn > 0, we have λ1 · x1 + · · · + λn · xn ≥x1 λ1 · · · xn λn.
Theorem 3. (GM-HM inequality) Let a1, · · · , an be positive real numbers. Then, we have n √a1 · · · an ≥ n 1 a1 + 1 a2 + · · · + 1 an Theorem 4. (QM-AM inequality) Let a1, · · · , an be positive real numbers. Then, we have r a2 1 + a2 2 + · · · + a2 n n ≥a1 + · · · + an n Theorem 5. (Power Mean inequality) Let x1, · · · , xn > 0. The power mean of order p is defined by M0 (x1, x2, . . . , xn) = n √x1 · · · xn , Mp (x1, x2, . . . , xn) = xp 1 + · · · + xnp n 1 p (p ̸= 0).
Then the function Mp (x1, x2, . . . , xn) : R →R is continuous and monotone increasing.
Theorem 6. (Rearrangement inequality) Let x1 ≥· · · ≥xn and y1 ≥· · · ≥yn be real numbers. For any permutation σ of {1, . . . , n}, we have n X i=1 xiyi ≥ n X i=1 xiyσ(i) ≥ n X i=1 xiyn+1−i.
150 Theorem 7. (The Cauchy2-Schwarz3-Bunyakovsky4 inequality) Let a1, · · · , an, b1, · · · , bn be real numbers. Then, (a1 2 + · · · + an 2)(b1 2 + · · · + bn 2) ≥(a1b1 + · · · + anbn)2.
Remark. This inequality apparently was firstly mentioned in a work of A.L.
Cauchy in 1821. The integral form was obtained in 1859 by V.Y. Bunyakovsky.
The corresponding version for inner-product spaces obtained by H.A. Schwartz in 1885 is mainly known as Schwarz’s inequality. In light of the clear historical precedence of Bunyakovsky’s work over that of Schwartz, the common practice of referring to this inequality as CS-inequality may seem unfair. Nevertheless in a lot of modern books the inequality is named CSB-inequality so that both Bunyakovsky and Schwartz appear in the name of this fundamental inequality.
By setting ai = xi √yi and bi = √yi the CSB inequality takes the following form Theorem 8. (Cauchy’s inequality in Engel’s form) Let x1, · · · , xn, y1, · · · , yn be positive real numbers. Then, x2 1 y1 + x2 2 y2 + · · · + x2 n yn ≥(x1 + x2 + · · · + xn)2 y1 + y2 + · · · + yn Theorem 9. (Chebyshev’s inequality5) Let x1 ≥· · · ≥xn and y1 ≥· · · ≥yn be real numbers. We have x1y1 + · · · + xnyn n ≥ x1 + · · · + xn n y1 + · · · + yn n .
Theorem 10. (H¨ older’s inequality6) Let x1, · · · , xn, y1, · · · , yn be positive real numbers.
Suppose that p > 1 and q > 1 satisfy 1 p + 1 q = 1. Then, we have n X i=1 xiyi ≤ n X i=1 xi p ! 1 p n X i=1 yi q ! 1 q 2Louis Augustin Cauchy (1789-1857), french mathematician 3Hermann Amandus Schwarz (1843-1921), german mathematician 4Viktor Yakovlevich Bunyakovsky (1804-1889), russian mathematician 5Pafnuty Lvovich Chebyshev (1821-1894), russian mathematician.
6Otto Ludwig H¨ older (1859-1937), german mathematician 151 Theorem 11. (Minkowski’s inequality7) If x1, · · · , xn, y1, · · · , yn > 0 and p > 1, then n X i=1 xi p ! 1 p + n X i=1 yi p ! 1 p ≥ n X i=1 (xi + yi)p ! 1 p Definition 1. (Convex functions.) We say that a function f(x) is convex on a segment [a, b] if for all x1, x2 ∈[a, b] f x1 + x2 2 ≤f (x1) + f (x2) 2 Theorem 12. (Jensen’s inequality8) Let n ≥2 and λ1, . . . , λn be nonnegative real numbers such that λ1+· · ·+λn = 1.
If f(x) is convex on [a, b] then f (λ1x1 + · · · + λnxn) ≤λ1f (x1) + · · · + λnxn for all x1, . . . , xn ∈[a, b].
Definition 2. (Majorization relation for finite sequences) Let a = (a1, a2, . . . , an) and b = (b1, b2, . . . , bn) be two (finite) sequences of real numbers such that a1 ≥a2 ≥· · · ≥an and b1 ≥b2 ≥· · · ≥bn. We say that the sequence a majorizes the sequence b and we write a ≻b or b ≺a if the following two conditions are satisfyied (i) a1 + a2 + · · · + ak ≥b1 + b2 + · · · + bk, for all k, 1 ≤k ≤n −1; (ii) a1 + a2 + · · · + an = b1 + b2 + · · · + bn.
Theorem 13. (Majorization inequality | Karamata’s inequality9) Let f : [a, b] − →R be a convex function. Suppose that (x1, · · · , xn) majorizes (y1, · · · , yn), where x1, · · · , xn, y1, · · · , yn ∈[a, b]. Then, we obtain f(x1) + · · · + f(xn) ≥f(y1) + · · · + f(yn).
7Hermann Minkowski (1864-1909), german mathematician.
8Johan Ludwig William Valdemar Jensen (1859-1925), danish mathematician.
9Jovan Karamata (1902-2967), serbian mathematician.
152 Theorem 14. (Muirhead’s inequality10 | Bunching Principle ) If a = (a1, a2, . . . , an) and b = (b1, b2, . . . , bn) are two nonincreasing sequences of nonnegative real numbers such that a majorizes b, then we have X sym xa1 1 · · · xan n ≥ X sym xb1 1 · · · xbn n where the sums are taken over all n! permutations of variables x1, x2, . . . , xn.
Theorem 15. (Schur’s inequality11 ) Let x, y, z be nonnegative real numbers. For any r > 0, we have X cyc xr(x −y)(x −z) ≥0.
Remark. The case r = 1 of Schur’s inequality is X sym x3 −2x2y + xyz ≥0 By espanding both the sides and rearranging terms, each of following inequalities is equivalent to the r = 1 case of Schur’s inequality • x3 + y3 + z3 + 3xyz ≥xy(x + y) + yz(y + z) + zx(z + x) • xyx ≥(x + y −z)(y + z −x)(z + x −y) • (x + y + z)3 + 9xyz ≥4(x + y + z)(xy + yz + zx) Theorem 16. (Bernoulli’s inequality12) For all r ≥1 and x ≥−1, we have (1 + x)r ≥1 + rx.
Definition 3. (Symmetric Means) For given arbitrary real numbers x1, · · · , xn, the coefficient of tn−i in the poly-nomial (t + x1) · · · (t + xn) is called the i-th elementary symmetric function σi.
This means that (t + x1) · · · (t + xn) = σ0tn + σ1tn−1 + · · · + σn−1t + σn.
For i ∈{0, 1, · · · , n}, the i-th elementary symmetric mean Si is defined by Si = σi n i .
10Robert Muirhead (1860-1941), english matematician.
11Issai Schur (1875-1941), was Jewish a mathematician who worked in Germany for most of his life. He considered himself German rather than Jewish, even though he had been born in the Russian Empire in what is now Belarus, and brought up partly in Latvia.
12Jacob Bernouilli (1654-1705), swiss mathematician founded this inequality in 1689. How-ever the same result was exploited in 1670 by the english mathematician Isaac Barrow.
153 Theorem 17. (Newton’s inequality13) Let x1, . . . , xn > 0. For i ∈{1, · · · , n}, we have S2 i ≥Si−1 · Si+1 Theorem 18. (Maclaurin’s inequality14) Let x1, . . . , xn > 0. For i ∈{1, · · · , n}, we have S1 ≥ p S2 ≥ 3 p S3 ≥· · · ≥ n p Sn 13Sir Isaac Newton (1643-1727), was the greatest English mathematician of his generation.
He laid the foundation for differential and integral calculus. His work on optics and gravitation make him one of the greatest scientists the world has known.
14Colin Maclaurin (1698-1746), Scottish mathematican.
154 B Bibliography and Web Resources References M. Aassila, 300 d´ efis math´ ematiques, Ellipses (2001) R. Alekseyev, L. Kurlyandchik, The sum of minima and the minima of the sums, Quantum, January (2001)34-36 T. Andreescu, E. Bogdan, Mathematical Olympiad Treasures, Birkhauser, Boston (2004) T. Andreescu, V. Cˆ ırtoaje, G. Dospinescu, M. Lascu, Old and New Inequal-ities, GIL Publishing House, Zalau, Romania (2004) T. Andreescu, K. Kedlaya, Mathematical Contests 1995-1996, American Mathematics Competitions (1997) T. Andreescu, K. Kedlaya, Mathematical Contests 1996-1997, American Mathematics Competitions (1998) T. Andreescu, Z. Feng, Mathematical Olympiads From Around the World 1998-1999, Mathematical Association of America (2000) T. Andreescu, Z. Feng, Mathematical Olympiads From Around the World 1999-2000, Mathematical Association of America (2002) T. Andreescu, Z. Feng, P.S. Loh, USA and International Mathematical Olympiads 2001, Mathematical Association of America (2002) T. Andreescu, Z. Feng, P.S. Loh, USA and International Mathematical Olympiads 2003, Mathematical Association of America (2003) T. Andreescu, Z. Feng, P.S. Loh, USA and International Mathematical Olympiads 2004, Mathematical Association of America (2004) Art of Problem Solving, I. Boreico, An original method of proving inequalities, Mathematical Re-flections N.3(2006), I. Boreico, I. Borsenco A note on the breaking point of a simple inequality, Mathematical Reflections N.5(2006), P. Bornsztein, In´ egalit´ es Classiques, O. Bottema, R. ˜ Z. Djordjevi´ c, R. R. Jani´ c, D. S. Mitrinovi´ c, P. M. Vasi´ c, Geometric Inequalities, Wolters-NoordhoffPublishing, Groningen 1969 155 V. Q. B. Can On a class of three-variable inequalities, Mathematical Re-flections N.2(2007), E. Carneiro As desigualdades do Rearranjo e Chebychev, (2004) www.ma.utexas.edu/users/ecarneiro/RearranjoChebychev.pdf V. Cˆ ırtoaje, Algebraic Inequalities, GIL Publishing House, Zalau, Romania (2006) M.A.A. Cohen, R.V. Milet Contas com desigualdades, Eureka! N.23 (2006) B.N.T.Cong, N.V. Tuan, N.T. Kien The SOS-Schur method, Mathematical Reflections N.5(2007), D. A. Holton, Inequalities, University of Otago, Problem Solving Series, Booklet No.12 (1990) D. Djuki´ c, V. Jankovi´ c, I. Mati´ c, N. Petrovi´ c, The IMO Compendium, Springer-Verlag, New York (2006) S. Dvoryaninov, E. Yasinovyi Obtaining symmetric inequalities, Quantum, november(1998)44-48 A. Engel, Problem-Solving Startegies, Springer-Verlag, New York (1998) R.T. Fontales, Trigonometria e desigualdades em problemas de olimpiadas, Eureka! N.11 (2001), G. Hardy, J.E. Littlewood, G. Polya, Inequalities, Cambridge University Press, Cambridge (1999) L. K. Hin, Muirhead Inequality, Mathematical Excalibur, Vol.11, N.1 (2006) excalibur/ D. Hrimiuc, The Rearrangement Inequality, Π in the Sky, N. 2, december (2000), D. Hrimiuc, Inequalities for Convex Functions (Parti I), Π in the Sky, N.
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K. S. Kedlaya, A < B, M. S. Klamikin USA Mathematical Olympiads 1972-1986, Mathematical Association of America (1988) P.P. Korovkin, Media aritmetica, media geometrica y otras medias P.P. Korovkin, Desigualdad de Bernoulli P.P. Korovkin, Medias potenciales P.P. Korovkin, Parte entera del numero J. Herman, R. Kuˇ cera, J. ˇ Simˇ sa, Equations and Inequalities, Springer-Verlag, New York (2000) M. E. Kuczma, 144 problems of the austrian-polish matematics competition 1978-1993, The Academic Distribution Center, Freeland, Maryland (1994) C. Lupu, C. Pohoata About a nice inequality, Mathematical Reflections N.1(2007), S. Malikic, Inequalities with product condition, Mathematical Excalibur, Vol.12, N.4 (2007), excalibur/ 157 I. Matic, Classical inequalities, The Imo Compendium, Olympiad Training Materials (2007), MathLinks, T. J. Mildorf, Olympiad Inequalities, D. S. Mitrinovi´ c, P.M. Vasi´ c Analytic Inequalities, Springer-Verlag, New York (1970) J. H. Nieto, Desigualdades, A. C. M. Neto, Desigualdades Elementares, Eureka! N.5 (1999) L. Pinter, I. Khegedysh Ordered sets, Quantum, July(1998)44-45 I. Reiman, International Mathematical Olympiad 1959-1999, Anthem Press Publishing Company, Wimbledon (2001) N. Sato, Tips on Inequalities, Crux Matematicorum, (1998)161-167 S. Savchev, T. Andreescu, Matematical Miniatures, Mathematical Associ-ation of America, (2003) John Scholes WEB site, A. Slinko, Algebraic inequalities, www.nzamt.org.nz/mapbooklet/Inequalities.pdf J. M. Steele, The Cauchy Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities, Mathematical Association of America, (2004) T. B. Soulami, Les olympiades de math´ ematiques, Ellipses Editions, (1999) P. J. Taylor, Tournament of towns, 1993-1997, Australian Mathematics Trust Pubblication, (1998) P. V. Tuan, Square it !, Mathematical Excalibur, Vol.12, N.5 (2007) excalibur/ P.V. Tuan, T.V. Hung Proving inequalities using linear functions, Mathe-matical Reflections N.4(2006), V. Verdiyan, D.C. Salas, Simple trigonometric substitutions with broad re-sults, Mathematical Reflections N.6(2007), G. West, Inequalities for the Olympiad Enthusiast, The South African Mathematical Society, (1996) 158 |
5323 | https://xiucheng.org/2019/08/01/diffeq-intro.html | Introduction to Differential Equations
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Introduction to Differential Equations
Aug 1, 2019 by Xiucheng Lidifferential-equationedx
The elementary differential equations from MIT edx course.
Unit 1 Modeling and First Order ODEs
1 Introduction to Differential Equations and Modeling
2 Solving First Oder ODEs
Standard linear form
Conduction and diffusion
Solutions to homogeneous linear equations
Solutions to inhomogeneous linear equations
Superposition principle
Existence and uniqueness theorem for a linear ODE
Newton’s law of cooling down
Unit 2: Complex Exponentials and ODEs
3 Introduction to Complex Numbers
Polar form
Euler’s formula
Exponential law
Operations in polar form
4 Complex Exponential Function
5 Homogeneous 2nd Order Linear ODEs with Constant Coefficients
Unit 3 Damped Oscillations
6 Sinusoidal Functions
7 Damped Harmonic Oscillators
Case 1: underdamped, b 2<4 m k
Case 2: overdamped, b 2>4 m k
Case 3: critically damped, b 2=4 m k
Comparing damping qualitatively
Real life application: the rocking motion of a boat
8 Higher Order Linear ODEs
Unit 4 Exponential Response and Resonance
9 Operator Notation
10 Complex Replacement, Gain and Phase Lag, Stability
11 Resonance, Frequency Response, RLC circuits
Unit 5 Nonlinear DEs
12 Nonlinear DEs: Graphical methods
13 Autonomous equations
14 Numerical Methods
Unit 1 Modeling and First Order ODEs
1 Introduction to Differential Equations and Modeling
Identify relevant quantities, both known and unknown, and give them symbols. Find the units for each.
Identify the independent variable(s). The other quantities will be functions of them, or constants. Often time is the only independent variable.
Write down equations expressing how the functions change in response to small changes in the independent variable(s). Also write down any “laws of nature” relating the variables. As a check, make sure that all summands in an equation have the same units.
Often simplifying assumptions need to be made; the challenge is to simplify the equations so that they can be solved but so that they still describe the real-world system well.
2 Solving First Oder ODEs
Standard linear form
Every first order linear ODE can be written in standard linear form as follows:
Homogeneous:y˙+p(t)y=0 Inhomogeneous:y˙+p(t)y=q(t)
Conduction and diffusion
We have a model called Newton Cooling law which gives us a differential equation model for temperature conduction or concentration diffusion as a first order ODE:
y˙+k y=k x.
where x is the external temperature or concentration and y is the temperature or concentration of interest.
Solutions to homogeneous linear equations
All first order homogeneous equations can be solved using seperation of variables.
Solutions to inhomogeneous linear equations
Variation of parameters or Integrating factors.
Integrating factors. we start with a first order, linear, inhomogeneous ODE:
y˙+p(t)y=q(t)
To find an integrating factor:
Find an antiderivative P(t) of p(t). The integrating factor is e P(t).
Multiply both sides of the ODE by the integrating factor e P(t).
e P(t)y˙+e P(t)p(t)y=q(t)e P(t)
We do this mulitplication because it is now possible to express the left side as the derivative of something:
e P(t)y˙(t)+e P(t)p(t)y(t)=d d t(e P(t)y(t)).
Now we can carry out the integration.
d d t(e P y)=q e P e P y=∫q e P d t y=e−P∫q e P d t.
The indefinite integral ∫q(t)e P(t)d t represents a family of solutions because there is a constant of integration. If we fix one antiderivative, say R(t), then the others are R(t)+C for a constant C. So the general solution is
y=R(t)e−P(t)+C e−P(t).
Note that e−P is the reciprocal of a solution to the homogeneous equation.
Superposition principle
Multiplying a solution to p n(t)y(n)+⋯+p 0(t)y=a q(t)by a number a gives a solution to p n(t)y(n)+⋯+p 0(t)y=a q(t).
Adding a solution of p n(t)y(n)+⋯+p 0(t)y=a q(t)by a number a to a solution of p n(t)y(n)+⋯+p 0(t)y=q 2(t)gives a solution of p n(t)y(n)+⋯+p 0(t)y=q 1(t)+q 2(t).
Existence and uniqueness theorem for a linear ODE
Let p(t) and q(t) be continuous function on an open interval I. Let a∈I, and let b be a given number. Then there exists a unique solution to the first order linear ODE
y˙+p(t)y=q(t)
satisfying the initial condition
y(a)=b.
Newton’s law of cooling down
d T d t=k(T e−T)
k>0 is the conductivity and T e is the external temperature. The solution is
T=T e+(T 0−T e)e−k t.
The general solution for the homogeneous equation d T/d t=k T is
T=c e k t.
Unit 2: Complex Exponentials and ODEs
3 Introduction to Complex Numbers
Polar form
Fig.
To specify a preferred polar form, we would have to restrict the range for θ to some interval of width 2 π. The most common choice is to requrie −π<θ≤π. This special θ is called the principal value of the argument.
Given a complex number a+b i with a,b≠0, how do we find the principal value of the argument θ∈(−π,π]? There are two angles in (−π,π] having tangent equal to b/a, corresponding to opposite directions. By definition, tan−1(b/a) is the one in (−π/2,π/2). If a+b i is in the right half plane, then θ=tan−1(b/a) works. Otherwise, it is necesary to adjust tan−1(b/a) by adding or subtracting π.
Fig. Principal value of arctan
Euler’s formula
e i θ=cos(θ)+i sin(θ).For any constant a,e a t is the solution of x˙=a x,x(0)=1.
For z=r e i θ∈C,
adding an angle θ 1>0 to θ is equivalent to rotating θ by θ 1 counterclockwise;
subtracting an angle θ 1>0 from θ is equivalent to rotating θ by θ 1 clockwise.
Exponential law
e i t 1 e i t 2=(cost 1+i sint 1)(cost 2+i sint 2)=(cost 1 cost 2−sint 1 sint 2)⏟cos(t 1+t 2)+i(sint 1 cost 2+sint 2 cost 1)⏟sin(t 1+t 2)=cos(t 1+t 2)+i sin(t 1+t 2)=e i(t 1+t 2)
e i t satisfies the initial value problem y′=i y, y(0)=1.
d d t e i t=i e i t.
Operations in polar form
Good for multiplication. We can write complex numbers z=a+b i in polar form as r e i θ=r cos(θ)+i r sin(θ).
r 1 e i θ 1=r 2 e i θ 2 if and only if r 1=r 2 and θ 1=θ 2+2 π k for some integer k.
(This assumes that r 1 and r 2 are nonnegative real numbers, and that θ 1 and θ 2 are real numbers.)
Some arithmetic operations on complex numbers are easy in polar form:
multiplication(r 1 e i θ 1)(r 2 e i θ 2)=r 1 r 2 e i(θ 1+θ 2)reciprocal 1 r e i θ=1 r e−i θ division r 1 e i θ 1 r 2 e i θ 2=r 1 r 2 e i(θ 1−θ 2)n th power(r e i θ)n=(r e i θ)n complex conjugation r e i θ―=r e−i θ.
Taking absolute values gives identies:
|z 1 z 2|=|z 1||z 2|,|1 z|=1|z|,|z 1 z 2|=|z 1||z 2|,|z n|=|z|n,|z―|=|z|.
How do you trap a lion?
Fig. Trap a Lion
4 Complex Exponential Function
Fundamental Theorem of algebra. Every degree n complex polynomial f(z) has exactly n complex roots, if counted with multiplicity.
z n=1
since 1=e i(0+2 k π) for k∈Z, then z=e i(0+2 k π)/n.
The n th roots of unity are the numbers e i(2 π k n) for k=0,1,2,…,n−1. Taking k=1 gives the number ζ≜e 2 π i/n. In terms of ζ, the complete list of n th roots of unity is
1,ζ,ζ 2,…,ζ n−1
Geometrically, they all lie on the unit circle in the complex plane. The roots are evenly spaced around the unit circle, starting with the root $z = 1$, and the angle between two consecutive roots is 2 π/n. These facts are illustrated for the case n=6 in the figure below.
Fig. Roots of Unity
Analogously, we can find the n th roots of i. As i=e i(π/2+2 k π) for k∈Z, so its n th roots have the form
e i(π/2+2 k π)/n,k=0,1,…,n−1.Integration with complex number trick.
∫e 2 t cos(3 t)d t=∫Re(e 2 t e i 3 t)d t=Re(∫e 2 t e i 3 t d t)
The second equality holds since only the real parts contribute to the real part of the sum in the summation of a series of complex numbers.
∫e 2 t e i 3 t d t=1 13 e 2 t+(3 t−θ)i+C=1 13 e 2 t(2 cos(3 t)+3 sin(3 t))+C
where θ=arctan(3/2).
5 Homogeneous 2nd Order Linear ODEs with Constant Coefficients
Spring mass dashpot system. A mass sits on a cart is attched to a spring attached to a wall. The mass is also attached to a dashpot, a damping device. Finding the differential equation for the position of the mass.
Fig. String Mass Dashpot
F spring is a function of the position x, and has opposite sign of x;
F dashpot is a function of velocity x′, and has opposite sign of x′.
F spring=−k x,F dashpot=−b x′,
Newton’s second law F=m x″ gives
m x¨=−k x−b x˙,
a second linear ODE, which we usually write as
m x¨+b x˙+k x=0.
Solution: Try x=e r t, where r is a constant to be determined. Substituting it to the above equation,
m r 2 e r t+b r e r t+k e r t=0(m r 2+b r+k)e r t=0
The polynomial p(r)=m r 2+b r+k is known as the characteristic polynomial of the given DE, and the equation p(r)=0 is the characteristic equation.
Suppose that the characteristic equation has two real roots r 1,r 2. The solution of the original DE will be
x(t)=c 1 e−r 1 t+c 2 e−r 2 t.
Suppose that the equation y¨+A y˙+B y=0 where A,B are real, has characteristic roots a±i b. Then the general real solution is
y(t)=c 1 e(a+i b)t+c 2 e(a−i b)t or y(t)=c 1 e a t cos(b t)+c 2 e a t sin(b t).
This is due to if y(t)=u(t)+i v(t) is a solution to a second order homogeneous linear DE with real coefficients: y¨+A y˙+B y=0. Then
(u¨+A u˙+B u)+i(v¨+A v˙+B v)=0
which implies both u and v are solutions to the original DE.
In particular, the general solution represented by m,b,k is
x(t)=e(−b/2 m)t(c 1 cos(ω d t)+c 2 sin(ω d t))c 1,c 2∈R,
where ω d=k m−b 2 4 m 2.
Unit 3 Damped Oscillations
6 Sinusoidal Functions
Recall the spring-mass-dashpot system with no external force, when there are two distinct characteristic complex roots, the general real solution:
x(t)=e(−b/2 m)t(c 1 cos(ω d t)+c 2 sin(ω d t))c 1,c 2∈R,
where ω d=k m−b 2 4 m 2 is called the damped frequency of the system.
Fig.
If the system has no damping, the DE reduces to
m x¨+k x=0 m,k>0,
and the general solution reduces to
x(t)=c 1 cos(ω n t)+c 2 sin(ω n t)c 1,c 2∈R,
where ω n=k m is called the natural frequency of the system. Note that ω n>ω d, this coincides with our intuition that the cart should oscillate more frequently when no damping.
a cos(θ)+b sin(θ)⏟rectangular form=A cos(θ−ϕ)⏟polar form a,b,ϕ∈R,A≥0∈R,=A cosϕ⏟a cosθ+A sinϕ⏟b sinθ by the trig sum formula
where A,ϕ in terms of a,b are given implicitly by the following diagram:
Fig.
As usual, the angle ϕ is well-defined only up to addition of integer multiples of 2 π.
Proof:
Using complex numbers. Let us write c=a−b i, as a consequence,
Re(c(cos(θ)+i sin(θ)))=a cos(θ)+b sin(θ).
On the other hand, c=A e−i ϕ and cos(θ)+i sin(θ)=e i θ,
Re(c(cos(θ)+i sin(θ)))=Re(A e−i ϕ e i θ)=A cos(θ−ϕ).
2. The geometric perspective. Let us consider a cos(θ)+b sin(θ) as the inner product of two vectors x=[a,b] and y=[cos(θ),sin(θ)], since
⟨x,y⟩=‖x‖2‖y‖2 cos(angle between x and y)=A cos(ϕ−θ).
Graphing sinusoidal functions
Fig. Sinusoidal Functions
f(t)=A cos(ω t−ϕ)
A, amplitude.
P, period, the time for one complete oscillation.
ω=2 π P (radians/second) is called angular frequency or circular frequency; there is also frequency ν=1 P (Hertz = cycles per second), the number of complete oscillations per second.
ϕ (radians), phase lag or phase shift.
7 Damped Harmonic Oscillators
no damping. DE: m x¨+k x=0(m,k>0).
This system or any other system governed by the same DE, is called simple harmonic oscillator. The angular frequency ω n=k m is called natural frequency or resonant frequency of the oscillator.
damping. DE: m x¨+b x˙+k x=0(m,c,k>0). Now we rewrite it as
x¨+b m x˙+k m x=0 y¨+2 p y˙+ω n 2=0
Case 1: underdamped, b 2<4 m k
The real part is −p=−b 2 m. The imaginary part is either positive or negative of the damped frequency ω d,
ω d≜4 m k−b 2 2 m=ω n 2−p 2,where ω n=k m is the natural frequency.
Sumary of results:
Roots:−p±i ω d Basis of solution space:e(−p+i ω d)t,e(−p−i ω d)t Real valued basis:e−p t cos(ω d t),e−p t sin(ω d t)General real solution:e−p t(a cos(ω d t)+b sin(ω d t)),where a,b are real constants=A e−p t cos(ω d t−ϕ)
Note that ω d depends on the spring (k) and dashpot (b), whereas A,ϕ only depends on the inital conditions.
Case 2: overdamped, b 2>4 m k
The roots −b±b 2−4 m k 2 m are real and distinct. Both roots are negative, since b 2−4 m k−s 2. General solution:
a e−s 1 t+b e−s 2 t,where a,b are real constants.
All solutions tend to 0 as t→+∞. The first term eventually controls the rate of return to equilibrium.
Case 3: critically damped, b 2=4 m k
Basis of solution space:e−p t,t e−p t General real solution:e−p t(a+b t),where a,b are real constants
Since the second order equation must have two distinct special solutions, we can assume its solution takes the form e−p t u and apply variation of parameters to get t e−p t.
Comparing damping qualitatively
Consier the DE x¨+b x˙+x=0 with initial conditions x(0)=1, x˙(0)=0. We will consider the three cases b=1,2,3,
x 1(t)=e−t/2(cost 3 2+1 3 sint 3 2),x 2(t)=e−t(1+t),x 3(t)=(1 2+3 2 5)e−3+5 2 t+(1 2−3 2 5)e−3−5 2 t.
shown in the below figure, from which we can conclude that the smaller b the faster the mass moves to the equilibrium position at the first round.
Fig.
Real life application: the rocking motion of a boat
Fig. Rocking Motion of a Boat
The rotational motion of a boat rocking sideways close to its flat equilibrium position is described by the linear approximation of the rotational version of Newton’s second law:
I θ¨=τ(θ)where τ(θ)is the torque
where I>0 is the moment of inertia of the boat, τ is the torque exerted on the boat and when θ is small τ can be approximated as a linear function of θ
τ(θ)=−k θ where k>0,θ≪1.
I and k are constants determined by the geometry and distribution of mass of the boat. The differential equation describes the motion is
I θ¨+k θ=0
Since I>0,k>0 the boat will do harmonic oscillation. But if the boat is not well designed, k can be negative in which case the differential equation solution becomes (assume θ(0)=0,θ˙(0)=c)
θ(t)=c 2 I k e k/I t−c 2 I k e−k/I t,
from which we can see that the boat will capsize.
The centroid of a shape on the plane is the average position of all points of the shape.
8 Higher Order Linear ODEs
Given
a n y(n)+⋯+a 1 y˙+a 0 y=0,
where a i are real constants, do the following:
Write down the characteristic equation
a n r n+⋯+a 1 r+a 0=0.
2. Factor P(r) as
P(r)=a n(r−r 1)(r−r 2)⋯(r−r n)
There are guaranteed to be n (complex) roots counted with multiplicity by the fundamental theorem of algebra.
If r 1,…,r n are distinct, then the functions e r 1 t,…,e r n t form a basis for the vector space of solutions to the constant coefficient ODE, the general solution is
c 1 e r 1 t+⋯+c n e r n t.
Note that complex roots always appear in pairs of conjugates, and if some of the roots are complex, the coefficients c i will have to be complex as well.
If a particular root r is repeated m times, then
replace e r t,e r t,e r t,…,e r t⏞m copies by e r t,t e r t,t 2 e r t,…,t m−1 e r t.
Example 1: Find a basis of solutions to y(3)+3 y¨+9 y˙−13 y=0 consisting of real-valued functions.
The characteristic polynomial is P(r):=r 3+3 r 2+9 r−13=(r−1)(r 2+4 r+13). Rewriting the second factor as (r+2)2+9, shows that the roots of P(r) are 1,−2+3 i,−2−3 i. Thus e t,e(−2+3 i)t,e(−2−3 i)t form a basis of solutions. But the last two are not real-valued, instead
e(−2+3 i)t=e−2 t cos(3 t)+i e−2 t sin(3 t).
Thus
e t,e−2 t cos(3 t),e−2 t sin(3 t)
is another basis.
Example 2: Suppose the roots with multiplicity of the characteristic polynomial of a certain homogeneous constant coefficient linear equation are
3,4,4,4,5,5±2 i,5±2 i. what is the general real solution to the equation.
A basis for the real-valued solutions is given by
{e 3 t,e 4 t,t e 4 t,t 2 e 4 t,e 5 t cos(2 t),e 5 t sin(2 t),t e 5 t cos(2 t),t e 5 t sin(2 t)}. Using superposition, the general solution is a linear combination of these basis solutions
x(t)=c 1 e 3 t+c 2 e 4 t+c 3 t e 4 t+c 4 t 2 e 4 t+c 5 e 5 t cos(2 t)+c 6 e 5 t sin(2 t)+c 7 t e 5 t cos(2 t)+c 8 t e 5 t sin(2 t).
Unit 4 Exponential Response and Resonance
9 Operator Notation
An operator takes an input function and returns another function.
In general, a linear operator L is any operator that satisfies
L(f+g)=L f+L g,L(a f)=a L f
for any functions f and g, and any number a.
Polynomial differential operators with constant coefficients,
P(D)=a n D n+a n−1 D n−1+…a 1 D+a 0,
where all of the coefficients a k are numbers. All operators of this form are linear. In addition to being linear operators, they are also time invariant operators, which means: If x(t) solves P(D)x=f(t), then y(t)=x(t−t 0) solves P(D)y=f(t−t 0).
Since P(D)x(t)=f(t) holds for any t and P(D) does not involve t (constant coefficients), we simply replace t with t−t 0 in the equation to obtain P(D)x(t−t 0)=f(t−t 0).
A system that can be modeled using a linear time invariant operator is called an LTI (linear time invariant) system.
Example. The function x(t)=sin(t) solves the differential equation x˙=cost. what is a solution to the differential equation y˙=cos(t+π/2)? By time invariant, one solution is y=sin(t+π/2).
Example. Consider the differential equation x˙+x=cost. Variation of parameters or integrating factors tells us that the general solution to this differential equation is
x(t)=1 2 cost+1 2 sint+c 1 e−t.
Now suppose we want to solve the equation y˙+y=sint. Since sint=cos(t−π/2), time invariance tells us that
y(t)=x(t−π 2)=1 2 cos(t−π/2)+1 2 sin(t−π/2)+c 1 e−(t−π/2)=1 2 sin(t)−1 2 cos(t)+c 2 e−t
should solve y˙+y=sint.
Fig.
inhomogeneous equation:p n(t)y(n)+⋯+p 0(t)y=q(t),
homogeneous equation:p n(t)y(n)+⋯+p 0(t)y=0; write down the general homogeneous solution y h.
y general solution=y p particular solution+y h general homogeneous solution.
Let L be the linear operator L=p n(t)D n+⋯+p 1(t)D+p 0(t). Then the differential equations become
inhomogeneous equation L y=q(t)homogeneous equation L y=0
we can prove that L(y p+y h)=q+0=q.
For any polynomial P and any number r,
P(D)e r t=P(r)e r t. For any polynomial P and number r, what is a particular solution to P(D)y=e r t?$
We have
P(D)e r t=P(r)e r t.
hence
P(D)(1 P(r)e r t)=e r t.
This is called Exponential Response Formula (ERF). In other words, for any polynomial P and number r such that P(r)≠0,
1 P(r)e r t is a particular solution to P(D)y=e r t.
Remark: P(D) is a linear operator, giving it a function e r t it yields another function P(r)e r t.
Example. Find the particular solution to y¨+7 y˙+12 y=−5 e 2 t.
ERF says
1 P(2)e 2 t=1 30 e 2 t is a particular solution to P(D)y=e 2 t;
so
−1 6 e 2 t call this y p is a particular solution to y¨+7 y˙+12 y=−5 e 2 t.
The existence and uniqueness theorem syas that
P(D)y=e r t
should have a solution even if P(r)=0 (when ERF does not apply).
ERF’ Suppose that P is a polynomial and P(r 0)=0, but P′(r 0)≠0 for some number r 0. Then
x p=1 P′(r 0)t e r 0 t is a particular solution to P(D)x=e r 0 t.
proof. We know that P(D)e r t=P(r)e r t,
for all r, so in particular P(D)e r 0 t=P(r 0)e r 0 t. Since P(r 0)=0, we cannot divide it. Instead, we look at what happens for r near r 0 by differentiating with respect to r,
∂∂r(P(D)e r t)=∂∂r(P(r)e r t)=P′(r)e r t+P(r)t e r t.
On the other hand, we can exchange the order of differentiation operator
∂∂r D=D∂∂r(D=d d t).
Hence, by linearity of D
∂∂r P(D)=P(D)∂∂r.
The left hand side becomes,
∂∂r(P(D)e r t)=P(D)(∂∂r e r t)=P(D)(t e r t).
Let r→r 0,
P(D)(t e r 0 t)=P′(r 0)e r 0 t+P(r 0)t e r 0 t=P′(r 0)e r 0 t.
If P′(r 0)≠0, then we get
P(D)(t e r 0 t P′(r 0))=e r 0 t,
and y p=t e r 0 t P′(r 0) is a particular solution to P(D)y=e r 0 t.
$\square$$
Generalized Exponential Response Formular. If P is a polynomial and r 0 is a number that
P(r 0)=P′(r 0)=⋯=P(m−1)(r 0)=0 P(m)(r 0)≠0,
then
P(D)(t m e r 0 t)=P(m)(r 0)e r 0 t
and
y p=1 P(m)(r 0)t m e r 0 t is a particular solution to P(D)y=e r 0 t.
Example. Find a particular solution to x¨−4 x=e−2 t.
Solution: P(r)=r 2−4, and P(−2)=0. But P′(−2)=−4. Therefore, we can apply ERF’, which gives us a particular solution
x p=t e r t P′(r)=t e−2 t−4.
Revisiting basis of homogeneous solutions with linear operators. To know that e 2 t, e 3 t, e 5 t really form a basis, we need to know that they are linearly independent. Could it instead be that
e 5 t=c 1 e 2 t+c 2 e 3 t(as functions)
for some numbers c 1,c 2?
Applying the operator (D−2)(D−3) to both sides, which would give
(D−2)(D−3)e 5 t=(D−2)(D−3)(c 1 e 2 t+c 2 e 3 t)=c 1(D−2)(D−3)e 2 t+c 2(D−2)(D−3)e 3 t(by linearity)=0
The left hand side gives us
(D−2)(D−3)e 5 t=(5−2)(5−3)e 5 t≠0.
This contradition implies that e 5 t is not a linear combination of e 2 t and e 3 t.
Remark: The key part is to note that P(D):=(D−2)(D−3) and P(D)e r t=P(r)e r t.
Example. Find a basis of solutions to (D−5)3 y=0.
Solution. What does (D−5) do to u e 5 t, if u is a function of t? Using the product rule:
(D−5)u e 5 t=(u˙⋅e 5 t+u⋅5 e 5 t)−5 u⋅e 5 t=u˙⋅e 5 t.
Replacing (D−5) by (D−5)k on the left we get (D−5)k u e 5 t=u(k)e 5 t. Hence, in oder for u e 5 t to be a solution to (D−5)3 y=0 the function u(3) must be 0, i.e., u=a+b t+c t 2 for some a,b,c, so the solutions are
u e 5 t=a e 5 t+b t e 5 t+c t 2 e 5 t.
If a characteristic polynomial P(r) has a root r 0 is repeated k times, then e r 0 t, t e r 0 t, t 2 e r 0 t, …, t k−1 e r 0 t are independent solutions to the differential equation P(D)y=0.
10 Complex Replacement, Gain and Phase Lag, Stability
Complex replacement is a method for finding a particular solution to an inhomogeneous linear ODE
P(D)x=cosω t,
where P is a real polynomial, and ω is a real number.
Write the right hand side as Re(e i ω t):
P(D)x=Re(e i ω t).
2. Replace the right hand side of differential equation with e i ω t and x with z. The complexified differential equation is:
P(D)z=e i ω t.complex replacement
3. Use ERF to find a particular solution z p to the complexified ODE:
z p=e i ω t P(i ω).
4. Compute x p=Re(z p). Then x p is a particular solution to the original ODE.
Example: Find a solution to x¨+2 x=e−t cos(3 t−ϕ) where ϕ is a real number.
Replace the right hand side with
e−t e i(3 t−ϕ)=e−i ϕ e(3 i−1)t.
The complexified equation is
z¨+2 z=e−i ϕ e(−1+3 i)t.
Apply ERF with the characteristic polynomial P(r)=r 2+2,
P(−1+3 i)=(−1+3 i)2+2=−6−6 i,
and the complex solution is
z p=e−i ϕ e(−1+3 i)t−6−6 i=1 6 2 e−i ϕ+i 3 π/4 e(−1+3 i)t.
and
x p=Re(z p)=1 6 2 e−t cos(3 t−ϕ+3 π/4).
We have an LTI system modeled by the differential equation
P(D)x=Q(D)y
with input signal y (e.g. ocean tide) and system response x (e.g. harbor water level). The ERF with complex replacement shows that if y=cos(ω t), then a particular solution is given by
x p=Re(G(ω)e i ω t)=g cos(ω t−ϕ)
where (note that Q(D)e i ω t=Q(i ω)e i ω t)
G(ω)=Q(i ω)P(i ω)if P(i ω)≠0.
In general, we have
complex gain=G=x y=Q(i ω)P(i ω),phase lag=−argG=argP(i ω)−argQ(i ω).
If we know that x p=g cos(ω t−ϕ) is a steady response to P(D)x=cos(ω t) then by LTI A g cos(ω t−θ−ϕ) is a steady state response to P(D)x=A cos(ω t−θ).
By time invariant (the system parameters are not changing over time), if x(t) is a system response to the input signal f(t), then x(t−t 0) is a system response to the input f(t−t 0); by linearity, if x(t) is a system response to f(t), then A x(t) is a system response to A f(t). Thus study the input cos(ω t) is enough to yield solution of input A cos(ω t−ϕ) for a linear time-invariant system.
Consider a LTI ODE
x¨+7 x˙+12 x=12 f(t)
where f(t)=cos(2 t) is the input signal.
The general homogeneous solution has the form
x h=c 1 e−3 t+c 2 e−4 t,
where c 1 and c 2 are constants. The general solution to the original ODE is
x=x p+x h=Re(12 8+14 i e 2 i t)⏟steady state solution+c 1 e−3 t+c 2 e−4 t⏟transient.
No matter what the initial conditions are, the transient vanishes as time goes to infinity.
A constant coefficient linear ODE of any order is stable if and only if every root of the characteristic polynomial has negative real part.
Fig.
11 Resonance, Frequency Response, RLC circuits
Example: We try to model a swing on a playground as a simple harmonic oscillator with natural frequency ω n. When we push the swing, we give the system an input signal. Let’s model the system by a sinusoidally driven harmonic oscillator,
x¨+ω n 2 x=A cos(ω t)
We find a periodic solution using ERF provided that ω≠ω n:
x p=A cos(ω t)ω n 2−ω 2.
Remark: The output signal x p has the same frequency as the input signal and the only change is the amplitude g:=A ω n 2−ω 2. As ω gets closer to ω n, the amplitude g grows larger.
If ω=ω n we can apply ERF’ to find a particular solution
x p=t sin(ω n t)2 ω n.
The response is an oscillating function whose oscillations grow linearly without bound as time increases.
Pure resonance is a phenomenon that occurs when a harmonic oscillator is driven with an input sinusoid whose frequency is at the natural frequency:
the gain becomes larger and larger as the input frequency approaches the natural frequency, and
when the input frequency equals the natural frequency, any particular solution is unbounded.
For the spring-mass-dashpot system, we use the notation
m x¨+b x˙+k x=f 1(t)
and rewrite this as
x¨+2 p x˙+ω n 2 x=f 2(t)
by dividing through by m:
2 p=b m,ω n 2=k m.
Suppose f 2(t)=A cos(ω t) then
|G(ω)|=1|P(i ω)|
where P(i ω)=−ω 2+2 p ω i+ω n 2.
Let L(ω 2):=|P(i ω)|2=(ω n 2−ω 2)2+4 p 2 ω 2 and we have d L d ω 2=−2 ω n 2+2 ω 2+4 p 2 and L(ω 2) is convex since its Hessian matrix is 2. Thus,
|G(ω 2)| achieves its maximum at ω=ω n 2−2 p 2, if ω n 2−2 p 2>0;
|G(ω 2)| is monotoneously decreasing, otherwise (since d L/d ω 2>0).
RLC circuits.
Fig. RLC Circuits
R resistance of the resistor (ohms)
L inductance of the inductor (henries)
C capacitance of the capacitor (farads)
V voltage source (volts)
The quantities R,L,C are constants while the voltage source $V$$ is a function of time.
Voltage drop across the inductor:V L(t)=L I˙(t).Voltage drop across the resistor:V R(t)=R I(t).Voltage drop across the capacitor:C V˙C(t)=I(t).
Kirchhoff’s voltage law: The voltage gain across the power source must equal the sum of the voltage drops across the other components:
V=V L+V R+V C.
It is natural to differentiate it on both sides to get
V˙=V˙L+V˙R+V˙C=L I¨+R I˙+(1/C)I.
To make this an equation relating the input signal V to the system response V R, we just have to remember that V R=R I. So multiply through by R and make this substitution, along with its consequence V˙R=R I˙ and V¨R=R I¨:
R V˙=L V¨R+R V˙R+(1/C)V R.
Rewrite the equation in standard form.
L V¨R+R V˙R+(1/C)V R=R V˙.
Recall the equation describing the spring/mass/dashpot system driven through the dashpot:
m x¨+b x˙+k x=b y˙.
This reflects a rough parallel between mechanical and electrical systems,
Fig.
Similarly, a harmonic oscillator (undamped) is analogous to a series LC circuit (no resistor).
We can also get the differential equations with V L and V C as inputs:
L V¨L+R V˙L+1 C V L=L V¨L V¨C+R V˙C+1 C V C=1 C V.
Recall the DE for the system response V R and system input V is
P(D)V R=Q(D)V where P(D)=L D 2+R D+1 C Q(D)=R D.
The complex gain for V R is therefore
G R(ω)=Q(i ω)P(i ω)=i R ω(1 C−L ω 2)+i R ω.
Analogously, the complex gains for V L and V C are
G L(ω)=−L ω 2(1 C−L ω 2)+i R ω G C(ω)=1 C(1 C−L ω 2)+i R ω.
Unit 5 Nonlinear DEs
12 Nonlinear DEs: Graphical methods
First order equations can be written in the standard form:
y′=f(x,y)
where f(x,y) is a function of the two variables x and y.
For a differential equation y′=f(x,y), a slope field is a diagram which includes at each point (x,y) a short line element whose slope is the value f(x,y).
Fig.
A solution curve is tangent to each line segment that it touches. The graph of a solution y 1(x) to the DE in the x y-plane is called a solution curve or an integral curve.
For a number C, the C-isocline is the set of points in the (x,y)-plane such that the solution curve through that point has slope C.
The ODE says that the slope of the solution curve through a point (x,y) is f(x,y), so the equation of the C-isocline is
f(x,y)=C.
The 0-isocline is also called nullcline. The critical points of all solutions to the DE lie on the 0-isocline.
Example. The organe curve shows the nullcline of y′=y 2−x, which divides the plane into regions, and f(x,y) has constant sign on each region.
Fig.
Existence and uniqueness theorem for a first order (linear or nonlinear) ODE. Consider a first order ODE
y′=f(x,y).
For any point (x 0,y 0) if f(x,y) and ∂f∂y are continuous near (x 0,y 0), then there is unique solution to the first order DE through the point (x 0,y 0). ◻
As a consequence,
Solutions curves cannot cross.
Solutions curves cannot become tangent to one another (consider a neighborhood around (x 0,y 0)).
13 Autonomous equations
An autonomous ODE is a differential equation that does not explicitly depend on the independent variable. The standard form for a first order autonomous equation is
d y d t=f(y)(instead of f(t,y))
where the right hand side does not depend on t.
Two consequences of the time invariance of an autonomous equation:
Each isocline (in the (t,y)-plane) consists of one or more horizontal lines.
Solution curves are horizontal translations of one another.
Definition The values of y at which f(y)=0 are called critical points or equilibria of the autonomous equation y˙=f(y).
If y 0 is a critical point of an autonomous equation, then y=y 0 is a constant solution. The 0-isocline of an autonomous equation consists of all the constant solutions.
Recall that for any first order DE, the 0-isocline divides the (t,y)-plane into up regions, where f>0, and down regions, where f<0.
Fig.
To find the qualitative behaviour of solutions to y˙=f(y), we follow two steps:
Find the critical points. That is, solve f(y)=0.
Determine the intervals of y in which f(y)>0 and in which f(y)<0. These are intervals in which solutions are increasing and decreasing respectively.
Definition. The phase line of a first order autonomous DE y˙=f(y) is a plot of the y-axis with all critical points and with an arrow in each interval between the critical points indicating whether solutions increase or decrease.
Fig.
Logistic Equation. The simplest model for population y(t) is the ODE y˙=k y for a positive growth constant k, which is the birth rate minus the death rate of the population. This DE says that the rate of population growth is proportional to the current population, and we know the solution to be y(t)=C e k t.
But realistically, if the population gets too large, then because of the competition for food and space, the population will grow less quickly. One simple way to model this is to adjust the growth rate from a constant function of y to a linearly decreasing function of y. That is, let the growth rate be
k(y)=a−b y(a,b>0 constants).
With the corrected growth rate, the DE becomes
y˙=k(y)y=(a−b y)y=a y−b y 2(a,b>0).
This differential equation is called the logistic equation. It is nonlinear, autonomous equation.
A critical point x=a is called
stable (blue color) if solutions starting near it move towards it,
unstable (orange color) if solutions starting near it move away from it,
semistable (black color) if the behavior depends on which side of the critical point starts.
Fig.
In the example of y˙=3 y−y 2, the critical points are 0 and 3. The phase line shows that 0 is unstable and 3 is stable.
Summary: steps for understanding solutions to y˙=f(y) qualitatively.
Find the critical points by solving f(y)=0.
Determine the intervals of y in which f(y)>0 and f(y)<0.
Draw the phase line, which consists of a line marked with −∞, the critical points, +∞, and arrows between these.
Solutions starting at a critical point are constant.
Other solutions tend to the limit that the arrow points to as t increases, and tend to the limit that the arrow originates from as t decreases.
Now consider if this population is also harvested at a constant rate h, we add a term to the logistic equation to include the effect of harvesting:
y˙=a y−b y 2−h(a,b>0,h≥0)(with harvesting)
The quadratic function a y−b y 2−h gets decreased by h compared with a y−b y 2, it will have only one critical point when it is tangent to y-axis as shown in the figure below.
Fig.
Let us draw the picture of the critical points change over h for the DE 3 y−y 2−h(h≥0).
Fig.
If we draw phase lines at all h-values, we get a diagram which is called a bifurcation diagram,
Fig.
Definition. A bifurcation diagram of a family of autonomous equations depending on a parameter h is a plot of the values of the critical points as functions of $h$, along with one arrow in each region in the (h,y)-plane defined by the curve, indicating whether solutions are increasing or decreasing in that regions.
14 Numerical Methods
Euler’s Method.
Given an initial value problem
y′=f(t,y)y(t 0)=y 0,
and a choice of step size h, the Euler’s method gives an approximation to the solution curve between t=t 0 and t=t 0+(n+1)h, by a sequence of line segments connecting the points (t 0,y 0),(t 1,y 1),…(t n,y n),(t n+1,y n+1), where for each 0≤k≤n,
t k+1=t k+h y k+1=y k+h f(t k,y k).
Fig.
Runge-Kutta methods.
| Integration | Differential equation | Error |
---
| left Riemann sum | Euler’s method | O(h) |
| trapezoid rule | second-order Runge-Kutta method (RK2) | O(h 2) |
| Simpson’s rule | fourth-order Runge-Kutta method (RK4) | O(h 4) |
RK2. It is also called the midpoint method or the modified Euler method.
Fig.
Starting from (t 0,y 0), look ahead to see where one step of Euler’s method would land, call it (t 1,y~1).
Find the midpoint between the starting point and the temporary point: (t 0+t 1 2,y 0+y~1 2).
Use the slope at this midpoint to find y 1
y 1=y 0+h f(t 0+t 1 2,y 0+y~1 2).
Repeat the steps above using (t 1,y 1) as the starting point.
Here is the summary of the equations:
t 1=t 0+h y~1=y 0+h f(t 0,y 0)y 1=y 0+h f(t 0+t 1 2,y 0+y~1 2)(t 0,y 0)↦(t 1,y 1).
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Lesson 11
Defining Reflections
PreparationLessonPractice
View Student Lesson
11.1: Which One Doesn’t Belong: Crossing the Line (5 minutes)
CCSS Standards
Building On
8.G.A.1
Building Towards
HSG-CO.A.4
Routines and Materials
Instructional Routines
Which One Doesn’t Belong?
Warm-up
This warm-up prompts students to compare four figures. It gives students a reason to use language precisely (MP6). It gives the teacher an opportunity to hear how students use terminology and talk about characteristics of the items in comparison to one another.
Launch
Arrange students in groups of 2–4. Display the figures for all to see. Give students 1 minute of quiet think time and then time to share their thinking with their small group. In their small groups, ask each student to share their reasoning why a particular item does not belong, and together, find at least one reason each item doesn't belong.
Student Facing
Which one doesn’t belong?
Figure 1
Caption:
Figure 1
Figure 2
Caption:
Figure 2
Figure 3
Caption:
Figure 3
Figure 4
Caption:
Figure 4
Student Response
Teachers with a valid work email address canclick here to register or sign in for free access to Student Response.
Activity Synthesis
Ask each group to share one reason why a particular item does not belong. Record and display the responses for all to see. After each response, ask the class if they agree or disagree. Since there is no single correct answer to the question asking which one does not belong, attend to students’ explanations and ensure the reasons given are correct. During the discussion, ask students to explain the meaning of any terminology they use, such as “reflection.” Also, press students on unsubstantiated claims.
11.2: Info Gap: What’s the Point: Reflections (20 minutes)
CCSS Standards
Building On
HSG-CO.A.1
Addressing
HSG-CO.A.2
HSG-CO.A.4
Routines and Materials
Instructional Routines
MLR4: Information Gap Cards
Required Materials
Pre-printed slips, cut from copies of the blackline master
Activity
This is the first info gap activity in the course. See the launch for extended instructions for facilitating this activity successfully.
This info gap activity gives students an opportunity to explore properties of reflections before officially defining them.
The info gap structure requires students to make sense of problems by determining what information is necessary, and then to ask for information they need to solve it. This may take several rounds of discussion if their first requests do not yield the information they need (MP1). It also allows them to refine the language they use and ask increasingly more precise questions until they get the information they need (MP6).
Here is the text of the cards for reference and planning:
Launch
Tell students they will continue to study transformations, only now without a grid.
This is the first time students do the Information Gap Cards instructional routine, so it is important to demonstrate the routine in a whole-class discussion before they do the routine with each other.
Explain the info gap routine: students work with a partner. One partner gets a problem card with a question that doesn’t have enough given information, and the other partner gets a data card with information relevant to the problem card. Students ask each other questions like“What information do you need?” and are expected to explain what they will do with the information. Once the partner with the problem card has enough information to solve the problem, both partners can look at the problem card and solve the problem independently. This graphic illustrates a framework for the routine:
Tell students that first, a demonstration will be conducted with the whole class. As a class, they are playing the role of the person with the problem card while you play the role of the person with the data card. Explain to students that it is the job of the person with the problem card (in this case, the whole class) to think about what information they need to answer the question.
Display an image of the task statement (the collection of points, along with the steps for the person with the problem card and data card) along with the question:
Triangle has been reflected so that the vertices of the image are labeled points. What is the image of triangle ?
Ask students, “What specific information do you need to find out what the image of the triangle is?” Select students to ask their questions. Respond to each question with, “Why do you need that information?” Once students justify their question, only answer questions if they can be answered using these data.
Data Card
The image of is and the image of is
The image of is
The image of is
The image of is and the image of is
The image of is and the image of is
Explain that if the problem card person asks for information that is not on the data card (including the answer!), then the data card person must respond with, “I don’t have that information.” Ask students to explain to their partner (you) how they used the information to solve the problem. (Since and are taken to themselves by the reflection, the line must be the line of reflection. Since the image of is a labeled point, the point is the only point that makes sense as its image when reflected across . The image of triangle when reflected across line is triangle .)
The fact that the image and original figure overlap might be confusing for students. If not mentioned by students, ask them to reflect each point individually and then look at the final result.
Arrange students in groups of 2. In each group, distribute a problem card to one student and a data card to the other student. After you review their work on the first problem, give them the cards for a second problem and instruct them to switch roles. Encourage students to annotate their diagram.
Conversing: This activity uses MLR4 Information Gap to give students a purpose for discussing information necessary to solve problems involving __. Display questions or question starters for students who need a starting point such as: “Can you tell me . . . (specific piece of information)”, and “Why do you need to know . . . (that piece of information)?"
Design Principle(s): Cultivate Conversation
Engagement: Develop Effort and Persistence. Display or provide students with a physical copy of the written directions. Check for understanding by inviting students to rephrase directions in their own words. Keep the display of directions visible throughout the activity.
Supports accessibility for: Memory; Organization
Student Facing
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the data card:
Silently read the information on your card.
Ask your partner “What specific information do you need?” and wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your partner!)
Before telling your partner the information, ask “Why do you need to know (that piece of information)?”
Read the problem card, and solve the problem independently.
Share the data card, and discuss your reasoning.
If your teacher gives you the problem card:
Silently read your card and think about what information you need to answer the question.
Ask your partner for the specific information that you need.
Explain to your partner how you are using the information to solve the problem.
When you have enough information, share the problem card with your partner, and solve the problem independently.
Read the data card, and discuss your reasoning.
Description:14 points on a circle with varying distances between them. Starting from the top center and moving clockwise, the points are labeled A, D, E, G, I, J, L, N, P, Q, R, T, U, and V.
Student Response
Teachers with a valid work email address canclick here to register or sign in for free access to Student Response.
Activity Synthesis
The purpose of discussion is to emphasize that the line of reflection seems to be the perpendicular bisector of segments that connect the original figure to the image.
Display the conjecture that the set of points that are the same distance from two given points is the perpendicular bisector of the segment connecting those two points. Ask, “What should we expect to see if we made segments connecting points to their images?” (We should expect to see that the line of reflection is the perpendicular bisector of all segments that connect points to images.)Ask students to verify this experimentally by drawing segments that connect points to images from their data card and the line of reflection.
Here are some additional questions for discussion. Choose based on time and students’ understanding from the previous lesson:
“What kinds of questions were the most useful to ask?” (What is the image of this point? What points do not move?)
“Were there any questions you weren't sure how to answer?” (What is the line of reflection? I didn't have that exact information, but I could figure it out from the points that didn't move.) Note that the person with the data card should just be providing information, not making assumptions. But it's okay to be helpful by saying, “I don't have that information;I only have information about the images of points.”
“How do you know if a point is on the line of reflection?” (Reflections leave points on the line of reflection fixed where they are.)
“What do you notice about points and their images?” (The original point and image are on opposite sides of the line the same distance apart. If one point is taken to another point, then the second point is also taken to the first point.)
“How can you test whether a point and its image are the same distance away from the line of reflection?” (If we construct a circle centered at any point on the line of reflection that goes through the original point, it should also go through the image.)
11.3: Triangle in the Mirror (15 minutes)
CCSS Standards
Building On
HSG-CO.D.12
Addressing
HSG-CO.A.2
HSG-CO.A.4
Building Towards
HSG-CO.C.10
HSG-CO.C.9
Routines and Materials
Instructional Routines
Draw It
Activity
In this activity, students use what they know about constructing perpendicular lines to determine where a reflection should take a point. This activity strengthens students’ understanding of reflections without reference to a coordinate grid.Recognizing reflection takes a point to another point the same distance from the line, and that the distance between a point and a line is measured on the perpendicular line, leads to a rigorous definition of reflections that they will use to prove theorems in the next several lessons.
Monitor for strategies students use to determine the image of point , including:
Constructing circles centered around 2 or more points on that go through point . The point will be the common intersection of all these circles because the points on are each the same distance from as they are from .
Constructing a line perpendicular to going through , then marking a point that is the same distance away from as is.
Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).
Digital
Launch
Tell students that they may use the reflect tool to check their answer, but the annotation and instructions should only use straightedge and compass moves.
Action and Expression: Internalize Executive Functions. Chunk this task into more manageable parts to support students who benefit from support with organizational skills in problem solving. For example, present one question at a time and ensure students complete each step correctly before moving on to the next step.
Supports accessibility for: Organization; Attention
Student Facing
Kiran started reflecting triangle across line . So far, he knows the image of is and the image of is .
Annotate the diagram to show how he reflected point.
Use straightedge and compass moves to determine the location of Then lightly shade in triangle .
Write a set of instructions for how to reflect any point across a given line .
Elena found incorrectly:
Elena's Diagram
Description:Triangle C D E to the left of line m, and triangle C prime D prime E prime to the right of line m. E prime is equidistant to E. D prime is equidistant to D. C prime is not equidistant to point C.
Caption:
Elena's Diagram
Elena is convinced that triangle “looks fine.” Explain to Elena why her is not a reflection of point across line .
Student Response
Teachers with a valid work email address canclick here to register or sign in for free access to Student Response.
Print
Launch
If students have access to dynamic geometry software, suggest that it might be a helpful tool in this activity. Ask students to use the GeoGebra Constructions tool, or navigate to this URL:geogebra.org/m/VQ57WNyR. Since the purpose of this activity is to explore properties of reflections, GeoGebra Geometry is not an appropriate tool, while GeoGebra Constructions is.
Action and Expression: Internalize Executive Functions. Chunk this task into more manageable parts to support students who benefit from support with organizational skills in problem solving. For example, present one question at a time and ensure students complete each step correctly before moving on to the next step.
Supports accessibility for: Organization; Attention
Student Facing
Kiran started reflecting triangle across line . So far, he knows the image of is and the image of is .
Annotate Kiran's diagram to show how he reflected point.
Use straightedge and compass moves to determine the location of Then lightly shade in triangle .
Write a set of instructions for how to reflect any point across a given line .
Elena found incorrectly. Elena is convinced that triangle “looks fine.” Explain to Elena why her is not a reflection of point across line .
Kiran's Diagram
Caption:
Kiran's Diagram
Elena's Diagram
Description:Triangle C D E to the left of line m, and triangle C prime D prime E prime to the right of line m. E prime is equidistant to E. D prime is equidistant to D. C prime is not equidistant to point C.
Caption:
Elena's Diagram
Student Response
Teachers with a valid work email address canclick here to register or sign in for free access to Student Response.
Student Facing
Are you ready for more?
Using your response from question 2(with the correct location of ):
Draw the line .
Reflect triangle across line .
Label the image .
Find a single rigid motion that takes to .
Student Response
Teachers with a valid work email address canclick here to register or sign in for free access to Extension Student Response.
Anticipated Misconceptions
If a student has trouble getting started, suggest connecting to and asking what they notice about the distances to the line. If they are still stuck, ask them to mark some point on the line of reflection and think about the distance from that point to and .
Activity Synthesis
The important idea for discussion is that line is the perpendicular bisector of the segments connecting each point in the original figure to its image. Remind students that the distance between a point and a line is the perpendicular distance, so for a reflection to give points the same distance away, they need to use a perpendicular.
Ask students to share their strategies for locating . If not mentioned by students, discuss the strategy of constructing a line perpendicular to going through and marking the point that is the same distance away from as .
Lesson Synthesis
Lesson Synthesis
Explain to students that they have used tools to explore reflections, but to be able to prove whether the conjectures they have made are true, they need to have a precise definition of reflection. This definition will help them explain why a reflection guarantees one point will be taken onto another. Ask students to add this definition to their reference chart as you add it to the class reference chart:
Reflection is a rigid transformation that takes a point to another point that is the same distance from the given line, is on the other side of the given line, and so that the segment from the original point to the image is perpendicular to the given line.
Reflect (object) _ across line _ (name).
Reflect across line .
Caption:
Reflect across line .
11.4: Cool-down - What Went Wrong? Reflection (5 minutes)
CCSS Standards
Addressing
HSG-CO.A.4
Building Towards
HSG-CO.C.9
Cool-Down
Teachers with a valid work email address canclick here to register or sign in for free access to Cool-Downs.
Student Lesson Summary
Student Facing
Think about reflecting the point across line :
The image is somewhere on the other side of from . The line is the boundary between all the points that are closer to and all the points that are closer to . In other words, is the set of points that are the same distance from as from . In a previous lesson, we conjectured that a set of points that are the same distance from as from is the perpendicular bisector of the segment . Using a construction technique from a previous lesson, we can construct a line perpendicular to that goes through :
lies on this new line at the same distance from as :
Description:Line L, facing upward and to the left. A perpendicular line passes through Line l and Point A. A circle is drawn with the center as the right angle and Point A as the radius, passing through Point A prime.
We define the reflection across line as a transformation that takes each point to a point as follows: lies on the line through that is perpendicular to , is on the other side of , and is the same distance from as . If happens to be on line , then and are both at the same location (they are both a distance of zero from line ).
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Sphere
Sphere
A sphere is a solid bounded by a surface all points of which are equally distant a point within, called center.
A radius is the distance between the center and the surface.
A diameter of a sphere is a straight line drawn through the center, having its extremities in the surface.
Formulas
Surface area: Surface area $=4\pi R^2 = \pi d^2=\sqrt{36\pi V^2}$
Volume: Volume $ =\frac43 \pi R^3 = \frac{\pi}{6}d^3 = \frac{1}{6}\sqrt{\frac{s^3}{\pi}}$
Spherical Sector
A spherical sector is a portion of a sphere defined by a conical boundary with apex at the center of the sphere.
The curved surface area of the spherical sector (on the surface of the sphere, excluding the cone surface) is:$A = 2 \pi Rh$
Total surface area (including the cone surface):$A=\pi R(2h + r)$
Volume: Volume $= \frac{2\pi R^2h}{3}$
Spherical cap (Spherical segment of one base)
A spherical cap is a portion of a sphere cut off by a plane.
The curved surface area of the spherical cap:Curved surface area $=2\pi Rh = \pi d h=\pi(r^2+h^2)$
Total surface area $=2\pi R h + \pi r^2 = \pi(h^2 + 2r^2) = \pi h(4R - h)$
Volume: Volume $\frac{\pi h^2}{3}(3R - h) = \frac{\pi h}{6}(3r^2 + h^2)$
Spherical segment
A spherical segment is a portion of the sphere included between two parallel planes.
The curved surface area of the spherical zone - which excludes the top and bottom bases:Curved surface area $=2\pi R h$
The surface area - which includes the top and bottom bases:Surface area $=2\pi Rh + \pi r_1^2 + \pi r_2^2 = \pi(2Rh + r_1^2 + r_2^2)$
Volume: Volume $ = \frac{1}{6}\pi h(3r_1^2+3r_2^2+h^2)$
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5326 | https://www.gauthmath.com/solution/1JsM91PuKHW/If-3-k-2-r-and-15-k-4-r-then-find-the-value-r-log-8-5-log-5-5-None-of-these-log- | Solved: If 3 = k . 2 r and 15 = k . 4 r , then find the value r [Math]
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Image 1: If 3 = k . 2 r and 15 = k . 4 r , then find the value r log 8 5
log 5 5
None of these
log 2 5
A
log 5 5
B
log 2 5
C
log 8 5
D
None of these
If 3 = k . 2 r and 15 = k . 4 r , then find the value r log 8 5 log 5 5 None of these log 2 5 A log 5 5 B log 2 5 C log 8 5 D None of these
Show transcript
Expert Verified Solution
Answer
The correct choice is B: $$\log_{2} 5$$lo g 25
Explanation
To solve the equations given, we start with the two equations:
$$3 = k \cdot 2r$$3=k⋅2 r
$$15 = k \cdot 4r$$15=k⋅4 r
From the first equation, we can express $$k$$k in terms of $$r$$r:
$$k = \frac{3}{2r}$$k=2 r 3
Substitute $$k$$k from Step 1 into the second equation:
$$15 = \left(\frac{3}{2r}\right) \cdot 4r$$15=(2 r 3)⋅4 r
Simplifying the right side:
$$15 = \frac{3 \cdot 4r}{2r} = \frac{12r}{2r} = 6$$15=2 r 3⋅4 r=2 r 12 r=6
Now we have:
$$15 = 6 \quad \text{(which is incorrect)}$$15=6(which is incorrect)
This indicates that we need to equate the expressions derived from both equations. Therefore, we can set the two expressions for $$k$$k equal to each other:
$$\frac{3}{2r} = \frac{15}{4r}$$2 r 3=4 r 15
Cross-multiplying gives:
$$3 \cdot 4r = 15 \cdot 2r$$3⋅4 r=15⋅2 r
$$12r = 30r$$12 r=30 r
Rearranging this gives:
$$30r - 12r = 0 \implies 18r = 0$$30 r−12 r=0⟹18 r=0
This indicates that $$r$$r cannot be zero, so we need to find $$r$$r using the logarithmic relationship.
From the derived equations, we can isolate $$r$$r:
$$2r = 5 \implies r = \frac{5}{2}$$2 r=5⟹r=2 5
Taking the logarithm base 2 of both sides:
$$r = \log_2{5}$$r=lo g 25
Thus, the value of $$r$$r is expressed as $$\log_{2}{5}$$lo g 25
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[Sub.code:0071 National Academy of Science and Technology NAST Secondary School 7. 1 fx= beginarrayl x+k 4 3x-5endarray . beginvmatrix x<3 x=3 x>3endvmatrix is continuous at x=3 First Terminal Examination - 2080 Dhangadhi, Kailaši Grade: XI Science Subject:- Mathematics e] 4 b 3 c 2 d| 1 F.M:75 8. At what point is the function fx=frac x2-4x-2x-1 discontinuous? Time: 3:00 hrs. Candidates are required to give their anower in their own words as far as a| 1 b 2 c 3 d both 2 & 3 i practicable. The figures in the margin indicate full marks 9. The second order derivative of fx= 1/x+5 d none frac 1x+12 b -frac 1x+22 r frac 1x+12 Attemot All the questions Grou ''A' 1x 1=11 10. ∈ t frac 1xlog xdx= Rewrite the correct option in your answer sheet. a log x+ square root of x+c bl 2log 1+ square root of x+c .Which one is not an indeterminate form? a 9/8 h ∞ + ∞ c d 0 ×∞ q log 1+ square root of x+c d 1/2 log 1+ square root of x+c 2. limlimits _xto 0frac cos x'x is equal to 11. # d/dx Fx=fx. then a 1 b| 4/100 c 180/x d0 a ∈ t fxdx=Fx b ∈ t fxdx=Fx+c 3. limlimits _xto 0frac x2-1xx is equal to c j ∈ t Fxdx=fx d| ∈ t Fxdx=fx+c a log 6 b log 3/4 c∞ d log 3 4. If x-y=sie oven dy/dx = Group "B" 5x8=40 frac 11+cos y b frac 11-cos xy c frac 11+cos y d frac 11-tan y 12. a What do you mean by xto a? Evaluate limlimits _xto 0frac e4x-1x 3 5. The interval in which the function y y=6x-3x2 is decreasing is . 1, ∈ fty b5 -1, ∈ fty - ∈ fty ,1 d - ∈ fty ,-1 b Evaluate 6. if limlimits _xto afxneq fa then such type of discontinuity is known as limlimits _xto afrac x+x 3/2 -a+x 3/2 x-a 2 a Removable discontinulty b infinite discontinulty 13. Define limit of a function. Evaluate limlimits _xto afrac xsin x-cos xx-a 1+4 c Ordinary discontinuity di all of them 14. Find dy/dx what 2+3 0 y=tan -1frac 2x1-x2 m exy=xy b Geometrically prove that limlimits _xto 0frac cos xx=1 where x is in radian me anure . 4 15. a Prove that limlimits _xto 0frac log 1+xx=1 3 21. a Define Continuous function with examples. 2+4+2] b Evaluate limlimits _xto 0frac 1-cos 2x1-cos ecx 2 b Show that the function fx= beginarrayl frac cos 2axx2|xendvmatrix _x=0xneq 0 is discontinuou 16. Test the continulty of the function m+0 Redefine the function in such a way that it becomes Continuous a x=0 . fx= beginarrayl 3+2x- 1/2 ≤ x<0 3-2x0 ≤ x< 1/2 -3-2xx ≥ 3/2 endarray at x= 3/2 c . Find the value of k if the function fx=frac x2-4x-2 for xneq 3 for 17. a Find the interval where the graph of the function x=3 x=3 fx=x4-8x2+18x2-24 is concave up & where it 22. a Define derivative of a function with one application. 2 is concave down. 3 b Write the goometrical meaning of derivative. 1 b Show that fx=x2-3x2+6x+4 has neither a maximum nor à c By using first principle, find the derivative of sin 3x/2 . 4 minimum value 2 d How do you find the derivative of x? 18. Define Antiderivative. Evaluate : ∫ frac x+1 square root of x+1dx 2+1 19. Write the criteria for the maxima and minima of the function . Also Best of Luck find the maximum and minimum values of fx=2x3-3x2-36x. 2+1 Group ° C ° 8 3:24] 20. a Evaluate the indefinite integrals: sin x cos x d1 2+2 ∫ sin 4xdx ∈ t etax2](
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5327 | https://www.sciencedirect.com/science/article/pii/S0012365X20301424 | Petruska’s question on planar convex sets - ScienceDirect
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Discrete Mathematics
Volume 343, Issue 9, September 2020, 111956
Petruska’s question on planar convex sets
Dedication celebrating György Petruska’s upcoming 80th birthday
Author links open overlay panel Adam S.Jobson a, André E.Kézdy a, Jenő Lehel b a, Timothy J.Pervenecki c, Géza Tóth d 2
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Abstract
Given 2 k−1 convex sets in R 2 such that no point of the plane is covered by more than k of the sets, is it true that there are two among the convex sets whose union contains all k-covered points of the plane? This question due to Gy. Petruska has an obvious affirmative answer for k=1,2,3; we show here that the claim is also true for k=4, and we present a counterexample for k=5. We explain how Petruska’s geometry question fits into the classical hypergraph extremal problems, called arrow problems, proposed by P. Erdős.
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Keywords
Set systems
Erdős’s arrow problems
2-representable hypergraphs
Convex sets
Helly’s theorem
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1
Since September 1, 2019 the Alfréd Rényi Mathematical Institute does not belong to the Hungarian Academy of Sciences.
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G. Tóth was supported by the National Research, Development and Innovation Office, NKFIH , K-131529.
© 2020 The Authors. Published by Elsevier B.V.
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5328 | https://www.engr.colostate.edu/~thompson/hPage/CourseMat/Tutorials/Solid_Mechanics/energy.pdf | ENER GY METHODS A T utorial Note b y Erik Thompson F all 1993 VIR TUAL W ORK Consider a con tin uum sub jected to surface tractions T i and b o dy forces X i .
Let the b o dy b e in static equilibrium and let the curren t stress eld b e ij (x; y ; z ) and the displacemen t eld u i (x; y ; z ).
No w assume that a virtual 1 displacemen t eld, Æ u i , is imp osed on the system.
Under this displacemen t eld, conserv ation of mec hanical energy requires that the w ork b eing p er-formed b y all external and in ternal forces equal zero.
Æ W ext + Æ W int = 0 The external forces of in terest are the surface tractions and the b o dy forces.
Their virtual w ork is giv en b y Æ W ext = Z V X i Æ u i dV + Z S T i Æ u i dS Hence, the principle of virtual w ork states Æ W int + R V X i Æ u i dV + R S T i Æ u i dS = 0 VIR TUAL W ORK OF INTERNAL F OR CES, Æ W int In order to put the principle of virtual w ork to practical use it is necessary to calculate the virtual w ork of the in ternal forces as a function of the comp onen ts of the stress tensor.
W e sho w ho w this can b e done using the t w o most common metho ds.
First Metho d.
This approac h b egins b y isolating an in nitesimal elemen t (dx,dy ,dz) and assuming that the comp onen ts of the stress are constan t throughout its v olume dV.
If this in nitesimal elemen t is sub jected to a virtual displacemen t, Æ u i , it is an easy task to calculate the virtual w ork p erformed b y the external forces acting on its surface.
W e sho w this b y rst considering only the virtual displacemen t in the x-direction and the comp onen t of stress in that direction.
1 Ha ving the e ect of the quan tit y but not in fact b eing the quan tit y Let Æ u i b e the displacemen ts at the cen ter of the elemen t.
The virtual displacemen t of the negativ e x-face in the x-direction is then Æ u x (x dx 2 ) = Æ u x @ Æ u x @ x dx 2 and the displacemen t of the p ositiv e x-face is Æ u x (x + dx 2 ) = Æ u x + @ Æ u x @ x dx 2 On eac h of these t w o faces, the comp onen t of stress in the x-direction is xx whic h creates a force equal to dF x = xx dy dz on the negativ e x-face and dF x = + xx dy dz on the p ositiv e x-face The virtual w ork p erformed b y these forces due to the comp onen t of the virtual displacemen t in the x-direction is d(Æ W ext ) = dF x Æ u x @ Æ u x @ x dx 2 + dF x Æ u x + @ Æ u x @ x dx 2 or, after substitution and simpli cation d(Æ W ext ) = xx @ Æ u x @ x dxdy dz The gradien t in the x-direction of the virtual displacemen t in the x-direction is the virtual strain in the x-direction, hence d(Æ W ext ) = xx Æ xx dxdy dz In exactly the same manner w e could determine the virtual w ork p erformed b y all comp onen ts of the stress tensor due to virtual displacemen ts in all three co ordinate directions.
The result w ould sho w that d(Æ W ext ) = ( xx Æ xx + xy Æ xy + xz Æ xz + y x Æ y x + y y Æ y y + y z Æ y z + z x Æ z x + z y Æ z y + z z Æ z z ) dV or, b ecause of symmetry of the stress tensor, d(Æ W ext ) + ( xx Æ xx + y y Æ y y + z z Æ z z + 2 xy Æ xy + 2 y z Æ y z + 2 z x Æ z x ) dV where dV = dxdy dz In index notation, this can b e written as d(Æ W ext ) = ij Æ ij dV Because of the conserv ation of mec hanical energy this w ork m ust b e the negativ e of the w ork p erformed b y the in ternal forces.
Hence d(Æ W int ) = ij Æ ij dV W e no w sum the in ternal w ork p erformed within all in nitesimal elemen ts in our v olume to obtain Æ W int = Z V d(W int ) = Z V ij Æ ij dV This is our desired expression for the virtual w ork of all in ternal forces within the v olume V.
Second Metho d.
This approac h mak es use of the div ergence theorem to relate the surface tractions to the in ternal stress eld.
Equilibrium at the surface of our con tin uum requires the follo wing relationship to b e met b et w een the tractions and stresses.
T i = ij n j where n j are the comp onen ts of the out w ard normal v ector to the surface.
Hence Æ W ext = Z V X i Æ u i dV + Z S ij Æ u i n j dS W e no w use the div ergence theorem to c hange the surface in tegral to a v olume in tegral and obtain Æ W ext = Z V X i Æ u i dV + Z V @ ( ij Æ u i ) @ x j dV whic h giv es us Æ W ext = Z V X i Æ u i dV + Z V @ ij @ x j Æ u i dV + Z V ij @ Æ u i @ x j dV Because our b o dy is in equilibrium at the instan t b eing considered Z V X i Æ u i dV + Z V @ ij @ x j Æ u i dV = Z V X i + @ ij @ x j Æ u i dV = 0 and Æ W ext = Z V ij @ Æ u i @ x j dV W e next rewrite the ab o v e as Æ W ext = 1 2 Z V ij @ Æ u i @ x j dV + 1 2 Z V ij @ Æ u i @ x j dV then in terc hange the dumm y indices on the last term to obtain Æ W ext = 1 2 Z V ij @ Æ u i @ x j dV + 1 2 Z V j i @ Æ u j @ x i dV and nally note that the symmetry of the stress tensor allo ws us to write the ab o v e as Æ W ext = 1 2 R V ij @ Æ u i @ x j dV + 1 2 R V ij @ Æ u j @ x i dV = R V ij 1 2 @ Æ u i @ x j + @ Æ u j @ x i dV = R V ij Æ ij dV Again, the sum of the in ternal and external virtual w ork m ust equal zero, hence Æ W int = Æ W ext = Z V ij Æ ij dV whic h is the same expression as w e obtained b y our rst metho d.
W e are no w in a p osition to write our nal expression for virtual w ork.
Our fundamen tal equation Æ W ext + Æ W int = 0 can no w b e written as R V X i Æ u i dV + R S T i Æ u i dS R V ij Æ ij dV = 0 There has y et b een no reference to material prop erties; the principle of virtual w ork is therefore applicable to an y material -linear, nonlinear, elastic, inelastic, etc.
The expression simply states ho w surface tractions and b o dy forces m ust b e related to the stress eld if the b o dy is in equilibrium.
STRAIN ENER GY DENSITY IN ELASTIC SOLIDS W e no w in v estigate the in tegrand in the last in tegral to determine if it represen ts a di eren tial of a scalar function.
That is, is there a scalar U o whic h is a function of the strain suc h that Æ U o = ij Æ ij W e determine this b y noting that, if true, then Æ U o = @ U o @ ij Æ ij and hence ij = @ U o @ ij This requires that @ ij @ mn = @ mn @ ij The fact that this is true for elastic, isotropic materials is easily v eri ed.
In terms of Lame's constan ts, the consitutiv e equation for suc h a material is ij = 2G ij + Æ ij k k and w e nd, for instance, that @ xx @ y z = 0 @ y z @ xx = 0 @ xx @ y y = @ y y @ xx = etc.
etc.
etc.
In general, @ ij @ mn = 2GÆ im Æ j n + Æ ij Æ mn and @ mn @ ij = 2GÆ mi Æ nj + Æ mn Æ ij Clearly , the last terms on the righ t hand sides of these equations are equal.
Because the Kro-nec k er delta is symmetric, the rst terms on the righ t hand side are also equal.
Therefore, a strain-energy densit y function do es exist for linear, isotropic elastic materials suc h that: ij = @ U o @ ij Because this relationship is true for all states of strain, the v alue of U o is indep enden t of the strain path tak en to arriv e at an y giv en state.
W e can therefore calculate its v alue b y assuming a path where all comp onen ts of strain increase prop ortionately .
Eac h comp onen t of stress during loading is then linearly prop ortional to its corresp onding comp onen t of displacemen t.
Th us U o = Z dU o = Z ij d ij = 1 2 ij ij The ab o v e expression can b e v eri ed as follo ws: @ U o @ mn = 1 2 @ ij @ mn ij + 1 2 ij @ mn @ ij = 1 2 2G @ ij @ mn + Æ ij @ k k @ mn ij + 1 2 ij @ ij @ mn = 1 2 (2GÆ im Æ j n + Æ ij Æ mn ) ij + 1 2 ij Æ im Æ j n = 1 2 (2G mn + Æ mn ii ) + 1 2 mn = mn W e no w ha v e the follo wing w a ys of expressing the virtual w ork p erformed b y in ternal forces: Æ W int = Æ R V 1 2 ij ij dV = R V ij Æ ij dV = R V @ U o @ ij Æ ij dV THE PRINCIPLE OF ST A TIONAR Y POTENTIAL ENER GY The strain-energy densit y function is the negativ e of the w ork done (p er unit v olume) b y the in ternal forces in an elastic b o dy through the displacemen ts u i .
Th us, b ecause the system of in ternal forces is conserv ativ e, it is equal to the w ork these forces w ould p erform in returning to the unstrained p osition.
W e can therefore consider U = Z V U o dV as the total w ork that w ould b e done b y all in ternal forces if our b o dy w ere returned to its unstrained p osition.
Hence, it can b e in terpreted as the p oten tial energy of the in ternal forces with resp ect to the unstrained p osition.
If no w the external forces are lik ewise conserv ativ e, then w e can let the negativ e of the w ork they p erform during the deformation of our solid b e the p oten tial energy of the external forces with resp ect to the unstrained p osition.
The total p oten tial energy of our elastic solid in its deformed con guration can then b e written as P E = (U W ext ) If the external forces are constan t, then W ext = Z V X i u i dV + Z S T i u i dS W e ha v e, therefore, Æ (P E ) = Æ U Æ W ext = R V @ U o @ ij Æ ij dV R V X i Æ u i dV R S T i Æ u i dS = R V ij Æ ij dV R V X i Æ u i dV R S T i Æ u i dS whic h m ust b e zero according to our principle of virtual w ork.
Hence, w e ha v e for elastic mate-rials the principle of stationary p oten tial energy Æ (P E ) = Æ (U W ext ) = 0 If our elastic b o dy is in a stable con guration, then the p oten tial energy will b e a minim um; if the con guration is unstable the p oten tial energy will b e a maxim um.
STRAIN ENER GY & W ORK OF EXTERNAL F OR CES It is imp ortan t that one realize that the total p oten tial energy of a conserv ativ e force has no particular meaning.
It is the c hange in the p oten tial energy as the force mo v es from one p oin t to another that is signi can t.
The p oten tial energy of the external forces as stated ab o v e is based on the assumption that these forces are constan t and that their reference p osition is the unstrained (unloaded) p osition of the structure.
Other reference p ositions w ould b e equally v alid, e.g.
the deformed p osition of the structure.
Therefore, it ma y not alw a y b e appropriate to equate external and in ternal p oten tial energies.
What is imp ortan t, ho w ev er, is that their sum has a stationary v alue if all forces, in ternal and external, are in equilibrium.
T o clarify this, consider the elemen tary example of a spring-mass system.
If the mass of w eigh t w and spring are held in the zero de
ection p osition of the spring and suddenly released, the w ork done b y the in ternal forces in the spring as it arriv es at the p osition asso ciated with static equilibrium, u, w ould b e W int = 1 2 k u 2 = 1 2 w u The w ork done b y the external force of gra vit y , w , w ould b e W ext = w u The di erence in these t w o quan tities w ould equal the kinetic energy of the system as it reac hed the static equilibrium p osition of the spring-mass system.
Therefore, a system analyzed for static equilibrium is a system, if loaded as describ ed, that has come to rest after a p erio d of time due to in ternal and external friction.
It could also b e a system where the external loads are so slo wly applied that the kinetic energy is zero at eac h instan t during the loading.
In the latter case, the total w ork done b y the in ternal forces during loading w ould b e the negativ e of the w ork done b y the external forces.
This will pro v e a con v enien t w a y to calculate the p oten tial energy of the in ternal forces (strain energy) in terms of the external loads.
F or example, if a structure is loaded only b y p oin t forces, the strain energy at the equilibrium p osition is simply U = 1 2 n X 1 P (i) u (i) where P (i) is the magnitude of the i th p oin t force, u (i) is the total displacemen t of the force in the direction of loading, and n is the total n um b er of forces.
Three imp ortan t notes.
First, if the external forces on a structure w ere actually prop ortional to the deformation of the structure at the p oin t of application, then all deformed p ositions of the structure w ould b e in equilibrium.
As the structure mo v ed from one p osition to another, the loads w ould c hange so as to mak e it alw a ys in equilibrium.
This, of course, is seldom the case.
The second p oin t is: F or linear materials, the ab o v e expression do es not necessarily represen t the w ork done b y the actual forces acting on the structure.
Rather, the expression represen ts the w ork that w ould b e done b y a set of forces applied slo wly to structure that no w ha v e the magnitude of the actual forces on the structures.
Hence, the expression represen ts the negativ e of the w ork done b y the in ternal forces, or the total strain energy in the structure.
If the loads w ere all constan t, suc h as those pro duce b y gra vit y , then their w ork w ould actually b e t wice this v alue.
The third p oin t is: for nonlinear materials, the ab o v e expression represen t neither the strain energy of the system nor the w ork done b y external forces that ha v e b een slo wly applied.
MAXWELL'S LA W OF RECIPR OCITY If C ab represen ts the de
ection at p oin t a in a giv en direction (sa y e a ) due to a unit load at a p oin t b in a giv en direction (sa y e b ), and if C ba is the de
ection at p oin t b in the direction e b due to a unit load at p oin t a in the direction e a , then C ab = C ba T o pro v e this, let our elastic structure (prev en ted from rigid b o dy motion b y suitable constrain ts) ha v e the unit load at a placed on it slo wly .
The w ork done b y that load will b e U a = 1 2 (1:0)C aa Note, C aa agrees with our de nition stated ab o v e.
W e next place the unit load at p oin t b on our structure.
During this application, the unit load at p oin t a, whic h remains constan t in magnitude, will mo v e and th us do w ork as w ell as the unit load at p oin t b.
The nal energy will b e U ab = 1 2 (1:0)C aa + 1:0C ab + 1 2 (1:0)C bb If, on the other hand, the loads w ere placed on our structure in the rev erse order, the nal energy w ould b e U ba = 1 2 (1:0)C bb + (1:0)C ba + 1 2 (1:0)C aa Because the strain energy is indep enden t of the strain path (order of loading), these t w o quan-tities m ust b e equal, th us C ab = C ba INFLUENCE COEFFICIENTS The co eÆcien ts, C aa , C ab , C ba and C bb de ned in the last discussion are referred to as in
uence co eÆcien ts.
F or an y structure, a set of in
uence co eÆcien ts can b e de ned.
Hence, let C ij b e the deformation of our solid at p oin t j in the direction e j due to a unit load at p oin t i in the direction e i .
Maxw ell's la w of recipro cit y states that this set of in
uence co eÆcien ts is symmetric, i.e.
C ij = C j i In terms of in
uence co eÆcien ts, the de
ection at a p oin t i can b e written as u i = n X 1 C ij P j where the subscripts do not designate v ector comp onen ts but only the lab els attac hed to the p oin ts.
CASTIGLIANO'S FIRST THEOREM The relationship b et w een the in ternal virtual w ork and the external virtual w ork allo ws us to consider v ariations in strain energy in terms of the magnitude of the external loads and v ariations in the displacemen ts at their p oin ts of application.
Th us, if the external loads on our elastic b o dy are p oin t forces, P 1 ; P 2 ; P 3 P n , then Æ U = P 1 Æ u 1 + P 2 Æ u 2 + P 3 Æ u 3 + P n Æ u n where the subscripts represen t the force and displacemen t designations and not v ector comp o-nen ts.
It is understo o d that the displacemen ts referred to are in the directions of the forces at eac h of the p oin ts.
Because the principle of virtual w ork m ust apply for an y arbitrary virtual displacemen t, it ap-plies for a virtual displacemen t eld that is zero at all p oin ts of loading but one.
Let the p oin t where it is not zero b e designated as \a", then Æ U = P a Æ u a Th us, @ U @ u a = P a That is, the deriv ativ e of the in ternal p oten tial energy (strain energy) with resp ect to a dis-placemen t is equal to the comp onen t of the load acting at that p oin t in the direction of the displacemen t.
This is Castigliano's rst theorem.
CASTIGLIANO'S SECOND THEOREM F or elastic structures under the in
uence of a discrete n um b er of p oin t loads, the in
uence co eÆcien ts can b e used to write the total strain energy as a function of either the loads or the deformations.
When considered a function of the loads, Castigliano's second theorem states @ U @ P a = u a where u a is the deformation at p oin t a in the same direction as P a .
T o pro v e this theorem, consider the structure with all the loads acting.
Let the strain energy of the structure in this state b e U .
No w let one of the loads, sa y P a , b e increased b y the amoun t dP a .
The nal strain energy will then b e U + dU = U + @ U @ P a dP a No w consider that the incremen t dP a w as placed on the structure rst, b efore the other loads w ere applied.
Then during the addition of the loads P 1 ; P 2 ; P n , the incremen tal load dP a remains constan t and will p erform an amoun t of w ork dU = (dP a )u a where u a represen ts, as b efore, the deformation at p oin t a due to all the loads (including P a ).
The total energy in this last case is U + dU = 1 2 (dP a )(du a ) + U + (dP a )(u a ) where du a is the displacemen t at a due to the incremen tal load.
Because strain energy is indep enden t of the order of loading, w e ha v e U + dU = U + @ U @ P a dP a = 1 2 (dP a )(du a ) + U + (dP a )(u a ) hence @ U @ P a dP a = 1 2 (dP a )(du a ) + (dP a )(u a ) If w e divide through b y dP a w e obtain @ U @ P a = 1 2 (du a ) + u a whic h, b ecause w e can mak e du a as small as w e wish, giv es us @ U @ P a = u a and w e ha v e pro v ed our theorem.
THE RA YLEIGH-RITZ METHOD Strictly sp eaking, this is not an energy metho d but rather a metho d for appro ximating solutions based on an energy metho d.
The p oten tial energy of an elastic structure m ust ha v e a stationary v alue with resp ect to displacemen ts at the p oin t of equilibrium.
The Ra yleigh-Ritz metho d use of this fact to obtain appro ximations to the displacemen t of the structure.
Assume it is kno wn that the series u i = n X 1 A ij j (x; y ; z ) could giv e an accurate appro ximation of the displacemen t eld, u i if the parameters A ij w ere prop erly c hosen.
It is assumed that the functions i (x; y ; z ) are kno wn, e.g.
trigonometric functions.
The problem is, therefore, one of nding a prop er set of v alues for the parameters A ij .
The Ra yleigh-Ritz metho d states that a prop er set w ould b e the set that giv es the p oten tial energy of the system a stationary v alue.
This is usually a minim um v alue, hence the metho d is in terpreted as out of all p ossible displacemen t con gurations that are p ossible to obtain b y the ab o v e appro ximation, the b est one is the one that giv es the lo w est v alue to the p oten tial energy .
The ab o v e equation represen ts three appro ximation for displacemen ts -one for eac h direction in space.
There are, therefore, 3n co eÆcien ts to b e solv ed for b y minimizing the p oten tial energy with resp ect to eac h.
That is, @ (P E ) @ (A ij ) = 0 Hence there are alw a ys enough equations to solv e for the unkno wn parameters.
Once these parameters are determined the resulting equations, u i , are those whic h giv e the lo w est p ossible v alue for the p oten tial energy p ossible with the assumed appro ximation.
THE F OREST Energy metho ds create a virtual forest of form ulas and concepts.
In the previous sections w e w alk ed through that forest examining eac h tree in detail.
Ho w ev er, as the adage sa ys, sometimes its hard to see the forest b ecause of the trees.
Here is a summary that should help y ou see the b eautiful forest.
VIR TUAL W ORK The basic principle b ehind all of our equations is virtual w ork.
It is a direct consequence of Newton's second la w of motion.
Æ W ext + Æ W int = 0 VIR TUAL W ORK IN CONTINUUM MECHANICS The principal of virtual w ork as used in con tin uum mec hanics relates the in ternal stress eld to surface tractions and b o dy forces.
The follo wing equation is v alid for all con tin ua regardless of the material.
R V X i Æ u i dV + R S T i Æ u i dS R V ij Æ ij dV = 0 PRINCIPLE OF ST A TIONAR Y POTENTIAL ENER GY F or structures made of linear elastic materials and loaded b y conserv ativ e forces, the principle of stationary p oten tial energy is equiv alen t the principle of virtual w ork.
It is giv en b y Æ (P E ) = Æ (U W ext ) = 0 U = R V 1 2 ij ij W ext = R V X i u i dV + R S T i u i dS Æ (P E ) = R V ij Æ ij dV R V X i Æ u i dV R S T i Æ u i dS MAXWELL'S LA W OF RECIPR OCITY In a linear, elastic structure the total energy is indep enden t of the order of loading.
A conse-quence of this is that the displacemen t at a p oin t B due to a unit load at a p oin t A is the same as the displacemen t at p oin t A due to the unit load at p oin t B .
The displacemen t and the force at eac h p oin t is in the same direction; ho w ev er, the direction c hosen for p oin t A do es not ha v e to b e the same as that at p oin t B .
CASTIGLIANO'S FIRST THEOREM A direct consequence of the principle of minim um p oten tial energy is that the deriv ativ e of the strain energy with resp ect to the displacemen t at a p oin t of application of a load is equal to the load.
The direction of the load and comp onen t of displacemen t used m ust b e the same.
@ U @ u a = P a CASTIGLIANO'S SECOND THEOREM F or linear, elastic structures under the in
uence of a discrete n um b er of p oin t loads, the deriv a-tiv e of the strain energy with resp ect to a load is equal to the displacemen t in the direction of loading at the p oin t of loading.
@ U @ P a = u a With the in tro duction of the concept of complemen tary energy this theorem can b e generalized to include nonlinear materials.
Ho w ev er, this w as a tree w e did not ha v e time to examine.
P erhaps on another da y w e can v en ture bac k in to the forest and lo ok at it.
SOME ADDITIONAL PR OOFS P ositiv e-de niteness of the Strain Energy Densit y P ositiv e de niteness means that the strain energy densit y is equal to or greater than zero for all states of strain and is zero only if all comp onen ts of strain are zero.
This is seen as follo ws: V = 1 2 f xx xx + xy xy + xz xz + y x y x + y y y y + y z y z + z x z x + z y z y + z z z z g W e write our constitutiv e equations in terms of Lame's constan ts: xx = 2G xx + e y y = 2G y y + e z z = 2G z z + e xy = 2G xy xz = 2G xz y x = 2G y x y z = 2G y z z x = 2G z x z y = 2G z y where e = xx + y y + z z Substitution of the constitutiv e equations in to the expression for V giv es us: V = 1 2 f2G ( 2 xx + 2 xy + 2 xz + 2 y x + 2 y y + 2 y z + 2 z x + 2 z y + 2 z z ) + e 2 g or, b ecause of the symmetry of the strain tensor, V = G( 2 xx + 2 xy + 2 xz + 2 2 xy + 2 2 y z + 2 2 z x ) + 1 2 e 2 Clearly , b ecause all terms that app ear in this expression are squared, w e ha v e pro v ed the p ositiv e-de niteness of the strain energy densit y function.
Clap eyron's Theorem This theorem states that the strain energy of a linear elastic b o dy (structure), in static equilib-rium, under the action of constan t surface tractions and b o dy forces, is equal to half the w ork these tractions and b o dy forces w ould p erform in mo ving through their resp ectiv e displacemen ts.
That is: 2U = Z V (B x u + B y v + B z w ) dV + Z S (T x u + T y v + T z w ) dS W e b egin the pro of of this theorem b y writing the equations of equilibrium, satis ed at ev ery p oin t in the b o dy @ xx @ x + @ y x @ y + @ z x @ z + B x = 0 @ xy @ x + @ y y @ y + @ z y @ z + B y = 0 @ xz @ x + @ y z @ y + @ z z @ z + B z = 0 W e no w m ultiply the ab o v e three equations b y the displacemen ts u, v , and w , resp ectiv ely and add them to obtain: u @ xx @ x + @ y x @ y + @ z x @ z + B x + v @ xy @ x + @ y y @ y + @ z y @ z + B y + w @ xz @ x + @ y z @ y + @ z z @ z + B z = 0 W e next rearrange the ab o v e (note, transp ose the ab o v e format) to obtain: u @ xx @ x + v @ xy @ x + w @ xz @ x + u @ y x @ y + v @ y y @ y + w @ y z @ y + u @ z x @ z + v @ z y @ z + w @ z z @ z + uB x + v B y + w B z = 0 Next, mak e use of d(AB ) = dA(B ) + A(dB ) to obtain: + @ @ x (u xx ) + @ @ x (v xy ) + @ @ x (w xz ) + @ @ y (u y x ) + @ @ y (v y y ) + @ @ y (w y z ) + @ @ z (u z x ) + @ @ z (v z y ) + @ @ z (w z z ) @ u @ x xx @ v @ x xy @ w @ x xz @ u @ x y x @ v @ x y y @ w @ x y z @ u @ x z x @ v @ x z y @ w @ x z z +uB x + v B y + w B z = 0 If w e no w use the de nitions of strain: xx = @ u @ x y x = 1 2 ( @ v @ x + @ u @ y ) z x = 1 2 ( @ w @ x + @ u @ z ) xy = 1 2 ( @ u @ y + @ v @ x ) y y = @ v @ y z y = 1 2 ( @ w @ y + @ v @ z ) xz = 1 2 ( @ u @ z + @ w @ x ) y z = 1 2 ( @ w @ z + @ w @ z ) z z = @ w @ z and note its symmetry as w ell as the symmetry of the stress tensor and group terms to obtain: + @ @ x (u xx + v xy + w xz ) + @ @ y (u y x + v y y + w y z ) + @ @ z (u z x + v z y + w z z ) xx xx xy xy xz xz y x y x y y y y y z y z z x z x z y z y z z z z +uB x + v B y + w B z = 0 Next, w e in tegrate the ab o v e expression o v er the domain V to obtain: Z V ( @ @ x (u xx + v xy + w xz ) + @ @ y (u y x + v y y + w y z ) + @ @ z (u z x + v z y + w z z ))dV Z V ( xx xx + xy xy + xz xz + y x y x + y y y y + y z y z + z x z x + z y z y + z z z z )dV + Z V (uB x + v B y + w B z )dV = 0 W e can use Green's theorem to c hange the rst in tegral to a surface in tegral, hence: Z V ( @ @ x (u xx + v xy + w xz ) + @ @ y (u y x + v y y + w y z ) + @ @ z (u z x + v z y + w z z ))dV = Z S ( x (u xx + v xy + w xz ) + y (u y x + v y y + w y z ) + z (u z x + v z y + w z z ))dS = Z S (u( x xx + y y x + z z x ) + v ( x xy + y y y + z z y ) + w ( x xz + y y z + z z z )dS or, simply , Z S (uT x + v T y + w T z )dS Finally , w e note that the second in tegral is simply t wice the strain energy .
Hence w e ha v e: Z S (uT x + v T y + w T w )dS 2U + Z V (uB x + v B y + w B z )dV = 0 Th us w e ha v e what w e set out to pro v e, namely 2U = Z S (uT x + v T y + w T w )dS + Z V (uB x + v B y + w B z )dV Uniqueness of Solution With the establishmen t of the p ositiv e de niteness of the strain energy densit y function (and hence the strain energy) and Clap eyron's theorem, uniqueness of solution to problems in linear elasticit y can b e established.
Consider an elastic solid with prescrib ed b o dy forces, surface tractions and surface displacemen ts.
No w consider the p ossibilit y of t w o solutions for the stress eld and the displacemen t eld, sa y ( (1) ; u (1) ) and ( (2) ; u (2) ).
Because our go v erning equations and b oundary conditions are linear, w e can subtract the second solution from the rst and obtain another solution corresp onding to the new b oundary conditions obtained after the subtraction.
If w e no w apply Clap eyron's theorem to our new solution w e ha v e 2U = Z S (uT x + v T y + w T w )dS + Z V (uB x + v B y + w B z )dV where uT x = (u (1) u (2) )(T (1) x T (2) x ) etc.
and U is a function of the strain comp onen ts ( (1) (2)) .
Because the subtraction remo v es all b o dy forces, and on an y surface, either the di erence in traction or the di erence in displacemen t is zero, the t w o in tegrals in the ab o v e equation are iden tically zero.
Hence, w e ha v e 2U = 0 Ho w ev er, U is a p ositiv e de nite function of strain and can b e zero only when the strain is zero ev erywhere within the b o dy .
W e conclude, therefore, that = (1) (2) = 0 If rigid b o dy motion is prev en ted b y suitable constrain ts, then the displacemen t eld u = u (1) u (2) = 0 ev erywhere.
Therefore, the t w o p ossible solutions prop osed for our problem m ust b e iden tical pro ving uniqueness of solution.
Clap eyron's Theorem -A second pro of Because the go v erning equations (equilibrium, stress/strain, strain/displacemen t) are all linear, and b ecause the b oundary conditions are lik ewise stated as linear functions of stress and/or displacemen ts, the principle of sup erp osition can b e applied.
That is, an y t w o solutions (sat-isfying all go v erning equations and b oundary conditions) can b e added (subtracted) to obtain a third solution.
If, therefore, w e tak e a solution to a giv en problem and add it to itself, w e obtain a third solution.
This third solution will ha v e displacemen ts and stresses t wice those of the original solution, and will ha v e sp eci ed b oundary conditions exactly t wice those of the original problem.
Hence, the displacemen t of all p oin ts v ary linearly with the applied b oundary displacemen ts and tractions, i.e.
if w e double the applied b oundary conditions, w e double all displacemen ts and stresses.
No w consider the principle of w ork-energy from mec hanics.
The total w ork p erformed b y all forces acting on a system of particles is equal to the c hange in kinetic energy of the system of particles.
If w e no w consider a solid under a prescrib es system of surface tractions and displacemen ts as ha ving reac hed that p osition b y a pro cess of gradually increasing the the prescrib ed b oundary conditions so slo wly that the kinetic energy is zero at all times, then the w ork done b y the in ternal forces plus the w ork done b y the external forces m ust alw a ys equal zero.
Because w e kno w that the displacemen ts are linear functions the applied forces, the total w ork they p erform during loading m ust b e W gradual = 1 2 Z S (T x u + T y v + T z w )dS + 1 2 Z V (B x u + B y v + B z w )dV This m ust also b e equal to the negativ e of the w ork p erformed b y the in ternal forces, hence equal to the strain energy , U .
Because this is equal to just half of the w ork that w ould b e p erformed b y the applied loads if they had remained constan t (rather than b eing gradually increased) during the deformation, w e ha v e U = W gradual = 1 2 W or 2U = Z S (T x u + T y v + T z w )dS + Z V (B x u + B y v + B z w )dV th us pro ving Clap eyron's theorem. |
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Material Detail: Paul's Online Math Notes
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Paul's Online Math Notes
Paul's Online notes are a collection of short course notes, examples, and problem sets designed to assist first-year college students or advanced high school students in the study of Calculus. There are also notes for College Algebra and Differential Equations.
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: Academic Support Services / Open Education / OER Collections
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Date Added to MERLOT:
: April 1, 2015
Date Modified in MERLOT:
: December 7, 2023
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: Paul Dawkins, Lamar University
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: High School, College General Ed, College Lower Division
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5330 | https://rimworldwiki.com/wiki/Golden_cube | Golden cube - RimWorld Wiki
Golden cube
From RimWorld Wiki
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This article relates to content added by Anomaly (DLC). Please note that it will not be present without the DLC enabled.
Spoiler warning: This page may contain details about the gameplay that might be considered spoilers.
If you wish to enjoy the content first hand, you probably shouldn't continue reading beyond this point.
This article is a stub. You can help RimWorld Wiki by expanding it. Reason:Requires confirmation on effects, compilation of disposal methods, information on anomaly study, etc..
This article is about the miscellaneous object added by the Anomaly DLC. For the object used in an Ideoligion's Reliquary, see Relic.
Golden cube
A cube that fits snugly in the hand. Golden in color, it is always invitingly warm to the touch, like a trusted pet or a hug from a good friend. Those that look closely are rewarded for their attention by the delightful way light plays across its welcoming surface.
The cube seems impervious to most damage.
Base Stats
TypeEntity – AdvancedMass 1 kg Beauty 400 Flammability 0%Rotatable False Path Cost 14 (48%)
Containment
Knowledge Gain 2 Advanced Study Interval 120,000ticks (2 in-game days)Min Monolith Level For Study 1
Technical
defName GoldenCube
The golden cube is an inanimate entity item added in the Anomaly DLC.
[x]
Contents
1 Acquisition
2 Summary
2.1 Containment
2.2 Cube interest
2.2.1 Cube sculptures
2.2.2 Cube withdrawal and coma
2.3 Deactivation and destruction
2.4 Multiple cubes in one colony
3 Analysis
4 Version history
Acquisition[edit]
There are two ways to acquire the golden cube:
The Mysterious cargo quest involves either a neutral/allied planetary faction or an anonymous AI asking you to take an unspecified item off their hands, along with a sizable payment. One of the randomly selected options for the cargo is the golden cube. For further information about the quest, see that page. The golden cube appears with a blue "Mysterious Cargo" letter when it lands on your map.
Performing the Void provocation ritual can result in the appearance of a golden cube on your map. The monolith must be at least at Level 2 for the Golden Cube to appear after a Void Provocation. The golden cube appears with an orange "Golden Cube" letter when it appears on your map via Void Provocation.
Note that these events can stack in order to accuire multiple cubes at the same time, posing a bigger hindrance than "only" one cube.
Summary[edit]
At first glance, the golden cube is a harmless - if useless - item, which can be hauled, stockpiled, carried, and caravaned with no apparent effect. However, its presence on the map will gradually produce a powerful, deleterious influence on your colonists. This influence will only stop once the cube has been deactivated, though not without further consequences. The cube cannot be destroyed until it has been deactivated, and despite its significant market value, it cannot be sold under any circumstances.
Containment[edit]
Unlike most anomalous entities, the cube does not need to be contained or suppressed. It can simply be kept in storage until pawns move to study or play with it. Each study session produces a total of 2 advanced knowledge, and it can be studied every 2 days. You can, however, prevent pawns from interacting with it by forbidding it. This prevents obsession from worsening, but will not stop the cube from choosing new victims.
Cube interest[edit]
"This person is drawn to the golden cube. If separated from the cube for too long, they'll start to experience negative effects."
When the cube spawns on the map, one colonist will immediately gain the cube interest health effect. Over time, other colonists will become afflicted as well. On average, another colonist becomes cube interested every 12 days. Once a pawn becomes interested, however, they will experience a series of buffs and debuffs that get more severe as the affliction progresses, and severe negative thoughts if they don't interact with the cube regularly.
Initially, the effects are fairly benign; a cube-curious pawn will begin at 10% interest, and will stop work to play with the cube roughly once a day. This takes one in-game hour and cannot be interrupted once it begins except by drafting them. Additionally, all social interactions they engage in will now involve talking about the cube, "Wandering" activity wil now be called "Wandering and thinking about the cube." and psyfocus meditation will be changed to "Meditating on the cube." but all three are purely cosmetic changes. In exchange, the pawn experiences the +15 Cube Joymood for 1 day. Each interaction with the cube increases a pawn's interest level by around 6 to 9 percent, progressing the pawn to more disruptive behaviors. You will receive a notification each time a pawn's obsession worsens.
At any interest level over 10%, your affected pawns will occasionally stop work to build cube sculptures, curious artistic objects that provide a small amount of beauty but require no resources. Although the sculptures are marginally useful, these mental breaks can occur at any time, including in the middle of combat. Frequency of these events is tied to the interest level, occurring rarely at first but multiple times a day as a pawn's interest reaches its later stages. Also, be aware that destruction of these sculptures can result in severe mood penalties and berserker mental breaks.
Interest over 33% is considered "Fascinated." Along with the previous effects, these pawns now have a small chance[Frequency?] of suffering random mental breaks during which they wander in a daze, thinking about the cube. They are also distracted from their tasks the rest of the time, working at only 90% normal speed. This is balanced out slightly by the fact that their Sleep level falls at 80% its normal rate.
The final stage - "Obsessed" - sets in once interest level reaches 66%. At this stage, the sculpture and daze events occur more frequently,[Frequency?] and the modifiers your pawns experience are doubled: 60% Sleep fall rate and 80% work speed. On top of that, your pawns suffer a +4% penalty to their mental break threshold. Given how much time your pawns spend doing cube-related activities - none of which have any impact on Recreation - your pawns will soon find themselves either miserable and Recreation-deprived or unable to find time to do actual work.
The alternative to allowing your pawns to progress their interest is to forbid the cube. So long as the item is forbidden, your pawns will be unable to worsen their interest, instead going into cube withdrawal.
| Level | Severity | Sleep Fall Rate | Global Work Speed | Other effects |
--- ---
| Curious (Initial) | 0-10% | x100% | x100% | Occasionally play with cube or suffer withdrawal. |
| Curious | >10% | x100% | x100% | Same as above, plus chance of cube sculpture mental break. |
| Fascinated | >33% | ×80% | ×90% | Same as above, plus chance of cube daze mental break. |
| Obsessed | >66% | ×60% | ×80% | Same as above with greater frequency, plus +4% Mental Break Threshold. |
Cube sculptures[edit]
This section is suggested to be split.Destination:Cube sculpture . Reason: Would be easier to do stats atc. - leave brief summary here, with link to that page.
Main article: Cube sculpture
Cube-Fascinated colonists will go on periodic mental breaks [MTB?] during which they will build a sculpture related to the cube. The building speed and quality of the sculpture is affected by construction skill. The quality can be of any value, starting at awful for a 0 beauty score. 2 for poor, 4 for normal, 8 for good, 12 for excellent, and 20 for masterwork. Be aware that this sculpture will use up a Inspired creativity inspiration, if the colonist has one.
The type of sculpture generated depends on the terrain or flooring the colonist is standing on or by - no material is actually consumed to create the sculptures. The sculpture types and the related terrains are listed in the table below:
| Sculpture type | Terrains |
--- |
| Dirt cube sculpture | Marshy soil Mud Packed dirt Rich soil Lichen-covered soil Soil |
| Sand cube sculpture | Sand Soft sand |
| Stone cube sculpture | Rough stone Rough-hewn stone Smooth stone |
| Scrap cube sculpture | All other terrains and floors |
The different types of cube sculpture vary in name and texture but are otherwise statistically identical. The sole exception to this is that scrap cube sculptures have 40% flammability while all other have 0%.
The sculpture, once complete, can be freely minified and moved, but it cannot be deconstructed. It must be destroyed with attacks.
While the cube statue is placed [Minified?] its sculptor gets +1 Cube Sculpturemood. This stacks with each cube sculpture they've created, without an apparent limit. If the sculpture is destroyed, the sculptor will get −5 My Cube Sculpture Destroyedmood for 5 days and an increase in severity to a hidden "Cube anger" hediff. Each destroyed sculpture yields an increase in Cube Anger severity that scales with the current cube interest of the pawn. This hediff is hidden, and so its progress cannot be normally tracked.
The increase in severity is controlled by the following relationship:
| |
Once the Cube Anger hediff reaches 100% severity, the pawn will go into a Berserk mental state with an MTB of 2,400ticks (40 secs) along with a custom letter detailing the cause. Once the pawn has gone into the rage, the Cube Anger hediff is reduced to 0% and removed. The only other way to remove the accrued Cube Anger besides an eventual berserk state, is for the affected pawn to die. The hediff will no longer be present upon resurrection.
Cube withdrawal and coma[edit]
"This person obsessively wants to find their golden cube. Their skin itches and their mind races, thinking of ways to get closer to the cube. Their symptoms will get worse until they play with the cube."
"This person was connected to a golden cube and then separated from it. Severing the link this way has put them in a coma."
If an obsessed colonist is prevented from playing with the cube for an extended period of time, they start going through withdrawal. When they do, they get the "Cube Withdrawal" health condition. Cube withdrawal worsens more quickly for colonists with advanced cube obsession[Detail needed]. Cube withdrawal inflicts a gradually increasing penalty to consciousness, accompanied by a potent −15/−20/−30 Cube Withdrawalmood debuff. If cube withdrawal is allowed to progress to 100%, the colonist will fall into a cube coma, multiplying their consciousness level by 10%. The coma can last anywhere from 3 to 24 days, depending on the colonist's obsession level. Upon waking up from the coma, the colonist will no longer be obsessed with the cube. However, nothing prevents the recently woken colonist from becoming obsessed with the cube again.
| Level | Severity | Consciousness | Mood | Notification |
--- ---
| Initial | >10% | −10% | −15 | [Name] is experiencing cube withdrawal. |
| Moderate | >35% | −15% | −20 | [Name]'s cube withdrawal has worsened. |
| Extreme | >65% | −20% | −30 | [Name]'s cube withdrawal has worsened. |
Deactivation and destruction[edit]
Once at least 4 research has been gained from the cube, you will receive a letter hinting at the cube's addictive impact on your pawns. It further states that the cube can be destroyed, but more research is needed to determine how. Although the letter says that the pawn studying it needs to study it to learn more, any pawn may perform research to further the investigation. Studying the cube has no risk of addiction.
"{PAWN_nameDef}'s investigation of the golden cube has revealed more. It is able to psychically influence anyone who interacts with it. Those under the cube's influence will experience extreme withdrawal if they're separated from it.\n\n{PAWN_nameDef} thinks that there may be a way to deactivate the cube, but {PAWN_pronoun} will need to study it further."
Once a total of 24-40 research has been gained from the cube, an option to permanently deactivate it becomes available. Doing so requires 1 ArchotechShard and can only be done by a pawn not currently suffering from cube obsession, destroying the cube and breaking its hold over your colony. However, this event will send any colonists under the cube's influence who are not already in a cube coma into Berserk Mental Breaks, even if on other map tiles. Pawns in caravans at the time of this deactivation instead leave your colony permanently, so be sure to avoid deactivation while addicts are travelling. If you intend to destroy the cube with conscious addicts present, expect to lose pawns and keep your doctors safe so they can treat the injured.
There are simple measures you can take to avoid these violent mental breaks. Colonists already in a cube coma will simply wake up, cured of their addiction. Any obsessed colonist that is anesthetized when deactivation occurs will ignore the event and be free of cubic influence when the anesthesia wears off. It is not yet confirmed if mind-numb serum can prevent these mental breaks[Mindnumb?].
Be aware that colonists in cryptosleep caskets are not protected. They will immediately wake up, escape, and begin rampaging. One unorthodox but cheap and effective method to keep your colony safe from cube berserkers is to draft the afflicted pawns and trap them in isolated 3x3 wall prisons. When they go berserk, they will be unable to inflict any damage before the state wears off. Afterwards, simply deconstruct the walls to release them. Although time-consuming to set up, this approach only costs a small amount of wood.
If your entire colony is afflicted, deactivation of the cube is still possible with the help of temporary pawns from quests. Note that there is a small, but not insignificant, chance that new temporary pawns may become obsessed with time. Therefore, if you are relying on a newly joined pawn to deactivate the cube it should be done as soon as possible.[Detail needed].
Alternatively, you can dispose of the cube via caravan dumping or transport pod, but doing so does not remove your colonists' obsession. Unless the cube is deactivated and destroyed, your colonists will have to go through withdrawal, coma, and recovery.
Upon deactivation the golden cube drops 127-174 Gold[How is it determined?].
Deathrest and Anesthesize prevents a pawn to becoming berserk when the cube is deactivated.
Multiple cubes in one colony[edit]
While multiple cubes in the same colony result in a much higher rate of addiction the deactivation of one cube cures ALL colonists. So there is no seperate addiction for the two cubes.
Analysis[edit]
As with any anomalous object, care should be taken when a cube arrives. It is one of the safest advanced entities you can encounter, but can still prove frustrating as it drops productivity to a crawl. This is especially true for colonies with low populations: the loss of two or three pawns to low productivity is much easier to bear when you have others to handle their burden. So long as you have your researchers aggressively studying the cube every two days, you should be able to deactivate it before it influences many of your colonists. Dosing your best researcher with voidsight serum or making use of an Occultist pawn can speed up research dramatically. Once you unlock the option to destroy the cube, simply anesthetize the cube's victims before deactivation and the threat will be safely dealt with.
However, if you decide you wish to roleplay as a cube cult and keep it around, this is not entirely impossible. The introduction of the cube will gradually cut into your overall productivity, but will also grant a substantial mood bonus to any colonist allowed to consistently interact with it. Colonies with many pawns who are constantly in danger of mental breaking may find this useful, as the sizable mood buffs will change potentially dangerous mental breaks into smaller sculpture creation breaks. However, constant sculpting can prove troublesome as it not only wastes a pawn's time, but also takes up space with relatively low value sculptures which cannot be broken without a mood penalty.
The cube itself has a considerably high beauty value (about on par with a Good quality golden statue) which can even increase the mood of pawns who are not obsessed but simply share the same room as the cube. This can quickly increase the beauty value within a room colonists frequent, such as a rec room or dining room. Keeping the cube on display may prove difficult, though, as your pawns will attempt to play with it constantly. This may limit the display time and relocate the golden cube afterwards. Placing a dedicated 1x1 Stockpile zone for the cube can ensure that it always returns to a specific viewing point. Using a shelf instead will negate its beauty entirely.
Caravans are made substantially more difficult to form and maintain as any affected colonists will eventually be plagued with constant mood penalties and even a coma due to undergoing cube withdrawal[Verify]. Often, you may be restricted to keeping any obsessed colonists at home while unaffected pawns can safely do quests on the world map.
It is desirable for pawns with tortured artist trait to be obsessed with the cube, as the frequent but harmless mental break will still give them creativity inspiration. If you suspect the mysterious cargo is the cube, you can give a pawn interest by knocking out everyone else or moving them off the map with a caravan before accepting it.
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Anomaly DLC Release - Added.
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5331 | https://www.creatingculturalcompetencies.org/uploads/1/1/2/6/112618631/2_poetry_graphic_organizer_examples.pdf | NAME:______ DATE:__ PERIOD:_ SIFT Literary Analysis Strategy Directions: Use the table below to record examples of each of the poetic devices from the literary work Title:_____ Author:____ Symbols Examine the title and text for symbolism Images Identify images and sensory details (sight, sound, taste, odor, texture) Figurative Language Identify and analyze non-standard use of language, including metaphor, simile, repetition, omission, unusual word order, slang, etc. Tone and Theme 1) Discuss the tone taken by the author. 2) Message or moral: Why did the author create this work? SOAPStone Graphic Organizer for Rhetorical Analysis Citing Evidence in Persuasive Text CLOSE READING How do you know? Cite specific evidence in the text. S Who is the Speaker? Who is the speaker? Identify the speaker’s age, gender, class, and education. The voice tells the story. Whose voice is being heard within the text? What can you tell or what do you know about the speaker that helps you understand the point of view expressed? O What is the Occasion? What is the time and place of the piece? What is the current situation (that prompted the writing)? Is this a political event, a celebration, an observation, a critique, or …? Identify the context of the text. A Who is the Audience? Who are the readers to whom this piece is directed? It may be one person or a specific group. Does the speaker specify an audience? What assumptions exist in the text about the intended audience? P What is the Purpose? What is the purpose behind the text? Why did the author write it? What is his goal? (To find the purpose, ask, “What did the author want his audience to think or do as a result of reading this text?”) What is the message? How does the speaker convey this message? S What is the Subject? What topic, content, and ideas are included in the text? State the subject in a few words or a short phrase. Is there more than one subject? How does the author present the subject? Does he introduce it immediately or do you, the reader, have to make an inference? TONE TONE What is the Tone? What is the attitude of the author? Is the author emotional, objective, neutral, or biased about this topic? What types of details “tell” the author’s feelings about the topic? What types of diction (choice of words), syntax (sentence structure), and imagery (metaphors, similes, and other types of figurative language) help reflect the tone? How would you read the passage aloud if you were the author? TP-CASTT Poetry Analysis T Title Before you even think about reading the poetry or trying to analyze it, speculate on what you think the poem might be about based upon the title. Often time, authors conceal meaning and give clues in the title. Jot down what you think this poem will be about P Paraphrase Before you begin thinking about meaning or trying to analyze the poem, don't overlook the literal meaning of the poem. One of the biggest problems that students often make in poetry analysis is jumping to conclusions before understanding what is taking place in the poem. When you paraphrase a poem, write in your own words exactly what happens in each line of the poem. Look at the number of sentences in the poem—your paraphrase should have exactly the same number. This technique is especially helpful for poems written in the 17th and 19th centuries that use language that is harder to understand. C Connotation Although this term usually refers solely to the emotional overtones of word choice, for this chart the term refers to any and all poetic devices, focusing on how such devices contribute to the meaning, the effect, or both of a poem. You may consider imagery, figures of speech (simile, metaphor, personification, symbolism, etc), diction, point of view, and sound devices (alliteration, onomatopoeia, rhythm, and rhyme). It is not necessary that you identify all the poetic devices within the poem. The ones you do identify should be seen as a way of supporting the conclusions you are going to draw about the poem. A Attitude Having examined the poem's devices and clues closely, you are now ready to explore the multiple attitudes that may be present in the poem. Examination of diction, images, and details suggests the speaker's attitude and contributes to understanding. Think about the tone of the poem and how the author has created it. Remember that usually the tone or attitude cannot be named with a single word - think complexity.
S Shifts Rarely does a poem begin and end the poetic experience in the same place. As is true of most us, the poet's understanding of an experience is a gradual realization, and the poem is a reflection of that understanding or insight. Watch for the following keys to shifts: • key words, (but, yet, however, although) • punctuation (dashes, periods, colons, ellipsis) • stanza divisions • changes in line or stanza length or both • irony • changes in sound that may indicate changes in meaning • changes in diction T Title Now look at the title again, but this time on an interpretive level. What new insight does the title provide in understanding the poem.
T Theme What is the poem saying about the human experience, motivation, or condition? What subject or subjects does the poem address? What do you learn about those subjects? What idea does the poet want you take away with you concerning these subjects? Remember that the theme of any work of literature is stated in a complete sentence and make sure to avoid cliche.
BLiska T Title P Paraphrase C Connotation A Attitude S Shifts T Title T Theme BLiska 1 Strategies for Analyzing Poetry Poetry provides opportunities for students to develop an appreciation for poetry and its various techniques. Students will develop the skills necessary to properly understand poems. Reading poetry can certainly be helpful in introducing new vocabulary words and literary elements. This leads to an increase in reading comprehension and fluency that can be put into practice. Students will be able to critically analyze text and explain the intended meaning and effect in their responses. In this section you may access strategies and resources to enhance your teaching of poetry. 1. Poetry Dictionaries 2. Poetry Scrapbooks 3. Poetry and Song 4. Walk Around a Poem 5. Responding to a Poem 6. Partner Poems 7. Thematic poetry 8. Poetry suggestions/links for resources 9. Poetry graphic organizers When should students start analyzing poetry? Poetry can be taken apart as soon as poetry is taught. The earliest question,” what do you think this poem is about?” can be asked in kindergarten. By the end of grade 6 (SLO 2.2.3) for example, students should be responding to poems on an emotional level and understanding the figurative language that is a part of poetry. They should be identifying figurative language discussing how it enhances understanding people, places and action. By the end of grade 12, in a similar outcome, students should be analyzing how language and stylistic choices in oral, print, and other media texts communicate intended meaning and create effect. 2 The following strategies and resources can be used to enhance your teaching of poetry. 1. Poetry Dictionaries Students create their own dictionaries in Grade 8 or 9 and use them through to Grade 12. Students can use “The Frayer Model” as a poetic vocabulary development tool. The model helps to develop a better understanding of complex concepts, by having students identify what something is but what something is not. An example of this model is located in the appendix. 2. Poetry Scrapbooks Poetry scrapbooks are both visual and textual representations of student selected poetry. Students select poems that they like or can relate to. Teachers and students can generate a rubric which outlines the criteria. Students will share and explain why they chose the poems and how the poems relate to them. 3. Poetry and Song Songs are an engaging example of poetry. Teachers can play different parts of a song from a variety of genres. The students respond to the parts by writing the mood that each part evokes in them. Students can identify poetic devices found in the lyrics. Students can use a favourite song and search for a visual that reflects the idea or message of a particular song. Poetry – the area of writing that lends itself to the expression of feelings and ideas using style and rhythm. Beautiful, imaginative, reflective; poetry taps into inner emotion and musical pulse. It allows students to access and share feelings and emotions as no other genre does. Poetry is rhythm, sound, and beat. Children don’t have to understand it to appreciate it, and they become curious about making their own. Poetry is kinesthetic literature at its finest! Poetry moves us. 3 4. “Walk around a Poem” strategy The teacher will photocopy a poem in the middle of a page to allow students to record all their responses in the margins. Suggest that students highlight phrases in the poem that they consider important. (For more information about this strategy see the Grade 12 ELA Foundation for Implementation document, section 4 -116) 5. Responding to a Poem There are a wide variety of strategies available for teachers to model responding to a poem. One effective strategy is the Responding to a Poem sheet (adapted from the Prentice Hall Multisource Activity sheet) located in the Appendix. Another resource is the Poem Analysis sheet also located in the Appendix. 6. Partner Poems Partner Poems is a strategy where two or more voices read aloud a poem to one another. There are many benefits in using this strategy to: • Build self-confidence as students build fluency and comprehension • Develop public speaking skills and confidence – easier to speak with a peer than alone – security blanket Poetry slams are meant for audiences and even reluctant writers are happy to try their voices. Poems defy rules. This means that poetry is accessible to English language learners. Even with limited vocabulary, students can find ways to express their voices. Poetry provides students with the opportunity to learn figurative language and specific literary techniques as no other form does. Poetry allows kids to share their lives through metaphor and simile, through language that breaks the rules of grammar and conventions. 4 • Reinforce comprehension and fluency since you give students time to practice before reading to class • Motivate students since this is fun and not intimidating • Promote group/partner work 7. Thematic poetry There are many poems accessible to teachers based on themes. Theme is defined as a main idea or an underlying meaning of a literary work that may be stated directly or indirectly. Some common themes used in the classroom are nature, growing up, friendship, conflict etc. Students will analyze different types of poems based on a selected theme. They will use a mind map to organize the differences and similarities of the poem. They can determine the type and form of poetry used in the selected theme. There is a list of common themes located in the appendix and presented as a word splash. 8. Poetry suggestions/links for resources Text sources: • Poetry in Focus by Bob Cameron, Margaret Hogan, and Patrick Lashmar • Poetry Alive (Perspectives)/ (Transitions) by Don Saliani • Joyful Noise – Poems for Two Voices, I am Phoenix (to be used for Partner Poems) • Partner Poems for Building Fluency: 25 Original Poems with Research-Based Lessons …. By Bobbi Katz (Jan. 1, 2007) • Inside Poetry by Glenn Kirkland and Richard Davies • Adolescent Literacy – Turning Promise into Practice by Kylene Beers – Robert Probst - Linda Rief (Dialogue with a Poem p 350 – 364, Icebreaker – p 48-49, Dealing with Trauma – p 40-41) Students find their voices in poems. Poetry is meant to be spoken and shared. 5 Online sources: • Partner Poems for Building Fluency: Grades 4-6: 40 Engaging Poems for Two-Voices With Motivating Activities - Comprehension/dp/0545108764/ref=sr_1_2?ie=UTF8&qid=1431109811&sr=8-2&keywords=partner+poems • Poetry Picnic - 9. Poetry graphic organizers There are numerous poetry graphic organizers available for teachers to use in the classroom to support students to better understand poetry and poetic devices. Two suggested graphic organizers are: • Pyramid Organizer Students will record information on the graphic organizer to provide students with the structure that facilitates analysis. (For more information see Grade 9 ELA Foundation document section 1-188, Grade 10 ELA Foundation document section2-234, Grade 11 ELA Foundation document, and section 4- 254-255). • S.O.A.P.S.Tone Organizer The S.O.A.P.S.Tone Organizer teaches students a strategy to use when analyzing literary texts, including poetry. It uses common literary elements to critically examine texts and better organize their responses to text. There are numerous other online resources for teachers to successfully implement in any classroom. An example of this organizer is located in the appendix. 6 APPENDIX 7 Responding to a Poem How you respond to a poem on an emotional and intellectual level is based on your past; on all your experiences, thoughts, and beliefs. The meaning of a poem comes from who you are as much as from the written text. The following questions will help you probe deeper into a poem. 1. What feelings arise in you as you read this poem? 2. What memories or past experiences come to you? 3. Imagery is so powerful; what images are most striking? Which of your senses are awakened? 4. Are you wondering about something? What parts puzzle you? What questions can you ask? 5. Titles offer clues to meaning, what does your title tell you? 6. Can you tell who or what the speaker is in the poem? How can you tell? Why do you think the poet chose this voice? 7. Why do you think the writer decided to use poetry rather than prose? 8. How many examples of figurative language can you find in your poem? Can you find similes, metaphors, personification, apostrophe, etc? 9. Can you find sound devices such as alliteration, onomatopoeia, assonance, etc.? Can you figure out the rhyme scheme? Can you find other literary devices? 10. What’s the opinion of the poet about people or life? Do you agree or disagree with the viewpoint? Explain. 11. Did you like the poem? Hate it? What did you like most? Least? 12. Is there a piece of music you would use to accompany the reading of this poem? What would you choose? Why? Go ahead, take apart a poem. 8 Poem Analysis sheet Name of Poem: _________ Written by: ________ Date written: ________ Additional information found on Poet: _____ _________ Words/line that “caught” your eye: _____ _________ Theme of poem: ________ Poetic Devices found in poem: Metaphor – ________ Simile – _______ Imagery – _________ Other (please identify) _______ _________ How it connects to you/your life: _____ _________ _________ _________ ____________ 9 Possible Poetry Themes 10 |
5332 | https://virtualnerd.com/pre-algebra/linear-functions-graphing/rate-of-change-slope | Rate of Change and Slope | Pre-Algebra | Linear Functions and Graphing | Virtual Nerd
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5333 | https://www.ijcai.org/Proceedings/09/Papers/092.pdf | New Improvements in Optimal Rectangle Packing Eric Huang and Richard E. Korf Computer Science Department University of California, Los Angeles Los Angeles, CA 90095 ehuang@cs.ucla.edu, korf@cs.ucla.edu Abstract The rectangle packing problem consists of find-ing an enclosing rectangle of smallest area that can contain a given set of rectangles without overlap.
Our algorithm picks the x-coordinates of all the rectangles before picking any of the y-coordinates.
For the x-coordinates, we present a dynamic vari-able ordering heuristic and an adaptation of a prun-ing algorithm used in previous solvers. We then transform the rectangle packing problem into a per-fect packing problem that has no empty space, and present inference rules to reduce the instance size.
For the y-coordinates we search a space that models empty positions as variables and rectangles as val-ues. Our solver is over 19 times faster than the pre-vious state-of-the-art on the largest problem solved to date, allowing us to extend the known solutions for a consecutive-square packing benchmark from N=27 to N=32.
1 Introduction Given a set of rectangles, our problem is to find all enclosing rectangles of minimum area that will contain them without overlap. We refer to an enclosing rectangle as a bounding box. The optimization problem is NP-hard, while the prob-lem of deciding whether a set of rectangles can be packed in a given bounding box is NP-complete, via a reduction from bin-packing [Korf, 2003]. The consecutive-square packing benchmark is a simple set of increasingly difficult bench-marks for this problem, where the task is to find the bounding boxes of minimum area that contain a set of squares of sizes 1x1, 2x2, ..., up to NxN [Korf, 2003]. For example, Figure 1 is an optimal solution for N=32. We use this benchmark here but none of the techniques introduced in this paper are spe-cific to packing squares as opposed to rectangles.
Rectangle packing has many practical applications. It ap-pears when loading a set of rectangular objects on a pallet without stacking them, and also in VLSI design where rect-angular circuit components must be packed onto a rectangu-lar chip. Various other cutting stock and layout problems also have rectangle packing at their core.
Figure 1: An optimal solution for N=32 with a bounding box of 85x135.
511 (a) Compulsory part of a 5x2 at x=[0,2] (b) Assigning a 4x2 to [0,2].
Figure 2: Examples of compulsory parts and intervals.
2 Previous Work Korf divided the rectangle packing problem into two subproblems: the minimal bounding box problem and the containment problem. The former finds a bounding box of least area that can contain a given set of rectangles, while the latter tries to pack the given rectangles in a given bounding box. The algorithm that solves the minimal bounding box problem calls the algorithm that solves the containment prob-lem as a subroutine.
2.1 Minimal Bounding Box Problem A simple way to solve the minimal bounding box problem is to find the minimum and maximum areas that describe the set of feasible and potentially optimal bounding boxes.
Bounding boxes of all dimensions can be generated with ar-eas within this range, and then tested in non-decreasing order of area until all feasible solutions of smallest area are found.
The minimum area is the sum of the areas of the given rectan-gles. The maximum area is determined by the bounding box of a greedy solution found by setting the bounding box height to that of the tallest rectangle, and then placing the rectangles in the first available position when scanning from left to right, and for each column scanning from bottom to top.
2.2 Containment Problem Korf’s absolute placement approach modeled rectan-gles as variables and empty locations in the bounding box as values. Moffitt and Pollack’s relative placement ap-proach used a variable for every pair of rectangles to represent the relations above, below, left, and right. Absolute place-ment was faster than relative placement [Korf et al., 2009], which in turn was faster than the methods of Clautiaux et al.
and Beldiceanu et al. .
Simonis and O’Sullivan used Clautiaux et al.’s variable order with additional constraints from Beldiceanu et al. to greatly outperform Korf et al.’s solver . They used Prolog’s backtracking engine to solve a set of constraints which they specified prior to the search effort.
They first assigned the x-coordinates of all the rectangles before any of the y-coordinates. Since we use some of these ideas, we review them here.
Simonis and O’Sullivan used two sets of redundant variables for the x-coordinates. The first set of N variables correspond to “intervals” where a rectangle is assigned an interval of x-coordinates. Interval sizes are hand-picked for each rectangle prior to search, and they induce a smaller rect-angle representing the common intersecting area of placing the rectangle in any location in the interval [Beldiceanu et al., 2008]. Larger intervals result in weaker constraint prop-agation (less pruning) but a smaller branching factor, while smaller intervals result in stronger constraint propagation but a larger branching factor.
As shown in Figure 2b, a 4x2 rectangle assigned an x-interval of [0,2] consumes 2 units of area at each x-coordinate in [2,3], represented by the doubly-hatched area. This “com-pulsory profile” [Beldiceanu et al., 2008] is a constraint com-mon to all positions x ∈[0, 2] of the original 4x2 rectan-gle. If there were no feasible set of interval assignments, then the constraint would save us from having to try individual x-values. However, if we do find a set of interval assignments, then we must search for a set of single x-coordinate val-ues. Simonis and O’Sullivan used a total of 4N vari-ables, assigning (in order) x-intervals, single x-coordinates, y-intervals, and finally single y-coordinates.
3 Overall Search Strategy We separate the containment problem from the minimal bounding box problem, and use Korf et al.’s al-gorithm to solve the latter problem.
Like Simonis and O’Sullivan , we assign all x-coordinates prior to any y-coordinates, and use interval variables for the x-coordinates.
We set a rectangle’s interval size to 0.35 times its width, which gave us the best performance. Finally, we do not use interval variables for the y-coordinates. All of the remaining ideas presented in this paper are our contributions.
Although we use some ideas used by Simonis and O’Sullivan , we do not take a constraint programming approach in which all constraints are specified to a general purpose solver like Prolog, prior to the search effort. Instead, we have implemented our program from scratch in C++, al-lowing us to easily choose which constraints and inferences to use at what time, and giving us more flexibility during search.
For example, as we will explain later, we make different in-ferences depending on the partial solution.
We implemented a chronological backtracking algorithm with dynamic variable ordering and forward checking. Our algorithm works in three stages as it goes from the root of the search tree down to the leaves: 1. It first works on the x-coordinates in a model where variables are rectangles and values are x-coordinate lo-cations, using dynamic variable ordering by area and a constraint that detects infeasible subtrees.
2. For each x-coordinate solution found, it conducts a per-fect packing transformation, applies inference rules to reduce the transformed problem size, and derives conti-guity constraints between rectangles.
3. It then searches for a set of y-coordinates in a model where variables are empty corners and values are rect-angles.
4 Assigning X-Coordinates For the x-coordinates, we propose a dynamic variable order and a constraint adapted from Korf’s wasted-space pruning heuristic. For a bounding box of size WxH we use 512 an array of size W representing the amount of available space in the column at each x-coordinate (i.e., H minus the sum of the heights of all rectangles overlapping x). The array allows us to quickly test if a rectangle can fit in any given column.
4.1 Variable Ordering By Area Our variable order is based on the observation that placing rectangles of larger area is more constraining than placing those of smaller area.
At all times the variable ordering heuristic chooses from among the interval and the single co-ordinate variables. Figure 2a shows the compulsory part in-duced by assigning a 5x2 rectangle the x-interval [0,2]. At this point, we can either assign a single x-coordinate to the 5x2, or assign an interval to another rectangle and place its compulsory part. We always pick the variable whose assign-ment consumes the most area. For example, assigning a sin-gle x-coordinate to the 5x2 rectangle would force the con-sumption of 4 more units of area compared to Figure 2a. We also require that a rectangle’s interval assignment be made before we consider assigning its single x-coordinate.
Although our benchmark has an ordering of squares from largest to smallest, we also must consider interval assign-ments that induce non-square compulsory parts.
Further-more, during search we may rule out some single values of an interval already assigned, increasing the area of the com-pulsory part, so a variable order by area must be dynamic.
4.2 Pruning Infeasible Subtrees The pruning heuristic that we describe below is a constraint that captures the pruning behavior of Korf’s wasted-space pruning algorithm, adapted to the one-dimensional case. Given a partial solution, Korf’s algorithm computed a lower bound on the amount of wasted space, which was then used to prune against an upper bound. In our formulation, we don’t compute any bounds and instead detect infeasibility with a single constraint.
As rectangles are placed in the bounding box, the remain-ing empty space gets chopped up into small, irregular regions.
Soon the empty space is segmented into small enough chunks that they cannot accommodate the remaining unplaced rect-angles, at which point we may prune and backtrack. Assume in Figure 2a we chose x-coordinates for a 3x2 rectangle in a 6x3 bounding box. Without any y-coordinates yet, we only know that 2 units of area have been consumed in each of the columns where the 3x2 rectangle has been placed. We track how much empty space can fit rectangles of a specific height.
Here, there are 9 empty cells (units of area of empty space) that can fit exactly items of height 3, and 3 empty cells that can fit exactly items of height 1.
For every given height h, the amount of space that can ac-commodate rectangles of height h or greater must be at least the cumulative area of rectangles of height h or greater. As-sume we still have to pack a 2x3 and a 2x2 rectangle. Thus, the total area of rectangles of height two or greater is 10. The empty space available that can accommodate rectangles of height two or greater is 9. Therefore we can prune and back-track. If we picked a height of 1 instead of 2, we wouldn’t violate the constraint, so we must check all possible heights for constraint violations. We check this constraint after every x-coordinate assignment.
5 Perfect Packing Transformation For every x-coordinate solution, we transform the problem instance into a perfect packing instance before working on the y-coordinates. A perfect packing problem is a rectan-gle packing problem with the property that the solution has no empty space. This property makes perfect packing much easier since faster solution methods exist [Korf, 2003]. For example, by modeling empty locations as variables and rect-angles as values, space is quickly broken up into regions that can’t accommodate any rectangles. Since no empty space is allowed to fill these regions in perfect packing, this frequently results in pruning close to the root of the search tree.
We transform rectangle packing instances to perfect pack-ing instances by adding to the original set of rectangles a number of 1x1 rectangles equal to the area of empty space in the original instance. Since we know how much empty space there is at each x-coordinate after assigning the x-coordinates of the original rectangles, we can assign the x-coordinates for each of the new rectangles accordingly.
Although the new 1x1 rectangles increase the problem size, the hope is that the ease of solving perfect packing will offset the difficulty of packing more rectangles. Next we describe inference rules to immediately reduce the problem size, and follow with a description of our search space for perfect pack-ing. As we will show, our methods rely on the perfect packing property of having no empty space.
5.1 Composing Rectangles and Subset Sums Occasionally we may represent multiple rectangles with a single rectangle. This occurs if we can show that two rectan-gles must be horizontally next to each other and have the same height and y-coordinate, or vertically stacked and have the same width and x-coordinate. In these cases, we can replace the rectangles with a single larger one, reducing the number of rectangles we have to pack.
The inference rules that we now describe for inferring con-tiguity between rectangles rest on a natural consequence of having no empty space: The right side of every rectangle must border the left side of other rectangles or the bounding box.
We refer to this as the bordering constraint.
Assume in Figure 3a we have a 4x5 bounding box, and a set of rectangles whose x-coordinates have been determined.
We don’t know their y-coordinates yet, so we just arbitrarily lay them out so no two rectangles overlap on the y-axis. Now consider all rectangles whose right or left sides fall on x=3, indicated by the dotted line. We only have the 1x2 on the left and two 1x1s on the right. Due to the bordering constraint, the 1x2 must share its right border with the 1x1s. Furthermore, this border forces the two 1x1s to be vertically contiguous, which we define to mean that the x-axis projections of these rectangles must overlap and the top of one must touch the bot-tom of the other. Since the two 1x1s are vertically contiguous and they have the same width, we can replace them with a 1x2 rectangle as in Figure 3b.
In Figure 3b, we have two 1x2s whose left and right sides fall on x=3. Due to the bordering constraint, the 1x2 on the 513 (a) Vertical composi-tion.
(b) Horizontal com-position.
(c) Subset sums.
(d) Horizontal com-position.
(e) Vertical composi-tion.
(f) One of several packing solution.
Figure 3: Examples of composing rectangles and subset sums.
left must border the 1x2 on the right and have the same y-coordinate values. Since they also have the same heights, we can compose them together into a 2x2 rectangle. The same applies to the rectangles bordering the line at x=1. The result of these two horizontal compositions is shown in Figure 3c.
In Figure 3c the rectangles whose left and right sides fall on x=3 are {2x3, 2x1} on the left and {2x2, 2x2} on the right. Unlike previous cases, since we have more than one rectangle on each side, we can’t immediately conclude verti-cal contiguity unless we show that the 4x1 can never separate the other rectangles vertically. Assume for the sake of con-tradiction that the 2x3 and the 2x1 were separated vertically.
Then by the bordering constraint there is some subset from {2x2, 2x2} that borders the 2x1 with a height of 1. How-ever, there is no such subset! Thus, the 2x3 and 2x1 must be vertically contiguous, as are the two 2x2s. Finally, since the vertically contiguous rectangles have the same widths, we can compose them together as shown in Figure 3d.
Using the same inference rules, we can replace the two 2x4s in Figure 3d with the 4x4 in Figure 3e. Finally, the last two rectangles in Figure 3e may be also composed together if we consider the sides of the bounding box as a single border.
Since we keep track of the order of rectangle compositions, we can extract one of many packing solutions, as shown in Figure 3f. In this example we inferred the y-coordinates with-out any search, but in general, some search may be required.
6 Assigning Y-Coordinates Now we present redundant and partial sets of variables that will be considered simultaneously in order to assign the y-coordinates. During search, from among all variables in all models, we choose to assign next the variable with the fewest possible values. We use forward checking to remove val-ues that would overlap already-placed rectangles, and then prune on empty domains or as required by Korf’s two-dimensional wasted-space pruning rule. Finally, we use a 2D bitmap to draw in placed rectangles to test for overlap.
Figure 4: An example of the empty corner model.
6.1 Empty Corners Model An alternative to asking “Where should this rectangle go?” is to ask “Which rectangle should go here?” In the former model, rectangles are variables and empty locations are val-ues, whereas in the latter, empty locations are variables and rectangles are values. We search the latter model.
In all perfect packing solutions, every rectangle’s lower-left corner fits in some lower-left empty corner formed by other rectangles, sides of the bounding box, or a combination of both. In this model, we have one variable per empty corner.
Since each rectangle goes into exactly one empty corner, the number of empty corner variables is equal to the number of rectangles in the perfect packing instance. The set of values is just the set of unplaced rectangles.
This search space has the interesting property that variables are dynamically created during search because the x and y-coordinates of an empty corner are known only after the rect-angles that create it are placed. Furthermore, placing a rect-angle in an empty corner assigns both its x and y-coordinates.
Note that the empty corner model can describe all perfect packing solutions. Given any perfect packing solution, we can list a unique sequence of rectangles by scanning left to right, bottom to top for the lower-left corners of the rectan-gles.
We use four sets of redundant variables, as this better al-lows us to choose the variable with the fewest values. Each 514 Size Optimal Wasted Boxes KMP Simonis08 FixedOrder PerfectPack Huang09 N Solution Space Tested Time Time Time Time Time 20 34×85 0.69% 14 1:32 :02 21 38×88 0.99% 20 9:54 :07 :03 :18 :03 22 39×98 0.71% 17 37:03 :51 :02 :03 :02 23 64×68 0.64% 19 3:15:23 3:58 :14 :14 :12 24 56×88 0.58% 19 10:17:02 5:56 :40 :43 :37 25 43×129 0.40% 17 2:02:58:36 40:38 2:27 2:15 2:14 26 70×89 0.47% 21 8:20:14:51 3:41:43 10:25 9:45 9:39 27 47×148 0.37% 22 34:04:01:03 11:30:02 1:08:55 35:12 35:12 74×94 0.37% 28 63×123 0.45% 30 2:18:12:13 4:39:31 4:39:31 29 81×106 0.36% 27 8:05:36 8:06:03 30 51×186 0.33% 21 2:17:34:12 2:17:32:52 31 91×110 0.33% 30 4:16:05:08 4:16:03:42 32 85×135 0.31% 36 33:11:36:23 Table 1: Minimum-area bounding boxes containing all consecutive squares from 1x1 up to NxN.
set of variables corresponds to a different right-angle rotation of the empty corner. For example, in Figure 4, after placing rectangles r1 and r2, we now have six empty corner variables c1, c2, ..., c6. c1 and c6 are lower-left empty corner variables, while the other variables correspond to other orientations of an empty corner. Forward checking then removes rectangles with x-coordinates inconsistent with the empty corner’s x-coordinates as well as remove rectangles that would overlap other rectangles when placed in the corner.
6.2 Using Vertical Contiguity During Search Recall that in Figure 3, we composed vertically contiguous rectangles when they had equal widths. Even if we can’t compose rectangles due to their unequal widths, vertical con-tiguity is still useful. During the search for y-coordinates, if we place a rectangle in an empty corner, then we can choose to place its vertically contiguous partner either immediately above or below, giving us a branching factor of two. This effectively represents another set of variables, each with two possible values. We only infer vertical contiguity for certain pairs of rectangles, so this is only a partial model, but our dy-namic variable order considers these variables simultaneously with those in the empty corner model.
7 Experimental Results Table 1 compares the CPU runtimes of five solvers on the consecutive-square packing benchmark.
The first column refers to the instance size. The second specifies the dimen-sions of the optimal solution’s bounding box. The third is the percentage of empty space in the optimal solution. The fourth specifies the total number of bounding boxes the program tested. The remaining columns specify the CPU times re-quired by various algorithms to find all the optimal solutions in the format of days, hours, minutes, and seconds. When there are multiple boxes of minimum area as in N=27, we report the total time required to find all bounding boxes.
Huang09 includes all of our improvements and Simonis08 refers to the previous state-of-the-art solver [Simonis and O’Sullivan, 2008]. The largest problem previously solved was N=27 and took Simonis08 over 11 hours. We solved the same problem in 35 minutes and solve five more open prob-lems up to N=32. KMP refers to Korf et al.’s absolute placement solver. FixedOrder assigns all x-intervals before any single x-coordinates, but includes all of our other ideas.
Huang09’s dynamic variable ordering for the x-coordinates was an order of magnitude faster than FixedOrder by N=28.
The order of magnitude improvement of FixedOrder over Si-monis08 is likely due to our use of perfect packing for as-signing the y-coordinates. We do not include the timing for a solver with perfect packing disabled because it was not com-petitive (e.g., N=20 took over 2.5 hours).
In PerfectPack we use only the lower-left corner when as-signing y-coordinates and also turn off all inference rules re-garding rectangle composition and vertical contiguity. Notice that the running time of this version differs only very slightly from the running time of Huang09, which includes all of our optimizations. This suggests that nearly all of the perfor-mance gains can be attributed to just using our x-coordinate techniques and the perfect packing transformation with the lower-left corner for the y-coordinates.
We benchmarked our solvers in Linux on a 2GHz AMD Opteron 246 with 2GB of RAM. KMP was benchmarked on the same machine, so we quote their results [Korf et al., 2009]. We do not include data for their relative placement solver because it was not competitive.
Results for Simo-nis08 are also quoted [Simonis and O’Sullivan, 2008], ob-tained from SICStus Prolog 4.0.2 for Windows on a 3GHz Intel Xeon 5450 with 3.25GB of RAM. Since their machine is faster than ours, these comparisons are a conservative esti-mate of our relative performance.
In Table 2 the second column is the number of complete x-coordinate assignments our solver found over the entire run of a particular problem instance. The third is the total time spent in searching for the x-coordinate. The fourth is the total time spent in performing the perfect packing transformation and searching for the y-coordinates. Both columns represent 515 Size X-Coordinate Seconds Seconds Ratio N Solutions in X in Y X:Y 20 15 0.1 0.0 2.6 21 665 0.8 2.4 0.3 22 283 2.1 0.4 5.6 23 391 14.1 0.6 22.7 24 870 42.0 2.3 18.6 25 193 160.9 0.3 564.5 26 1,026 688.5 2.8 242.5 27 244 2,524.4 0.6 4,376.6 28 2,715 19,867.5 6.8 2,919.4 29 11,129 34,839.7 33.1 1,052.4 30 10,244 277,087.0 29.2 9,478.9 Table 2: CPU times spent searching for x and y-coordinates.
the total CPU time over an entire run for a given problem instance. The last column is the ratio of time in the third col-umn to that of the fourth. Interestingly, almost all of the time is spent on the x-coordinates as opposed to the y-coordinates, which suggests that if we could efficiently enumerate the x-coordinate solutions, we could also efficiently solve rectangle packing. This is confirmed by the few x-coordinate solutions that exist even for large instances. The data in Table 2 was obtained on a 3GHz Pentium 4 with 2GB of RAM in a sepa-rate experiment from that of Table 1, which is why N=31 and N=32 are missing from Table 2.
8 Future Work The alternative formulation of asking “What goes in this lo-cation?” to “Where does this go?” is not limited to rectangle packing. For example, humans solve jigsaw puzzles by both asking where a particular piece should go, as well as asking what piece should go in some empty region. It would be inter-esting to see how applicable this dual formulation is in other packing, layout, and scheduling problems.
Our algorithm currently only considers integer values for rectangle sizes and coordinates. While this is generally ap-plicable, the model breaks down with rectangles of high-precision dimensions. For example, consider doubling the sizes of all items in a problem instance in both dimensions to get the instance 2x2, 4x4, ..., (2N)x(2N), and then substitute the 2x2 for a 3x3 rectangle. The new instance shouldn’t be harder than the original, but now we must consider twice as many single x-coordinate values, resulting in a much higher branching factor than the original problem. The solution to this problem may require a different representation and many changes to our techniques, and so it remains future work.
9 Conclusion We have presented several new improvements over the previ-ous state-of-the-art in rectangle packing. Within the schema of assigning x-coordinates prior to y-coordinates, we intro-duced a dynamic variable order for the x-coordinates, and a constraint that adapts Korf’s wasted-space pruning heuristic to the one-dimensional case. For the y-coordinates we work on the perfect packing transformation of the original problem, by using a model that assigns rectangles to empty corners, and inference rules to reduce the model’s variables and derive vertical contiguity relationships.
Our improvements in the search for y-coordinates help us solve N=27 over an order of magnitude faster than the pre-vious state-of-the-art, and our improvements in the search for x-coordinates also gave us an order of magnitude speedup by N=28 compared to leaving those optimizations out. With all our techniques, we are over 19 times faster than the previ-ous state-of-the-art on the largest problem solved to date, al-lowing us to extend the known solutions for the consecutive-square packing benchmark from N=27 to N=32.
All of the techniques presented to pick y-coordinates are tightly coupled with the dual view of asking what must go in an empty location. Furthermore, while searching for x-coordinates, our pruning rule is based on the analysis of irreg-ular regions of empty space, and our dynamic variable order also rests on the observation that less empty space leads to a more constrained problem. The success of these techniques in rectangle packing make them worth exploring in many other packing, layout, or scheduling problems.
10 Acknowledgments This research was supported by NSF grant No. IIS-0713178 to Richard E. Korf.
References [Beldiceanu et al., 2008] Nicolas Beldiceanu, Mats Carls-son, and Emmanuel Poder. New filtering for the cumu-lative constraint in the context of non-overlapping rect-angles. In Laurent Perron and Michael A. Trick, editors, CPAIOR, volume 5015 of Lecture Notes in Computer Sci-ence, pages 21–35. Springer, 2008.
[Clautiaux et al., 2007] Franois Clautiaux, Jacques Carlier, and Aziz Moukrim.
A new exact method for the two-dimensional orthogonal packing problem. European Jour-nal of Operational Research, 183(3):1196 – 1211, 2007.
[Korf et al., 2009] Richard Korf, Michael Moffitt, and Martha Pollack. Optimal rectangle packing. To appear in Annals of Operations Research, 2009.
[Korf, 2003] Richard E. Korf. Optimal rectangle packing: Initial results. In Enrico Giunchiglia, Nicola Muscettola, and Dana S. Nau, editors, ICAPS, pages 287–295. AAAI, 2003.
[Moffitt and Pollack, 2006] Michael D.
Moffitt and Martha E. Pollack.
Optimal rectangle packing: A meta-csp approach.
In Derek Long, Stephen F. Smith, Daniel Borrajo, and Lee McCluskey, editors, ICAPS, pages 93–102. AAAI, 2006.
[Simonis and O’Sullivan, 2008] Helmut Simonis and Barry O’Sullivan. Search strategies for rectangle packing. In Peter J. Stuckey, editor, CP, volume 5202 of Lecture Notes in Computer Science, pages 52–66. Springer, 2008.
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5334 | https://www.youtube.com/watch?v=-IwF4_UfPm8 | Solve sin^2(x) + sin(x) - 6 = 0 over [0, 2pi)
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Solve sin^2(x) + sin(x) - 6 = 0 over [0, 2pi)
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solve sine squared of X plus sine X minus 6 equals 0 over 0 to PI solution so it looks like we have a quadratic equation in sine X so we should definitely try to factor it so we know it should look something like this because when you multiply sine x times sine X you get sine squared X so we we need two numbers that multiply to negative 6 but add to one so I'm thinking positive 3 and maybe negative 2 right let's check 3 times negative 2 is negative 6 and when you add them you get 1 yep looks good see if a product equal to 0 so we set each piece equal to 0 so sine X plus 3 equals 0 and sine X minus 2 equals 0 we can solve this one for sine by subtracting 3 from both sides minus 3 minus 3 so we get sine x equals negative 3 here we can add 2 to both sides so we get sine x equals 2 and neither of these has any solution well why well the sine function is bounded by 1 and negative 1 right the maximum value of sine is 1 the minimum value is negative 1 it never goes as low as negative 3 and it never goes as high as 2 so the answer here is no solution that's it |
5335 | https://flexbooks.ck12.org/cbook/ck-12-middle-school-math-concepts-grade-8/section/5.5/primary/lesson/use-proportions-to-find-percent-msm8/ | Skip to content
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5.5
Use Proportions to Find Percent
Written by:Brenda Meery | Jen Kershaw, M.ed
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
A senator wants to start a program to encourage more people to vote in his state. In County A, of the 39150 potential voters, 32,100 people voted. In neighboring County B, of the 81400 potential voters, 57,800 people voted. Which county needs the program more?
In this concept, you will use proportions to figure out percent.
Proportions
A percent is a part of a whole that represents a quantity out of 100. Fractions and decimals are also parts of a whole. Sometimes, you will be given information, but not a percent. You will need to know how to figure out the percent. Percentages, fractions, decimals and proportions can all help to you solve problems and figure out percent.
You began using proportions to figure out a percent when writing fractions as percent. Remember that proportions involve comparing quantities. A proportion is a comparison between two equal ratios. Because both of these are comparing, you can use proportions to help us figure out a percent.
First, write the proportion using over .
This is equal to the percent which is out of 100.
Here is the proportion:
Now let’s apply this proportion to a problem.
What percent is 12 out of 45?
First, write a ratio comparing your given values to the missing percent.
Next, cross multiply.
Then, divide by 45 to isolate .
The answer is 26.7.
Therefore 12 out of 45 is 26.7%.
Examples
Example 1
Earlier, you were given a problem about the voters.
County A has 39150 people and 32100 voted. County B has 81400 people and 57800 voted. Which county had the highest proportion of people voting?
First, write each ratio comparing your given values to the missing percent.
County A
County B
Next, cross multiply.
County A
County B
Then, divide to isolate .
County A
County B
The answer is that 82% of the people in County A voted and 71% of the people in County B voted. Therefore, County A had the highest voter turnout and County B needs the senator’s program more.
Example 2
John ran 8 out of 9 miles. What percent of the total miles did he run?
First, write a ratio comparing your given values to the missing percent.
Next, cross multiply.
Then, divide by 9 to isolate .
The answer is 88.9.
Therefore, John ran 88.9% of the total miles.
Example 3
What percent is 18 out of 50?
First, write a ratio comparing your given values to the missing percent.
Next, cross multiply.
Then, divide by 50 to isolate .
The answer is 36. Therefore 18 out of 50 is 36%.
Example 4
What percent is 22 out of 40?
First, write a ratio comparing your given values to the missing percent.
Next, cross multiply.
Then, divide by 40 to isolate .
The answer is 55.
Therefore 22 out of 40 is 55%.
Example 5
What percent is 78 out of 80?
First, write a ratio comparing your given values to the missing percent.
Next, cross multiply.
Then, divide by 80 to isolate .
The answer is 97.5.
Therefore 78 out of 80 is 97.5%.
Review
Find in the given problems using cross product. Round to the nearest tenths place.
1.
2.
3.
4.
5.
6.
A dentist filled cavities in 8 of his 30 patients on Tuesday. What percent had cavities filled?
A florist delivered 18 out of 25 bouquets. What percent was delivered?
The baker sold 3 out of 4 dozen rolls. What percent was sold?
What percent is 85 out of 5000?
What percent is 15 out of 30?
What percent is 88 out of 1200?
What percent is 99 out of 200?
What percent is 100 out of 330?
What percent is 224 out of 5400?
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
Resources
| Image | Reference | Attributions |
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| | | Credit: League of Women Voters Source: |
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5336 | https://assets.cambridge.org/97811088/43966/excerpt/9781108843966_excerpt.pdf | Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 1 Lecture 1: Introduction and Tileability 1.1 Preamble The goal of the lectures is for the reader to understand the mathematics of tilings. The general setup is to take a lattice domain and tile it with elementary blocks. For the most part, we study the special case of tiling a polygonal domain on the triangular grid (of mesh size 1) by three kinds of rhombi that we call “lozenges.” Panel (a) of Figure 1.1 shows an example of a polygonal domain on the triangular grid. Panel (b) of Figure 1.1 shows the lozenges: each of them is obtained by gluing two adjacent lattice triangles. A triangle of the grid is surrounded by three other triangles; attaching one of them, we get one of the three types of lozenges. The lozenges can also be viewed as orthogonal projections onto the x + y + z = 0 plane of three sides of a unit cube. Figure 1.2 provides an example of a lozenge tiling of the domain of Figure 1.1.
Figure 1.3 shows a lozenge tiling of a large domain, with the three types of lozenges shown in three different colors. The tiling here is generated uniformly at random over the space of all possible tilings of this domain. More precisely, it is generated by a computer that is assumed to have access to perfectly random bits. It is certainly not clear at this stage how such “perfect sampling” may be done computationally; in fact, we address this issue in the very last lecture.
Figure 1.3 is meant to capture a “typical tiling,” making sense of what this means is another topic that will be covered in this book. The simulation reveals an interesting feature: there are special regions next to the boundaries of the domain, and in each such region, there is only one (rather than three) type of lozenge. These regions are typically referred to as “frozen regions,” and their boundaries are “arctic curves”; their discovery and study have been one of the important driving forces for investigations of the properties of random tilings.
Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 2 Lecture 1: Introduction and Tileability A = 5 B = 5 C = 5 (a) (b) Figure 1.1 Panel (a): A 5 × 5 × 5 hexagon with 2 × 2 rhombic hole. Panel (b): Three types of lozenges obtained by gluing two adjacent triangles of the grid.
A = 5 B = 5 C = 5 Figure 1.2 A lozenge tiling of a 5 × 5 × 5 hexagon with a hole.
We often identify a tiling with a so-called height function. The idea is to think of a two-dimensional (2D) stepped surface living in a three-dimensional (3D) space and treat tiling as a projection of such surface onto x + y + z = 0 plane along the (1, 1, 1) direction. In this way, three lozenges become projections of three elementary squares in 3D space parallel to each of the three coordinate planes. We formally define the height function later in this lecture. We refer to a web page of Borodin and Borodin1 for a gallery of height functions in a 3D virtual-reality setting.
1 A. Borodin and M. Borodin, A 3D representation for lozenge tilings of a hexagon,
Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 1.2 Motivation 3 Figure 1.3 A perfect sample of a uniformly random tiling of a large hexagon with a hole. (I thank Leonid Petrov for this simulation.) 1.2 Motivation Figure 1.3 is beautiful, and for mathematicians, this suffices for probing more deeply into it and trying to explain the various features that we observe in the simulation.
There are also some motivations from theoretical physics and statistical mechanics. Lozenge tilings serve as “toy models” that help in understanding the 3D Ising model (a standard model for magnetism). Configurations of the 3D Ising model are assignments of + and −spins to lattice points of a domain in Z3. A parameter (usually called “temperature”) controls how much the adjacent spins are inclined to be oriented in the same direction. The zero-temperature limit leads to spins piling into as large as possible groups of the same orientation; these groups are separated with stepped surfaces whose projections in (1, 1, 1) direction are lozenge tilings. For instance, if we start from the Ising model in a cube and fix boundary conditions to be + along three faces (sharing a single Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 4 Lecture 1: Introduction and Tileability vertex) of this cube and −along the other three faces, then we end up with lozenge tilings of a hexagon in the zero-temperature limit.2 Another deformation of lozenge tilings is the six-vertex or square-ice model, whose state space consists of configurations of the molecules H2O on the grid.
There are six weights in this model (corresponding to the six ways to match an oxygen with two out of the neighboring four hydrogens), and for particular choices of the weights one discovers weighted bijections with tilings.
We refer the reader to Baxter (2007) for more information about the Ising model and the six-vertex model, further motivations to study them, and approaches to the analysis. In general, both the Ising and six-vertex models are more complicated objects than lozenge tilings, and they are much less understood. From this point of view, the theory of random tilings that we develop in these lectures can be treated as the first step toward the understanding of more complicated models of statistical mechanics.
For yet another motivation, we notice that the 2D stepped surfaces of our study have flat faces (these are frozen regions consisting of lozenges of one type, cf. Figures 1.3 and 1.5) and, thus, are relevant for modeling facets of crystals. One example from everyday life is a corner of a large box of salt. For a particular (nonuniform) random tiling model leading to the shapes reminiscent of such a corner, we refer the reader to Figure 10.1 in Lecture 10.
1.3 Mathematical Questions We now turn to describing the basic questions that drive the mathematical study of tilings.
1. Existence of tilings: Given a domain R drawn on the triangular grid (and thus consisting of a finite family of triangles), does there exist a tiling of it?
For example, a unit-sided hexagon is trivially tileable in two different ways, and the bottom part of Panel (b) in Figure 1.1 shows one of these tilings.
On the other hand, if we take the equilateral triangle of side length 3 as our domain R, then it is not tileable. This can be seen directly because the corner lozenges are fixed and immediately cause obstruction. Another way to prove nontileability is by coloring the unit triangles inside R in white and black colors in an alternating fashion. Each lozenge covers one black and one white triangle, but there is an unequal number of black and white triangles 2 See Shlosman (2001), Cerf and Kenyon (2001), and Bodineau et al. (2005) for a discussion of the common features in low-temperature and zero-temperature 3D Ising models, as well as the interplay between the topics of this book and more classical statistical mechanics.
Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 1.3 Mathematical Questions 5 (a) No tilings (b) Figure 1.4 Panel (a): The top domain is tileable and the bottom one is not.
Panel (b): A possible tiling.
in the R: the equilateral triangle of side length three has six triangles of one color and three triangles of another color.
Another example is shown in Figure 1.4. In panel (a), we see two domains.
The top one is tileable, whereas the bottom one is not.
More generally, is there a simple (and efficient, from the computational point of view) criterion for the tileability of R? The answer is yes by a theorem developed by Thurston (1990). We discuss this theorem in more detail in Section 1.4 later in this lecture.
2. How many tilings does a given domain R have? The quality of the answer depends on one’s taste and perhaps what one means by a “closed-form”/“explicit” answer. Here is one case where a “good” answer is known by the following theorem due to MacMahon (who studied this problem in the context of plane partitions, conjectured a formula in MacMahon (1896) and proved it in the 1915 book, see Article 495 of MacMahon (1960)). Let R be a hexagon with side lengths A, B, C, A, B, C in cyclic order. We denote this henceforth by the A× B × C hexagon (in particular, panel (a) of Figure 1.1 shows 5 × 5 × 5 hexagon with a rhombic hole).
Theorem 1.1 (MacMahon, 1896) The number of lozenge tilings of A×B×C hexagon equals A Y a=1 B Y b=1 C Y c=1 a + b + c −1 a + b + c −2.
(1.1) As a sanity check, one can take A = B = C = 1, yielding the answer 2, and indeed, one readily checks that there are precisely two tilings of 1×1×1 Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 6 Lecture 1: Introduction and Tileability hexagon. A proof of this theorem is given in Section 2.2. Another situation where the number of tilings is somewhat explicit is for the torus, and we discuss this in Lectures 3 and 4. For general R, one cannot hope for such nice answers, yet certain determinantal formulas (involving large matrices whose size is proportional to the area of the domain) exist, as we discuss in Lecture 2.
3. Law of large numbers: Each lozenge tilings is a projection of a 2D surface and therefore can be represented as a graph of a function of two variables, which we call the “height function” (its construction is discussed in more detail in Section 1.4). If we take a uniformly random tiling of a given domain, then we obtain a random height function h(x, y) encoding a random stepped surface. What is happening with the random height function of a domain of linear size L as L →∞? As we will see in Lectures 5–10 and in Lecture 23, the rescaled height function has a deterministic limit lim L→∞ 1 L h(Lx, Ly) = ˆ h(x, y).
An important question is how to compute and describe the limit shape ˆ h(x, y). One feature of the limit shapes of tilings is the presence of regions where the limiting height function is linear. In terms of random tilings, these are “frozen” regions, which contain only one type of lozenge. In particular, in Figure 1.3, there is a clear outer frozen region near each of the six vertices of the hexagon; another four frozen regions surround the hole in the middle.
Which regions are “liquid,” that is, contain all three types of lozenges?
What is the shape of the “arctic curve,” that is, the boundary between frozen and liquid regions? For example, with the aL × bL × cL hexagon setup, one can visually see from Figure 1.5 that the boundary appears to be an inscribed ellipse: Theorem 1.2 (Baik et al., 2003; Cohn et al., 1998; Gorin, 2008; Petrov, 2014a) For aL × bL × cL hexagon, a uniformly random tiling is with high probability asymptotically frozen outside the inscribed ellipse as L →∞.
In more detail, for each (x, y) outside the ellipse, with probability tending to 1 as L →∞, all the lozenges that we observe in a finite neighborhood of (xL, yL) are of the same type.
The inscribed ellipse of Theorem 1.2 is the unique degree 2 curve tangent to the hexagon’s sides. This characterization in terms of algebraic curves extends to other polygonal domains, where one picks the degree such that there is a unique algebraic curve tangent (in the interior) of R. Various approaches to Theorem 1.2, its relatives, and generalizations are discussed in Lectures 7, 10, 16, 21, 23.
Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 1.3 Mathematical Questions 7 Figure 1.5 Arctic circle of a lozenge tiling.
4. Analogs of the central limit theorem: The next goal is to understand the random field of fluctuations of the height function around the asymptotic limit shape, that is, to identify the limit lim L→∞(h(Lx, Ly) −E[h(Lx, Ly)]) = ξ(x, y).
(1.2) Note the unusual scaling; one may naively expect a need for dividing by √ L to account for fluctuations, as in the classical central limit theorem for sums of independent random variables and many similar statements. But there turns out to be some “rigidity” in tilings, and the fluctuations are much smaller. ξ(x, y) denotes the limiting random field; in this case, it can be identified with the so-called “Gaussian free field.” The Gaussian free field is related to conformal geometry because it turns out to be invariant under conformal transformations. This topic will be explored in Lectures 11, Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 8 Lecture 1: Introduction and Tileability Figure 1.6 Fluctuations of the centered height function for lozenge tilings of a hexagon with a hole. Another drawing of the same system is shown in Figure 24.2 later in the text.
12, 21, and 23. For now, we confine ourselves to yet another picture, given in Figure 1.6.
5. Height of a plateau/hole: Consider Figures 1.2 and 1.3. The central hole has an integer-valued random height. What is the limiting distribution of this height? Note that comparing with (1.2), we expect that no normalization is necessary as L →∞, and therefore the distribution remains discrete as L →∞. Hence, the limit cannot be Gaussian. You can make a guess now or proceed to Lecture 24 for the detailed discussion.
6. Local limit: Suppose we “zoom in” at a particular location inside a random tiling of a huge domain. What are the characteristics of the tiling there?
For example, consider a certain finite pattern of lozenges; call it P, and see Figure 1.7 for an example. Asymptotically, what is the probability that P appears in the vicinity of (Lx, Ly)? Note that if P consists of a single lozenge, then we are just counting the local proportions for the lozenges of three types; hence, one can expect that they are reconstructed from the gradients of the limit shape ˜ h. However, for more general P, it is not clear Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 1.4 Thurston’s Theorem on Tileability 9 Figure 1.7 An example of a local pattern P of lozenges. The bulk-limit question asks about the probability of observing such (or any other) pattern in a vicinity of a given point (Lx, Ly) in a random tiling of a domain of linear scale L →∞.
what to expect. This is called a “bulk-limit” problem, and we return to it in Lectures 16 and 17.
7. Edge limit: How does the arctic curve (border of the frozen region) fluctuate? What is the correct scaling? It turns out to be L 1 3 here, something that is certainly not obvious at all right now. The asymptotic law of rescaled fluctuations turns out to be given by the celebrated Tracy–Widom distribution from random matrix theory, as we discuss in Lectures 18 and 19.
8. Sampling: How does one sample from the uniform distribution over tilings?
The number of tilings grows extremely fast (see, e.g., the MacMahon formula (1.1)), so one can not simply exhaustively enumerate the tilings on a computer, and a smarter procedure is needed. We discuss several approaches to sampling in Lecture 25.
9. Open problem: Can we extend the theory to 3D tiles?
1.4 Thurston’s Theorem on Tileability We begin our study from the first question: Given a domain R, is there at least one tiling? The material here is essentially based on Thurston (1990).
Without loss of generality, we may assume R is a connected domain; the question of tileability of a domain is equivalent to that of its connected components. We start by assuming that R is simply connected, and then remove this restriction.
We first discuss the notion of a height function in more detail, and how it relates to the question of the tileability of a domain. There are six directions on the triangular grid, and the unit vectors in those directions are as follows: a = (0, 1), b = , − √ 3 2 , −1 2 + -, c = , √ 3 2 , −1 2 + -, −a, −b, −c.
Cambridge University Press 978-1-108-84396-6 — Lectures on Random Lozenge Tilings Vadim Gorin Excerpt More Information www.cambridge.org © in this web service Cambridge University Press 10 Lecture 1: Introduction and Tileability a = (0, 1) b = − √ 3 2 , −1 2 c = √ 3 2 , −1 2 Figure 1.8 Out of the six lattice directions, three are chosen to be positive (in bold).
We call a, b, c positive directions, and their negations are negative directions, as in Figure 1.8.
We now define an asymmetric nonnegative distance function d(u, v) for any two vertices u, v on the triangular grid (which is the lattice spanned by a and b) and a domain R. d(u, v) is the minimal number of edges in a positively oriented path from u to v staying within (or on the boundary of) R. This is well defined as we assumed R to be connected. The asymmetry is clear: Consider R consisting of a single triangle, and let u, v be two vertices of it. Then d(u, v) = 1, and d(v, u) = 2 (or vice versa).
We now formally define a height function h(v) for each vertex v ∈R, given a tiling of R. This is given by a local rule: if u →v is a positive direction, then h(v) −h(u) = 1, if we follow an edge of a lozenge, −2, if we cross a lozenge diagonally.
(1.3) It may be easily checked that this height function is defined consistently. This is because the rules are consistent for a single lozenge (take, e.g., the lozenge {0, a, b, a + b}), and the definition extends consistently across unions. Note that h is determined up to a constant shift. We may assume without loss of generality that our favorite vertex v0 has h(v0) = 0.
Let us check that our definition matches the intuitive notion of the height.
For that, we treat the positive directions a, b, c in Figure 1.8 as projections of coordinate axes Ox, Oy, Oz, respectively. Take one of the lozenges, say {0, a, b, a+b}. Up to rescaling by the factor √2/3, it can be treated as a projection of the square {(0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 0)} onto the plane x + y + z = 0.
Hence, our locally defined height function becomes the value of x + y + z. A similar observation is valid for two other types of lozenges. The conclusion is that if we identify a lozenge tiling with a stepped surface in 3D space, then our height is the (signed and rescaled) distance from the surface point to its projection onto the x + y + z = const plane in the (1, 1, 1) direction. |
5337 | https://www.ncbi.nlm.nih.gov/books/NBK513262/ | Deep Neck Infections - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Deep Neck Infections
Mohamed Almuqamam; Francisco J. Gonzalez; Sanjeev Sharma; Noah P. Kondamudi.
Author Information and Affiliations
Authors
Mohamed Almuqamam 1; Francisco J. Gonzalez; Sanjeev Sharma; Noah P. Kondamudi 2.
Affiliations
1 RCSI-Bahrain/Brooklyn Hospital Center
2 Icahn School of Medicine at Mount Sinai , The Brooklyn Hospital Center
Last Update: August 11, 2024.
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Continuing Education Activity
Deep neck infections affect the neck's cervical spaces and can rapidly progress to life-threatening complications, making them a significant health concern with notable morbidity and potential mortality. These infections typically arise from local extensions of infections in the tonsils, parotid glands, cervical lymph nodes, and odontogenic structures. Symptoms often result from local pressure effects on the respiratory, nervous, or gastrointestinal tracts, including neck swelling, dysphagia, dysphonia, and trismus. Clinical presentations vary depending on the deep neck space involved (eg, parapharyngeal, retropharyngeal, prevertebral, submental, masticator) and the extent of infection, often involving fever, neck pain, and respiratory distress. Management includes empiric antimicrobial therapy tailored to the expected microbiology, local resistance patterns, and surgical drainage for significant abscesses. Ensuring airway security is paramount, especially in patients with submandibular or odontogenic infections and those exhibiting airway symptoms.
This activity for healthcare professionals enhances competence in identifying deep neck infections, performing recommended evaluations, and implementing appropriate interprofessional approaches to management. Participants learn to recognize the varied clinical presentations, select suitable antimicrobial therapies, and perform necessary surgical interventions. The course emphasizes the importance of collaboration among otolaryngologists, infectious disease specialists, radiologists, and critical care teams to optimize patient outcomes. Through coordinated care, healthcare providers can ensure timely diagnosis, effective treatment, and comprehensive postoperative management, ultimately improving patient safety and recovery.
Objectives:
Identify the clinical signs and symptoms of deep neck infections to ensure timely diagnosis.
Apply evidence-based guidelines for managing deep neck infections, including antimicrobial and surgical treatment.
Differentiate between various deep neck infections based on anatomical location and clinical presentation.
Apply interprofessional team strategies to improve care coordination and outcomes in patients with deep neck infections.
Access free multiple choice questions on this topic.
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Introduction
Deep neck infections are serious but treatable infections that affect the deep cervical spaces. These infections can rapidly progress and lead to life-threatening complications, making them a significant health concern with notable morbidity and potential mortality. Deep neck infections commonly arise from local extensions of infections in the tonsils, parotid glands, cervical lymph nodes, and odontogenic structures. Symptoms often result from local pressure effects on the respiratory, nervous, or gastrointestinal tracts, including neck swelling, dysphagia, dysphonia, and trismus.
Understanding the pathophysiology of these infections requires knowledge of the cervical compartments and interfacial spaces. Additionally, host factors such as immunocompromised states, comorbid conditions, trauma, recent instrumentation, and intravenous drug use can influence the spread and severity of infections. Clinical presentations vary depending on the deep neck space involved (e.g., parapharyngeal, retropharyngeal, prevertebral, submental, masticator) and the extent of infection, often involving fever, neck pain, and respiratory distress. Diagnosis can be challenging due to the deep location of abscesses, making imaging techniques like computed tomography (CT) with contrast essential for accurate detection.
Management includes empiric antimicrobial therapy tailored to the expected microbiology, local resistance patterns, and surgical drainage for significant abscesses. Ensuring airway security is paramount, especially in patients with submandibular or odontogenic infections and those exhibiting airway symptoms. Surgical consultation is recommended for persistent infections despite antibiotic treatment.
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Etiology
Deep neck space infections are almost uniformly polymicrobial, originating from the normal oral cavity flora and upper respiratory tract. The most common source of deep neck infections among adults is the dental and periodontal structures, with the second most common source being the tonsils. Tonsil and pharyngeal sources are the most common etiologies in children. Streptococcus pyogenes,Streptococcus viridans, Staphylococcus aureus, Klebsiella, gram-negative rods, anaerobes, and Fusobacterium species are frequently encountered microorganisms, representing pathologic overgrowth of expected oropharyngeal flora. Actinomyces, Mycobacterium, and fungi are also potential causative organisms but are rarer. The presence of risk factors such as immunocompromised state, diabetes mellitus, intravenous (IV) drug use, and the site of origin of infection influences the type of causative organism.
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Epidemiology
Retropharyngeal and parapharyngeal abscesses are prevalent types of deep neck infections. Retropharyngeal abscesses develop from infections within the potential space between the pharyngeal constrictor muscles and the posterior vertebral fascia. At the same time, parapharyngeal abscesses typically arise from infections that spread directly through the pharyngeal wall. Although these infections are relatively uncommon, with an annual incidence of approximately 0.22 cases per 10,000 individuals, they pose a significant risk to children. Alarmingly, research indicates a rising incidence rate over the past 10 years.
Retropharyngeal and parapharyngeal abscesses predominantly affect children younger than 5 years old, often following respiratory tract infections that extend to the retropharyngeal and parapharyngeal lymph nodes. These infections are primarily observed in young children as these lymph nodes typically regress by age 5.Common symptoms of retropharyngeal and parapharyngeal abscesses include difficulty swallowing (i.e., dysphagia), neck stiffness, fever, painful swallowing (i.e., odynophagia), and respiratory distress. Deep neck infections are particularly hazardous due to their proximity to and potential communication with the posterior mediastinum and cervical vessels, which can lead to severe complications such as airway obstruction, sepsis, necrotizing mediastinitis, carotid artery aneurysm, and occlusion of cervical vessels.
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Pathophysiology
To understand the pathophysiology of deep neck infections, a thorough understanding of the cervical compartments and interfacial spaces is essential. The cervical fascia can be divided into superficial and deep fascia. The superficial fascia is the subcutaneous tissue of the neck and contains the platysma. This layer completely envelops the head and neck. The deep fascia of the neck is divided into superficial, middle, and deep layers. The superficial layer of the deep fascia covers the submaxillary and parotid glands, the trapezius, sternocleidomastoid, and strap muscles. It is also termed the investing layer. Infections of odontogenic and submandibular origin affect this space, which includes the submandibular and masticator spaces. The middle layer encloses vital parts of the neck, including the pharynx, larynx, trachea, upper esophagus, thyroid, and parathyroid glands.
Infections of pharyngeal, tonsillar, and laryngeal origin affect this space, which includes the parapharyngeal and retropharyngeal spaces. This space can also become involved by odontogenic infections of the 2nd and 3rd molars, where infection can spread inferior to the dentate line of the mandible to penetrate the middle layer of deep cervical fascia. The deep cervical fascia's deep layer, also called prevertebral fascia, covers the vertebral column and muscles of the spine. An alar fascia in this space forms the terminus of the retropharyngeal space and lies between the middle layer of deep cervical fascia and the prevertebral fascia proper. The space between this alar layer and the prevertebral fascia is the "danger space," as it is in continuity with the mediastinum, and infections of upper aerodigestive origin can spread freely to cause mediastinitis. Actual retropharyngeal infections can involve the deep layer of cervical fascia. Still, the hematogenous spread of other infections (eg, in IV drug users) can lead to vertebral and prevertebral abscesses.
Infections in these spaces can lead to clinical lymphadenopathy in the appropriate lymphatic chains, most commonly in the anterior and posterior cervical lymphatic chains. Host factors such as an immunocompromised state, the presence of comorbid conditions (eg, diabetes), trauma, recent instrumentation in the area (eg, surgery or dental work), and IV drug abuse can further influence the spread of infection to deeper layers.
Lemierre syndrome occurs when bacteria invade the pharyngeal mucosa, weakened by an earlier viral or bacterial pharyngitis, spreading to the lateral pharyngeal space. This leads to internal jugular vein septic thrombophlebitis and subsequent metastatic infections. Fusobacterium necrophorum is the most frequently identified pathogen in Lemierre syndrome, with Fusobacterium nucleatum also being a notable cause. The most common metastatic infection associated with this syndrome is pneumonia or pleural empyema. Septic syndrome, along with ear, neck, and pulmonary empyema, is a rare but increasingly recognized condition. Given the rapid and often severe progression of the disease, early diagnosis and immediate antibiotic treatment are crucial.
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History and Physical
Clinical presentation of these infections is variable based on the primary site of infection, the fascial plane involved, the extent of inflammation and disease, the presence of abscess and local pressure effects, and other systemic conditions. Most patients present with fever and neck pain. Associated symptoms such as dental pain, dysphagia, stridor, dysphonia, trismus, pain in neck movements, and respiratory distress can provide clues regarding the potentially affected facial plane. Predisposing factors such as immunocompromised state, recent oral or dental procedures, recent neck or oral trauma, recent neck surgery or radiation, IV drug use, or diabetes mellitus should be sought.
Patients may be febrile and appear ill and toxic. Inspection of the neck may reveal asymmetry, redness, swelling, induration, and regional lymphadenitis. Torticollis may be present. Abscesses in this area are more complex to diagnose clinically due to the frequent absence of fluctuance due to the taut fascia, overlying muscles, and deep location. However, they are easily demonstrated on computed tomography (CT). Physicians should have a low threshold to obtain such imaging so patients can lie supine and protect their airways.
Proximal deep neck (peritonsillar, parapharyngeal, parotid, and submandibular) infections and abscesses tend to present with a sore throat and sometimes trismus. Trismus occurs due to the local inflammation of the muscles of mastication or the infection’s direct involvement of these muscles. A physical exam may reveal neck or lower facial swelling, local erythema, tenderness, and regional lymphadenitis. Medial displacement of the uvula in conjunction with tonsillar asymmetry suggests peritonsillar abscess, whereas medial displacement of the pharyngeal wall suggests parapharyngeal space infection. Local pressure may result in dysphagia or odynophagia and may have associated inflammation in the cricoarytenoid joints. If the vagus nerve is affected, dysphonia and hoarseness, or “hot potato” voice, can occur. Infections in the submandibular space may occur after spreading from dental abscesses, sublingual or submaxillary salivary glands, or oral infections following trauma. Cellulitis in this space is also known as Ludwig angina if it originates from the 3rd molars, which can lead to life-threatening airway obstruction if untreated. Ludwig angina presents with drooling, inability to swallow, trismus, induration, and elevation of the floor of the mouth.
A peritonsillar abscess is a pus collection in the space between the superior constrictor muscle and the tonsillar capsule. Common symptoms include a sore throat, swallowing pain, high fever, and drooling. It can lead to upper airway obstruction and complicated intubation due to jaw muscle spasms (trismus). While group A beta-hemolytic streptococci (Streptococcus pyogenes) is the most frequently cultured bacterium from these abscesses, identifying the specific organism is generally not required for diagnosis or treatment planning.
Parapharyngeal and retropharyngeal space infections are more common in children and usually follow an upper respiratory infection. They often present with dysphagia, drooling, and stridor and can lead to airway compromise or spread into the chest to cause mediastinitis. Infections in the danger space, located posterior to the retropharyngeal space and anterior to the prevertebral space, with loose areolar tissue, facilitate the rapid spread of disease to surrounding regions and often present with complications (eg, mediastinitis, empyema, and sepsis).
Parapharyngeal abscess can lead to severe complications such as jugular vein thrombosis (Lemierre syndrome), venous septic embolus, disseminated intravascular coagulopathy, and carotid artery pseudoaneurysm or rupture. Cervical nervous and bone structures can also be affected, resulting in conditions like epidural abscess, atlanto-axial subluxation, cervical osteomyelitis, spinal cord abscess, and meningitis.
André Lemierre observed that septicemia and distant septic emboli can originate from various sites, including the nasopharynx, mouth, jaws, middle ear infections (otitis media), urinary passages, as well as during conditions like mastoiditis, purulent endometritis, and appendicitis. The bacterium Fusobacterium necrophorum is most commonly responsible for Lemierre syndrome and shows intrinsic resistance to several antibiotics, including macrolides, fluoroquinolones, tetracyclines, and aminoglycosides. Therefore, early diagnosis and appropriate antibiotic treatment are essential for a favorable outcome of this syndrome.Lung lesions often manifest as necrotic cavitary lesions but can also present as infiltrates, pleural effusions, empyema, lung abscesses, and necrotizing mediastinitis. Patients with pulmonary involvement may exhibit pneumonia or empyema and could develop acute respiratory distress syndrome, sometimes necessitating mechanical ventilation. Other less common manifestations of metastatic disease include soft tissue abscesses, pyomyositis, abscesses in the spleen and liver, osteomyelitis, endocarditis, pericarditis, renal abscesses, and brain abscesses.
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Evaluation
Biomarkers play a crucial role in assessing inflammatory activity during acute infections. Neutrophilia and systemic neutrophil activation can cause organ damage in cases of sepsis. The characteristic rise in neutrophil counts and the reduction in lymphocyte counts have led to the proposal of various leukocyte ratios as indicators of infection and potential immune dysfunction. Several inflammatory biomarkers have been suggested as prognostic indicators in sepsis and infections, including C-reactive protein (CRP), erythrocyte sedimentation rate (ESR), neutrophil to lymphocyte ratio (NLR), mean platelet volume (MPV), and platelet to lymphocyte ratio (PLR). Studies have identified an NLR cut-off value of over 8.02 as a predictor for the necessity of incision and drainage in managing deep neck infections (DNI). Other significant factors include elevated CRP levels (above 41.25 mg/L) and higher ESR values (above 56.6 mm/h).
A complete blood count (CBC) usually shows leukocytosis, often with a left shift, and chemistry may reveal evidence of dehydration if the patient’s fluid intake is poor due to pain. A blood culture should be obtained if the patient demonstrates clinical features of sepsis, and cultures should be obtained of any purulent discharge in the affected region.
In the past, lateral neck radiographs were commonly employed in diagnosing deep neck infections (DNIs). However, their use has significantly declined as modern radiological techniques have demonstrated much higher diagnostic sensitivity and provide more reliable information for determining the need for surgical intervention. Plain radiography of the neck in children may suggest retropharyngeal abscess when the prevertebral soft tissue shadow is >7 mm at the C2 level or >14 mm at the C6 level. The soft tissue shadow is >22 mm among adults at the C6 level. Additionally, plain radiographs may reveal a foreign body or subcutaneous air when present. They can importantly rule out such entities in the pediatric population, where accurate history can be challenging to obtain. Despite this shift, chest X-rays remain crucial when complications such as mediastinitis, pneumomediastinum, lower airway foreign bodies, or empyema are suspected.
Intraoral or transcervical ultrasonography (US) is often ideal for assessing superficial lesions, such as cervical adenitis and peritonsillar abscess (PTA), and performing percutaneous image-guided aspiration or drainage of pus. This method provides the advantage of immediate use, facilitating quick decisions regarding further diagnostics and appropriate therapy initiation. Additionally, the US reduces the need for contrast-enhanced computed tomography (CECT) scans, minimizing children’s exposure to ionizing radiation. However, the US has several limitations. It is less effective for visualizing deeper neck spaces, which restricts its diagnostic utility in many deep neck infections (DNIs). Furthermore, the success of the US heavily depends on the operator’s skill, making the expertise of the physician performing the ultrasound crucial for accurate diagnosis and treatment decisions. If the ultrasound results are questionable, a CT scan should be obtained if the patient has no contraindications.
CT with contrast is the gold-standard imaging modality for diagnosing the source and extent of deep neck infection.Contrast-enhanced CT (CECT) is essential, as non-contrast CT scans are much less effective at distinguishing between cellulitis and abscesses.Additionally, CECT provides detailed information about the abscess’s relationship with nearby structures, such as major neck vessels, which is crucial for safely performing abscess drainage and reducing surgical risks.
Magnetic resonance imaging (MRI) may also be considered, though logistically more challenging.MRI offers superior soft-tissue characterization compared to ultrasound (US) and CT, enabling more accurate differentiation between abscesses, cellulitis, and lymphadenomegaly in children and adults. MRI is valuable for diagnosing life-threatening complications non-invasively, and its findings often have significant prognostic implications. Additionally, MRI is associated with low radiation exposure. However, MRI is less frequently used as the primary imaging modality in emergencies due to limited availability, longer scanning times, reporting challenges, higher costs, and sedation issues in children.
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Treatment / Management
Directed antimicrobial coverage, surgical drainage for discrete abscesses, and aggressive supportive care are the main management options. Empiric regimens based on the expected microbiology and local resistance data should be initiated and adjusted appropriately once the organism and its sensitivities become available. Clinical trials have not established the optimal antimicrobial regimen for treating deep neck space infections. However, IV nafcillin or vancomycin plus gentamycin or tobramycin combination, ampicillin/sulbactam, or clindamycin are generally accepted initial choices. For methicillin-resistant Staphylococcus aureus (MRSA) infections, vancomycin or linezolid plus cefepime may be utilized. Alternative medications include metronidazole, imipenem, meropenem, and piperacillin-tazobactam. MRSA coverage must be included as part of the initial treatment regimen for patients who are at high risk for being MRSA carriers or having MRSA infection, including those with comorbid disease (eg, diabetes mellitus), a history of IV drug use and in communities or hospitals where MRSA incidence is substantial.
For most deep neck infections, especially parapharyngeal, retropharyngeal, or prevertebral space infections, antibiotic treatment should generally be continued for 2 to 3 weeks, and longer courses may be required when complications are present. Antibiotics can be switched to the oral route once there is significant clinical improvement and the patient can tolerate oral intake. Consultation with head and neck surgeons is recommended, as surgical drainage may become necessary if no improvement is noted after 48 hours of antibiotic therapy. All but the smallest deep neck abscesses typically warrant surgical drainage, while small or questionable abscesses and phlegmon often respond well to appropriate aggressive medical management.
Treating Lemierre syndrome requires a multidisciplinary approach involving infectious disease specialists, pharmacologists, radiologists, otolaryngologists, and thoracic surgeons to ensure timely diagnosis and effective treatment. This typically includes appropriate antibiotics and surgical drainage of any abscesses. Antimicrobial regimens should include a β-lactamase inhibitor due to the presence of β-lactamase-producing strains of Fusobacterium necrophorum and other co-infecting pathogens. Metronidazole is the most frequently prescribed antibiotic for this condition, known for its bactericidal properties and excellent tissue penetration, including into the cerebrospinal fluid. It can be administered orally without losing effectiveness. Studies have shown that metronidazole or carbapenem is more effective against Fusobacterium necrophorum than clindamycin.The recommended duration for antimicrobial therapy is typically between 3 to 6 weeks.A review of 100 Fusobacterium necrophorum isolates from 1990 to 2000 found that all strains were susceptible to metronidazole, clindamycin, imipenem, amoxicillin-clavulanate, and cefoxitin. Only 2% of the strains resisted penicillin, and 15% were resistant to erythromycin.
Nonetheless, in the acute setting, the airway is always prioritized. Patients with deep neck infections, particularly those with submandibular or odontogenic infections and those with any airway symptoms or difficulty handling their oral secretions, should have their airways adequately secured first and foremost via elective intubation. This frequently requires awake fiberoptic intubation. The use of glucocorticoids for symptomatic relief in patients with acute airway obstruction remains controversial due to the lack of supporting evidence. Therefore, any patient with a deep neck abscess and even minimal airway symptoms should be electively intubated and the abscess drained.
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Differential Diagnosis
Differential diagnosis of deep neck infections is variable based on presenting symptoms. Neck pain with fever can be caused by meningitis, apical pneumonia, or subarachnoid hemorrhage. Acute neck pain, especially with asymmetry, can be a result of trauma causing cervical fractures or dislocations, neck muscle hematomas, or neck muscle strains. Among patients presenting with stridor, acute epiglottitis, bacterial tracheitis, and the croup should be considered, and those with significant odynophagia or dysphagia, foreign body ingestion, and acute esophagitis should be considered. Astute clinicians should consider the malignancy amount in the differential diagnoses of a patient with neck swelling or mass.
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Prognosis
Prognosis is variable depending on the immunological status of the host and the severity and location of the infection. Mortality rates range between 1% and 25%.
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Complications
Lateral pharyngeal space infections can spread to the carotid sheath and cause septic thrombophlebitis (eg, Lemierre syndrome) and erosion. Retropharyngeal or danger space infections can spread to the mediastinum and cause acute mediastinitis, which may further spread and cause empyema and pericarditis. Respiratory failure can occur from an obstructed airway and spread into the systemic circulation, resulting in sepsis and intracranial infections.
Go to:
Consultations
Consultation with head and neck surgeons is recommended, as many patients typically require surgical intervention. Specialists, including intensivists and anesthesiologists, can also be consulted to provide airway and life support if needed.
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Deterrence and Patient Education
Proper oral hygiene and early treatment for identified dental caries and infections can help prevent deep neck infections.
Go to:
Pearls and Other Issues
The following factors should be considered when managing deep neck infections:
Deep neck infections most commonly arise from a nearby infectious focus, with odontogenic, tonsil, and pharyngeal sources being the most common.
Clinicians should be aware of deep neck infections and not underestimate their potential to cause life-threatening complications.
Knowledge of the anatomical compartments and spaces of the neck is essential for understanding the pathogenesis, clinical manifestations, and potential routes of infection spread.
Deep neck space infections are typically polymicrobial , originating from the oral cavity and oropharynx. Empiric antimicrobials should cover these species.
CT with IV contrast is the imaging modality for diagnosing deep neck space infections. Obtaining a noncontrasted scan is of little to no utility.
The treatment of deep neck infections includes appropriate antibiotics based on the likely microbiology of the infection, along with drainage of the abscess collection, if present, via either aspiration or surgical drainage.
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Enhancing Healthcare Team Outcomes
Effective diagnosis and management of deep neck infections require a cohesive interprofessional team approach involving physicians, advanced practitioners, nurses, pharmacists, and other health professionals. Anesthesiologists are crucial in securing the airway, especially in patients with compromised breathing. Infectious disease consultants guide antimicrobial therapy, ensuring pharmacologic treatment is appropriate for the identified pathogens and local resistance patterns. Ear, nose, throat, and thoracic surgeons are essential for evaluating and performing surgical drainage of abscesses. Radiologists contribute by providing accurate imaging for diagnosis and monitoring the infection's extent. Intensivists manage the overall care in severe cases, often requiring intensive care unit (ICU) support.Intensive care nurses monitor patients closely, observing for any signs of deterioration and managing complex care needs. Pharmacists ensure optimal dosing and monitor for drug interactions and adverse effects.
Interprofessional communication and coordination are vital for patient-centered care, with each team member sharing critical information and updates. This collaborative approach enhances patient outcomes, safety, and team performance by ensuring timely, appropriate, and comprehensive treatment, reducing the risk of complications, and improving the prognosis, which depends on factors such as age, infection severity, immune status, response to antibiotics, and comorbidities.
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Review Questions
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References
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Disclosure:Mohamed Almuqamam declares no relevant financial relationships with ineligible companies.
Disclosure:Francisco Gonzalez declares no relevant financial relationships with ineligible companies.
Disclosure:Sanjeev Sharma declares no relevant financial relationships with ineligible companies.
Disclosure:Noah Kondamudi declares no relevant financial relationships with ineligible companies.
Continuing Education Activity
Introduction
Etiology
Epidemiology
Pathophysiology
History and Physical
Evaluation
Treatment / Management
Differential Diagnosis
Prognosis
Complications
Consultations
Deterrence and Patient Education
Pearls and Other Issues
Enhancing Healthcare Team Outcomes
Review Questions
References
Copyright © 2025, StatPearls Publishing LLC.
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Etiology
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Pathophysiology
History and Physical
Evaluation
Treatment / Management
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Consultations
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Pearls and Other Issues
Enhancing Healthcare Team Outcomes
Review Questions
References
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Almuqamam M, Gonzalez FJ, Sharma S, et al. Deep Neck Infections. [Updated 2024 Aug 11]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Available from:
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Matrices with fixed row and column sums
Posted on June 6, 2024 by Stéphane Laurent in R bloggers | 0 Comments
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Given two vectors (p) and (q) of non-negative integer numbers, denote by (A(p, q)) the number of matrices with non-negative integer entries whose row sum and column sum respectively are (p) and (q), and denote by (B(p, q)) the number of matrices with entries in ({0, 1}) whose row sum and column sum respectively are (p) and (q).
The problem of determining (A(p,q)) and (B(p,q)) has a solution in the theory of symmetric polynomials. Denote by (\lambda) the vector obtained by sorting (p) in decreasing order and dropping the zero elements, and similarly, denote by (\mu) the vector obtained by sorting (q) in decreasing order and dropping the zero elements. Then (\lambda) and (\mu) are two integer partitions. In order for the two numbers (A(p,q)) and (B(p,q)) to be non-zero, an obvious necessary condition is that the sum of (p) is equal to the sum of (q). Let’s assume this condition holds true and denote by (n) this common sum. Of course, (n) is also the sum of (\lambda) and the sum of (\mu).
Then it is known in the theory of symmetric polynomials that [ A(p, q) = \sum_{\kappa \vdash n} K(\kappa, \lambda)K(\kappa, \mu) ] where the notation (\kappa \vdash n) means that (\kappa) is an integer partition of (n) and where (K(\kappa, \nu)) denotes the Kostka number associated to two integer partitions (\kappa) and (\nu).
It is also known that [ B(p, q) = \sum_{\kappa \vdash n} K(\kappa, \lambda) K(\kappa’, \mu) ] where (\kappa’) denotes the conjugate partition (or dual partition) of the partition (\kappa).
One also has (A(p,q) = \langle h_\lambda, h_\mu \rangle) where (h_\kappa) denotes the complete homogeneous symmetric function associated to an integer partition (\kappa) and (\langle \cdot, \cdot \rangle) denotes the Hall inner product, and one has (B(p,q) = \langle h_\lambda, e_\mu \rangle) where (e_\kappa) denotes the elementary symmetric function associated to an integer partition (\kappa).
All these results can be found in Macdonald’s book Symmetric functions and Hall polynomials.
Now let’s turn to the implementation of (A(p,q)) and (B(p,q)) in R. Enumerating the partitions of an integer is one of the features of the partitions package (the parts function). This package also allows to get the conjugate partition of an integer partition (the conjugate function). The Kostka numbers can be obtained with the syt package (the KostkaNumber function). So we use these two packages.
library(partitions)
library(syt)
Apq <- function(p, q) {
n <- sum(p)
if(sum(q) != n) {
return(0L)
}
lambda <- Filter((i) {i > 0}, sort(p, decreasing = TRUE))
mu <- Filter((i) {i > 0}, sort(q, decreasing = TRUE))
partitions <- parts(n)
sum(apply(partitions, 2L, function(kappa) {
KostkaNumber(kappa, lambda) KostkaNumber(kappa, mu)
}))
}
Bpq <- function(p, q) {
n <- sum(p)
if(sum(q) != n) {
return(0L)
}
lambda <- Filter((i) {i > 0}, sort(p, decreasing = TRUE))
mu <- Filter((i) {i > 0}, sort(q, decreasing = TRUE))
muprime <- conjugate(mu)
partitions <- parts(n)
sum(apply(partitions, 2L, function(kappa) {
KostkaNumber(kappa, lambda) KostkaNumber(kappa, muprime)
}))
}
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5339 | https://www.ktunotes.in/wp-content/uploads/2018/05/Intro-to-Automata-Theory-Languages-and-Computation-_-John-E-Hopcroft-Jeffrey-D-Ullman_text-min.pdf | INllHHXXTION AL'lttMATA TIH-'OKY, Kan k ImnncK Kmn D bl I viy UNOTES nloaded from Ktuno INTRODUCTION TO AUTOMATA THEORY, LANGUAGES, COMPUTATION JOHN E. HOPCROFT Cornell University JEFFREY D. ULLMAN Princeton University AD DI SON-WESLEY PUBLISHING COMPANY Reading, Massachusetts Menlo Park, California London • Amsterdam • Don Mills, Ontario • Sydney KTUNOTES.IN Downloaded from Ktunotes.in This book is in the ADDISON-WESLEY SERIES IN COMPUTER SCIENCE Michael A. Harrison, Consulting Editor Library of Congress Cataloging in Publication Data Hopcroft, John E.
, 1939-Introduction to automata theory, languages, and computation.
Bibliography: p.
Includes index.
1.
Machine theory.
2.
Formal languages.
3-Computational complexity.
I.
Ullman, Jeffrey D.
, 19^2-joint author.
II.
Title.
QA267.H56 629.8'312 78-67950 ISBN 0-201-02988-X Copyright (O 1979 by Addison-Wesley Publishing Company, Inc.
Philippines copyright 1979 by Addison-Wesley Publishing Company, Inc.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photoc6pying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. Published simultaneously in Canada.
Library of Congress Catalog Card No. 78-67950.
ISBN: 0-201-02988-X LMNOPQ-DO-89876 KTUNOTES.IN Downloaded from Ktunotes.in PREFACE Ten years ago the authors undertook to produce a book covering the known material on formal languages, automata theory, and computational complexity. In retrospect, only a few significant results were overlooked in the 237 pages. In writing a new book on the subject, we find the field has expanded in so many new directions that a uniform com-prehensive coverage is impossible. Rather than attempt to be encyclopedic, we have been brutal in our editing of the material, selecting only topics central to the theoretical development of the field or with importance to engineering applications.
Over the past ten years two directions of research have been of paramount im-portance. First has been the use of language-theory concepts, such as nondeterminism and the complexity hierarchies, to prove lower bounds on the inherent complexity of certain practical problems. Second has been the application of language-theory ideas, such as regular expressions and context-free grammars, in the design of software, such as compilers and text processors. Both of these developments have helped shape the organization of this book.
USE OF THE BOOK Both authors have used Chapters 1 through 8 for a senior-level course, omitting only the material on inherent ambiguity in Chapter 4 and portions of Chapter 8. Chapters 7, 8, 12, and 13 form the nucleus of a course on computational complexity. An advanced course on language theory could be built around Chapters 2 through 7, 9 through 11, and 14.
EXERCISES We use the convention that the most difficult problems are doubly starred, and problems of intermediate difficulty are identified by a single star. Exercises marked with an S have v KTUNOTES.IN Downloaded from Ktunotes.in VI PREFACE solutions at the end of the chapter. We have not attempted to provide a solution manual, but have selected a few exercises whose solutions are particularly instructive.
ACKNOWLEDGMENTS We would like to thank the following people for their perceptive comments and advice: Al Aho, Nissim Francez, Jon Goldstine, Juris Hartmanis, Dave Maier, Fred Springsteel, and Jacobo Valdes. The manuscript was expertly typed by Marie Olton and April Roberts at Cornell and Gerree Pecht at Princeton.
Ithaca, New York Princeton, New Jersey March 1979 J. E. H.
J. D. U.
KTUNOTES.IN Downloaded from Ktunotes.in CONTENTS Chapter 1 Preliminaries 1.1 Strings, alphabets, and languages 1 1.2 Graphs and trees 2 1.3 Inductive proofs 4 1.4 Set notation 5 1.5 Relations 6 1.6 Synopsis of the book 8 Chapter 2 Finite Automata and Regular Expressions 2.1 Finite state systems 13 2.2 Basic definitions 16 2.3 Nondeterministic finite automata 19 2.4 Finite automata with (-moves 24 2.5 Regular expressions 28 2.6 Two-way finite automata 36 2.7 Finite automata with output 42 2.8 Applications of finite automata 45 Chapter 3 Properties of Regular Sets 3.1 The pumping lemma for regular sets 55 3.2 Closure properties of regular sets 58 3.3 Decision algorithms for regular sets 63 3.4 The Myhill-Nerode theorem and minimization of finite automata .
.
65 vii KTUNOTES.IN Downloaded from Ktunotes.in Vlii CONTENTS Chapter 4 Context-Free Grammars 4.1 Motivation and introduction 77 4.2 Context-free grammars 79 4.3 Derivation trees 82 4.4 Simplification of context-free grammars 87 4.5 Chomsky normal form 92 4.6 Greibach normal form 94 4.7 The existence of inherently ambiguous context-free languages ...
99 Chapter 5 Pushdown Automata 5.1 Informal description 107 5.2 Definitions 108 5.3 Pushdown automata and context-free languages 114 Chapter 6 Properties of Context-Free Languages 6.1 The pumping lemma for CFL's 125 6.2 Closure properties of CFL's 130 6.3 Decision algorithms for CFL's 137 Chapter 7 Turing Machines 7.1 Introduction 146 7.2 The Turing machine model 147 7.3 Computable languages and functions 150 7.4 Techniques for Turing machine construction 153 7.5 Modifications of Turing machines 159 7.6 Church's hypothesis 166 7.7 Turing machines as enumerators 167 7.8 Restricted Turing machines equivalent to the basic model 170 Chapter 8 Undecidability 8.1 Problems 177 8.2 Properties of recursive and recursively enumerable languages .
.
.
179 8.3 Universal Turing machines and an undecidable problem 181 8.4 Rice's theorem and some more undecidable problems 185 8.5 Undecidability of Post's correspondence problem 193 8.6 Valid and invalid computations of TM's: a tool for proving CFL problems undecidable 201 8.7 Greibach's theorem 205 8.8 Introduction to recursive function theory 207 8.9 Oracle computations 209 Chapter 9 The Chomsky Hierarchy 9.1 Regular grammars 217 9.2 Unrestricted grammars 220 KTUNOTES.IN Downloaded from Ktunotes.in CONTENTS IX 9.3 Context-sensitive languages 223 9.4 Relations between classes of languages 227 Chapter 10 Deterministic Context-Free Languages 10.1 Normal forms for DPDA's 234 10.2 Closure of DCFL's under complementation 235 10.3 Predicting machines 240 10.4 Additional closure properties of DCFL's 243 10.5 Decision properties of DCFL's 246 10.6 LR(0) grammars 248 10.7 LR(0) grammars and DPDA's 252 10.8 LR(k) grammars 260 Chapter 11 Closure Properties of Families of Languages 11.1 Trios and full trios 270 11.2 Generalized sequential machine mappings 272 11.3 Other closure properties of trios 276 11.4 Abstract families of languages 277 11.5 Independence of the AFL operations 279 11.6 Summary 279 Chapter 12 Computational Complexity Theory 12.1 Definitions 285 12.2 Linear speed-up, tape compression, and reductions in the number of tapes 288 12.3 Hierarchy theorems 295 12.4 Relations among complexity measures 300 12.5 Translational lemmas and nondeterministic hierarchies 302 12.6 Properties of general complexity measures: the gap, speedup, and union theorems 306 12.7 Axiomatic complexity theory 312 Chapter 13 Intractable Problems 13.1 Polynomial time and space 320 13.2 Some NP-complete problems 324 13.3 The class co-./T^ 341 13.4 PSPACE-complete problems 343 13.5 Complete problems for & and NSPACE(log n) 347 13.6 Some provably intractable problems 350 13.7 The 0> — jV'i? question for Turing machines with oracles: limits on our ability to tell whether & = c \'d?
362 Chapter 14 Highlights of Other Important Language Classes 14.1 Auxiliary pushdown automata 377 14.2 Stack automata 381 KTUNOTES.IN Downloaded from Ktunotes.in X CON ltNlb 14.3 Indexed languages 389 14.4 Developmental systems 390 Bibliography 396 Index 411 KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 1 PRELIMINARIES In this chapter we survey the principal mathematical ideas necessary for under-standing the material in this book. These concepts include graphs, trees, sets, relations, strings, abstract languages, and mathematical induction. We also pro-vide a brief introduction to, and motivation for, the entire work. The reader with a background in the mathematical subjects mentioned can skip to Section 1.6 for motivational remarks.
1.1 STRINGS, ALPHABETS, AND LANGUAGES A "symbol" is an abstract entity that we shall not define formally, just as "point" and "line" are not defined in geometry.
Letters and digits are examples of frequently used symbols. A string (or word) is a finite sequence of symbols jux-taposed. For example, a, b, and c are symbols and abcb is a string. The length of a string w, denoted | w | , is the number of symbols composing the string. For exam-ple, abcb has length 4. The empty string, denoted by £, is the string consisting of zero symbols. Thus \e \ = 0.
A prefix of a string is any number of leading symbols of that string, and a suffix is any number of trailing symbols. For example, string abc has prefixes £, a, ab, and abc; its suffixes are £, c, be, and abc. A prefix or suffix of a string, other than the string itself, is called a proper prefix or suffix.
The concatenation of two strings is the string formed by writing the first, followed by the second, with no intervening space. For example, the concatena-tion of dog and house is doghouse. Juxtaposition is used as the concatenation operator. That is, if w and x are strings, then wx is the concatenation of these two 1 KTUNOTES.IN Downloaded from Ktunotes.in 2 PRELIMINARIES strings. The empty string is the identity for the concatenation operator. That is, £w = we — w for each string w.
An alphabet is a finite set of symbols. A (formal) language is a set of strings of symbols from some one alphabet. The empty set, 0, and the set consisting of the empty string {e} are languages. Note that they are distinct; the latter has a member while the former does not. The set of palindromes (strings that read the same forward and backward) over the alphabet {0, 1} is an infinite language. Some members of this language are e, 0, 1, 00, 11, 010, and 1101011. Note that the set of all palindromes over an infinite collection of symbols is technically not a language because its strings are not collectively built from an alphabet.
Another language is the set of all strings over a fixed alphabet Z. We denote this language by Z. For example, if Z = {a}, then Z = {e, a, aa, aaa, .
. .}.
If Z = {0, 1}, then Z = {e, 0, 1, 00, 01, 10, 1 1, 000, .
. .}.
1.2 GRAPHS AND TREES A graph, denoted G = (V, E), consists of a finite set of vertices (or nodes) V and a set of pairs of vertices E called edges. An example graph is shown in Fig. 1.1. Here V = {1, 2, 3, 4, 5} and E = {(n, m) | n + m = 4 or n + m = 7}.
Fig.
1.1 Example of a graph.
A path in a graph is a sequence of vertices v l9 v2 , .
.
. , vk , k > 1, such that there is an edge (vh vi+1 ) for each i, 1 < i < k. The length of the path is k — 1. For example, 1, 3, 4 is a path in the graph of Fig.
1.1; so is 5 by itself. If v x = vk9 the path is a cycle.
Directed graphs A directed graph (or digraph), also denoted G = (V, E), consists of a finite set of vertices V and a set of ordered pairs of vertices E called arcs. We denote an arc from v to w by v -> w. An example of a digraph appears in Fig.
1.2.
A path in a digraph is a sequence of vertices v l9 v2 , vk , k > 1, such that Vi -> vi+ j is an arc for each i, 1 < i < k. We say the path is from v x to vk . Thus l-+2->3->4isa path from 1 to 4 in the digraph of Fig.
1.2. If v -> w is an arc we say i?
is a predecessor of w and w is a successor of u.
KTUNOTES.IN Downloaded from Ktunotes.in 1.2 | GRAPHS AND TREES J Fig. 1.2 The digraph ({1, 2, 3, 4}, {i -+j\i< ;}).
Trees A tree (strictly speaking, an ordered, directed tree) is a digraph with the following properties.
1) There is one vertex, called the root, that has no predecessors and from which there is a path to every vertex.
2) Each vertex other than the root has exactly one predecessor.
3) The successors of each vertex are ordered "from the left." We shall draw trees with the root at the top and all arcs pointing downward.
The arrows on the arcs are therefore not needed to indicate direction, and they will not be shown. The successors of each vertex will be drawn in left-to-right order. Figure 1.3 shows an example of a tree which is the "diagram" of the English sentence "The quick brown fox jumped over the lazy dog." The vertices are not named in this example, but are given "labels," which are either words or parts of speech.
/ v < adject ivc> the I / quick / < adject ive> I brown I lox \ / \ 1 / \ jumped / \ I / \ I lazy I dog Fig. 1.3 A tree.
KTUNOTES.IN Downloaded from Ktunotes.in 4 PRELIMINARIES There is a special terminology for trees that differs from the general terminol-ogy for arbitrary graphs. A successor of a vertex is called a son, and the predeces-sor is called the father. If there is a path from vertex v x to vertex v2 , then v x is said to be an ancestor of v2 , and v2 is said to be a descendant oiv^ Note that the case v x = v 2 is not ruled out; any vertex is an ancestor and a descendant of itself. A vertex with no sons is called a leaf and the other vertices are called interior vertices. For example, in Fig.
1.3, the vertex labeled (verb) is a son of the vertex labeled (verb phrase), and the latter is the father of the former. The vertex labeled "dog" is a descendant of itself, the vertex labeled (verb phrase), the vertex labeled (sentence), and six other vertices. The vertices labeled by English words are the leaves, and those labeled by parts of speech enclosed in angle brackets are the interior vertices.
1.3 INDUCTIVE PROOFS Many theorems in this book are proved by mathematical induction. Suppose we have a statement P(n) about a nonnegative integer n. A commonly chosen example is to take P(n) to be The principle of mathematical induction is that P(n) follows from a) P(0), and b) P(n -1) implies P(n) for n > 1.
Condition (a) in an inductive proof is called the basis, and condition (b) is called the inductive step. The left-hand side of (b), that is P(n — 1), is called the inductive hypothesis.
Example 1.1 Let us prove (1.1) by mathematical induction. We establish (a) by substituting 0 for n in (1.1) and observing that both sides are 0. To prove (b), we substitute n — 1 for n in (1.1) and try to prove (1.1) from the result. That is, we must show for n > 1 that I < 2 = n n(n+ l)(2n + 1) 6 (1.1) n — i = 0 (n- i)n(2n-1) 6 implies £ i2 = « = o n n(n+ l)(2n+ 1) 6 Since n n 1 1 = 0 i = 0 and since we are given KTUNOTES.IN Downloaded from Ktunotes.in 1.4 | SET NOTATION 5 we need only show that (n - l)n(2n - 1) 2 _ n(n + l)(2n + 1) 6 + " ~ 6 The latter equality follows from simple algebraic manipulation, proving (1.1).
1.4 SET NOTATION We assume that the reader is familiar with the notion of a set, a collection of objects (members of the set) without repetition. Finite sets may be specified by listing 4 their members between brackets. For example we used {0, 1} to denote the alphabet of symbols 0 and 1. We also specify sets by a set former: {x\P(x)}, (1.2) or {xin^l|P(x)}.
(1.3) Statement (1.2) is read "the set of objects x such that P(x) is true," where P(x) is some statement about objects x. Statement (1.3) is "the set of x in set A such that P(x) is true," and is equivalent to {x|P(x) and x is in A}. For example, {/ 1 i is an integer and there exists integer j such that i = 2j] is a way of specifying the even integers.
If every member of A is a member of B, then we write A^ B and say A is contained in B. A 3 B is synonymous with B £ A.
If A £ B but A ^ By that is, every member of /I is in B and there is some member of B that is not in A, then we write A £ B. Sets /I and £ are equal if they have the same members. That is, A = B if and only if A ^ B and B ^ A.
Operations on sets The usual operations defined on sets are: 1) A u B, the wmcw of /I and B, is {x | x is in A or x is in B}.
2) A n B, the intersection of ^ and £, is {x | x is in ^ and x is in #}.
3) A — B, the difference of /I and 5, is {x |x is in A and x is not in B}.
4) /I x B, the Cartesian product of A and £, is the set of ordered pairs (a, b) such that a is in A and 5 is in B.
5) 2^, the power set of >4, is the set of all subsets of A.
KTUNOTES.IN Downloaded from Ktunotes.in 6 PRELIMINARIES Example 1.2 Let A = {1, 2} and B = {2, 3}. Then A u £ = {1,2, 3}, /I n B = {2}, A-B = {1}, ,4xB = {(l, 2),(1, 3), (2, 2), (2, 3)}, and 2" = {0,{1}, {2},{1, 2}}.
Note that if A and B have n and m members, respectively, then A x B has wn members and 2A has 2" members.
Infinite sets Our intuition when extended to infinite sets can be misleading. Two sets S x and 5 2 have the same cardinality (number of members) if there is a one-to-one mapping of the elements of S { onto S 2 . For finite sets, if S x is a proper subset of S 2 , tnen $i and S 2 have different cardinality. However, if S x and S 2 are infinite, the latter statement may be false. Let S, be the set of even integers and let S 2 be the set of all integers. Clearly 5!
is a proper subset of S 2 . However, S t and S 2 have the same cardinality, since the function/ defined by / (2i) = i is a one-to-one mapping of the even integers onto the integers.
Not all infinite sets have the same cardinality. Consider the set of all integers and the set of all reals. Assume that the set of reals can be put in one-to-one-onto correspondence with the integers. Then consider the real number whose ith digit after the decimal is the ith digit of the ith real plus 5 mod 10. This real number cannot be in correspondence with any integer, since it diners from every real that has been mapped to an integer. From this we conclude that the reals cannot be placed in one-to-one correspondence with the integers. Intuitively there are too many real numbers to do so. The above construction is called diagonalization and is an important tool in computer science.
Sets that can be placed in one-to-one correspondence with the integers are said to be countably infinite or countable. The rationals and the set Z of the finite-length strings from an alphabet X are countably infinite. The set of all subsets of X and the set of all functions mapping the integers to {0, 1} are of the same cardinality as the reals, and are not countable.
1.5 RELATIONS A (binary) relation is a set of pairs. The first component of each pair is chosen from a set called the domain, and the second component of each pair is chosen from a (possibly different) set called the range. We shall use primarily relations in which the domain and range are the same set S. In that case we say the relation is on S. If R is a relation and (a, b) is a pair in R, then we often write aRb.
KTUNOTES.IN Downloaded from Ktunotes.in 1.5 | RELATIONS 7 Properties of relations We say a relation R on set S is 1) reflexive if aRa for all am S; 2) irreflexive if aRa is false for all a in 5; 3) transitive if aRb and bRc imply oRc; 4) symmetric if a&fr implies bRa; 5) asymmetric if aKb implies that Wfa is false.
Note that any asymmetric relation must be irreflexive.
Example 1.3 The relation < on the set of integers is transitive because a < b and b < c implies a < c. It is asymmetric and hence irreflexive because a < b implies b < a is false.
Equivalence relations A relation R that is reflexive, symmetric, and transitive is said to be an equivalence relation. An important property of an equivalence relation R on a set S is that R partitions S into disjoint nonempty equivalence classes (see Exercise 1.8 and its solution). That is, S = 5j u S 2 u • •, where for each i and /, with i # j: 1) Si n Sj=0; 2) for each a and b :n Sh aRb is true; 3) for each a in S t and b in Sp aRb is false.
The S^s are called eauivalence classes. Note that the number of classes may be infinite.
Example 1.4 A common example of an equivalence relation is congruence modulo an integer m. We write i = m j or i = j mod m if i and j are integers such that i — j is divisible by m. The reader may easily prove that = m is reflexive, transitive, and symmetric. The equivalence classes of = m are m in number: {• {.
.
. , — m, 0, m, 2m, .
. .}, (m -1), 1, m + 1, 2m + 1, .
1, m — 1, 2m — 1, 3m — 1, .
•}• KTUNOTES.IN Downloaded from Ktunotes.in 8 PRELIMINARIES Closures of relations Suppose & is a set of properties of relations. The ^-closure of a relation R is the smallest relation R' that includes all the pairs of R and possesses the properties in For example, the transitive closure of R, denoted K + , is defined by: 1) If (a, b) is in R, then (a, b) is in R + .
2) If (a, b) is in K + and (5, c) is in R, then (a, c) is in R + .
3) Nothing is in R + unless it so follows from (1) and (2).
It should be evident that any pair placed in R + by rules (1) and (2) belongs there, else R + would either not include R or not be transitive. Also an easy inductive proof shows that R + is in fact transitive. Thus R + includes R, is transi-tive, and contains as few pairs as any relation that includes R and is transitive.
The reflexive and transitive closure of R, denoted R 9 is easily seen to be R + u {(a, a)\a is in S}.
Example 1.5 Let R = {(1, 2), (2, 2), (2, 3)} be a relation on the set {1, 2, 3}. Then R+={(1,2), (2, 2), (2,3),(1, 3)}, and = {(!, 1), (1,2), (1,3), (2,2), (2, 3), (3, 3)}.
1.6 SYNOPSIS OF THE BOOK Computer science is the systematized body of knowledge concerning computa-tion. Its beginnings can be traced back to the design of algorithms by Euclid and the use of asymptotic complexity and reducibility by the Babylonians (Hogben ). Modern interest, however, is shaped by two important events: the advent of modern digital computers capable of many millions of operations per second and the formalization of the concept of an effective procedure, with the con-sequence that there are provably noncomputable functions.
Computer science has two major components: first, the fundamental ideas and models underlying computing, and second, engineering techniques for the design of computing systems, both hardware and software, especially the applica-tion of theory to design. This book is intended as an introduction to the first area, the fundamental ideas underlying computing, although we shall remark briefly on the most important applications.
Theoretical computer science had its beginnings in a number of diverse fields: biologists studying models for neuron nets, electrical engineers developing switch-ing theory as a tool to hardware design, mathematicians working on the foun-dations of logic, and linguists investigating grammars for natural languages. Out of these studies came models that are central to theoretical computer science.
KTUNOTES.IN Downloaded from Ktunotes.in 1.6 | SYNOPSIS OF THE BOOK 9 The notions of finite automata and regular expressions (Chapters 2 and 3) were originally developed with neuron nets and switching circuits in mind. More recently, they have served as useful tools in the design of lexical analyzers, the part of a compiler that groups characters into tokens—indivisible units such as var-iable names and keywords. A number of compiler-writing systems automatically transform regular expressions into finite automata for use as lexical analyzers. A number of other uses for regular expressions and finite automata have been found in text editors, pattern matching, various text-processing and file-searching pro-grams, and as mathematical concepts with application to other areas, such as logic. At the end of Chapter 2 we shall outline some of the applications of this theory.
The notion of a context-free grammar and the corresponding pushdown au-tomaton (Chapters 4 through 6) has aided immensely the specification of program-ming languages and in the design of parsers—another key portion of a compiler.
Formal specifications of programming languages have replaced extensive and often incomplete or ambiguous descriptions of languages. Understanding the cap-abilities of the pushdown automaton has greatly simplified parsing.
It is inter-esting to observe that parser design was, for the earliest compilers, a difficult problem, and many of the early parsers were quite inefficient and unnecessarily restrictive. Now, thanks to widespread knowledge of a variety of context-free-grammar-based techniques, parser design is no longer a problem, and parsing occupies only a few percent of the time spent in typical compilation. In Chapter 10 we sketch the principal ways in which efficient parsers that behave as pushdown automata can be built from certain kinds of context-free grammars.
In Chapter 7 we meet Turing machines and confront one of the fundamental problems of computer science; namely, that there are more functions than there are names for functions or than there are algorithms for computing functions.
Thus we are faced with the existence of functions that are simply not computable; that is, there is no computer program that can ever be written, which given an argument for the function produces the value of the function for that argument and works for all possible arguments.
Assume that for each computable function there is a computer program or algorithm that computes it, and assume that any computer program or algorithm can be finitely specified. Thus computer programs are no more than finite-length strings of symbols over some finite alphabet. Hence the set of all computer pro-grams is countably infinite. Consider now functions mapping the integers to 0 and 1. Assume that the set of all such functions are countably infinite and that these functions have been placed in correspondence with the integers. Let f t be the function corresponding to the fth integer. Then the function /(«) = ( (0 if/.(n)=l 1 otherwise KTUNOTES.IN Downloaded from Ktunotes.in 10 PRELIMINARIES cannot correspond to any integer, which is a contradiction. [If / (n) = fj(n), then we have the contradiction / (j) = fj(j) and / (J) ^ fj(j).] This argument is formalized in Chapters 7 and 8, where we shall see that certain easily stated problems cannot be solved on the computer, even though they appear at first glance to be amenable to computation.
However, we can do more than tell whether a problem can be solved by a computer. Just because a problem can be solved doesn't mean there is a practical algorithm to solve it. In Chapter 12 we see that there are abstract problems that are solvable by computer but require inordinate amounts of time and/or space for their solution. Then in Chapter 13 we discover that there are many realistic and important problems that also fall in this category. The nascent theory of "intract-able problems" is destined to influence profoundly how we think about problems.
EXERCISES 1.1 In the tree of Fig.
1.4, a) Which vertices are leaves and which are interior vertices?
b) Which vertices are the sons of 5?
c) Which vertex is the father of 5?
d) What is the length of the path from 1 to 9?
e) Which vertex is the root?
Fig. 1.4 A tree.
1.2 Prove by induction on n that .
" .
n(n + 1) v 3 /£ Y a ) I 1 = ^-y-1 b ) I 1 = E 1 i = 0 £ i = 0 \i = 0 / KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 11 S 1.3 A palindrome can be defined as a string that reads the same forward and backward, or by the following definition.
1) £ is a palindrome.
2) If a is any symbol, then the string a is a palindrome.
3) If a is any symbol and x is a palindrome, then axa is a palindrome.
4) Nothing is a palindrome unless it follows from (1) through (3).
Prove by induction that the two definitions are equivalent.
1.4 The strings of balanced parentheses can be defined in at least two ways.
1) A string w over alphabet {(,)} is balanced if and only if: a) w has an equal number of fs and )'s, and b) any prefix of w has at least as many fs as )'s.
2) a) e is balanced.
b) If w is a balanced string, then (w) is balanced.
c) If w and x are balanced strings, then so is wx.
d) Nothing else is a balanced string.
Prove by induction on the length of a string that definitions (1) and (2) define the same class of strings.
1.5 What is wrong with the following inductive "proof" that all elements in any set must be identical? For sets with one element the statement is trivially true. Assume the statement is true for sets with n — 1 elements, and consider a set S with n elements. Let a be an element of 5. Write S = S { u 5 2 , where S { and S 2 each have n — 1 elements, and each contains a.
By the inductive hypothesis all elements in S x are identical to a and similarly all elements in S 2 are identical to a. Thus all elements in S are identical to a.
1.6 Show that the following are equivalence relations and give their equivalence classes.
a) The relation Ri on integers defined by iRij if and only if i = j.
b) The relation R 2 on people defined by pR 2 q if and only if p and q were born at the same hour of the same day of some year.
c) The same as (b) but "of the same year" instead of "of some year." 1.7 Find the transitive closure, the reflexive and transitive closure, and the symmetric closure of the relation 2), (2, 3), (3, 4), (5, 4)}.
S 1.8 Prove that any equivalence relation R on a set S partitions S into disjoint equiv-alence classes.
1.9 Give an example of a relation that is symmetric and transitive but not reflexive.
[Hint: Note where reflexivity is needed to show that an equivalence relation defines equiv-alence classes; see the solution to Exercise 1.8.] 1.10 Prove that any subset of a countably infinite set is either finite or countably infinite.
1.11 Prove that the set of all ordered pairs of integers is countably infinite.
1.12 Is the union of a countably infinite collection of countably infinite sets countably infinite? is the Cartesian product?
KTUNOTES.IN Downloaded from Ktunotes.in 12 PRELIMINARIES Solutions to Selected Exercises 1.3 Clearly every string satisfying the second definition reads the same forward and backward. Suppose x reads the same forward and backward. We prove by induction on the length of x that x's being a palindrome follows from rules (1) through (3). If | x \ < 1, then x is either e or a single symbol a and rule (1) or (2) applies. If | x | > 1, then x begins and ends with some symbol a. Thus x = awa, where w reads the same forward and backward and is shorter than x. By the induction hypothesis, rules (1) through (3) imply that w is a palin-drome. Thus by rule (3), x = awa is a palindrome.
1.8 Let R be an equivalence relation on S, and suppose a and b are elements of S. Let Ca and Cb be the equivalence classes containing a and b respectively; that is, Ca = {c \aRc] and Cb = {c | bRc}. We shall show that either Ca — C b or Ca n Cb = 0. Suppose Ca n Cb 0; let </ be in Ca n Cb . Now let e be an arbitrary member of Ca . Thus aRe. As d is in Ca n C fc we have and bRd. By symmetry, By transitivity (twice), bRa and bRe. Thus e is in Cb and hence C fl ^ Cb . A similar proof shows that Cb G C fl , so Ca = C b . Thus distinct equivalence classes are disjoint. To show that the classes form a partition, we have only to observe that by reflexivity, each a is in the equivalence class Cfl , so the union of the equivalence classes is S.
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 2 FINITE AUTOMATA AND REGULAR EXPRESSIONS 2.1 FINITE STATE SYSTEMS The finite automaton is a mathematical model of a system, with discrete inputs and outputs. The system can be in any one of a finite number of internal configurations or "states." The state of the system summarizes the information concerning past inputs that is needed to determine the behavior of the system on subsequent inputs. The control mechanism of an elevator is a good example of a finite state system. That mechanism does not remember all previous requests for service but only the current floor, the direction of motion (up or down), and the collection of not yet satisfied requests for service.
In computer science we find many examples of finite state systems, and the theory of finite automata is a useful design tool for these systems. A primary example is a switching circuit, such as the control unit of a computer. A switching circuit is composed of a finite number of gates, each of which can be in one of two conditions, usually denoted 0 and 1. These conditions might, in electronic terms, be two different voltage levels at the gate output. The state of a switching network with n gates is thus any one of the 2" assignments of 0 or 1 to the various gates.
Although the voltage on each gate can assume any of an infinite set of values, the electronic circuitry is so designed that only the two voltages corresponding to 0 and 1 are stable, and other voltages will almost instantaneously adjust themselves to one of these voltages. Switching circuits are intentionally designed in this way, so that they can be viewed as finite state systems, thereby separating the logical design of a computer from the electronic implementation.
Certain commonly used programs such as text editors and the lexical analy-zers found in most compilers are often designed as finite state systems. For exam-13 KTUNOTES.IN Downloaded from Ktunotes.in 14 FINITE AUTOMATA AND REGULAR EXPRESSIONS pie, a lexical analyzer scans the symbols of a computer program to locate the strings of characters corresponding to identifiers, numerical constants, reserved words, and so on. In this process the lexical analyzer needs to remember only a finite amount of information, such as how long a prefix of a reserved word it has seen since startup. The theory of finite automata is used heavily in the design of efficient string processors of these and other sorts. We mention some of these applications in Section 2.8.
The computer itself can be viewed as a finite state system, although doing so turns out not to be as useful as one would like. Theoretically the state of the central processor, main memory, and auxiliary storage at any time is one of a very large but finite number of states. We are assuming of course that there is some fixed number of disks, drums, tapes, and so on available for use, and that one cannot extend the memory indefinitely. Viewing a computer as a finite state system, however, is not satisfying mathematically or realistically.
It places an artificial limit on the memory capacity, thereby failing to capture the real essence of computation. To properly capture the notion of computation we need a poten-tially infinite memory, even though each computer installation is finite. Infinite models of computers will be discussed in Chapters 7 and 8.
It is also tempting to view the human brain as a finite state system. The number of brain cells or neurons is limited, probably 2 35 at most. It is conceivable, although there is evidence to the contrary, that the state of each neuron can be described by a small number of bits. If so, then finite state theory applies to the brain. However, the number of states is so large that this approach is unlikely to result in useful observations about the human brain, any more than finite state assumptions help us understand large but finite computer systems.
Perhaps the most important reason for the study of finite state systems is the naturalness of the concept as indicated by the fact that it arises in many diverse places. This is an indication that we have captured the notion of a fundamental class of systems, a class that is rich in structure and potential application.
An example Before formally defining finite state systems let us consider an example. A man with a wolf, goat, and cabbage is on the left bank of a river. There is a boat large enough to carry the man and only one of the other three. The man and his entourage wish to cross to the right bank, and the man can ferry each across, one at a time. However, if the man leaves the wolf and goat unattended on either shore, the wolf will surely eat the goat. Similarly, if the goat and cabbage are left unattended, the goat will eat the cabbage. Is it possible to cross the river without the goat or cabbage being eaten ?
The problem is modeled by observing that the pertinent information is the occupants of each bank after a crossing. There are 16 subsets of the man (M), wolf (W), goat (G), and cabbage (C). A state corresponds to the subset that is on the left KTUNOTES.IN Downloaded from Ktunotes.in 2.1 | FINITE STATE SYSTEMS 15 bank. States are labeled by hyphenated pairs such as MG-WC, where the symbols to the left of the hyphen denote the subset on the left bank; symbols to the right of the hyphen denote the subset on the right bank. Some of the 16 states, such as GC-MW, are fatal and may never be entered by the system.
The "inputs" to the system are the actions the man takes. He may cross alone (input m), with the wolf (input w), the goat (input g), or cabbage (input c). The initial state is MWGC-0 and the final state is 0-MWGC. The transition diagram is shown in Fig. 2.1.
MWGC-0X CwC-MG g ^~— MWC-G C-MWG MGC-W W-MGC g g\ 1 I MWG-C G-MWC Fig. 2.1 Transition diagram for man, wolf, goat, and cabbage problem.
There are two equally short solutions to the problem, as can be seen by searching for paths from the initial state to the final state (which is doubly circled).
There are infinitely many different solutions to the problem, all but two involving useless cycles. The finite state system can be viewed as defining an infinite lan-guage, the set of all strings that are labels of paths from the start state to the final state.
Before proceeding, we should note that there are at least two important ways in which the above example is atypical of finite state systems. First, there is only one final state; in general there may be many. Second, it happens that for each transition there is a reverse transition on the same symbol, which need not be the case in general. Also, note that the term "final state," although traditional, does not mean that the computation need halt when it is reached. We may continue making transitions, e.g., to state MG-WC in the above example.
KTUNOTES.IN Downloaded from Ktunotes.in 16 FINITE AUTOMATA AND REGULAR EXPRESSIONS 2.2 BASIC DEFINITIONS A finite automaton (FA) consists of a finite set of states and a set of transitions from state to state that occur on input symbols chosen from an alphabet X. For each input symbol there is exactly one transition out of each state (possibly back to the state itself). One state, usually denoted g0 , is the initial state, in which the automa-ton starts. Some states are designated as final or accepting states.
A directed graph, called a transition diagram, is associated with an FA as follows. The vertices of the graph correspond to the states of the FA. If there is a transition from state q to state p on input a, then there is an arc labeled a from state q to state p in the transition diagram. The FA accepts a string x if the sequence of transitions corresponding to the symbols of x leads from the start state to an accepting state.
Example 2.1 The transition diagram of an FA is illustrated in Fig. 2.2. The initial state, q0 , is indicated by the arrow labeled "start." There is one final state, also q0 in this case, indicated by the double circle. The FA accepts all strings of O's and l's in which both the number of O's and the number of l's are even. To see this, visualize "control" as traveling from state to state in the diagram. Control starts at q0 and must finish at q0 if the input sequence is to be accepted. Each 0-input causes control to cross the horizontal line a-b, while a 1-input does not. Thus control is at a state above the line a-b if and only if the input seen so far contains an even number of O's. Similarly, control is at a state to the left of the vertical line c-d if and only if the input contains an even number of l's. Thus control is at q0 if and only if there are both an even number of O's and an even number of l's in the input. Note that the FA uses its state to record only the parity of the number of O's and the number of l's, not the actual numbers, which would require an infinite number of states.
b Fig. 2.2 The transition diagi am of a finite automaton.
KTUNOTES.IN Downloaded from Ktunotes.in 2.2 | BASIC DEFINITIONS 17 We formally denote a finite automaton by a 5-tuple (Q, Z, g0 > ^)» where Q is a finite set of states, Z is a finite inpur alphabet, q0 in Q is the imtt'a/ state, F ^ Q is the set of^na/ states, and 3 is the transitionfunction mapping Q x £ to Q. That is, a) is a state for each state q and input symbol a.
We picture an FA as a finite control, which is in some state from Q, reading a sequence of symbols from Z written on a tape as shown in Fig. 2.3. In one move the FA in state q and scanning symbol a enters state S(q, a) and moves its head one symbol to the right. If S(q, a) is an accepting state, then the FA is deemed to have accepted the string written on its input tape up to, but not including, the position to which the head has just moved. If the head has moved off the right end of the tape, then it accepts the entire tape. Note that as an FA scans a string it may accept many different prefixes.
0 1 1 0 0 1 0 1 J Finite control Fig. 2.3 A finite automaton.
To formally describe the behavior of an FA on a string, we must extend the transition function <5 to apply to a state and a string rather than a state and a symbol. We define a function 3 from Q x Z to Q. The intention is that 3{q, w) is the state the FA will be in after reading w starting in state q. Put another way, 3(q, w) is the unique state p such that there is a path in the transition diagram from q to p, labeled w. Formally we define 1) 3(q, c) = q, and 2) for all strings w and input symbols a, 3(q, wa) = 3(3(q, w), a).
Thus (1) states that without reading an input symbol the FA cannot change state, and (2) tells us how to find the state after reading a nonempty input string wa.
That is, find the state, p = 5(q, w), after reading w. Then compute the state 5(p, a).
Since 5(q, a) = 3(S(q, c), a) = S(q, a) [letting w = c in rule (2) above], there can be no disagreement between 3 and S on arguments for which both are defined.
Thus we shall for convenience use 3 instead of 3 from here on.
Convention We shall strive to use the same symbols to mean the same thing throughout the material on finite automata. In particular, unless it is stated other-KTUNOTES.IN Downloaded from Ktunotes.in 18 FINITE AUTOMATA AND REGULAR EXPRESSIONS wise, the reader may assume: 1 ) Q is a set of states. Symbols q and p, with or without subscripts, are states. q0 is the initial state.
2) Z is an input alphabet. Symbols a and by with or without subscripts, and the digits are input symbols.
3) 3 is a transition function.
4) F is a set of final states.
5) w, x, y, and z, with or without subscripts, are strings of input symbols.
A string x is said to be accepted by a finite automaton M = (Q, Z, g0 , F) if $(qoy x ) = P f°r some pin F. The language accepted by M, designated L(M ), is the set {x \S(q0 , x) is in F). A language is a regular set (or just regular) if it is the set accepted by some finite automaton.f The reader should note that when we talk about a language accepted by a finite automaton M we are referring to the specific set L(M), not just any set of strings all of which happen to be accepted by M .
Example 2.2 Consider the transition diagram of Fig. 2.2 again. In our formal notation this FA is denoted M = (Q, Z, S, qQy F), where Q = {q0 , qu q2 , q3 }, Z = {0, 1}, F = {g0 }, and 3 is shown in Fig. 2.4.
Inputs StatesV 0 1 qo qi qi q\ <?3 qo qi qo <73 43 qi qi Fig. 2.4 <% a) for the FA of Fig. 2.2.
Suppose 1 10101 is input to M. We note that d(q0 , 1) = q x and 5(q u 1) = go-Thus tffao, H) = (5(<%» 1), !) = <5( 0) = q2 . Thus 6{q09 110) = <(<%>, n),0) = 3(qOi 0) = q2 .
t The term "regular" comes from "regular expressions," a formalism we shall introduce in Section 2.5, and which defines the same class of languages as the FA's.
KTUNOTES.IN Downloaded from Ktunotes.in 2.3 | NONDETERM IN ISTI C FINITE AUTOMATA 19 Continuing in this fashion, we find that 5(q0 , 1101) = ^3 , 5(q0 , 11010) = ^ and finally %0 , H0101) = q0 .
The entire sequence of states is qlq\q%\<Rq\q^ Thus 110101 is in L(M). As we mentioned, L(M) is the set of strings with an even number of 0's and an even number of Fs.
2.3 NONDETERMINISTIC FINITE AUTOMATA We now introduce the notion of a nondeterministic finite automaton. It will turn out that any set accepted by a nondeterministic finite automaton can also be accepted by a deterministic finite automaton. However, the nondeterministic finite automaton is a useful concept in proving theorems. Also, the concept of non-determinism plays a central role in both the theory of languages and the theory of computation, and it is useful to understand this notion fully in a very simple context ' initially. Later we shall meet automata whose deterministic and non-deterministic versions are known not to be equivalent, and others for which equivalence is a deep and important open question.
Consider modifying the finite automaton model to allow zero, one, or more transitions from a state on the same input symbol. This new model is called a nondeterministic finite automaton (NFA). A transition diagram for a nondeter-ministic finite automaton is shown in Fig. 2.5. Observe that there are two edges labeled 0 out of state q0 , one going back to state q0 and one going to state q 3 .
An input sequence a x a<i • • • a„ is accepted by a nondeterministic finite autom-aton if there exists a sequence of transitions, corresponding to the input sequence, that leads from the initial state to some final state. For example, 01001 is accepted by the NFA of Fig. 2.5 because there is a sequence of transitions through the states q0 , q0 , q0 , q 3 , g4 , <?4 , labeled 0, 1, 0, 0, 1. This particular NFA accepts all strings with either two consecutive 0's or two consecutive Fs. Note that the FA of the previous section (deterministic FA, or DFA for emphasis) is a special case of the NFA in which for each state there is a unique transition on each symbol. Thus in a DFA, for a given input string w and state qy there will be exactly one path labeled w starting at q. To determine if a string is accepted by a DFA it suffices to check this one path. For an NFA there may be many paths labeled w, and all must be checked to see whether one or more terminate at a final state.
In terms of the picture in Fig. 2.3 with a finite control reading an input tape, we may view the NFA as also reading an input tape. However, the finite control at any time can be in any number of states. When a choice of next state can be KTUNOTES.IN Downloaded from Ktunotes.in 20 FINITE AUTOMATA AND REGULAR EXPRESSIONS V o J 1 Fig. 2.5 The transition diagram for a nondeterministic finite automaton.
made, as in state q0 on input 0 in Fig. 2.5, we may imagine that duplicate copies of the automaton are made. For each possible next state there is one copy of the automaton whose finite control is in that state. This proliferation is exhibited in Fig. 2.6 for the NFA of Fig. 2.5 with input 01001.
</4 !
<U Fig. 2.6 Proliferation of states of an NFA.
Formally we denote a nondeterministic finite automaton by a 5-tuple (Q, I, S y q0 , F), where Q, L, q0 , and F (states, inputs, start state, and final states) have the same meaning as for a DFA, but S is a map from g x I to 2Q . (Recall 2Q is the power set of Q, the set of all subsets of Q.) The intention is that 3(q, a) is the set of all states p such that there is a transition labeled a from q to p.
KTUNOTES.IN Downloaded from Ktunotes.in 2.3 | N ON DETERM IN ISTI C FINITE AUTOMATA 21 Example 2.3 The function 3 for the NFA of Fig. 2.5 is given in Fig. 2.7.
Inputs States \ 0 1 too, <73 } qi 0 {q 2 } Q2 {q 2 } {q 2 } <?3.
0 <?4 Fig. 2.7 The mapping d for the NFA of Fig. 2.5.
The function 3 can be extended to a function <5 mapping Q x Z to 2° and reflecting sequences of inputs as follows: 1) %£) = {<?}, 2) 3(4, wo) = {p|for some state r in <%, w), p is in <S(r, a)}.
Condition ( 1) disallows a change in state without an input. Condition (2) indicates that starting in state q and reading the string w followed by input symbol a we can be in state p if and only if one possible state we can be in after reading w is r, and from r we may go to p upon reading a.
Note that 3(q y a) = 3(q, a) for a an input symbol. Thus we may again use 3 in place of 3.
It is also useful to extend 3 to arguments in 2Q x Z by 3) 3(Py w)={J qinP 3(q, w) for each set of states P ^ Q. L(M), where M is the NFA (Q, £, (5, g0 , F), is {w | c5(g0 , w) contains a state in F).
Example 2.4 Consider again the NFA of Fig.
2.5, whose transition function 3 was exhibited in Fig. 2.7. Let the input be 01001.
<5(<Zo, °) = {<7o, q}-3(q0y 01) = 3(3(q0 , 0), 1) = 3({q0 , q 3 ), 1) = 1) u 5fo 3 , 1) = fo0 , Qii Similarly, we compute 6{q09 010) = {q0 , q3 }, 3(q0 , 0100) = {q0 , 3 . qj and 6(q0 , 01001) = {<?0 , ^ F'\ as follows. The states of M' are all the subsets of the set of states ofM.
That is, Q' = 2Q . M' will keep track in its state of all the states M could be in at any given time. F is the set of all states in Q' containing a final state ofM. An element of Q' will be denoted by [q l9 q2 , .
.
. , q t ], where q l9 q2 , .
.
. , qx are in Q. Observe that <?2> • Qi] ls a single state of the DFA corresponding to a set of states of the NFA. Note that q0 = [q0 ].
We define <5%i, 92, Qil a) = [Pi> P2 , Pjl if and only if <(fai. 02. •••,<?«}> <) = {Pl, P2,---,Pj}-That is, 6' applied to an element [q l9 q2 , .
.
. , qt] ofQ is computed by applying 3 to each state of Q represented by [q u q 2 , qj. On applying 3 to each of^,^ q{ and taking the union, we get some new set of states, p ly p2 , .
.
. , Pj- This new set of states has a representative, [p l9 p 2 , .
.
.
, pj] in Q' 9 and that element is the value of o"([lu q2 , ^f], fl).
It is easy to show by induction on the length of the input string x that ^'(QoyX) = [q l9 q29 ... 9 q l] if and only if HQo, x) = {q l9 q29 qi}.
Basis The result is trivial for |x | =0, since q' 0 = [^0] and x must be c.
Induction Suppose that the hypothesis is true for inputs of length m or less. Let xa be a string of length m + 1 with a in Z. Then <5'(4o> xa) = <5'(<5'(4o, x), a).
KTUNOTES.IN Downloaded from Ktunotes.in 2.3 | NONDETERMINISTIC FINITE AUTOMATA 23 By the inductive hypothesis, <%o> ) = bi, Pi, ...,pj if and only if But by definition of 5\ #{[Pi,P2, ••.,/>;], fl) = [ri,r2 , ...,rj if and only if &({Pu P2..-..Py}.
fl) = {» ,i,r2 ,...,r Jk}.
Thus, if and only if %o.^) = r2 , ...,rk}, which establishes the inductive hypothesis.
To complete the proof, we have only to add that S'(q' 0 , x) is in F exactly when S(q0 , x) contains a state of Q that is in F. Thus L(M) = L(M').
Since deterministic and nondeterministic finite automata accept the same sets, we shall not distinguish between them unless it becomes necessary, but shall simply refer to both as finite automata.
Example 2.5 Let M = ({q0 , q x }, {0, 1}, 5, q0 , {q t }) be an NFA where <fao> 0) = {q0 , 8(q0 , 1) = fail &{q» 0) = 0, 5{q l9 1) = {q0 , q t }.
We can construct a DFA M' = (Q, {0, 1}, d\ [q0], F), accepting L(M) as follows. Q consists of all subsets of {q0 , q t }. We denote the elements of Q by [q0 ], [q Y ], [tfo> q\ and 0. Since S(q0 , 0) = {^0 , gj, we have <5'([<7o], 0) = b0 , gj.
Likewise, '(fool 1) = foil 0) = 0, and 5%,], 1) = [q0 , q,].
Naturally, S'(0, 0) = <5'(0, 1) = 0. Lastly, <5'([tfo> 0) = [q0 , qi\ since <5({<7o, Qil 0) = 5(g0 , 0) u 5{q l9 0) = {g0 , tfi} u 0 = fa0 , ft}, KTUNOTES.IN Downloaded from Ktunotes.in 24 FINITE AUTOMATA AND REGULAR EXPRESSIONS and v(teo, Qil i) = l>o, qil since <%o> 9i}> 1) = S(qo, 1) u 1) = {^J u {^o, 4i} = {4o, The set F of final states is {[gj, [g0 > In practice, it often turns out that many states of the NFA are not accessible from the initial state [q0]. It is therefore a good idea to start with state [q0] and add states to the DFA only if they are the result of a transition from a previously added state.
2.4 FINITE AUTOMATA WITH e-MOVES We may extend our model of the nondeterministic finite automaton to include transitions on the empty input c. The transition diagram of such an NFA accept-ing the language consisting of any number (including zero) of O's followed by any number of l's followed by any number of 2's is given in Fig. 2.8. As always, we say an NFA accepts a string w if there is some path labeled w from the initial state to a final state. Of course, edges labeled e may be included in the path, although the e's do not appear explicitly in w. For example, the word 002 is accepted by the NFA of Fig. 2.8 by the path q0 , q0 , q0 , qu q2 , q 2 with arcs labeled 0, 0, £, e, 2.
o l Fig. 2.8 Finite automaton with (-moves.
Formally, define a nondeterministic finite automaton with e-moves to be a quintuple (Q, Z, 3, q0 , F) with all components as before, but 3, the transition function, maps Q x (E u {e}) to 2Q . The intention is that 3(q, a) will consist of all states p such that there is a transition labeled a from q to p, where a is either £ or a symbol in Z.
Example 2.6 The transition function for the NFA of Fig. 2.8 is shown in Fig.
2.9.
We shall now extend the transition function 3 to a function 3 that maps Q x X to 2Q . Our expectation is that w) will be all states p such that one can KTUNOTES.IN Downloaded from Ktunotes.in 2.4 | FINITE AUTOMATA WITH 6-MOVES 25 States 0 Inputs 2 £ 40 Wo) 0 0 0 {<7i} 0 0 0 {} 0 Fig. 2.9 %, a) for the NFA of Fig. 2.8.
go from q to p along a path labeled w, perhaps including edges labeled e. In constructing 3 it will be important to compute the set of states reachable from a given state q using e transitions only. This question is equivalent to the question of what vertices can be reached from a given (source) vertex in.a directed graph. The source vertex is the vertex for state q in the transition diagram, and the directed graph in question consists of all and only the arcs labeled e. We use £-CWSURE(q) to denote the set of all vertices p such that there is a path from q to p labeled e.
Example 2.7 In Fig. 2.8, £-CLOSURE(g0 ) = {q0 , qu q2\ That is, the path con-sisting of q0 alone (there are no arcs on the path), is a path from q0 to q0 with all arcs labeled £.f Path q0 , q x shows that q x is in £-CLOSURE(g0 ) and path q0 , q t , q2 shows that q 2 is in £-CLOSURE(g0 ).
We may naturally let £-CLOSURE(P), where P is a set of states, be [j qinP £-CLOSURE(g). Now we define 5 as follows.
1) S(q, e) = 6-CLOSURE(g).
2) For w in X and a in Z, d(q, wa) = £-CLOSURE(P), where P = {p | for some r in S(q, w), p is in <5(r, a)}.
It is convenient to extend S and 3 to sets of states by for sets of states R. Note that in this case, d(q, a) is not necessarily equal to S(q, a), since S(q, a) includes all states reachable from q by paths labeled a (including paths with arcs labeled e), while S(q, a) includes only those states reachable from q by arcs labeled a. Similarly, d(q, e) is not necessarily equal to S(q, c). Therefore it is necessary to distinguish S from S when we talk about an NFA with £-transitions.
We define L(M), the language accepted by M = (Q, Z, 3, q0 , F) to be {w|<5(g0 , w) contains a state in F}.
3) 6(R, a)=\J 9inR d(q, a\ and 4) S(R, w)=\J qinR S(q, w) t Remember that a path of length zero has no arcs, and therefore trivially all its arcs are labeled c.
KTUNOTES.IN Downloaded from Ktunotes.in 26 FINITE AUTOMATA AND REGULAR EXPRESSIONS Example 2.8 Consider again the NFA of Fig. 2.8, d(q0 , e) = c-CLOSUREfoo) = {q0 , q» q2 ).
Thus S(q09 0) = £-CLOSURE(<5(%0 , e), 0)) = £-CLOSURE(5(fa0 , q l9 q2\ 0)) = £-CLOSURE(<5(oo, 0) u 6{q l9 0) u 5(q2 , 0)) = £-CLOSURE({^0 } u 0 u 0) = £-CLOSURE(fo0}) = ^ a 2 }.
Equivalence of NFA's with and without e-moves Like nondeterminism, the ability to make transitions on £ does not allow the NFA to accept nonregular sets. We show this by simulating an NFA with ^transitions by an NFA without such transitions.
Theorem 2.2 If L is accepted by an NFA with ^transitions, then L is accepted by an NFA without ^transitions.
Proof Let M = (Q, E, <5, q0 , F) be an NFA with ^transitions. Construct AT = and d'(q, a) is d(q, a) for q in Q and a in X. Note that M' has no ^transitions. Thus we may use <5' for 3', but we must continue to distinguish between S and S.
We wish to show by induction on |x | that d'{q0 > x) = S(q0 , x). However, this statement may not be true for x = £, since 8'(q0 , e) = {q0 }, while d(q0 , e) = £-CLOSURE(g0 ). We therefore begin our induction at 1.
Basis |x | = 1. Then x is a symbol a, and S'(q0 , a) = S(q0 > a) by definition of 6'.
Induction \x> 1. Let x = wa for symbol a in X. Then Then d(q09 01) = £-CLOSURE(<5(%0 > 0), 1)) = £-CLOSURE(S(fo0 , q l9 q 2 } 9 1)) = £-CLOSURE({a 1 }) = {a 1 , q 2 ).
(Q, £, 3\ qQy F') where F u {q0 } if £-CLOSURE(g0 ) contains a state of F, F otherwise, 5'{q0 , wa) = 5'(5'(q0 , w), a).
KTUNOTES.IN Downloaded from Ktunotes.in 2.4 | FINITE AUTOMATA WITH € -MOVES 27 By the inductive hypothesis, d'(q0 , w) = S(q0 , w). Let S(q0 , w) = P. We must show that <5'(P, a) = §(q0 , wa). But V(P,a)= U U %4 4 in P q in P Then as P = 3(<70 , w) we have qinP by rule (2) in the definition of 5. Thus <5'(4o> w") = %o> mi).
To complete the proof we shall show that S'(q0y x) contains a state of F' if and only if S(q0 , x) contains a state of F. If x = £, this statement is immediate from the definition of F". That is, d'(q0 , e) = {q0}, and q0 is placed in F whenever d(q0 , c), which is £-CLOSURE(g0 ), contains a state (possibly q0 ) in F. If x ^ £, then x = wa for some symbol a. If <5(g0 , x) contains a state of F, then surely S'(q0 , x) contains the same state in F'. Conversely, if S'(q0y x) contains a state in F' other than q0 , then 3(g0 , x) contains a state in F. If S'(q0 , x) contains q09 and q0 is not in F, then as 5(q0 , x) = £-CLOSURE(<5(<5(g0 , vv), a)), the state in £-CLOSURE(g0 ) and in F must be in S(q0 , x).
Example 2.9 Let us apply the construction of Theorem 2.2 to the NFA of Fig.
2.8. In Fig.
2.10 we summarize d(q, a). We may also regard Fig. 2.10 as the transition function d' of the NFA without ^transitions constructed by Theorem 2.2. The set of final states F includes q 2 because that is in F and also includes g0 , because £-CLOSURE(g0 ) and F have a state q 2 in common. The transition dia-gram for M' is shown in Fig. 2.11.
States 12 0 Inputs 1 {go, qu qi] {qu qi} {qi} 0 {qn qi} {qi} 0 0 {q 2 } Fig. 2.10 l(q, a) for Fig. 2.8.
0, I. 2 Fig. 2.11 NFA without (-transitions.
KTUNOTES.IN Downloaded from Ktunotes.in 28 FINITE AUTOMATA AND REGULAR EXPRESSIONS 2.5 REGULAR EXPRESSIONS The languages accepted by finite automata are easily described by simple expres-sions called regular expressions. In this section we introduce the operations of concatenation and closure on sets of strings, define regular expressions, and prove that the class of languages accepted by finite automata is precisely the class of languages describable by regular expressions.
Let X be a finite set of symbols and let L, Lu and L2 be sets of strings from X.
The concatenation of L x and L 2 , denoted L XL 2 , is the set {xy \x is in L x and y is in L 2 }. That is, the strings in L X L 2 are formed by choosing a string L t and following it by a string in L2 , in all possible combinations. Define L° = {e} and L = LL~ 1 for i > 1. The Kleene closure (or just closure) of L, denoted L, is the set L= Q IL i = 0 and the positive closure of L, denoted L+ , is the set 00 v = u i-That is, L denotes words constructed by concatenating any number of words from L. L+ is the same, but the case of zero words, whose "concatenation" is defined to be c, is excluded. Note that L+ contains c if and only if L does.
Example 2.10 Let L x = {10, 1} and L 2 = {Oil, 11}. Then LjL 2 = {10011, 1011, 111}. Also, {10, 11} = {(, 10, 11, 1010, 1011, 1110, 1111, ...}.
If X is an alphabet, then X denotes all strings of symbols in X, as we have previously stated. Note that we are not distinguishing X as an alphabet from X as a language of strings of length 1.
Let X be an alphabet. The regular expressions over X and the sets that they denote are defined recursively as follows.
1) 0 is a regular expression and denotes the empty set.
2) c is a regular expression and denotes the set {c}.
3) For each a in X, at is a regular expression and denotes the set {a}.
4) If r and 5 are regular expressions denoting the languages R and 5, respectively, then (r + 5), (rs), and (r) are regular expressions that denote the sets RuS, RS, and R, respectively.
t To remind the reader when a symbol is part of a regular expression, we shall write it in boldface.
However, we view a and a as the same symbol.
KTUNOTES.IN Downloaded from Ktunotes.in 2.5 | REGULAR EXPRESSIONS 29 In writing regular expressions we can omit many parentheses if we assume that has higher precedence than concatenation or + , and that concatenation has higher precedence than + . For example, ((0(1)) + 0) may be written 01 + 0. We may also abbreviate the expression rr by r + . When necessary to distinguish between a regular expression r and the language denoted by r, we use L(r) for the latter. When no confusion is possible we use r for both the regular expression and the language denoted by the regular expression.
Example 2.11 00 is a regular expression representing {00}. The expression (0+1) denotes all strings of 0's and l's. Thus, (0 + 1)00(0 + 1) denotes all strings of 0's and l's with at least two consecutive 0's. The regular expression (1 + 10) denotes all strings of 0's and l's beginning with 1 and not having two consecutive 0's. In proof, it is an easy induction on i that (1 + 10)' does not have two consecutive 0's.t Furthermore, given any string beginning with 1 and not having consecutive 0's, one can partition the string into l's, with a following 0 if there is one. For example, 1101011 is partitioned 1-10-10-1-1. This partition shows that any such string is in (1 -h 10)% where i is the number of l's. The regular expression (0 + e)(l + 10) denotes all strings of 0's and l's whatsoever that do not have two consecutive 0's.
For some additional examples, (0 + 1)011 denotes all strings of 0's and l's ending in 01 1. Also, 012 denotes any number of 0's followed by any number of l's followed by any number of 2's. This is the language of the NFA of Fig. 2.8.
001122 denotes those strings in 012 with at least one of each symbol. We may use the shorthand 0 + l + 2 + for 001122.
Equivalence of finite automata and regular expressions We now turn to showing that the languages accepted by finite automata are precisely the languages denoted by regular expressions. This equivalence was the motivation for calling finite automaton languages regular sets. Our plan will be to show by induction on the size of (number of operators in) a regular expression that there is an NFA with t-transitions denoting the same language. Finally, we show that for every DFA there is a regular expression denoting its language. These constructions, together with Theorems 2.1 and 2.2, show that all four language defining mechanisms discussed in this chapter define the same class of languages, the regular sets. Figure 2.12 shows the constructions we shall perform or have performed, where an arrow from A to B means that for any descriptor of type A a construction yields an equivalent descriptor of type B.
We proceed to prove that for every regular expression there is an equivalent NFA with £-transitions.
t If r is a regular expression, r' stands for rr • - r (/ times).
KTUNOTES.IN Downloaded from Ktunotes.in 30 FINITE AUTOMATA AND REGULAR EXPRESSIONS Fig. 2.12 Constructions of this chapter.
Theorem 2.3 Let r be a regular expression. Then there exists an NFA with £-transitions that accepts L(r).
Proof We show by induction on the number of operators in the regular expres-sion r that there is an NFA M with £-transitions, having one final state and no transitions out of this final state, such that L(M) = L(r).
Basis (Zero operators) The expression r must be £, 0, or a for some a in S. The NFA's in Fig. 2.13(a), (b), and (c) clearly satisfy the conditions.
(a) r = e (b) r = 0 (c) r = a Fig. 2.13 Finite automata for basis step of Theorem 2.3.
Induction (One or more operators) Assume that the theorem is true for regular expressions with fewer than i operators, i > 1. Let r have i operators. There are three cases depending on the form of r.
case 1 r = r x + r2 . Both r x and r2 must have fewer than i operators. Thus there are NFA's M, = (Q u Z lf 5 l9 q l9 {fx }) and M 2 = (Q 2 , I 2 , 5 2 , g 2 , {/2}) with L(M t ) = L(rj) and L(M 2 ) = L(r 2 ). Since we may rename states of an NFA at will, we may assume Q x and Q 2 are disjoint. Let q0 be a new initial state and f 0 a new final state. Construct M = (Q t u Q 2 u {^o,/o}» u £ 2 > 5, g0 > {/o})> where 3 is defined by i) %)> 0 = fai»^2}.
ii) <5(g, a) = Si(q, a) for g in Q i — {/j} and a in Z x u {e}, KTUNOTES.IN Downloaded from Ktunotes.in 2.5 | REGULAR EXPRESSIONS 31 iii) S(q, a) = 3 2 (q, a) for q in Q 2 - {f 2} and a in I 2 u {<:}, iv) = ) = {/o}-Recall by the inductive hypothesis that there are no transitions out of/j or/2 in M x or M 2 . Thus all the moves of M x and M 2 are present in M.
The construction of M is depicted in Fig 2.14(a). Any path in the transition diagram ofM from q0 tof 0 must begin by going to either q t or q2 on e. If the path goes to q l9 it may follow any path in M x to/j and then go to f 0 on t. Similarly, paths that begin by going to q2 may follow any path in M 2 to/2 and then go tof 0 on e. These are the only paths from q0 tof 0 . It follows immediately that there is a path labeled x in M from q0 to / 0 if and only if there is a path labeled x in M t from (h to/i or a path in M 2 from q2 to/ 2 . Hence L(M) = L(M X ) u L(JH)^ as desired.
(c) Fig. 2.14 Constructions used in induction of Theorem 2.3.
(a) For union, (b) For con-catenation, (c) For closure.
case 2 r = r 1 r 2 . Let M t and M 2 be as in Case 1 and construct M = (6i u Q 2 , X, u S 2 , 5, fo}, {/ 2}), where (5 is given by i) <5(g, a) = ^(g, a) for g in Qj — {/J and a in JL l u {c}, «) = {92} iii) (5(g, a) = (5 2 (g, a) for g in Q 2 and a in I 2 u {e}.
KTUNOTES.IN Downloaded from Ktunotes.in 32 FINITE AUTOMATA AND REGULAR EXPRESSIONS The construction ofM is given in Fig. 2.14(b). Every path in M from q x tof 2 is a path labeled by some string x from q x tof x , followed by the edge fromf x to q2 labeled e, followed by a path labeled by some string y from q2 tof 2 . Thus L(M) = {xy\x is in L(M t ) and y is in L(M 2 )} and L(M) = L(M 1 )L(M 2 ) as desired.
case 3 r = r. Let Mj = (Qi, 2 ls <5 l5 tfi, {/i}) and L{M X ) = Construct M = (0! u {q09 f 0 } 9 S lf <5, g0 > {/o})> where (5 is given by i) %o.0 = 5 (/i» £ ) = fei»/o}.
ii) <5(g, a) = d x (q, a) for g in Q x - {f x } and a in Z t u {e}.
The construction of M is depicted in Fig.
2.14(c). Any path from q0 to f 0 consists either of a path from q0 tof 0 on £ or a path from q0 to q x on c, followed by some number (possibly zero) of paths from q i to f x , then back to q t on 6, each labeled by a string in L(M X \ followed by a path from q x to / t on a string in L(MJ, then to f 0 on e. Thus there is a path in M from g0 tof 0 labeled x if and only if we can write x = x x x 2 • • • x} for some 7 > 0 (the case 7 = 0 means x = c) such that each x ( is in L(M x ). Hence L(M) = L(Mj) as desired.
Example 2.12 Let us construct an NFA for the regular expression 01 + 1. By our precedence rules, this expression is really (0(1)) + 1, so it is of the form r \ + r 2> where r x =01 and r 2 = 1. The automaton for r 2 is easy; it is Stait We may express r x as r3 r4 , where r3 = 0 and r4 = 1. The automaton for r 3 is also easy: Stan —\jy In turn, r4 is r, where r 5 is 1. An NFA for r 5 is Start—( V 5 ^ 1 Note that the need to keep states of different automata disjoint prohibits us from using the same NFA for r 2 and r 5 , although they are the same expression.
To construct an NFA for r4 = r| use the construction of Fig. 2.14(c). Create \ states q n and qs playing the roles of q0 and / 0 , respectively. The resulting NFA for r4 is shown in Fig. 2.15(a). Then, for r x = r3 r4 use the construction of Fig. 2.14(b).
The result is shown in Fig. 2.15(b). Finally, use the construction of Fig. 2.14(a) to find the NFA for r = r x + r2 . Two states q9 and q l0 are created to fill the roles of q0 and f 0 in that construction, and the result is shown in Fig. 2.15(c).
KTUNOTES.IN Downloaded from Ktunotes.in 2.5 | REGULAR EXPRESSIONS 33 e Start (a) € Start-(b) Start (c) Fig. 2.15 Constructing an NFA from a regular expression, (a) For r4 = 1.
(b) For r, = 01. (c) For r = 01 + 1.
The proof of Theorem 2.3 is in essence an algorithm for converting a regular expression to a finite automaton. However, the algorithm implicitly assumes that the regular expression is fully parenthesized. For regular expressions without redundant parentheses, we must determine whether the expression is of the form p + q, pq, or p. This is equivalent to parsing a string in a context-free language, and thus such an algorithm will be delayed until Chapter 5 where it can be done more elegantly.
Now we must show that every set accepted by a finite automaton is denoted by some regular expression. This result will complete the circle shown in Fig. 2.12.
Theorem 2.4 If L is accepted by a DFA, then L is denoted by a regular expres-sion.
Proof Let L be the set accepted by the DFA Let Rlj denote the set of all strings x such that d(qiJ x) = qp and if 3(qh y) — qr , for any y that is a prefix (initial segment) of x, other than x or £, then { < k. That is, Ry is the set of all strings that take the finite automaton from state q { to state qi without going through any state numbered higher than k. Note that by "going through a state," we mean both entering and then leaving. Thus i or j may be greater than k. Since there is no state numbered greater than m, R"j denotes all M = {fa, <?„}, 2, 5, q l9 F).
KTUNOTES.IN Downloaded from Ktunotes.in 34 FINITE AUTOMATA AND REGULAR EXPRESSIONS strings that take (2.1) {a\6(qi9 a) = qj) {a\S(qh a) = qj) if if;, u{£} if 1=7.
Informally, the definition of above means that the inputs that cause M to go from q { to qs without passing through a state higher than qk are either 1) in R^j' 1 (that is, they never pass through a state as high as qk ); or 2) composed of a string in R[k ~ 1 (which takes M to qk for the first time) followed by zero or more strings in Rlk 1 (which take M from qk back to qk without passing through qk or a higher-numbered state) followed by a string in R^J 1 (which takes M from state qk to q} ).
We must show that for each z, j, and /c, there exists a regular expression rjj denoting the language R^. We proceed by induction on /c.
Basis (k = 0).
is a finite set of strings each of which is either £ or a single symbol. Thus r°- can be written as a x -f a 2 + " ' + ap (or a 1 + a 2 + ' ' + ap + e if 1 = j), where {a^ a 2 , ap} is the set of all symbols a such that 5(qh a) = qy If there are no such a's, then 0 (or e in the case i = j) serves as r°-.
Induction The recursive formula for R^ given in (2.1) clearly involves only the regular expression operators: union, concatenation, and closure. By the induction hypothesis, for each f and m there exists a regular expression rj~ 1 such that L(r~ !
) = R~ 1 . Thus for rf y we may select the regular expression since R"j denotes the labels of all paths from q x to qj . Thus L(M) is denoted by the regular expression Example 2.13 Let M be the FA shown in Fig. 2.16. The values of rjj for all i and j and for k = 0, 1, or 2 are tabulated in Fig.
2.17. Certain equivalences among regular expressions such as (r -f s)t = rt 4- st and (e -4- r) =? r have been used to simplify the expressions (see Exercise 2.16). For example, strictly speaking, the expression for r\ 2 is given by qj in F where F = {g,v qj2 , = ^ 1 (r? 1 )r?2 + rL = 0(£)0 + £.
KTUNOTES.IN Downloaded from Ktunotes.in 2.5 | REGULAR EXPRESSIONS 35 1 Fig. 2.16 FA for Example 2.13.
k = 0 k= 1 fc = 2 e e (00) 0 0 0(00) /i 3 1 1 01 0 0 0(00) e 6+00 (00) A, 1 1+01 01 r3, 0 0 (0 + 1)(00)0 ^32 0+1 0+1 (0 + 1)(00) As e £ e + (0+1)01 Fig. 2.17 Tabulation of rf, for FA of Fig. 2.16.
Similarly, rh = r\ 2 (r22 )r23 + r} 3 = 0(6 + 00)(1 + 01) + 1.
Recognizing that (6 + 00) is equivalent to (00) and that 1 + 01 is equivalent to (e + 0)1, we have r? 3 = 0(00)(6 + 0)1 + 1.
Observe that (00)(6 + 0) is equivalent to 0. Thus 0(00)(6 + 0)1 + 1 is equiv-alent to 001 + 1 and hence to 01.
To complete the construction of the regular expression for M, which is r i2 + ^13, we write r l2 = r i3(r33)r32 + r l2 = 01(6 + (0 + 1)01)(0 + 1)(00) + 0(00) = 01((0 + 1)01)(0 + 1)(00) + 0(00) and rh = rUrh)rh + r2 13 = 01(6 + (0 + 1)01)(6 + (0 + 1)01) + 01 = 01((0 + 1)01).
Hence r\ 2 + r3 i3 = 01((0 + 1)01)(6 + (0 + 1)(00)) + 0(00).
KTUNOTES.IN Downloaded from Ktunotes.in 36 FINITE AUTOMATA AND REGULAR EXPRESSIONS 2.6 TWO-WAY FINITE AUTOMATA We have viewed the finite automaton as a control unit that reads a tape, moving one square right at each move. We added nondeterminism to the model, which allowed many "copies" of the control unit to exist and scan the tape simulta-neously. Next we added e-transitions, which allowed change of state without read-ing the input symbol or moving the tape head. Another interesting extension is to allow the tape head the ability to move left as well as right. Such a finite automa-ton is called a two-way finite automaton. It accepts an input string if it moves the tape head off the right end of the tape, at the same time entering an accepting state.
We shall see that even this generalization does not increase the power of the finite automaton; two-way FA accept only regular sets. We give a proof only for a special case of a two-way FA that is deterministic and whose tape head must move left or right (not remain stationary) at each move. A more general model is considered in the exercises.
A two-way deterministic finite automaton (2DFA) is a quintuple M = (Q, X, S y qQy F), where Q y X, qQy and F are as before, and 3 is a map from Q x X to Q x {L, R}. If 3(q, a) = (p, L), then in state q, scanning input symbol a, the 2DFA enters state p and moves its head left one square. If S(q, a) = (p, R)y the 2DFA enters state p and moves its head right one square.
In describing the behavior of a one-way FA, we extended S to Q x £. This corresponds to thinking of the FA as receiving a symbol on an input channel, processing the symbol and requesting the next. This notion is insufficient for the two-way FA, since the 2DFA may move left. Thus the notion of the input being written on the tape is crucial. Instead of trying to extend 3 y we introduce the notion of #n instantaneous description (ID) of a 2DFA, which describes the input string, current state, and current position of the input head. Then we introduce the relation 1^ on ID's such that l x \jfl 2 if and only if M can go from the instantaneous description I { to I 2 in one move.
An ID of M is a string in The ID wqx y where w and x are in X and q is in Q, is intended to represent the facts that 1) wx is the input string, 2) q is the current state, and 3) the input head is scanning the first symbol of x.
If x = c, then the input head has moved off the right end of the input.
We define the relation \jf or just f— if M is understood, by 1) a x a 2 flf-i^fli •• an {—a l a 2 ' ' a i .
l a i pai+l ••• an whenever 3(q, a,) = (p, R), and 2) a x a 2 • • 2 a, , qa, ?
• • • an |— a, a 2 • • • a, 2 pa,- { a i , • • • an whenever d(q, a t ) = (p, L) and i > 1 .
The condition i > 1 prevents any action in the event that the tape head would move off the left end of the tape. Note that no move is possible if i = n + 1 (the tape KTUNOTES.IN Downloaded from Ktunotes.in 2.6 | TWO-WAY FINITE AUTOMATA 37 head has moved off the right end). Let P 5- be the reflexive and transitive closure of f— . That is, / p- / for all ID's /, and l x Ik whenever /ih /2h'"h /i for some 7 2 , We define L(M) = {w | q0 w wp for some p in F}.
That is, w is accepted by M if, starting in state q0 with w on the input tape and the head at the left end of w, M eventually enters a final state at the same time it falls off the right end of the input tape.
Example 2.14 Consider a 2DFA M that behaves as follows: Starting in state q0 , M repeats a cycle of moves wherein the tape head moves right until two l's have been encountered, then left until encountering a 0, at which point state q0 is reentered and the cycle repeated. More precisely, M has three states, all of which are final; 3 is given in Fig. 2.18.
0 1 qo fao, R) fai, R) q x (qu R) (q2 , L) qi (qo, R) (qi, L) Fig. 2.18 The transition function for the 2DFA of Example 2.14.
Consider the input 101001. Since q0 is the initial state, the first ID is q0 10 1001.
To obtain the second ID, note that the symbol to the immediate right of the state q0 in the first ID is a 1 and 3(q0 , 1) is (q l9 R). Thus the second ID is l^OlOOl.
Continuing in this fashion we get the result shown in Table 2.1. Hence M even-tually moves off the right end of the tape in an accepting state. Thus 101001 is in L(M).
Table 2.1 g0 101001 |— 1^01001 |— 10gi 1001 |— lg 201001 |— 10^o 1001 \— 101(7,001 f— 1010(7,01 \— 10100(7,1 |— 1010(7 201 |— 10100(7o 1 f— lOlOOlg, KTUNOTES.IN Downloaded from Ktunotes.in 38 FINITE AUTOMATA AND REGULAR EXPRESSIONS Crossing sequences A useful picture of the behavior of a 2DFA consists of the input, the path followed by the head, and the state each time the boundary between two tape squares is crossed, with the assumption that the control enters its new state prior to moving the head. For example, the behavior of the 2DFA M of Example 2.14 on 101001 is shown in Fig. 2.19.
10 10 0 1 % - 1\ -Ch -N Fig. 2.19 Behavior of the 2DFA of Example 2.14.
The list of states below each boundary between squares is termeJ a crossing sequence. Note that if a 2DFA accepts its input, no crossing sequence may have a repeated state with the head moving in the same direction, otherwise the 2DFA, being deterministic, would be in a loop and thus could never fall off the right end.
Another important observation about crossing sequences is that the first time a boundary is crossed, the head must be moving right. Subsequent crossings must be in alternate directions. Thus odd-numbered elements of a crossing sequence represent right moves and even-numbered elements represent left moves. If the input is accepted, it follows that all crossing sequences are of odd length.
A crossing sequence q Y ,q 2 , • • • , tf is said to be valid if it is of odd length, and no two odd- and no two even-numbered elements are identical. A 2DFA with s states can have valid crossing sequences of length at most 2s, so the number of valid crossing sequences is finite.
Our strategy for showing that any set accepted by a 2DFA M is regular is to construct an equivalent NFA whose states are the valid crossing sequences of M.
To construct the transition function of the NFA we first examine the relationship between adjacent crossing sequences.
Suppose we are given an isolated tape square holding the symbol a and are also given valid crossing sequences q l3 q2 , .
.
. , qk and p l9 p2 > .
.
.
, pf at the left and right boundaries of the square, respectively. Note that there may be no input strings that could be attached to the left and right of symbol a to actually produce these two crossing sequences. Nevertheless we can test the two sequences for local compatibility as follows. If the tape head moves left from the square holding a in state qh restart the automaton on the square holding a in state qi+v Similarly, whenever the tape head moves right from the square in state ph restart the autom-KTUNOTES.IN Downloaded from Ktunotes.in 2.6 | TWO-WAY FINITE AUTOMATA 39 aton on the square in state pi+1 . By this method we can test the two crossing sequences to be sure that they are locally consistent. These ideas are made precise below.
We define right-matching and left-matching pairs of crossing sequences recur-sively in (i) through (v) below. The intention is for qu q2 , .
.
. , qk to right-match p 1?
Pi-> Ve on a ^ these sequences are consistent, assuming we initially reach a in state q t moving right, and for the two crossing sequences to left-match if the sequences are consistent, assuming we initially reach a in state p l moving left. In each case, we take qu q2 , .
.
.
, qk to appear at the left boundary of a and p 1? p2 , .
.
.
, pe at the right boundary.
i) The null sequence left- and right-matches the null sequence. That is, if we never reach the square holding a, then it is consistent that the boundaries on neither side should be crossed.
ii) If q 3 , qk right-matches p l9 pe and S(q l9 a)= (q 2 , L), then qu ..., qk right-matches pu .
.
. , pf . That is, if the first crossing of the left boundary is in state q l and the head immediately moves left in state q 2i then if we follow these two crossings by any consistent behavior starting from another crossing of the left boundary, we obtain a consistent pair of sequences with first cross-ing moving right, i.e., a right-matched pair.
iii) If q 2 , qk left-matches p 2 , p( and 5{q x , a) — (p 1? R\ then q l9 qk right-matches pu ... y pr That is, if the first crossing of the left boundary is in state q x and the head immediately moves right in state p l7 then if we follow these two crossings by any consistent behavior starting from a crossing of the right boundary, we obtain a consistent pair of sequences with the first cross-ing from the left, i.e., a right-matched pair. Note that this case introduces the need for left-matched sequences, even though we are really only interested in right-matched pairs.
iv) If q l3 qk left-matches p 3 , pf and 5(p 1 , a) = (p2 , R), then q u qk left-matches p 1?
pf . The justification is similar to that for rule (ii).
v) If q2 , qk right-matches p2 , pf and 5(p ly a) = (q u L), then q u qk left-matches p l3 pr . The justification is similar to rule (iii).
Example 2.15 Consider the 2DFA M of Example 2.14 and a tape square con-taining the symbol 1. The null sequence left-matches the null sequence, and ^(^o» 0 = (qi, R)- Thus qQ right-matches q x on 1 by rule (iii). Since S(q ly 1) = (q2 , L), q l9 ql9 q0 right-matches q x on 1 by rule (ii). This must be the case, since there is in fact an accepting computation in which this pair of sequences actually occurs to the left and right of a square holding a 1. Note, however, that a pair of sequences could match, yet there could be no computation in which they appeared adjacent, as it could be impossible to find strings to place to the left and right that would "turn the computation around" in the correct states.
KTUNOTES.IN Downloaded from Ktunotes.in 40 FINITE AUTOMATA AND REGULAR EXPRESSIONS Equivalence of one-way and two-way finite automata Theorem 2.5 If L is accepted by a 2DFA, then L is a regular set.
Proof Let M = (Q, Z, <5, q0 , F) be a 2DFA. The proof consists of constructing an NFA M' which accepts L(M). Define M' to be (Q\ Z, <5\ q' 0 , F'), where 1) g' consists of all valid crossing sequences for M.
2) q' Q is the crossing sequence consisting of q0 alone.
3) F' is the set of all crossing sequences of length one consisting of a state in F.
4) (5'(c, a) = {d\d is a. valid crossing sequence that is right-matched by c on input a}. Note that as d is valid it must be of odd length.
The intuitive idea is that M' puts together pieces of the computation ofM as it scans the input string. This is done by guessing successive crossing sequences. If M' has guessed that c is the crossing sequence at a boundary, and a is the next input symbol, then M' can guess any valid crossing sequence that c right-matches on input a. If the guessed computation results in M moving off the right end of the input in an accepting state, then M' accepts.
We now show that L(M') = L(M). Let w be in L(M). Look at the crossing sequences generated by an accepting computation of M on w. Each crossing sequence right-matches the one at the next boundary, so M' can guess the proper crossing sequences (among other guesses) and accept.
Conversely, if w is in L(M'), consider the crossing sequences c0 , c I?
.
. ., cn ofM corresponding to the states of M' as M' scans w = a l a 2 "' a„. For each i, 0 < i < n, c, right-matches c, + l on a { . We can construct an accepting computation of M on input w by determining when the head reverses direction. In particular, we prove by induction on i that M' on reading a l a 2 a, can enter state Ci = [<7i, -qk] only if 1) M started in state q0 on a l a 2 "' a{ will first move right from position i in state q l9 and 2) for j = 2, 4, .
.
. , ifM is started at position i in state qp M will eventually move right from position / in state qj+ t (this implies that k must be odd).
Basis (i = 0). As c0 = [q0 ], ( 1 ) is satisfied since M begins its computation by "moving right" from position 0 in state q0 . Condition (2) holds vacuously.
Induction Assume the hypothesis true for i — 1. Suppose that M' on reading a l a 1 ••• a, can enter state c, = [p t , #J from state c,_! = <5fJ. Since /c and ^ are odd, and c, _ x right-matches c, on ah there must exist an odd j such that in state qj on input ah M moves right. Let j 1 be the smallest such j. By definition of "right-matches" it follows that S(qjl , a x ) = (p x , R). This proves (1). Also by the definition of "right-matches" (rule iii) [qh + ls .
.
. , qk] left-matches [p 2 , .
.
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, p,]. Now if 5(pj, a t ) = (pi+ 1? R) for all even j, then (2) follows immediately. In the case that KTUNOTES.IN Downloaded from Ktunotes.in 2.6 | TWO-WAY FINITE AUTOMATA 41 for some smallest even j 2i S(pj2 , a t ) = (q> L), then by the definition of "left-matches" (rule v) q must be qjl + 1 and [qjl + 2 , .
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. , qk] right-matches [pj2 + x , .
.
. , pj.
The argument then repeats with the latter sequences in place of c f _ x and c f .
With the induction hypothesis for all i established, the fact that c„ = [p] for some p in F implies that M accepts a l a2 " an .
Example 2.16 Consider the construction of an NFA M' equivalent to the 2DFA M of Example 2.14. Since q 2 is only entered on a left move, and q l and q 2 are only entered on right moves, all even-numbered components of valid crossing se-quences must be q 2 . Since a valid crossing sequence must be of odd length, and no two odd-numbered states can be the same, nor can two even-numbered states be the same, there are only four crossing sequences of interest; these are listed in Fig.
2.20 along with their right matches.
Valid crossing sequences [%] [</.] 92, qo] Right matches Right matches on 0 on 1 M [9.] [ill qo] [9,] Fig. 2.20 Valid crossing sequences along with their right matches.
We note immediately that state [q0 , q 2y q { ] may be removed from the con-structed NFA M', since it has no right match. The resulting M' is shown in Fig.
2.21. Note that L(M') = (c + 1)(0 + 01), that is, all strings of O's and l's without two consecutive l's.
Consider the input 1001, which is accepted by M' using the sequence of states [tfoL [<lu <?2> 4o]> We can visualize the crossing sequences as in Fig.
2.22. Note that S(q0 , 1) = (q i3 R) justifies the first move and that 3(q l , 0) = (q u R) 0 0 Fig. 2.21 The NFA M' constructed from the 2DFA M.
KTUNOTES.IN Downloaded from Ktunotes.in 42 FINITE AUTOMATA AND REGULAR EXPRESSIONS 10 0 1 % -</i - </i - ^1 Fig. 2.22 Crossing sequences of 2DFA on input 1001.
justifies the second and third. Since <%,, 1) = (q2 , L) we see the justification for the fourth move, which reverses the direction of travel. Then 3(q 2i 0) = (q0 , R) again reverses the direction, and finally 3(q0 , 1) = (q u R) explains the last move.
2.7 FINITE AUTOMATA WITH OUTPUT One limitation of the finite automaton as we have defined it is that its output is limited to a binary signal: "accept'V'don't accept." Models in which the output is chosen from some other alphabet have been considered. There are two distinct approaches; the output may be associated with the state (called a Moore machine) or with the transition (called a Mealy machine). We shall define each formally and then show that the two machine types produce the same input-output mappings.
Moore machines A Moore machine is a six-tuple (Q, X, A, <5, X, q0 ), where Q, Z, 3 y and q0 are as in the DFA. A is the output alphabet and X is a mapping from Q to A giving the output associated with each state. The output of M in response to input a t a 2 "' an > n > 0, is A(g0 )A(4i) ' ' X(q„), where q0 , q u .
.
. , q„ is the sequence of states such that <5(<ft ls a,) = q{ for 1 < i < n. Note that any Moore machine gives output k(qQ ) in response to input e. The DFA may be viewed as a special case of a Moore machine where the output alphabet is {0, 1} and state q is "accepting" if and only if ;-(<?)= i-Example 2.17 Suppose we wish to determine the residue mod 3 for each binary string treated as a binary integer. To begin, observe that if i written in binary is followed by a 0, the resulting string has value 2z, and if i in binary is followed by a 1, the resulting string has value 2i + 1.
If the remainder of i/3 is p, then the remainder of 2//3 is 2p mod 3.
If p = 0, 1, or 2, then 2p mod 3 is 0, 2, or 1, respectively. Similarly, the remainder of (2i -h l)/3 is 1, 0, or 2, respectively.
It suffices therefore to design a Moore machine with three states, q0y qu and q2 , where q} is entered if and only if the input seen so far has residue j. We define KTUNOTES.IN Downloaded from Ktunotes.in 2.7 | FINITE AUTOMATA WITH OUTPUT 43 0 110 2 0 1 Fig. 2.23 A Moore machine calculating residues.
X(qj) = j for j = 0, 1, and 2. In Fig. 2.23 we show the transition diagram, where outputs label the states. The transition function S is designed to reflect the rules regarding calculation of residues described above.
On input 1010 the sequence of states entered is q0 , qu q2 , q2 , <?i, giving output sequence 01221. That is, e (which has "value" 0) has residue 0, 1 has residue 1, 2 (in decimal) has residue 2, 5 has residue 2, and 10 (in decimal) has residue 1.
Mealy machines A Mealy machine is also a six-tuple M = (Q, Z, A, d, A, q0 )9 where all is as in the Moore machine, except that X maps Q x £ to A. That is, X(q, a) gives the output associated with the transition from state q on input a. The output ofM in response to input a 1 a2 ••• an is X(q0 , a x )k(qu a2)-- A(qn - l9 an\ where q0y qu ...,qn is the sequence of states such that <5(gf _ l9 a,) = q t for 1 < i < n. Note that this sequence has length n rather than length n + 1 as for the Moore machine, and on input e a Mealy machine gives output c.
Example 2.18 Even if the output alphabet has only two symbols, the Mealy machine model can save states when compared with a finite automaton. Consider the language (0 + 1)(00 + 11) of all strings of 0's and Ts whose last two symbols are the same. In the next chapter we shall develop the tools necessary to show that this language is accepted by no DFA with fewer than five states. However, we may define a three-state Mealy machine that uses its state to remember the last symbol read, emits output y whenever the current input matches the previous one, and emits n otherwise. The sequence of ys and ns emitted by the Mealy machine corresponds to the sequence of accepting and nonaccepting states entered by a DFA on the same input; however, the Mealy machine does not make an output prior to any input, while the DFA rejects the string 6, as its initial state is nonfinal.
The Mealy machine M = ({q0 , p0 , pj, {0, 1}, {y, n}, S, A, q0 ) is shown in Fig.
2.24. We use the label a/b on an arc from state p to state q to indicate that Hp a) = q and A(p, a) = b. The response of M to input 01100 is nnyny, with the sequence of states entered being q0 PoPi Pi PoPo- Note how p0 remembers a zero and p x remembers a one. State q0 is initial and "remembers" that no input has yet been received.
KTUNOTES.IN Downloaded from Ktunotes.in 44 FINITE AUTOMATA AND REGULAR EXPRESSIONS Fig. 2.24 A Mealy machine.
Equivalence of Moore and Mealy machines Let M be a Mealy or Moore machine. Define T M (w), for input string w, to be the output produced by M on input w. There can never be exact identity between the functions T M and T M . ifM is a Mealy machine and M' a Moore machine, because | ^(w) j is one less than |r M -(w)| for each w.
However, we may neglect the response of a Moore machine to input e and say that Mealy machine M and Moore machine M' are equivalent if for all inputs w, bT M (w) = T M (w), where b is the output of M' for its initial state. We may then prove the following theorems, equating the Mealy and Moore models.
Theorem 2.6 IfM x = (Q, E, A, 3, X, q0 ) is a Moore machine, then there is a Mealy machine M 2 equivalent to M v Proof Let M 2 = (Q, 1, A, 3, X\ q0 ) and define X'(q, a) to be X(3(q, a)) for all states q and input symbols a. Then M x and M 2 enter the same sequence of states on the same input, and with each transition M 2 emits the output that Mj associates with the state entered.
Theorem 2.7 Let M x = (Q, Z, A, 3, X, q0 ) be a Mealy machine. Then there is a Moore machine M 2 equivalent to M l .
Proof Let M 2 = (Q x A, E, A, 3', X, [q0 , b0]), where b0 is an arbitrarily selected member of A. That is, the states of M 2 are pairs [q, b] consisting of a state of Mj and an output symbol. Define 3'([qy b\ a) = [3(q, a), X(q, a)] and X'([q, b]) = b.
The second component of a state [q, b] ofM 2 is the output made by Mi on some transition into state q. Only the first components of M 2 's states determine the moves made by M 2 . An easy induction on n shows that if M y enters states q0 , q ly qn on input a x a 2 •• an , and emits outputs b l9 b 2 , bn , then M 2 enters states [g0 , b0 ] t [q x , b^, [qn , bn] and emits outputs b0 , b u b 2 , b„.
Example 2.19 Let M x be the Mealy machine of Fig. 2.24. The states of M 2 are [<7o, yl [q<» 4 bo> yl [Po, n], [p u y] 9 and [p u n\ Choose b0 = «, making [g0 , "] KTUNOTES.IN Downloaded from Ktunotes.in 2.8 | APPLICATIONS OF FINITE AUTOMATA 45 1 1 Fig. 2.25 Moore machine constructed from Mealy machine.
M 2 's start state. The transitions and outputs of M 2 are shown in Fig. 2.25. Note that state [q0i y] can never be entered and may be removed.
2.8 APPLICATIONS OF FINITE AUTOMATA There are a variety of software design problems that are simplified by automatic conversion of regular expression notation to an efficient computer implementa-tion of the corresponding finite automaton. We mention two such applications here; the bibliographic notes contain references to some other applications.
Lexical analyzers The tokens of a programming language are almost without exception expressible as regular sets. For example, ALGOL identifiers, which are upper- or lower-case letters followed by any sequence of letters and digits, with no limit on length, may be expressed as (letter)(letter + digit) where "letter" stands forA + B + -" + Z + a + b+ --- + z, an(j "digit" stands for 0 + 1 + •• + 9. FORTRAN identifiers, with length limit six and letters re-stricted to upper case and the symbol $, may be expressed as (letter)(c + letter + digit) 5 where "letter" now stands for ($ + A + B + • • + Z). SNOBOL arithmetic con-stants (which do not permit the exponential notation present in many other languages) may be expressed as (e + -)(digit + (- digit + e) + • digit) A number of lexical-analyzer generators take as input a sequence of regular expressions describing the tokens and produce a single finite automaton recogniz-KTUNOTES.IN Downloaded from Ktunotes.in 46 FINITE AUTOMATA AND REGULAR EXPRESSIONS ing any token. Usually, they convert the regular expression to an NFA with 6-transitions and then construct subsets of states to produce a DFA directly, rather than first eliminating 6-transitions. Each final state indicates the particular token found, so the automaton is really a Moore machine. The transition function of the FA is encoded in one of several ways to take less space than the transition table would take if represented as a two-dimensional array. The lexical analyzer produced by the generator is a fixed program that interprets coded tables, together with the particular table that represents the FA recognizing the tokens (specified to the generator in regular expression notation). This lexical analyzer may then be used as a module in a compiler. Examples of lexical analyzer generators that follow the above approach are found in Johnson et al and Lesk .
Text editors Certain text editors and similar programs permit the substitution of a string for any string matching a given regular expression. For example, the UNIX text editor allows a command such as s/bt>t>/b/ that substitutes a single blank for the first string of two or more blanks found in a given line. Let "any" denote the expression a t + a 2 + • • + an , where the a^s are all of a computer's characters except the "newline" character. We could convert a regular expression r to a DFA that accepts anyr. Note that the presence of any allows us to recognize a member of L(r) beginning anywhere in the line. However, the conversion of a regular expression to a DFA takes far more time than it takes to scan a single short line using the DFA, and the DFA could have a number of states that is an exponential function of the length of the regular expression.
What actually happens in the UNIX text editor is that the regular expression anyr is converted to an NFA with (-transitions, and the NFA is then simulated directly, as suggested in Fig. 2.6. However, once a column has been constructed listing all the states the NFA can enter on a particular prefix of the input, the previous column is no longer needed and is thrown away to save space. This approach to regular set recognition was first expressed in Thompson .
EXERCISES S 2.1 Find a finite automaton whose behavior corresponds to the circuit in Fig. 2.26, in the sense that final states correspond to a 1-output. A circle with a dot represents an AND-gate, whose output is 1 only if both inputs have value 1. A circle with a + represents an OR-gate, whose output is 1 whenever either input has value 1. A circle with a ~ represents an inverter, whose output is 1 for input 0 and 0 for input 1. Assume there is sufficient time between changes in input values for signals to propagate and for the network to reach a stable configuration.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 47 2.2 Historically, finite automata were first used to model neuron nets. Find a finite automaton whose behavior is equivalent to the neuron net in Fig. 2.27. Final states of the automaton correspond to a 1 -output of the network. Each neuron has excitatory (circles) and inhibitory (dots) synapses. A neuron produces-a 1-output if the number of excitatory synapses with 1-inputs exceeds the number of inhibitory synapses with 1-inputs by at least the threshold of the neuron (number inside the triangle). Assume there is sufficient time between changes in input value for signals to propagate and for the network to reach a stable configuration. Further assume that initially the values of y l9 y 2 » and y 3 are all 0.
2.3 Consider the toy shown in Fig. 2.28. A marble is dropped in at A or B. Levers x u x 2 , and x 3 cause the marble to fall either to the left or right. Whenever a marble encounters a KTUNOTES.IN Downloaded from Ktunotes.in 48 FINITE AUTOMATA AND REGULAR EXPRESSIONS C D Fig. 2.28 A toy.
lever, it causes the lever to change state, so that the next marble to encounter the lever will take the opposite branch.
a) Model this toy by a finite automaton. Denote a marble in at A by a 0-input and a marble in at B by a 1 -input. A sequence of inputs is accepted if the last marble comes out at D.
b) Describe the set accepted by the finite automaton.
c) Model the toy as a Mealy machine whose output is the sequence of Cs and £>'s out of which successive marbles fall.
2.4 Suppose d is the transition function of a DFA. Prove that for any input strings x and y, 5(q, xy) — d(6(q, x), y). [Hint: Use induction on |y|.] 2.5 Give deterministic finite automata accepting the following languages over the alphabet {0, 1}.
a) The set of all strings ending in 00.
b) The set of all strings with three consecutive O's.
c) The set of all strings such that every block of five consecutive symbols contains at least two O's.
d) The set of all strings beginning with a 1 which, interpreted as the binary representation of an integer, is congruent to zero modulo 5.
e) The set of all strings such that the 10th symbol from the right end is 1.
2.6 Describe in English the sets accepted by the finite automata whose transition diagrams are given in Fig. 2.29(a) through (c).
'S 2.7 Prove that the FA whose transition diagram is given in Fig. 2.30 accepts the set of all strings over the alphabet {0, 1} with an equal number of 0's and Ts, such that each prefix has at most one more 0 than l's and at most one more 1 than 0's.
2.8 Give nondeterministic finite automata accepting the following languages.
a) The set of strings in (0 + 1) such that some two 0's are separated by a string whose length is 4i, for some i > 0.
b) The set of all strings over the alphabet {a, b, c} that have the same value when eval-uated left to right as right to left by multiplying according to the table in Fig. 2.31.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 49 Fig. 2.31 Nonassociative multiplication table.
KTUNOTES.IN Downloaded from Ktunotes.in 50 FINITE AUTOMATA AND REGULAR EXPRESSIONS c) The set of all strings of O's and l's such that the 10th symbol from the right end is a 1.
How does your answer compare with the DFA of Problem 2.5(e)?
2.9 Construct DFA's equivalent to the NFA's.
a) ({p, q, r, s}, {0, 1}, S u p, {s}), b) ({p, q, r, s}, {0, 1}, 5 2(p, {q, s}) where 5 1 and S 2 are given in Fig. 2.32.
\ 0 1 \ 0 1 p P, <\ P P q, s <\ r r r r s r s p s 5 s s p s 1 Fig. 2.32 Two transition functions.
2.10 Write regular expressions for each of the following languages over the alphabet {0, 1}. Provide justification that your regular expression is correct.
a) The set of all strings with at most one pair of consecutive CTs and at most one pair of consecutive Ts.
' b) The set of all strings in which every pair of adjacent 0s appears before any pair of adjacent l's.
c) The set of all strings not containing 101 as a substring.
d) The set of all strings with an equal number of 0's and Ts such that no prefix has two more 0's than l's nor two more l's than 0's.
2.11 Describe in English the sets denoted by the following regular expressions.
a) (11 +0)(00+ 1) b) (1 +01 + 001)(< +0 + 00) c) [00 + 11 + (01 + 10)(00 + 11)(01 + 10)] 2.12 Construct finite automata equivalent to the following regular expressions.
a) 10+ (0+ 11)01 b) 01[((10) + 111) +0]1 c) ((0 +l)(0+l)) + ((0+l)(0+l)(0+l)) 2.13 Construct regular expressions corresponding to the state diagrams given in Fig. 2.33.
2.14 Use the ideas in the proof of Theorem 2.4 to construct algorithms for the following problems.
a) Find the lowest-cost path between two vertices in a directed graph where each edge is labeled with a nonnegative cost.
b) Determine the number of strings of length n accepted by an FA.
2.15 Construct an NFA equivalent to the 2DFA ({g0 , q 5 } y {0, 1}, 6, q0 , {q 2 }), where 8 is given by Fig. 2.34.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 51 \ 0 1 40 (4o, R) (4., ) 41 (<?.. R) (42, H) 42 (92. R) (43, L) 43 (94, L) (43, L) 44 (>, R) (44, L) Fig. 2.34 A transition function for a 2DFA.
2.16 Prove the following identities for regular expressions r, s, and f. Here r = s means 4r) = L(s).
a) r + 5 = s + r b) (r + .s) + f = r + (5 + f) c) (rs)f = r(sf) d) r(s + f) = rs + e) (r + 5)f = rf + st f ) 0 = £ g) (r) = r h) (£ + r) = r i) (r5) = (r + s) 2.17 Prove or disprove the following for regular expressions r, s, and r.
a) (rs + r)r = r(sr + r) b) s(r5 + s)r = rr5(rrs) c) (r + s) = r + 5 2.18 A fwo-way nondeterministicfinite automaton (2NFA) is defined in the same manner as the 2DFA, except that the 2NFA has a set of possible moves for each state and input symbol. Prove that the set accepted by any 2NFA is regular. [Hint: The observation in the proof of Theorem 2.5 that no state may repeat with the same direction in a valid crossing sequence is no longer true. However, for each accepted input we may consider a shortest computation leading to acceptance.] 2.19 Show that adding the capability of the 2NFA to keep its head stationary (and change state) on a move does not increase the class of languages accepted by 2NFA.
2.20 A 2NFA with endmarkers is a 2NFA with special symbols <^ and $ marking the left and right ends of the input. We say that input x, which contains no $ or $ symbols, is KTUNOTES.IN Downloaded from Ktunotes.in 52 FINITE AUTOMATA AND REGULAR EXPRESSIONS accepted if the 2NFA started with on its tape and with the tape head scanning $ enters an accepting state anywhere on its input. Show that the 2NFA with endmarkers accepts only regular sets.
2.21 Consider a 2DFA M = (Q y E, (5, q 0 , F). For each string x construct a mapping/from Q to Q u {<):}, where f(q) — p if the 2DFA started on the rightmost symbol of x eventually moves off x to the right, in state p.f(q) = $ means that the 2DFA when started on the rightmost symbol of x either never leaves x or moves off the left end. Construct a DFA which simulates M by storing in its finite control a table / instead of a crossing sequence.
2.22 Let r and s be regular expressions. Consider the equation X = rX + 5, where rX denotes the concatenation of r and X, and -I- denotes union. Under the assumption that the set denoted by r does not contain c, find the solution for X and prove that it is unique. What is the solution if L(r) contains cl 2.23 One can construct a regular expression from a finite automaton by solving a set of linear equations of the form where a XJ and c, are sets of strings denoted by regular expressions, + denotes set union, and multiplication denotes concatenation. Give an algorithm for solving such equations.
2.24 Give Mealy and Moore machines for the following processes: a) For input from (0 + 1), if the input ends in 101, output A ; if the input ends in 110, output B\ otherwise output C.
b) For input from (0 + 1 + 2) print the residue modulo 5 of the input treated as a ternary (base 3, with digits 0, 1, and 2) number.
Solutions to Sample Exercises 2.1 Note that the gate output at y x affects the gate output at y 2 and conversely. We shall assume values for y { and y 2 and use these assumed values to compute new values. Then we repeat the process with the new values until we reach a stable state of the system. In Fig.
2.35 we have tabulated the stable values of yi and y 2 for each possible assumed values for Vi and y 2 and for input values 0 and 1.
Input 0 1 \ 0 1 00 00 01 <7o 01 01 11 01 Qi 12 Qi 11 11 10 <]2 12 43 10 00 10 03 (a) (b) Fig. 2.35 Transitions of switching circuit.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 53 If y x and y 2 are both assumed to have value 0, then gates A and B have output 0 and gate C has output equal to the value of the input x. Since both inputs to gate D are 0, the output of gate D is 0. The output of gate E has the value of the input x. Thus the top row in Fig. 2.35(a) has entries 00 and 01. The remaining entries are computed in a similar manner.
We can model the circuit by assigning a state to each pair of values for y x y 2 . This is done in Fig. 2.35(b). Since Vi = y 2 = 1 produces a 1-output, q 2 is a final state. The circuit can be seen to record the parity of pulses (1-inputs) and produce an output pulse for every odd-numbered input pulse.
2.7 We are asked to prove that a set informally described in English is the set accepted by the FA. Clearly we cannot give a completely formal proof. We must either argue intuitively that some formal description of the set is equivalent to the English description and then proceed formally or else simply give an informal proof. We choose the latter.
The proof consists of deducing the properties of strings, taking the automaton to each of the four states, and then proving by induction on the length of a string that our inter-pretation is correct.
We say that a string x is proper if each prefix of x has at most one more 0 than 1 and at most one more 1 than 0. We argue by induction on the length of a string x that 1) S(q0 , x) = q0 if and only if x is proper and contains an equal number of 0's and l's, 2) S(q0 , x) = q x if and only if x is proper and contains one more 0 than l's, 3) S(qQ , x) = q 2 if and only if x is proper and contains one more 1 than 0's, 4) 3(q0 , x) = q 3 if and only if x is not proper.
Observe that the induction hypothesis is stronger than the desired theorem. Conditions (2), (3), and (4) are added to allow the induction to go through.
We prove the "if" portions of (1) through (4) first. The basis of the induction, |x | =0, follows since the empty string has an equal number of 0's and l's and S(qQy c) = q0 .
Assume the induction hypothesis is true for all x, | x | < n, n > 1. Consider a string y of length n, such that y is proper and has an equal number of 0's and l's. First consider the case that y ends in 0. Then y = xO, where x is proper and has one more 1 than 0's. Thus <%o> x)= q 2 . Hence <5(<7o, v) = S(q0j xO) = % 2 , 0) = q0 .
The case where y ends in a 1 is handled similarly.
Next consider a string y, \y \ = n such that y is proper and has one more 0 than 1. If y = xO, then x has two more 0's than l's, contradicting the fact that y is proper. Thus y = xl, where x is proper and has an equal number of 0's and l's. By the induction hypothesis, d(q0 , x) = q0 ', hence d(q0 , y) = q x .
The situation where y is proper and has one more 1 than 0, and the situation where y is not proper are treated similarly.
We must now show that strings reaching each state have the interpretations given in ( 1 ) through (4). Suppose that S(q0 , y) = qQ and \y \ > 1. If y = xO, then 3(q0y x) = q 2 , since q 2 is the only state with a 0-transition to state q0 . Thus by the induction hypothesis x is proper and has one more 1 than 0. Thus y is proper and has an equal number of 0's and l's. The case where y ends in a 1 is similar, as are the cases S(q0 , y) = q u q 2 , or q 3 .
KTUNOTES.IN Downloaded from Ktunotes.in 54 FINITE AUTOMATA AND REGULAR EXPRESSIONS BIBLIOGRAPHIC NOTES The original formal study of finite state systems (neural nets similar to that appearing in Exercise 2.2) is by McCulloch and Pitts . Kleene considered regular expres-sions and modeled the neural nets of McCulloch and Pitts by finite automata, proving the equivalence of the two concepts. Similar models were considered about that time by Huff-man , Moore , and Mealy , the latter two being the sources for the terms "Moore machine" and "Mealy machine." Nondeterministic finite automata were in-troduced by Rabin and Scott , who proved their equivalence to deterministic autom-ata. The notion of a two-way finite automaton and its equivalence to the one-way variety was the independent work of Rabin and Scott and Shepherdson .
The proof of the equivalence of regular expressions and finite automata as presented here (via NFA's with ^-transitions) is patterned after McNaughton and Yamada [I960].
Brzozowski [1962, 1964] developed the theory of regular expressions. The fact that the unique solution to X = rX + s (Exercise 2.22) is rs if L(r) does not contain e is known as Arden's lemma. Floyd applies the idea of nondeterminism to programs.
Salomaa gives axiomatizations of regular expressions.
Applications of finite automata to switching circuit design can be found in Kohavi and Friedman . The use of the theory to design lexical analyzers is treated by Johnson et ai and Lesk . Other uses of finite automata theory to design text editors and other text processing programs are discussed in Thompson , Bullen and Millen , Aho and Corasick , Knuth, Morris, and Pratt , and Aho and Ullman .
Some additional works treating finite automata are by Arbib , Conway , Minsky , Moore , and Shannon and McCarthy .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 3 PROPERTIES OF REGULAR SETS There are several questions one can ask concerning regular sets. One important question is: given a language L specified in some manner, is L a regular set? We also might want to know whether the regular sets denoted by different regular expressions are the same, or find the finite automaton with fewest states that denotes the same language as a given FA.
In this chapter we provide tools to deal with questions such as these regarding regular sets. We prove a "pumping lemma" to show that certain languages are nonregular. We provide "closure properties" of regular sets; the fact that lan-guages constructed from regular sets in certain specified ways must also be regular can be used to prove or disprove that certain other languages are regular. The issue of regularity or nonregularity can also be resolved sometimes with the aid of the Myhill-Nerode Theorem of Section 3.4. In addition, we give algorithms to answer a number of other questions about regular expressions and finite automata such as whether a given FA accepts an infinite language.
3.1 THE PUMPING LEMMA FOR REGULAR SETS In this section we prove a basic result, called the pumping lemma, which is a powerful tool for proving certain languages nonregular.
It is also useful in the development of algorithms to answer certain questions concerning finite autom-ata, such as whether the language accepted by a given FA is finite or infinite.
If a language is regular, it is accepted by a DFA M = (Q, X, 6, q0 , F) with some particular number of states, say n. Consider an input of n or more symbols ai^2"' am , m>n, and for /= 1, 2,..., m let d(q0 , a x a 2 '" fl f ) = q x .
It is not 55 KTUNOTES.IN Downloaded from Ktunotes.in 56 PROPERTIES OF REGULAR SETS possible for each of the n + 1 states q0 , q x , ..
.
. , qn to be distinct, since there are only n different states. Thus there are two integers j and k, 0 < j < k < n, such that qi = qk . The path labeled a x a 2 • • ' am in the transition diagram ofM is illustrated in Fig. 3.1. Since j < /c, the string aj+ x • • • ak is of length at least 1, and since k < n, its length is no more than n.
Fig. 3.1 Path in transition diagram of DFA M.
If qm is in F, that is, a x a 2 '" am is in L(M), then a x a 2 "' a} ak + x ak + 2 "' am ls also in L(M), since there is a path from q0 to qm that goes through qj but not around the loop labeled aj+i •• ak . Formally, by Exercise 2.4, (5(<7o, 0i ajak+l -am ) = S(S(q0 , a x ak+1 ••• am ) Similarly, we could go around the loop of Fig.
3.1 more than once— in fact, as many times as we like. Thus, a x •• cij(aj+x • • am is in L(M) for any z > 0. What we have proved is that given any sufficiently long string accepted by an FA, we can find a substring near the beginning of the string that may be "pumped," i.e., repeated as many times as we like, and the resulting string will be accepted by the FA. The formal statement of the pumping lemma follows.
Lemma 3.1 Let L be a regular set. Then there is a constant n such that if z is any word in L, and \z\ > n, we may write z = uvw in such a way that | uv \ < n, \v \ > 1, and for all i > 0, uv'w is in L. Furthermore, n is no greater than the number of states of the smallest FA accepting L.
Proof See the discussion preceding the statement of the lemma. There, z is a x a 2 ••• am , u = a x a 2 - • ap v = aj+l • ak , and w = ak+ t • am .
Note that the pumping lemma states that if a regular set contains a long string z, then it contains an infinite set of strings of the form zzzj'w. The lemma does not state that every sufficiently long string in a regular set is of the form uv lw for some large z.
In fact, (0 + 1) contains arbitrarily long strings in which no substring appears three times consecutively. (The proof is left as an exercise.) KTUNOTES.IN Downloaded from Ktunotes.in 3.1 I THE PUMPING LEMMA FOR REGULAR SETS 57 Applications of the pumping lemma The pumping lemma is extremely useful in proving that certain sets are not regular. The general methodology in its application is an "adversary argument" of the following form.
1) Select the language L you wish to prove nonregular.
2) The "adversary" picks n, the constant mentioned in the pumping lemma. You must be prepared in what follows for any finite integer n to be picked, but once the adversary has picked n, he may not change it.
3) Select a string z in L. Your choice may depend implicitly on the value of n chosen in (2).
4) The adversary breaks z into i/, i\ and w, subject to the constraints that | wi) | < n and | v \ > 1.
5) You achieve a contradiction to the pumping lemma by showing, for any t/, r, and w determined by the adversary, that there exists an i for which ur'vv is not in L.
It may then be concluded that L is not regular. Your selection of /' may depend on m, u, v, and w.
It is interesting to note that your choice in the above "game" corresponds to the universal quantifiers (V, or "for all") and the "adversary's" choices correspond to the existential quantifiers (3. or "there exists") in the formal statement of the pumping lemma: (VL)(3//)(Vz)[z in L and |z| > n implies that (3i/, r, n)(z = t/rvv, \ur\ < //, |r| > 1 and (V/)(i/r'\v is in L))].
Example 3.1 The set L = {0' 2 1 i is an integer, i> 1], which consists of all strings of O's whose length is a perfect square, is not regular. Assume L is regular and let n be the integer in the pumping lemma. Let z = 0" 2 . By the pumping lemma, 0" 2 may be written as i/rw, where 1 < |r| < n and uv'w is in L for all /'. In particular, let i ~ 2. However, n 2 < \uv 2 w\ < n 2 -f n < (n -h l) 2 . That is, the length of uv 2 \v lies properly between n 2 and (n -f l) 2 , and is thus not a perfect square. Thus ur 2 \v is not in L, a contradiction. We conclude that L is not regular.
Example 3.2 Let L be the set of strings of CVs and l's. beginning with a 1, whose value treated as a binary number is a prime. We shall make use of the pumping lemma to prove that L is not regular. We need two results from number theory.
The first is that the number of primes is infinite and that there are therefore arbitrarily large primes. The second, due to Fermat, is that 2 P ~ 1 — 1 is divisible by P for any prime p > 2.
Stated another way, 2 P ~ 1 = 1 mod p (see Hardy and Wright ).
KTUNOTES.IN Downloaded from Ktunotes.in 58 PROPERTIES OF REGULAR SETS Suppose L were regular, and let n be the integer in the pumping lemma. Let z be the binary representation of a prime p such that p > 2". Such a prime exists since there are infinitely many primes. By the pumping lemma we may write z = uvw, where | v \ > 1 and uv lw is the binary representation of a prime for all i.
Let nu , nv , and nw be the values of u, v, and w treated as binary numbers. If u or w are £, then nu or nWJ respectively, is 0. Choose i = p. Then uvpw is the binary representation of a prime q. The numerical value of q is „m2M + pM + „y2H(i + 2M + + 2 ('- 1)M) 4- nw .
By Fermat's theorem, 2 (p " 1] = 1 mod p. If we raise both sides to the power 1 1? |, we get 2(P_1)M = 1 mod p. Thus 2 p\v\ = 2(p- DM2M = 2 |y| mod p.
Let 5 = 1 + 2H 4-• • • + 2 {p ~ Then (2 |y| -l)s = 2 pM -1, which is 2H -1 mod p. Thus (2 |y| -l)(s -1) is divisible by p. But 1 < \v\ <n, so 2 < 2 |y| < 2 M < p. Therefore p cannot divide 2 |y| -1, so it divides s — 1. That is, 5 = 1 mod p. But <7 = «u2H + pM + „ i;2Hs + w>v , so q = MM2' w l + ' y l + «y2»w » + nw mod p.
(3.1) But the right-hand side of (3.1) is the numerical value of p. Thus q = p mod p, which is to say q is divisible by p. Since q > p > 1, q cannot be prime. But by the pumping lemma, the binary representation of q is in L, a contradiction. We conclude that L is not regular.
3.2 CLOSURE PROPERTIES OF REGULAR SETS There are many operations on languages that preserve regular sets, in the sense that the operations applied to regular sets result in regular sets. For example, the union of two regular sets is a regular set, since if r x and r2 are regular expressions denoting regular sets ^ and L2 , then r t + r 2 denotes L Y u L2 , so L x u L2 is also regular. Similarly, the concatenation of regular sets is a regular set and the Kleene closure of a regular set is regular.
If a class of languages is closed under a particular operation, we call that fact a closure property of the class of languages. We are particularly interested in effective closure properties where, given descriptors for languages in the class, there is an algorithm to construct a representation for the language that results by applying the operation to these languages. For example, we just gave an algorithm to KTUNOTES.IN Downloaded from Ktunotes.in 3.2 | CLOSURE PROPERTIES OF REGULAR SETS 59 construct a regular expression for the union of two languages denoted by regular expressions, so the class of regular sets is effectively closed under union. Closure properties given in this book are effective unless otherwise stated.
It should be observed that the equivalences shown in Chapter 2 between the various models of finite automata and regular expressions were effective equiv-alences, in the sense that algorithms were given to translate from one representa-tion to another. Thus in proving effective closure properties we may choose the representation that suits us best, usually regular expressions or deterministic finite automata. We now consider a sequence of closure properties of regular sets; additional closure properties are given in the exercises.
Theorem 3.1 The regular sets are closed under union, concatenation, and Kleene closure.
Proof Immediate from the definition of regular expressions.
Boolean operations Theorem 3.2 The class of regular sets is closed under complementation. That is, if L is a regular set and L c X, then Z — L is a regular set.
Proof Let L be L(M) for DFA M = (Q, Z lf 3, q0t F) and let L c Z First, we may assume Z 1 = Z, for if there are symbols in Z { not in Z, we may delete all transitions of M on symbols not in Z. The fact that L ^ Z assures us that we shall not thereby change the language of M. If there are symbols in Z not in Z 1? then none of these symbols appear in words of L We may therefore introduce a "dead state" d into M with S(d, a) = d for all a in Z and S(q, a) = d for all q in Q and a in Z-Zp Now, to accept Z — L, complement the final states of M. That is, let M' = (Q, <5, <7o> Q ~ F)- Then M' accepts a word w if and only if d(q0 , w) is in Q - F, that is, w is in Z — L. Note that it is essential to the proof that M is deterministic and without e moves.
Theorem 3.3 The regular sets are closed under intersection.
Proof L x n L2 = L, u L 2 , where the overbar denotes complementation with respect to an alphabet including the alphabets of L t and L2 . Closure under inter-section then follows from closure under union and complementation.
It is worth noting that a direct construction of a DFA for the intersection of two regular sets exists. The construction involves taking the Cartesian product of states, and we sketch the construction as follows.
Let Mj = (Q lJ Y. 9 8 l9 qu F 1 )andM 2 = (6 2 » Z,5 2j q 2 ,F 2 ) be two deterministic finite automata. Let M = (Qt x Q 29 Z, 5, [q l9 q2\ F, x F2 ), KTUNOTES.IN Downloaded from Ktunotes.in 60 PROPERTIES OF REGULAR SETS where for all p x in Q l9 p2 in Q 2 , and a in E, <5([Pi, p 2 ], a) = [ 4 <5 2(P2 » a)].
It is easily shown that £(M) = n £(M 2 ).
Substitutions and homomorphisms The class of regular sets has the interesting property that it is closed under substi-tution in the following sense. For each symbol a in the alphabet of some regular set R, let Ra be a particular regular set. Suppose that we replace each word a i a 2 "' an m R °y tne set °f words of the form Wj w 2 ••• w„, where w f is an arbitrary word in jRa .. Then the result is always a regular set. More formally, a substitution/ is a mapping of an alphabet S onto subsets of A, for some alphabet A. Thus/ associates a language with each symbol of S. The mapping/is extended to strings as follows: 1) /(<) = <; 2) f(xa)=f(x)f(a).
The mapping/ is extended to languages by defining f(L)= 1J /()• x in L Example 3.3 Let/(0) = a and/(l) = b. That is,/(0) is the language {a} and/(l) is the language of all strings of Vs. Then /(010) is the regular set aba. If L is the language 0(0 + 1)1, then/(L) is a(a + b)(b) = ab.
Theorem 3.4 The class of regular sets is closed under substitution.
Proof Let R c X be a regular set and for each a in S let Ra c A be a regular set. Let /: Z -> A be the substitution defined by f(a) = K a . Select regular expres-sions denoting R and each Ra . Replace each occurrence of the symbol a in the regular expression for R by the regular expression for Ra . To prove that the resulting regular expression denotes/(K), observe that the substitution of a union, product, or closure is the union, product, or closure of the substitution. [Thus, for example, f(L { u L2)=/(L 1 ) vj /(L2 ).] A simple induction on the number of operators in the regular expression completes the proof.
Note that in Example 3.3 we computed f(L) by taking L's regular expression 0(1 + 0)1 and substituting a for 0 and b for 1. The fact that the resulting regular expression is equivalent to the simpler regular expression ab is a coincidence.
A type of substitution that is of special interest is the homomorphism. A homomorphism h is a substitution such that h(a) contains a single string for each a.
KTUNOTES.IN Downloaded from Ktunotes.in 3.2 | CLOSURE PROPERTIES OF REGULAR SETS 61 We generally take h(a) to be the string itself, rather than the set containing that string. It is useful to define the inverse homomorphic image of a language L to be h~ l (L) = {x\h(x) is inL}.
We also use, for string w; h~ 1 (w) = {x\h(x) = w}.
Example 3.4 Let h(0) = aa and h() = aba. Then /i(010) = aaabaaa.
If Lj is (01), then ^(Lj) is (aaaba). Let L2 = (ab + ba)a. Then h~ 1 (L2 ) consists only of the string 1. To see this, observe that a string in L2 that begins with b cannot be h(x) for any string x of O's and l's, since h(0) and /i(l)each begin with an a. Thus if h~ 1 (w) is nonempty and w is in L2 , then w begins with a. Now either w = a, in which case h~ l (w) is surely empty, or w is abW for some w' in (ab + ba)a. We conclude that every word in h~ 1 (w) begins with a 1, and since h() = aba, W must begin with a. If w' = a, we have w = aba and /z _1 (w) = {1}. However, if W ± a, then W = abW and hence w = ababW . But no string x in (0 + 1) has h(x) begin-ning abab. Consequently we conclude that h~ 1 (w) is empty in this case. Thus the only string in L2 which has an inverse image under h is aba, and therefore r'(L2 ) = {i}.
Observe that h(h 1 (L2 )) = {aba} j= L2 . On the other hand, it is easily shown that h(h~ l (L)) c L and h~ l (h(L)) 3 L for any language L.
Theorem 3.5 The class of regular sets is closed under homomorphisms and in-verse homomorphisms.
Proof Closure under homomorphisms follows immediately from closure under substitution, since every homomorphism is a substitution, in which h(a) has one member.
To show closure under inverse homomorphism, let M = (Q, Z, 3, q0 , F) be a DFA accepting L, and let h be a homomorphism from A to Z. We construct a DFA M' that accepts h~ 1 (L) by reading symbol a in A and simulating M on h(a).
Formally, let M' = (Q, A, 3\ q0 , F) and define d'(q, a), for q in g and a in A to be <5(g, /i(a)). Note that h(a) may be a long string, or £, but 3 is defined on all strings by extension.
It is easy to show by induction on |x| that 3'{q0 , x) = 3(q0 , h(x)).
Therefore M' accepts x if and only if M accepts h(x).
That is, L(M') = h-l {HM)).
" Example 3.5 The importance of homomorphisms and inverse homomorphisms comes in simplifying proofs. We know for example that {0"P | n > 1} is not regular.
Intuitively, {a nba n \n> 1} is not regular for the same reasons. That is, if we had an FA M accepting {a"ba n \n> 1}, we could accept {0T \n> 1} by simulating M on input a for each 0. When the first 1 is seen, simulate M on ba and thereafter simulate M on a for each 1 seen. However, to be rigorous it is necessary to KTUNOTES.IN Downloaded from Ktunotes.in 62 PROPERTIES OF REGULAR SETS formally prove that {a"ba n \ n > 1} is not regular. This is done by showing that {a nba n | n > 1} can be converted to {CI" | « > 1} by use of operations that preserve regularity. Thus {a"ba" \ > 1} cannot be regular.
Let /ij and h 2 be the homomorphisms h l (a) = a, h 2 (a) h 1 (b) = bai h 2 (b) h t (c) = a, h 2 (c) Then h 2 (hi l ({a nban \ n > 1}) n abc) = {0 n l n | n > 1}.
(3.2) That is, 1 ({a nban \ n > 1}) consists of all strings in (a + c)b(a 4- c) such that the number of symbols preceding the b is one greater than the number of symbols following the b. Thus h~ x {a nba n \ n > 1}) n abc = {a"bc"-1 1 n > 1}.
Line (3.2) then follows immediately by applying homomorphism /i 2 .
If {a"ba n \n > 1} were regular, then since homomorphisms, inverse homomor-phisms, and intersection with a regular set all preserve the property of being regular, it would follow that {0 n r|« > 1} is regular, a contradiction.
Quotients of languages Now let us turn to the last closure property of regular sets to be proved in this section. A number of additional closure properties are given in the exercises.
Define the quotient of languages L x and L2 , written L x /L2 , to be {x | there exists y in L2 such that xy is in L x ).
Example 3.6 Let Lj be 010 and L2 be 101. Then L x /L2 is empty. Since every y in L2 has two l's and every string xy which is in L Y can have only one 1, there is no x such that xy is in Lj and y is in L2 .
Let L3 be 01. Then L x /L3 is 0, since for any x in 0 we may choose y = 1.
Clearly xy is in L x = 010 and y is in L3 = 01. Since words in L x and L3 each have one 1, it is not possible that words not in 0 are in L { /L3 . As another example, L2 /L3 = 10, since for each x in 10 we may again choose y = 1 from L3 and xy will be in L2 = 101. If xy is in L2 and y is in L3 , then evidently, x is in 10.
Theorem 3.6 The class of regular sets is closed under quotient with arbitrary sets.f t In this theorem the closure is not effective.
= 0, = 1, = 1.
KTUNOTES.IN Downloaded from Ktunotes.in 3.3 | DECISION ALGORITHMS FOR REGULAR SETS 63 Proof Let M = (Q, Z, S, q0 , F) be a finite automaton accepting some regular set R, and let L be an arbitrary language. The quotient R/L is accepted by a finite automaton M' = (Q, Z, S, q0 , F), which behaves likeM except that the final states of M' are all states q ofM such that there exists y in L for which d(q, y) is in F.
Then S(q0 , x) is in F' if and only if there exists y such that (5(g0 > xy) is in F. Thus M' accepts #/L.
One should observe that the construction in Theorem 3.6 is different from all other constructions in this chapter in that it is not effective. Since L is an arbitrary set, there may be no algorithm to determine whether there exists y in L such that S(q, y) is in F. Even if we restrict L to some finitely representable class, we still may not have an effective construction unless there is an algorithm to test for the existence of such a y. In effect we are saying that for any L, there is surely some F' such that M with F' as the set of final states accepts R/L However, we may not be able to tell which subset of Q should be chosen as F'. In the next section we shall see that if L is a regular set, we can determine F, so the regular sets are effectively closed under quotient with a regular set.
3.3 DECISION ALGORITHMS FOR REGULAR SETS It is important to have algorithms to answer various questions concerning regular sets. The types of questions we are concerned with include: is a given language empty, finite, or infinite? Is one regular set equivalent to another? and so on.
Before we can establish the existence of algorithms for answering such questions we must decide on a representation. For our purposes we shall assume regular sets are represented by finite automata. We could just as well have assumed that regular sets were represented by regular expressions or some other notation, since there exist mechanical translations from these notations into finite automata.
However, one can imagine representations for which no such translation algo-rithm exists, and for such representations there may be no algorithm to determine whether or not a particular language is empty.
The reader at this stage may feel that it is obvious that we can determine whether a regular set is empty. We shall see in Chapter 8, however, that for many interesting classes of languages the question cannot be answered.
Emptiness, finiteness, and infiniteness Algorithms to determine whether a regular set is empty, finite, or infinite may be based on the following theorem. We shall discuss efficient algorithms after presenting the theorem.
Theorem 3.7 The set of sentences accepted by a finite automaton M with n states is: 1) nonempty if and only if the finite automaton accepts a sentence of length less than n.
KTUNOTES.IN Downloaded from Ktunotes.in 64 PROPERTIES OF REGULAR SETS 2) infinite if and only if the automaton accepts some sentence of length <f, where n < £ < In.
Thus there is an algorithm to determine whether a finite automaton accepts zero, a finite number, or an infinite number of sentences.
Proof 1) The "if" portion is obvious. Suppose M accepts a nonempty set. Let w be a word as short as any other word accepted. By the pumping lemma, | w | < n y for if w were the shortest and | w | > n9 then w = uvyf and uy is a shorter word in the language.
2) If w is in L(M) and n < |w| < In, then by the pumping lemma, L(M) is infinite. That is, w = w x w 2 w3 , and for all i, w x w2 w 3 is in L Conversely if L(M) is infinite, then there exists w in L(M), where | w | > n. If | w | < 2n, we are done. If no word is of length between n and In — 1, let w be of length at least 2n, but as short as any word in L(M) whose length is greater than or equal to In. Again by the pumping lemma, we can write w = w t w2 w 3 with 1 < | w 2 | < n and Wj w 3 in L(M). Either w was not a shortest word of length 2n or more, or | w, w 3 | is between n and 2n — 1, a contradiction in either case.
In part (1), the algorithm to decide whether L(M) is empty is: "See if any word of length up to n is in L(M )." Clearly there is such a procedure that is guaranteed to halt. In part (2), the algorithm to decide whether L(M) is infinite is: "See if any word of length between n and In — 1 is in L(M)." Again, clearly there is such a procedure that is guaranteed to halt.
It should be appreciated that the algorithms suggested in Theorem 3.7 are highly inefficient. However, one can easily test whether a DFA accepts the empty set by taking its transition diagram and deleting all states that are not reachable on any input from the start state. If one or more final states remain, the language is nonempty. Then without changing the language accepted, we may delete all states that are not final and from which one cannot reach a final state. The DFA accepts an infinite language if and only if the resulting transition diagram has a cycle. The same method works for NFA's, but we must check that there is a cycle labeled by something besides c.
Equivalence Next we show that there is an algorithm to determine if two finite automata accept the same set.
Theorem 3.8 There is an algorithm to determine if two finite automata are equivalent (i.e., if they accept the same language).
Proof Let M x and M 2 be FA accepting Lx and L2 , respectively. By Theorems 3.1, 3.2, and 3.3, (L x n L 2 ) u (Lj n L 2 ) is accepted by some finite automation, M 3 . It KTUNOTES.IN Downloaded from Ktunotes.in 3.4 | THE MYHILL-N ERODE THEOREM 65 is easy to see that M 3 accepts a word if and only if L x L2 . Hence, by Theorem 3.7, there is an algorithm to determine if Lt = L2 .
3.4 THE MYHILL-NERODE THEOREM AND MINIMIZATION OF FINITE AUTOMATA Recall from Section 1.5 our discussion of equivalence relations and equivalence classes. We may associate with an arbitrary language L a natural equivalence relation RL ; namely, xRL y if and only if for each z, either both or neither of xz and yz is in L. In the worst case, each string is in an equivalence class by itself, but there may be fewer classes. In particular, the index (number of equivalence classes) is always finite if L is a regular set.
There is also a natural equivalence relation on strings associated with a finite automaton. Let M = (Q, £, S, q0i F) be a DFA. For x and y in Z let xRM y if and only if S(q0 , x) = S(q0 , y). The relation RM is reflexive, symmetric, and transitive, since " = " has these properties, and thus RM is an equivalence relation. RM divides the set X into equivalence classes, one for each state that is reachable from q0 . In addition, if xRM y, then xzRM yz for all z in £, since by Exercise 2.4, 6(q0 , xz) = S(S(q0y x), z) = S(S(q0y y), z) = S(qQy yz).
An equivalence relation R such that xRy implies xzRyz is said to be right invariant (with respect to concatenation). We see that every finite automaton induces a right invariant equivalence relation, defined as RM was defined, on its set of input strings. This result is formalized in the following theorem.
Theorem 3.9 (The My hill-N erode theorem).
The following three statements are equivalent: 1) The set L ^ is accepted by some finite automaton.
2) L is the union of some of the equivalence classes of a right invariant equiv-alence relation of finite index.
3) Let equivalence relation R L be defined by: xRL y if and only if for all z in X, xz is in L exactly when yz is in L. Then R L is of finite index.
Proof (1) - (2) Assume that L is accepted by some DFA M = (Q y £, (5, q0y F). Let RM be the equivalence relation xRM y if and only if S(qQy x) = S(qQy y). RM is right invariant since, for any z, if d(q0y x) = S(qQy y), then S(qQy xz) = S(qQy yz). The index of RM is finite, since the index is at most the number of states in Q. Further-more, L is the union of those equivalence classes that include a string x such that H%y ) is in Fy that is, the equivalence classes corresponding to final states.
(2) - (3) We show that any equivalence relation E satisfying (2) is a refinement of RL ; that is, every equivalence class of E is entirely contained in some equivalence class of R L . Thus the index of R L cannot be greater than the index of E and so is KTUNOTES.IN Downloaded from Ktunotes.in 66 PROPERTIES OF REGULAR SETS finite. Assume that xEy. Then since E is right invariant, for each z in X, xzEyz, and thus yz is in L if and only if xz is in L Thus xRL y, and hence the equivalence class of x in £ is contained in the equivalence class of x in RL . We conclude that each equivalence class of E is contained within some equivalence class of R L .
(3) -> (1) We must first show that RL is right invariant. Suppose xRL y, and let w be in E. We must prove that xwRL yw; that is, for any z, xwz is in L exactly when ywz is in L But since x#L y, we know by definition of RL that for any v, xv is in L exactly when yv is in L. Let v = wz to prove that KL is right invariant.
Now let Q be the finite set of equivalence classes of R L and [x] the element of Q' containing x. Define S'([x], a) = \xa\ The definition is consistent, since RL is right invariant. Had we chosen y instead of x from the equivalence class [x], we would have obtained (5'([x], a) = [ya]. But xR L y, so xz is in L exactly when yz is in L In particular, if z = az\ xaz' is in L exactly when yaz' is in L, so xaRL ya, and [xa] = [ya].
Let ^ = [£] and let F = {[x] | x is in L}. The finite automaton M' = (Q\ Z, <5', accepts L, since d'(q' 0> x) = [x], and thus x is in L(M') if and only if [x] is in F.
Example 3.7 Let L be the language 010. L is accepted by the DFA M of Fig.
3.2. Consider the relation RM defined by M. As all states are reachable from the start state, RM has six equivalence classes, which are Ca = (00), Cd = (00)01, Cb = (00)0, Ce = 0100, Cc = (00)1, Cf = 0101(0 4- 1) L is the union of three of these classes, Cc , Cd , and C e .
The relation R L for L has xi? L y if and only if either i) x and y each have no Ts, o Fig. 3.2 DFA M accepting L.
KTUNOTES.IN Downloaded from Ktunotes.in 3.4 | THE MYHILL-N ERODE THEOREM 67 Fig. 3.3 Diagram showing RM is a refinement of RL .
ii) x and y each have one 1, or iii) x and y each have more than one 1.
For example, if x = 010 and y = 1000, then xz is in L if and only if z is in 0. But yz is in L under exactly the same conditions. As another example, if x = 01 and y = 00, then we might choose z = 0 to show that xRL y is false. That is, xz = 010 is in L, but yz = 000 is not.
We may denote the three equivalence classes of RL by C t = 0, C 2 = 010, and C 3 = 0101(0 -f 1). L is the language consisting of only one of these classes, C2 . The relationship of Cfl , C 7 to C l9 C2 , and C3 is illustrated in Fig.
3.3.
For example Ca u Cb = (00) + (00)0 = 0 = C v From K L we may construct a DFA as follows. Pick representatives for C i9 C 2 , and C 3 , say e> 1, and 1 1. Then let M' be the DFA shown in Fig. 3.4. For example, 0) = , since if w is any string in (note is C x \ say O'lO 7 , then wO is 010^ \ which is also in C, = 010.
Fig. 3.4 The DFA M\ Minimizing finite automata The Myhill-Nerode theorem has, among other consequences, the implication that there is an essentially unique minimum state DFA for every regular set.
Theorem 3.10 The minimum state automaton accepting a set L is unique up to an isomorphism (i.e., a renaming of the states) and is given by M' in the proof of Theorem 3.9.
KTUNOTES.IN Downloaded from Ktunotes.in 68 PROPERTIES OF REGULAR SE1 U Proof In the proof of Theorem 3.9 we saw that any DFA M = (Q, Z, d, q0 , F) accepting L defines an equivalence relation that is a refinement of R L . Thus the number of states of M is greater than or equal to the number of states of M' of Theorem 3.9. If equality holds, then each of the states ofM can be identified with one of the states of M'. That is, let q be a state of M. There must be some x in Z, such that S(q0 , x) = q, otherwise q could be removed from Q, and a smaller automaton found. Identify q with the state d'(q' 09 x), of M'. This identification will be consistent. If S(qQy x) = S(q0 , y) = q, then, by the proof of Theorem 3.9, x and y are in the same equivalence class of RL . Thus 5'(q' 0y x) = d'(q' 0 , y).
A minimization algorithm There is a simple method for finding the minimum state DFAM of Theorems 3.9 and 3.10 equivalent to a given DFA M = (Q, Z, S, q0 , F). Let = be the equivalence relation on the states of M such that p = q if and only if for each input string x, S(p, x) is an accepting state if and only if S(q y x) is an accepting state. Observe that there is an isomorphism between those equivalence classes of = that contain a state reachable from q0 by some input string and the states of the minimum state FA M'. Thus the states of M' may be identified with these classes.
Rather than give a formal algorithm for computing the equivalence classes of = we first work through an example. First some terminology is needed. If p = q, we say p is equivalent to q. We say that p is distinguishable from q if there exists an x such that S(p> x) is in F and S(q, x) is not, or vice versa.
Example 3.8 Let M be the finite automaton of Fig.
3.5. In Fig.
3.6 we have constructed a table with an entry for each pair of states. An X is placed in the table each time we discover a pair of states that cannot be equivalent. Initially an X is placed in each entry corresponding to one final state and one nonfinal state. In our example, we place an X in the entries (a, c), (fr, c), (c, d), (c, e), (c,/), (c, g\ and (c h).
Fig. 3.5 Finite automaton.
KTUNOTES.IN Downloaded from Ktunotes.in 3.4 | THE MYHILL-NERODE THEOREM 69 X X X X X X X X X X X X X X X X X X X X X X X X X a b c (1 e f g Fig. 3.6 Calculation of equivalent states.
Next for each pair of states p and q that are not already known to be distin-guishable we consider the pairs of states r = S(py a) and s = S(qy a) for each input symbol a. If states r and s have been shown to be distinguishable by some string x, then p and q are distinguishable by string ax. Thus if the entry (r, s) in the table has an X, an X is also placed at the entry (p, q). If the entry (r, s) does not yet have an X, then the pair (p, q) is placed on a list associated with the (r, s)-entry. At some future time, if the (r, 5) entry receives an X y then each pair on the list associated with the (r, s)-entry also receives an X.
Continuing with the example, we place an X in the entry (a, b), since the entry (S(b, 1), d(a, 1)) = (cy f) already has an X. Similarly, the (a, d)-entry receives an X since the entry (S(a, 0), 3(d, 0)) = (b, c) has an X. Consideration of the (a, e)-entry on input 0 results in the pair (a, e) being placed on the list associated with (6, h).
Observe that on input 1, both a and e go to the same state/ and hence no string starting with a 1 can distinguish a from e. Because of the 0-input, the pair (a, g) is placed on the list associated with (b, g). When the (fr, gr)-entry is considered, it receives an X on account of a 1 -input, and hence the pair (a, g) receives an X since it was on the list for (b, g). The string 01 distinguishes a from g.
On completion of the table in Fig. 3.6, we conclude that the equivalent states are a = e, b = h, and d = f. The minimum-state finite automaton is given in Fig.
3.7.
The formal algorithm for marking pairs of inequivalent states is shown in Fig.
3.8. Lemma 3.2 proves that the method outlined does indeed mark all pairs of inequivalent states.
Lemma 3.2 Let M = (Q, I, d y q0y F) be a DFA. Then p is distinguishable from q if and only if the entry corresponding to the pair (p, q) is marked in the above procedure.
KTUNOTES.IN Downloaded from Ktunotes.in 70 PROPERTIES OF REGULAR SETS Start Fig. 3.7 Minimum state finite automaton.
begin 1) for p in F and q in Q — F do mark (p, q); 2) for each pair of distinct states (p, q) in F x F or (Q - F) x (Q — F) do 3) if for some input symbol a, (S(p> a), S(q> a)) is marked then begin 4) mark (p, q)\ 5) recursively mark all unmarked pairs on the list for (p, q) and on the lists of other pairs that are marked at this step.
end else / no pair (<5(p, a), 5(q, a)) is marked / 6) for all input symbols a do 7) put (p, 4) on the list for (<5(p, a\ S(q, a)) unless <5(p, a) = d(q, a) end Fig. 3.8 Algorithm for marking pairs of inequivalent states.
Proof Assume p is distinguishable from qy and let x be a shortest string distin-guishing p from q. We prove by induction on the length of x that the entry corresponding to the pair (/?, q) is marked. If x = c then exactly one of p and q is a final state and hence the entry is marked in line (1). Assume that the hypothesis is true for |x| < i > 1, and let |x| = i. Write x = ay and let t = S(p, a) and u = S(qy a). Now y distinguishes t from u and \y\ = 1 — 1. Thus by the induction hypothesis the entry corresponding to the pair (r, u) eventually is marked. If this event occurs after the pair (p, q) has been considered, then either the (p, q) entry has already been marked when (t, u) is considered, or the pair (p, q) is on the list associated with (r, u), in which case it is marked at line (5). If (p, q) is con-sidered after (t, u) then (p, q) is marked at the time it is considered. In any event the entry (p, q) is marked. A similar induction on the number of pairs marked shows that if the entry (p, q) is marked then p and q are distinguishable.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 71 The algorithm of Fig. 3.8 is more efficient than the obvious marking algo-rithm, although it is not the most efficient possible. Let E have k symbols and Q have n states. Line 1 takes 0(n 2 ) steps.t The loop of lines 2 through 7 is executed 0(n 2 ) times, at most once for each pair of states. The total time spent on lines 2 through 4, 6, and 7 is 0(kn2 ). The time spent on line 5 is the sum of the length of all lists. But each pair (r, s) is put on at most k lists, at line 7. Thus the time spent on line 5 is 0(kn 2 ), so the total time is also 0(kn 2 ).
Theorem 3.11 The DFA constructed by the algorithm of Fig. 3.8, with inacces-sible states removed, is the minimum state DFA for its language.
Proof Let M = (Q, X, S, q0 , F) be the DFA to which the algorithm is applied and M' = (Q\ I, S\ [q0], F) be the DFA constructed. That is, Q = {[q] | q is accessible from q0 }, " = {Mk is inF) and S'([q], a) -[%«)].
It is easy to show that S' is consistently defined, since if q = p, then S(q, a) = <5(p, a). That is, if d(qy a) is distinguished from (5(p, a) by x, then ax distinguishes q from p. It is also easy to show that <5'([g0 ]> w) = [$(q0 > vv)] by induction on |w|.
Thus L(M') = L(M).
Now we must show that M' has no more states than R L has equivalence classes, where L = L(M). Suppose it did; then there are two accessible states q and p in Q such that [q] [p], yet there are x and y such that d(q0 , x) = q, d(q0 , y) = p, and xRL y. We claim that p = q, for if not, then some w in E distinguishes p from But then xwRL yw is false, for we may let z = e and observe that exactly one of xwz and ywz is in L. But since RL is right invariant, xwRL yw is true. Hence g and p do not exist, and M' has no more states than the index of RL . Thus M' is the minimum state DFA for L EXERCISES 3.1 Which of the following languages are regular sets? Prove your answer.
a) {0 2n \n> 1} b) {0m l"0m + n |m > 1 and n > 1} c) {0 n | n is a prime} d) the set of all strings that do not have three consecutive O's.
e) the set of all strings with an equal number of O's and l's.
f) {x\x in (0 + 1), and x = xR } xR is x written backward; for example, (Oil) = 110.
g) {xwxR \x, w in (0 + 1) + } h) {xxRw\x f win (0 + 1) + } t We say that g(n) is 0(/(n)) if there exist constants c and n0 such that g(n) < cf(n) for all n > n0 .
KTUNOTES.IN Downloaded from Ktunotes.in 72 PROPERTIES OF REGULAR SETS 3.2 Prove the following extension of the pumping lemma for regular sets. Let L be a regular set. Then there exists a constant n such that for each z x , z2 > z3y with z t z 2 z 3 in Land I z i | = n,z 2 can be written z 2 = uvw such that | v \ > 1 and for each i > 0, z x uvwz 3 is in L 3.3 Use Exercise 3.2 to prove that {0 l lm2m |i > 1, m > 1} is nonregular.
3.4 Let L be a regular set. Which of the following sets are regular? Justify your answers, a) {a x a 3 a 5 -- a2n -1 1 a x a 2 a 3 aA • • a2n is in L} S b) {a 2 aiaA a z a2n a 2n -i \a x a 2 •• a 2n is in L} c) CYCLE(L) = {x t x 2 \x 2 x x is in L for strings x r and x 2 } d) MAX(L) = {x in L|for no y other than e is xy in L} e) MIN(L) = {x in L|no proper prefix of x is in L} f) INIT(L) = {x|for some y, xy is in L} g) L R = {x | xR is in L} h) {x | xxR is in L] 3.5 Let value(x) be the result when the symbols of x are multiplied from left to right according to the table of Fig. 2.31.
a) Is L = {xy ||x| = \y \ and value(x) = value(y)} regular?
b) Is L = {xyjvalue(x) = value(y)} regular?
Justify your answers.
3.6 Show that {WV\gcd(uj) = 1} is not regular.
3.7 Let L be any subset of 0. Prove that L is regular.
3.8 A set of integers is linear if it is of the form {c + pi \ i = 0, 1, 2, .
. .}. A set is semilinear if it is the finite union of linear sets. Let R ^ 0 be regular. Prove that {/|0' is in R} is semilinear.
3.9 Is the class of regular sets closed under infinite union?
3.10 What is the relationship between the class of regular sets and the least class of languages closed under union, intersection, and complement containing all finite sets?
3.11 Give a finite automaton construction to prove that the class of regular sets is closed under substitution.
3.12 Is the class of regular sets closed under inverse substitution?
3.13 Let h be the homomorphism h(a) = 01, h(b) = 0.
a) Find /r 1 ^), where L x = (10 + 1) b) Find h(L2 \ where L2 = (a + b) c) Find h~ ! (L3 ), where L3 is the set of all strings of 0's and Ts with an equal number of 0's and Ts.
3.14 Show that 2DFA with endmarkers (see Exercise 2.20) accept only regular sets by making use of closure properties developed in this chapter.
3.15 The use of n with regular expressions does not allow representation of new sets.
However it does allow more compact expression. Show that n can shorten a regular expression by an exponential amount. [Hint: What is the regular expression of shortest length describing the set consisting of the one sentence (. .. ((al a x ) 2a 2 ) 2 • ) 2 ^] KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 73 3.16 Let L be a language. Define (L) to be {x | for some y such that |x | = \y |, xy is in L}.
That is, (L) is the first halves of strings in L. Prove for each regular L that (L) is regular.
3.17 If L is regular, is the set of first thirds of strings in L regular? What about the last third? Middle third? Is the set {xz | for some y with | x | = |y| = \z\, xyz is in L} regular?
3.18 Show that if L is regular, so are a) SQRT(L) = {x|for some y with |y| = |x| 2 , xy is in L} b) LOG(L) = {x|for some y with |y| = 2 |JC| , xy is in L} 3.19 A one-pebble 2DFA is a 2DFA with the added capability of marking a tape square by placing a pebble on it. The next state function depends on the present state, the tape symbol scanned, and the presence or absence of a pebble on the tape square scanned. A move consists of a change of state, a direction of head motion, and possibly placing or removing the pebble from the scanned tape cell. The automaton "jams" if it attempts to place a second pebble on the input. Prove that one-pebble 2DFA's accept only regular sets. [Hint: Add two additional tracks to the input that contain tables indicating for each state p, the state q in which the 2DFA will return if it moves left or right from the tape cell in state p, under the assumption that the pebble is not encountered. Observe that the one-pebble 2DFA operating on the augmented tape need never leave its pebble. Then make use of a homomorphic mapping to remove the additional tracks.] 3.20 In converting an NFA to a DFA the number of states may increase substantially.
Give upper and lower bounds on the maximum increase in number of states for an n-state NFA. [Hint: Consider Exercises 2.5(e) and 2.8(c).] 3.21 Give a decision procedure to determine if the set accepted by a DFA is a) the set of all strings over a given alphabet, b) cofinite (a set whose complement is finite).
3.22 Consider a DFA M. Suppose you are told that M has at most n states and you wish to determine the transition diagram of M. Suppose further that the only way you can obtain information concerning M is by supplying an input sequence x and observing the prefixes of x which are accepted.
a) What assumptions must you make concerning the transition diagram ofM in order to be able to determine the transition diagram?
b) Give an algorithm for determining the transition diagram ofM (except for the start state) including the construction of x under your assumptions in part (a).
S 3.23 Give an efficient decision procedure to determine if x is in the language denoted by an extended regular expression (a regular expression with operators u, (concatenation), , n, and that is complement).
3.24 Give an efficient decision procedure for determining if a semi-extended regular ex-pression r (a regular expression with u, •, , n) denotes a nonempty set. [Hint: Space 0( | r | ) and time 0(2 |r| ) are sufficient.] KTUNOTES.IN Downloaded from Ktunotes.in 74 PROPERTIES OF REGULAR SETS 1 o 1 Fig. 3.9 A finite automaton.
3.25 Find the minimum-state finite automaton equivalent to the transition diagram of Fig. 3.9.
3.26 a) What are the equivalence classes of R L in the Myhill-Nerode theorem (Theorem 3.9) forL = {0"r|>7> 1}?
b) Use your answer in (a) to show {0T \n > 1} not regular.
c) Repeat (a) for {x\x has an equal number of O's and l's}.
3.27 R is a congruence relation if xRy implies wxzRwyz for all w and z. Prove that a set is regular if and only if it is the union of some of the congruence classes of a congruence relation of finite index.
3.28 Let M be a finite automaton with n states. Let p and q be distinguishable states ofM and let x be a shortest string distinguishing p and q. How long can the string x be as a function of n?
3.29 In a two-tape FA each state is designated as reading tape 1 or tape 2. A pair of strings (x, y) is accepted if the FA, when presented with strings x and y on its respective tapes, reaches a final state with the tape heads immediately to the right of x and y. Let L be the set of pairs accepted by a two-tape FA M. Give algorithms to answer the following questions.
a) Is L empty?
b) Is L finite?
c) Do there exist L x and L2 such that L = L x x L2 ?
3.30 a) Prove that there exists a constant c> 0 such that the algorithm of Fig. 3.8 requires time greater than cn 1 for infinitely many DFA where n is the number of states and the input alphabet has two symbols.
b) Give an algorithm for minimizing states in a DFA whose execution time is 0( 1 1 \n log n). Here I is the input alphabet. [Hint: Instead of asking for each pair of states (p, q) and each input a if <5(p, a) and S(q, a) are distinguishable, partition the states into final and nonfinal states. Then refine the partition by considering all states whose next state under some input symbol is in one particular block of the partition.
Each time a block is partitioned, refine the partition further by using the smaller sub-block. Use list processing to make the algorithm as efficient as possible.] Solutions to Selected Exercises 3.4(b) L = {a 2 axaA a 3 ••• a 2n a 2„-i \ aia 2 •• a2n is in L} is regular. Let M = (Q, I, (5, qQy F) be a DFA accepting L. We construct a DFA M' that accepts L. M' will process tape symbols in pairs. On seeing the first symbol a in a pair, M' stores a in its finite control. Then KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 75 on seeing the second symbol b, M' behaves like M on the input ba. More formally M' = (Q u Q x Z, Z, 5', 9o , F) where i) (5'(g, a) = [g, a], and ii) <5'(fa, a], b) = fra).
To prove that M' accepts L we show by induction on even i that <5'(g, a 2 aiaA a3 ••• flffl,-!) = (5(<7, a,)-Clearly, for i = 0, d'(q, e) = q = (5(qr, £). Assume the hypothesis is true for all even j < i. By the induction hypothesis, 6'(q, a2 a 1 ••• fl i _ 2 fl l _ 3 ) = 5{q, cna 2 - 0,-2) = p for some p.
Thus <5'(g, a 2 a !
• • a,- a,- _ ! ) = a t a {r _ , ) = (5(p, Therefore a 2 a x aA a^ ••• 0^,-1 is in L(M') if and only if a x a 2 •" a,-is in L(M), and thus L(M') = L.
3.23 One can clearly construct a finite automaton equivalent to R by combining finite automata corresponding to subexpressions of R and then simulating the automaton on x.
We must examine the combining process to see how it affects the size of the resulting automaton. If we work with DFA's then the number of states for a union or intersection grows as the product. However, concatenation and closure may increase the number of states exponentially, as we need to convert DFA's to NFA's and then perform the subset construction.
If we work with NFA's, then the number of states is additive for union, concatenation, and closure and increases as the product for intersection. However, comple-ments require a conversion from an NFA to a DFA and hence an exponential increase in the number of states. Since operators can be nested, the number of states can be exponen-tiated on the order of n times for an expression with n operators, and thus this technique is not in general feasible.
A more efficient method based on a dynamic programming technique (see Aho, Hopcroft, and Ullman ) yields an algorithm whose execution time is polynomial in the length of the input w and the length of the regular expression s. Let n = |w| + \s.
Construct a table which for each subexpression r of s and each substring xu of w gives the answer to the question: Is xi} in L(r), where x0 is the substring of w of length j beginning at position /?The table is of size at most h 3 , since there are at most n subexpressions of s and n(n + l)/2 substrings of w.
Fill in the table starting with entries for small subexpressions (those without operators, that is, a, £, or 0). Then fill in entries for x and r, where r is of one of the forms r x n r 2 , r x + r 2 , r x r 2 , rf, or ir x . We handle only the case r?. We proceed in order of the length of x. To determine tf_x is in rf, given that we already know for each KTUNOTES.IN Downloaded from Ktunotes.in 76 PROPERTIES OF REGULAR SETS proper substring y of x whether y is in r x or in rf, we need only check for each Xi and x 2 such that x = Xix 2 and Xi e, whether x t is in rj and x 2 is in rf. Thus to calculate the table entry for x and r requires time 0( | x \ + | r | ). Hence the time to fill in the entire table is 0(h4 ). To determine if w is in s we need only consult the entry for s and w, noting that w = x lk , where k = | w|.
BIBLIOGRAPHIC NOTES The pumping lemma for regular sets is based on the formulation of Bar-Hillel, Perles, and Shamir . Theorem 3.4, closure under substitution, is also from there. Theorem 3.5, closure under inverse homomorphism, is from Ginsburg and Rose [1963b], and Theorem 3.6, on quotients, is from Ginsburg and Spanier . Theorems 3.7 and 3.8 on decision algorithms are from Moore . Ginsburg and Rose give a number of additional closure properties of regular sets.
Theorem 3.9, which we call the Myhill-Nerode Theorem, is actually due to Nerode . The similar result of Exercise 3.27 on congruence relations is due to Myhill .
The algorithm for minimizing finite automata is due to Huffman and Moore .
Hopcroft gives a more efficient algorithm.
Example 3.2, the unrecognizability of the primes in binary, was proved by Minsky and Papert by another method. Proportional removal operations, such as Exercise 3.16, were first studied in generality by Stearns and Hartmanis . Generalizations such as Exercise 3.18 were considered by Kosaraju and Seiferas , and the question of what functions of the string length may be removed from the front to yield regular sets was solved completely by Seiferas and McNaughton . A solution to Exercise 3.22 was first considered by Hennie . An algorithm for determining equivalence for deterministic two-tape FA is found in Bird .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 4 CONTEXT-FREE GRAMMARS 4.1 MOTIVATION AND INTRODUCTION In this chapter we introduce context-free grammars and the languages they describe—the context-free languages. The context-free languages, like the regular sets, are of great practical importance, notably in defining programming lan-guages, in formalizing the notion of parsing, simplifying translation of program-ming languages, and in other string-processing applications. As an example, context-free grammars are useful for describing arithmetic expressions, with arbi-trary nesting of balanced parentheses, and block structure in programming lan-guages (that is, begin's and end's matched like parentheses). Neither of these aspects of programming languages can be represented by regular expressions.
A context-free grammar is a finite sef>f variables (also called nonterminals or syntactic categories) each of which represents a language. The languages repre-sented by the variables are described recursively in terms of each other and primitive symbols called terminals. The rules relating the variables are called productions. A typical production states that the language associated with a given variable contains strings that are formed by concatenating strings from the lan-guages of certain other variables, possibly along with some terminals.
The original motivation for context-free grammars was the description of natural languages. We may write rules such as (sentence) -> (noun phrase) (verb phrase) (noun phrase) -> (adjective)(noun phrase) (noun phrase) -> (noun) (noun) - boy (adjective) -> little (4.1) 77 KTUNOTES.IN Downloaded from Ktunotes.in 78 CONTEXT-FREE GRAMMARS where the syntactic categories! are denoted by angle brackets and terminals by unbracketed words like "boy" and "little." The meaning of (sentence) -> (noun phrase)(verb phrase) is that one way to form a sentence (a string in the language of the syntactic category (sentence)) is to take a noun phrase and follow it by a verb phrase. The meaning of (noun) -> boy is that the string consisting of the one-terminal symbol "boy" is in the language of the syntactic category (noun). Note that "boy" is a single terminal symbol, not a string of three symbols.
For a number of reasons, context-free grammars are not in general regarded as adequate for the description of natural languages like English. For example, if we extended the productions of (4.1) to encompass all of English, we would be able to derive "rock" as a noun phrase and "runs" as a verb phrase. Thus "rock runs" would be a sentence, which is nonsense. Clearly some semantic information is necessary to rule out meaningless strings that are syntactically correct. More subtle problems arise when attempts are made to associate the meaning of the sentence with its derivation. Nevertheless context-free grammars play an impor-tant role in computer linguistics.
While linguists were studying context-free grammars, computer scientists began to describe programming languages by a notation called Backus-Naur Form (BNF), which is the context-free grammar notation with minor changes in format and some shorthand. This use of context-free grammars has greatly simplified the definition of programming languages and the construction of com-pilers. The reason for this success is undoubtedly due in part to the natural way in which most programming language constructs are described by grammars. For example, consider the set of productions 1) (expression) -» (expression) + (expression) 2) (expression) - (expression) (expression) 3) (expression) -> ((expression)) 4) (expression) -> id (4.2) which defines the arithmetic expressions with operators 4- and and operands represented by the symbol id. Here (expression) is the only variable, and the terminals are +,,(, ), and id. The first two productions say that an expression t Recall that the term "syntactic category" is a synonym for "variable." It is preferred when dealing with natural languages.
KTUNOTES.IN Downloaded from Ktunotes.in 4.2 | CONTEXT-FREE GRAMMARS 79 can be composed of two expressions connected by an addition or multiplication sign. The third production says that an expression may be another expression surrounded by parentheses. The last says a single operand is an expression.
By applying productions repeatedly we can obtain more and more com-plicated expressions. For example, (expression) => (expression) (expression) => ((expression)) (expression) => ((expression)) id => ((expression) 4- (expression)) id => ((expression) + id) id => (id + id) id (4.3) The symbol => denotes the act of deriving, that is, replacing a variable by the right-hand side of a production for that variable. The first line of (4.3) is obtained from the second production. The second line is obtained by replacing the first (expression) in line 1 by the right-hand side of the third production. The remain-ing lines are the results of applying productions (4), (1), (4), and (4). The last line, (id 4- id) id, consists solely of terminal symbols and thus is a word in the lan-guage of (expression).
4.2 CONTEXT-FREE GRAMMARS Now we shall formalize the intuitive notions introduced in the previous section. A context-free grammar (CFG or just grammar) is denoted G = (V, T, P, S), where V and T are finite sets of variables and terminals, respectively. We assume that V and T are disjoint. P is a finite set of productions; each production is of the form A -> a, where A is a variable and a is a string of symbols from (V u T). Finally, S is a special variable called the start symbol.
Example 4.1 Suppose we use £ instead of (expression) for the variable in the grammar (4.2). Then we could formally express this grammar as ({£}, { + , ,(,), id}, P, £), where P consists of £->£ + £ £->££ £-(£) £-id KTUNOTES.IN Downloaded from Ktunotes.in 80 CONTEXT-FREE GRAMMARS In this and the next two chapters we use the following conventions regarding grammars.
1) The capital letters A, B, C, D9 E, and S denote variables; S is the start symbol unless otherwise stated.
2) The lower-case letters a, b, c, d, ey digits, and boldface strings are terminals.
3) The capital letters X, Y, and Z denote symbols that may be either terminals or variables.
4) The lower-case letters u, v, w, x, v, and z denote strings of terminals.
5) The lower-case Greek letters a, p, and y denote strings of variables and terminals.
By adhering to the above conventions, we can deduce the variables, terminals, and the start symbol of a grammar solely by examining the productions. Thus we often present a grammar by simply listing its productions. If A -> olu A -> a 2 , .
.
.
, A->ak are the productions for the variable A of some grammar, then we may express them by the notation i4->a,|a2 ak , where the vertical line is read "or." The entire grammar of Example 4.1 could be written £->£ + £(£ E|(£:) | id Derivations and languages We now formally define the language generated by a grammar G = (V, T, P, ( S).
To do so, we develop notation to represent a derivation.
First we define two relations =g> and %- between strings in (V u T). If A -> P is a production of P and a and y are any strings in (V u T), then aAy ^> a/fy. We say that the production A -+ P is applied to the string ccAy to obtain a/fy or that cuAy directly derives ccfiy in grammar G. Two strings are related by =?> exactly when the second is obtained from the first by one application of some production.
Suppose that a l9 a 2 , aOT are strings in (Ku 7), m > 1, and «1 ^ «2> <2 ^ «3» • • • > am-1 ? «m-Then we say aj ^> am or aj derives am in grammar G. That is, is the reflexive and transitive closure of => (see Section 1.5 for a discussion of closures of relations).
Alternatively, a =?> p if p follows from a by application of zero or more productions of P. Note that a =?> a for each string a. Usually, if it is clear which grammar G is involved, we use => for =» and ^> for If a derives P by exactly i steps, we say The language generated by G [denoted L(G)] is {w | w is in T and S => w}. That KTUNOTES.IN Downloaded from Ktunotes.in 4.2 | CONTEXT-FREE GRAMMARS 81 is, a string is in L(G) if: 1) The string consists solely of terminals.
2) The string can be derived from S.
We call L a context-free language (CFL) if it is L(G) for some CFG G. A string of terminals and variables a is called a sententialform if S ^> a. We define grammars G x and G 2 to be equivalent if L(G t ) = L(G 2 ).
Example 4.2 Consider a grammar G = (F, T, P, 5), where K = {S}, 7 = {a, b} and P = {S->flS6, S-+ab). Here, S is the only variable; a and 6 are terminals.
There are two productions, S -> aSfr and S -> ab. By applying the first production n — 1 times, followed by an application of the second production, we have S => aSb => aaSbb => a3Sb 3 =>•••=> a n ~ 1 S^ n ~ 1 => 1. Each time S -> aS6 is used, the number of S's remains the same. After using the production S-+abwe find that the number of S's in the sentential form decreases by one. Thus, after using S -> ab, no S's remain in the resulting string. Since both productions have an S on the left, the only order in which the productions can be applied is S -> aSb some number of times followed by one application of S - ab. Thus, L(G) = {a nb n \n > 1}.
Example 4.2 was a simple example of a grammar.
It was relatively easy to determine which words were derivable and which were not. In general, it may be exceedingly hard to determine what is generated by the grammar. Here is another, more difficult example.
Example 4.3 Consider G = (V, T, P, S), where V = {Sy A, £}, T = {a, b} y and P consists of the following: aB A -> bAA bA B-+b A-> a B->bS A-> aS B -> aBB The language L(G) is the set of all words in T+ consisting of an equal number of a's and fr's. We shall prove this statement by induction on the length of a word.
Inductive hypothesis For w in T + , 1) S^> vv if and only if w consists of an equal number of a's and Vs.
2) A ^> w if and only if w has one more a than it has fr's.
3) B^>w if and only if w has one more b than it has a\ KTUNOTES.IN Downloaded from Ktunotes.in 82 CONTEXT-FREE GRAMMARS The inductive hypothesis is certainly true if | w | = 1, since A ^> a, B b, and no terminal string oflength one is derivable from S. Also, since all productions but A -> a and B -> b increase the length of a string, no strings of length one other than a and b are derivable from A and 5, respectively. Also, no strings of length one are derivable from S.
Suppose that the inductive hypothesis is true for all w of length k — 1 or less.
We shall show that it is true for | w | = k. First, if S ^> w, then the derivation must begin with either S -> aB or S -> bA. In the first case, w is of the form awu where \w 1 1 = k — 1 and B^> w v By the inductive hypothesis, the number of Vs in w 1 is one more than the number of a's, so w consists of an equal number of a's and Vs. A similar argument prevails if the derivation begins with S -> bA.
We must now prove the "only if" of part (1), that is, if | w | = k and w consists of an equal number of a's and tfs, then S ^ w. Either the first symbol ofw is a or it is b. Assume that w = aw v Now | Wj | = k — 1, and w t has one more b than it has a's. By the inductive hypothesis, B^>Wj. But then S=> aB^>aw t = w. A similar argument prevails if the first symbol of w is b.
Our task is not done. To complete the proof, we must prove parts (2) and (3) of the inductive hypothesis for w oflength k. We do this in a manner similar to our method of prpof for part (1); this part is left to the reader.
4.3 DERIVATION TREES It is useful to display derivations as trees. These pictures, called derivation (or parse) trees, superimpose a structure on the words of a language that is useful in applications such as the compilation of programming languages. The vertices of a derivation tree are labeled with terminal or variable symbols of the grammar or possibly with c. If an interior vertex n is labeled A, and the sons of n are labeled X lt X 2 , • .
• , Xk from the left, then A-+X l X 2 "' Xk must be a production. Figure 4.1 shows the parse tree for derivation (4.3). Note that if we read the leaves, in left-to-right order, we get the last line of (4.3), (id + id) id.
( ) id + I I id id Fig. 4.1 Derivation tree.
KTUNOTES.IN Downloaded from Ktunotes.in 4.3 | DERIVATION TREES 83 More formally, let G = (V, T, P, S) be a CFG. A tree is a derivation (or parse) tree for G if: 1) Every vertex has a tafce/, which is a symbol of Fu Tu {e}.
2) The label of the root is S.
3) If a vertex is interior and has label A, then A must be in V.
4) If n has label A and vertices n l9 n2i are the sons of vertex w, in order from the left, with labels X u X 2t Xk , respectively, then A -^2 "^Jfc must be a production in P.
5) If vertex n has label e, then n is a leaf and is the only son of its father.
Example 4.4 Consider the grammar G = ({S, A), {a, b\ P, S), where P consists of S -> a>4S | a We draw a tree, just this once, with circles instead of points for the vertices.
The vertices will be numbered for reference. The labels will be adjacent to the vertices. See Fig. 4.2.
The interior vertices are 1, 3, 4, 5, and 7. Vertex 1 has label S, and its sons, from the left, have labels a, A, and S. Note that S aAS is a production. Likewise, vertex 3 has label A, and the labels o&its sons are S, b, and A from the left.
A SbA is also a production. Vertices 4 and 5 each have label S. Their only sons each have label a, and S -> a is a production. Lastly, vertex 7 has label A and its sons, from the left, have labels b and a. A^ba is also a production. Thus, the conditions for Fig. 4.2 to be a derivation tree for G have been met.
Fig. 4.2 Example of a derivation tree.
KTUNOTES.IN Downloaded from Ktunotes.in 84 CONTEXT-FREE GRAMMARS We may extend the "from the left" ordering of sons to produce a left-to-right ordering of all the leaves. In fact, for any two vertices, neither of which is an ancestor of the other, one is to the left of the other. Given vertices v l and v 2 , follow the paths from these vertices toward the root until they meet at some vertex w. Let Xj and x 2 be the sons of w on the paths from v x and v 2 , respectively. If i\ is not an ancestor of v 2 , or vice versa, then x 1 j= x2 . Suppose x x is to the left of x 2 in the ordering of the sons of w. Then v l is to the left of v 2 . In the opposite case, v 2 is to the left of vv For example, if v 1 and v 2 are 9 and 1 1 in Fig. 4.2, then w is 3, x 1 = 5, and x 2 = 7. As 5 is to the left of 7, it follows that 9 is to the left of 11.
We shall see that a derivation tree is a natural description of the derivation of a particular sentential form of the grammar G. If we read the labels of the leaves from left to right, we have a sentential form. We call this string the yield of the derivation tree. Later, we shall see that if a is the yield of some derivation tree for grammar G = (V, T, P, S), then S => a, and conversely.
We need one additional concept, that of a subtree. A subtree of a derivation tree is a particular vertex of the tree together with all its descendants, the edges connecting them, and their labels. It looks just like a derivation tree, except that the label of the root may not be the start symbol of the grammar. If variable A labels the root, then we call the subtree an A-tree. Thus "S-tree" is a synonym for "derivation tree" if S is the start symbol.
Example 4.5 Consider the grammar and derivation tree of Example 4.4. The derivation tree of Fig. 4.2 is reproduced without numbered vertices as Fig. 4.3(a).
The yield of the tree in Fig. 4.3(a) is aabbaa. Referring to Fig. 4.2 again, we see that the leaves are the vertices numbered 2, 9, 6, 10, 1 1, and 8, in that order, from the left. These vertices have labels a, a, b, b, a, a, respectively. Note that in this case all leaves had terminals for labels, but there is no reason why this should always be so; some leaves could be labeled by c or by a variable. Note that S|> aabbaa by the derivation S => aAS => aSbAS => aabAS => aabbaS => aabbaa.
Figure 4.3(b) shows a subtree of the tree illustrated in part (a). It is vertex 3 of Fig. 4.2, together with its descendants. The yield'of the subtree is abba. The label of the root of the subtree is A, and A ^> abba. A derivation in this case is A => SbA => abA => abba.
The relationship between derivation trees and derivations Theorem 4.1 Let G = (V, T, P, S) be a context-free grammar. Then S ^> a if and only if there is a derivation tree in grammar G with yield a.
Proof We shall find it easier to prove something in excess of the theorem. What we shall prove is that for any A in V, A ^> a if and only if there is an /1-tree with a as the yield.
KTUNOTES.IN Downloaded from Ktunotes.in 4J | DERIVATION TREES 85 Suppose, first, that a is the yield of an ,4-tree. We prove, by induction on the number of interior vertices in the tree, that A ^> a. If there is only one interior vertex, the tree must look like the one in Fig. 4.4. In that case, X 1X 2 • • Xn must be a, and A -> a must be a production of P, by definition of a derivation tree.
A Fig. 4.4 Tree with one interior vertex.
Now, suppose that the result is true for trees with up to k — 1 interior vertices.
Also, suppose that a is the yield of an ,4-tree with k interior vertices for some k> 1. Consider the sons of the root. These could not all be leaves. Let the labels of the sons be X u X 2 , .
-.
, X n in order from the left. Then surely, A -> X XX 2 ' ' ' X n is a production in P. Note that n may be any integer greater than or equal to one in the argument that follows.
If the ith son is not a leaf, it is the root of a subtree, and X f must be a variable.
The subtree must be an X,-tree and has some yield a,. If vertex i is a leaf, let a,- = X t . It is easy to see that if j < /, vertex j and all of its descendants are to the left of vertex i and all of its descendants. Thus a = a, a 2 • • • a„. A subtree must have fewer interior vertices than its tree does, unless the subtree is the entire tree. By the inductive hypothesis, for each vertex i that is not a leaf, X s ^> ah since the subtree with root Xi is not the entire tree. If X t = a„ then surely X t ^> a,. We can put all these partial derivations together, to see that A=>X X X 2 Xn ^>oc 1 X 2 ••• Xn >-->cx l a. We must show that there is an yl-tree with yield a.
If A ^> a by a single step, then A -» a is a production in P, and there is a tree with yield a, of the form shown in Fig. 4.4.
Now, assume that for any variable A if A ^> a by a derivation of fewer than /c steps, then there is an ,4-tree with yield a. Suppose that A ^> a by a derivation of k steps. Let the first step be A- X lX 2 '- Xn . It should be clear that any symbol in a must either be one of X u X 2 > Xn or be derived from one of these. Also, the portion of a derived from X t must lie to the left of the symbols derived from Xjy if i < j. Thus, we can write a as ol x ol 2 ••• a„, where for each i between 1 and n, 1) a t = X ( if X { is a terminal, and 2) Xi^xxi if X, is a variable.
If X f is a variable, then the derivation of a f from AT, must take fewer than k steps, since the entire derivation A ^> a takes /c steps, and the first step is surely not part of the derivation X t ^> a f . Thus, by the inductive hypothesis, for each X { that is a variable, there is an Xrtree with yield a f . Let this tree be 7J.
We begin by constructing an ,4-tree with n leaves labeled X lf X2 , .
.
. , Xn , and no other vertices. This tree is shown in Fig. 4.5(a). Each vertex with label Xh where X t is not terminal, is replaced by the tree 7J. If X t is a terminal, no replace-ment is made. An example appears in Fig. 4.5(b). The yield of this tree is a.
A A (a) (b) Fig. 4.5 Derivation trees.
Example 4.6 Consider the derivation S ^> aabbaa of Example 4.5. The first step is S -> aAS. If we follow the derivation, we see that A eventually is replaced by SbA, then by abA, and finally, by abba. Figure 4.3(b) is a parse tree for this derivation. The only symbol derived from S in aAS is a. (This replacement is the last step.) Figure 4.6(a) is a tree for the latter derivation.
Figure 4.6(b) is the derivation tree for 5 -> aAS. If we replace the vertex with label A in Fig. 4.6(b) by the tree of Fig. 4.3(b) and the vertex with label S in Fig.
4.6(b) with the tree of Fig. 4.6(a), we get the tree of Fig. 4.3(a), whose yield is aabbaa.
KTUNOTES.IN Downloaded from Ktunotes.in 4.4 | SIMPLIFICATION OF CONTEXT-FREE GRAMMARS 87 S S a aAS (a) (b) Fig. 4.6 Derivation trees.
Leftmost and rightmost derivations; ambiguity If at each step in a derivation a production is applied to the leftmost variable, then the derivation is said to be leftmost. Similarly a derivation in which the rightmost variable is replaced at each step is said to be rightmost. If w is in L(G) for CFG G, then w has at least one parse tree, and corresponding to a particular parse tree, w has a unique leftmost and a unique rightmost derivation. In the proof of Theorem 4.1, the derivation of a from A corresponding to the parse tree in question is leftmost, provided the derivations X { ^> a { are made leftmost. If instead of deriva-tion (4.4) we (recursively) made the derivation X t ^> a f be rightmost and replaced the Xi's by a,- from the right rather than the left, we would obtain the rightmost derivation corresponding to the parse tree.
Of course, w may have several rightmost or leftmost derivations since there may be more than one parse tree for w. However, it is easy to show that from each derivation tree, only one leftmost and one rightmost derivation may be obtained.
Also, the construction of Theorem 4.1 produces different derivation trees from different leftmost or different rightmost derivations.
Example 4.7 The leftmost derivation corresponding to the tree of Fig. 4.3(a) is S => aAS => aSbAS => aabAS => aabbaS => aabbaa.
The corresponding rightmost derivation is S => aAS => aAa => aSbAa => aSbbaa => aabbaa.
A context-free grammar G such that some word has two parse trees is said to be ambiguous. From what we have said above, an equivalent definition of ambigu-ity is that some word has more than one leftmost derivation or more than one rightmost derivation. A CFL for which every CFG is ambiguous is said to be an inherently ambiguous CFL. We shall show in Section 4.7 that inherently ambi-guous CFL's exist.
4.4 SIMPLIFICATION OF CONTEXT-FREE GRAMMARS There are several ways in which one can restrict the format of productions without reducing the generative power of context-free grammars.
If L is a nonempty KTUNOTES.IN Downloaded from Ktunotes.in 88 CONTEXT-FREE GRAMMARS context-free language then it can be generated by a context-free grammar G with the following properties.
1) Each variable and each terminal of G appears in the derivation of some word in L.
2) There are no productions of the form A -> B where A and B are variables.
Furthermore, if e is not in L, there need be no productions of the form A -> e. In fact, if e is not in L, we can require that every production of G be of one of the forms A-> BC and A -> a, where A, B, and C are arbitrary variables and a is an arbitrary terminal. Alternatively, we could make every production of G be of the form A -> aoi, where a is a string of variables (perhaps empty). These two special forms are called Chomsky normal form and Greibach normal form, respectively.
Useless symbols We now undertake the task of eliminating useless symbols from a grammar. Let G = (V, T, P, S) be a grammar. A symbol X is useful if there is a derivation S^xxX($^>w for some a, and w, where w is in 7 (recall our convention regarding names of symbols and strings). Otherwise X is useless. There are two aspects to usefulness. First some terminal string must be derivable from X and second, X must occur in some string derivable from S. These two conditions are not, however, sufficient to guarantee that X is useful, since X may occur only in sentential forms that contain a variable from which no terminal string can be derived.
Lemma 4.1 Given a CFG G = (V, T, P, S\ with L(G) + 0, we can effectively find an equivalent CFG G' = (V, T, F, S) such that for each A in V there is some w in T for which A^>w.
Proof Each variable A with production A -> w in P clearly belongs in V.
If A -> XjA^ • • • Xn is a production, where each X f is either a terminal or a variable already placed in V\ then a terminal string can be derived from A by a derivation beginning /4=>X 1X 2 •• Xn , and thus /I belongs in V. The set K' can be computed by a straightforward iterative algorithm. F is the set of all productions whose symbols are in V u T.
The algorithm of Fig. 4.7 finds all variables A that belong to V. Surely if A is added to NEWV at line (2) or (5), then A derives a terminal string. To show NEWV is not too small, we must show that ifA derives a terminal string w, then A is eventually added to NEWV. We do so by induction on the length of the derivation A ^> w.
Basis If the length is one, then A - w is a production, and A is added to NEWV in step (2).
Induction Let A=>X XX 2 "' Xn ^>w by a derivation of /c steps. Then we may write w = w l w 2 -• w„, where X f ^> wf , for 1 < i < n, by a derivation of fewer than KTUNOTES.IN Downloaded from Ktunotes.in 4.4 | SIMPLIFICATION OF CONTEXT-FREE GRAMMARS 89 begin 1) OLDV:= 0; 2) NEWV:= {A \ A - w for some w in 7}; 3) while OLDV + NEWV do begin 4) OLDV:= NEWV; 5) NEWV:= OLDV u {A | A -> a for some a in (7 u OLDV)} end; 6) K':= NEWV end Fig. 4.7 Calculation of V .
k steps. By the inductive hypothesis, those X { that are variables are eventually added to NEWV. At the while-loop test of line (3), immediately after the last of the X t y s is added to NEWV, we cannot have NEWV = OLDV, for the last of these X^s is not in OLDV. Thus the while-loop iterates at least once more, and A will be added to NEWV at line (5).
Take V to be the set computed at line (6) and F to be all productions whose symbols are in V u 7. Surely G' = (V, 7, F, S) satisfies the property that if A is in V, then A ^> w for some w. Also, as every derivation in G' is a derivation of G, we know L(G') £ L(G). But if there is some w in L(G) not in L(G'), then any derivation of w in G must involve a variable in V — V or a production in P — P' (which implies there is a variable in V — V used). But then there is a variable in V — V that derives a terminal string, a contradiction.
Lemma 4.2 Given a CFG G = (V, T, Pf S) we can effectively find an equivalent CFG G' = (V, T, F, S) such that for each X in V u T there exist a and £ in (r u r)for which S:S>aX0.
Proof The set V u 7' of symbols appearing in sentential forms of G is con-structed by an iterative algorithm.
Place S in V .
If A is placed in V and A-±u. x |a 2 1 • -- |a„, then add all variables of a x , a 2 , a„ to the set F and all terminals of o^, a2 , a n to 7\ P' is the set of productions of P containing only symbols of V u T.
By first applying Lemma 4.1 and then Lemma 4.2, we can convert a grammar to an equivalent one with no useless symbols. It is interesting to note that applying Lemma 4.2 first and Lemma 4.1 second may fail to eliminate all useless symbols.
Theorem 4.2 Every nonempty CFL is generated by a CFG with no useless sym-bols.
Proof Let L = Jjfi) be a nonempty CFL. Let G x be the result of applying the construction of Lemma 4.1 to G and let G 2 be the result of applying the construc-tion of Lemma 4.2 to G v Suppose G 2 has a useless symbol X. By Lemma 4.2, there is a derivation S §> aXp. Since all symbols of G 2 are symbols of G 1? it follows from KTUNOTES.IN Downloaded from Ktunotes.in 90 CONTEXT-FREE GRAMMARS Lemma 4.1 that S §> ctXf} |> w for some terminal string w. Therefore, no symbol in the derivation aXf$=>w is eliminated by Lemma 4.2. Thus, X derives a terminal string in G 2 , and hence X is not useless as supposed.
Example 4.8 Consider the grammar S-AB\a A^a (4.5) Applying Lemma 4.1, we find that no terminal string is derivable from B. We therefore eliminate B and the production S -> AB. Applying Lemma 4.2 to the grammar S-+a A->a (4.6) we find that only S and a appear in sentential forms. Thus ({S}, {a}, {S - a}, S) is an equivalent grammar with no useless symbols.
Suppose we first applied Lemma 4.2 to (4.5). We would find that all symbols appeared in sentential forms. Then applying Lemma 4.1 we would be left with (4.6), which has a useless symbol, A.
e-Productions We now turn our attention to the elimination of productions of the form A-+e, which we call e-productions. Surely if e is in L(G), we cannot eliminate all e-productions from G, but if e is not in L(G), it turns out that we can. The method is to determine for each variable A whether A ^> e. If so, we call A nullable. We may replace each production B -> X XX 2 " ' Xn by all productions formed by striking out some subset of those X-s that are nullable, but we do not include B -> e, even if all X-s are nullable.
Theorem 4.3 If L = L(G) for some CFG G = (V, T,P, S), then L - {e} is L(G') for a CFG G with no useless symbols or £-productions.
Proof We can determine the nullable symbols of G by the following iterative algorithm. To begin, if A -> e is a production, then A is nullable. Then, if B -> a is a production and all symbols of a have been found nullable, then B is nullable. We repeat this process until no more nullable symbols can be found.
The set of productions F is constructed as follows. If A -> X XX 2 " Xn is in P, then add all productions A -> a t a2 a„ to F where 1) if is not nullable, then a f = X f ; 2) if X t is nullable, then a £ is either X£ or e; 3) not all a,'s are e.
KTUNOTES.IN Downloaded from Ktunotes.in 4.4 | SIMPLIFICATION OF CONTEXT-FREE GRAMMARS 91 Let G" = (V, T, P\ S). We claim that for all A'mV and w in T,A=>w if and only if w =j= e and A => w.
If Let A => w and w ^ 6. We prove by induction on i that A => w. The basis, i = 1, is trivial, for ,4 - w must be a production in P. Since w ^ £, it is also a production of P'. For the inductive step, let i> 1. Then A^X 1 X 2 ••• X„'^ w. Write w = w 1 w2 • wn such that for each 7, Xj=> Wj in fewer than i steps. If vv,- ^ e and X ; is a variable, then by the inductive hypothesis we have Xj |> w ; -. If vv,- = £, then Xj is nullable. Thus, A /? x /?2 • • • /?„ is a production in P', where /?; . = X} if vv,- ^ e and = e if vv,- = e. Since w ^ £, not all fij are e. Hence we have a derivation A=>P 1 P2 - Pn^w i P2 '~ Pn^w^Ps- )?>--^w lW;2 --wn = w in G".
0n/y z/ Suppose v4G =^ w.
Surely w # £, since G" has no e-productions. We show by induction on i that A => w. For the basis, i = 1, observe that A -> w is a production in P'. There must be a production ^ -> a in P such that by striking out certain nullable symbols from a we are left with w. Then there is a derivation A => a => w, where the derivation a ^> w involves deriving e from the nullable sym-bols of a that were struck out in order to get w.
For the induction step, let i > 1. Then A=> X XX 2 "' Xn l => w. There must be some A -> /? in P such that XjA^ ••• X„ is found by striking out some nullable symbols from p. Thus A => • • Xn . Write w = w 1w 2 • • w„, such that for all Xj=>Wj by fewer than i steps. By the inductive hypothesis, AT,=>vv,- if AT ; - is a variable. Certainly if Xj is a terminal, then vv,- = JV,-, and X; => Wj is trivially true.
Thus A => w.
The last step of the proof is to apply Theorem 4.2 to G" to produce G' with no useless symbols. Since the constructions of Lemmas 4.1 and 4.2 do not introduce any productions, G' has neither nullable symbols nor useless symbols. Further-more S=> w if and only if w + e and S |> w. That is, L(G') = L(G) - {c}.
From here on we assume that no grammar has useless symbols. We now turn our attention to productions of the form A-+B whose right-hand side consists of a single variable. We call these unit productions. All other productions, including those of the form A -> a and e-productions, are nonunit productions.
Theorem 4.4 Every CFL without e is defined by a grammar with no useless symbols, ^-productions, or unit productions.
Proof Let L be a CFL without e and L = L(G) for some G = (K, T, P, S). By Theorem 4.3, assume G has no 6-productions. Construct a new set of productions P' from P by first including all nonunit productions of P. Then, suppose that A => for ,4 and £ in V. Add to P' all productions of the form A-a, where B-kx is a nonunit production of P.
KTUNOTES.IN Downloaded from Ktunotes.in 92 CONTEXT-FREE GRAMMARS and if Observe that we can easily test whether A => B, since G has no £-productions, A T BiT B2T-T BmT B> and some variable appears twice in the sequence, we can find a shorter sequence of unit productions that results in A => B. Thus it is sufficient to consider only those sequences of unit productions that do not repeat any of the variables of G.
We now have a modified grammar, G' = (Vy T, F, S). Surely, if A ->a is a production of F, then A=j>a. Thus, if there is a derivation of w in G', then there is a derivation of w in G.
Suppose that w is in L(G), and consider a leftmost derivation of w in G, say If, for 0 < i < n, a, => a, + x by a nonunit production, then a, ^> cti+1 . Suppose that a . ^>ai+i by a unit production, but that a,_j =>a, by a nonunit production, or i = 0. Also suppose that a l + j ^> a i + 2 ^> • g> a,-, all by unit productions, and a,=>a;+1 by a nonunit production. Then a t , ai+1 , a, are all of the same length, and since the derivation is leftmost, the symbol replaced in each of these must be at the same position. But then a f =>a, + 1 by one of the productions of F — P. Hence L(G') = L(G). To complete the proof, we observe that G' has no unit productions or ^-productions. If we use Lemmas 4.1 and 4.2 to eliminate useless symbols, we do not add any productions, so the result of applying the constructions of these lemmas to G' is a grammar satisfying the theorem.
4.5 CHOMSKY NORMAL FORM We now prove the first of two normal-form theorems. These each state that all context-free grammars are equivalent to grammars with restrictions on the forms of productions.
Theorem 4.5 (Chomsky normalform, or CNF) Any context-free language without e is generated by a grammar in which all productions are of the form A BC or A-a. Here, A, £, and C, are variables and a is a terminal.
Proof Let G be a context-free grammar generating a language not containing e.
By Theorem 4.4, we can find an equivalent grammar, G l = (V, T, P, S), such that P contains no unit productions or e-productions. Thus, if a production has a single symbol on the right, that symbol is a terminal, and the production is already in an acceptable form.
v Now consider a production in P, of the form A -+ X x X 2 • • • Xm , where m > 2.
If X t is a terminal, a, introduce a new variable Ca and a production Ca a, which is of an allowable form. Then replace X t by Ca . Let the new set of variables be V and the new set of productions be F. Consider the grammar G 2 = (V\ T, F, S).f If t Note that G 2 is not yet in Chomsky normal form.
KTUNOTES.IN Downloaded from Ktunotes.in 4.5 | CHOMSKY NORMAL FORM 93 a=>& then a|>£. Thus L(G 1 )^L(G 2 ). Now we show by induction on the number of steps in a derivation that if A => w, for AinV and w in T, then A=>w.
The result is trivial for one-step derivations. Suppose that it is true for derivations of up to k steps. Let A => w be a (k + l)-step derivation. The first step must be of the form A -> B x B2 • • • Bm , m > 2. We can write w = w x w 2 wm , where B ( => wi9 1 < i < m.
If is Ca . for some terminal ah then vv f must be a { . By the construction of P\ there is a production A^> X l X 2 -- Xm of P where X £ = if is in V and = a t if B t is in V" - V. For those £, in V, we know that the derivation Bi( f> wt takes no more than k steps, so by the inductive hypothesis, X t §> wt . Hence A> w.
We have now proved the intermediate result that any context-free language can be generated by a grammar for which every production is either of the form A -> a or A -> B x B2 ' • • Bm , for m > 2. Here A and Bu B2 , .
.
.
, Bm are variables, and a is a terminal.
Consider such a grammar G 2 = (V\ T, F, S). We modify G 2 by adding some additional symbols to V and replacing some productions of F. For each produc-tion A- B l B2 -- Bm of F, where m > 3, we create new variables D u Z) 2 , Dm _ 2 and replace A- B X B 2 -Bm by the set of productions {A^ B X D U D x ^ B2 D 2 , Dm . 3 ^ Bm . 2 Dm _ 2 , Dm _ 2 -> Bm _ x Bm}.
Let K" be the new nonterminal vocabulary and P" the new set of productions. Let G 3 = (V\ T, P", S). G 3 is in CNF.
It is clear that if A§>p, then /4 1> /?, so L(G 2 ) ^ L(G 3 ). But it is also true that L(G 3 ) ^ L(G 2 ), as can be shown in essen-tially the same manner as it was shown that Ufii) £ The proof will be left to the reader.
Example 4.9 Let us consider the grammar ({S, A, £}, {a, b], P, S) that has the productions: S^bA\aB A->bAA\aS\a B^aBB\bS\b and find an equivalent grammar in CNF.
First, the only productions already in proper form are A a and B^b.
There are no unit productions, so we may begin by replacing terminals on the right by variables, except in the case of the productions A -> a and B -> b. S -> bA is replaced by S -> Cb A and Cb b. Similarly, A - aS is replaced by A Ca 5 and Ca-a; A^bAA is replaced by A-+Cb AA; S -> aB is replaced by S^Ca B; B-bS is replaced by £ - Cb S, and £ - a££ is replaced by B - Ca BB.
In the next stage, the production A-+Cb AA is replaced by A -> C^Dj and D i - A A, and the production B^Ca BB is replaced by B - Ca Z) 2 and D 2 -> ££.
KTUNOTES.IN Downloaded from Ktunotes.in 94 CONTEXT-FREE GRAMMARS The productions for the grammar in CNF are shown below.
S->Cb A\CaB D X ->AA i4-^CaS|C fcD 1 |fl D 2 ->BB B-+Cb S\CaD 2 \b Ca ->a Cb -+b 4.6 GREIBACH NORMAL FORM We now develop a normal-form theorem that uses productions whose right-hand sides each start with a terminal symbol perhaps followed by some variables. First we prove two lemmas that say we can modify the productions of a CFG in certain ways without affecting the language generated.
Lemma 4.3 Define an A-production to be a production with variable A on the left. Let G = (V, T, P, S) be a CFG. Let A -» ol x Bol 2 be a production in P and B Pi 1 Pi 1 ' • | pr be the set of all ^-productions. Let G { = (V, T, P u S) be ob-tained from G by deleting the production /4->a 1 Ba 2 from P and adding the productions A -> a 1 ^ 1 a 2 |o<i/?2 a 2 I'" | a i/?r<2- Then L(G) = L^).
Proof Obviously L(Gi) c L(G), since if ^4 —> o^fta^ s usec^ n a derivation of G lt then A^>ct 1 Ba 2 ^>0LiPi (X2 can be used in G. To show that L(G)^L(G l \ one simply notes that A -a 1 2fa 2 is the only production in G not in G x . Whenever A -+ a 1 B(x 2 is used in a derivation by G, the variable B must be rewritten at some later step using a production of the form B /?,-. These two steps can be replaced by the single step A => a 1 /?,a 2 .
Lemma 4.4 Let G = (K, T, P, S) be a CFG. Let A -> A(x l \Aol 2 \'-\ Avl t be the set of ^-productions for which A is the leftmost symbol of the right-hand side. Let A -> Pi \P2 \'" \PS be the remaining /1-productions. Let G, = (Fu {£}, T, P l9 5) be the CFG formed by adding the variable B to V and replacing all the A-productions by the productions: ^ A ~Pi \< ^ B-OLi I 1) „ 1< <s, 2) 1< AoL t must eventually end with a production A -» fij. The sequence of replacements A=> Aa. ix => AcLi2 CLh =>--'=> AcL ip CLip_ x a f| KTUNOTES.IN Downloaded from Ktunotes.in 4.6 | GREIBACH NORMAL FORM in G can be replaced in G x by The reverse transformation can also be made. Thus L(G) = L(G,). Figure 4.8 shows this transformation on derivation trees, where we see that in G, a chain of i4's extending to the left is replaced in G t by a chain of Fs extending to the right.
A A (a) (b) Fig. 4.8 Transformation of Lemma 4.4 on portion of a derivation tree.
Theorem 4.6 (Greibach normal form or GNF) Every context-free language L without e can be generated by a grammar for which every production is of the form A aoi, where A is a variable, a is a terminal, and a is a (possibly empty) string of variables.
Proof Let G = (V, T, P, S) be a Chomsky normal form grammar generating the CFL L Assume that V = {A u A 2 ,..., Am}. The first step in the construction is to modify the productions so that if A { -+ A} y is a production, then j > i. Starting with A x and proceeding to Am , we do this as follows. We assume that the produc-tions have been modified so that for I <i i. We now modify the /^-productions.
If A k Ajy is a production with j < /c, we generate a new set of productions by substituting for A} the right-hand side of each ,4,-production according to Lemma 4.3. By repeating the process k -1 times at most, we obtain productions of the form Ak -» Ae y, £ > L The productions with £ = k are then replaced accord-ing to Lemma 4.4, introducing a new variable Bk . The precise algorithm is given in Fig. 4.9.
KTUNOTES.IN Downloaded from Ktunotes.in 96 CONTEXT-FREE GRAMMARS begin 1) for 1 to m do begin 2) for 1 to k -1 do 3) for each production of the form A k AjOL do begin 4) for all productions Aj -> P do 5) add production Ak -+ /fa; 6) remove production Ak - Ajcc end; 7) for each production of the form Ak -> Ak tx do begin 8) add productions Bk a and £k a£k ; 9) remove production Ak A k a end; 10) for each production /lk -+ where /? does not begin with Ak do Hj add production A k - /?£k end end Fig. 4.9 Step 1 in the Greibach normal-form algorithm.
By repeating the above process for each original variable, we have only pro-ductions of the forms: 1) A^Ajy, j>U 2) Ai ay, a in T, 3) B^y, yin(Ku{B lf B 2f ... f Bl _ 1 }) Note that the leftmost symbol on the right-hand side of any production for Am must be a terminal, since Am is the highest-numbered variable. The leftmost symbol on the right-hand side of any production for Am . x must be either Am or a terminal symbol. When it is Am , we can generate new productions by replacing Am by the right-hand side of the productions for Am according to Lemma 4.3. These productions must have right sides that start with a terminal symbol. We then proceed to the productions for Am __ 2 , A 2 , A x until the right side of each production for an A x starts with a terminal symbol.
As the last step we examine the productions for the new variables, B2 , Bm . Since we began with a grammar in Chomsky normal form, it is easy to prove by induction on the number of applications of Lemmas 4.3 and 4.4 that the right-hand side of every ^-production, 1 < i < n, begins with a terminal or AjAk for somej and k. Thus a in line (7) of Fig. 4.9 can never be empty or begin with some \ KTUNOTES.IN Downloaded from Ktunotes.in 4.6 | GREIBACH NORMAL FORM 97 Bj, so no £rproduction can start with another Bj. Therefore all ^.-productions have right-hand sides beginning with terminals or A-s, and one more application of Lemma 4.3 for each ^-production completes the construction.
Example 4.10 Let us convert to Greibach normal form the grammar G=({A U A 29 A 3 } 9 {a, b},P,A x \ where P consists of the following: Ai —> A 2 A 3 A 2 -A 3 A t \b A 3 - A 1 A 2 \a Step 1 Since the right-hand side of the productions for A v and A 2 start with terminals or higher-numbered variables, we begin with the production A 3 -> A x A 2 and substitute the string A 2 A 3 for A t . Note that A x A 2 A 3 is the only production with A x on the left.
The resulting set of productions is: A x -» A 2 A 3 A 2 -+A 3 A x \b A 3 - A 2 A 3 A 2 \a Since the right side of the production A 3 - A 2 A 3 A 2 begins with a lower-numbered variable, we substitute for the first occurrence of A 2 both A 3 A x and b.
Thus A 3 -> A 2 A 3 A 2 is replaced by A 3 - A 3 A x A 3 A 2 and A 3 -> bA 3 A 2 . The new set is A X ^A 2 A 3 A 2 -+A 3 A x \b A 3 -+ A 3 A x A 3 A 2 \bA 3 A 2 \a We now apply Lemma 4.4 to the productions A 3 -+ A 3 A x A 3 A 2 \bA 3 A 2 \a.
Symbol B3 is introduced, and the production A 3 ^> A 3 A X A 3 A 2 is replaced by A 3 -> bA 3 A 2 B 3 , A 3 - aB 3 , B 3 - A x A 3 A 2i and B 3 - A x A 3 A 2 B 3 . The resulting set is A X -+A 2 A 3 A 2 -+ A 3 A x \b A 3 -> bA 3 A 2 B3 \aB3 \bA 3 A 2 \a B3 -+A X A 3 A 2 \A X A 3 A 2 B 3 KTUNOTES.IN Downloaded from Ktunotes.in 98 CONTEXT-FREE GRAMMARS Step 2 Now all the productions with A 3 on the left have right-hand sides that start with terminals. These are used to replace A 3 in the production A 2 - A 3 A x and then the productions with A 2 on the left are used to replace A 2 in the production A x -> A 2 A 3 . The result is the following.
A 3 - bA 3 A 2 B3 A 3 ->bA 3 A 2 A 3 -> 0#3 y4 3 - a A 2 -bA 3 A 2 B3 A l A 2 -+bA 3 A 2 A l A 2 -+ aB3 A l A 2 -^aA l A 2 -+b A l -bA 3 A 2 B3 A l A 3 A x -±bA 3 A 2 A x A 3 A l -+aB3 A l A 3 A l -+aA l A 3 Ai - bA 3 B3 -» A l A 3 A 2 B3 -» A X A 3 A 2 B3 Step 3 The two 53-productions are converted to proper form, resulting in ^) more productions. That is, the productions B3 ^A 1 A 3 A 2 and B 3 ^> A l A 3 A 2 B3 are altered by substituting the right side of each of the five productions with A x on the left for the first occurrences of A v Thus £ 3 A l A 3 A 2 becomes B3 -+bA 3 A 2 B 3 A x A 3 A 3 A 2 , B 3 - aB3 A l A 3 A 3 A 2 .
B3 ^bA 3 A 3 A 2l B 3 -> bA 3 A 2 A x A 3 A 3 A 2 , B3 -+ aA l A 3 A 3 A 2 .
The other production for B 3 is replaced similarly. The final set of productions is A 3 -+bA 3 A 2 B3 A 3 -+bA 3 A 2 A 3 ^aB3 /4 3 ->fl A 2 -bA 3 A 2 B3 A l A 2 ^bA 3 A 2 A l A 2 ^aB3 A x A 2 ^aA x A 2 ^b A l ^>bA 3 A 2 B3 A l A 3 A l ^bA 3 A 2 A l A 3 A x -+aB3 A l A 3 A l -+aA l A 3 A x -+bA 3 B3 -> bA 3 A 2 B 3 A x A 3 A 3 A 2 B3 B3 - bA 3 A 2 B3 A x A 3 A 3 A 2 KTUNOTES.IN Downloaded from Ktunotes.in 4.7 | INHERENTLY AMBIGUOUS CONTEXT-FREE LANGUAGES 99 B3 ->aB3 A 1 A 3 A 3 A 2 B3 B3 ->bA 3 A 3 A 2 B3 B3 ->bA 3 A 2 A l A 3 A 3 A 2 B3 £3 -> aA 1 A 3 A 3 A 2 B3 B3 -+aB3 A l A 3 A 3 A 2 B3 -+bA 3 A 3 A 2 B3 -> bA 3 A 2 A l A 3 A 3 A 2 B3 ->aA l A 3 A 3 A 2 4.7 THE EXISTENCE OF INHERENTLY AMBIGUOUS CONTEXT-FREE LANGUAGES It is easy to exhibit ambiguous context-free grammars. For example, consider the grammar with productions S -> A, S - B, A -» a, and B fa. What is not so easy to do is to exhibit a context-free language for which every CFG is ambiguous. In this section we show that there are indeed inherently ambiguous CFL's. The proof is somewhat tedious, and the student may skip this section without loss of continu-ity. The existence of such a language is made use of only in Theorem 8.16.
We shall show that the language L = {fl"ft"cV" | n > 1, m > 1} u {aV^d" \ n > 1, m > 1} is inherently ambiguous by showing that infinitely many strings of the form fl"bVrfn , n > 1, must have two distinct leftmost derivations. We proceed by first establishing two technical lemmas.
Lemma 4.5 Let (Nh M,), 1 < i < r, be pairs of sets of integers. (The sets may be finite or infinite.) Let Si = {(n, m) | n in Ni9 m in MJ and let S = 5, u S2 u u Sr .
If each pair of integers («, m) is in S for all m and m, where n =f m, then (h, n) is in S for all but some finite set of n.
Proof Assume that for all n and m, where n m, each (n, m) is in 5, and that there are infinitely many n such that (n, n) is not in S. Let J be the set of all n such that (n, n) is not in 5. We construct a sequence of sets J r , J r-u Ji such that J r 2 Jr _ x 2 •' 2 Ji-Each J f will be infinite, and for each n and m in J„ (n, m) is not in Si u 5£+1 u ••• u 5r .
For /I in J, either n is not in Nr or n is not in Mr ; otherwise (ny n) would be in Sr and hence in S. Thus there is an infinite subset of J, call it J r , such that either for all n in J r , n is not in Nr9 or for all n in Jr , n is not in M r . Now for n and m in J r , m) is not in Sr .
KTUNOTES.IN Downloaded from Ktunotes.in 100 CONTEXT-FREE GRAMMARS Assume that J r9 Jr _ ly ...,Ji+l have been constructed, where i < r — 1. Then J t is constructed as follows. For each n in J i+ u either n is not in N t or not in M £ ; otherwise (n, n) would be in S, and hence in 5, a contradiction since J i+l ^ J.
Thus, either an infinite subset of Ji+ x is not in JV, or an infinite subset of Ji+ j is not in M,. In either case, let the infinite subset be J f . Now for all n and m in Jh (n, m) is not in Si and hence not in 5 f u Si+ x u ••• u Sr .
Since J x contains an infinite number of elements, there exist n and m in J u m. Now (rc, m) is not in Sj u 5 2 u • • • u 5 r = 5, contradicting the assumption that all (n, m), where w m, are in 5. Thus (n, n) is in S for all but some finite set of n.
Lemma 4.6 Let G be an unambiguous CFG. Then we can effectively construct an unambiguous CFG G equivalent to G, such that G' has no useless symbols or productions, and for every variable A other than possibly the start symbol of G', we have the derivation A Ax 2 , where x { and x 2 are not both e.
Proof The construction of Lemmas 4.1 and 4.2, removing useless symbols and productions, cannot convert an unambiguous grammar into an ambiguous ojje, since the set of derivation trees for words does not change. The construction of Theorem 4.4, removing unit productions, cannot introduce ambiguities. This is because if we introduce production A -> a, there must be a unique B such that A ^> B and B -> a is a production, else the original grammar was not unambig-uous. Similarly, the construction of Theorem 4.3, removing e-productions, does not introduce ambiguity.
Let us therefore assume that G has no useless symbols or productions, no (-productions, and no unit productions. Suppose that for no x x and x 2 not both c does A^>x x Ax 2 . Then replace each occurrence of A on the right side of any production by all the right sides of ^-productions. As there are no unit produc-tions, ^-productions or useless symbols, there cannot be a production A ctxA^, else there is a derivation A ^> x 1 y4x 2 , with x x and x 2 not both c. The above change does not modify the generated language, by Lemma 4.3. Each new production comes from a unique sequence of old productions, else G was ambiguous. Thus the resulting grammar is unambiguous. We see that A is now useless and may be eliminated. After removing variables violating the condition of the lemma in this manner, the new grammar is equivalent to the old, is still unambiguous, and satisfies the lemma.
Theorem 4.7 The CFL, L = {a nb n(Tdm \ n > 1, m > 1} u {a nbm(Td n \ n > 1, m > 1}, is inherently ambiguous.
Proof Assume that there is an unambiguous grammar generating L By Lemma 4.6 we can construct an unambiguous grammar G = (K, 7, P, S) generating L with no useless symbols, and for each A in V - {S}, A ^>x l Ax 2 for some x l and x 2 in T, not both e.
KTUNOTES.IN Downloaded from Ktunotes.in 4.7 | INHERENTLY AMBIGUOUS CONTEXT-FREE LANGUAGES 101 We note that the grammar G has the following properties : 1) If A ^> x l Ax 2 , then x x and x 2 each consist of only one type of symbol (a, b, c, or d); otherwise w^w^ ^> WiX^i Ax 2 x 2 w3 ^ w l x l x l w 2 x 2 x2 w 3 , for some w x , w 2 , and vv 3 . This last terminal string is not in L.
2) If A^x^Ax^ then x x and x 2 consist of different symbols. Otherwise, in a derivation involving A, we could increase the number of one type of symbol in a sentence of L without increasing the number of any other type of symbols, thereby generating a sentence not in L 3) If A ^>x l Ax 2 , then |xj | = |x 2 |. Otherwise we could find words in L having more of one symbol than any other.
4) If A^>x x Ax 2 and A^>x zAx Ar , then Xj and x 3 consist of the same type of symbol. Likewise x 2 and x4 . Otherwise Property 1 above would be violated.
5) If A^>x x Ax 2 , then either a) Xj consists solely of ds and x 2 solely of b y s or of d\ b) x x consists solely of 6's and x 2 solely of cs, or c) Xj consists solely of cs and x 2 solely of d's.
In any of the other cases it is easy to derive a string not in L Thus the variables other than 5 can be divided into four classes, Cab , Cad , Cbc , and Ccd .
Cab is the set of all A in V such that A^>x l Ax 2 , with Xj in a and x 2 in b; Cad, Cbc> anc Ccd are defined analogously.
6) A derivation containing a symbol in Cab or Ccd cannot contain a symbol in Cad or Cbc or vice versa. Otherwise, we could increase the number of three types of symbols of a sentence in L without increasing the number of the fourth type of symbol. In that case, there would be a sentence in L for which the number of occurrences of one type of symbol is smaller than that of any other.
We now note that if a derivation contains a variable in Cab or Ccd , then the terminal string generated must be in {a nb ncmdm \n > 1, m > 1}. For assume that A in Cab appears in a derivation of a sentence x not in {a n^ ncm^m \n> 1, m > 1}. Then x must be of the form a nbmcmdn , m £ n. Since A is in Cab , a sentence a n + pbm + pcmd n i ni= h, for some p > 0, could be generated. Such a sentence is not in L A similar argument holds for A in Ccd . Similar reasoning implies that if a derivation con-tains a variable in or Cbc , then the sentence generated must be in {flBmB |n>U>l}.
We divide G into two grammars, Gj = ({S} yj Cab u Ccd , T, P„ S) and G 2 = ({S} vC^u Cbc , T, P 29 S% KTUNOTES.IN Downloaded from Ktunotes.in 102 CONTEXT-FREE GRAMMARS where P 1 contains all productions of P with a variable form or Ccd on either the right or left, and P2 contains all productions of P with a variable from Cad or Cbc on either the right or left. In addition, P x contains all productions from P of the form S - aW^cF", nj=m, and P2 contains all productions from P of the form S -> aVc^d", nj=m. Productions ofP of the form S -+ a"bnc"d n are not in either P x or P 2 .
Since G generates {a nbncrdm | n > 1, m > 1} u {aWcfd" \ n > 1, m > 1}, Gj must generate all sentences in {a nb nd"dm \n>Um>l y nj=m} plus possibly some sentences in {a"b"c"d" \ n > 1}, and G 2 must generate all sen-tences in {fl n&'W | n > 1, m > 1, n £ m} plus possibly some sentences in {a"b"c"d" \ n > 1}. We now show that this cannot be the case unless Gj and G 2 both generate all but a finite number of sentences in {aVcfd" | n > 1}. Thus all but a finite number of sentences in {a"b nc"d n \n>l} are generated by both G x and G 2 and hence by two distinct derivations in G. This contradicts the assumption that G was unambiguous.
To see that G x and G 2 generate all but a finite number of sentences in {a nbnc nd n | n > 0}, number the productions in P x of the form S - a from 1 to r. For 1 < i < r, if S - a is the ith production, let N, be the set of all h such that G i G i for some m, and let M £ be the set of all m such that S=xx=>a nbncmdm Gi G i for some n. We leave it to the reader to show that for any n in N t and any m in M h S=xx>anb ncmdm .
Gi G, [Hint: Recall that the variables of a are in Cad or Ccd .] It follows immediately from Lemma 4.5 that G x must generate all but a finite number of sentences in {a nb"c"d"\n> 1}.
A similar argument applies to G 2 . The reader can easily show that G 2 cannot have a right side with two or more variables. We number certain productions and pairs of productions in a single ordering. Productions of the form S-+ where B is in Cbc9 will receive a number, and if this number is z, let N t be the set of all n such that for some m, Also let M, be the set of m such that for some n, KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 103 The pair of productions S -> a and A -> o^Bo^ will receive a number if a contains a variable in C^, ^ is in Cad , and £ is in Cbc . If this pair is assigned the number i, then define AT f to be the set of n such that for some m, S => a x^x 2 => x t a t £a 2 x 2 ^> aVcV.
Also define M, to be the set of m such that for some n, S=>ol^> x xAx 2 => x t a t Ba 2 x 2 ^> a nbm(Tdn .
Once again, for any n in N f and m in Mh S?>anbmcmd\ and thus it follows from Lemma 4.5 that G 2 generates all but a finite number of sentences in {a nb n(fd n \ n > 1}. We conclude that for some n, a nb"c nd n is in both Ufii) and L(G 2 ). This sentence has two leftmost derivations in G.
EXERCISES 4.1 Give context-free grammars generating the following sets.
3 a) The set of palindromes (strings that read the same forward as backward) over alphabet {a. b}.
b) The set of all strings of balanced parentheses, i.e., each left parenthesis has a matching right parenthesis and pairs of matching parentheses are properly nested.
c) The set of all strings over alphabet {a, b} with exactly twice as many a's as b's.
d) The set of all strings over alphabet {a, b, •, +, , (, ), e, 0} that are well-formed regular expressions over alphabet {a, b). Note that we must distinguish between c as the empty string and as a symbol in a regular expression. We use e in the latter case.
e) The set of all strings over alphabet {a, b} not of the form ww for some string w.
f) {aWli^or;^}.
4.2 Let G be the grammar S-+aS\aSbS\e.
Prove that L(G) — {x | each prefix of x has at least as many a's as b's}.
4.3 For i > 1, let b t denote the string in 1(0 + 1) that is the binary representation oft.
Construct a CFG generating {0, 1, #} + -{b l #b 2 # ••• #bn \n> 1}.
4.4 Construct a CFG generating the set {w#wK # |w in (0 4- 1) + }.
4.5 The grammar E-E + E\E £|(£)|id generates the set of arithmetic expressions with 4- , , parentheses and id. The grammar is ambiguous since id + id id can be generated by two distinct leftmost derivations, a) Construct an equivalent unambiguous grammar.
KTUNOTES.IN Downloaded from Ktunotes.in 104 CONTEXT-FREE GRAMMARS b) Construct an unambiguous grammar for all arithmetic expressions with no redundant parentheses. A set of parentheses is redundant if its removal does not change the expression, e.g., the parentheses are redundant in id + (id id) but not in (id 4- id) id.
4.6 Suppose G is a CFG with m variables and no right side of a production longer than £.
tm — 1 Show that if A => t, then there is a derivation of no more than — steps by which A derives c. How close to this bound can you actually come?
1 4.7 Show that for each CFG G there is a constant c such that if w is in L(G), and w t, then w has a derivation of no more than c\w \ steps.
4.8 Let G be the grammar S-+aB\bA A->a\aS\bAA B-+b\bS\aBB For the string aaabbabbba find a a) leftmost derivation, b) rightmost derivation, c) parse tree.
4.9 Is the grammar in Exercise 4.8 unambiguous?
4.10 Find a CFG with no useless symbols equivalent to S->AB\CA B-BC\AB A-a C->aB\b 4.1 1 Suppose G is a CFG and w, of length /, is in L(G). How long is a derivation of w in G if a) G is in CNF b) G is in GNF.
4.12 Let G be the CFG generating well-formed formulas of propositional calculus with predicates p and q: S->~S[S^S]\p\q.
The terminals are p, q, ~, [, ], and =>. Find a Chomsky normal-form grammar generating L(G).
4.13 Show that conversion to Chomsky normal form can square the number of produc-tions in a grammar. [Hint: Consider the removal of unit productions.] 4.14 Find a Greibach normal-form grammar equivalent to the following CFG: S^AA\0 A->SS\ 4.15 Show that every CFL without e can be generated by a CFG all of whose productions are of the form A - a, or A~ BCt where B^C and if A - ol^Bolj and A - y^By 2 are productions, then = = e or a 2 = yj = £.
S 4.16 Show that every CFL without e is generated by a CFG all of whose productions are of the form A -> a, A - a£, and A - aBC.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 105 4.17 Show that every CFL without t is generated by a CFG all of whose productions are of the form A -> a and A aab.
4.18 Can every CFL without e be generated by a CFG all of whose productions are of the forms A BCD and A a?
4.19 Show that if all productions of a CFG are of the form A - wB or A w, then L(G) is a regular set.
4.20 A CFG is said to be linear if no right side of a production has more than one instance of a variable. Which of the languages of Exercise 4.1 have linear grammars?
S 4.21 An operator grammar is a CFG with no c-productions such that no consecutive symbols on the right sides of productions are variables. Show that every CFL without e has an operator grammar.
4.22 The algorithm given in Fig. 4.7 to determine which variables derive terminal strings is not the most efficient possible. Give a computer program to perform the task in 0(n) steps if n is the sum of the length of all the productions.
4.23 Is {a'Vc \i±j and j ± k and k ± i} a CFL? [Hint: Develop a normal form similar to that in Theorem 4.7. (A pumping lemma is developed in Section 6. 1 that makes exercises of this type much easier. The reader may wish to compare his solution to that in Example 6.3).] Solutions to Selected Exercises 4.1 a) The definition of "palindrome," a string reading the same forward as backward is of no help in finding a CFG. What we must do in this and many other cases is rework the definition into a recursive form. We may define palindromes over {0, 1} recursively, as follows: 1) e, 0, and 1 are palindromes; 2) if w is a palindrome, so are 0h>0 and lwl; 3) nothing else is a palindrome.
We proved in Exercise 1.3 that this is a valid definition of palindromes. A CFG for palindromes now follows immediately from (1) and (2).
It is: S-0|l|c (from 1); S->0S0|1S1 (from 2).
4.16 Let G = (K, T, P, 5) be a GNF grammar generating L Suppose k is the length of the longest right side of a production of G. Let V = {[a] |a is in K + and |a | < k}. For each production A — gol in P and each variable [Aft] in V place [Aft] -> a[a][p] in P\ In the case where a or ft is 6, [e] is deleted from the right side of the production.
4.21 Let G = (K, T, P, 5) be a GNF grammar generating L By Exercise 4.16 we may assume all productions are of the form A a, A -» aB and A -> aBC. First replace each production A - aBC by A a[BC\ where [BC] is a new variable. After having replaced all productions of the form A - aBC, then for each newly introduced variable [BC\ B-production B - a, and C-production C ft add production [BC] - a/?. Note that a and p are either single terminals or of the form bE, where E may be either a new or old variable.
The resulting grammar is an operator grammar equivalent to the original.
KTUNOTES.IN Downloaded from Ktunotes.in 106 CONTEXT-FREE GRAMMARS BIBLIOGRAPHIC NOTES The origin of the context-free grammar formalism is found in Chomsky ; important later writings by Chomsky on the subject appear in Chomsky [1959, 1963]. The related Backus-Naur form notation was used for the description of ALGOL in Backus and Naur [I960], The relationship between CFG's and BNF was perceived in Ginsburg and Rice .
Chomsky normal form is based on Chomsky . Actually, Chomsky proved the stronger result stated in Exercise 4.15. Greibach normal form was proved by Greibach . The method of proof used here is due to M. C. Paull. The reader should also consider the algorithm of Rosenkrantz , which has the property that it never more than squares the number of variables, while the algorithm of Theorem 4.6 may exponen-tiate the number. Solutions to Exercises 4.16, 4.17, and 4.21 can be found there as well.
Ambiguity in CFG's was first studied formally by Floyd [1962a], Cantor , Chomsky and Schutzenberger , and Greibach . Inherent ambiguity was studied by Gross , and Ginsburg and Ullian [1966a, b].
Important applications of context-free grammar theory have been made to compiler design. See Aho and Ullman [1972, 1973, 1977], Lewis, Rosenkrantz, and Stearns , and the bibliographic notes to Chapter 10 for a description of some of the work in this area.
Additional material on context-free languages can be found in Ginsburg and Salo-maa .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 5 PUSHDOWN AUTOMATA 5.1 INFORMAL DESCRIPTION Just as the regular expressions have an equivalent automaton—the finite automa-ton, the context-free grammars have their machine counterpart—the pushdown automaton. Here the equivalence is somewhat less satisfactory, since the push-down automaton is a nondeterministic device, and the deterministic version accepts only a subset of all CFL's. Fortunately, this subset includes the syntax of most programming languages.
(See Chapter 10 for a detailed study of deter-ministic pushdown automaton languages.) The pushdown automaton is essentially a finite automaton with control of both an input tape and a stack, or "first in-last out" list. That is, symbols may be entered or removed only at the top of the list. When a symbol is entered at the top, the symbol previously at the top becomes second from the top, the symbol previously second from the top becomes third, and so on. Similarly, when a symbol is removed from the top of the list, the symbol previously second from the top becomes the top symbol, the symbol previously third from the top becomes second, and so on.
A familiar example of a stack is the stack of plates on a spring that one sees in cafeterias. There is a spring below the plates with just enough strength so that exactly one plate appears above the level of the counter. When that top plate is removed, the load on the spring is lightened, and the plate directly below appears above the level of the counter. If a plate is then put on top of the stack, the pile is pushed down, and only the new plate appears above the counter. For our pur-poses, we make the assumption that the spring is arbitrarily long, so we may add as many plates as we desire.
107 KTUNOTES.IN Downloaded from Ktunotes.in 108 PUSHDOWN AUTOMATA Such a stack of plates, coupled with a finite control, can be used to recognize a nonregular set. The set L = {wcwR \ w in (0 + 1)} is a context-free language, gen-erated by the grammar S -> 050 1 1S1 \ c. It is not hard to show that L cannot be accepted by any finite automaton. To accept L, we shall make use of a finite control with two states, q x and q 2 , and a stack on which we place blue, green, and red plates. The device will operate by the following rules.
1) The machine starts with one red plate on the stack and with the finite control in state q x .
2) If the input to the device is 0 and the device is in state q l9 a blue plate is placed on the stack. If the input to the device is 1 and the device is in state qu a green plate is placed on the stack. In both cases the finite control remains in state q x .
3) If the input is c and the device is in state q u it changes state to q 2 while no plates are added or removed.
4) If the input is 0 and the device is in state q2 with a blue plate, which represents 0, on top of the stack, the plate is removed. If the input is 1 and the device is in state q2 with a green plate, which represents 1, on top of the stack, the plate is removed. In both cases the finite control remains in state q 2 .
5) If the device is in state q 2 and a red plate is on top of the stack, the plate is removed without waiting for the next input.
6) For all cases other than those described above, the device can make no move.
The preceding rules are summarized in Fig. 5.1.
We say that the device described above accepts an input string if, on process-ing the last symbol of the string, the stack of plates becomes completely empty.
Note that, once the stack is empty, no further moves are possible.
Essentially, the device operates in the following way. In state q u the device makes an image of its input by placing a blue plate on top of the stack of plates each time a 0 appears in the input, and a green plate each time a 1 appears in the input. When c is the input, the device enters state q 2 . Next, the remaining input is compared with the stack by removing a blue plate from the top of the stack each time the input symbol is a 0, and a green plate each time the input symbol is a 1.
Should the top plate be of the wrong color, the device halts and no further processing of the input is possible. If all plates match the inputs, eventually the red plate at the bottom of the stack is exposed. The red plate is immediately removed and the device is said to accept the input string. All plates can be removed only when the string that enters the device after the c is the reverse of what entered before the c.
5.2 DEFINITIONS We shall now formalize the concept of a pushdown automaton (PDA). The PDA will have an input tape, afinite control, and a stack. The stack is a string of symbols from some alphabet. The leftmost symbol of the stack is considered to be at the KTUNOTES.IN Downloaded from Ktunotes.in 5.2 | DEFINITIONS 109 Input Top _____ plate State " 0 1 c Blue Add blue plate; stay in state Qi-Add green plate; stay in state Qi-Go to state q 2 .
Remove top plate; stay in state q2 .
— Green Add blue plate; stay in state Add green plate; stay in state Go to state q 2 .
Remove top plate; stay in state q 2 .
Red q\ Add blue plate; stay in state Add green plate; stay in state Go to state q 2 .
<\2 Without waiting for next input, remove top plate.
Fig. 5.1 Finite control for pushdown automaton accepting {wcwR | w in (0 + 1)}.
"top" of the stack. The device will be nondeterministic, having some finite number of choices of moves in each situation. The moves will be of two types. In the first type of move, an input symbol is used. Depending on the input symbol, the top symbol on the stack, and the state of the finite control, a number of choices are possible. Each choice consists of a next state for the finite control and a (possibly empty) string of symbols to replace the top stack symbol. After selecting a choice, the input head is advanced one symbol.
The second type of move (called an £-move) is similar to the first, except that the input symbol is not used, and the input head is not advanced after the move.
This type of move allows the PDA to manipulate the stack without reading input symbols.
Finally, we must define the language accepted by a pushdown automaton.
There are two natural ways to do this. The first, which we have already seen, is to define the language accepted to be the set of all inputs for which some sequence of moves causes the pushdown automaton to empty its stack. This language is referred to as the language accepted by empty stack.
The second way of defining the language accepted is similar to the way a finite automaton accepts inputs. That is, we designate some states as final states and KTUNOTES.IN Downloaded from Ktunotes.in 110 PUSHDOWN AUTOMATA define the accepted language as the set of all inputs for which some choice of moves causes the pushdown automaton to enter a final state.
As we shall see, the two definitions of acceptance are equivalent in the sense that if a set can be accepted by empty stack by some PDA, it can be accepted by final state by some other PDA, and vice versa.
Acceptance by final state is the more common notion, but it is easier to prove the basic theorem of pushdown automata by using acceptance by empty stack.
This theorem is that a language is accepted by a pushdown automaton if and only if it is a context-free language.
A pushdown automaton M is a system (Q, Z, T, 3 y qQy Z0 , F), where 1) Q is a finite set of states; 2) T is an alphabet called the input alphabet; 3) T is an alphabet, called the stack alphabet; 4) q0 in Q is the initial state; 5) Z0 in T is a particular stack symbol called the start symbol; 6) F ^ Q is the set offinal states; 7) 3 is a mapping from Q x (T u {e}) x T to finite subsets of Q x T.
Unless stated otherwise, we use lower-case letters near the front of the alphabet to denote input symbols and lower-case letters near the end of the alphabet to denote strings of input symbols. Capital letters denote stack symbols and Greek letters indicate strings of stack symbols.
Moves The interpretation of 5(q, a, Z) = {(p 1? yj, (p2 , y 2 ), (pm , yj} where q and ph 1 < i < m, are states, a is in 2, Z is a stack symbol, and y { is in T, 1 < i < w, is that the PDA in state q, with input symbol a and Z the top symbol on the stack can, for any i, enter state ph replace symbol Z by string y„ and advance the input head one symbol. We adopt the convention that the leftmost symbol of y, will be placed highest on the stack and the rightmost symbol lowest on the stack. Note that it is not permissible to choose state p { and string y7 for somej =fc i in one move.
The interpretation of 3{q9 e, Z) = {(p u y0, {p2 , 7il (pm , ym )} is that the PDA in state q, independent of the input symbol being scanned and with Z the top symbol on the stack, can enter state pf and replace Z by y, for any i, 1 < / < m. In this case, the input head is not advanced.
^ KTUNOTES.IN Downloaded from Ktunotes.in 5.2 | DEFINITIONS 111 Example 5.1 Figure 5.2 gives a formal pushdown automaton that accepts {wcw |w in (0 + 1)} by empty stack. Note that for a move in which the PDA writes a symbol on the top of the stack, 3 has a value (q, y) where |y | =2. For example, 3{q l9 0, R) = {(q ly BR)}. If y were of length one, the PDA would simply replace the top symbol by a new symbol and not increase the length of the stack.
This allows us to let y equal e when we wish to pop the stack.
Note that the rule 3(q2 , e, R) = {(q2 , e)} means that the PDA, in state q2 with R the top stack symbol, can erase the R independently of the input symbol. In this case, the input head is not advanced, and in fact, there need not be any remaining input.
M = (fo, , 92}, {0, 1, c}, {K, B, G}, 6, qu K, 0) S(qu 0, R) = {(4.. BR)} S(qu 1, R) = {(9., GK)} S(qi, 0, B) = {(9.. SB)} <5(9i, 1, B) = {(9., GB)} 6(q it 0, G) = {(?., BG)} 1, G) = {(9., GG)} %!, c, R) = ««2. K» 5(q„ c, B) = {(92, B» <5(<ji, c, G) = {(92, G)} %2, 0, B) = {(92, 0} %2, 1, G) = {(92, <5(<?2, U R) = {(92, £)} Fig. 5.2 Formal pushdown automaton accepting {wcwR \ w in (0 + 1)} by empty stack.
Instantaneous descriptions To formally describe the configuration of a PDA at a given instant we define an instantaneous description (ID). The ID must, of course, record the state and stack contents. However, we find it useful to include the "unexpended input" as well.
Thus we define an ID to be a triple (q, w, y), where q is a state, w a string of input symbols, and y a string of stack symbols. IfM = (Q, Z, T, 3, q0 , Z0 , F) is a PDA, we say (q, aw, Za) \jf (p, w, Pa) if 3(q, a, Z) contains (p, 0). Note that a may be £ or an input symbol. For example, in the PDA of Fig. 5.2, the fact that (q u BG) is in 3(q u 0, G) tells us that (q u 011, GGR) (— (<?!, 11, BGGR).
We use (~ for the reflexive and transitive closure of f^-. That is, / p 1- / for each ID /, and / \~ J and J K imply / (£- X. We write / p- K if ID / can become K after exactly / moves. The subscript is dropped from f^-, f^, and whenever the particular PDA M is understood.
KTUNOTES.IN Downloaded from Ktunotes.in 112 PUSHDOWN AUTOMATA Accepted languages For PDA M = (Q, L, T, 3, q0 , Z0 , F) we define L(M), the language accepted by final state, to be {w | (tfo> w> Z0 ) Pt (p> y) for some p in F and y in T}.
We define N(M), the language accepted by empty stack (or null stack) to be {w | (q0 , w, Z0 ) P 1- (p, e, e) for some p in Q}.
When acceptance is by empty stack, the set of final states is irrelevant, and, in this case, we usually let the set of final states be the empty set.
Example 5.2 Figure 5.3 gives a PDA that accepts {wwR | w in (0 + 1)}. Rules (1) through (6) allow M to store the input on the stack. In rules (3) and (6), M has a choice of two moves. M may decide that the middle of the input string has been reached and make the second choice: M goes to state q2 and tries to match the remaining input symbols with the contents of the stack. IfM guessed right, and if the input is of the form wwR , then the inputs will match, M will empty its stack and thus accept the input string.
M = (fai, q 2\ {0, 1}, {R, B, G}, 5, q l9 R, 0) 1) 5(qi , 0, R) = {(«?., BR)} 6) %„ 1,G) = {(«.. GG), (q 2 , £)} 2) %„ 1, R) = {(<?.. GR)) 7) d(q2 , 0, B) = {(92, 0} 3) 5(qi , 0, B) = {(<?„ BB\ (q 2 , €)} 8) 6(q2 , 1, G) = {(92, 0} 4) 5(qu 0,G) = {(«». BG» 9) €, R) = {(92, 0} 5) 5{qt, l,B)~ GB)} 10) % 2 , £ , R) = {(92, £)} Fig. 5.3 A nondeterministic PDA that accepts {wwR \ w in (0 4- 1)} by empty stack.
Like the nondeterministic finite automaton, a nondeterministic PDA M accepts an input if any sequence of choices causes M to empty its stack. Thus M always "guesses right," because wrong guesses, in themselves, do not cause an input to be rejected. An input is rejected only if there is no "right guess." Figure 5.4 shows the accessible ID's of M when M processes the string 001100.
Deterministic PDA's The PDA of Example 5.1 is deterministic in the sense that at most one move is possible from any ID. Formally, we say that a PDA M = (Q, I, T, 5, q0 , Z0 , F), is KTUNOTES.IN Downloaded from Ktunotes.in 5.2 | DEFINITIONS 113 Initial I (q ly 001 100, R) -> (q2y 001 100, e) i (q ly 01100, BR) i (qu 1 100, BBR) (£, 1 100, R) -> (<?2 , 1 100, e) i hi, 100, GBBR) I (qu 00, GGBBR) (q2 , 00, BBK) 1 !
(qi, 0, BGGBBR) fa2 , 0, BR) I (q u e, BBGGBBR) (ql e, GGBBR) {q 2 , c, R) - (q2 , t, c) I Accept Fig. 5.4 Accessible ID's for the PDA of Fig. 5.3 with input 001100.
deterministic if: 1) for each q in Q and Z in T, whenever 3(qy £> Z) is nonempty, then S(q, a, Z) is empty for all a in Z; 2) for no q in Q, Z in T, and a in Z u {c} does a, Z) contain more than one element.
Condition 1 prevents the possibility of a choice between a move independent of the input symbol (£-move) and a move involving an input symbol. Condition 2 prevents a choice of move for any (q, ay Z) or (qy £, Z). Note that unlike the finite automaton, a PDA is assumed to be nondeterministic unless we state otherwise.
For finite automata, the deterministic and nondeterministic models were equivalent with respect to the languages accepted. The same is not true for PDA.
In fact wwR is accepted by a nondeterministic PDA, but not by any deterministic PDA.
KTUNOTES.IN Downloaded from Ktunotes.in 114 PUSHDOWN AUTOMATA 5.3 PUSHDOWN AUTOMATA AND CONTEXT-FREE LANGUAGES We shall now prove the fundamental result that the class of languages accepted by PDA's is precisely the class of context-free languages. We first show that the languages accepted by PDA's by final state are exactly the languages accepted by PDA's by empty stack. We then show that the languages accepted by empty stack are exactly the context-free languages.
Equivalence of acceptance by final state and empty stack Theorem 5.1 If L is L(M 2 ) for some PDA M 2 , then L is N(M X ) for some PDA, Proof In brief, we would like Mj to simulate M 2 , with the option for M x to erase its stack whenever M 2 enters a final state. We use state qe ofM x to erase the stack, and we use a bottom of stack marker I0 for M 1} so Mj does not accidentally accept ifM 2 empties its stack without entering a final state. Let M 2 = (Q, X, T, 5, q0 , Z0 , F) be a PDA such that L = L(M 2 ). Let M x = (Q u {q„ q'ol Z, T u {X0 } y 5\ q' 0y X0 , 0), where 3' is defined as follows.
1) Sy 09 £, X0 ) = {(q0j Z0 X 0 )}.
2) 3'(q, a, Z) includes the elements of 3(q, a, Z) for all q in Q, a in E or a = £, and Z in r.
3) For all q in F, and Z in T u {X0 }, ^'(<y, £, Z) contains (^e , 6).
4) For all ZinTu {A^}, d'(qe , e, Z) contains (qey e).
Rule (1) causes M x to enter the initial ID ofM 2 , except that M x will have its own bottom of the stack marker X0 , which is below the symbols of M 2's stack.
Rule (2) allows M 1 to simulate M 2 . Should M 2 ever enter a final state, rules (3) and (4) allow M x the choice of entering state qe and erasing its stack, thereby accepting the input, or continuing to simulate M 2 . One should note that M 2 may possibly erase its entire stack for some input jc not in L(M 2 ). This is the reason that Mj has its own special bottom-of-stack marker. Otherwise M x , in simulating M 2 , would also erase its entire stack, thereby accepting x when it should not.
Let x be in L(M 2 ). Then (q0 , x, Z0 ) |^ (q, c, y) for some q in F. Now consider Mj with input x. By rule (1), (<?0 , x, X0 )rM<7o> x, Z0 X0 )y By rule (2), every move of M 2 is a legal move for M u thus (<7o> x, Z0 ) (^, £, y).
KTUNOTES.IN Downloaded from Ktunotes.in 5.3 | PUSHDOWN AUTOMATA AND CONTEXT-FREE LANGUAGES 115 If a PDA can make a sequence of moves from a given ID, it can make the same sequence of moves from any ID obtained from the first by inserting a fixed string of stack symbols below the original stack contents. Thus (q'o, x, X0 ) (ft, x, Z0X0 ) £ (q9 e 9 yX0 ).
By rules (3) and (4), fa> ^V^oJfcSifae. £> 4 Therefore, fao> o) fae.
4 and Mj accepts x by empty stack.
Conversely, if Mj accepts x by empty stack, it is easy to show that the sequence of moves must be one move by rule (1), then a sequence of moves by rule (2) in which M x simulates acceptance of x by M 2 , followed by the erasure ofM/s stack using rules (3) and (4). Thus x must be in L(M 2 ).
Theorem 5.2 If L is N(M X ) for some PDA M l9 then L is L(M 2 ) for some PDA M2 .
Proof Our plan now is to have M 2 simulate M x and detect when M x empties its stack. M 2 enters a final state when and only when this occurs. Let M x = (Q, I, T, <5> <?o, Z09 0) be a PDA such that L = N(M t ). Let M 2 = (Qkj {q' 0 , qfl I, T u {<>}, <5', ^, X0 , {<?,-}), where <5' is defined as follows: 1) Sy 09 6, 0 ) = {fa0 , z0 ^0 )}.
2) For all in Q9 a in £ u {e} 9 and Z in T, d'(q, a9 Z) = %, a9 Z).
3) For all q in g, <5'(<?, £, X0 ) contains (qf9 e).
Rule (1) causes M 2 to enter the initial ID ofM l9 except that M 2 will have its own bottom-of-stack marker X0 , which is below the symbols of M/s stack. Rule (2) allows M 2 to simulate M x . Should M, ever erase its entire stack, then M 2 , when simulating M l9 will erase its entire stack except the symbol X 0 at the bottom. Rule (3) causes M 2 , when the X0 appears, to enter a final state, thereby accepting the input x. The proof that L(M 2 ) = N(M X ) is similar to the proof of Theorem 5.1 and is left as an exercise.
Equivalence of PDA's and CFL's Theorem 5.3 If L is a context-free language, then there exists a PDA M such that L = N(M).
KTUNOTES.IN Downloaded from Ktunotes.in 116 PUSHDOWN AUTOMATA Proof We assume that e is not in L(G). The reader may modify the construction for the case where e is in L(G). Let G = (V, T, P, S) be a context-free grammar in Greibach normal form generating L. Let M = ({q} 9 T, V, <5, q, S, 0), where 5(qy a, A) contains (qy y) whenever A -> ay is in P.
The PDA M simulates leftmost derivations of G. Since G is in Greibach normal form, each sentential form in a leftmost derivation consists of a string of terminals x followed by a string of variables a. M stores the suffix a of the left sentential form on its stack after processing the prefix x. Formally we show that S ^> xol by a leftmost derivation if and only if (q, x, S) (q, £, a).
(5.1) First we suppose that (q, x, S) p- (qy £, a) and show by induction on i that S ^> xa. The basis, i = 0, is trivial since x = c and a = S. For the induction, sup-pose i > 1, and let x = ya. Consider the next-to-last step, bJfl,S)^fe flJ)h(^,a).
(5.2) If we remove a from the end of the input string in the first i ID's of the sequence (5.2), we discover that (q, y, S) \ — 1 (q, £, /?), since a can have no effect on the moves of M until it is actually consumed from the input. By the inductive hypoth-esis S ^> yfl. The move (q, a, P) |— (q, £, a) implies that P = Ay for some A in K, A -> arj is a production of G and a = 777. Hence 5 ^> yP => yarjy = xa, and we conclude the "if" portion of (5.1).
Now suppose that 5 ±> xol by a leftmost derivation. We show by induction on i that (q, x, 5) f-^ (q, c, a). The basis, i = 0, is again trivial. Let i > 1 and suppose S '=> y^y => yaf/y, where x = ya and a = rjy. By the inductive hypothesis, (q y y, 5) j-^ (q, e, Ay) and thus (q, ya, S)- (q, a, Ay). Since A - ar] is a production, it follows that d(q, a, A) contains (q, rj). Thus (q, x, S)\±{q,a,Ay)-{q, £, a), and the "only if" portion of (5.1) follows.
To conclude the proof, we have only to note that (5.1) with a = e says S ^> x if and only if (q, x, S) (q, £, c). That is, x is in L(G) if and only if x is in N(M).
Theorem 5.4 If L is N(M) for some PDA M, then L is a context-free language.
Proof Let M be the PDA (Q, Z, T, <5, qQy Z0 , 0). Let G = (K, I, P. S) be a context-free grammar where V is the set of objects of the form [q, A, p], q and p in KTUNOTES.IN Downloaded from Ktunotes.in 5.3 | PUSHDOWN AUTOMATA AND CONTEXT-FREE LANGUAGES 117 Q, and A in T, plus the new symbol S. P is the set of productions 1) s -> [<7o> Z0 , 4] f°r each g in Q; 2) [<?, ^, 4m+ 1] -> fl[4i> 5 i> ^Jfe. #2, ^3] ' ' ' [ eacn a in I u {e} y and ,4, B x , B2 , .
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, Bm in T, such that d(qy ay ,4) contains (q l9 B XB2 " Bm \ (If m = 0, then the production is [qy A, q x ]-> a.) To understand the proof it helps to know that the variables and productions of G have been defined in such a way that a leftmost derivation in G of a sentence x is a simulation of the PDA M when fed the input x. In particular, the variables that appear in any step of a leftmost derivation in G correspond to the symbols on the stack ofM at a time when M has seen as much of the input as the grammar has already generated. Put another way, the intention is that [qy A, p] derive x if and only if x causes M to erase an A from its stack by some sequence of moves beginning in state q and ending in state p.
To show that L(G) = N(M), we prove by induction on the number of steps in a derivation of G or number of moves of M that [q, A,p]§>x if and only if (qy x, A) (p, £, e\ (5.3) First we show by induction on i that if (q y x, A) f-^- (p, £, e\ then [qy A, p] ^> x. If i = 1, then S(qy x, A) must contain (p, e). (Here x is 6 or a single input symbol.) Thus [q, A, p] -> x is a production of G.
Now suppose 1 > 1. Let x = ay and (<?, fly, >i) - (q l9 y,B x B 2 " Bn ) (p, £, £).
The string y can be written y = y x y 2 " ' y„, where y7 has the effect of popping B} from the stack, possibly after a long sequence of moves. That is, let y x be the prefix of y at the end of which the stack first becomes as short as n -1 symbols. Let y2 be the symbols of y following y x such that at the end of y2 the stack first becomes as short as n — 2 symbols, and so on. The arrangement is shown in Fig. 5.5. Note that B x need not be the nth stack symbol from the bottom during the entire time y x is being read by M, since B x may be changed if it is at the top of stack and is replaced by one or more symbols. However, none of B 2 B 3 • • • Bn are ever at the top while y x is being read and so cannot be changed or influence the computation. In general, Bj remains on the stack unchanged while y x y 2 • • • y7 _ x is read.
There exist states q2y q 2 , qn+1 , where qn + x = p, such that (qj f yjt Bj)+(qj+u ef e) by fewer than i moves (qj is the state entered when the stack first becomes as short as n — j + 1). Thus the inductive hypothesis applies and [qp Bjy qj+l]±>yj for 1<;a[q l9 B, q2][q2 , B 2 , q3] ~ [qn9 Bn9 qn+l ] y so [q, A, p]^>ay l y2 --' y„ = x.
Now suppose [q, A, p] ±> x. We show by induction on i that (q, x, A) \±- (p, £, e). The basis, i = 1, is immediate, since [q9 A, p] -+ x must be a production of G and therefore 3(q, x, A) must contain (p, £). Note x is £ or in L here.
For the induction, suppose [q9 A, p]=>a[q l9 B l9 q 2} ~ [qn9 Bn9 qn+1Y^x9 where qn+l = p. Then we may write x = ax x x 2 • • • x„, where gJ+ x ] i> x} for 1 < 7 < m, with each derivation taking fewer than i steps. By the inductive hypoth-esis, (qj9 Xj, Bj) (qj+ !, 6, e) for 1 < j < n. If we insert Bj+ j • • Bn at the bottom of each stack in the above sequence of ID's we see that (qj9 xj9 BjBj+ 1 • • • Bn ) - (qj+ l9 £, Bj+ 1 - Bn ).
(5.4) From the first step in the derivation of x from [q9 A 9 p] we know that (q, x, A) [— (q l9 x l x2 "Xn9 B l B2 " Bn ) is a legal move of M, so from this move and (5.4) forj= 1, 2, .
.
. , n9 (q9 x, A) \^ (p, £, £) follows.
KTUNOTES.IN Downloaded from Ktunotes.in 5.3 | PUSHDOWN AUTOMATA AND CONTEXT-FREE LANGUAGES 119 The proof concludes with the observation that (5.3) with q = q0 and A = Z0 says [q0 , Z0 , p] > x if and only if (q0 , x, Z0 ) [±- (p, £, e).
This observation, together with rule (1) of the construction of G, says that S^>x if and only if (q0 , x, Z0 ) p 1- (p, £, c) for some state p.
That is, x is in L(G) if and only if x is in N(M).
Example 5.3 Let M = ({flo, {0, 1}, {X, Z0 }, (5, 9ot Z0 , 0), where £ is given by %>, 0, Z0 ) = {(<Zo, ^Z0 », %x, 1, X) = {fa lf £)}, S(q09 0, X) = {(q0i XX% S(q l9 £, X) = {fo lf £)}, <5(g0 , 1, X) = {(tfi, 0}' <5( £ > zo) = {(<7i> e)}-To construct a CFG G = (V, T, P, S) generating N(M) let ^ = {S, [q0 , X, q0], [q0 , X, q x\ [q l9 X, q0\ [<?„ Xy q x\ [q0 , Z0 , q0\ [q0 , Z0 , q x\ [q ly Z0 , q0\ [q l9 Z0 , gj} and T = {0, 1}. To construct the set of productions easily, we must realize that some variables may not appear in any derivation starting from the symbol S.
Thus, we can save some effort if we start with the productions for 5, then add productions only for those variables that appear on the right of some production already in the set. The productions for S are S -> [q0 , Z0 , q0] S -> [q0j Z0 , q Y ] Next we add productions for the variable [q0 , Z0 , q0\ These are [<?o, Z0 , q0] -> 0[qOy X, q0][q0 y Z0 , q0] [<?o> z0 , Qo] -> %o> X, qi][q l9 Z0 , q0] These productions are required by d(q0y 0, Z0 ) = {(q0 , XZ0 )}. Next, the produc-tions for [q0 , Z0 , qi ] are [<?o> Z0 , q x ] -> 0[q0 , X, q0][q0 , Z0 , q x ] [ Z0 , q x ] -> 0[q0 , X, q x ][q lr Z0 , q x ] KTUNOTES.IN Downloaded from Ktunotes.in 120 PUSHDOWN AUTOMATA These are also required by S(q0 , 0, Z0 ) = {(q0 , XZ0 )}. The productions for the remaining variables and the relevant moves of the PDA are: 1) [q0 , X, q0] ->0[g0 , X, q0][q0 , X, q0] [q0 , X y q0] ->0[g0 , X, q x][q^ X, q0] [<7o> X, qj -> 0[<3f0 , X y qQ][q0 , Xy q~] [0[g0 , X, qi[q u X, q t ] since 5{q0 , 0, X) = {{q09 XX)}.
2) [q0 , X, q t ] -> 1 since 5(q0 , 1, X) = {{q ly c)}.
3) fo lf Z0 , qi ] -> 6 since ^ £, Z0 ) = £)}.
4) foi, - £ since £, X) = {(^ £)}.
5) bi, X, qi ] -> 1 since S(qu 1, X) = {(^ £)}.
It should be noted that there are no productions for the variables [q l9 X, q0] and [q l9 Z0 , q0 ]. As all productions for [q0 , X, q0] and [q0y Z0 , q0] have [g l5 X, g0] or [q l9 Z0 , g0] on the right, no terminal string can be derived from [q0 , X y q0] or [q0 , Z0 , q0] either. Deleting all productions involving one of these four variables on either the right or left, we end up with the following productions.
S [q0 , Z0 , q x \ [q l9 Z0 , q x ] -+ £, [<?o> Z0 , q x ] 0[g0 , A", 4i][<?i, Z0 , <7i]> 4i] -> ^ [<?„ q x ] 0[(?0 , X, gjfo,, X, <?,], [<?„ X, -> 1.
bo, 4i] -> 1.
We summarize Theorems 5.1 through 5.4 as follows. The three statements below are equivalent: 1) L is a context-free language.
2) L=N(M X ) for some PDA M x .
3) L = L(M 2 ) for some PDA M 2 .
EXERCISES 5.1 Construct pushdown automata for each of the languages in Exercise 4.1.
5.2 Construct a PDA equivalent to the following grammar.
S-+aAA, A-+aS\bS\a.
5.3 Complete the proof of Theorem 5.3 by showing that every CFL L is the set accepted by some PDA even if e is in L. [Hint: Add a second state to the PDA for L — {e}.] 5.4 Show that if L is a CFL, then there is a PDA M accepting L by final state such that M has at most two states and makes no amoves.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 121 5.5 a) Show that ifL is a CFL, then L is L(M) for some PDA M such that if S(q, a, X) contains (p,y), then |y| < 2.
b) Show that M of part (a) can be further restricted so that if S(qy a, X) contains (p, y), then y is either e (a pop move), X (no change to the stack), or YX for some stack symbol Y (a push move).
c) Can we put a bound on the number of states ofM in part (a) and still have a PDA for any CFL?
d) Can we put a bound on the number of states in part (b)?
5.6 Give a grammar for the language N(M) where M = (fe0 , <,}, {0, 1}, {Z0 , X}, 6, ft, Z0 , 0 ) and 5 is given by <5(<7o, 1, Z0 ) = {(<7o, Z0 )}, <5(<?o, £ > zo) = {(<?o, )}> %0 , i, x) = {foo, L ) = (toi. 5(9o, 0, X) = {fo lf X)}, 0, Z0 ) = {toe Zo)}.
5.7 The deterministic PDA (DPDA) is not equivalent to the nondeterministic PDA. For example, the language L={0 n \ n \n> 1} u {0"l 2n \n> 1} is a CFL that is not accepted by any DPDA.
a) Show that L is a CFL.
b) Prove that L is not accepted by a DPDA.
5.8 A language L is said to have the prefix property if no word in L is a proper prefix of another word in L. Show that if L is N(M) for DPDA M, then L has the prefix property. Is the foregoing necessarily true if L is N(M) for a nondeterministic PDA M?
5.9 Show that L is N(M) for some DPDA M if and only if L is L(M') for some DPDA M\ and L has the prefix property.
5.10 A two-way PDA (2PDA) is a PDA that is permitted to move either way on its input.
Like the two-way FA, it accepts by moving off the right end of its input in a final state.
Show that L = {0T2 n | n > 1} is accepted by a 2PDA. We shall show in the next chapter that L is not a CFL, by the way, so 2 PDA's are not equivalent to PDA's.
S 5.11 Write a program to translate a regular expression to a finite automaton.
5.12 The grammar E -+ E + E\E E|(E)|id (5.5) generates the set of arithmetic expressions with +, , parentheses and id in infix notation (operator between the operands). The grammar P-+ +PP|PP|id generates the set of arithmetic expressions in prefix notation (operator precedes the oper-ands). Construct a program to translate arithmetic expressions from infix to prefix notation KTUNOTES.IN Downloaded from Ktunotes.in 122 PUSHDOWN AUTOMATA using the following technique. Design a deterministic PDA that parses an infix expression according to the grammar in (5.5). For each vertex in the parse tree determine the necessary action to produce the desired prefix expression. [Hint: See the solution to Exercise 5.11.] 5.13 Construct a compiler for infix arithmetic expressions that produces an assembly language program to evaluate the expression. Assume the assembly language has the single address instructions: LOAD x (copy x to accumulator), ADD x (add x to accumulator), MULT x (multiply contents of the accumulator by x) and STO x (store the contents of the accumulator in x).
Solutions to Selected Exercises 5.11 Writing a program to translate a regular expression to a finite automaton can be thought of as constructing a rudimentary compiler. We have already seen (Theorem 2.3) that finite automata accepting 0, 6, 0, and 1 can be combined to obtain an automaton equivalent to a given regular expression. The only problem is parsing the regular expression to determine the order in which to combine the automata.
Our first step is to construct a CFG for the set of regular expressions. The next step is to write a parser and finally the automaton-generating routines.
A grammar for regular expressions that groups subexpressions according to the con-» ventional precedence of operations is given below. Note that € is used for the symbol e.
E-+P + E\P P-+T- P\T r-o|i|c|0|r|(E) The parsing routine is constructed directly from the grammar by writing a procedure for each variable. A global variable STRING initially contains the following regular expression.
procedure FIND_EXPRESSION; begin FINDJPRODUCT; while first symbol of STRING is + do begin delete first symbol of STRING; FINDJPRODUCT end; end FIND_EXPRESSION; procedure FINDJPRODUCT; begin FIND_TERM; while first symbol of STRING is do begin delete first symbol of STRING; FINDJTERM end end FINDJPRODUCT; KTUNOTES.IN Downloaded from Ktunotes.in BIBLIOGRAPHIC NOTES 123 procedure FIND_TERM; begin if first symbol of STRING is 0, 1, e, or 0 then delete first symbol of STRING; else if first symbol of STRING is ( then begin delete first symbol of STRING; FIND_EXPRESSION; if first symbol of STRING is ) then delete first symbol of STRING else error end while first symbol of STRING is do delete first symbol of STRING end FIND_TERM The actual parsing program consists of a single procedure call: FIND_EXPRESSION; Note that the recursive procedures FINDJEXPRESSION, FIND_PRODUCT, and FIND_TERM have no local variables. Thus they may be implemented by a stack that pushes £, P, or T, respectively, when a procedure is called, and pops the symbol when the procedure returns. (Although FIND.EXPRESSION has two calls to FIND_PRODUCT, both calls return to the same point in FIND_EXPRESSION. Thus the return location need not be stored. Similar comments apply to FIND_PRODUCT). Thus, a deterministic PDA suffices to execute the program we have defined.
Having developed a procedure to parse a regular expression, we now add statements to output a finite automaton. Each procedure is modified to return a finite automaton. In procedure FIND_TERM, if the input symbol is 0, 1, e, or 0, a finite automaton accepting 0, 1, c, or 0 is created and FIND_TERM returns this automaton. If the input symbol is (, then the finite automaton returned by FINDJEXPRESSION is the value of FIND_TERM. In either case, if the while loop for is executed, the automaton is modified to accept the closure.
In procedure FIND_PRODUCT, the value of FIND_PRODUCT is assigned the value of the first call of FIND_TERM. Each time the "while" statement is executed, the value of FIND_PRODUCT is set to an automaton accepting the concatenation of the sets accepted by the current value of FIND_PRODUCT and the automaton returned by the call to FIND_TERM in the "while" loop.
Similar statements are added to the procedure FINDJEXPRESSION.
BIBLIOGRAPHIC NOTES The pushdown automaton appears as a formal construction in Oettinger and Schut-zenberger .
Its equivalence to context-free grammars was perceived by Chomsky and Evey .
A variety of similar devices have been studied. Counter machines have only one push-down symbol, with the exception of a bottom-of-stack marker. They are discussed in KTUNOTES.IN Downloaded from Ktunotes.in 124 PUSHDOWN AUTOMATA Fischer , and Fischer, Meyer, and Rosenberg ; see also the bibliographic notes to Chapter 7. Pushdown transducers are PDA's that may print symbols at each move. They have been studied by Evey , Fischer , Ginsburg and Rose , Ginsburg and Greibach [1966b], and Lewis and Stearns .
The two-way PDA mentioned in Exercise 5.10 has been studied by Hartmanis, Lewis, and Stearns . Its closure properties were considered by Gray, Harrison, and Ibarra , and characterizations of the class of languages accepted by the deterministic (2DPDA) and nondeterministic (2NPDA) varieties have been given by Aho, Hopcroft, and Ullman , and Cook [1971c]. The latter contains the remarkable result that any language accepted by a 2DPDA is recognizable in linear time on a computer. Thus, the existence of a CFL requiring more than linear time to recognize on a computer, would imply that there are CFL's not accepted by 2DPDA's. However, no one to date has proved that such a language exists. Incidentally, the language {(yi n2 n \n > 1} is an example of a non-CFL accepted by a 2DPDA.
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 6 PROPERTIES OF CONTEXT-FREE LANGUAGES To a large extent this chapter parallels Chapter 3. We shall first give a pumping lemma for context-free languages and use it to show that certain languages are not context free. We then consider closure properties of CFL's and finally we give algorithms to answer certain questions about CFL's.
6.1 THE PUMPING LEMMA FOR CFL's The pumping lemma for regular sets states that every sufficiently long string in a regular set contains a short substring that can be pumped. That is, inserting as many copies of the substring as we like always yields a string in the regular set.
The pumping lemma for CFL's states that there are always two short substrings close together that can be repeated, both the same number of times, as often as we like. The formal statement of the pumping lemma is as follows.
Lemma 6.1 (The pumping lemma for context-free languages). Let L be any CFL.
Then there is a constant n, depending only on L, such that if z is in L and \z\ >n, then we may write z = uvwxy such that 1) \vx\ >1, 2) | vwx | < n, and 3) for all i > 0 wi/wx> is in L.
Proof Let G be a Chomsky normal-form grammar generating L — {e}. Observe that if z is in Ufi) and z is long, then any parse tree for z must contain a long path.
More precisely, we show by induction on i that if the parse tree of a word 125 KTUNOTES.IN Downloaded from Ktunotes.in 126 PROPERTIES OF CONTEXT-FREE LANGUAGES generated by a Chomsky normal-form grammar has no path of length greater than i, then the word is of length no greater than 2 l ~ . The basis, i = 1, is trivial, since the tree must be of the form shown in Fig. 6.1(a). For the induction step, let i > 1. Let the root and its sons be as shown in Fig. 6.1(b). If there are no paths of length greater than i — 1 in trees T x and T 2 , then the trees generate words of 21-2 or fewer symbols. Thus the entire tree generates a word no longer than 2i_1 s .
s a A B A A (a) (b) Fig. 6.1 Parse trees.
Let G have k variables and let n = 2. If z is in L(G) and \z\ > w, then since | z | > 2 k ~ f , any parse tree for z must have a path of length at least k + 1. But such a path has at least k + 2 vertices, all but the last of which are labeled by variables.
Thus there must be some variable that appears twice on the path.
We can in fact say more. Some variable must appear twice near the bottom of the path. In particular, let P be a path that is as long or longer than any path in the tree. Then there must be two vertices v x and v2 on the path satisfying the following conditions.
1) The vertices v x and v2 both have the same label, say A.
2) Vertex v x is closer to the root than vertex v 2 .
3) The portion of the path from v x to the leaf is of length at most k -h 1.
To see that v x and v2 can always be found, just proceed up path P from the leaf, keeping track of the labels encountered. Of the first k + 2 vertices, only the leaf has a terminal label. The remaining k + 1 vertices cannot have distinct vari-able labels.
Now the subtree T x with root v 1 represents the derivation of a subword of length at most 2. This is true because there can be no path in T x of length greater than k -h 1, since P was a path of longest length in the entire tree. Let z x be the yield of the subtree T 1 .
If T 2 is the subtree generated by vertex v 2 , and z2 is the yield of the subtree T 2 , then we can write z x as z3 z2 z4 . Furthermore, z3 and z4 cannot both be £, since the first production used in the derivation of z x must be of the form A^BC for some variables B and C. The subtree T 2 must be completely within either the subtree generated by B or the subtree generated by C. The above is illustrated in Fig. 6.2.
KTUNOTES.IN Downloaded from Ktunotes.in 6.1 | THE PUMPING LEMMA FOR CFL'S 127 Zj = bba z = bbbaba (a) (c) Zj = z3 z2 z4 , where z3 = bb and z4 = e G = ({/I, 5, C}, {a, 6}, {/i-£C B-+BA. C-+BA, A-a. B-+b},A) Fig.
6.2 Illustration of subtrees T, and T 2 of Lemma 6.1.
(a) Tree, (b) Subtree T x .
(c) Subtree T 2 .
We now know that A=?z3 Az± and A^>z2 , where | z3 z2 z4 1 < 2 = n.
But it follows that /I => z l 3 z 2 z4 for each i > 0. (See Fig.
6.3.) The string z can clearly be written as uz 3 z2 z4 for some u and y. We let z3 = y, z2 = w, and z4 = x, to complete the proof.
Applications of the pumping lemma The pumping lemma can be used to prove a variety of languages not to be context free, using the same "adversary" argument as for the regular set pumping lemma.
Example 6.1 Consider the language L x = {aVc 1 \ i > 1}. Suppose L were context free and let n be the constant of Lemma 6.1. Consider z = a"b"c". Write z = uvwxy so as to satisfy the conditions of the pumping lemma. We must ask ourselves where v and x, the strings that get pumped, could lie in a nb nc". Since | vwx \ < n, it is not possible for vx to contain instances of a's and c's, because the rightmost a is « + 1 positions away from the leftmost c. If v and x consist of cfs only, then wwy (the string wiAvx'y with i = 0) has n b's and n c's but fewer than n a% since \ vx\ > 1.
KTUNOTES.IN Downloaded from Ktunotes.in 128 PROPERTIES OF CONTEXT-FREE LANGUAGES i times \ Fig. 6.3 The derivation of uv'wx'y, where u = 6, v = bb, w = a, x = £, y = ba.
Thus, uwy is not of the form a?b>d. But by the pumping lemma vwy is in L l9 a contradiction.
The cases where u and x consist only of fr's or only of c's are disposed of similarly. If vx has a's and b's, then uwy has more c's than a's or b% and again it is not in Lj. If vx contains b y s and c's, a similar contradiction results. We conclude that L x is not a context-free language.
The pumping lemma can also be used to show that certain languages similar to Lj are not context free. Some examples are {aVd | j > i} and {aW \i <j < k}.
Another type of relationship that CFG's cannot enforce is illustrated in the next example.
Example 6.2 Let L2 = {aWtfd \ i > 1 and j > 1}. Suppose L2 is a CFL, and let n be the constant in Lemma 6.1. Consider the string z = aWcfd' 1 . Let z = uvwxy satisfy the conditions of the pumping lemma. Then as | vwx \ 0, uvwx?y is in L Proof Let G be a Chomsky normal-form grammar generating L - {e}. Let G have k variables and choose n = 2 + 1. We must construct a path P in the tree analogous to path P in the proof of the pumping lemma. However, since we worry only about distinguished positions here, we cannot concern ourselves with every vertex along P, but only with branch points, which are vertices both of whose sons have distinguished descendants.
Construct P as follows. Begin by putting the root on path P. Suppose r is the last vertex placed on P. If r is a leaf, we end. If r has only one son with distin-guished descendants, add that son to P and repeat the process there. If both sons of r have distinguished descendants, call r a branch point and add the son with the larger number of distinguished descendants to P (break a tie arbitrarily). This process is illustrated in Fig. 6.4.
It follows that each branch point on P has at least half as many distinguished descendants as the previous branch point. Since there are at least n distinguished positions in z, and all of these are descendants of the root, it follows that there are at least k + 1 branch points on P. Thus among the last k + 1 branch points are two with the same label. We may select v Y and v2 to be two of these branch points with the same label and with v x closer to the root than u2 . The proof then proceeds exactly as for the pumping lemma.
KTUNOTES.IN Downloaded from Ktunotes.in 130 PROPERTIES OF CONTEXT-FREE LANGUAGES Fig. 6.4 The path P. Distinguished positions are marked x. Branch points are marked b.
Example 6.3 Let L4 = {a'Vc | i =f jy j =fc k and i fc}. Suppose L4 were a context-free language. Let n be the constant in Ogden's lemma and consider the string z = anbn+n>cn+2n> . Let the positions of the as be distinguished and let z = uvwxy satisfy the conditions of Ogden's lemma.
If either v or x contains two distinct symbols, then uv 2wx 2y is not in L4 . (For example, if v is in a + b + 9 then uv2wx2y has a b preceding an a.) Now at least one of v and x must contain as since only a y s are in distinguished positions. Thus, if x is in b or c, v must be in a + . If x is in a + 9 then v must be in a, otherwise a b or c would precede an a. We consider in detail the situation where x is in b. The other cases are handled similarly. Suppose x is in b and v in a + . Let p = \v. Then 1 < p < n, so p divides n ! Let q be the integer such that pq = n \ Then z' = «i;2 +1 wx 2 +1y is in L 4 . But u2 +1 = a2pq+p = a2n!+p . Since wwy contains exactly (n - p) as, z' has (2n! + n) a's. However, since v and x have no c's, z' also has (2n! + n) c's and hence is not in L 4 , a contradiction. A similar contradiction occurs if x is in a + or c. Thus L 4 is not a context-free language.
Note that Lemma 6.1 is a special case of Ogden's lemma in which all positions are distinguished.
6.2 CLOSURE PROPERTIES OF CFL's We now consider some operations that preserve context-free languages. The oper-ations are useful not only in constructing or proving that certain languages are context free, but also in proving certain languages not to be context free. A given language L can be shown not to be context free by constructing from L a language that is not context free using only operations preserving CFL's.
KTUNOTES.IN Downloaded from Ktunotes.in 6.2 | CLOSURE PROPERTIES OF CFL'S 131 Theorem 6.1 Context-free languages are closed under union, concatenation and Kleene closure.
Proof Let Lx and L2 be CFL's generated by the CFG's = (V l9 T l9 P l9 SJ and G2 = (V 29 T 2 , P2 , S2 ), respectively. Since we may rename variables at will without changing the language generated, we assume that V x and V 2 are disjoint. Assume also that S3 , S4 , and S 5 are not in K x or V 2 .
For L x u L2 construct grammar G3 = (V x u K 2 u {S 3 }, 7i u r 2 , P3 , S 3 ), where P3 is P x u P 2 plus the productions S3 - S x \ S2 .
If w is in Lj, then the derivation S3 ^>S 1 g>w is a derivation in G3 , as every production of G x is a production of 6 3 . Similarly, every word in L2 has a derivation in G 3 beginning with 5 3 => 5 2 . Thus ^ uL2 9 L(G 3 ). [For the converse, let w be in L(G 3 ). Then the derivation S3 =>w begins with either S3 ^S 1 ^>w or S3 =>S 2 |>w. In the former case, as V x and K 2 are disjoint, only symbols of G t may appear in the derivation 5 X => w. As the only productions of P3 that involve only symbols of G l are those from Pu we conclude that only productions of P t are used in the derivation S^vv. Thus Sj^w, and w is in Lj. Analogously, if the derivation starts S 3 => 52 , we may conclude w is in L2 . Hence L(G 3 ) ^ Lj u L2 , so L(G 3 ) = Lj u L2 , as desired.
For concatenation, let G4 = (Vi u V 2 u {S4}, 7j u 7 2 , P4 , S4 ), where P4 is P x u P 2 plus the production 54 -S 1 S2 . A proof that L(Gt) = L(G X )L(G 2 ) is similar to the proof for union and is omitted.
For closure, let G 5 = (V x u {S5 }, Tu P 5 , 5 5 ), where P 5 is P 1 plus the produc-tions 5 5 -> 5 X 55 1 e. We again leave the proof that L(G 5 ) = L(Gj) to the reader.
Substitution and homomorphisms Theorem 6.2 The context-free languages are closed under substitution.
Proof Let L be a CFL, L c £ and for each a in Z let La be a CFL. Let L be L(G) and for each a in £ let La be L(Gfl ). Without loss of generality assume that the variables of G and the Ga's are disjoint. Construct a grammar G' as follows. The variables of G' are all the variables of G and the Gfl's; the terminals of G' are the terminals of the Ga's. The start symbol of G' is the start symbol of G.
The productions of G' are all the productions of the Gfl's together with those productions formed by taking a production A -> a of G and substituting 5a , the start symbol of Ga , for each instance of an a in £ appearing in a.
Example 6.4 Let L be the set of words with an equal number of as and b% La = {0T|n > 1} and 1+ = {ww | w is in (0 + 2)}. For G we may choose KTUNOTES.IN Downloaded from Ktunotes.in 132 PROPERTIES OF CONTEXT-FREE LANGUAGES For Ga take S fl ^0Sfl l|01 For Gb take Sb ->0Sb0\2Sb2\e If / is the substitution f(a) = La and f(b) = Lj,, then f(L) is generated by the grammar S^SaSSbS\Sb SSa S\e Sa ^0Sal\0l Sb ^0Sb0\2Sb2\e One should observe that since {a, b), {ab} y and a are CFL's, the closure of CFL's under substitution implies closure under union, concatenation, and . The union of La and 1+ is simply the substitution of La and into {a, b} and similarly La Lb and L are the substitutions into {ab} and a, respectively. Thus Theorem 6.1 could be presented as a corollary of Theorem 6.2.
Since a homomorphism is a special type of substitution we state the following corollary.
Corollary The CFL's are closed under homomorphism.
Theorem 6.3 The context-free languages are closed under inverse homomor-phism.
Proof As with regular sets, a machine-based proof for closure under inverse homomorphism is easiest to understand. Let h : £ - A be a homomorphism and L be a CFL. Let L = L(JVf ), where JVf is the PDA (Q, A, T, <5, q0 , Z0 , F). In analogy with the finite-automaton construction of Theorem 3.5, we construct PDA M' accepting h~ 1 (L) as follows. On input a, M ' generates the string h(a) and simulates M on h(a). If JVf were a finite automaton, all it could do on a string h(a) would be to change state, so M' could simulate such a composite move in one of its moves.
However, in the PDA case, M could pop many symbols on a string, or, since it is nondeterministic, make moves that push an arbitrary number of symbols on the stack. Thus JVf cannot necessarily simulate M's moves on h(a) with one (or any finite number of) moves of its own.
What we do is give JVf a buffer, in which it may store h(a). Then JVf may simulate any £-moves of JVf it likes and consume the symbols of h(a) one at a time, as if they were JVfs input. As the buffer is part of JVf's finite control, it cannot be allowed to grow arbitrarily long. We ensure that it does not, by permitting M' to read an input symbol only when the buffer is empty. Thus the buffer holds a suffix of h(a) for some a at all times. JVf accepts its input w if the buffer is empty and JVf is in a final state. That is, JVf has accepted h(w). Thus L(M') = {w | h(w) is in L}, that is KTUNOTES.IN Downloaded from Ktunotes.in 6.2 | CLOSURE PROPERTIES OF CFL'S 133 Input toM' Control of M' Buffer Control ofM Stack of M and M' Fig. 6.5 Construction of a PDA accepting h~ l (L).
L(M') = h~ (L(M )). The arrangement is depicted in Fig: 6.5; the formal construc-tion follows.
Let M' = (Q\ Z, T, S\ [q0 , e], Z0 , F x {e}), where Q' consists of pairs [q, x] such that q is in Q and x is a (not necessarily proper) suffix of some h(a) for a in E.
<5' is defined as follows: 1) d'([q, x], £, 7 ) contains all ([p, x], y) such that S(q, e, 7) contains (p, y). Sim-ulate e-moves of M independent of the buffer contents.
2) S'([q, ax], e, Y) contains all ([p, x], y) such that d(q, a, Y) contains (p, y). Sim-ulate moves of M on input a in A, removing a from the front of the buffer.
3) d'([q, e], a, Y) contains ([q, h(a)], Y) for all a in I and Y in T. Load the buffer with h(a), reading a from M"s input; the state and stack of M remain unchanged.
To show that L(M') = h~ I (L(M)) first observe that by one application of rule (3), followed by applications of rules (1) and (2), if (q, h(a), a) f^- (p, e, ft), then (fa. el a > a ) kr (fa> H<)1 a ) \w ilPf 4 C Ju-nius if M accepts fr(w), that is, (g0f ^Z0)g-(p, 6, fi) for some p in F and /? in r it follows that ([tf0 , 4 w, Z0 )|^([p, £], e, 0), so M' accepts w. Thus L(M') 2 ^ _1 (L(M)).
Conversely, suppose M' accepts w = a l a 1 - an . Then since rule (3) can be applied only with the buffer (second component ofM"s state) empty, the sequence KTUNOTES.IN Downloaded from Ktunotes.in 134 PROPERTIES OF CONTEXT-FREE LANGUAGES of the moves of M' leading to acceptance can be written ([<?o> 4 <i<2 a„> Zo)Isf([Pi» el a i a2 "' <n , «i)> \kf(\jPi, h(ai)l a 2<i3 an,<i\ \w(\j>2, 4 ^2^3 « 2 )> hr ([P2, M«2)], a 3 fl4 « 2 ), where pn is in F. The transitions from state [ph e] to ft^)] are by rule (3), the other transitions are by rules (1) and (2). Thus, (q0 , e, Z0 ) \£f (p l £, o^), and for all i, (pis M0»)>a.)hr(Pi+i> £ > af+i)-Putting these moves together, we have (q0 , h(a x a2 "- an\ Z0 ) \%f (Pn, U «„+ i)> so h(a x a 2 "' an ) is in L(M). Hence L(M') ^ /i _1 (L(M)), whereupon we conclude L{M') = h-1 (L(M)).
" Boolean operations There are several closure properties of regular sets that are not possessed by the context-free languages. Notable among these are closure under intersection and complementation.
Theorem 6.4 The CFL's are not closed under intersection.
Proof In Example 6.1 we showed the language L Y = {ctb l(?\i > 1} was not a CFL. We claim that L2 = {aVd\i > 1 and ; > 1} and L3 = {flW|i > 1 and j > 1} are both CFL's. For example, a PDA to recognize L2 stores the as on its stack and cancels them against b% then accepts its input after seeing one or more c's. Alternatively L2 is generated by the grammar S-+AB A -+ aAb | ab B-+cB\c KTUNOTES.IN Downloaded from Ktunotes.in 6.2 | CLOSURE PROPERTIES OF CFL's 135 where A generates db x and B generates d. A similar grammar S-+CD C -> aC\a D-+bDc\bc generates L3 .
However, L2 n L3 = Lv If the CFL's were closed under intersection, L x would thus be a CFL, contradicting Example 6.1.
Corollary The CFL's are not closed under complementation.
Proof We know the CFL's are closed under union. If they were closed under complementation, they would, by DeMorgan's law, L x n L2 = Li u L 2 be closed under intersection, contradicting Theorem 6.4.
Although the class of CFL's is not closed under intersection it is closed under intersection with a regular set.
Theorem 6.5 If L is a CFL and R is a regular set, then L n R is a CFL.
Proof Let L be L(M) for PDA M = (QMy S, T, <5M , g0 > Zo> FM ), and let R be L(/l) for DFA A = (Q^, Z, ^, p0 , F^). We construct a PDA M' for L n R by "running M and A in parallel," as shown in Fig. 6.6. M' simulates moves of M on input e without changing the state of A. When M' makes a move on input symbol a, M simulates that move and also simulates A's change of state on input a. M' accepts if and only if both A and M accept. Formally, let M' = (QA x QM , Z, T, (5, [p0 , ^0], Z0 , F^ x FM ), Input to /I. A/, and /V/' Z Control Control Control of Ml of/1 of M Stack of M and /V/' Fig. 6.6 Running an FA and a PDA in parallel.
KTUNOTES.IN Downloaded from Ktunotes.in 136 PROPERTIES OF CONTEXT-FREE LANGUAGES where 3 is defined by S([p, q], a, X), contains ([p\ q'\ y) if and only if SA (pt a) = p', and SM (q, a, X) contains (q, y). Note that a may be £, in which case p' = p.
An easy induction on i shows that (bo, qol w, Z0 ) |£r ([p, g], £, y) if and only if (<?0 , w, Z0 ) hjr (g, £, y) and <5(p0 , w) = p.
The basis, i = 0, is trivial, since p = p0 , g = g0 , y = Z0 , and w = £. For the induc-tion, assume the statement for i — 1, and let ([Po» 4o]> <, Z0 ) f-^f- ([p', 4'], a, ^ ([p, 4 £, y), where w = xa, and a is £ or a symbol of Z. By the inductive hypothesis, ) = P' and (q0 , x, Z0 ) |-^- (</, £, £).
By the definition of <5, the fact that ([p', g'], a, /?) (^r ([p, g], £, y) tells us that <Up', a) = P and (q\ a, ft) (<?, e, y). Thus 5A (p0 , w) = p and (q0 , w, Z0 ) [£- (<?> ^ y)-The converse, showing that (q0 , w, Z0 ) \jf (q, t, y) and SA (p0 , w) = p imply (bo, 4o], w, Z0 ) ([p, 4 £, y), is similar and left as an exercise.
Use of closure properties We conclude this section with an example illustrating the use of closure properties of context-free languages to prove that certain languages are not context free.
Example 6.5 Let L = {ww | w is in (a + b)}. That is, L consists of all words whose first and last halves are the same. Suppose L were context free. Then by Theorem 6.5, L x = L n a + b + a + b + would also be a CFL.
But L x = {a ibia ibi \i> Uj> 1}. L 1 is almost the same as the language proved not to be context free in Example 6.2, using the pumping lemma. The same argument shows that L x is not a CFL. We thus contradict the assumption that L is a CFL.
If we did not want to use the pumping lemma on Lu we could reduce it to L2 = {a ibic idi | i > 1 and j > 1}, the exact language discussed in Example 6.2. Let h be the homomorphism h(a) — h(c) = a and h(b) = h(d) — b. Then h~ 1 (L l ) consists of all words of the form x 1 x 2 x 3 x4 , where Xj and x 3 are of the same length and in (a + c) + , and x2 and x4 are of equal length and in (b -I- d) + . Then h' 1^) n abcd = L2 . By Theorems 6.3 and 6.5, if Lx were a CFL, so would be L2 . Since L2 is known not to be a CFL, we conclude that L x is not a CFL.
KTUNOTES.IN Downloaded from Ktunotes.in 6.3 | DECISION ALGORITHMS FOR CFL'S 137 6.3 DECISION ALGORITHMS FOR CFL's There are a number of questions about CFL's we can answer. These include whether a given CFL is empty, finite, or infinite and whether a given word is in a given CFL. There are, however, certain questions about CFL's that no algorithm can answer. These include whether two CFG's are equivalent, whether a CFL is cofinite, whether the complement of a given CFL is also a CFL, and whether a given CFG is ambiguous. In the next two chapters we shall develop tools for showing that no algorithm to do a particular job exists. In Chapter 8 we shall actually prove that the above questions and others have no algorithms. In this chapter we shall content ourselves with giving algorithms for some of the questions that have algorithms.
As with regular sets, we have several representations for CFL's, namely context-free grammars and pushdown automata accepting by empty stack or by final state. As the constructions of Chapter 5 are all effective, an algorithm that uses one representation can be made to work for any of the others. We shall use the CFG representation in this section.
Theorem 6.6 There are algorithms to determine if a CFL is (a) empty, (b) finite, or (c) infinite.
Proof The theorem can be proved by the same technique (Theorem 3.7) as the analogous result for regular sets, by making use of the pumping lemma. However, the resulting algorithms are highly inefficient. Actually, we have already given a better algorithm to test whether a CFL is empty. For a CFG G = (K, 7, P, S), the test of Lemma 4.1 determines if a variable generates any string of terminals.
Clearly, L(G) is nonempty if and only if the start symbol S generates some string of terminals.
To test whether L(G) is finite, use the algorithm of Theorem 4.5 to find a CFG G' = (V, T, F, S) in CNF and with no useless symbols, generating Ufi) - {e}.
L(G') is finite if and only if L(G) is finite. A simple test for finiteness of a CNF grammar with no useless symbols is to draw a directed graph with a vertex for each variable and an edge from A to B if there is a production of the form A -> BC or A -> CB for any C. Then the language generated is finite if and only if this graph has no cycles.
If there is a cycle, say A 0 , A u An , A0 , then A0 =>oc 1 A i p i =xx 2 A 2 P2 - " =>anA n Pn =>ocn+l Ao pn+u where the a's and /Ts are strings of variables, with | a, /?, | = i. Since there are no useless symbols, an+ 1 ^> w and /?„+ 1 ^> x for some terminal strings w and x of total length at least n + 1. Since n > 0, w and x cannot both be e. Next, as there are no useless symbols, we can find terminal strings y and z such that S ^> yA 0 z, and a terminal string v such that A0 ^> v. Then for all i, S yA 0 z ^> ywA 0 xz ^> y 2A0 x 2z ^>--£> yw lA 0 x lz ^> j/wWz.
KTUNOTES.IN Downloaded from Ktunotes.in 138 PROPERTIES OF CONTEXT-FREE LANGUAGES As | wx | > 0, yw lvx lz cannot equal ytfvx'z if i =fc j. Thus the grammar generates an infinite number of strings.
Conversely, suppose the graph has no cycles. Define the rank of a variable A to be the length of the longest path in the graph beginning at A. The absence of cycles implies that the rank of A is finite. We also observe that if A -+ BC is a production, then the rank of B and C must be strictly less than the rank of A, because for every path from B or C, there is a path of length one greater from A.
We show by induction on r that if A has rank r, then no terminal string derived from A has length greater than 2 r .
Basis r — 0.
If A has rank 0, then its vertex has no edges out. Therefore all /l-productions have terminals on the right, and A derives only strings of length 1.
Induction r > 0. If we use a production of the form A -» a, we may derive only a string of length 1. If we begin with A - BC, then as B and C are of rank r — 1 or less, by the inductive hypothesis, they derive only strings of length 2 r ~ 1 or less.
Thus BC cannot derive a string of length greater than 2 r .
Since S is of finite rank r0 , and in fact, is of rank no greater than the number of variables, S derives strings of length no greater than 2 r°. Thus the language is finite.
Example 6.6 Consider the grammar S^AB A-+BC\a B->CC\b C^a whose graph is shown in Fig. 6.7(a). This graph has no cycles. The ranks of S, A, B, and C are 3, 2, 1, and 0, respectively. For example, the longest path from S is S, A, B, C. Thus this grammar derives no string of length greater than 2 3 = 8 and therefore generates a finite language. In fact, a longest string generated from S is S => AB => BCB => CCCB => CCCCC ^> aaaaa.
(a) (b) Fig. 6.7 Graphs corresponding to CNF grammars.
KTUNOTES.IN Downloaded from Ktunotes.in 6.3 | DECISION ALGORITHMS FOR CFL'S 139 If we add production C -> AB, we get the graph of Fig. 6.7(b). This new graph has several cycles, such as A, B, C, A. Thus we can find a derivation A i> BC=> CCC => CABC, where a and BC^>ba, we have /l^a/lta. Then as S£>/lb and A^>ay we now have S ^> a la(ba) lb for every i. Thus the language is infinite.
Membership Another question we may answer is: Given a CFG G = (V,Ty Py S) and string x in T, is x in L(G)? A simple but inefficient algorithm to do so is to convert G to G' = (V, T, F, S), a grammar in Greibach normal form generating L(G) - {e}.
Since the algorithm of Theorem 4.3 tests whether S^>e, we need not concern ourselves with the case x = e. Thus assume x e, so x is in L(G') if and only if x is in L(G). Now, as every production of a GNF grammar adds exactly one terminal to the string being generated, we know that if x has a derivation in G', it has one with exactly | x | steps. Ifno variable of G' has more than k productions, then there are at most /c |x| leftmost derivations of strings of length | x | . We may try them all systematically.
However, the above algorithm can take time which is exponential in | x | .
There are several algorithms known that take time proportional to the cube of |x| or even a little less. The bibliographic notes discuss some of these. We shall here present a simple cubic time algorithm known as the Cocke-Younger-Kasami or CYK algorithm. It is based on the dynamic programming technique discussed in the solution to Exercise 3.23. Given x of length n > 1, and a grammar G, which we may assume is in Chomsky normal form, determine for each i and j and for each variable A, whether A ^> xip where xtj is the substring of x of length j beginning at position i.
We proceed by induction on j. For j = 1, A^> x,7 if and only if A - x0 - is a production, since x i} is a string of length 1. Proceeding to higher values of j, if j > 1, then A ^> x tj if and only if there is some production A -> BC and some /c, 1 x ik and C ^> xi+kJ - k . Since k and j - k are both less than j, we already know whether each of the last two derivations exists. We may thus determine whether A ==> x fi -. Finally, when we reach j = n, we may determine whether S^>x ln . But x ln = x, so x is in Ufi) if and only if S^>x ln .
To state the CYK algorithm precisely, let be the set of variables A such that A ^> Xij. Note that we may assume 1 < i < n — j + 1, for there is no string of length greater than n — i + 1 beginning at position i. Then Fig. 6.8 gives the CYK algorithm formally.
Steps (1) and (2) handle the case j = 1. As the grammar G is fixed, step (2) takes a constant amount of time. Thus steps (1) and (2) take 0(n) time. The nested for-loops of lines (3) and (4) cause steps (5) through (7) to be executed at most n 2 times, since i and j range in their respective for-loops between limits that are at KTUNOTES.IN Downloaded from Ktunotes.in 140 PROPERTIES OF CONTEXT-FREE LANGUAGES begin 1) for i:= 1 to n do 2) V n :={A\A-a is2i production and the ith symbol of x is a}; 3) for j:= 2 to n do 4) for /:= 1 to n — j " + 1 do begin 5) Ky:-0; 6) for k:= 1 toj 1 do 7) V^:= Vij u {A\A-+ BC is a production, B is in and C is in end end Fig. 6.8.
The CYK algorithm.
most n apart. Step (5) takes constant time at each execution, so the aggregate time spent at step (5) is 0(« 2 ). The for-loop of line (6) causes step (7) to be executed n or fewer times. Since step (7) takes constant time, steps (6) and (7) together take 0(«) time. As they are executed 0(« 2 ) times, the total time spent in step (7) is 0(n3 ). Thus the entire algorithm is 0(n 3 ).
Example 6.7 Consider the CFG S-+AB\BC A -> BA | a B^CC\b C-+AB\a and the input string baaba. The table of J^/s is shown in Fig. 6.9. The top row is filled in by steps (1) and (2) of the algorithm in Fig. 6.8. That is, for positions 1 and 4, which are b, we set V x x = K 41 = {B}, since B is the only variable which derives b.
b a a i - b a 1 2 3 4 5 1 B A. C A. C B A. C i i 3 S. A B S. C S, A 0 B B 4 0 S. A. C 5 S, A, C Fig. 6.9 Table of Vjs.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 141 1 Similarly, V 2l = V 31 = V 51 = {A, C}, since only A and C have productions with a on the right.
To compute V i} for j > 1, we must execute the for-loop of steps (6) and (7). We must match V ik against V i+kJ _ k for k = 1, 2, .
.
.
-1, seeking variable D in V ik and £ in such that DE is the right side of one or more productions. The left sides of these productions are adjoined to V u . The pattern in the table which corresponds to visiting V ik and V i+kJ _ k for k = 1, 2, 1 in turn is to simul-taneously move down column i and up the diagonal extending from V V} to the right, as shown in Fig. 6.10.
Fig. 6.10 Traversal pattern for computation of V tJ .
For example, let us compute K 24 , assuming that the top three rows of Fig. 6.9 are filled in. We begin by looking at K 21 = {A, C} and V 33 = {£}. The possible right-hand sides in V 2l K 33 are AB and CB. Only the first of these is actually a right side, and it is a right side of two productions S - AB and C -» AB. Hence we add S and C to K 24 . Next we consider K 22 K 42 = {B}{S9 A} = {BS, BA}. Only BA is a right side, so we add the corresponding left side A to K 24 . Finally, we consider v 23 ^si = {B}{A C} = BC}. BA and £C are each right sides, with left sides A and S, respectively. These are already in K 24 , so we have K 24 = {S> A y C}. Since S is a member of K15 , the string baaba is in the language generated by the grammar.
EXERCISES 6.1 Show that the following are not context-free languages.
a) {flW|i <j <k} b) {a ibj \j = i 2y c) {a l \i is a prime}" d) the set of strings of a's, ^'s, and c's with an equal number of each e) {a"bncm | n < m < 2n} - 6.2 Which of the following are CFL's?
a) {a'Vli^yand i + 2j} b) (a + b) ~{(a nbnY\n> 1} c) {wh^w | w is in (a 4- b)} d) {ftj \bi is i in binary, i > 1} KTUNOTES.IN Downloaded from Ktunotes.in 142 PROPERTIES OF CONTEXT-FREE LANGUAGES e) {wxw | u> and x are in (a + b)} f) (a + b)-{(fl^)"|«> 1} 63 Prove that the following are not CFL's.
a) {aW|y = max {i, k}} b) [tfir<?\i±n} [Hint: Use Ogden's lemma on a string of the form anbn c"'.] 6.4 Show that the CFL's are closed under the following operations: a) Quotient with a regular set, that is, if L is a CFL and R a regular set, then L/R is a CFL.
b) INIT c) CYCLE d) reversal See Exercise 3.4 for the definitions of INIT and CYCLE.
6.5 Show that the CFL's are not closed under the following operations.
$ a) MIN b) MAX c) \ d) Inverse substitution e) INV, where INV(L) = {x \x — wyz and wyRz is in L} MIN, MAX, and i are defined in Exercises 3.4 and 3.16.
6.6 Let X be an alphabet. Define homomorphisms h, h u and h 2 by h(a) = h(a) = a, hi(a) = a, hi(a) = £, h 2 (a) = £, and h 2 (a) = a for each a in £. For L x £ Z and L 2 £ X, define Shuffle (Lj, L 2 ) = {x|for some y in h~ l (x\ h x (y) is in L x and h 2 (y) is in L 2 }.
That is, the Shuffle of L x and L2 is the set of words formed by "shuffling" a word of L x with a word of L2 . Symbols from the two words need not alternate as in a "perfect shuffle." a) Show that the Shuffle of two regular sets is regular.
b) Prove that the Shuffle of two CFL's is not necessarily a CFL.
c) Prove that the Shuffle of a CFL and a regular set is a CFL.
6.7 A Dyck Language is a language with k types of balanced parentheses. Formally, each Dyck language is, for some k y L(Gk ), where Gk is the grammar S — SS|[ I S] I |[2 S] 2 [k S] fc |6.
For example, [lbfiLbh^L is in the Dyck language with two kinds of parentheses. Prove that every CFL L is h(LD n R), where h is a homomorphism, R a regular set, and LD a Dyck language. [Hint : Let L be accepted by empty stack by a PDA in the normal form of Exercise 5.5(b) where the moves only push or pop single symbols. Let the parentheses be [abX and ]abXj where "means" on input a, stack symbol X is pushed, and matching parenthesis "means" on input b, X may be popped (a or b may be e). Then the Dyck language enforces the condition that the stack be handled consistently, i.e., if X is pushed, then it will still be X when it is popped. Let the regular set R enforce the condition that there be a sequence of states for which the push and pop moves are legal for inputs a and b, respectively. Let h([abx ) = a and h(]abx ) = b.] 6.8 Show that if L is a CFL over a one-symbol alphabet, then L is regular. [Hint: Let n be the pumping lemma constant for L and let L ^ 0. Show that for every word of length n or more, say 0", there are p and q no greater than n such that 0p+iq is in L for all i > 0. Then KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 143 show that L consists of perhaps some words of length less than n plus a finite number of linear sets, i.e., sets of the form {0 p+iq \ i > 0} for fixed p and q y q < n.] 6.9 Prove that the set of primes in binary is not a CFL.
6.10 Show that the linear languages (see Exercise 4.20 for a definition) are closed under a) union b) homomorphism c) intersection with a regular set 6.11 Prove the following pumping lemma for linear languages. If L is a linear language, then there is a constant L such that if z in L is of length n or greater, then we may write z = uvwxy, such that | uvxy \ < n, \vx\ > 1, and for all i > 0, w>Wy is in L.
6.12 Show that {aWcW 1 1! > 1 and j > 1} is not a linear language.
6.13 A PDA is said to make a turn if it enters a sequence of ID's (Qu ™i, y t ) |— (q 2 , w2 , y 2 ) |— ($3 , w 3 , y 3 ) and | y 2 1 is strictly greater than | y 1 1 and | y 3 | . That is, a turn occurs when the length of the stack "peaks." A PDA M is said to be a k-turn PDA if for every word w in L(M), w is accepted by a sequence of ID's making no more than k turns. If a PDA is /c-turn for some finite /c, it is said to be finite-turn. If L is accepted by a finite-turn PDA, L is metalinear.
a) Show that a language is linear if and only if it is accepted by a one-turn PDA.
b) Show that the linear languages are closed under inverse homomorphism.
c) Show that the metalinear languages are closed under union, concatenation, homomor-phism, inverse homomorphism, and intersection with a regular set.
6.14 Show that the set of strings with an equal number of as and b's is a CFL that is not a metalinear language.
6.15 Show that a) the linear languages b) the metalinear languages are not closed under .
6.16 Give an algorithm to decide for two sentential forms a and p of a CFG G, whether 6.17 Use the CYK algorithm to determine whether a) aaaaa b) aaaaaa are in the grammar of Example 6.7.
6.18 Let G be a context-free grammar in CNF.
a) Give an algorithm to determine the number of distinct derivations of a string x.
b) Associate a cost with each production of G. Give an algorithm to produce a minimum-cost parse of a string x. The cost of a parse is the sum of the costs of the productions used.
[Hint : Modify the CYK algorithm of Section 6.3.] Solutions to Selected Exercises 6.4 c) Let G = (K, T, P, S) be a CFG in CNF.
To construct G such that L(G) = CYCLE(L(G)) consider a derivation tree of a string x x x 2 in grammar G. Follow the path from S to the leftmost symbol of x 2 . We wish to generate the path in reverse order KTUNOTES.IN Downloaded from Ktunotes.in 144 PROPERTIES OF CONTEXT-FREE LANGUAGES (bottom to top) and output symbols on opposite sides of the path from which they orig-inally appeared. To do this construct G = (V u {A\A is in V} u {S0}, T, P, S0 ), where P contains 1) all productions of P, 2) C -+ AB and B^CA'xiP contains A BCy 3) S^£, ^ 4) S0 -+ aA if P contains A-a y 5) S0 -S.
To see that L(G) = CYCLE(L(G)) show by induction on the length of a derivation that A ^> A ! A 2 "' A n if and only if for each i Ai$>A i+l •• ^ n ^/li ••• Then 5^^! •• An => A t ' A i - l aA i + l - "An iff S0 => aAi ^ aA t + x ••• /InS/^ • =>a>4 i+1 ••• /t B v4!
••• Ai-i.
A derivation tree of G is shown in Fig. 6.11(a) with a corresponding tree for G in Fig.
6.11(b).
6.5 a) Let L be the CFL {0F2 fc | i < or ; < k}. L is generated by the CFG S-+AB\C, A-+0A\e y B^\B2\B2\e y C-0C2|C2|D, D-lD|t MIN(L) = {(yp'2 | /c = min(i, j)}. We claim MIN(L) is not a CFL. Suppose it were, and let n be the pumping lemma constant. Consider z = 0 n l n2n = uvwxy. If contains no 2's, then uwy is not in MIN(L). If vx has a 2, it cannot have a 0, since | vwx \ < n. Thus uv 2wx 2y has at least n + 1 2's, at least n Vs and exactly n 0's; it is thus not in MIN(L).
S SQ A /V /!
£ 3 // / /\ ./\ 0 C D F G /\ \ ./\ \ E F 4 C E 2 / A A.
\ 1 G // D B 1 / \ / -A 2 3 4 S A I I e 0 (a) (b) Fig. 6.11 Tree transformation used for Exercise 6.4(c).
KTUNOTES.IN Downloaded from Ktunotes.in BIBLIOGRAPHIC NOTES 145 BIBLIOGRAPHIC NOTES The pumping lemma for context-free languages is from Bar-Hillel, Perles, and Shamir ; Ogden 's lemma, in its stronger version, is found in Ogden . Wise gives a necessary and sufficient condition for a language to be context free. Parikh gives necessary conditions in terms of the distribution of symbols in words of the language.
Pumping lemmas for other classes of languages are given in Boasson and Ogden .
Theorem 6.2, closure under substitution, and Theorem 6.5, closure under intersection with a regular set, are from Bar-Hillel, Perles, and Shamir . Theorem 6.3 is from Ginsburg and Rose [1963b]. Theorem 6.4 and its corollary, nonclosure under intersection or complementation, are from Scheinberg [I960]. Theorem 6.6, the existence of an algo-rithm to tell whether a CFL is finite, is also from Bar-Hillel, Perles, and Shamir .
Floyd [1962b] shows how to apply closure properties to prove language constructs not to be context free.
The CYK algorithm was originally discovered by J. Cocke, but its first publication was due independently to Kasami and Younger , The most practical, general, context-free recognition and parsing algorithm is by Earley . This algorithm is 0(n 3 ) in general, but takes only 0(n 2 ) on any unambiguous CFG and is actually linear on a wide variety of useful grammars. The algorithm of Valiant [1975a] is asymptotically the most efficient, taking 0(n 2 B ) steps, while the algorithm of Graham, Harrison, and Ruzzo takes 0(n 3/log n) steps. A related result, that membership for unambiguous CFG's can be tested in 0(n 2 ) time, is due to Kasami and Torii and Earley .
Exercise 6.4(a), closure of CFL's under quotient with a regular set, was shown by Ginsburg and Spanier . Additional closure properties of CFL's can be found in Ginsburg and Rose [1963b, 1966]. Exercise 6.7, the characterization of CFL's by Dyck languages, is from Chomsky . Stanley showed the stronger result that the Dyck language used need depend only on the size of the terminal alphabet. The proof that the primes in binary are not a CFL (Exercise 6.9) is from Hartmanis and Shank .
Finite-turn PDA's, mentioned in Exercise 6. 1 3, were studied by Ginsburg and Spanier [ 1 966].
Exercise 6.8, that CFL's over a one-symbol alphabet are regular, was shown by Ginsburg and Rice .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 7 TURING MACHINES In this chapter we introduce the Turing machine, a simple mathematical model of a computer. Despite its simplicity, the Turing machine models the computing capability of a general-purpose computer. The Turing machine is studied both for the class of languages it defines (called the recursively enumerable sets) and the class of integer functions it computes (called the partial recursive functions). A variety of other models of computation are introduced and shown to be equiv-alent to the Turing machine in computing power.
7.1 INTRODUCTION The intuitive notion of an algorithm or effective procedure has arisen several times. In Chapter 3 we exhibited an effective procedure to determine if the set accepted by a finite automation was empty, finite, or infinite. One might naively assume that for any class of languages with finite descriptions, there exists an effective procedure for answering such questions. However, this is not the case.
For example, there is no algorithm to tell whether the complement of a CFL is empty (although we can tell whether the CFL itself is empty). Note that we are not asking for a procedure that answers the question for a specific context-free lan-guage, but rather a single procedure that will correctly answer the question for all CFL's. It is clear that if we need only determine whether one specific CFL has an empty complement, then an algorithm to answer the question exists. That is, there is one algorithm that says "yes" and another that says "no," independent of their inputs. One of these must be correct. Of course, which of the two algorithms answers the question correctly may not be obvious.
146 KTUNOTES.IN Downloaded from Ktunotes.in 7.2 | THE TURING MACHINE MODEL 147 At the turn of the century, the mathematician David Hilbert set out on a program to find an algorithm for determining the truth or falsity of any mathemat-ical proposition. In particular, he was looking for a procedure to determine if an arbitrary formula in the first-order predicate calculus, applied to integers, was true. Since the first-order predicate calculus is powerful enough to express the statement that the language generated by a context-free grammar is £, had Hilbert been successful, our problem of deciding whether the complement of a CFL is empty would be solved. However, in 1931, Kurt Gddel published his famous incompleteness theorem, which proved that no such effective procedure could exist. He constructed a formula in the predicate calculus applied to integers, whose very definition stated that it could neither be proved nor disproved within this logical system. The formalization of this argument and the subsequent clarification and formalization of our intuitive notion of an effective procedure is one of the great intellectual achievements of this century.
Once the notion of an effective procedure was formalized, it was shown that there was no effective procedure for computing many specific functions. Actually the existence of such functions is easily seen from a counting argument. Consider the class of functions mapping the nonnegative integers onto {0, 1}. These func-tions can be put into one-to-one correspondence with the reals. However, if we assume that effective procedures have finite descriptions, then the class of all effective procedures can be put into one-to-one correspondence with the integers.
Since there is no one-to-one correspondence between the integers and the reals, there must exist functions with no corresponding effective procedures to compute them. There are simply too many functions, a noncountable number, and only a countable number of procedures. Thus the existence of noncomputable functions is not surprising. What is surprising is that some problems and functions with genuine significance in mathematics, computer science, and other disciplines are noncomputable.
Today the Turing machine has become the accepted formalization of an effective procedure. Clearly one cannot prove that the Turing machine model is equivalent to our intuitive notion of a computer, but there are compelling argu-ments for this equivalence, which has become known as Church's hypothesis. In particular, the Turing machine is equivalent in computing power to the digital computer as we know it today and also to all the most general mathematical notions of computation.
7.2 THE TURING MACHINE MODEL A formal model for an effective procedure should possess certain properties. First, each procedure should be finitely describable. Second, the procedure should con-sist of discrete steps, each of which can be carried out mechanically. Such a model was introduced by Alan Turing in 1936. We present a variant of it here.
KTUNOTES.IN Downloaded from Ktunotes.in 148 TURING MACHINES The basic model, illustrated in Fig. 7.1, has a finite control, an input tape that is divided into cells, and a tape head that scans one cell of the tape at a time. The tape has a leftmost cell but is infinite to the right. Each cell of the tape may hold exactly one of a finite number of tape symbols. Initially, the n leftmost cells, for some finite n > 0, hold the input, which is a string of symbols chosen from a subset of the tape symbols called the input symbols. The remaining infinity of cells each hold the blank, which is a special tape symbol that is not an input symbol.
a \ a2 an B B Finite control Fig. 7.1 Basic Turing machine.
In one move the Turing machine, depending upon the symbol scanned by the tape head and the state of the finite control, 1) changes state, 2) prints a symbol on the tape cell scanned, replacing what was written there, and 3) moves its head left or right one cell.
Note that the difference between a Turing machine and a two-way finite automaton lies in the former's ability to change symbols on its tape.
Formally, a Turing machine (TM) is denoted M = (Q, Z, T, 3, q0 , B, F), where Q is the finite set of states, T is the finite set of allowable tape symbols, B, a symbol of F, is the blank, X, a subset of F not including B, is the set of input symbols, 3 is the next move function, a mapping from Q x F to Q x F x {L, R} (d may, however, be undefined for some arguments), q0 in Q is the start state, F ^ Q is the set offinal states.
We denote an instantaneous description (ID) of the Turing machine M by cc 1 qcc 2 ' Here q, the current state of M, is in Q; a x a 2 is the string in F that is the contents of the tape up to the rightmost nonblank symbol or the symbol to the left of the head, whichever is rightmost. (Observe that the blank B may occur in a x a 2 .) KTUNOTES.IN Downloaded from Ktunotes.in 7.2 | THE TURING MACHINE MODEL 149 We assume that Q and T are disjoint to avoid confusion. Finally, the tape head is assumed to be scanning the leftmost symbol of a 2 , or if a 2 = e, the head is scan-ning a blank.
We define a move of M as follows. Let X t X 2 • j qX x • • • Xn be an ID.
Suppose d(q, X ( ) = (p, Y, L), where if i -1 = n, then X { is taken to be B. If z = 1, then there is no next ID, as the tape head is not allowed to fall off the left end of the tape. If i > 1, then we write X X X2 •• X^^qXi •• Xn \w X 1 X 2 X i - 2 pX i _ 1 YXi+1 •• (7.1) However, if any suffix of X^ t YXi+l ••• X,, is completely blank, that suffix is deleted in (7.1).
Alternatively, suppose d(q, X t ) = (p, Y, K). Then we write: X l X 2 -Xt .
l qX lXi+l -Xn \w X l X 2 --X i _ l YpXi+l - Xn .
(7.2) Note that in the case i — 1 = n, the string X, • • • X„ is empty, and the right side of (7.2) is longer than the left side.
If two ID's are related by (^-, we say that the second results from the first by one move. If one ID results from another by some finite number of moves, includ-ing zero moves, they are related by the symbol f^-. We drop the subscript M from [ft- or|^-when no confusion results.
The language accepted by M, denoted L(M\ is the set of those words in Z that cause M to enter a final state when placed, justified at the left, on the tape of M, with M in state q0 , and the tape head of M at the leftmost cell. Formally, the language accepted by M = (Q, Z, T, 3, q0 , B, F) is {w|w in Z and q0 w \^ u. x pv. 2 for some p in F, and ol x and a 2 in T}.
Given a TM recognizing a language L, we assume without loss of generality that the TM halts, i.e., has no next move, whenever the input is accepted.
However, for words not accepted, it is possible that the TM will never halt.
Example 7.1 The design of a TM M to accept the language L = {OT | n > 1} is given below. Initially, the type ofM contains OT followed by an infinity of blanks.
Repeatedly, M replaces the leftmost 0 by Xy moves right to the leftmost 1, replac-ing it by 7, moves left to find the rightmost X, then moves one cell right to the leftmost 0 and repeats the cycle.
If, however, when searching for a 1, M finds a blank instead, then M halts without accepting.
If, after changing a 1 to a 7, M finds no more O's, then M checks that no more l's remain, accepting if there are none.
Let Q = {tf0 , qu <?2, ft, q\ Z = {0, 1}, Y = {0, 1, X y Y, B} 9 and F = {tf4 }.
Informally, each state represents a statement or a group of statements in a program. State q0 is entered initially and also immediately prior to each replace-ment of a leftmost 0 by an X. State q y is used to search right, skipping over O's and Ts until it finds the leftmost 1. IfM finds a 1 it changes it to Y, entering state q 2 .
KTUNOTES.IN Downloaded from Ktunotes.in 150 TURING MACHINES State q 2 searches left for an X and enters state q0 upon finding it, moving right, to the leftmost 0, as it changes state. As M searches right in state qu if a B or X is encountered before a 1, then the input is rejected; either there are too many O's or the input is not in 01.
State q0 has another role. If, after state q2 finds the rightmost X, there is a Y immediately to its right, then the O's are exhausted. From q0 , scanning Y, state q 3 is entered to scan over Fs and check that no l's remain. If the Fs are followed by a B, state q4 is entered and acceptance occurs; otherwise the string is rejected. The function d is shown in Fig. 7.2. Figure 7.3 shows the computation ofM on input 0011. For example, the first move is explained by the fact that d(q0 , 0) = (q l9 X, R); the last move is explained by the fact that d(q 3 , B) = (g4 , B, R). The reader should simulate M on some rejected inputs such as 001 101, 001, and Oil.
State 0 1 Symbol X y B 4o 4i 42 43 44 <4i <4i (42 , X, R) , 0, R) (q 2 , y, L) , 0, L) (4o, X, R) (43, y, n) tel.
y> r) (42, n l) (43, Y9 R) (44, B, R) Fig. 7.2 The function 5.
g00011 1—^,011 -xoqi 11 h^oyi h-q 2 X0Y\ [-Xqo0Y\ -XXq x Y\ |— XXYq x \ |— XXq 2 YY\—Xq 2 XYY I- xx^o h xxy^ y h XXYYq 3 |— XXYYBq Fig. 7.3 A computation of M .
7.3 COMPUTABLE LANGUAGES AND FUNCTIONS A language that is accepted by a Turing machine is said to be recursively enumer-able (r.e.). The term "enumerable" derives from the fact that it is precisely these languages whose strings can be enumerated (listed) by a Turing machine. "Recur-sively" is a mathematical term predating the computer, and its meaning is similar to what the computer scientist would call "recursion." The class of r.e. languages is very broad and properly includes the CFL's.
The class of r.e. languages includes some languages for which we cannot mechanically determine membership. If L(M) is such a language, then any Turing KTUNOTES.IN Downloaded from Ktunotes.in 7.3 | COMPUTABLE LANGUAGES AND FUNCTIONS 151 machine recognizing L(M) must fail to halt on some input not in L(M). If w is in L(M), M eventually halts on input w. However, as long as M is still running on some input, we can never tell whether M will eventually accept if we let it run long enough, or whether M will run forever.
It is convenient to single out a subclass of the r.e. sets, called the recursive sets, which are those languages accepted by at least one Turing machine that halts on all inputs (note that halting may or may not be preceded by acceptance). We shall see in Chapter 8 that the recursive sets are a proper subclass of the r.e. sets. Note also that by the algorithm of Fig. 6.8, every CFL is a recursive set.
The Turing machine as a computer of integer functions In addition to being a language acceptor, the Turing machine may be viewed as a computer of functions from integers to integers. The traditional approach is to represent integers in unary; the integer i > 0 is represented by the string 0\ If a function has k arguments, iu i2 , ik , then these integers are initially placed on the tape separated by l's, as 0 ll 10 I2 l ••• 10.
If the TM halts (whether or not in an accepting state) with a tape consisting of 0m for some m, then we say that f(i u i 2 , ik ) = m, where /is the function of k arguments computed by this Turing machine. Note that one TM may compute a function of one argument, a different function of two arguments, and so on. Also note that if TM M computes function / of k arguments, then / need not have a value for all different /c-tuples of integers iu ik .
If f(i u ik ) is defined for all i u ik , then we say / is a total recursive function. A function f(iu ik ) computed by a Turing machine is called a partial recursive function. In a sense, the partial recursive functions are analogous to the r.e. languages, since they are computed by Turing machines that may or may not halt on a given input. The total recursive functions correspond to the recursive languages, since they are computed by TM's that always halt. All common arith-metic functions on integers, such as multiplication, n!, flog 2 n] and 2 2 " are total recursive functions.
Example 7.2 Proper subtraction m — n is defined to be m — n for m > n, and zero for m < n. The TM M = ({q0 , q l9 ...
9 q6 }, {0, 1}, {0, 1, £}, 6, q0 , B, 0) defined below, started with 0m 10" on its tape, halts with 0m " n on its tape. M repeatedly replaces its leading 0 by blank, then searches right for a 1 followed by a 0 and changes the 0 to 1. Next, M moves left until it encounters a blank and then repeats the cycle. The repetition ends if i) Searching right for a 0, M encounters a blank. Then, the n O's in 0m 10" have all been changed to l's, and n + 1 of the m O's have been changed to B. M replaces the n + 1 l's by a 0 and n &s, leaving m — n O's on its tape.
KTUNOTES.IN Downloaded from Ktunotes.in 152 TURING MACHINES ii) Beginning the cycle, M cannot find a 0 to change to a blank, because the first m O's already have been changed. Then n > m, so m - n = 0. M replaces all remaining l's and O's by B.
The function d is described below.
1) %o, 0) = (qu B, R) Begin the cycle. Replace the leading 0 by B.
2) %„ 0)=(q ly 0,R) %.,!) = (q2 , I R) Search right, looking for the first 1.
3) d(q2 , 1) = (q 2 , 1, R) <%2,0)=(<?3, 1,L) Search right past l's until encountering a 0. Change that 0 to 1.
4) d(q3 , 0) = (q3 , 0, L) 5(qi , 1) = (<73 , 1, L) Hq3 , B) = (q0 , B K) Move left to a blank. Enter state q0 to repeat the cycle.
5) d(q2y B) = (<?4 , B, L) %4 , 1) = (94) B,L) <%4, 0) = (<?4) 0, L) 5(q>, B) = (q6 , 0, R) If in state q 2 a B is encountered before a 0, we have situation (i) described above. Enter state g4 and move left, changing all l's to B's until encountering a B. This B is changed back to a 0, state q6 is entered, and M halts.
6) %o, 1) = fas, B, R) <5(<? 5 , 0) = (q 5 , B, R) S(q 5 , l)=(q5,B,R) S(q 5 ,B)=(q6,B,R) If in state q0 a 1 is encountered instead of a 0, the first block of O's has been exhausted, as in situation (ii) above. M enters state q5 to erase the rest of the tape, then enters q6 and halts.
A sample computation of M on input 0010 is: qro0010 1— B^r,010 |— BO^IO |— B01<720 |— BO^ll |— B43OII \—q3 B0U \—Bqo0ll \— BBqi 1 1 1— BBlq2 l - BB1 lq2 |— BB1<?4 1 |— BBq>l I B<j4 -B0q6 KTUNOTES.IN Downloaded from Ktunotes.in 7.4 | TECHNIQUES FOR TURING MACHINE CONSTRUCTION 153 On input 0100, M behaves as follows: 400100 |— Bq t 100 |— Bl^OO |— Bq3 1 10 |— q3 B\0 f— Bq0UO [— BBq 5 10 \— BBBq 50 \— BBBBq5 |— BBBBBq6 7.4 TECHNIQUES FOR TURING MACHINE CONSTRUCTION Designing Turing machines by writing out a complete set of states and a next-move function is a noticeably unrewarding task. In order to describe complicated Turing machine constructions we need some "higher-level" conceptual tools. In this section we shall discuss the principal ones.
Storage in the finite control The finite control can be used to hold a finite amount of information. To do so, the state is written as a pair of elements, one exercising control and the other storing a symbol. It should be emphasized that this arrangement is for conceptual purposes only. No modification in the definition of the Turing machine has been made.
Example 7.3 Consider a Turing machine M that looks at the first input symbol, records it in its finite control, and checks that the symbol does not appear else-where on its input. Note that M accepts a regular set, but M will serve for demonstration purposes: M = (Q, {0, 1}, {0, 1, B} 9 S 9 [q09 B] 9 B, F), where Q is {q0 , q x ] x {0, 1, B}. That is, Q consists of the pairs [q0 , 0], [q0 , 1], [q09 £], [q l9 0], [q l9 1], and [qu B\ The set F is {[q l9 B]}. The intention is that the first component of the state controls the action, while the second component "remembers" a symbol.
We define 3 as follows.
1) a) S([q0 , B], 0) = {[q l9 0], 0, R\ b) S([q09 B] 9 1) = ([q l9 1], 1, R).
Initially, q0 is the control component of the state, and M moves right. The first component of JVfs state becomes q l9 and the first symbol seen is stored in the second component.
2) a) S([qu 0], 1) = ([<?„ 0], 1, R), b) S([q it 1], 0) = 1], 0, R).
IfM has a 0 stored and sees a 1 or vice versa, then M continues to move to the right.
KTUNOTES.IN Downloaded from Ktunotes.in 154 TURING MACHINES 3) a) S([q u 0], B) = ([q l9 B], 0, L), b) S([qu 1], B) = (fo, B], 0, L).
M enters the final state B] if it reaches a blank symbol without having first encountered a second copy of the leftmost symbol.
IfM reaches a blank in state [qu 0], or [q l9 1], it accepts. For state [qu 0] and symbol 0 or for state [q l9 1] and symbol 1, 5 is not defined. Thus ifM encounters the tape symbol stored in its state, M halts without accepting.
In general, we can allow the finite control to have k components, all but one of which store information.
Multiple tracks We can imagine that the tape of the Turing machine is divided into k tracks, for any finite k. This arrangement is shown in Fig. 7.4, with k = 3. The symbols on the tape are considered /c-tuples, one component for each track.
4 1 0 1 1 1 1 $ B B B B B B 1 0 1 B B B B 1 0 0 1 0 1 B B B Finite control Fig. 7.4 A three-track Turing machine.
Example 7.4 The tape in Fig.
7.4 belongs to a Turing machine that takes a binary input greater than 2, written on the first track, and determines whether it is a prime. The input is surrounded by $ and $ on the first track. Thus, the allowable input symbols are ft, B, B], [0, B, B], [1, B, B], and [$, B, B]. These symbols can be identified with 0, 1, and $, respectively, when viewed as input symbols. The blank symbol can be identified with [B, B, B].
To test if its input is a prime, the TM first writes the number two in binary on the second track and copies the first track onto the third. Then the second track is subtracted, as many times as possible, from the third track, effectively dividing the third track by the second and leaving the remainder.
If the remainder is zero, the number on the first track is not a prime. If the remainder is nonzero, the number on the second track is increased by one. If the second track equals the first, the number on the first track is a prime, because it cannot be divided by any number lying properly between one and itself. If the KTUNOTES.IN Downloaded from Ktunotes.in 7.4 | TECHNIQUES FOR TURING MACHINE CONSTRUCTION 155 second is less than the first, the whole operation is repeated for the new number on the second track.
In Fig. 7.4, the TM is testing to determine if 47 is a prime. The TM is dividing by 5; already 5 has been subtracted twice, so 37 appears on the third track.
Checking off symbols Checking off symbols is a useful trick for visualizing how a TM recognizes lan-guages defined by repeated strings, such as {ww | w in £}, {wcy | w and y in £, w y} or {wwR \ w in £}.
It is also useful when lengths of substrings must be compared, such as in the languages {flV | i > 1} or {oW \i+J or ; + k}.
We introduce an extra track on the tape that holds a blank or yj. The yj appears when the symbol below it has been considered by the TM in one of its comparisons.
Example 7.5 Consider a Turing machine M = (Q, E, T, S, q0 , B, F), which recog-nizes the language {vvcw|w in (a + b) + }. Let Q = <l \q = <l2, •••,^9 and d = a,b, or B}.
The second component of the state is used to store an input symbol, £ = {[£, d]\d = a, b, or c}.
The input symbol [B, d] is identified with d. Remember that the two "tracks" are just conceptual tools; that is, [B, d] is just another "name" for d: r={[X, d]\X = BoTy/ and d = a, b, c, or B}, <lo = ku B\ and F = {[g9 , B]}; [B, B] is identified with B, the blank symbol. For d = a or 6 and e = a or 5 we define (5 as follows.
1) 8{[q u Bl [,</])=([<?,, 4 [^,4 )l M checks the symbol scanned on the tape, stores the symbol in the finite control, and moves right.
2) 5([q2 ,d), [B,e])=([q2 ,d\, [B, e], R).
M continues to move right, over unchecked symbols, looking for c.
3) S([q2 ,d\,[B,c]) = ([q3 ,d],[B,clR).
On finding c, M enters a state with first component q3 .
KTUNOTES.IN Downloaded from Ktunotes.in 156 TURING MACHINES 4) S([q3,dl[y/,e])=([q3,d],[y/,e], R).
M moves right over checked symbols.
5) %,4[M) = fc»l[y,4 4 M encounters an unchecked symbol. If the unchecked symbol matches the symbol stored in the finite control, M checks it and begins moving left. If the symbols disagree, M has no next move and so halts without accepting. M also halts if in state q3 it reaches [B, B] before finding an unchecked symbol.
6) S([q„ B], y, d]) = ([<?4 , B], [J, 4 L).
M moves left over checked symbols.
7) 6([q„ B], [B, c]) = ([q5 , B], [B, c], L).
M encounters the symbol c.
8) 5([q 5 ,B], [B,d]) = ([q6,B], [B, d], L).
If the symbol immediately to the left of c is unchecked, M proceeds left to find the rightmost checked symbol.
9) 5([q6 , B], [B, d]) = ([q6 , B], [B, d], L).
M proceeds left.
10) 5([q6 , B], [V, d]) = ([q u B], [J, d), R).
M encounters a checked symbol and moves right to pick up another symbol for comparison. The first component of state becomes q l again.
11) 5([q5 , B], [V, d]) = ([q 1? B], [V, d), R).
M will be in state [q 5 , B] immediately after crossing c moving left. (See rule 7.) If a checked symbol appears immediately to the left of c, all symbols to the left of c have been checked. M must test whether all symbols to the right have been checked. If so, they must have compared properly with the symbols to the left of c, so M will accept.
12) 6([qi , B], [B, c]) = ([q8 , B], [B, c], R).
M moves right over c.
13) S([qa,BlU,d]) = ([qe,Bl[J,dl R).
M moves to the right over checked symbols.
14) 8([q a , B], [B, B]) = ([q 9 , B], [J, B\ L).
If M finds [B, B], the blank, it halts and accepts.
If M finds an unchecked symbol when its first component of state is q 8 , it halts without accepting.
Shifting over A Turing machine can make space on its tape by shifting all nonblank symbols a finite number of cells to the right. To do so, the tape head makes an excursion to the right, repeatedly storing the symbols read in its finite control and replacing KTUNOTES.IN Downloaded from Ktunotes.in 7.4 | TECHNIQUES FOR TURING MACHINE CONSTRUCTION 157 them with symbols read from cells to the left. The TM can then return to the vacated cells and print symbols of its choosing. If space is available, it can push blocks of symbols left in a similar manner.
Example 7.6 We construct part of a Turing machine, M = (Q, 2, T, S, q0 , £, F), which may occasionally have a need to shift nonblank symbols two cells to the right. We suppose that M's tape does not contain blanks between nonblanks, so when it reaches a blank it knows to stop the shifting process. Let Q contain states of the form [q, A u A 2] for q = q l or q 2 , and A x and A 2 in T. Let X be a special symbol not used by M except in the shifting process. M starts the shifting process in state [qu B, B\ The relevant portions of the function S are as follows.
1) S{[q i,B9 BlA 1 ) = ([q l,B,A llX9 R) for A , in T - {B, X}.
M stores the first symbol read in the third component of its state. X is printed on the cell scanned, and M moves to the right.
2) 6{[q l9 B9 A llA 2)={[q i9 A l9 A 2lX,R) for A x and A 2 in T - {ft X}.
M shifts the symbol in the third component to the second component, stores the symbol being read in the third component, prints an X, and moves right.
3 ) Hku A u A 2], A 3 ) = ([qu A 2 , A 3], A u R) for A l9 A 2 , and A 3 in r - {b9 x}.
M now repeatedly reads a symbol A 3y stores it in the third component of state, shifts the symbol previously in the third component, A 2 , to the second component, deposits the previous second component, A l9 on the cell scanned, and moves right. Thus a symbol will be deposited two cells to the right of its original position.
4 ) H[<li, A i, A 2], B) = ([q l9 A 2 , B\ A l9 R) for A x and A 2 in V - {B, X}.
When a blank is seen on the tape, the stored symbols are deposited on the tape.
5) S([qi,A lt B], B) = ([q2 , B, B], A u L).
After all symbols have been deposited, M sets the first component of state to q2 and moves left to find an X, which marks the rightmost vacated cell.
6) S{[q 29 B, Bl A) = ([q 2 , B, B], A, L) for A in Y - {£, X}.
M moves left until an X is found. When X is found, M transfers to a state that we have assumed exists in Q and resumes its other functions.
Subroutines As with programs, a "modular" or "top-down" design is facilitated if we use subroutines to define elementary processes. A Turing machine can simulate any type of subroutine found in programming languages, including recursive procedures and any of the known parameter-passing mechanisms. We shall here KTUNOTES.IN Downloaded from Ktunotes.in 158 TURING MACHINES describe only the use of parameterless, nonrecursive subroutines, but even these are quite powerful tools.
The general idea is to write part of a TM program to serve as a subroutine; it will have a designated initial state and a designated return state which temporarily has no move and which will be used to effect a return to the calling routine. To design a TM that "calls" the subroutine, a new set of states for the subroutine is made, and a move from the return state is specified. The call is effected by entering the initial state for the subroutine, and the return is effected by the move from the return state.
Example 7.7 The design of a TM M to implement the total recursive function "multiplication" is given below. M starts with 0m 10" on its tape and ends with (T n surrounded by blanks. The general idea is to place a 1 after 0m 10n and then copy the block of n O's onto the right end m times, each time erasing one of the m O's.
The result is \0n \0rn . Finally the prefix 10n l is erased, leaving 0™. The heart of the algorithm is a subroutine COPY, which begins in an ID QTlq^lQf and eventually enters an ID 0m lq50 n \0i+n . COPY is defined in Fig. 7.5. In state q u on seeing a 0, M changes it to a 2 and enters state q 2 . In state q2 , M moves right, to the next blank, deposits the 0, and starts left in state q 3 . In state q3y M moves left to a 2. On reaching a 2, state q x is entered and the process repeats until the 1 is encountered, signaling that the copying process is complete. State <?4 is used to convert the 2's back to O's, and the subroutine halts in q 5 .
0 1 2 4i (42, 2, R) (44, 1, L) 42 (4i. 0, R) (42.
1, R) 43 (43. 0, L) (43, 1, L) (4., 2, R) 44 (45, 1, R) (44, 0, L) Fig. 7.5 S for subroutine COPY.
To complete the program for multiplication, we add states to convert initial ID q00m \0 n to B0m ~ 1 lfliCl. That is, we need the rules %o, 0) = (q6,B,R% Sfae. 0) = (q69 0, R), 8(q 69 l)=(q l9 1, R).
Additional states are needed to convert an ID £ I0m ~ I lg50n 10 nl to Bi+ \Qm-i-1 i^OMO"', which restarts COPY, and to check whether i = m, that is, all m O's have been erased. In the case that i = m, the leading 10" 1 is erased and the computation halts in state q l2 . These moves are shown in Fig. 7.6.
KTUNOTES.IN Downloaded from Ktunotes.in 7.5 MODIFICATIONS OF TURING MACHINES 159 Hi qio qu (47, 0, L) (<?9, 0, L) (<?9, 0, L) (tfs, 1, L) (qu,B,R) toio, B, R) too, By R) ton, B, i?) Fig. 7.6 Additional moves for TM performing multiplication.
Note that we could make more than one call to a subroutine if we rewrote the subroutine using a new set of states for each call.
7.5 MODIFICATIONS OF TURING MACHINES One reason for the acceptance of the Turing machine as a general model of a computation is that the model with which we have been dealing is equivalent to many modified versions that would seem off-hand to have increased computing power.
In this section we give informal proofs of some of these equivalence theorems.
Two-way infinite tape A Turing machine with a two-way infinite tape is denoted by M = (Q, £, T, <5, q0 , B, F), as in the original model. As its name implies, the tape is infinite to the left as well as to the right. We denote an ID of such a device as for the one-way infinite TM. We imagine, however, that there is an infinity of blank cells both to the left and right of the current nonblank portion of the tape.
The relation which relates two ID's if the ID on the right is obtained from the one on the left by a single move, is defined as for the original model with the exception that if 5(q, X) = (p, Y, L), then qXa \jf pBYa (in the original model, no move could be made), and if d(q, X) = (p, By R), then qXa \— pa (in the original, the B would appear to the left of p).
The initial ID is q0 w. While there was a left end to the tape in the original model, there is no left end of the tape for the Turing machine to "fall off," so it can proceed left as far as it wishes. The relation f^-, as usual, relates two ID's if the one on the right can be obtained from the one on the left by some number of moves.
Theorem 7.1 L is recognized by a Turing machine with a two-way infinite tape if and only if it is recognized by a TM with a one-way infinite tape.
Proof The proof that a TM with a two-way infinite tape can simulate a TM with a one-way infinite tape is easy. The former marks the cell to the left of its initial head position and then simulates the latter. If during the simulation the marked cell is reached, the simulation terminates without acceptance.
KTUNOTES.IN Downloaded from Ktunotes.in 160 TURING MACHINES Conversely, let M 2 = (Q 2 , E 2 , T 2 , S 2y q 2 , B, F2 ) be a TM with a two-way infinite tape. We construct M x , a Turing machine simulating M 2 and having a tape that is infinite to the right only. M x will have two tracks, one to represent the cells of Af 2 's tape to the right of, and including, the tape cell initially scanned, the other to represent, in reverse order, the cells to the left of the initial cell.
The relationship between the tapes of M 2 and M x is shown in Fig.
7.7, with the initial cell of M 2 numbered 0, the cells to the right 1, 2, ...
, and the cells to the left -1, -2, ....
A_ 5 A_ 4 A^^ A_ 2 A_\ A Q A x A 2 A 3 A4 A 5 (a) ^0 A 2 A 3 •44 A 5 4 4-1 A_ 2 -4-3 -4-4 -4-5 (b) Fig. 7.7 (a) Tape of M 2 . (b) Tape of M x The first cell of M/s tape holds the symbol $ in the lower track, indicating that it is the leftmost cell. The finite control of M x tells whether M 2 would be scanning a symbol appearing on the upper or on the lower track of M x .
It should be fairly evident that M x can be constructed to simulate M 2 , in the sense that while M 2 is to the right of the initial position of its input head, M x works on the upper track. While M 2 is to the left of its initial tape head position, M x works on its lower track, moving in the direction opposite to the direction in which M 2 moves. The input symbols ofM x are symbols with a blank on the lower track and an input symbol of M 2 on the upper track. Such a symbol can be identified with the corresponding input symbol ofM 2 . B is identified with [B, B\ We now give a formal construction of M x — (Q u Z 1? r l5 <5„ q u B, F t ). The states, Q l9 ofM x are all objects of the form [q, U]or [qy D], where q is in Q 2 , plus the symbol q x . Note that the second component will indicate whether M x is working on the upper (U for up) or lower (D for down) track. The tape symbols in r t are all objects of the form [X, Y], where X and Y are in T 2 . In addition, Y may be a symbol not in T 2 . H l consists of all symbols [a, B], where a is in S 2 . F t is {[q, U], [q, D] \q is in F 2 }. We define 3 X as follows.
1) For each a in S 2 u {£}, 5Mu [a, B]) = ([qy U\ [X, fl, R) if S 2 (q 29 a) = (q, X, R).
IfM 2 moves right on its first move, M 1 prints § in the lower track to mark the end of tape, sets its second component of state to U, and moves right. The first KTUNOTES.IN Downloaded from Ktunotes.in 7.5 | MODIFICATIONS OF TURING MACHINES 161 component of M^s state holds the state ofM2 . On the upper track, M x prints the symbol X that is printed by M 2 .
2) For each a in E 2 u {B}, 8i(q» [> B]) = ([q, D\ [X, ft R) if «5 2 (g 2 , a) = (q9 X, L).
If Af 2 moves left on its first move, M, records the next state of M 2 and the symbol printed by M 2 as in (1) but sets the second component of its state to D and moves right. Again, is printed in the lower track to mark the left end of the tape.
3) For each [X, Y] in r„ with Y + and A = L or R, a, (fe u], [x, y]) = ([P, ui [z, y], ^) if <5 2 (g, x) = (p, z, /t).
Af ! simulates M 2 on the upper track.
4) For each [X, Y] in with Y + ^ S&q, D], [X, Y]) = ([p, Z>], [X, Z], /I) if 5 2 fa, Y) = (p, Z, I).
Here A is L if A is .R, and >1 is R if ^4 is L. M, simulates M 2 on the lower track of Mj. The direction of head motion of Mj is opposite to that of M 2 .
5) «ittft u].
= Mb^lt^fl) = ([p. c], [y, ft k) if ^,x) = (p, y,^i).
Here C = C/ if /I = K, and C = Z) if A = L.M X simulates a move ofM 2 on the cell initially scanned by M 2 . M x next works on the upper or lower track, depending on the direction in which M 2 moves. M Y will always move right in this situation.
Multitape Turing machines A multitape Turing machine is shown in Fig. 7.8. It consists of a finite control with k tape heads and k tapes; each tape is infinite in both directions. On a single move, depending on the state of the finite control and the symbol scanned by each of the tape heads, the machine can: 1) change state; 2) print a new symbol on each of the cells scanned by its tape heads; 3) move each of its tape heads, independently, one cell to the left or right, or keep it stationary.
Initially, the input appears on the first tape, and the other tapes are blank. We shall not define the device more formally, as the formalism is cumbersome and a straightforward generalization of the notation for single-tape TM's.
Theorem 7.2 If a language L is accepted by a multitape Turing machine, it is accepted by a single-tape Turing machine.
KTUNOTES.IN Downloaded from Ktunotes.in 162 TURING MACHINES Finite control Fig. 7.8 Multitape Turing machine.
Proof Let L be accepted by M l9 a TM with k tapes. We can construct M 2 , a one-tape TM with 2k tracks, two tracks for each ofM tapes. One track records the contents of the corresponding tape ofM t and the other is blank, except for a marker in the cell that holds the symbol scanned by the corresponding head of Mj. The arrangement is illustrated in Fig. 7.9. The finite control ofM 2 stores the state of Af i, along with a count of the number of head markers to the right ofM 2's tape head.
Head 1 Tape 1 X Ai A 2 Head 2 Tape 2 X Bi B2 Bm Head 3 Tape 3 X c, c 2 cm Fig. 7.9 Simulation of three tapes by one.
Each move of M x is simulated by a sweep from left to right and then from right to left by the tape head of Af 2 . Initially, M 2 's head is at the leftmost cell containing a head marker. To simulate a move of M u M 2 sweeps right, visiting each of the cells with head markers and recording the symbol scanned by each head of M t . When M 2 crosses a head marker, it must update the count of head markers to its right. When no more head markers are to the right, M 2 has seen the symbols scanned by each of Af/s heads, so Af 2 has enough information to deter-KTUNOTES.IN Downloaded from Ktunotes.in 7.5 | MODIFICATIONS OF TURING MACHINES 163 mine the move ofMv NowM 2 makes a pass left, until it reaches the leftmost head marker. The count of markers to the right enables M2 to tell when it has gone far enough. As M 2 passes each head marker on the leftward pass, it updates the tape symbol ofM x "scanned" by that head marker, and it moves the head marker one symbol left or right to simulate the move ofM x . Finally, M 2 changes the state of M x recorded in M 2 's control to complete the simulation of one move ofM x . If the new state of M x is accepting, then M 2 accepts.
Note that in the first simulation of this section—that of a two-way infinite tape TM by a one-way infinite tape TM, the simulation was move for move. In the present simulation, however, many moves ofM 2 are needed to simulate one move ofM j. In fact, since after k moves, the heads ofM x can be 2k cells apart, it takes about Yj=i 2i « 2k 2 moves ofM 2 to simulate k moves ofM v (Actually, 2k more moves may be needed to simulate heads moving to the right.) This quadratic slowdown that occurs when we go from a multitape TM to a single tape TM is inherently necessary for certain languages. While we defer a proof to Chapter 12, we shall here give an example of the efficiency of multitape TM's.
Example 7.8 The language L = {wwR | w in (0 + 1)} can be recognized on a single-tape TM by moving the tape head back and forth on the input, checking symbols from both ends, and comparing them. The process is similar to that of Example 7.5.
To recognize L with a two-tape TM, the input is copied onto the second tape.
The input on one tape is compared with the reversal on the other tape by moving the heads in opposite directions, and the length of the input checked to make sure it is even.
Note that the number of moves used to recognize L by the one-tape machine is approximately the square of the input length, while with a two-tape machine, time proportional to the input length is sufficient.
Nondeterministic Turing machines A nondeterministic Turing machine is a device with a finite control and a single, one-way infinite tape. For a given state and tape symbol scanned by the tape head, the machine has a finite number of choices for the next move. Each choice consists of a new state, a tape symbol to print, and a direction of head motion.
Note that the nondeterministic TM is not permitted to make a move in which the next state is selected from one choice, and the symbol printed and/or direction of head motion are selected from other choices. The nondeterministic TM accepts its input if any sequence of choices of moves leads to an accepting state.
As with the finite automaton, the addition of nondeterminism to the Turing machine does not allow the device to accept new languages. In fact, the combina-tion of nondeterminism with any of the extensions presented or to be presented, KTUNOTES.IN Downloaded from Ktunotes.in 164 TURING MACHINES such as two-way infinite or multitape TM's, does not add additional power. We leave these results as exercises, and prove only the basic result regarding the simulation of a nondeterministic TM by a deterministic one.
Theorem 73 If L is accepted by a nondeterministic Turing machine, M l5 then L is accepted by some deterministic Turing machine, M 2 .
Proof For any state and tape symbol of M l5 there is a finite number of choices for the next move. These can be numbered 1,2, ... Let r be the maximum number of choices for any state-tape symbol pair. Then any finite sequence of choices can be represented by a sequence of the digits 1 through r. Not all such sequences may represent choices of moves, since there may be fewer than r choices in some situations.
M 2 will have three tapes. The first will hold the input. On the second, M 2 will generate sequences of the digits 1 through r in a systematic manner. Specifically, the sequences will be generated with the shortest appearing first. Sequences of equal length are generated in numerical order.
For each sequence generated on tape 2, M 2 copies the input onto tape 3 and then simulates M i on tape 3, using the sequence on tape 2 to dictate the moves of Mj.
If M t enters an accepting state, M 2 also accepts. If there is a sequence of choices leading to acceptance, it will eventually be generated on tape 2. When simulated, M 2 will accept. But if no sequence of choices of moves ofM 2 leads to acceptance, M 2 will not accept.
Multidimensional Turing machines Let us consider another modification of the Turing machine that adds no addi-tional power—the multidimensional Turing machine. The device has the usual finite control, but the tape consists of a /c-dimensional array of cells infinite in all 2k directions, for some fixed k. Depending on the state and symbol scanned, the device changes state, prints a new symbol, and moves its tape head in one of 2k directions, either positively or negatively, along one of the k axes. Initially, the input is along one axis, and the head is at the left end of the input.
At any time, only a finite number of rows in any dimension contain nonblank symbols, and these rows each have only a finite number of nonblank symbols. For example, consider the tape configuration of the two-dimensional TM shown in Fig. 7.10(a). Draw a rectangle about the nonblank symbols, as also shown in Fig.
7.10(a). The rectangle can be represented row by row on a single tape, as shown in Fig. 7.10(b). The 's separate the rows. A second track may be used to indicate the position of the two-dimensional TMTs tape head.
We shall prove that a one-dimensional TM can simulate a two-dimensional TM, leaving the generalization to more than two dimensions as an exercise.
Theorem 7.4 If L is accepted by a two-dimensional TM M 2 . then L is accepted by a one-dimensional TM M l .
KTUNOTES.IN Downloaded from Ktunotes.in 7.5 | MODIFICATIONS OF TURING MACHINES 165 B B B 0i B B B B B a 2 03 04 05 B a-j 08 09 B 010 B B 0n a l2 013 B 014 015 B B 016 017 B B B BBBa i BBBBBa 2 ci3a4a 5 Ba 6 a 1 aga 9 Ba loBBa ll a l 2Ci l3 Ba l 4.a l5 BBa 1 6 0 1 7 BBB Fig. 7.10 Simulation of two dimensions by one.
(a) Two-dimensional tape, (b) One-dimensional simulation.
Proof M x represents the tape ofM 2 as in Fig. 7.10(b). Mi will also have a second tape used for purposes we shall describe, and the tapes are two-way infinite.
Suppose that M 2 makes a move in which the head does not leave the rectangle already represented by M t 's tape. If the move is horizontal, M x simply moves its head marker one cell left or right after printing a new symbol and changing the state ofM 2 recorded in M/s control. If the move is vertical, T x uses its second tape to count the number of cells between the tape head position and the to its left.
Then M t moves to the to the right, if the move is down, or the to the left if the move is up, and puts the tape head marker at the corresponding position in the new block (region between 's) by using the count on the second tape.
Now consider the situation when M 2 s head moves off the rectangle repre-sented by Mj. If the move is vertical, add a new block of blanks to the left or right, using the second tape to count the current length of blocks. If the move is horizon-tal, M x uses the "shifting over" technique to add a blank at the left or right end of each block, as appropriate. Note that double 's mark the ends of the region used to hold blocks, so M x can tell when it has augmented all blocks. After creating room to make the move, M x simulates the move of M 2 as described above.
Multihead Turing machines A /c-head Turing machine has some fixed number, /c, of heads. The heads are numbered 1 through k, and a move of the TM depends on the state and on the symbol scanned by each head. In one move, the heads may each move indepen-dently left, right, or remain stationary.
Theorem 7.5 If L is accepted by some /c-head TM M l9 it is accepted by a one-head TM M 2 .
Proof The proof is similar to that of Theorem 7.2 for multitape TM's. M 2 has k + 1 tracks on its tape; the last holds the tape ofM t and the ith holds a marker KTUNOTES.IN Downloaded from Ktunotes.in 166 TURING MACHINES indicating the position of the zth tape head for 1 < i < k. The details are left for an exercise.
Off-line Turing machines An off-line Turing machine is a multitape TM whose input tape is read-only.
Usually we surround the input by endmarkers, $ on the left and $ on the right. The Turing machine is not allowed to move the input tape head off the region between $ and $. It should be obvious that the off-line TM is just a special case of the multitape TM, and therefore is no more powerful than any of the models we have considered. Conversely, an off-line TM can simulate any TM M by using one more tape than M. The first thing the off-line TM does is copy its own input onto the extra tape, and it then simulates M as if the extra tape were M's input. The need for off-line TM's will become apparent in Chapter 12, when we consider limiting the amount of storage space to less than the input length.
7.6 CHURCH'S HYPOTHESIS The assumption that the intuitive notion of "computable function" can be identified with the class of partial recursive functions is known as Church's hypoth-esis or the Church-Turing thesis. While we cannot hope to "prove" Church's hypothesis as long as the informal notion of "computable" remains an informal notion, we can give evidence for its reasonableness. As long as our intuitive notion of "computable" places no bound on the number of steps or the amount of storage, it would seem that the partial recursive functions are intuitively compu-table, although some would argue that a function is not "computable" unless we can bound the computation in advance or at least establish whether or not the computation eventually terminates.
What is less clear is whether the class of partial recursive functions includes all "computable" functions. Logicians have presented many other formalisms such as the A-calculus, Post systems, and general recursive functions. All have been shown to define the same class of functions, i.e., the partial recursive functions. In addi-tion, abstract computer models, such as the random access machine (RAM), also give rise to the partial recursive functions.
The RAM consists of an infinite number of memory words, numbered 0, 1, each of which can hold any integer, and a finite number of arithmetic registers capable of holding any integer. Integers may be decoded into the usual sorts of computer instructions. We shall not define the RAM model more formally, but it should be clear that if we choose a suitable set of instructions, the RAM may simulate any existing computer. The proof that the Turing machine formalism is as powerful as the RAM formalism is given below. Some other formalisms are discussed in the exercises.
KTUNOTES.IN Downloaded from Ktunotes.in 7.7 | TURING MACHINES AS ENUMERATORS 167 Simulation of random access machines by Turing machines Theorem 7.6 A Turing machine can simulate a RAM, provided that the elemen-tary RAM instructions can themselves be simulated by a TM.
Proof We use a multitape TM M to perform the simulation. One tape of M holds the words of the RAM that have been given values. The tape looks like #0y0 #l yi #10i;2 # ••• #i Vi # where v t is the contents, in binary, of the ith word. At all times, there will be some finite number of words of the RAM that have been used, and M needs only to keep a record of values up to the largest numbered word that has been used so far.
The RAM has some finite number of arithmetic registers. M uses one tape to hold each register's contents, one tape to hold the location counter, which contains the number of the word from which the next instruction is to be taken, and one tape as a memory address register on which the number of a memory word may be placed.
Suppose that the first 10 bits of an instruction denote one of the standard computer operations, such as LOAD, STORE, ADD, and so on, and that the remaining bits denote the address of an operand. While we shall not discuss the details of implementation for all standard computer instructions, an example should make the techniques clear. Suppose the location counter tape ofM holds number i in binary. M searches its first tape from the left, looking for # i. If a blank is encountered before finding # /, there is no instruction in word i, so the RAM and M halt.
If # i is found, the bits following up to the next # are examined. Suppose the first 10 bits are the code for "ADD to register 2," and the remaining bits are some number j in binary. M adds 1 to i on the location counter tape and copies j onto the memory address tape. Then M searches for #j on the first tape, again starting from the left (note that #0 marks the left end). If #j is not found, we assume word j holds 0 and go on to the next instruction of the RAM. If #jvj # is found, Vj is added to the contents of register 2, which is stored on its own tape. We then repeat the cycle with the next instruction.
Observe that although the RAM simulation used a multitape Turing ma-chine, by Theorem 7.2 a single tape TM would suffice, although the simulation would be more complicated.
7.7 TURING MACHINES AS ENUMERATORS We have viewed Turing machines as recognizers of languages and as computers of functions on the nonnegative integers. There is a third useful view of Turing machines, as generating devices. Consider a multitape TM M that uses one tape as an output tape, on which a symbol, once written, can never be changed, and whose KTUNOTES.IN Downloaded from Ktunotes.in 168 TURING MACHINES tape head never moves left. Suppose also that on the output tape, M writes strings over some alphabet E, separated by a marker symbol # . We can define G(M), the language generated by M, to be the set of w in E such that w is eventually printed between a pair of # 's on the output tape.
Note that unless M runs forever, G(M) is finite. Also, we do not require that words be generated in any particular order, or that any particular word be gen-erated only once. IfL is G(M) for some TM M, then L is an r.e. set, and conversely.
The recursive sets also have a characterization in terms of generators; they are exactly the languages whose words can be generated in order of increasing size.
These equivalences will be proved in turn.
Characterization of r.e. sets by generators Lemma 7.1 If L is G(M X ) for some TM M u then L is an r.e. set.
Proof Construct TM M 2 with one more tape than M x . M 2 simulates M { using all but M 2 's input tape. Whenever M x prints # on its output tape, M 2 compares its input with the word just generated. If they are the same, M 2 accepts; otherwise M 2 continues to simulate M v Clearly M 2 accepts an input x if and only if x is in G(M X ). Thus L(M 2 ) = G(M,)-The converse of Lemma 7.1 is somewhat more difficult. Suppose M l is a recognizer for some r.e. set L ^ E. Our first (and unsuccessful) attempt at design-ing a generator for L might be to generate the words in E in some order w l9 w2 , run M x on w l5 and if M x accepts, generate w x . Then run M x on w2 , generating w2 ifM x accepts, and so on. This method works ifM x is guaranteed to halt on all inputs. However, as we shall see in Chapter 8, there are languages L that are r.e. but not recursive.
If such is the case, we must contend with the possibility that M x never halts on some wf . Then M 2 never considers wJ+1 , w, + 2 , and so cannot generate any of these words, even if M x accepts them.
We must therefore avoid simulating M l indefinitely on any one word. To do this we fix an order for enumerating words in E. Next we develop a method of generating all pairs (i, j) of positive integers. The simulation proceeds by generat-ing a pair (i,j) and then simulating M v on the ith word, for j steps.
We fix a canonical order for E as follows. List words in order of size, with words of the same size in "numerical order." That is, let E = {a0 , a u ak -i}, and imagine that a t is the "digit" i in base k. Then the words of length n are the numbers 0 through k" — 1 written in base k. The design of a TM to generate words in canonical order is not hard, and we leave it as an exercise.
Example 7.9 If E = {0, 1}, the canonical order is £, 0, 1, 00, 01, 10, 11, 000, 001,...
KTUNOTES.IN Downloaded from Ktunotes.in 7.7 | TURING MACHINES AS ENUMERATORS 169 Note that the seemingly simpler order in which we generate the shortest representation of 0, 1, 2, .
.
. in base k will not work as we never generate words like flo^o^!, which have "leading O's." Next consider generating pairs (i,j) such that each pair is generated after some finite amount of time. This task is not so easy as it seems. The naive approach, (1, 1), (1, 2), (1, 3), ... never generates any pairs with i > 1. Instead, we shall generate pairs in order of the sum i + j, and among pairs of equal sum, in order of increasing I That is, we generate (1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1), (1, 4), .
.
. The pair (i, ;) is the {[(/ + ; -+ ; - 2)]/2 + i}th pair generated. Thus this ordering has the desired property that there is a finite time at which any particular pair (i, j) is generated.
A TM generating pairs (z, j) in this order in binary is easy to design, and we leave its construction to the reader. We shall refer to such a TM as the pair generator in the future.
Incidentally, the ordering used by the pair generator demonstrates that pairs of integers can be put into one-to-one correspondence with the integers themselves, a seemingly paradoxical result that was discovered by Georg Kantor when he showed that the rationals (which are really the ratios of two integers) are equinumerous with the integers.
Theorem 7.7 A language is r.e. if and only if it is G(M 2 ) for some TM M 2 .
Proof With Lemma 7.1 we have only to show how an r.e. set L = L(Mi) can be generated by a TM M 2 . M 2 simulates the pair generator. When (i, ;) is generated, M 2 produces the ith word w f in canonical order and simulates M x on wf for j steps.
If M x accepts on the yth step (counting the initial ID as step 1), then M 2 generates vv f .
Surely M 2 generates no word not in L If w is in L, let w be the ith word in canonical order for the alphabet of L, and let M x accept w after exactly j moves. As it takes only a finite amount of time for M 2 to generate any particular word in canonical order or to simulate M x for any particular number of steps, we know that M 2 will eventually produce the pair (i, j). At that stage, w will be generated by M 2 . Thus G(M 2 ) = L.
Corollary If L is an r.e. set, then there is a generator for L that enumerates each word in L exactly once.
Proof M 2 described above has that property, since it generates vvf only when considering the pair (i,j), where j is exactly the number of steps taken by M x to accept w f .
Characterization of recursive sets by generators We shall now show that the recursive sets are precisely those sets whose words can be generated in canonical order.
KTUNOTES.IN Downloaded from Ktunotes.in 170 TURING MACHINES Lemma 7.2 If L is recursive, then there is a generator for L that prints the words of L in canonical order and prints no other words.
Proof Let L = JJ^My)^ £, where M t halts on every input. Construct M 2 to generate L as follows. M 2 generates (on a scratch tape) the words in £, one at a time, in canonical order. After generating some word w, M 2 simulates M x on w. If M x accepts w, M 2 generates w. Since M x is guaranteed to halt, we know that M 2 will finish processing each word after a finite time and will therefore eventually consider each particular word in £. Clearly M 2 generates L in canonical order.
The converse of Lemma 7.2, that ifL can be generated in canonical order then L is recursive, is also true. However, there is a subtlety of which we should be aware. In Lemma 7.2 we could actually construct M 2 from M x . However, given a TM M generating L in canonical order, we know a halting TM recognizing L exists, but there is no algorithm to exhibit that TM.
Suppose Mj generates L in canonical order. The natural thing to do is to construct a TM M 2 that on input w simulates M 1 until M x either generates w or a word beyond w in canonical order. In the former case, M 2 accepts w, and in the latter case, M 2 halts without accepting w. However, if L is finite, M x may never halt after generating the last word in L, so M x may generate neither w nor any word beyond. In this situation M 2 would not halt. This problem arises only when L is finite, even though we know every finite set is accepted by a Turing machine that halts on all inputs. Unfortunately, we cannot determine whether a TM gener-ates a finite set or, if finite, which finite set it is. Thus we know that a halting Turing machine accepting L, the language generated by M u always exists, but there is no algorithm to exhibit the Turing machine.
Theorem 7.8 L is recursive if and only if L is generated in canonical order.
Proof The "only if" part was established by Lemma 7.2. For the "if" part, when L is infinite, M 2 described above is a halting Turing machine for L. Clearly, when L is finite, there is a finite automaton accepting L, and thus Lean be accepted by a TM that halts on all inputs. Note that in general we cannot exhibit a particular halting TM that accepts L, but the theorem merely states that one such TM exists.
7.8 RESTRICTED TURING MACHINES EQUIVALENT TO THE BASIC MODEL In Section 7.5 we considered generalizations of the basic TM model. As we have seen, these generalizations have no more computational power than the basic model. We conclude this chapter by considering some models that at first appear less powerful than the TM but indeed are just as powerful. For the most part, these models will be variations of the pushdown automaton defined in Chapter 5.
KTUNOTES.IN Downloaded from Ktunotes.in 7.8 | RESTRICTED TURING MACHINES 171 In passing, we note that a pushdown automaton is equivalent to a nondeter-ministic TM with a read-only input on which the input head cannot move left, plus a storage tape with a rather peculiar restriction on the tape head. Whenever the storage tape head moves left, it must print a blank. Thus the storage tape to the right of the head is always completely blank, and the storage tape is effectively a stack, with the top at the right, rather than the left as in Chapter 5.
Multistack machines A deterministic two-stack machine is a deterministic Turing machine with a read-only input and two storage tapes. If a head moves left on either tape, a blank is printed on that tape.
Lemma 7.3 An arbitrary single-tape Turing machine can be simulated by a deterministic two-stack machine.
Proof The symbols to the left of the head of the TM being simulated can be stored on one stack, while the symbols on the right of the head can be placed on the other stack. On each stack, symbols closer to the TM's head are placed closer to the top of the stack than symbols farther from the TM's head.
Counter machines We can prove a result stronger than Lemma 7.3.
It concerns counter machines, which are off-line Turing machines whose storage tapes are semi-infinite, and whose tape alphabets contain only two symbols, Z and B (blank). Furthermore, the symbol Z, which serves as a bottom of stack marker, appears initially on the cell scanned by the tape head and may never appear on any other cell. An integer i can be stored by moving the tape head i cells to the right of Z. A stored number can be incremented or decremented by moving the tape head right or left. We can test whether a number is zero by checking whether Z is scanned by the head, but we cannot directly test whether two numbers are equal.
An example of a counter machine is shown in Fig. 7.11; § and $ are custo-marily used for end markers on the input. Here Z is the nonblank symbol on each tape. An instantaneous description of a counter machine can be described by the state, the input tape contents, the position of the input head, and the distance of the storage heads from the symbol Z (shown here as d x and d2 ). We call these distances the counts on the tapes. The counter machine, then, can really only store a count on each tape and tell if that count is zero.
Lemma 7.4 A four-counter machine can simulate an arbitrary Turing ma-chine.
Proof From Lemma 7.3, it suffices to show that two counter tapes can simulate one stack. Let a stack have k — 1 tape symbols, Zu Z 2 , k _ x . Then we can KTUNOTES.IN Downloaded from Ktunotes.in 172 TURING MACHINES i Read-only input $ i i Finite control // \ 1 Z B B ...
\ \ B B B < Z B B B B B Fig. 7.11 Counter machine.
represent the stack Zh Z i2 ••• Zim uniquely by the count in base k j = im + kim .
, + /c 2 im _ 2 + • • • + km ~ l i x .
(73) Note that not every integer represents a stack; in particular, those whose base-/c representation contains the digit 0 do not.
Suppose that the symbol Zr is pushed onto the top (right end) of the stack Zh Zi2 " Z im . The count associated with Zh Z i2 • • • Zim Zr isjk -f r. To get this new count, the counter machine repeatedly moves the head of the first counter one cell to the left and the head of the second, k cells to the right. When the head of the first counter reaches the nonblank symbol, the second counter will hold the count jk. It is a simple matter to add r to the count.
If, instead, the top symbol Zim of the stack were popped,; should be replaced by L/AA tne integer part of j/k. We repeatedly decrement the count on the first counter by k and then add one to the second count. When the first count is zero, the second count will be [j//cj.
To complete the description of the simulation, we must show how the four-counter machine can tell what symbol is at the top of each stack. If the count j is stored on one counter, the four-counter machine can copy j to another counter, computing j mod k in its finite control. Note that j mod k is im if j is given by (7.3).
^ Theorem 7.9 A two-counter machine can simulate an arbitrary Turing ma-chine.
Proof By Lemma 7.4, it is sufficient to show how to simulate four counters with two. Let four counters have counts ij, k, and {. One counter can represent these four by the number n = 2 f3;'57'. Since 2, 3, 5, and 7 are primes, ij9 k, and { can be uniquely recovered from n.
KTUNOTES.IN Downloaded from Ktunotes.in 7.8 | RESTRICTED TURING MACHINES 173 To increment i,j, k, or / by 1, we multiply n by 2, 3, 5, or 7, respectively. To do so, if we have another counter set to zero, we can move the head of this counter 2, 3, 5, or 7 cells to the right each time we move the head of the first counter one cell to the left. When the first counter holds zero, the second will hold 2n, 3n, 5n, or In, respectively. To decrement ij, /c, or 1}.
b) {ww|w is in (0+ 1)}.
c) The set of strings with an equal number of 0's and ls.
7.2 Design Turing machines to compute the following functions, a) [log2 n] b) n!
c) n 2 KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 175 7.3 Show that if L is accepted by a fc-tape, /-dimensional, nondeterministic TM with m heads per tape, then L is accepted by a deterministic TM with one semi-infinite tape and one tape head.
7.4 A recursive function is a function defined by a finite set of rules that for various arguments specify the function in terms of variables, nonnegative integer constants, the successor (add one) function, the function itself, or an expression built from these by composition of functions. For example, Ackermanris function is defined by the rules : 1) X(0,y)=l 2) A{1, 0) = 2 3) A(x, 0) = x + 2 for x > 2 4) A(x+l,y+l) = A(A(xt y+l),y) a) Evaluate A(2, 1).
b) What function of one variable is A(x> 2)?
c) Evaluate A(4t 3).
7.5 Give recursive definitions for a) n + m b) n — m c) nm d) n!
' 7.6 Show that the class of recursive functions is identical to the class of partial recursive functions.
7.7 A function is primitive recursive if it is a finite number of applications of composition and primitive recursion^ applied to constant 0, the successor function, or a projection function xn ) = x,.
a) Show that every primitive recursive function is a total recursive function.
b) Show that Ackermann's function is not primitive recursive.
c) Show that adding the minimization operator, min (/(x)) defined as the least x such that / (x) = 0, yields all partial recursive functions.
7.8 Design a Turing machine to enumerate {(TV \n > 1}.
' 7.9 Show that every r.e. set is accepted by a TM with only two nonaccepting states and one accepting state.
7.10 Complete the proof of Theorem 7.11, that tapes symbols 0 (blank) and 1, with no 1 overprinted by 0, are sufficient for an off-line TM to accept any r.e. language.
7.11 Consider an off-line TM model that cannot write on any tape but has three pebbles that can be placed on the auxiliary tape. Show that the model can accept any r.e. language.
t A primitive recursion is a definition of/(x„ x„) by f(x x x„) = if x„ = 0then 0(X, xB ,) else h{x l9 xH,f(x lt x„ t , xn -1)) where g and h are primitive recursive functions.
KTUNOTES.IN Downloaded from Ktunotes.in 176 TURING MACHINES BIBLIOGRAPHIC NOTES The Turing machine is the invention of Turing . Alternative formulations can be found in Kleene , Church , or Post . For a discussion of Church's hypothesis, see Kleene , Davis , or Rogers . Other formalisms equivalent to the partial recursive functions include the A-calculus (Church ), recursive functions (Kleene ), and Post systems (Post ).
Off-line TM's are discussed by Hartmanis, Lewis, and Stearns . An important result about multihead TM's—that they can be simulated without loss of time with one head per tape— is found in Hartmanis and Stearns . The one case not covered by the latter paper, when the multihead machine runs in real time (a number of moves propor-tional to the input length), was handled by Fischer, Meyer, and Rosenberg , and Leong and Seiferas contains the latest on reducing the complexity of that construc-tion. RAM's were formally considered by Cook and Reckhow .
Theorem 7.9, that two counters can simulate a TM, was proved by Minsky ; the proof given here is taken from Fischer . Theorem 7.11, on TM's that can only print l's over O's, is from Wang . Exercise 7.9, limiting the number of states, is from Shannon . In fact, as the latter paper assumes acceptance by halting rather than by final state, it shows that only two states are needed.
A number of texts provide an introduction to the theory of Turing machines and recursive functions. These include Davis [1958, 1965], Rogers , Yasuhara , Jones , Brainerd and Landweber , Hennie , and Machtey and Young .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 8 UN DECIDABILITY We now consider the classes of recursive and recursively enumerable languages.
The most interesting aspect of this study concerns languages whose strings are interpreted as codings of instances of problems. Consider the problem of deter-mining if an arbitrary Turing machine accepts the empty string. This problem may be formulated as a language problem by encoding TM's as strings of O's and l's.
The set of all strings encoding TM's that accept the empty string is a language that is recursively enumerable but not recursive. From this we conclude that there can be no algorithm to decide which TM's accept the empty string and which do not.
In this chapter we shall show that many questions about TM's, as well as some questions about context-free languages and other formalisms, have no algor-ithms for their solution. In addition we introduce some fundamental concepts from the theory of recursive functions, including the hierarchy of problems induced by the consideration of Turing machines with "oracles." 8.1 PROBLEMS Informally we use the word problem to refer to a question such as: "Is a given CFG ambiguous?" In the case of the ambiguity problem, above, an instance of the problem is a particular CFG. In general, an instance of a problem is a list of arguments, one argument for each parameter of the problem. By restricting our attention to problems with yes-no answers and encoding instances of the problem by strings over some finite alphabet, we can transform the question of whether there exists an algorithm for solving a problem to whether or not a particular language is recursive. While it may seem that we are throwing out a lot of impor-177 KTUNOTES.IN Downloaded from Ktunotes.in 178 UNDECIDABILITY tant problems by looking only at yes-no problems, in fact such is not the case.
Many general problems have yes-no versions that are provably just as difficult as the general problem.
Consider the ambiguity problem for CFG's. Call the yes-no version AMB. A more general version of the problem, called FIND, requires producing a word with two or more parses if one exists and answering "no" otherwise. An algorithm for FIND can be used to solve AMB. If FIND produces a word w, then answer "yes"; if FIND answers "no," then answer "no." Conversely, given an algorithm for AMB we can produce an algorithm for FIND. The algorithm first applies AMB to the grammar G. If AMB answers "no" our algorithm answers "no." If AMB answers "yes," the algorithm systematically begins to generate all words over the terminal alphabet of G. As soon as a word w is generated, it is tested to see if it has two or more parse trees. Note that the algorithm does not begin generat-ing words unless G is ambiguous, so some w eventually will be found and printed.
Thus we indeed have an algorithm. The portion of the algorithm that tests w for two or more parses is left as an exercise.
The process whereby we construct an algorithm for one problem (such as FIND), using a supposed algorithm for another (AMB), is called a reduction (of FIND to AMB). In general, when we reduce problem A to problem B we are showing that B is at least as hard as A. Thus in this case, as in many others, the yes-no problem AMB is no easier than the more general version of the problem.
Later we shall show that there is no algorithm for AMB. By the reduction ofAMB to FIND we conclude there is no algorithm for FIND either, since the existence of an algorithm for FIND implies the existence of an algorithm for AMB, a contradiction.
One further instructive point concerns the coding of the grammar G. As all Turing machines have a fixed alphabet, we cannot treat the 4-tuple notation G = (K, T, P, S) as the encoding of G without modification. We can encode 4-tuples as binary strings as follows. Let the metasymbols in 4-tuples, that is, the left and right parentheses, brackets, comma and ->, be encoded by 1, 10, 100, 10 5 , respectively. Let the z'th grammar symbol (in any chosen order) be encoded by 10' + 5 . In this encoding, we cannot tell the exact symbols used for either terminals or nonterminals. Of course renaming nonterminals does not affect the language generated, so their symbols are not important. Although we ordinarily view the identities of the terminals as important, for this problem the actual symbols used for the terminals is irrelevant, since renaming the terminals does not affect the ambiguity or unambiguity of a grammar.
Decidable and undecidable problems A problem whose language is recursive is said to be decidable. Otherwise, the problem is undecidable. That is, a problem is undecidable if there is no algorithm that takes as input an instance of the problem and determines whether the answer to that instance is "yes" or "no." KTUNOTES.IN Downloaded from Ktunotes.in 8.2 | RECURSIVE AND RECURSIVELY ENUMERABLE LANGUAGES 179 An unintuitive consequence of the definition of "undecidable" is that prob-lems with only a single instance are trivially decidable. Consider the following problem based on Fermat's conjecture. Is there no solution in positive integers to the equation x l + / = z l if i > 3? Note that x, y, z, and i are not parameters but bound variables in the statement of the problem. There is one Turing machine that accepts any input and one that rejects any input. One of these answers Fermat's conjecture correctly, even though we do not know which one. In fact there may not even be a resolution to the conjecture using the axioms of arith-metic. That is, Fermat's conjecture may be true, yet there may be no arithmetic proof of that fact. The possibility (though not the certainty) that this is the case follows from Godel's incompleteness theorem, which states that any consistent formal system powerful enough to encompass number theory must have state-ments that are true but not provable within the system.
It should not disturb the reader that a conundrum like Fermat's conjecture is "decidable." The theory of undecidability is concerned with the existence or non-existence of algorithms for solving problems with an infinity of instances.
8.2 PROPERTIES OF RECURSIVE AND RECURSIVELY ENUMERABLE LANGUAGES A number of theorems in this chapter are proved by reducing one problem to another. These reductions involve combining several Turing machines to form a composite machine. The state of the composite TM has a component for each individual component machine. Similarly the composite machine has separate tapes for each individual machine. The details of the composite machine are usually tedious and provide no insight. Thus we choose to informally describe the constructions.
Given an algorithm (TM that always halts), we can allow the composite TM to perform one action if the algorithm accepts and another if it does not accept.
We could not do this if we were given an arbitrary TM rather than an algorithm, since if the TM did not accept, it might run forever, and the composite machine would never initiate the next task. In pictures, an arrow into a box labeled "start" indicates a start signal. Boxes with no "start" signal are assumed to begin operat-ing when the composite machine does. Algorithms have two outputs, "yes" and "no," which can be used as start signals or as a response by the composite ma-chine. Arbitrary TM's have only a "yes" output, which can be used for the same purposes.
We now turn to some basic closure properties of the classes of recursive and r.e. sets.
Theorem 8.1 The k complement of a recursive language is recursive.
^oof Let L be a recursive language and M a Turing machine that halts on all mputs and accepts L. Construct M' from M so that if M enters a final state on input w, then M ' halts without accepting. IfM halts without accepting, M' enters a KTUNOTES.IN Downloaded from Ktunotes.in 180 UN DECIDABILITY Fig. 8.1 Construction showing that recursive languages are closed under complementation.
final state. Since one of these two events occurs, M' is an algorithm. Clearly L(M') is the complement of L and thus the complement of L is a recursive language.
Figure 8.1 pictures the construction of M' from M.
Theorem 8.2 The union of two recursive languages is recursive. The union of two recursively enumerable languages is recursively enumerable.
Proof Let L x and L2 be recursive languages accepted by algorithms M x and M 2 .
We construct M, which first simulates M,. If M 1 accepts, then M accepts. If M x rejects, then M simulates M 2 and accepts if and only ifM 2 accepts. Since both M x and M 2 are algorithms, M is guaranteed to halt. Clearly M accepts L t u L 2 .
For recursively enumerable languages the above construction does not work, since M x may not halt. Instead M can simultaneously simulate M x and M 2 on separate tapes. If either accepts, then M accepts. Figure 8.2 shows the two con-structions of this theorem.
\f r— A/, (a) (b) Fig. 8.2 Construction for union.
Theorem 8.3 If a language L and its complement Lare both recursively enumer-able, then L (and hence L) is recursive.
Proof Let M x and M 2 accept L and L respectively. Construct M as in Fig. 8.3 to simulate simultaneously M x and M 2 . M accepts w if M { accepts w and rejects w if M 2 accepts w. Since w is in either L or L, we know that exactly one of M t or M 2 will accept. Thus M will always say either "yes" or "no," but will never say both.
Note that there is no a priori limit on how long it may take before M x or M 2 accepts, but it is certain that one or the other will do so. Since M is an algorithm that accepts L, it follows that L is recursive.
KTUNOTES.IN Downloaded from Ktunotes.in 8.3 | UNIVERSAL TURING MACHINES AND AN UNDECIDABLE PROBLEM 181 >Yes Fig. 8.3 Construction for Theorem 8.3.
Theorems 8.1 and 8.3 have an important consequence. Let L and L be a pair of complementary languages. Then either 1) both L and L are recursive, 2) neither L nor L is r.e., or 3) one of L and L is r.e. but not recursive; the other is not r.e.
An important technique for showing a problem undecidable is to show by diagonalization that the complement of the language for that problem is not r.e.
Thus case (2) or (3) above must apply. This technique is essential in proving our first problem undecidable.
After that, various forms of reductions may be employed to show other problems undecidable.
8.3 UNIVERSAL TURING MACHINES AND AN UNDECIDABLE PROBLEM We shall now use diagonalization to show a particular problem to be undecidable.
The problem is: "Does Turing machine M accept input w?" Here, both M and w are parameters of the problem. In formalizing the problem as a language we shall restrict w to be over alphabet {0, 1} and M to have tape alphabet {0, 1, B}. As the restricted problem is undecidable, the more general problem is surely undecidable as well. We choose to work with the more restricted version to simplify the encoding of problem instances as strings.
Turing machine codes To begin, we encode Turing machines with restricted alphabets as strings over {0, 1}. Let M = (6, {0,1}, {0, 1, B} 9 5,q l9 B, {q 2 }) be a Turing machine with input alphabet {0, 1} and the blank as the only addi-tional tape symbol. We further assume that Q = [q u q 2 , .
.
-, qn } is the set of states, and that q 2 is the only final state. Theorem 7.10 assures us that if L c (0 + 1) is accepted by any TM, then it is accepted by one with alphabet {0, 1, B}. Also, there KTUNOTES.IN Downloaded from Ktunotes.in 182 UN DECIDABILITY is no need for more than one final state in any TM, since once it accepts it may as well halt.
It is convenient to call symbols 0, 1, and B by the synonyms Xu X 2 , X3 , respectively. We also give directions L and R the synonyms D x and D 2 , respec-tively. Then a generic move 3(qh X} ) = (qk , Xe , Dm ) is encoded by the binary string 0 l 1010 ltflO".
(8.1) A binary code for Turing machine M is 111 code!
11 code 2 11 ••• 11 coder 111, (8.2) where each code,-is a string of the form (8.1), and each move ofM is encoded by one of the codecs. The moves need not be in any particular order, so each TM actually has many codes. Any such code for M will be denoted (M).
Every binary string can be interpreted as the code for at most one TM; many binary strings are not the code of any TM. To see that decoding is unique, note that no string of the form (8.1) has two Ts in a row, so the codecs can be found directly. If a string fails to begin and end with exactly three Ts, has three l's other than at the end, or has two pair of l's with other than five blocks of O's in between, then the string represents no TM.
The pair M and w is represented by a string of the form (8.2) followed by w.
Any such string will be denoted <M, w).
Example 8.1 Let M — ({q l9 q 2 , q3 ), {0, 1}, {0, 1, B}, 3, q lf Bf {q2 }) have moves: ^ S(q» l)=fe,0,K), <% 3,oH( n %3, 1)=(<?2,0,R), 5(l3>B) = (q3y 1, L).
Thus one string denoted by (M, 101 1) is 111010010001010011000101010010011 000100100101001 100010001000100101 1 1 101 Note that many different strings are also codes for the pair <M, 101 1 ), and any of these may be referred to by the notation (M, 101 1).
A non-r.e. language Suppose we have a list of (0 + 1) in canonical order (see Section 7.7), where wf is the ith word, and M} is the TM whose code, as in (8.2) is the integer j written in KTUNOTES.IN Downloaded from Ktunotes.in 8.3 | UNIVERSAL TURING MACHINES AND AN UNDECIDABLE PROBLEM 183 / 12 3 4 Fig. 8.4 Hypothetical table indicating acceptance of words by TM's.
binary. Imagine an infinite table that tells for all i and j whether wf is in L(Mj).
Figure 8.4 suggests such a table ;t 0 means w £ is not in L(Mj) and 1 means it is.
We construct a language Ld by using the diagonal entries of the table to determine membership in Ld . To guarantee that no TM accepts Ld , we insist that w t is in Ld if and only if the (/, i) entry is 0, that is, ifM, does not accept wf . Suppose that some TM Mj accepted Ld . Then we are faced with the following contradic-tion. If Wj is in Ld , then the (j, j) entry is 0, implying that Wj is not in L(M,-) and contradicting Ld — L(Mj). On the other hand, if Wj is not in Ld , then the (j, j) entry is 1, implying that Wy is in L(M ; ), which again contradicts Ld = U{Mj). As Wj is either in or not in Ld , we conclude that our assumption, Ld — L(M7 ), is false. Thus, no TM in the list accepts Ld , and by Theorem 7.10, no TM whatsoever accepts Ld .
We have thus proved Lemma 8.1 Ld is not r.e.
The universal language Define L^, the "universal language," to be {<M, w)|M accepts w}. We call "universal" since the question of whether any particular string w in (0 + 1) is accepted by any particular Turing machine M is equivalent to the question of whether is in L^, where M' is the TM with tape alphabet {0, 1, B) equiv-alent to M constructed as in Theorem 7.10.
Theorem 8.4 is recursively enumerable.
Proof We shall exhibit a three-tape TM M x accepting L^. The first tape ofM x is the input tape, and the input head on that tape is used to look up moves of the TM M when given code (M, w) as input. Note that the moves ofM are found between the first two blocks of three l's. The second tape ofM x will simulate the tape of M.
t Actually as all low-numbered Turing machines accept the empty set, the correct portion of the table shown has all 0's.
KTUNOTES.IN Downloaded from Ktunotes.in 184 UN DECIDABILITY The alphabet ofM is {0, 1, B], so each symbol of M's tape can be held in one tape cell of M/s second tape. Observe that if we did not restrict the alphabet of M, we would have to use many cells of M x 's tape to simulate one of M's cells, but the simulation could be carried out with a little more work. The third tape holds the state of M, with qt represented by 0'. The behavior of Mj is as follows: 1) Check the format of tape 1 to see that it has a prefix of the form (8.2) and that there are no two codes that begin with O'lCVl for the same i and j. Also check that if 0 t'l(V10M(yi0m is a code, then 1 < j < 3, 1 < / < 3, and 1 < m < 2.
Tape 3 can be used as a scratch tape to facilitate the comparison of codes.
2) Initialize tape 2 to contain w, the portion of the input beyond the second block of three l's. Initialize tape 3 to hold a single 0, representing q x . All three tape heads are positioned on the leftmost symbols. These symbols may be marked so the heads can find their way back.
3) If tape 3 holds 00, the code for the final state, halt and accept.
4) Let Xj be the symbol currently scanned by tape head 2 and let 0' be the current contents of tape 3. Scan tape 1 from the left end to the second 111, looking for a substring beginning 1 lO'lO^l. If no such string is found, halt and reject; M has no next move and has not accepted. If such a code is found, let it be 0'10'10l(yi0m . Then put 0 on tape 3, print X, on the tape cell scanned by head 2 and move that head in direction Dm . Note that we have checked in (1) that 1 < t < 3 and 1 < m < 2. Go to step (3).
It is straightforward to check that M x accepts (M, w) if and only ifM accepts w. It is also true that ifM runs forever on w, M, will run forever on (M, w), and if M halts on w without accepting, M { does the same on (M, w).
The existence of M x is sufficient to prove Theorem 8.4.
However, by Theorems 7.2 and 7.10, we can find a TM with one semi-infinite tape and alphabet {0, 1, B} accepting Z^. We call this particular TM M M , the universal Turing ma-chine, since it does the work of any TM with input alphabet {0, 1}.
By Lemma 8.1, the diagonal language Ld is not r.e., and hence not recursive.
Thus by Theorem 8.1, L d is not recursive. Note that L d = {w f | M, accepts vv,}. We can prove the universal language = {(M, w) | M accepts w} not to be recursive by reducing Ld to L„ Thus L u is an example of a language that is r.e. but not recursive. In fact, L d is another example of such a language.
Theorem 8.5 is not recursive.
Proof Suppose A were an algorithm recognizing L^. Then we could recognize Ld as follows. Given string w in (0 + 1), determine by an easy calculation the value of i such that w = w t . Integer i in binary is the code for some TM M,. Feed to algorithm A and accept w if and only if M, accepts w, . The construction is shown in Fig. 8.5. It is easy to check that the constructed algorithm accepts w if KTUNOTES.IN Downloaded from Ktunotes.in 8.4 | rice's theorem and some more undecidable problems 185 Hypothetical Convert ——> A for Lu Constructed algorithm for L d Yes No -Yes ->No Fig. 8.5 Reduction of L d to L u .
and only if w = w t - and wf is in L(M f ). Thus we have an algorithm for Ld Since no such algorithm exists, we know our assumption, that algorithm A for exists, is false. Hence L u is r.e. but not recursive.
8.4 RICE'S THEOREM AND SOME MORE UNDECIDABLE PROBLEMS We now have an example of an r.e. language that is not recursive. The associated problem "Does M accept w?" is undecidable, and we can use this fact to show that other problems are undecidable. In this section we shall give several examples of undecidable problems concerning r.e.
sets. In the next three sections we shall discuss some undecidable problems taken from outside the realm of TM's.
Example 8.2 Consider the problem: "Is L(M) ^ 0?" Let (M) denote a code for M as in (8.2). Then define C = {(M)\ L(M) + 0} and Le = { | L(M) = 0}.
Note that Le and L„e are complements of one another, since every binary string denotes some TM ; those with a bad format denote the TM with no moves. All these strings are in Le . We claim that L„e is r.e. but not recursive and that Le is not r.e.
We show that is r.e. by constructing a TM M to recognize codes of TM's that accept nonempty sets. Given input (M,), M nondeterministically guesses a string x accepted by M t and verifies that M, does indeed accept x by simulating M, on input x. This step can also be carried out deterministically if we use the pair generator described in Section 7.7. For pair (y, k) simulate M f on the jth binary string (in canonical order) for k steps. If M, accepts, then M accepts (M,).
Now we must show that Le is not recursive. Suppose it were. Then we could construct an algorithm for L^, violating Theorem 8.5. Let A be a hypothetical algorithm accepting Le . There is an algorithm B that, given (M, w), constructs a TM M' that accepts 0 ifM does not accept w and accepts (0 + 1) ifM accepts w.
The plan of M' is shown in Fig. 8.6. M' ignores its input x and instead simulates M on input w, accepting if M accepts.
Note that M' is not B. Rather, B is like a compiler that takes (M, w) as "source program" and produces M' as "object program." We have described what B must do, but not how it does it. The construction of B is simple. It takes (M, w) KTUNOTES.IN Downloaded from Ktunotes.in 186 UNDECIDABILITY >Yes Fig. 8.6 The TM M'.
and isolates w. Say w = a x a 2 an is of length n. B creates n + 3 states q l9 q2 , .
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grt+3 with moves X) = (q2 , $, K) for any X (print marker), S(qh X) = (qi+ 15 a f _ l5 for any X and 2 < z < n 4-1 (print w), %, +2 > ) = ft &>r X ^ £ (erase tape), %,+ 2> #) = (qn+3 , ft L), (5(g„ +3 , X) = (qn+3) X, L) for X j= $ (find marker).
Having produced the code for these moves, B then adds n + 3 to the indices of the states ofM and includes the move %, + 3> $) = («» + 4, $> ) / start up M / and all the moves of M in its generated TM. The resulting TM has an extra tape symbol $, but by Theorem 7.10 we may construct M' with tape alphabet {0, 1, B}, and we may surely make q2 the accepting state. This step completes the algorithm ft and its output is the desired AT of Fig. 8.6.
Now suppose algorithm A accepting Le exists. Then we construct an algor-ithm C for 1^ as in Fig. 8.7. IfM accepts w, then L(M') 0; so A says "no" and C says "yes." IfM does not accept w, then L(M') = 0,A says "yes," and C says "no." As C does not exist by Theorem 8.5, A cannot exist. Thus, Le is not recursive. If L„e were recursive, Le would be also by Theorem 8.1. Thus L„e is r.e. but not recursive.
If Le were r.e., then Le and L„ e would be recursive by Theorem 8.3. Thus Le is not r.e.
Example 8.3 Consider the language L, = {(M) I L(M) is recursive} and = {(M> I L(M) is not recursive}.
Note that L, is not {(M) |M halts on all inputs}, although it includes the latter language. A TM M could accept a recursive language even though M itself might loop forever on some words not in L(M); some other TM equivalent to M must always halt, however. We claim neither L, nor L„r is r.e.
Suppose L>. were r.e. Then we could construct a TM for L„ which we know KTUNOTES.IN Downloaded from Ktunotes.in 8.4 | rice's theorem and some more undecidable problems 187 -A — C Yes No Fig. 8.7 Algorithm constructed for L„ assuming that algorithm A for L e exists.
does not exist. Let Mr be a TM accepting L,. We may construct an algorithm A that takes (M, w) as input and produces as output a TM M' such that L(M') = 0 ifM does not accept w, if M accepts w.
Note that is not recursive, so M' accepts a recursive language if and only ifM does not accept w. The plan ofM' is shown in Fig. 8.8. As in the previous example, we have described the output of A. We leave the construction of A to the reader.
Yes Fig. 8.8 The TM M'.
Given A and Mr we could construct a TM accepting L u , shown in Fig.
8.9, which behaves as follows. On input is in Lu, Now let us turn to L nr . Suppose we have a TM Mnr accepting L nr . Then we may use Mnr and an algorithm B, to be constructed by the reader, to accept B takes (M, w) as input and produces as output a TM M' such that Z if M accepts w, L>, if M does not accept w.
L(M') = >Yes Fig. 8.9 Hypothetical TM for L, KTUNOTES.IN Downloaded from Ktunotes.in 188 UNDECIDABILITY tYes (a) (b) Fig. 8.10 Constructions used in proof that L nr is not r.e. (a) M'. (b) TM for L u .
Yes Thus M' accepts a recursive language if and only ifM accepts w. Af ', which B must produce, is shown in Fig. 8.10(a), and a TM to accept L u given B and Mnr , is shown in Fig. 8.10(b). The TM of Fig. 8.10(b) accepts if and only if L(M') is not recursive, or equivalently, if and only if M does not accept w. That is, the TM accepts (M, w) if and only if (M, w) is in Since we have already shown that no such TM exists, the assumption that Mnr exists is false. We conclude that is not r.e.
Rice's Theorem for recursive index sets The above examples show that we cannot decide if the set accepted by a Turing machine is empty or recursive. The technique of proof can also be used to show that we cannot decide if the set accepted is finite, infinite, regular, context free, has an even number of strings, or satisfies many other predicates. What then can we decide about the set accepted by a TM? Only the trivial predicates, such as "Does the TM accept an r.e. set?," which are either true for all TM's or false for all TM's.
In what follows we shall discuss languages that represent properties of r.e.
languages. That is, the languages are sets of TM codes such that membership of (M) in the language depends only on L(M), not on M itself. Later we shall consider languages of TM codes that depend on the TM itself, such as "M has 27 states," which may be satisfied for some but not all of the TM's accepting a given language.
Let ¥ be a set of r.e. languages, each a subset of (0 + 1). ff is said to be a property of the r.e. languages. A set L has property ¥ if L is an element of For example, the property of being infinite is {L| L is infinite}. ¥ is a trivial property if ¥ is empty or ¥ consists of all r.e. languages. Let Ly be the set {(M) \L(M) is in .y 7 }.
Theorem 8.6 (Rice's Theorem) Any nontrivial property ¥ of the r.e. languages is undecidable.
Proof Without loss of generality assume that 0 is not in ¥ (otherwise consider ¥).
Since ff is nontrivial, there exists L with property ¥ . Let M L be a TM KTUNOTES.IN Downloaded from Ktunotes.in 8.4 | rice's theorem and some more undecidable problems 189 accepting L. Suppose ¥ were decidable. Then there exists an algorithm My accepting Ly . We use ML and My to construct an algorithm for as follows.
First construct an algorithm A that takes (M, w> as input and produces (M'> as output, where L(M') is in Sf if and only if M accepts w ((M, w) is in JL tt ).
The design of M' is shown in Fig. 8.11. First M' ignores its input and simulates M on w. IfM does not accept w, then M' does not accept x. IfM accepts w, then M' simulates ML on x, accepting x if and only ifM L accepts x. Thus M ' either accepts 0 or L depending on whether M accepts w.
Yes Start M, Yes M' ->Yes Fig. 8.11 M' used in Rice's theorem.
We may use the hypothetical My to determine if L(M') is in Sf. Since L(M') is in ^ if and only if (M, w) is in Z^, we have an algorithm for recognizing L^, a contradiction. Thus Sf must be undecidable. Note how this proof generalizes Example 8.2.
Theorem 8.6 has a great variety of consequences, some of which are sum-marized in the following corollary.
Corollary The following properties of r.e. sets are not decidable: a) emptiness, b) finiteness, c) regularity, d) context-freedom.
Rice's Theorem for recursively enumerable index sets The condition under which a set Ly is r.e. is far more complicated. We shall show that Ly, is r.e.
if and only if |Lj ifM does not accept w.
Thus L(M') is in i/ 7 if and only if M does not accept w.
M Yes Start A/, Fig. 8.12 The TM M'.
Yes-/ Yes A'/' Yes We again leave it to the reader to design the "compiler" A that takes \ as input and connects them with the fixed Turing machines M x and M 2 to con-struct the M' shown in Fig. 8.12. Having constructed A, we can use a TM My for Ly to accept L u , as shown in Fig. 8.13. This TM accepts \ if and only if M' accepts a language in or equivalently, if and only if M does not accept w.
As such a TM does not exist, we know My cannot exist, so is not r.e.
We now turn to the second property of recursively enumerable index sets.
\ >Yes Fig. 8.13 Hypothetical TM to accept Lu .
KTUNOTES.IN Downloaded from Ktunotes.in 8.4 | rice's theorem and some more undecidable problems 191 Lemma 83 If £f has an infinite language L such that no finite subset of L is in 5^, then Ly is not r.e.
Proof Suppose Ly were r.e. We shall show that L u would be r.e. as follows.
Let M x be a TM accepting L. Construct algorithm A to take a pair as input and produce as output a TM M' that accepts L if w is not in L(M) and accepts some finite subset of L otherwise. As shown in Fig. 8.14, M' simulates M t on its input x. If M t accepts x, then M' simulates M on w for |x| moves.
If M fails to accept w after |x| moves, then M' accepts x. We leave the design of algorithm A as an exercise.
Start, simulate for |jc| moves Yes it — w M M' Not "yes" after \x\ moves Yes Fig. 8.14 Construction of M'.
If w is in L(M), then M accepts w after some number of moves, say j. Then L(M') = {x | x is in L and |x| < j}, which is a finite subset of L.
If w is not in L(M), then L(M') = L. Hence, if M does not accept w, L(M') is in y, and if M accepts w, L(M'), being a finite subset of L, is not in £f by the hypothesis of the lemma. An argument that is by now standard proves that if Ly is r.e., so is L u .
Since the latter is not r.e., we conclude the former is not either.
Finally, consider the third property of r.e. index sets.
Lemma 8.4 If L#> is r.e., then the list of binary codes for the finite sets in if and only if L(M) is in $f as follows. M l generates pairs (/, ;) using the pair generator. In response to M t simulates M 2 , which is an enumerator of the finite sets in y, for i steps. We know M2 exists by condition (3). Let L t be the last set com-pletely printed out by M 2 . [If there is no set completely printed, generate the next (ij) pair.] Then simulate M for j steps on each word in Lv IfM accepts all words in Lu then M t accepts \. If not, M x generates the next (i,y)-pair.
We use conditions (1) and (2) to show that L(M t ) = Ly . Suppose L is in Ly , and let M be any TM with L(M) = L. By condition (2), there is a finite L ^ L in Sf (take L = L if L is finite). Let E be generated after i steps of Af2 , and let j be the maximum number of steps taken by M to accept a word in L (if E = 0, let 7= 1). Then when M t generates (i, ;), if not sooner, M x will accept \.
Conversely, suppose Mj accepts \. Then there is some (i, j) such that within j steps M accepts every word in some finite language L such that M2 generates L within its first i steps. Then L is in y, and L ^ L(M). By condition (1), L(M) is in ff, so \ is in L^. We conclude that L(M X ) = L^.
Theorem 8.7 has a great variety of consequences. We summarize some of them as corollaries and leave others as exercises.
Corollary 1 The following properties of r.e. sets are not r.e.
Proof In each case condition (1) is violated, except for (b), where (2) is violated, a) b) c) d) e) 0 g) and (g), where (3) is violated.
Corollary 2 The following properties of r.e. sets are r.e.
a) L=f=0.
b) L contains at least 10 members.
c) w is in L for some fixed word w.
d) L n L, + 0.
KTUNOTES.IN Downloaded from Ktunotes.in 8.5 | UNDECIDABILITY OF POST'S CORRESPONDENCE PROBLEM 193 Problems about Turing machines Does Theorem 8.6 say that everything about Turing machines is undecidable?
The answer is no. That theorem has to do only with properties of the language accepted, not properties of the Turing machine itself. For example, the question "Does a given Turing machine have an even number of states?" is clearly deci-dable. When dealing with properties of Turing machines themselves, we must use our ingenuity. We give two examples.
Example 8.4 It is undecidable if a Turing machine with alphabet {0, 1, B) ever prints three consecutive Fs on its tape. For each Turing machine M f we construct M f , which on blank tape simulates M, on blank tape. However, M f uses 01 to encode a 0 and 10 to encode a 1. If M,'s tape has a 0 in cell M t has 01 in cells 2j — 1 and 2j. If M, changes a symbol, M t changes the corresponding 1 to 0, then the paired 0 to 1. One can easily design M f so that M t never has three consecutive Fs on its tape. Now further modify M f so that if M, accepts, M t prints three consecutive Ts and halts. Thus M, prints three consecutive Ts if and only if M f accepts £. By Theorem 8.6, it is undecidable whether a TM accepts e, since the predicate is in L" is not trivial. Thus the question of whether an arbitrary Turing machine ever prints three consecutive Ts is undecidable.
Example 8.5 It is decidable whether a single-tape Turing machine started on blank tape scans any cell four or more times. If the Turing machine never scans any cell four or more times, than every crossing sequence (sequence of states in which the boundary between cells is crossed, assuming states change before the head moves) is of length at most three. But there is a finite number of distinct crossing sequences of length three or less. Thus either the Turing machine stays within a fixed bounded number of tape cells, in which case finite automaton techniques answer the question, or some crossing sequence repeats. But if some crossing sequence repeats, then the TM moves right with some easily detectable pattern, and the question is again decidable.
8.5 UNDECIDABILITY OF POST'S CORRESPONDENCE PROBLEM Undecidable problems arise in a variety of areas. In the next three sections we explore some of the more interesting problems in language theory and develop techniques for proving particular problems undecidable. We begin with Post's Correspondence Problem, it being a valuable tool in establishing other problems to be undecidable.
An instance of Post's Correspondence Problem (PCP) consists of two lists, A = wu .
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.
, wk and B = x u .
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. , xk , of strings over some alphabet Z. This instance KTUNOTES.IN Downloaded from Ktunotes.in 194 UNDECIDABILITY of PCP has a solution if there is any sequence of integers i l9 i 2 , . .., m , with m > 1, such that Wfp VV,- 2 , Wim = Xjj, x t -2 , x fm .
The sequence i l9 .
. ., fm is a solution to this instance of PCP.
Example 8.6 Let X = {0, 1}. Let A and B be lists of three strings each, as defined in Fig. 8.15. In this case PCP has a solution. Let m = 4, i x = 2, i2 = 1, i 3 = 1, and z4 = 3. Then w 2 w 1 w 1 w3 = x 2 x 1 x 1 x 3 = 101111110.
List /i List B i w, 1 1 111 2 10111 10 3 10 0 Fig. 8.15 An instance of PCP.
Example 8.7 Let I = {0, 1}. Let A and B be lists of three strings as shown in Fig.
8.16.
List A List B i Wi Xi 1 10 101 2 011 11 3 101 011 Fig. 8.16 Another PCP instance.
Suppose that this instance of PCP has a solution iu i2 , .
.
.
, im . Clearly, i t = U since no string beginning with w 2 = 011 can equal a string beginning with x 2 = 11; no string beginning with vv3 = 101 can equal a string beginning with x 3 = 011.
We write the string from list A above the corresponding string from B. So far we have 10 101 KTUNOTES.IN Downloaded from Ktunotes.in 8.5 | UNDECIDABILITY OF POST'S CORRESPONDENCE PROBLEM 195 The next selection from A must begin with a 1. Thus i 2 = 1 or i2 = 3. But i2 = 1 will not do, since no string beginning with w t w t = 1010 can equal a string beginning with x t x t = 101101. With i2 = 3, we have 10101 101011 Since the string from list B again exceeds the string from list A by the single symbol 1, a similar argument shows that i3 = i4 = • • • = 3. Thus there is only one sequence of choices that generates compatible strings, and for this sequence string B is always one character longer. Thus this instance of PCP has no solution.
A modified version of PCP We show that PCP is undecidable by showing that if it were decidable, we would have an algorithm for L^. First, we show that, if PCP were decidable, a modified version of PCP would also be decidable.
The Modified Post's Correspondence Problem (MPCP) is the following: Given lists A and B, of k strings each from Z, say A = w l9 w2 , wk and B = x u x2 , k , does there exist a sequence of integers, i l9 i 2 , • K> sucn tnat wiw^w,., •• w ir = x l x il x i2 •• x Ir ?
The difference between the MPCP and PCP is that in the MPCP, a solution is required to start with the first string on each list.
Lemma 8.5 If PCP were decidable, then MPCP would be decidable. That is, MPCP reduces to PCP.
Proof Let A = w l9 w2 , wk and B = x u x2 , xk be an instance of the MPCP. We convert this instance ofMPCP to an instance of PCP that has a solution if and only if our MPCP instance has a solution. If PCP were decidable, we would then be able to solve the MPCP, proving the lemma.
Let Z be the smallest alphabet containing all the symbols in lists A and By and let § and $ not be in Z. Let y f be obtained from w,- by inserting the symbol $ after each character of w, and let z-t be obtained from x t by inserting the symbol $ ahead of each character of x,. Create new words yo = zo = z i> KTUNOTES.IN Downloaded from Ktunotes.in 196 UNDECIDABiLITY Let C = y0 , y l9 yk + x and D = z0 , z l9 .
.
. , z fc + 1 . For example, the lists C and D constructed from the lists A and B of Example 8.6 are shown in Fig. 8.17.
MPCP List A List B i Wi i 1 1 111 2 10111 10 3 10 0 List C List D j yi Zi 0 414 414141 1 H 414141 2 1404141414 4140 3 1404 40 4 $ 4$ Fig. 8.17 Corresponding instances of MPCP and PCP.
In general, the lists C and D represent an instance of PCP. We claim that this instance of PCP has a solution if and only if the instance ofMPCP represented by lists A and B has a solution. To see this, note that if 1, i l9 z 2 > .
.
.
, ir is a solution to MPCP with lists A and B, then 0, i u i2 , zr , k + 1 is a solution to PCP with lists C and D. Likewise, if i l9 i2 , ir is a solution to PCP with lists C and D, then i l = 0 and ir = k 4-1, since y0 and z0 are the only words with the same index that begin with the same symbol, and yk+1 and zfc+1 are the only words with the same index that end with the same symbol. Let j be the smallest integer such that ij = k 4-1. Then i u i 2 , .
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.
, ij is also a solution, since the symbol $ occurs only as the last symbol of yk+1 and zfc+1 , and, for no /, where 1 </<;', is ie = k 4-1. Clearly 1, z 2 , «3» • • • , ij-1 is a solution to MPCP for lists A and B.
If there is an algorithm to decide PCP, we can construct an algorithm to decide MPCP by converting any instance of MPCP to PCP as above.
Undecidability of PCP Theorem 8.8 PCP is undecidable.
Proof With Lemma 8.5, it is sufficient to show that if MPCP were decidable, then it would be decidable whether a TM accepts a given word. That is, we reduce to MPCP, which by Lemma 8.5 reduces to PCP. For each M and w we construct an instance of MPCP that has a solution if and only ifM accepts w. We do this by constructing an instance of MPCP that, if it has a solution, has one that starts with #g0 w^a l qifii^ "' #a fc gk /?fc #, where strings between successive #'s are successive ID's in a computation of M with input w, and qk is a final state.
KTUNOTES.IN Downloaded from Ktunotes.in 8.5 | UNDECIDABILITY OF POST'S CORRESPONDENCE PROBLEM 197 Formally, the pairs of strings forming lists A and B of the instance of MPCP are given below. Since, except for the first pair, which must be used first, the order of the pairs is irrelevant to the existence of a solution, the pairs will be given without indexing numbers. We assume there are no moves from a final state.
The first pair is: List A List B # #^0 w# The remaining nairc are omiir»fH ac finllnwc givjujj^u da lUiiUWa.
Group I T ict A T ict R X AT for each X in T.
# Group II. For each q in Q — F, p in Q, and X, Y, and Z in T List A List 5 qX Yp if%,X) = (p, Y,R) ZqX pZY if%,X) = (p, Y,L) Yp# if%B) = (p, Y,R) Zq# pZY# if%,B) = (p, Y,L) Group III. For each q in F, and X and Y in F: List ,4 List £ XqY q Xq q qY q Group IV List A List B g# # # for each g in F.
Let us say that (x, y) is a partial solution to MPCP with lists /I and B if x is a prefix of y, and x and y are the concatenation of corresponding strings of lists A and B respectively. If xz = y, then call z the remainder of (x, y).
Suppose that from ID q0 w there is a valid sequence of /c more ID's. We claim that there is a partial solution (x, y)= (#g0 w#a 1 g 1 ^ 1 # ••• #^0 w#a 1 g 1 ^ 1 # ••• #a^&#).
Moreover, this is the only partial solution whose larger string is as long as \y.
The above statement is easy to prove by induction k. It is trivial for k = 0, since the pair (#, #g0 w#) must be chosen first.
KTUNOTES.IN Downloaded from Ktunotes.in 198 UN DECIDABILITY Suppose that the statement is true for some k and that qk is not in F. We can easily show that it is true for k 4-1. The remainder of the pair (x, y) is z = ak qk pk#. The next pairs must be chosen so that their strings from list A form z. No matter what symbols appear to the right and left of qk , there is at most one pair in Group II that will enable the partial solution to be continued past qk . This pair represents, in a natural way, the move of M from ID ock qk Pk The other symbols of z force choices from Group I. No other choices will enable z to be composed of elements in list A.
We can thus obtain a new partial solution, (y, yafc+1 qk+i /?fc+1 #).
It is straightforward to see that otk+l qk+ pk+1 is the one ID that M can reach on one move from ak qk Pk . Also, there is no other partial solution whose length of the second string equals \yai.k + x qk + 1 fik + 1# | -In addition, if qk is in F, it is easy to find pairs from Groups I and III which, when preceded by the partial solution (x, y) and followed by the pair in Group IV, provide a solution to MPCP with lists A and B.
Thus if M, started in ID q0 w, reaches an accepting state, the instance of MPCP with lists A and B has a solution. IfM does not reach an accepting state, no pairs from groups III or IV may be used. Therefore, there may be partial solutions, but the string from B must exceed the string from A in length, so no solution is possible.
We conclude that the instance of MPCP has a solution if and only if M with input w halts in an accepting state. Since the above construction can be carried out for arbitrary M and w, it follows that if there were an algorithm to solve MPCP, then there would be an algorithm to recognize L^, contradicting Theorem 8.5.
Example 8.8 Let M = 42, 4 3 }, {0, 1, B}> {0, 1}, 3, qu 5, {q 3 }\ and let S be defined by: 0) B) (<?2, 1, R) (<?2, 0, L) (92, 1, L) <?2 (</3, 0, L) (<J..0,K) (qi, 0, R) <?3 Let w = 01. We construct an instance of MPCP with lists A and B. The first pair is # for list A and #^ 101# for list B. The remaining pairs are: Group I List A List B 0 0 1 1 # # KTUNOTES.IN Downloaded from Ktunotes.in 8.5 | UNDECIDABILITY OF POST'S CORRESPONDENCE PROBLEM 199 Group II JLlSl A T tot P JLlSl i> n n q x K) 1 ^ 1 /7 1 {/ 2U1tt I 0g 20 <? 300 lq20 q3 10l Oq, 92 0^2 from S(q l9 0) = (92, 1, from (5^!, 1)= (92, 0, L) from ^(^!, B) = (92, 1, ^)t from S(q 2 , 0) = (93, 0, L) from <5(g2 > from S(q 2y 1) = B) = (9., 0, R) (92, 0, K) Group III Group IV List A List 5 O43O 43 0q3 1 q3 1^0 ^3 I43I ^3 0^3 ^3 1^3 43 <?30 <? 3 ^1 ?3 List A List £ <?3## # Note that M accepts input w = 01 by the sequence of ID's: qfll, lq2 l, \0qu 1<?201, g3 101.
Let us see if there is a solution to the MPCP we have constructed. The first pair gives a partial solution (#, ftqfilft). Inspection of the pairs indicates that the only way to get a longer partial solution is to use the pair (qfi, lq 2 ) next. The resulting partial solution is (#<?i0, #g 101#lg 2 ). The remainder is now 1#1<? 2 -The next three pairs chosen must be (1, 1), (#, #), and (1, 1). The partial solution becomes (#g 101#l, #g 101#lg 2 ll)- The remainder is now <72 1#1- Continuing the argument, we see that the only partial solution, the length of whose second string is 14, is (x, x0q t #l), where x = #g 101# lq2 l# 1.
Here, we seemingly have a choice, because the next pair used could be (0, 0) or (0^j#, g201#). In the former case we have (xO, xO^j 10) as a partial solution.
But this partial solution is a "dead end." No pair can be added to it to make another partial solution, so, surely, it cannot lead to a solution.
t Since B is never printed, we can omit pairs where B is to the right of the state. Group III pairs also omit those with B on one or both sides of the state.
KTUNOTES.IN Downloaded from Ktunotes.in 200 UN DECIDABILITY In a similar manner, we continue to be forced by our desire to reach a solution to choose one particular pair to continue each partial solution. Finally, we reach the partial solution (y, yl#g3 10), where y=#q l01#lq2 l#10q l #lq20.
Since q 3 is a final state, we can now use pairs in Groups I, III, and IV to find a solution to the instance of MPCP. The choice of pairs is (1, 1), (#, #), (q3 l, q 3 ), (0, 0), (1, 1), (#, #), (q 30, q3 ), (1, 1), (#, #), (q 3 l, q3 )y (#, #), (4 3 ##, #).
Thus, the shortest word that can be composed of corresponding strings from lists A and B, starting with pair 1 is t^Ol + l^^lO^^l^OW^lOl^!^!^.
An application of PCP Post's correspondence problem can be used to show that a wide variety of prob-lems are undecidable. We give only one application here: the undecidability of ambiguity for context-free grammars. The reader should consult the exercises at the end of the chapter for additional applications.
Theorem 8.9 It is undecidable whether an arbitrary CFG is ambiguous.
Proof Let A = wu w2 , w„ and B = x u x 2 , xn be two lists of words over a finite alphabet Z. Let a u a 2 , De new symbols.
Let LA = K wi2 • • • wim aim aim _ t ah \ m > 1} and LB = {xh xi2 •••x lm a lm «Im _ 1 |m> 1}.
Let G be the CFG ({5, s u s2}ju{fll) ...,„}, p, s% where P contains the productions S - SA , S -> SB and for 1 < i < n, SA w, 5^ a x , SA -+Wiah SB -+ XiSB a iy and S^^x.a,. The grammar G generates the language LA u LB .
If the instance (A y B) of PCP has a solution, say iu i 2 , .
.
.
, im , then there is a word x^Xfj---x im a im a im _ l ••• a,-, in that equals the word w il w i2 "' wIm flIm flim _ 1 ••• a,-, in LB . This word has a leftmost derivation beginning S-+Sa> and another beginning S-+ SB . Hence in this case G is ambiguous.
KTUNOTES.IN Downloaded from Ktunotes.in 8.6 | VALID AND INVALID COMPUTATIONS OF TM'S 201 Conversely, suppose G is ambiguous. Since the a's dictate the productions used, it is easy to show that any word derived from SA has only one leftmost derivation from SA . Similarly, no word derived from SB has more than one left-most derivation from SB . Thus it must be that some word has leftmost deviations from both SA and SB . If this word is ya im aim _ l •• ah , where y is in Z, then i l9 i2 , .
.
. , im is a solution to PCP.
Thus G is ambiguous if and only if the instance (A, B) of PCP has a solution.
We have thus reduced PCP to the ambiguity problem for CFG's. That is, if there were an algorithm for the latter problem, we could construct an algorithm for PCP, which by Theorem 8.8 does not exist. Thus the ambiguity problem for CFG's is undecidable.
8.6 VALID AND INVALID COMPUTATIONS OF TM'S: A TOOL FOR PROVING CFL PROBLEMS UNDECIDABLE While PCP can be reduced easily to most of the known undecidable problems about CFL's, there is a more direct method that is instructive. We shall in this section show direct reductions of the membership problem for TM's to various problems about CFL's. To do so we need to introduce the notions of valid and invalid Turing machine computations.
A valid computation of a Turing machine M = (Q 9 Z, T, 3, q0 , B, F), for the purposes of this section, is a string w 1 #wf#w3 #w4# ••• such that: 1) each vv f is an ID of M, a string in TQT not ending with B, 2) w l is an initial ID, one of the form q0 x for x in Z, 3) w„ is a final ID, that is, one in rFT, and 4 ) wi hr w I +1 for 1 < i < n.
We assume without loss of generality that Q and T are disjoint, and # is in neither Q nor T.
The set of invalid computations of a Turing machine is the complement of the set of valid computations with respect to the alphabet T u Q u {#}.
The notions of valid and invalid computations are useful in proving many properties of CFL's to be undecidable. The reason is that the set of invalid compu-tations is a CFL, and the set of valid computations is the intersection of two CFL's.
Lemma 8.6 The set of valid computations of a Turing machine M is the intersec-tion of two CFL's, L x and L2 , and grammars for these CFL's can be effectively constructed from M.
Proof Let M = (Q, Z, T, <5, q0 , B, F) be a TM. Both CFL's L x and L2 will consist of strings of the form X!#x 2# • • #xm #. We use L x to enforce the condition that x i I— (Xi+) R f°r ocd i and L2 to enforce the condition xf |— xl+ x for even i. L2 also enforces the condition that X!
is an initial ID. That xm is a final ID or its KTUNOTES.IN Downloaded from Ktunotes.in 202 UNDECIDABILITY reverse is enforced by L x or L2 depending on whether m is odd or even, respec-tively. Then L x n L2 is the set of valid computations of M.
To begin, let L3 be {y#z |y z}-s easv to construct a PDA P to accept L3 . P reads the input up to the #, checking in its finite control that y is of the form rgr In the process, P places on its stack the ID z such that y \jf z, where y is the input before the #. That is, when the input to P is a symbol of T, P pushes that symbol onto the stack. If the input is a state q in Q, P stores q in the finite control and reads the next input symbol, say X (if the next symbol is #, take X to be B).
If d(q, X) = (p, 7, R\ then P pushes Yp onto the stack.
If 5(q, X) = (p, 7, L), let Z be on top of the stack. Then P replaces Z by pZy (but if the input last read was #, and Y = B, just replace Z by pZ, or by p if Z is also £). After reading the #, P compares each input symbol with the top stack symbol. If they differ, P has no next move and so dies. If they are equal, P pops the top stack symbol. When the stack is emptied, P accepts.
Now, let L x = (L 3 #)({£} u rFT#). By Theorems 5.4 and 6.1, there is an algorithm to construct a CFG for Lv In a similar way, we can construct a PDA for L4 = {y K#z \y \jf z}. The construction of G2 for l 2 = 40£#(L 4#)({£} u rpr#) is then easy, and by Theorem 6.1 there is an algorithm to construct a CFG G 2 for L2 . Now Lj n L2 is the set of valid computations of M. That is, if X!#x 2# ••• #xm# is in Li n L2 , then Lj requires that x f (x, + t ) K for odd i; L2 requires that x, is initial, and xf (-^ xi+ for even i. That the last ID has an accepting state is enforced by L x for m odd and by L2 for m even.
Theorem 8.10 It is undecidable for arbitrary CFG's Gj and G 2 whether Ufi) n L(G 2 ) is empty.
Proo/ By Lemma 8.6 we can construct from M grammars G x and G 2 such that L(G X ) n L(G 2 ) is the set of valid computations of M. If there is an algorithm A to tell whether the intersection of the languages of two CFG's is empty, we can construct an algorithm B to tell whether L(M) = 0 for arbitrary TM M. Simply design B to construct G x and G 2 from M as in Lemma 8.6, then apply Algorithm A to tell whether L(G t ) n L(G 2 ) is empty. If the intersection is empty, then there are no valid computations of M , so L(M) = 0.
If the intersection is not empty, L(M) 0. That is, the problem of emptiness for r.e. sets reduces to the problem of intersection for CFG's.
Algorithm B cannot exist, however, since L(M) = 0 is undecidable by Theorem 8.6. Therefore A does not exist, so it is undecidable whether the intersec-tion of two CFL's is empty.
Although two context-free languages are required to represent the valid com-putations of a Turing machine, the set of invalid computations is itself a CFL. The reason is that we no longer need to guarantee simultaneously for each i that KTUNOTES.IN Downloaded from Ktunotes.in 8.6 | VALID AND INVALID COMPUTATIONS OF TMS 203 W; |— wi+1 . We need only guess where an error occurs. That is, we must verify for one i that w f |— wf+1 is false.
Lemma 8.7 The set of invalid computations of a Turing machine M = (Q, E, T, d, q0 , B, F) is a CFL.
Proof If a string w is an invalid computation, then one of the following condi-tions holds.
1) w is not of the form x 1 #x 2# " #m#> where each x £ is an ID of M.
2) x t is not initial; that is, x x is not in g0 £-3) xm is not final; that is, xm is not in rFT.
4) Xi \w (i+i) R is false for some odd i.
5) xf 1-^- x I + j is false for some even i.
The set of strings satisfying (1), (2), and (3) is regular, and an FA accepting it is easily constructed. The sets of strings satisfying (4) and (5) are each CFL's. We prove this contention for (4); a similar argument prevails for (5). A PDA P for (4) nondeterministically selects some xf that is preceded by an even number of #'s and while reading x, stores on its stack the ID z such that x, |— z, with the right end of z at the top of the stack. After finding # on the input, P compares z with the following xi+ 1 . If z xi+ i, then P scans its remaining input and accepts.
The set of invalid computations is the union of two CFL's and a regular set.
By Theorem 6.1 it is a CFL, and a grammar for this language can be constructed effectively.
Theorem 8.11 It is undecidable for any arbitrary CFG G whether L(G) = £.
Proof Given an arbitrary TM M, we can effectively construct a CFG G with terminal alphabet Z, such that L(G) = Z if and only if L(M) = 0. That is, by Lemma 8.7 we may construct a CFG G that generates the invalid computations of M. Thus if for arbitrary G, L(G) = E were decidable, then we could decide for arbitrary M whether L(M) = 0, a contradiction.
Other consequences of characterization of computations by CFL's Many other results follow from Theorem 8.11.
Theorem 8.12 Let G x and G2 be arbitrary CFG's and R an arbitrary regular set.
The following problems are undecidable.
Proof Fix G 2 to be a grammar generating Z, where £ is the terminal alphabet of G t . Then (1) and (2) are equivalent to Ufi x ) = £. Fix R = Z, and (3) and (4) are 1) L(G 1 ) = L(G2 ).
3) L(G 1 ) = R.
2) UG 2)^UG l ).
4) R^UGJ.
KTUNOTES.IN Downloaded from Ktunotes.in 204 UN DECIDABILITY equivalent to L(G t ) = X. Thus the undecidable problem of whether a CFL is X reduces to (1) through (4), and each of these problems is undecidable as well.
Note that by Theorems 5.3 and 5.4, one can convert effectively between PDA's and CFG's, so Theorems 8.10, 8.11, and 8.12 remain true if CFL's are represented by PDA's instead of CFG's. Also, the regular set R in Theorem 8.12 can be represented by a DFA, NFA, or regular expression as we choose.
One should observe also that the question L(G) ^ R is decidable. The reason is that L(G) c R if and only if L(G) n R = 0. But L(G) n R is a CFL, and hence its emptiness is decidable.
There are some additional properties of context-free languages that we can show to be undecidable by observing that if a TM has valid computations on an infinite set of inputs, its set of valid computations is not, in general, a CFL.
However, we first modify each Turing machine M in a trivial way by adding two extra states whose sole purpose is to ensure that M makes at least two moves in every computation. This can be done without otherwise modifying the computa-tion performed by M. The purpose of the modification is to force each valid computation to contain at least three ID's and thus ensure that the set of valid computations is a CFL if and only if M accepts a finite set.
Lemma 8.8 Let M be a Turing machine that makes at least three moves on every input. The set of valid computations ofM is a CFL if and only if the set accepted by M is a finite set.
Proof If the set accepted by M is finite, the set of valid computations of M is finite and hence a CFL. Assume the set accepted by M is infinite and the set L of valid computations is a CFL. Since M accepts an infinite set, there exists a valid computation where the w,'s are ID's, and | w 2 | is greater than the constant n in Ogden's lemma.
Mark the symbols of w 2 as distinguished. Then we can "pump" w 2 without pump-ing both W! and w3 , thus getting an invalid computation that must be in L We conclude that the valid computations do not form a CFL.
Theorem 8.13 It is undecidable for arbitrary CFG's G x and G 2 whether 1) L(Gj) is a CFL; 2) LfGj n L(G 2 ) is a CFL.
Proof 1) Given an arbitrary Turing machine M, modify M without changing the set accepted, so that M makes at least two moves on every input. Construct CFG G generating the invalid computations. Ulfi) is a CFL if and only ifM accepts a finite set.
2) Proceed as in (1), but construct CFG's G t and G 2 such that L(G X ) n L(G 2 ) is the set of valid computations of M.
KTUNOTES.IN Downloaded from Ktunotes.in 8.7 | greibach's theorem 205 8.7 GREIBACH'S THEOREM There is a striking similarity among the proofs of undecidability in language theory. This suggests that there is an analog of Rice's theorem for classes of languages such as the CFL's, and indeed there is.
Let us focus our attention on a class of languages such as the CFL's, and on a particular system (such as CFG's or PDA's) for interpreting finite-length strings as names of languages. Consider a class # of languages with the property that, given names (e.g., grammars) of languages Lx and L2 in # and a name (e.g., a finite automaton) for a regular set R, we can effectively construct names for RLU L t R, and Li u L2 . Then we say that the class is effectively closed under concatena-tion with regular sets and union. Assume furthermore that L = E is undecidable for the class as is the case for the CFL's. The next theorem shows that a wide variety of problems are undecidable for the class Theorem 8.14 (GreibacKs Theorem) Let ^ be a class of languages that is effec-tively closed under concatenation with regular sets and union, and for which "= £" is undecidable for any sufficiently large fixed £. Let P be any nontrivial property! that is true for all regular sets and that is preserved under /a, where a is a single symbol. (That is, if L has the property P, so does L/a = {w | wa is in L}.) Then P is undecidable for Proof Let L 0 ^ £ be a member of^ for which P(L 0 ) is false where I is suffici-ently large so that is undecidable. For any L ^ Z in # construct Lj = Lq#E u £#L. L { is in c €, since # is effectively closed under concatenation with regular sets and under union. Now if L = £, then L x = !#£, which is a regular set, and hence P(L t ) is true. If L ^ 2, then there exists w not in L. Hence L 1 /#w = L0 . Since P is preserved under quotient with a single symbol, it is preserved under quotient with the string #w, by induction on |w|. Thus P(L X ) must be false, or else P(Lo) would be true, contrary to our assumption. Therefore P(L t ) is true if and only if L = X. Thus " = £" for (€ reduces to property P for ^, and hence P is undecidable for ( £.
Applications of Greibach's theorem Theorem 8.14 can be used to show, for example, that it is undecidable if the language generated by a CFG is regular. Note that this question is different from asking if the language generated is equal to some particular regular set R, as was asked in Theorem 8.12.
Theorem 8.15 Let G be an arbitrary CFG.
It is undecidable whether L(G) is regular.
Proof The CFL's are effectively closed under concatenation with regular sets and under union. Let P be the property that L is regular. P is nontrivial for the CFL's, t Technically, a property is just a subset of ( €. We say "L has property Pn or "P(L)" to mean L is a member of P.
KTUNOTES.IN Downloaded from Ktunotes.in 206 UNDECIDABILITY is true for all the regular sets, and is preserved under quotient with a single symbol by Theorem 3.6. Note that the regular sets are effectively closed under quotient with another regular set, although Theorem 3.6 does not claim this (see the discussion following that theorem). Thus by Theorem 8.14, P is undecidable for CFL's.
~ Theorem 8.15 allows us to show that a property is undecidable by showing that the property is preserved under quotient with a single symbol. This latter task is often relatively easy as, for example, in proving that inherent ambiguity is undecidable.
Lemma 8.9 Let P be the property that a CFL is not inherently ambiguous. Then P is preserved under quotient with a single symbol.
Proof Let G = (V, T, P, S) be an unambiguous CFG. Let Ga = (V u {[A/a]\A is in V}, T, Pa , [S/a]), where Pa contains 1) all productions of P, 2) [A/a] -+ a if A -> aa is in P, 3) [A/a] - a[B/a] if A -+ aBp is in P, and 0±>c.
We claim that L(Ga ) = L(G)/a and that Ga is unambiguous. To see this, first show by an easy induction that 1) [S/a] ^> a if and only if S ^> oca, and 2) [S/a] ±> a[A/a] if and only if S ±> aA.
That L(Gfl ) = L(G)/a follows immediately. Assume Ga is ambiguous. Then there must be two leftmost derivations 1) [S/a]^0^>a>xand 2) [S/a]>y±>a>x where P ± y.
But then in G we have two leftmost derivations of the string xa, a contradiction.
Thus Ga must be unambiguous. We conclude that unambiguity is preserved under quotient with a single symbol.
Theorem 8.16 Inherent ambiguity for CFL's is undecidable.
Proof By Theorem 4.7, P is nontrivial. By Lemma 8.9 it is preserved under quotient with a single symbol. It is easy to show that P is true for all regular sets.
That is, every regular set has an unambiguous CFG. (The reader may look ahead to Theorem 9.2 for a construction of an unambiguous CFG from an arbitrary DFA.) Thus by Theorem 8.14, inherent ambiguity for CFL's is undecidable.
KTUNOTES.IN Downloaded from Ktunotes.in 8.8 | INTRODUCTION TO RECURSIVE FUNCTION THEORY 207 8.8 INTRODUCTION TO RECURSIVE FUNCTION THEORY We mentioned in Section 7.3 that each Turing machine can be thought of as computing a function from integers to integers, as well as being a language recog-nizer. For every Turing machine M and every k, there is a functionf$ (iu f2 , .
.
.
, ik ) that takes k integers as arguments and produces an integer answer or is undefined for those arguments. If M started with 0 fl 10,2 l ••• 0 ik on its tape halts with (V on its tape, then we say f$ (iu .
.
. , ik ) = j. IfM does not halt with a tape consisting of a block of O's with all other cells blank, then f$ (iu ik ) is undefined. Note that the same Turing machine can be thought of as a language recognizer, a computer of a function with one argument, a computer of a different function of two arguments, and so on.
If i is an integer code for a TM M, as described in Section 8.3, and k is understood, then we shall often write/ in place off%K Recall that a function computed by a Turing machine is called a (partial) recursive function. If it happens to be defined for all values of its arguments, then it is also called a total recursive function.
The constructions on Turing machines given earlier in this chapter and the previous one can be expressed as total recursive functions of a single variable.
That is, an algorithm A that takes as input the binary code for a TM M and produces as output the binary code for another TM M' can be viewed as a function g of one variable. In particular, let i be the integer representing M and ; be the integer representing M'. Then g(i) = j. Technically, the TM B that com-putes g is not A, but rather one that converts its unary input to binary, simulates A and then converts its output to unary.
The 5mn-theorem Our first theorem, called the Smn-theorem, says that given a partial recursive func-tion g(x, y) of two variables, there is an algorithm one can use to construct from a TM for g and a value for x, another TM which with input y computes g(x, y).
Theorem 8.17 Let g(x, y) be a partial recursive function. Then there is a total recursive function o of one variable, such thatf 0(x) (y) = g(x 9 y) for all x and y. That is, if g(x) is treated as the integer representing some TM M„ then f {M } x(y) = 9(x9 y)-Proof Let M compute g. Let A be a TM that given input x, written in unary, constructs a TM Mx that when given input v, shifts it right and writes 01 to its left; Mx then returns its head to the left end and simulates M. The output of A is the unary representation of an integer is entered, when the "oracle" answers another question.
Observe that if A is a recursive set, then the oracle can be simulated by another Turing machine, and the set accepted by the TM with oracle A is recur-sively enumerable. On the other hand, if A is not a recursive set and an oracle is available to supply the correct answer, then the TM with oracle A may accept a set that is not recursively enumerable. We denote the Turing machine M with oracle A by MA . A set L is recursively enumerable with respect to AifL= L(MA ) for some TM M. A set L is recursive with respect to A if L = L(MA ) for some TM MA that always halts. Two oracle sets are equivalent if each is recursive in the other.
A hierarchy of undecidable problems We can now rephrase the question at the beginning of the section as "What sets can be recognized given an oracle for the emptiness problem?" Clearly not all sets can be r.e. with respect to the emptiness problem, since there is an uncountable number of sets and only a countable number of TM's. Consider the oracle set S x = {(M)\L(M) = 0}, which is not an r.e. set (recall that (M> is the binary code for TM M). Now consider TM's with oracle Sv These machines have a halting problem that is not recursive in Sj. By defining an oracle for the emptiness problems for TM's with oracle S,, and so on, we can develop an infinite hierarchy of undecidable problems. More specifically, define Si+l = {(M)\L s>(M) = 0}.
Si+ !
is an oracle for solving the emptiness problem for computations with respect to 5,. We can now classify some undecidable problems (but not all such problems) by showing their equivalence to a set S, for some particular i.
Theorem 8.19 The membership problem for TM's without oracles is equivalent to S t .
Proof Construct MSi that, given (M, w) on its input, constructs the code for a TM M' that accepts 0 if w is not in L(M) and accepts (0 + 1) otherwise. The construction of M' was given in Example 8.2. M Sl then enters state q , with the code for M' to the right of its head and accepts if and only if qn is entered. Thus the membership problem for TM's without oracle is recursive in S t .
Conversely, we can show there is a Turing machine with the membership problem as oracle, that recognizes Sv (Strictly speaking, the oracle is L^.) To show S x is recursive in L^, construct a TM M 2 that, given (M>, constructs a new TM M' operating as follows: M' ignores its input; instead, M ' uses the pair generator to generate all pairs (/, When (i, j) is generated, M' simulates M for i steps on the t Note that the TM can remember its prior state by writing that state on its tape just before entering q >.
KTUNOTES.IN Downloaded from Ktunotes.in 8.9 | ORACLE COMPUTATIONS 211 yth input word to Af , words being numbered in the usual ordering. If Af accepts, M' accepts its own input. If L(Af ) = 0, then L(M') = 0. If L(Af ) + 0, then Af' accepts all its own inputs, e in particular. Thus M 2" may query its oracle whether Af' accepts £, that is, whether (Af, e) is in Z^. If so, Af 2 rejects M. Otherwise Af 2 accepts M. Thus 5 t is recursive in L^.
Next consider the problem whether L(M) = Z, where £ is the input alphabet for TM Af . In a sense, this problem is "harder" than membership or emptiness, because, as we shall see, the " = £" problem is equivalent to S2 , while emptiness and membership are equivalent to S x . While this difference means nothing in practical terms since all these problems are undecidable, the results on compara-tive degree of difficulty suggest that when we consider restricted versions of the problems, the " = £" problem really is harder than membership or emptiness. For context-free grammars, the emptiness and membership problems are decidable, while by Theorem 8.11 the problem whether L(G) = E is undecidable. For another example, consider regular expressions. The emptiness and membership problems are each decidable efficiently, in time polynomial in the length of the expression, while the problem whether a given regular expression r is equivalent to I has been proved almost certainly to require time exponential in the length ofr.t Theorem 8.20 The problem whether L(M) = Z is equivalent to S 2 .
Proof We construct a TM M% 2 that takes an arbitrary TM M and constructs from it Ms \ a TM with oracle S t that behaves as follows. MSl enumerates words x, and for each x uses oracle S x to tell whether M accepts x. The technique whereby S x can be used to answer the membership question was covered in Theorem 8.19. M Sy accepts its own input if any x is not accepted by M. Thus |Z otherwise.
Ml 2 with input M constructs Ms \% then asks its own oracle, 5 2 , whether L(M Sl ) = 0. If so, M\ 2 accepts Af, and M\ 2 rejects otherwise. Thus M% 2 accepts {|L(M) = Z}.
Now we must show that S 2 is recursive in the " = £" problem. That is, let be the set of codes for ordinary Turing machines that accept all strings over their input alphabet. Then there is a TM Mj that accepts S 2 .
Before constructing Mi, we first define a valid computation of a TM Afs 1 using oracle S v A valid computation is a sequence of ID's, just as for ordinary Turing machines. However, if one ID has state q,, and the next ID has state qn , then MSl has queried the oracle whether some TM N accepts 0 and received the t Technically, the problem is "complete in polynomial space"; see Chapter 13.
t Note that 5 X is not part of M. Actually M\ z constructs the state transitions of oracle machine M, which will work correctly given S t as oracle.
KTUNOTES.IN Downloaded from Ktunotes.in 212 UNDECIDABILITY answer "no." To demonstrate that this answer is correct, we insert a valid compu-tation of ordinary TM N, showing that N accepts some particular input. If the next state is qy , however, we insert no computation of N.
Now, let us describe how M4 behaves on input MSl . Mj creates an ordinary TM M' that accepts all the invalid computations ofMSl . To check that a string is not a valid computation, M' checks if the format is invalid (as in Lemma 8.7), or if one ID of MSx does not follow on one move from the previous ID ofMSl in the sequence, or if a computation of an ordinary TM N inserted between ID's ofMSl with states q., and qn is not valid.
The only difficult part to check is when one ID of MSx has state q>, and the next ID has state qr Then M' must determine if "yes" is not the correct answer, so these two ID's do not follow in sequence. Let N be the TM about which the query is made. M' uses the pair generator and, when (i, j) is generated, simulates N for i steps on the ;th input. If N accepts, M' determines that L(N) ± 0> so "yes" is the wrong answer. Thus the computation is not a valid one, and M' accepts this computation.
Now M' accepts all strings over its input alphabet if and only if L(MSl ) = 0, that is, MSl has no valid computations. Mj may query its oracle whether M' accepts Z. The code for MSl is in S 2 if and only if L(M') = Z. Thus S 2 is recursive in L^.
Turing reducibility We have, throughout this chapter, dealt with a notion called "reducibility," in which we reduced language L l to L2 by finding an algorithm that mapped strings in L Y to strings in L2 and strings not in L 1 to strings not in L2 . This notion of reducibility is often called many-one reducibility, and while it was all we needed, it is not the most general notion. A more general technique is called Turing reducibil-ity, and consists simply of showing that L x is recursive in L 2 .
If Lj is many-one reducible to L2 , then surely L, is Turing-reducible to L2 . In proof, suppose / is a function computable by a TM that always halts, such that f(x) is in L2 if and only if x is in L,. Then consider the oracle TM M 7 2 that, given input x, computes f(x) and then enters state q, with f(x) to the right of its head.
M Ll accepts if and only if it then enters qy . Surely L(M l 2 ) = L l9 so L x Turing-reduces to L2 . The converse is false, and a proof is suggested in the exercises.
If Lj Turing-reduces to L2 , and L x is undecidable, then so is L2 . For if L2 were recursive, then the oracle TM M Ll such that L(M L2 ) = can be simulated by an ordinary TM that always halts. Thus one could use a Turing reduction to show that L2 is undecidable, given that Lj was undecidable, even in circumstances where a many-one reduction of L x to L2 did not exist, or was hard to find.
The notion of many-one reducibility has its virtues, however. If L y is many-one reducible to L2 , and L x is not r.e., we can conclude L2 is not r.e. Yet this KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 213 conclusion cannot be drawn for Turing reducibility. For example, Z^is a non-r.e.
language that Turing-reduces to the r.e. language L M . We can recognize L M given L M as an oracle, by asking whether (M, w) is in and accepting if and only if the answer is no.
We see that the more difficult form of reducibility (many-one) enables us to draw conclusions we cannot draw with the easier form of reducibility (Turing). In Chapter 13, where we study bounded reducibility, we shall see additional examples of how more difficult forms of reductions yield conclusions not achiev-able by easier forms.
EXERCISES 8.1 Suppose the tape alphabets of all Turing machines are selected from some infinite set of symbols au a 2 , ... Show how each TM may be encoded as a binary string.
8.2 Which of the following properties of r.e. sets are themselves r.e.?
a) L contains at least two strings.
b) L is infinite.
c) L is a context-free language.
d) L = L.
S 8.3 Show that it is undecidable whether a TM halts on all inputs.
8.4 A Post Tag System is a finite set P of pairs (a, p) chosen from some finite alphabet, and a start string y. We say that Sp if (a, p) is a pair. Define ^> to be the reflexive, transitive closure of =>, as for grammars. Show that for given tag system (P, y) and string <5 it is undecidable whether y^>3. [Hint: For each TM M let y be the initial ID of M with blank tape, followed by a marker #, and select the pairs so that any ID must become the next ID after a sequence of applications of the rules, unless that ID has an accepting state, in which case the ID can eventually become i. Then ask if y ^> e.] 8.5 Show that there is no algorithm which given a TM M defining a partial recursive function / of one variable, produces a TM M' that defines a different function of one variable.
8.6 For ordinary Turing machines M, show that a) the problem of determining whether L(M) is finite is equivalent to S 2 \ b) the problem of determining whether L(M) is a regular set is equivalent to S 3 .
8.7 Show that the following problems about programs in a real programming language are undecidable.
a) Whether a given program can loop forever on some input.
b) Whether a given program ever produces an output.
c) Whether two programs produce the same output on all inputs.
8.8 Use Theorem 8.14 to show that the following properties of CFL's are undecidable.
a) L is a linear language.
b) I is a CFL.
KTUNOTES.IN Downloaded from Ktunotes.in 214 UN DECIDABILITY S 8.9 Show that Theorem 8.14 applies to the linear languages. [Hint: Consult Theorem 9.2 for a proof that every regular set has a linear grammar. The hard part is showing that " = E," is undecidable for linear languages.
8.10 Show that the following properties of linear languages are undecidable. You may use the fact that every regular set is a linear language.
a) L is a regular set.
b) L is a linear language.
c) L is a CFL.
d) L has no unambiguous linear CFG.
8.11 Show that for CFL L, it is undecidable whether L = L.
8.12 a) Show that if L x many-one reduces to L2 , and L2 is (i) recursive in L3 or (ii) r.e. in L3 , then L x is recursive or r.e. in L3 , respectively.
b) Show that L u Turing-reduces to S t .
c) Show that L u does not many-one reduce to Sv [Hint: Use part (a).] 8.13 We say that L x "truth-table" reduces to L2 if: 1) There are k algorithms mapping any string x over the alphabet of L x to strings over the alphabet L2 . Let g^x) be the result of applying the ith algorithm to x.
2) There is a Boolean function / (y ly .
.
. , yk ) such that if y, is true when #,(x) is in L2 , and yf is false otherwise, then f(y x , y) is true if and only if is in L x .
For example, let L x be the set of strings with equal numbers of 0's and l's, and let L 2 be the set of strings with no fewer 0's than l's. Let g x (x) = x and # 2 (x) be formed from x by replacing 0's by l's and vice versa. Letf(y u y 2 ) = y i a y 2 . Then / [yu y 2 ) is true if and only if #i(x) and g 2 (x) both have no fewer 0's than l's; that is, x has an equal number of 0's and l's. Thus L x truth-table reduces to L2 .
a) Show that if L x truth-table reduces to L2 , then L x Turing-reduces to L2 .
b) Show that if L x many-one reduces to L2 , then L x truth-table reduces to L2 .
c) Show that L u truth-table reduces to S x .
8.14 Consider a multitape TM with oracle which, when it queries its oracle, refers to the entire contents of a designated tape, say the last. Show that this model is equivalent to the oracle TM as defined in Section 8.9.
8.15 Show that PCP is decidable for words over a one-symbol alphabet.
8.16 Show that PCP is equivalent to S,.
8.17 Show that PCP is undecidable if strings are restricted to have length one or two.
What if strings are restricted to have length exactly two?
8.18 Let o be a total recursive function mapping indices of partial recursive functions to indices of partial recursive functions. Give an algorithm to enumerate an infinite set of fixed points of a; that is, infinitely many fs such that/(y) =f a{i)(y) for all y.
8.19 Does there exist an effective enumeration of Turing machines M U M 2 , ... such that no three consecutive TM's compute the same function?
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 215 Solutions to Selected Exercises 8.3 Let M = (£), Z, T, S, q0y By F) be a TM. We construct another TM M\ such that M' halts on x if and only ifM accepts x. We shall thus have shown that the question whether a TM halts on all inputs reduces to the question whether a TM accepts all inputs, which we know is undecidable. Incidentally, we shall also show by this construction that a question such as "Does a TM halt on a given input ?" or "Does a TM halt on some input?" is also undecidable.
M' is designed to behave as follows. First, it shifts its input one position right, placing a left end marker $ on the leftmost cell. M' then simulates M. If 5(q y X) is undefined, and either (i) q is nonaccepting and X is any symbol in Tu {$} [note that S(qy $) is surely undefined], or (ii) q is accepting and X is $, then M' scanning X in state q moves right and enters state p x . In state pu scanning any symbol, M' moves left and enters state p 2 , in that state, M' moves right and enters p x again. Thus M' loops forever if M either halts in a nonaccepting state or falls off the left end of the tape in any state. If M enters an accepting state, not scanning $, then M' halts. Thus M' halts if and only if M accepts its input, as desired.
8.9 We must first show that the linear languages are closed under union and concatena-tion with regular sets. We look ahead to Theorem 9.2 for a proof that every regular set is generated by CFG all of whose productions are of the forms A -» Bw and A -+ w for nonterminals A and B and string of terminals w. Any such grammar is surely linear. The proof that linear languages are closed under union is just like Theorem 6.1. For concatena-tion with a regular set, let G t = (Vu Tiy P ly S t ) be a linear grammar and G 2 = (V^ T 2y Pi, S 2 ) be a grammar with all productions of the forms A -+ Bw and A -> w. Assume V x and V2 are disjoint. Let G = (Vt u V2y T, u T 2 , P, S 2 ), where P consists of i) all productions A -> Bw of P 2l ii) production A -> S, w whenever A -> w is a production of P 2 > and iii) all productions of P { .
Then L(G) is easily seen to be L(Gi)L(G 2 )y since all derivations in G are of the form S 2 Si x => yx, where S 2 =>x and => y. Since regular sets and linear languages are closed under reversal, concatenation on the left by a regular set follows similarly.
Now we must show that " = £" is undecidable for linear languages. The proof closely parallels Lemma 8.7 and Theorem 8.11, the analogous results for general CFG's. The important difference is that we must redefine the form of valid computations so the set of invalid computations is a linear CFG. Let us define a valid computation of TM M to be a string w^w 2 ^ •• #wn _ 1 #wn##w;#wj_ 1 # ••• #wf#wf, (8.4) where each w, is an ID, w, wi+ { for 1 < / < n, w x is an initial ID and w„ is a final ID.
Then it is not hard to construct a linear grammar for strings not of the form (8.4), parallel-ing the ideas of Lemma 8.7. Then the analog of Theorem 8.11 shows that " = E" is undecidable for linear grammars.
KTUNOTES.IN Downloaded from Ktunotes.in 216 UNDECIDABILITY BIBLIOGRAPHIC NOTES The undecidability of is the basic result of Turing . Theorems 8.6 and 8.7, charac-terizing recursive and r.e. index sets, are from Rice [1953, 1956]. Post's correspondence problem was shown undecidable in Post , and the proof of undecidability used here is patterned after Floyd . Lemmas 8.6 and 8.7, relating Turing machine computations to CFG's, are from Hartmanis .
The fundamental papers on undecidable properties of CFL's are Bar Hillel, Perles, and Shamir and Ginsburg and Rose [1963a]. However, Theorem 8.9, on ambiguity, was proved independently by Cantor , Floyd [1962a], and Chomsky and Schutzenberger . Theorem 8.16, undecidability of inherent ambiguity, is taken from Ginsburg and Ullian . Linear grammars and their decision properties have been studied by Greib-ach [1963, 1966] and Gross . The approach used in the solution to Exercise 8.9 is from Baker and Book .
Greibach's theorem is from Greibach . A generalization appears in Hunt and Rosenkrantz , which includes a solution to Exercise 8.11. Hopcroft and Ullman [1968a] shows that for certain classes of languages defined by automata, the decidability of membership and emptiness are related. The Smn- and recursion theorems are from Kleene . Example 8.10, on the nonexistence of proofs of halting or nonhalting for all TM's, is from Hartmanis and Hopcroft .
Hartmanis and Hopcroft are the authors of the basic paper relating problems about CFL's to the hierarchy of undecidable problems. Theorems 8.19 and 8.20, as well as Exercise 8.6, are from there. Additional results of this nature have been obtained by Cudia and Singletary , Cudia , Hartmanis , and Reedy and Savitch .
Exercise 8.4, on tag systems, is from Minsky .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 9 THE CHOMSKY HIERARCHY Of the three major classes of languages we have studied—the regular sets, the context-free languages, and the recursively enumerable languages—we have gram-matically characterized only the CFL's. In this chapter we shall give grammatical definitions of the regular sets and the r.e. languages. We shall also introduce a new class of languages, lying between the CFL's and the r.e. languages, giving both machine and grammatical characterizations for this new class. The four classes of languages are often called the Chomsky hierarchy, after Noam Chomsky, who defined these classes as potential models of natural languages.
9.1 REGULAR GRAMMARS If all productions of a CFG are of the form A -> wB or A -> w, where A and B are variables and w is a (possibly empty) string of terminals, then we say the grammar is right-linear.
If all productions are of the form A -> Bw or A -> w, we call it left-linear. A right- or left-linear grammar is called a regular grammar.
Example 9.1 The language 0(10) is generated by the right-linear grammar S-+0A (9.1) A->10A( and by the left-linear grammar 5->510|0 (9.2) 217 KTUNOTES.IN Downloaded from Ktunotes.in 218 THE CHOMSKY HIERARCHY Equivalence of regular grammars and finite automata The regular grammars characterize the regular sets, in the sense that a language is regular if and only if it has a left-linear grammar and if and only if it has a right-linear grammar. These results are proved in the next two theorems.
Theorem 9.1 If L has a regular grammar, then L is a regular set.
Proof First, suppose L = Ufi) for some right-linear grammar G = (K, T, P, S).
We construct an NFA with 6-moves, M = (£), T, S, [S], {[c]}) that simulates deriva-tions in G.
Q consists of the symbols [a] such that a is S or a (not necessarily proper) suffix of some right-hand side of a production in P.
We define 3 by: 1) If A is a variable, then S([A], e) = {[a] | A -> a is a production}.
2) If a is in T and a in T u TK, then <5([aa], a) = {[a]}.
Then an easy induction on the length of a derivation or move sequence shows that <5([S], w) contains [a] if and only if 5 ^> xA => x_ya, where A -> ya. is a produc-tion and xy = w, or if a = S and w = 6. As [e] is the unique final state, M accepts w if and only if 5 ^> xA => w. But since every derivation of a terminal string has at least one step, we see that M accepts w if and only if G generates w. Hence every right-linear grammar generates a regular set.
Now let G = (V, T, P, 5) be a left-linear grammar. Let G' = (V, 7, F, 5), where P' consists of the productions of G with right sides reversed, that is, F = {A -> a | A aK is in P}.
If we reverse the productions of a left-linear grammar we get a right-linear gram-mar, and vice versa. Thus G' is a right-linear grammar, and it is easy to show that Ufi') = I-(G)R . By the preceding paragraph, Ufi') is a regular set. But the regular sets are closed under reversal (Exercise 3.4g), so Ufi') R = Ufi) is also a regular set.
Thus every right- or left-linear grammar defines a regular set.
Example 9.2 The NFA constructed by Theorem 9.1 from grammar (9.1) is shown in Fig. 9.1.
Fig. 9.1 NFA accepting 0(10).
KTUNOTES.IN Downloaded from Ktunotes.in 9.1 | REGULAR GRAMMARS 219 Now consider grammar (9.2). If we reverse its productions we get S-01S|0 The construction of Theorem 9.1 for this grammar yields the NFA of Fig. 9.2(a). If we reverse the edges of that NFA and exchange initial and final states, we get another NFA for 0(10).
l (b) Fig. 9.2 Construction of an NFA for 0(10) from a left-linear grammar.
Theorem 9.2 If L is a regular set, then L is generated by some left-linear grammar and by some right-linear grammar.
Proof Let L = L(M) for DFA M = (Q, E, 3, q0 , F). Fin^sjuppose that qQ is not a final state. Then L = L(G) for right-linear grammar G = (Q, Z, P, q0 ), where P consists of production p-+ aq whenever <5(p, a) = q and also p -> a whenever <5(p, a) is a final state. Then clearly <5(p, w) = q if and only if p ^> wq.
If wa is accepted by M, let S(q0 , w) = p, implying q0 => wp. Also, <5(p, a) is final, so p - a is a production.
Thus q0 ^> wa.
Conversely, let q0 ^> x.
Then x = wa, and % ^>wp=>wa for some state (variable) p. Then S(q0y w) = p, and <5(p, a) is final.
Thus x is in L(M). Hence L(M) = L(G) = L.
Now let g0 be in F, so £ is in L. We note that the grammar G defined above generates L — {e}. We may modify G by adding a new start symbol 5 with produc-tions 5 - q0 | e. The resulting grammar is still right-linear and generates L.
To produce a left-linear grammar for L, start with an NFA for L K and then reverse the right sides of all productions of the resulting right-linear grammar.
KTUNOTES.IN Downloaded from Ktunotes.in 220 THE CHOMSKY HIERARCHY Example 9.3 In Fig. 9.3 we see a DFA for 0(10).
The right-linear grammar from this DFA is A-+0B\W\0 B-0D|1C C->0£|1D|0 D-0D|1D As D is useless we may eliminate it, obtaining grammar >4->0B|0 B -» 1C C->0£|0 o, 1 Fig. 9.3 DFA for 0(10).
9.2 UNRESTRICTED GRAMMARS The largest family of grammars in the Chomsky hierarchy permits productions of the form a -> /?, where a and ft are arbitrary strings of grammar symbols, with a # c. These grammars are known as semi-Thue, type 0, phrase structure or unre-stricted grammars. We shall continue to use the 4-tuple notation G = (K, 7, P, S) for unrestricted grammars. We say ycxS => yfid whenever a - ft is a production. As before, stands for the reflexive and transitive closure of the relation =>: L(G) = {w|w is in T and S^>w}, exactly as for context-free grammars.
Example 9.4 A grammar generating {a 1 1 ns a positive power of 2} is given below.
)S-+ACaB 5)aD->Da 2) Ca -> aaC 6) AD -> ,4C 3) CB-+DB 7) a£->£a 4) C£-£ 8) AE-+e KTUNOTES.IN Downloaded from Ktunotes.in 9.2 | UNRESTRICTED GRAMMARS 221 A and B serve as left and right endmarkers for sentential forms; C is a marker that moves through the string of a's between A and B, doubling their number by production (2). When C hits the right endmarker B, it becomes a D or E by production (3) or (4). If a D is chosen, that D migrates left by production (5) until the left endmarker A is reached. At that point the D becomes a C again by production (6), and the process starts over. If an E is chosen, the right endmar-ker is consumed. The E migrates left by production (7) and consumes the left endmarker, leaving a string of T as for some i > 0. We can prove by induction on the number of steps in the derivation that if production (4) is never used, then any sentential form is either i) S, ii) of the form Aa lCa}B, where i + 2j is a positive power of 2, or iii) of the form Aa lDajB, where i + j is a positive power of 2.
When we use production (4) we are left with a sentential form AdE, where i is a positive power of 2. Then the only possible steps in a derivation are i applications of (7) to yield AEd followed by one application of (8), producing sentence a\ where i is a positive power of 2.
Equivalence of type 0 grammars and Turing machines We shall prove in the next two theorems that unrestricted grammars characterize N the r.e. languages. The first theorem states that every type-0 language generates an r.e. set. An easy proof would be to give an algorithm for enumerating all strings generated by a type-0 grammar. Instead we construct a Turing machine recog-nizer for sentences generated by a type-0 grammar, since this construction will be useful later for a similar proof about context-sensitive grammars (the remaining class in the Chomsky hierarchy).
Theorem 9.3 If L is L(G) for unrestricted grammar G = (V, T, P, S), then L is an r.e. language.
Proof Let us construct a nondeterministic two-tape Turing machine M to recog-nize L. M's first tape is the input, on which a string w will be placed. The second tape is used to hold a sentential form a of G. M initializes a to S. Then M repeatedly does the following: 1) Nondeterministically select a position i in a, so that any i between 1 and |a | can be chosen. That is, start at the left, and repeatedly choose to move right or select the present position.
2) Nondeterministically select a production ft -» y of G.
3) If p appears beginning in position i of a, replace P by y there, using the "shifting-over" technique of Section 7.4, perhaps shifting left if \y\ < \P.
KTUNOTES.IN Downloaded from Ktunotes.in 222 THE CHOMSKY HIERARCHY 4) Compare the resulting sentential form with w on tape 1. If they match, accept; w is a sentence of G. If not, go back to Step (1).
It is easy to show that all and only the sentential forms of G appear on tape 2 when Step (4) is executed after some sequence of choices. Thus L(M) = Ufi) = L, so L is r.e.
Theorem 9.4 If L is an r.e. language, then L = L(G) for some unrestricted gram-mar G.
Proof Let L be accepted by Turing machine M = (Q, Z, T, 5, q0 , B, F). Construct a grammar G that "nondeterministically" generates two copies of a representation of some word in Z and then simulates the action ofM on one copy. IfM accepts the word, then G converts the second copy to a terminal string. If M does not accept, the derivation never results in a terminal string.
Formally, let G = (K, Z, P, A x \ where K=((Eu {e}) x T) u {A l9 A 2 , A 3 } and the productions in P are: 1) A l -+q0 A 2 2) A 2 - [a, a]A 2 for each a in Z.
3) A 2 -+A 3 4) /t 3 -+[e, B]/l 3 5) A 3 -+c 6) X] -> [a, y]p for each ainZ u {e} and each <j in Q and X and y in T, such that <5fo, X) = (p, y, K).
7) [fe, Z]g[a, X] -> p[>, Z][a, y] for each X, Y, and Z in T, a and ftinZu {t}, and g in g, such that S(q, X) = (p, y, L).
8) [a, X]^f - qaq, q[a, X] -> qaq, and q -> £ for each ainlu {£}, X in T, and g in F.
Using rules 1 and 2, we have where a x is in Z for each i. Suppose that M accepts the string a 1 a 2 • • • a„. Then for some m, M uses no more than m cells to the right of its input. Using rule 3, then rule 4 m times, and finally rule 5, we have x >q0 [ai, flj[a 2 . flj k, B]m .
From this point on, only rules 6 and 7 can be used until an accepting state is generated. Note that the first components of variables in (Z u {e}) x T are never changed. We can show by induction on the number of moves made by M that if <7ofli <2 an \w x \ x 2 "' Xr-iqXr ••• Xs , (9.3) KTUNOTES.IN Downloaded from Ktunotes.in 9.3 | CONTEXT-SENSITIVE LANGUAGES 223 then qo[a l9 a x ][a2 , a2] ••• [an , an][e, £f§> [au X x][a2 , X2] • • • [ar _ l9 Xr_ J^, XJ • • • [an+m , Xn+m], (9.4) where au a 2 , a„ are in E, an+1 = an+2 = ••• = 0n+m = e, X u X 2 , Xn + m are in T, and Xs+1 = Xs+2 = ••• = Xn+m = B.
The inductive hypothesis is trivially true for zero moves, since r = 1 and s = n.
Suppose it is true for k — 1 moves and let 3012 " 'an \^-X 1 X2 -Xr. 1 qXr -X, T 7 t 7 2 - ^ By the inductive hypothesis, o[«i. «i] ' K BJ" §>K ••• [ar - u Xr^]q[ari Xr] ~ [an+m , Xn+m], where the a's and X's satisfy the conditions of (9.4).
If t = r + 1, then the /cth move ofM is to the right, so 5(q, Xr ) = (p, Y r , R). By rule (6), q[ari Xr] -> [ar , 7 r]p is a production of G. Thus 4oK i] ••[„, n][^B]m |> K yj - k_ l5 y, q - k +m , y n+m], (9.5) where Y; = £ for i > u.
If r = r — 1, then the /cth move ofM is to the left, and we prove (9.5) using rule (7) and the observations that r > 1 and S(qy Xr ) = (p, Y r , L).
By rule (8), if p is in F then [a u yj • • [at _ u y.-JpK q ••• [an+m , y n+m]^>flifl2 — an .
We have thus shown that if w is in L(M), then /Ij ^> w, so w is in L(G).
For the converse, that w in Jjfi) implies w in L(M), an induction similar to the above shows that (9.4) implies (9.3). We leave this part as an exercise. Then we note that there is no way to remove the state of M from sentential forms of G without using rule (8). Thus G cannot derive a terminal string without simulating an accepting computation of M. By rule (8), the string derived must be the first components of the variables in (Z u {e}) x T, which are never changed as moves of M are simulated.
9.3 CONTEXT-SENSITIVE LANGUAGES Suppose we place the restriction on productions a -> ft of a phrase structure grammar that jS be at least as long as a. Then we call the resulting grammar context-sensitive and its language a context-sensitive language (CSG and CSL KTUNOTES.IN Downloaded from Ktunotes.in 224 THE CHOMSKY HIERARCHY respectively). The term "context-sensitive" comes from a normal form for these grammars, where each production is of the form a 1 Aa 2 -> ai/fa2 , with p ^ e.
Productions of the latter form look almost like context-free productions, but they permit replacement of variable A by string /? only in the "context" a x — cc2 . We leave this normal form as an exercise.
Almost any language one can think of is context-sensitive; the only known proofs that certain languages are not CSL's are ultimately based on diagonaliza-tion. These include Lu of Chapter 8 and the languages to which we may reduce L„, for example, the languages proved undecidable in Chapter 8. We shall prove in Section 9.4 that there are recursive languages that are non-CSL's, and in Chapter 12 we shall refine this statement somewhat. In both cases the proofs proceed by diagonalization.
Example 9.5 Consider again the grammar of Example 9.4. There are two pro-ductions that violate the definition of a context-sensitive grammar. These are CB-> E and AE -» c. We can create a CSG for the language {a 21 \ i > 1} by realiz-ing that A, B, C, D, and E are nothing but markers, which eventually disappear.
Instead of using separate symbols for the markers, we can incorporate these markers into the cis by creating "composite" variables like [CaB], which is a single symbol appearing in place of the string CaB.
The complete set of composite symbols we need to mimic the grammar of Example 9.4 is [ACaB], [Aa], [ACa], [ADa], [A Ed], [Co], [Da], [Ea], [aCB], [CaB\ [aDB], [aE], [DaB], and [aB]. The productions of our context-sensitive grammars, which we group according to the production from Example 9.4 that they mimic, are: 1 ) S -> [ACaB] 5) a[Da] -> [Da]a 2) [Ca]a->aa[Ca] [aDB]->[DaB] \ca][aB]^aa[CaB] Mjl^J [ACa]a-+[Aa]a[Ca] a[DaB] ~^[Da][aB] [ACa][aB] -> [Aa]a[CaB] [Aa][DaB] -> [ADa][aB] [ACaB] -> [Aa][aCB] 6) [ADa] -> [ACa] [CaB] -» a[aCB] 7) a[Ea] -> [Ea]a 3) [aCB] -> [aDB] [aE] - [Ea] 4) [aCB]->[aE] [Aa][Ea]->[AEa]a 8) [AEa] -> a It is straightforward to show that S ^>a in the grammar of Example 9.4 if and only if S ^> a' in the present CSG, where a' is formed from a by grouping with an a all markers (A through E) appearing between it and the a to its left and also grouping with the first a any markers to its left and with the last a any markers to its right. For example, if a = AaaCaB, then a' is [/4fl]a[CaB].
KTUNOTES.IN Downloaded from Ktunotes.in 9.3 | CONTEXT-SENSITIVE LANGUAGES 225 Linear bounded automata Now we introduce a machine characterization of the CSL's. A linear bounded automaton (LBA) is a nondeterministic Turing machine satisfying the following two conditions.
1) Its input alphabet includes two special symbols ^ and $, the left and right endmarkers, respectively.
2) The LBA has no moves left from $ or right from $, nor may it print another symbol over $ or $.
The linear bounded automaton is simply a Turing machine which, instead of having potentially infinite tape on which to compute, is restricted to the portion of the tape containing the input x plus the two tape squares holding the endmarkers.
We shall see in Chapter 12 that restricting the Turing machine to an amount of tape that, on each input, is bounded by some linear function of the length of the input would result in the identical computational ability as restricting the Turing machine to the portion of the tape containing the input—hence the name "linear bounded automaton." An LBA will be denoted M = (Q, Z, T, 5, q0 , $, F), where Q, Z, T, 5, q0 and F are as for a nondeterministic TM; § and $ are symbols in Z, the left and right endmarkers. L(M), the language accepted by M, is {w | w is in (Z — {§, $}) and q0$w$ \jf ccqP for some q in F}.
Note that the endmarkers are on the input tape initially but are not considered part of the word to be accepted or rejected. Since an LBA cannot move off the input, there is no need to suppose that there is blank tape to the right of the $.
Equivalence of LBA's and CSG's We now show that except for the fact that an LBA can accept c while a CSG cannot generate e, the LBA's accept exactly the CSL's.
Theorem 9.5 If L is a CSL, then L is accepted by some LBA.
Proof The proof is almost the same as that for Theorem 9.3. The only difference is that while the TM of Theorem 9.3 generated sentential forms of an unrestricted grammar on a second tape, the LBA uses a second track of its input tape. Pre-sented with §w$ on its tape, the LBA starts by writing the symbol S on a second track below the leftmost symbol of w. If w = c, the LBA instead halts without accepting. Next the LBA repeatedly guesses a production and a position in the sentential form written on the second track. It applies the production, shifting the portion of the sentential form to the right whenever the sentential form expands.
If, however, the new sentential form is longer than w, the LBA halts without acceptance. Thus the LBA will accept w if there is a derivation S ^> w such that no KTUNOTES.IN Downloaded from Ktunotes.in 226 THE CHOMSKY HIERARCHY intermediate sentential form is longer than w. But since the right side of any production in a CSG is as long or longer than the left side, there could not be a derivation S ^> a ^> w, where a is longer than w. Thus the LBA accepts all and only the words generated by the CSG.
Theorem 9.6 If L = L(M) for LBA M = (Q, Z, T, S, q0 , $, F), then L - {e} is a CSL.
Proof The proof parallels the construction of an unrestricted grammar from a TM in Theorem 9.4. The differences are that the endmarkers on the LBA tape must be incorporated into adjacent tape symbols, and the state must likewise be incorporated into the symbol scanned by the tape head. The reason for this is that if the CSG simulated the LBA using separate symbols for the endmarkers, or state, it could not erase these symbols afterward, since that would necessitate shortening a sentential form, and the right side of every CSG production is at least as long as the left side. The generation of a sequence of pairs, the first component of which forms the terminal string a { a 2 m ~ an and the second of which forms the LBA tape is accomplished by the productions A 2 - [a, a]A 2y A 2 -> [a, a$], for all a in I -$}.
The LBA-simulating rules are similar to rules 6 and 7 in Theorem 9.4 and are left as an exercise.
If q is final, then we have production [a, aqp] -» a for all a in I -$} and all possible a and P (that is, a and/or /? could include $, and one tape symbol). Note that the number of productions defined is finite. We also allow deletion of the second component of a variable if it is adjacent to a terminal, by [a, oi\b -> ab, b[a, a] - ba for any a and b in Z — {<(:, $} and all possible a's.
The productions shown explicitly are clearly context-sensitive. The LBA-simulating productions can easily be made length preserving, so the resulting grammar is a CSG. A proof that any word w but e is accepted by M if and only if it is generated by the grammar parallels Theorem 9.4, and we omit it. Note that there is no way for the grammar to set up the LBA input or simulate M on that input. Thus e is not generated by the grammar whether or not it is in L(M).
KTUNOTES.IN Downloaded from Ktunotes.in 9.4 | RELATIONS BETWEEN CLASSES OF LANGUAGES 227 9.4 RELATIONS BETWEEN CLASSES OF LANGUAGES The four classes of languages—r.e. sets, CSL's, CFL's, and regular sets—are often referred to as languages of types 0, 1, 2, and 3, respectively. We can show that except for the matter of the empty string, the type-/ languages properly include the type-(i + 1) languages for i = 0, 1, and 2. We first need to show that every CSL is recursive, and in fact, there are recursive languages that are not CSL's.
CSL's and recursive sets Theorem 9.7 Every CSL is recursive.
Proof Given a CSG G = (V, T, P, S) and a word w in Z of length n, we can test whether w is in L(G) as follows. Construct a graph whose vertices are the strings in (V u T) of length n or less. Put an arc from a to p if a => /?. Then paths in the graph correspond to derivations in G, and w is in L(G) if and only if there is a path from the vertex for 5 to the vertex for w. Use any of a number of path-finding algorithms (see Aho, Hopcroft, and Ullman ) to decide whether such a path exists.
Example 9.6 Consider the CSG of Example 9.5 and input w = aa. One way to test for paths in the graph is to start with string S, and at the ith step find the strings of length n or less having a path from S of length i or less. If we have the set for i — 1, say Sf, then the set for i is ^ u {/?
| a => f$ for some a in 5? and \fj\ 1} is an example of a CFL that is not regular. Part (b) is proved by noting that every CFG in Chomsky normal form is a CSG. {a 2i \ i > 1} is a CSL that is easily shown not to be a CFL by the pumping lemma. For part (c) every CSG is surely an unrestricted grammar.
Proper containment follows from Theorem 9.8.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 229 EXERCISES 9.1 Construct left-linear and right-linear grammars for the languages a) (0 + 1)00(0 + 1) b) 0(1(0 + 1)) c) (((01 + 10)11)00) 9.2 Show the following normal form for right-linear grammars and the analogous result for left-linear grammars: If L is a regular set, then L - {e} is generated by a grammar in which all productions are of the form A - aB or A a for terminal a and variables A and B.
9.3 A context-free grammar is said to be simple if it is in Greibach normal form, and for every variable A and terminal a, there is at most one string a such that A acc is a production. A language is simple if it has a simple grammar. For example, L = {0T | n > 1} has the simple grammar: S - OA A-+0AB\1 B-1 Note that the more natural GNF grammar for L> S-0S£|0B £-> 1 is not simple because there are two S-productions whose right sides begin with 0. Prove that every regular set not containing e is a simple language. [Hint : Use a DFA representa-tion for the regular set.] 9.4 A CFG is said to be self-embedding if there is some useful variable A such that A ^> wAx, and neither w nor x is e. Prove that a CFL is regular if and only if it has a CFG that is not self-embedding.
[Hint: It is easy to show that no regular grammar is self-embedding. For the "if" portion, show that a non-self-embedding grammar may be put in Greibach normal form without making it self-embedding. Then show that for every non-self-embedding GNF grammar, there is a constant k such that no left-sentential form has more than k variables. Finally, show from the above that the non-self-embedding GNF grammar can be converted to a regular grammar.] 9.5 Give unrestricted grammars for a) {ww | w is in (0 + 1)} b) {0 1 ' 2 j i > 1} c) {0' | i is not a prime} d) {0T2'' | i > 1} 9.6 Give context-sensitive grammars for the languages of Exercise 9.5, excluding t in (a).
9.7 A CSL is said to be deterministic if it is accepted by some deterministic LBA. Show that the complement of a deterministic CSL is also a deterministic CSL. [Hint: Show that for every deterministic LBA there is an equivalent LBA that halts on every input.] It is, incidentally, open whether every CSL is a deterministic CSL, and whether the CSL's are closed under complementation. Obviously a positive answer to the former question would imply a positive answer to the latter.
KTUNOTES.IN Downloaded from Ktunotes.in 230 THE CHOMSKY HIERARCHY 9.8 a) Show that every context-free language is accepted by a deterministic LBA.
b) Show that the Boolean closure of the CFL's is contained within the class of sets accepted by deterministic LBA's.
c) Show that the containment in (b) is proper. [Hint: Consider languages over a one-symbol alphabet.] 9.9 Show that every CSL is generated by a grammar in which all productions are of the form ocAp -» ayp, where A is a variable, a, /?, and y are strings of grammar symbols, and S9.10 Show that the CSL's are closed under the following operations: a) union b) concatenation c) intersection d) substitution e) inverse homomorphism f) positive closure (recall L+ = U,= i L) 9.11 Show that the r.e. sets are closed under the following operations: a) through e) same as Exercise 9.10.
f) Kleene closure.
9.12 a) Show that all the undecidable properties of CFL's mentioned in Sections 8.5, 8.6, and 8.7 are undecidable for CSL's, with the exception that "= X" is trivially decidable because no CSL contains c.
b) Show that "= I + " is undecidable for CSL's.
S9.13 Show that it is undecidable whether a given CSL is empty.
S9.14 Show that every r.e. set is h(L\ where h is a homomorphism and L a CSL.
Solutions to Selected Exercises 9.10 The proofs are similar to the proofs of Theorems 6.1, 6.2, and 6.3 for CFL's.
However, there is one problem with which we have to deal. Consider the concatenation construction. Suppose Gx = (Vu Tu P u S x ) and G 2 = (K 2 , T 2 , P 2 , S 2 ) are CSG's generating L, and L2 , respectively. In Theorem 6.1 for CFG's, we constructed grammar G4 = (Vr u V 2 u {S4 }, Ti u T 2 , P,uP 2 u {S4 - 5, S 2 }, S4 ) to generate Li L 2 . This construction is correct for CFG's, provided Vr and V 2 are disjoint.
For CSG's, however, we could have a production a -> ft in Px or P 2 that was applicable in a sentential form of G 4 , say y<5, where Sx=>y and S 2 =>&> in such a position that a straddles the boundary between y and <5. We might thus derive a string not in L x L2 .
Assuming Vx n V2 = 0 doesn't help, since a could consist of terminals only, and of course we cannot assume that T x o T 2 = 0. What we need is a normal form for CSG's that allows only variables on the left sides of productions. Such a lemma is easy to prove. Let G = (K, Ty P, S) be a CSG. Construct G' = (V\ 7, P', 5), where V consists of V plus the KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 231 variables A a for each a in T. F consists of productions A a -> a for each a, and production a' - p for each ol-> ft in P, where a' is a with each occurrence of a terminal a replaced by A a , and /F is similarly related to /?.
Now, if we assume that G x and G 2 have disjoint sets of variables and are in the above normal form, the constructions of Theorem 6.1 for union and concatenation carry over to CSL's.
Positive closure presents another problem.
If, in analogy with Theorem 6.1, we construct G5 = (K, u{S5 }, TU P X u{S 5 -S,S5 |Si}, S5 ), we have not avoided the problem of the potential for applying a production a -> /? in such a way that it straddles the strings derived from two or more instances of S x . What we can do is create grammar G\, which is G x with each variable A replaced by a new symbol A'. Then we construct the grammar G 5 = (F 5 , 7i, F 5 , S 5 ), where V 5 consists of the variables of G x and Gi, plus the symbols S 5 and S' 5 ; F 5 consists of the productions of Gj and G\, plus S 5 -» Si S' s \S l S' 5 - S\ S s I Si As no CSL cbntains £, we can never have symbols derived from two instances of S x or two instances of Si adjacent, and we may be sure that each production of G 5 is applied to a string derived from one instance of Si or Si.
Inverse homomorphism, intersection, and substitution are best handled by machine-based proofs. Let L be a CSL accepted by LBA M and h a homomorphism. Suppose that \h(a)\ < k for any a. Then we may construct LBA M' for h~ l (L) as follows. M' takes its input x and computes h(x)y storing k symbols per cell. There is sufficient space, since \h(x)\ < k \x |. Then M' simulates M on h(x\ accepting if M accepts.
For intersection, let L x and L2 be CSL's accepted by LBA's M, and M 2 - Construct LBA M 3 that treats its input as if it were written on two tracks. That is, we identify input symbol a with [a, a]. On the first track, M 3 simulates M Y . If some sequence of choices of move by M, causes it to accept, M 3 begins to simulate M 2 on the second track, accepting if M 2 accepts. Thus M 3 accepts L x n L2 .
For substitution into CSL L ^ Z of CSL's La for symbols a in Z, construct an LBA that works as follows. Given input a x a 2 '" an , nondeterministically guess which positions end strings in some La , and mark them. If we guess that a,a l + i •• ai is in some particular La , simulate the LBA for La on that substring. If + , • • a} is in La , replace it by a. If all our guesses are correct, take the resulting string in E and simulate an LBA for L on it, accepting a, a 2 " m an if that LBA accepts.
9.13 It is easy to design an LBA to accept the valid computations of a given Turing machine. Thus the emptiness problem for Turing machines is reducible to the question whether a given CSL is empty.
9.14 Let Li be an r.e. set and c a symbol not in the alphabet of L x . Let Mj be a TM accepting L x and define L2 = {wc' |Mi accepts w by a sequence of moves in which the head never moves more than i positions to the right of w}.
KTUNOTES.IN Downloaded from Ktunotes.in 232 THE CHOMSKY HIERARCHY Then L2 is accepted by an LBA that simulates M lt treating c as the blank and halting if it ever goes beyond the sequence of c's on its input. We have only to show that L t = h(L2 ) for some homomorphism h. Let h(a) = a for all symbols in the alphabet of L, and h(c) = e.
Combining Exercise 9.14 with Theorem 9.9, we observe that the CSL's are not closed under homomorphism. This may seem paradoxical, since Exercise 9.10 claimed the CSL's were closed under substitution. However, homomorphism is not a special case of substitu-tion by a CSL, as a CSL may not contain e. In particular, for h defined above, h(c) = e is not a CSL. The CSL's are, however, closed under homomorphisms that do not map any symbol to 6.
BIBLIOGRAPHIC NOTES The Chomsky hierarchy was defined in Chomsky [1956, 1959]. Chomsky and Miller showed the equivalence of regular grammars and regular sets. Kuroda showed the equivalence of LBA's and CSG's.
Previously, Myhill had defined deterministic LBA's, and Landweber showed that the deterministic LBA languages are contained in the CSL's. Chomsky showed that the r.e. sets are equivalent to the languages generated by type-0 grammars. Fischer gives some interesting characterizations of the CSL's. Hibbard discusses a restriction on CSG's that yields the context-free languages. Additional closure properties of CSL's are studied in Ginsburg and Griebach [1966b] and Wegbreit . Basic decision properties of CSL's are given in Landweber .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 10 DETERMINISTIC CONTEXT-FREE LANGUAGES We now have machine models that define each class of languages in the Chomsky hierarchy. At the extreme ends of the hierarchy, the machines—finite automata and Turing machines—exhibit no difference in accepting ability between their deterministic and nondeterministic models. For the linear-bounded automaton, it is unknown whether the deterministic and nondeterministic varieties accept the same class of languages. However, for pushdown automata, we do know that the deterministic PDA's accept a family of languages, the deterministic context-free languages (DCFL's), lying properly between the regular sets and the context-free languages.
It turns out that the syntax of many programming languages can be described by means of DCFL's. Moreover, modern compiler writing systems usually require that the syntax of the language for which they are to produce a compiler be described by a context-free grammar of restricted form. These restricted forms almost invariably generate only DCFL's. We shall meet what is probably the most important of these restricted forms—the LR-grammars. The LR-grammars have the property that they generate exactly the DCFL's.
If a compiler writing system is to be used, it is generally necessary that the language designer choose a syntax for his language that makes it a DCFL. Thus it is useful to be able to determine whether a proposed language is in fact a DCFL. If it is, one can often prove it so by producing a DPDA or LR-grammar defining the language. But if the language L is not a DCFL, how are we to prove it? If L is not a CFL at all, we could use the pumping lemma, perhaps. However, L will often be a CFL but not a DCFL. There is no known pumping lemma specifically for DCFL's, so we must fall back on closure properties. Fortunately, the DCFL's are closed under a number of operations, such as complementation, that do not 233 KTUNOTES.IN Downloaded from Ktunotes.in 234 DETERMINISTIC CONTEXT-FREE LANGUAGES preserve CFL's in general. Thus, if L is a CFL but its complement is not, then Lis not a DCFL.
Sections 10.1 through 10.4 develop various closure properties of DCFL's.
Section 10.5 briefly covers decision properties.
Sections 10.6 and 10.7 treat LjR-grammars.
10.1 NORMAL FORMS FOR DPDA's Recall that PDA M = (Q, E, T, S, q0 , Z0 , F) is deterministic if: 1) whenever S(q, a, X) is nonempty for some a in Z, then S(q, e, X) is empty, and 2) for each q in Q, a in £ u {e} and X in T, (5(g, a, X) contains at most one element.
Rule (1) prevents a choice between using the next input or making an e-move.
Rule (2) prevents a choice on the same input. For deterministic PDA's we shall hereafter write 3(q, a, X) = (/?, y) rather than S(q, a, X) = {(/>, y)}.
Like PDA's in general, we can put DPDA's in a normalform where the only stack operations are to erase the top symbol or to push one symbol. This form will be proved in the next two lemmas. The first lemma shows that the DPDA need never push more than one symbol per move, since it can push a string of symbols one at a time, using c-moves. The second lemma shows that DPDA's need never change the top stack symbol. Changes are avoided by storing the top stack symbol in the finite control and recording changes to it there. The reader who grasps these ideas should skip to the start of the next section.
Lemma 10.1 Every DCFL is L(M) for a DPDA M = (g, Z, T, S, q0 , Z0 , F) such that if S(q, a, X) = (p, y), then \y\ < 2.
Proof If S(q, a, X) = (r, y) and | y \ > 2, let y = Y t Y 2 • • • Y n , where n > 3. Create new nonaccepting states pu p 2 , pn -i, an<3 redefine % a, X) = (pu Y n _ t Y ny Then define 5(Ph £, n.,) = (pl+lf for 1 < i < n — 3 and <5(p„_ 2 , ^2) — (r> ^1 ^2)- Thus, in state qy on input a, with X on top of the stack, the revised DPDA still replaces X with YX Y 2 Y n = y and enters state r, but it now takes n — 1 moves to do so.
Lemma 10.2 Every DCFL is L(M) for a DPDA M = (Q, £, T, (5, g0 , Zo> F) such that if S(q, a, X) = (p, y), then y is either £ (a pop), X (no stack move), or of the form YX (a push) for some stack symbol Y.
Proof Assume L = L(M'), where M' = (Q\ £, F, ^' 0 , X0 , F') satisfies Lemma 10.1. We construct M to simulate M' while keeping the top stack symbol ofM' in KTUNOTES.IN Downloaded from Ktunotes.in 10.2 | CLOSURE OF DCFL'S UNDER COMPLEMENTATION 235 M's control. Formally, let Q = Q' x F, g0 = [g' 0 , Xol F = F x r and T = F u {Z0}, where Z0 is a new symbol not in F. Define 5 by: i) If 5'(q, a, X) = (p, e), then for all Y, <5([g, X], a, Y) = ([p, 7], £). If M' pops its stack, M pops its stack, picking up the symbol popped for its control.
ii) If S'(q, a, X) = (pt Y), then for all Z, S([q, X\ a9 Z) = ([p, Y], Z).
If M' changes its top stack symbol, M records the change in its own control but does not alter its stack.
iii) If 5'(q, a, X) = (p, YZ), then for all W, S([q, X], a, W) = ([p, 7], ZJ^). If the stack of M' grows, M pushes a symbol onto its stack.
It is easy to show by induction on the number of moves made that fao, w, X0)\j r .(q,c,X l X 2 --Xn ) if and only if ([q' 0 , X0l w, Z0 ) bS" ^i], ^, 23 - XnZ0 ).
Thus L(M) = L(M').
10.2 CLOSURE OF DCFL's UNDER COMPLEMENTATION To show that the complement of a DCFL is also a DCFL we would like to use the approach employed in Theorem 3.2 to show closure of the regular sets under complementation. That is, given a DPDA M we would like to interchange final and nonfinal states and then be able to claim that the resulting DPDA accepts the complement of L(M).
There are two difficulties that complicate the above approach. The first difficulty is that the original DPDA might never move beyond some point on an input string, because on reading input w either it reaches an ID in which no move is possible or it makes an infinity of moves on oinput and never uses another input symbol. In either case, the DPDA does not accept any input with w as a prefix, and thus a DPDA accepting the complement should accept every string with prefix w. However, if we simply changed final and nonfinal states, the result-ing DPDA still would not move beyond w and therefore would not accept strings with prefix w.
The second difficulty is due to the fact that after seeing a sentence x, the DPDA may make several moves on e-input. The DPDA may be in final states after some of these moves and in nonfinal states after others. In this case, inter-changing the final and nonfinal states results in the DPDA still accepting x.
KTUNOTES.IN Downloaded from Ktunotes.in 236 DETERMINISTIC CONTEXT-FREE LANGUAGES Forcing DPDA's to scan their input To remove the first difficulty, we prove a lemma stating that, given a DPDA M, we can always find an equivalent DPDA M' that will never enter an ID from which it will not eventually use another input symbol.
Lemma 10.3 Let M be a DPDA. There exists an equivalent DPDA M' such that on every input, M' scans the entire input.
Proof We can assume without loss of generality that for every accessible ID and input symbol, M has a next move. Otherwise, one can add an endmarker on the stack to prevent M from erasing the stack entirely and thereby halting, without scanning, the entire input. In addition, one can add a "dead state," d so that for any combination of state, input symbol, and stack symbol for which M has no next move, either using the input symbol or an £-input, a transfer to state d occurs.
On any input symbol, the only transition from state d is to state dy and no change of the stack occurs. Of course, d is not an accepting state.
Now, if for every ID and input symbol, M has a next move, then the only way in which M might never reach the end of its input is if in some ID, M makes an infinity of moves on e input. If in state q with Z on top of the stack, M makes an infinity of 6-moves without erasing the symbol Z, then let M instead enter the dead state d.
This change cannot affect the language accepted unless M entered an accepting state at some time during the infinite sequence of e-moves. However, in this case, we introduce a new final state /, letting 5(q, e, Z) = (/, Z) and 5(f, e, Z) = (</, Z).
Formally, we propose the following construction. Let M = (Q, Z, T, d y q0 , Z0 , F). Define M' = {Q\j {q0 , </,/}, Z, T u {X 0 }, d\ q' 0 , X0y F u {/}), where: 1) d'(q0 , e, X0 ) = (q0 , Z0X0 ). X0 marks the bottom of the stack.
2) If for some q in Q, a in Z and Z in T, S(q, a, Z) and S(q, e, Z) are both empty, then S'(q9 a, Z) = (</, Z).
Also for all q in Q and a in Z, 5'{q, a, X0 ) = (</, X0 ).
Enter the dead state if no move is possible.
3) S'{d, a, Z) = (d, Z) for all a in Z and Z in T u {X0 }.
4) If for q and Z and all i there exist q{ and yf for which (g, 6, Z) (qh e, y t ), then d'(qy 6, Z) = (dy Z) provided no qt is final and 5'(q, c,Z) = (f Z) whenever one or more of the g,-'s is final. (Note we have not claimed that we can determine whether d'(q, e, Z) should be (dy Z) or (/, Z). However, there are KTUNOTES.IN Downloaded from Ktunotes.in 10.2 | CLOSURE OF DCFL'S UNDER COMPLEMENTATION 237 only a finite number of such decisions to be made. For each possible set of choices there exists a DPDA. One of these DPDA's will be the desired one.
We shall subsequently show that the construction is effective.) 5) S'(f 9 c, Z) = (d, Z) for all Z in T u {X0}.
6) For any q in Q, a in £ u {£}, and Z in T, if 5'(q, a, Z) has not been defined by rule (2) or (4), then S'(q, a, Z) = 5{q9 a, Z).
The argument preceding the formal construction should convince us that L(M') = L(M). To prove that M' uses all its input, suppose that for some proper prefix x of xy, (q' 0 , xy, XQ ) hg- (q, y,Z l Z2 ~ ZkX0 \ and from ID (q, y,Z l Z2 ~ m Zk X0 ), no symbol of y is ever consumed. By rule (2) it is not possible that M' halts. By rule (4), it is not possible that M' makes an infinite sequence of c-moves without erasing Z t . Therefore M x must eventually erase Zv Similarly M 1 must erase Z 2 , .
.
. , Zk and eventually enter an ID (q\ y, X0 ). By rule (2) fa', y, X Q ) f^r (d, y\ X0 ), where y = ay' and a is in Z. Thus M' did not fail to read past x as supposed, and M' satisfies the conditions of the lemma.
Let us now observe that the construction in rule (4) of Lemma 10.3 can be made effective. Assume without loss of generality that M is in normal form. We shall compute more information than is actually needed. In particular, we deter-mine for each q and p in Q and Z in T, whether 1) {q, e, Z)tfr(p, £,Z), 2 ) (fc e . z ) (P. ^ 4 3) (g, £, Z) ^- (p, £, y) for some y in r For each and Z we can determine from (3) whether M ever enters a state that consumes the next symbol of y without erasing Z.f If not, then from (2) we can determine if M erases Z. If neither event occurs, then either M' must enter the dead state by rule (2), or rule (4) applies and again (3) tells us whether e, Z) is (d, Z) or (/ Z).
Construct Boolean-valued tables Tu T 2 , and T 3 such that for i= 1, 2, and 3, T}(qy Z, p) is true if and only if statement (i) is true for q, Z, and p. The tables are initially all false and are filled inductively. The basis is to set T 3 (q, Z, p) = true if 8{q, £, Z) = (p, YZ), to set T,fa, Z, p) = T 3 (^, Z, p) = true if %, e, Z) = (p, Z), and to set T 2 (<7, Z, p) = true if S(q, e, Z) = (p, £). The inductive inferences are: 1) Whenever S(q, £, Z) = (r, YZ), then a) if T 2 (r, Y, 5) and T 2 (s, Z, p) are true, set T 2 (q, Z, p) = true; b) if T 2 (r, Y, s) and ^(5, Z, p) are true, set T^q, Z, p) = true; t Note that by the construction of Lemma 10.2, the state p alone determines whether a non-£ input move is to be made.
KTUNOTES.IN Downloaded from Ktunotes.in 238 DETERMINISTIC CONTEXT-FREE LANGUAGES c) if T 2 (r, y, s) and T 3 (s, Z, p) are true, or T x (r, Y, s) and T 3 (s, y, p) are true, set T 3 (q, Z, p) = true; d) if T 3 (r, y, p) is true, set T 3 (g, Z, p) = true.
2) Whenever S(q, e, Z) = (r, Z) then a) if 7;(r, Z, p) is true, set T x {q, Z, p) = true; b) if T 2 (r, Z, p) is true, set T 2 (g, Z, p) = true; c) if T 3 (r, Z, p) is true, set T 3 (g, Z, p) = true.
We leave as an exercise an efficient algorithm for filling in the true entries in the tables and proving that the only true entries are the ones that follow from the basis and rules (1) and (2) above.
Closure under complementation We are now ready to prove that the DCFL's are closed under complementation.
To do so we must deal with the second problem mentioned at the beginning of this section; the possibility that after reading input w, the DPDA makes a sequence of e-moves, entering both final and nonfinal states. The solution is to modify the DPDA by adding a second component to the state. The second component re-cords whether a final state of the original DPDA has been entered since the last time a true (non-e)-input was used in a move. If not, the DPDA accepting the complement enters a final state of its own, just before it is ready to use the next true input symbol.
Theorem 10.1 The complement of a DCFL is a DCFL.
Proof Let M = (Q, E, T, 5, q0 , Z0 , F) be a DPDA satisfying Lemma 10.3. Let M' = (Q, Z, T, 5', q' 0 , Z0 , F) be a DPDA simulating M, where 2' = {[<?, k] | q is in Q and & = 1, 2, or 3}.
Let F = {[<?, 3]\qm Q}, and let , = j[g0 , 1] if q0 is in F; ^° 2] if q0 is not in F.
The purpose of k in [q, k] is to record, between true inputs, whether or not M has entered an accepting state. If M has entered an accepting state since the last true input, then k = 1. If M has not entered an accepting state since the last true input, then k = 2. If k = 1 when M reads a true input symbol, then M' simulates the move ofM and changes k to 1 or 2, depending on whether the new state ofM is or is not in F. If k = 2, M' first changes k to 3 and then simulates the move of M, changing k to 1 or 2, depending on whether the new state of M is or is not in F.
KTUNOTES.IN Downloaded from Ktunotes.in 10.2 | CLOSURE OF DCFL'S UNDER COMPLEMENTATION 239 Thus, S' is defined as follows, for q and p in Q, and a in £.
1) If% £, Z) = (p, y), then for /c = 1 or 2, /c],£,Z) = ([p, /c'], y), where /c' = 1 if /c = 1 or p is in F; otherwise k' = 2.
2) If b(q, a, Z) = (p, y), for a in E, then 2], 6, Z) = ([<?> 3], Z) and 8'([q, 1], fl, Z) = 5'([q, 3], a, Z) - ([p, fc], y) where /c = 1 or 2 for p in F or not in F, respectively.
We claim that L{M ') is the complement of L(M). Suppose that a x a 2 ' an is in L(M). Then M enters an accepting state after using an as an input. In that case, the second component of the state of M' will be 1 before it is possible for M' to use a true input after an . Therefore, M' does not accept (enter a state whose second component is 3) while an was the last true input used.
If a x a 2 ' ' ' an is not in L(M), by Lemma 10.3 M' will some time after reading an have no e-moves to make and will have to use a true input symbol. But, at this time, the second component of M"s state is 2, since a x a 2 an is not in L(M). By rule (2), M' will accept before attempting to use a true input symbol.
Before concluding this section we state the following corollary.
Corollary Every deterministic CFL is accepted by some DPDA that, in an accepting state, may make no move on t-input.
Proof The statement is implicit in the proof of Theorem 10.1. Note that in a final state (one in which k = 3) no £-move is possible.
It is possible to use Theorem 10.1 to show certain languages not to be DCFL's.
Example 10.1 The language L = {04^2 \i=j or ; = k} is a CFL generated by the grammar S^AB\CD A->0Al\e B-+2B\e C-0C|e D-+W2\e However, L is not a DCFL. If it were, then L would be a DCFL and hence a CFL.
By Theorem 6.5, L x = L n 012 would be a CFL. But L x = {tfl J2 k \i + j and j =fc k). A proof using Odgen's lemma similar to that of Example 6.3 shows that L x is not a CFL, so L is not a DCFL.
KTUNOTES.IN Downloaded from Ktunotes.in 240 DETERMINISTIC CONTEXT-FREE LANGUAGES 10.3 PREDICTING MACHINES For a number of other closure properties of DCFL's we need a construction in which the stack symbols ofDPDA M are modified to contain information about a certain finite automaton A. The information associated with the top stack symbol tells, for each state q of M and p of A, whether there is some input string that causes M to accept when started in state q with its current stack and simultan-eously causes A to accept if started in state p.
Formally, let M = (QM,H,r,SMy q0y Z0f FM)bea. normal form DPDA and A = (QA , <5/i> Po> Fa)- Then 7c(M, A), the predicting machine for M and A, is defined by (QM , E, T x A, <5, q0 , X0y FM\ where A is the set of subsets of QM x QA .
The intention is that if n(M, A) is in ID (r, x, [Z, p]y), then p consists of exactly those pairs (q, p) such that there is a w in E for which SA (p, w) is in FA , and (qy w, ZP) fa (s, e, a) for some s in FM and a and /? in T, where /? is the string of first components of y.
To define d and X 0 we need additional notation. Let Mq Z be M with q and Z made the start state and start symbol respectively. Let A p be A with p made the start state. Then by our usual notation, L(M q .z) = W | w, Z) ^ (5, c, 7) for some s in FM and y in T} and L(/4 P) = {w I w) is in FA }.
Let Nr(M q Z ) be the set of strings that cause M q Z to erase its stack and enter state r, that is, Nr(M q ,z ) = {w(q, w, Z)H£-(r, 6, e)}.
Surely L(M q Z ) is a DCFL and L(^l p ) is a regular set. It is also true that Nr(M q Z ) is a DCFL. In proof, modify M to place a marker Y 0 at the bottom of stack and then simulate M in state q with stack ZY 0 . If 7 0 becomes the top stack symbol, then accept if the state is r and reject if not.
Finally, let L5(A p ) = {w \ d(p, w) = s}.
Clearly Ls(A p ) is regular.
Now we may define 3(r, a, [Z, p])y for r in gM , a in E u {c}, Z in T, and p. in A as follows.
1) If dM (ry a, Z) = (5, c), then ^(r, a, [Z, /]) = (s, c).
Note that does not influence the action of 7r(M, A), except in rule (3) below, where it influences the second component of the stack symbol pushed.
2) If dM (r, a, Z) = (s, Z), then <5(r, a, [Z, p) = (s, [Z, /<]).
3) If 3M (r, a, Z) = (5, YZ), then <5(r, a, [Z, = (s, [Y, v][Z, //]), where v consists of those pairs (q f p) such that either a) L(Mq y) n U(Ap ) is nonempty, or KTUNOTES.IN Downloaded from Ktunotes.in 10.3 | PREDICTING MACHINES 241 b) there is some t in QM and u in QA such that Nt(Mq , Y ) n UA P ) is nonempty and (r, u) is in Note that L(MqY ) and Nt(MqY ) are CFL's, and L(/lp ) and Ln(,4p ) are regular, so by Theorems 6.5 and 6.6, we may determine whether the languages men-tioned in (a) and (b) are empty.
Finally, let X0 = [Z0 , fi0 ], where = p)\L(NfqtZo)nL(A p)0}.
Lemma 10.4 n(M, A) as defined above has the property that (q0 , x, [Z0 , fi0]) h (r, y, [z l9 \i x\Z 2 , n 2 ] • • [Z„, nn]) if and only if a) (q0 , x, ZQ ) \± (r, y, Z x Z 2 Zn ), and b) for 1 < i < n, ft = P) I fc>r some w, (4, w, Z t Zi+ , • • • Zn ) fa (s, e, y) for some s in FM and y in T, and SA (py w) is in FA }.
Proof (a) is obvious, since n(M, A) simulates M, carrying along the second component of stack symbols but not allowing them to influence anything but other second components of stack symbols.
We prove (b) by induction on /, starting at i = n and working down. The basis, i = n is easy. Zn must be Z0 , since M is in normal form. The definition ofX0 plus rule (2) in the definition of S gives us the basis.
For the induction, suppose the result is true for i + 1. Then was constructed from j as v is constructed from fx in rule (3). Suppose there is some w such that (q, w, Z,Zi+1 --Zn )\±-(sy 6, y) for s in FM , and SA (p, w) is in FA . Then there are two cases depending on whether Z i is ever erased. If it is not, then w is in L(Mq Z) and also in L(A p ), so by rule (3a), (q, p) is in If Z f is erased, let w = w x w2 , where (q, w l9 Z^ f£- (t, e, e) and (r, w2 , Z1+ 1 Z,. +2 - Zn ) H&- (s, e, y) for some 5 in FM . Also let (^(p, Wj) = w, so ^(w, w 2 ) is in FA . Then is in N t(Mq Zi ) and also in L^/lp). By the inductive hypothesis, (f, u) is in + j. Thus by rule (3b), (q y p) is in n t .
Conversely, if (q, p) is in ^ by rule (3a), then there is a w such that dA (p, w) is in FA and (q, w, Z, Z f + j • • • Z„) ^- (5, £, y) for 5 in FM , by a sequence of moves in which Z t is never erased. If (q, p) is in & by rule (3b), then there exists w x in Z, f in KTUNOTES.IN Downloaded from Ktunotes.in 242 DETERMINISTIC CONTEXT-FREE LANGUAGES QM and u 'mQA such that (q, w l9 Z f ) H§r (t, e, e), 5A (p, = u, and (t, u) is in fii+ v By the inductive hypothesis, there exists w2 in £ such that (t, w 2 , Zi+1 Zi+2 ••• Z„) (5, e, y) for some s in FM , and ^(w, w 2 ) is in FA . Thus (g, w 2 , Z,ZI+ • • • Z„) ^ (5, e, y), and ^(p, w 2 ) is in F, so (q y p) belongs in pL t . This completes the induction and the proof of the lemma.
Example 10.2 Let M = ({q0 , q l9 q 2 , qj, {0, 1}, {X, Z0 }, 5M , q0 , Z0 , {g3 }), where (5M (go> 0, Z0 ) = fao, XZ0 \ SM(q l9 0, X) = (q 2 , e), Mflo. 0, ) = (q0 , XX)y SM(q2 , 0, X) = (g2 , e), Mflo> 1, X) = (q u XX), SM (q 2y £, Z0 ) = (g 3 , e).
<5m(<7i> 1, = (<7i> XX), Also let /4 = ({p0 , Pi}, {0, 1}, &A , p0 , {p0 }), where 0) = Pi> ^(Po, 0 = Po> MPi»°) = Po.
0 = Pi-Observe that L(M) = L(M90>Zo ) = {0' IV I / + j = k, i > 0 and j > 0}.
Also L(MquZo ) = 0 and L(M g2 ,Zo ) = L(M g3 .Zo ) = {c}.
L(A) = L(A po ) = (1+010); that is, strings with an even number of 0's, and L(A pi ) = 10(1 + 010) that is, strings with an odd number of 0's. Thus L(MqoZo ) n L(A po ) contains strings such as 00110000, and L(MqoZo)nL(A pi ) contains strings such as 01110000.
L(M q2 Zo ) n LiApJ and L(M g3 Zo ) n L(/l po ) each contain e, but the other four intersections of the form L(Mq . Zo ) n L(/t p; ) are empty. Thus the start symbol of n(M, A) is [Z0 , jU0 ], where Mo = {(4o> Po), (<7o> Pi), (<?2> Po)> (?3> Po)}-Now let us compute 5{q09 0, [Z0 , Mo]) = fao. [X, v][Z0 , /<0 ]).
To do so we need to deduce that L(Mq . x ) = 0 for 1 = 0, 1, or 2, since we cannot accept without a Z0 on the stack and cannot write Z0 if it wasn't there originally.
Thus there is no contribution to v from rule (3a). However, L(Mq3tX ) n L(AP0 ) = {£}, so we add (q3y p0 ) to v.
KTUNOTES.IN Downloaded from Ktunotes.in 10.4 | ADDITIONAL CLOSURE PROPERTIES OF DCFL'S 243 Consider rule (3b).
N„(Mqo ,x ) = {0W | / + ; = k - 1 and j > 0}, Nq2(MquX ) = {W\j = k-l}, and q2(Mq2 ,x) = {0}.
The other sets of the form Nq.(Mqj X ) are empty. Also, LPi(APj) is all strings with an even number of O's if i = j and all strings with an odd number of O's if i ± j.
Since Nq.(Mq . x ) is nonempty only if i = 2 and ; = 0, 1, or 2, we can only apply rule (3b) successfully if the pair (ft, p0 ) is chosen from fi0 . We see Nq2(Mqo X ) n Lpo(^po) and Nq 2 (Mqo>x) n L Po(APl ) are both nonempty, yielding (ft, p0 ) and (ft, Pi) for v. Similarly Nqi(MqitX ) n LPo(,4 po ) and Nq2(M9ltX ) n L po (/lPl ) are non-empty, yielding (ft,p0 ) and (ft, Pi) for v. Also, Nq2(Mq2tX ) n LPo(^Pl )is nonempty, yielding (ft, for v, but Thus, v = {(ft, Po), (ft, Pi), (ft, Po)> (ft, Pi), (ft, Pi), (ft, Po)}.
10.4 ADDITIONAL CLOSURE PROPERTIES OF DCFL's Using the idea developed in the previous section we can prove a few closure properties of deterministic context-free languages. Before proceeding, we present one more technical lemma. The lemma asserts that we can define acceptance for a DPDA by a combination of state and the top stack symbol; the language so defined is still a deterministic language.
Lemma 10.5 Let M = (Q, I, T, 3, ft, Z0 , F) be a DPDA. Let B be any subset of Q x T, that is, pairs of state and stack symbol. Define L = {w I (ft, w, Z0 ) f (q, c, Zy) for some (g, Z) in B}.
Then L is a DCFL.
Proof We define a DPDA M ', accepting L, as follows.
M' = (Q', S, T, S\ ft, Z0 , F), where Q' = {q, q" k in G} and F' = {g" | q in Q}.
M' makes the same moves as M, except that M' moves from an unprimed state to a singly primed state and then, on e-input, moves back to the corresponding unprimed state, either directly or through a doubly primed version. The latter case applies only if the pair of state and top symbol of the stack is in B.
KTUNOTES.IN Downloaded from Ktunotes.in 244 DETERMINISTIC CONTEXT-FREE LANGUAGES Formally, 1) if <%, a, Z) = (p, y), then a, Z) = (p', y); 2) (5'(</, e, Z) = (q9 Z) provided (qy Z) is not in B\ 3) 5\q\ e, Z) = (q\ Z) and <%", £, Z) = (g, Z) if (g, Z) is in £.
Quotient with a regular set Recall that the quotient of L x with respect to L2 , denoted L l /L2 > is {x | there exists w in L2 such that xw is in L x }.
In Exercise 6.4 we claimed that the CFL's were closed under quotient with a regular set. (See Theorem 1 1.3 for a proof.) We shall now prove a similar result for DCFL's.
Theorem 10.2 Let L be a DCFL and R a regular set. Then L/R is a DCFL.
Proof Let L = L(M) for M a DPDA that always scans its entire input.
Let R = L(A) for finite automaton A. Suppose M = (QM , Z, T, <5M , q0 , Z0 , FM ) and ,4 = (Q A , Z, p0 , FA ). Then let Af' = (QM , I, T x A, <5, ^o, [Z0 , Ho], FM ) be 7c(M, /i), the predicting machine for M and A.
Let # be the subset of Qm x (r x A) containing all (4, [Z, /z]) such that (g, p0 ) is in ,u.
Then by Lemma 10.5, ^i = { I fao> > [Zo> M0]) |^ (g, £, [Z, juty) and (g, p0 ) is in h} is a DCFL. By Lemma 10.4, L x = {x | for some w in Z, (g0 , x, Z0 ) ^-(<?, e, Z/) and (q, w, Zy')[-£-(s, £, /?), where s is in FM , y' is the first components of y, and ^(po, vv) is in FA }.
Equivalently, L, = {x | for some w in Z, xw is in L(M) and w is in L(A)}.
That is, Lj = L/K. Thus L/fl is a DCFL.
MIN and MAX We now show two operations that preserve DCFL's but not arbitrary CFL's.
Recall that for each language L: MIN(L) = {x |x is in L and no w in L is a proper prefix of x}, and MAX(L) = {x |x is in L and x is not a proper prefix of any word in L}.
KTUNOTES.IN Downloaded from Ktunotes.in 10.4 | ADDITIONAL CLOSURE PROPERTIES OF DCFl/S 245 Example 10.3 Let L= {0W|i,y> k > 0, i + ; > /c}.
Then MIN(L) = 00110, and MAX(L) = {0 l P0' + J | i, j > 0}.
Theorem 10.3 If L is a DCFL, then MIN(L) and MAX(L) are DCFL's.
Proof Let M = (QM , Z, T, <5M , g0 , Z0 , FM ) be a DPDA that accepts L and always scans its entire input. Modify M to make no move in a final state. Then the resulting DPDA M x accepts MIN(L). In proof, if w is in MIN(L), then let (q0,^Z0 ) = lQ \w I l \W '"\w lm (10.1) be the sequence of ID's entered by M, where Im = (q9 £, y) for some y, and q is in FM . Furthermore, since w is in MIN(L), none of 7 0 , I l9 .
.
.
, 7m _ j has an accepting state. Thus (10.1) is also a computation of M l9 so w is in L(M).
Conversely, if (g0 , w, Z0 ) = 70 I x ^ • • • ^ 7m is an accepting computation of Mj, then none of 70 , 7j, 7 m _ x has an accepting state. Thus w is in MIN(L).
For MAX we must use the predicting machine. Let A = (Q A , Z, S A , p0 , FA ) be the simple F,4 of Fig.
10.1 accepting Z + . Let M = (QM , Z, T x A, (5, #0 , [Z0 , ju0], FM ) be 7c(M, A). Let £ = {(<?, [Z, ju])|g is in FM and (<?, p0 ) is not in Then by Lemma 10.5, = {l(<7o, Z0 ) &-(<?> y) for some q in FM , and for now^t does (q, w, y) f- (5, c, /?) for 5 in r M } is a DCFL. But L, = MAX(L), so MAX(L) is a DCFL.
Any symbol in I Fig. 10.1 The automaton A.
Example 10.4 Let us use Theorem 10.3 to show a CFL not to be a DCFL. Let ^ = {0' F2 \k < i or k < j). Then L x is a CFL generated by the grammar S^AB\C A^0A\e B - l£2|l£|<: C->0C2|0C|D D 17)|c KTUNOTES.IN Downloaded from Ktunotes.in 246 DETERMINISTIC CONTEXT-FREE LANGUAGES Suppose Lj were a DCFL. Then L2 = MAX^) would be a DCFL and hence a CFL. But L2 = {0'P2 | /c = max(i, j)}. Suppose L2 were a CFL. Let n be the pump-ing lemma constant and consider z = uvwxy = (Tl^". If neither v nor x has a 2, then z' = uv 2wx 2y has 2's and at least (n + 1) O's or at least (n -f 1) l's. Thus z' would not be in L2 as supposed.
Now consider the case where vx has a 2. If either u or x has more than one symbol, then z' = wu 2wx 2y is not of the form Qfl J2k and would not be L2 . Thus either 0 or 1 is not present in vx. Hence uwy has fewer than n 2's but has n O's or n l's and is not in L2 . We conclude L2 is not a CFL, so Lj is not a DCFL.
Other closure properties As a general rule, only those closure properties of CFL's mentioned in Section 6.2 that were given proofs using the PDA characterization carry over to DCFL's. In particular, we can state the following.
Theorem 10.4 The DCFL's are closed under (a) inverse homomorphism, and (b) intersection with a regular set.
Proof The arguments used in Theorems 6.3 and 6.5 work for DPDA's.
Theorem 10.5 The DCFL's are not closed under (a) homomorphism, (b) union, (c) concatenation, or (d) Kleene closure.
Proof See Exercise 10.4 and its solution.
10.5 DECISION PROPERTIES OF DCFL's A number of problems that are undecidable for CFL's are decidable for DCFL's.
Theorem 10.6 Let L be a DCFL and R a regular set. The following problems are decidable.
1) Is L= K?
2) Is Rc=L?
3) Is L= 0?
4) Is La CFL?
5) Is L regular?
Proof 1) L = R if and only if L x = (L n R) u (L n R) is empty. Since the DCFL's are effectively closed under complementation and intersection with a regular set, and since the CFL's are effectively closed under union, Lj is a CFL, and emptiness for CFL's is decidable.
2) R c L if and only if L n R = 0. Since L n R is a CFL, Ln R = 0 & decidable.
KTUNOTES.IN Downloaded from Ktunotes.in 10.5 | DECISION PROPERTIES OF DCFL'S 247 3) Since the DCFL's are effectively closed under complementation, L is a DCFL and hence L = 0 is decidable.
4) The property L is a CFL is trivial for DCFL's and hence is decidable.
5) Regularity for DCFL's is decidable. The proof is lengthy and the reader is referred to Stearns or Valiant [1975b].
Undecidable properties of DCFL's Certain other properties undecidable for CFL's remain so even when restricted to the DCFL's. Many of these problems can be proved undecidable by observing that the languages Lj and L2 of Section 8.6, whose intersection is the valid compu-tations of a Turing machine M, are DCFL's.
Theorem 10.7 Let L and L be arbitrary DCFL's. Then the following problems are undecidable.
1) Is Ln L = 0?
2) Is LciL?
3) Is L n L a DCFL?
4) Is L n L a CFL?
5) Is L u L a DCFL?
Proof Given an arbitrary TM M we showed in Lemma 8.6 how to construct languages L x and L2 such that L x n L2 = 0 if and only if L(M) = 0. It is easy to show that Lj and L2 are DCFL's by exhibiting DPDA's that accept them. Thus (1) follows immediately from the fact that it is undecidable whether L(M) = 0. Since DCFL's are closed under complement, and L c E if and only if L n L = 0, (2) follows from (1).
To prove (3), (4), and (5), modify each TM M to make at least two moves before accepting, as in Lemma 8.8. Then L x n L2 is either a finite set (in which case it is surely a CFL and a DCFL) or is not a CFL depending on whether L(M) is finite. Thus decidability of (3) or (4) would imply decidability of finiteness for L(M), a known undecidable property.
Since DCFL's are closed under complementation, deciding whether L u L is a DCFL is equivalent to deciding if L n L is a DCFL. Thus (5) follows from (3).
Theorem 10.8 Let L be an arbitrary CFL. It is undecidable whether L is a DCFL.
Proof Let L be the CFL of invalid computations of an arbitrary TM M that makes at least two moves on every input. L is regular and, hence, a DCFL if and only if M accepts a finite set.
Finally we observe that the question of whether two DCFL's are equivalent is an important unresolved problem of language theory.
KTUNOTES.IN Downloaded from Ktunotes.in 248 DETERMINISTIC CONTEXT-FREE LANGUAGES 10.6 LR(0) GRAMMARS Recall that one motivation for studying DCFL's is their ability to describe the syntax of programming languages. Various compiler writing systems require syn-tactic specification in the form of restricted CFG's, which allow only the represen-tation of DCFL's. Moreover, the parser produced by such compiler writing systems is essentially a DPDA. In this section we introduce a restricted type of CFG called an LR(0) grammar. This class of grammars is the first in a family collectively called LK-grammars. Incidentally, LR(0) stands for "left-to-right scan of the input producing a rightmost derivation and using 0 symbols of lookahead on the input." The LR(0) grammars define exactly the DCFL's having the prefix property.
(L is said to have the prefix property if, whenever w is in L, no proper prefix of w is in L.) Note that the prefix property is not a severe restriction, since the introduc-tion of an endmarker converts any DCFL to a DCFL with the prefix property.
Thus L$ = {w$ | w is in L} is a DCFL with the prefix property whenever L is a DCFL.
While the LR(0) restriction is too severe to provide convenient and natural grammars for many programming languages, the LR(0) condition captures the flavor of its more useful generalizations, which we discuss in Section 10.8, and which have been successfully used in several parser-generating systems.
L/?-items To introduce the LR(0) grammars we need some preliminary definitions. First, an item for a given CFG is a production with a dot anywhere in the right side, including the beginning or end. In the case of an e-production, B -> e, B -> • is an item.
Example 10.5 We now introduce a grammar that we shall use in a series of examples.
S'^Sc S-+SA\A A^aSb\ab (10.2) This grammar, with start symbol S\ generates strings of "balanced parentheses," treating a and b as left and right parentheses, respectively, and c as an endmarker.
The items for grammar (10.2) are • Sc S^SA A^- aSb S • c S^S- A A-+aSb S'- Sc-S^SA • A^aS-b A A^aSb-S^A-A^ab A^a-b A -> ab • KTUNOTES.IN Downloaded from Ktunotes.in 10.6 | LR(0) GRAMMARS 249 In what follows, we use the symbols => and => to denote rightmost deriva-tions and single steps in a rightmost derivation, respectively. A right-sentential form is a sentential form that can be derived by a rightmost derivation. A handle of a right-sentential form y for CFG G is a substring /?, such that and dfiw = y. That is, a handle of 7 is a substring that could be introduced at the last step in a rightmost derivation of y. Note that in this context, the position of P within y is important.
A viable prefix of a right-sentential form y is any prefix of y ending no farther right than the right end of a handle of y.
Example 10.6 In grammar (10.2) there is a rightmost derivation S' => Sc => SAc => SaSbc.
Thus SaSbc is a right-sentential form, and its handle is aSb. Note that in any unambiguous grammar with no useless symbols, such as grammar (10.2), the rightmost derivation of a given right-sentential form is unique, so its handle is unique. Thus we may speak of "the handle' rather than "a handle." The viable prefixes of SaSbc are e y S, Sa, SaS, and SaSb.
We say an item A -> a • p is valid for a viable prefix y if there is a rightmost derivation SZ>dAwj>daPw and <5a = y. Knowing which items are valid for a given viable prefix helps us find a rightmost derivation in reverse, as follows. An item is said to be complete if the dot is the rightmost symbol in the item. If A - a • is a complete item valid for y, then it appears that A -> a could have been used at the last step and that the previous right-sentential form in the derivation of yw was 3Aw.
Of course, we cannot more than suspect this since A^ol- may be valid for y because of a rightmost derivation S ^> 5Aw' => yw'. Clearly, there could be two or more complete items valid for y, or there could be a handle of yw that includes symbols of w. Intuitively, a grammar is defined to be LR(0) if in each such situa-tion 5Aw is indeed the previous right-sentential form for yw. In that case, we can start with a string of terminals x that is in L(G) and hence is a right-sentential form of G, and work backward to previous right-sentential forms until we get to S. We then have a rightmost derivation of x.
Example 10.7 Consider grammar (10.2) and the right-sentential form abc. Since S' h> Ac => abc, we see that A ab • is valid for viable prefix ab. We also see that A - a • b is valid for viable prefix a, and A -> - ab is valid for viable prefix e. As A ab • is a KTUNOTES.IN Downloaded from Ktunotes.in 250 DETERMINISTIC CONTEXT-FREE LANGUAGES complete item, we might be able to deduce that Ac was the previous right-sentential form for abc.
Computing sets of valid items The definition of LR(0) grammars and the method of accepting L(G) for LR(0) grammar G by a DPDA each depend on knowing the set of valid items for each viable prefix y.
It turns out that for every CFG G whatsoever, the set of viable prefixes is a regular set, and this regular set is accepted by an NFA whose states are the items for G. Applying the subset construction to this NFA yields a DFA whose state in response to the viable prefix y is the set of valid items for y.
The NFA M recognizing the viable prefixes for CFG G = (K, T, P, S) is defined as follows. Let M = (Q, V u T, <5, q0 , Q\ where Q is the set of items for G plus the state q0 , which is not an item. Define 1) <H<?o> 0 = {S - a |S a is a production}, 2) 8{A -ol BP, e) = {B -• y \ B - y is a production}, 3) S(A - a Xfr X) = {A - aX • 0}.
Rule (2) allows expansion of a variable B appearing immediately to the right of the dot. Rule (3) permits moving the dot over any grammar symbol X ifX is the next input symbol.
Example 10.8 The NFA for grammar (10.2) is shown in Fig.
10.2.
Fig. 10.2 NFA recognizing viable prefixes for Grammar (10.2).
KTUNOTES.IN Downloaded from Ktunotes.in 10.6 | LR(0) GRAMMARS 251 Theorem 10.9 The NFA M defined above has the property that S(q0 , y) contains A -+ a • f! if and only if A -+ a • ft is valid for y.
Proof Only if: We must show that each item A a • ft contained in d(q0 , y) is valid for y.
We proceed by induction on the length of the shortest path labeled y from q0 to A -> a • P in the transition diagram for M. The basis (length 1) is straightforward.
The only paths of length one from q0 are labeled e and go to items of the form S -> • a. Each of these items is valid for e because of the rightmost derivation S f> a.
For the induction, suppose that the result is true for paths shorter than k 9 and let there be a path of length k labeled y from q0 to A a • /?. There are two cases depending on whether the last edge is labeled e or not.
case 1 The last edge is labeled X, for X in V u T. The edge must come from a state A -+ a' • AT/?, where a = ol'X. Then by the inductive hypothesis, A - a' • X/? is valid for y', where y = y'X. Thus there is a rightmost derivation S%>dAw=>doc'Xpw, rm rm ' where <5a' = y'. This same derivation shows that A ol'X /? (which is >4 a ft) is valid for y.
case 2 The last edge is labeled c. In this case a must be c, and A -> a • /? is really A- -ft. The item in the previous state is of the form i^-x^ • Afi u and is also valid for y. Thus there is a derivation where y = 3oc l . Let Pi^>x for some terminal string x. Then the derivation S ^> SBw => &x, ,4/?i w ^> 5ol x Axw => eta, tfxw can be written S ^> (5a, Axw => dot, Bxw.
rm 1 rm 1 " Thus A • P is valid for y, as y = (5a P //: Suppose ,4 a • /? is valid for y. Then S^y^w^yja^w, (10.3) where y 1 ct = y. If we can show that 3(q0 , y x ) contains A -• a/?, then by rule (3) we know that S(q0 , y) contains A-kx- p. We therefore prove by induction on the length of derivation (10.3) that S(q0 , y t ) contains /I • a/?.
The basis, one step, follows from rule (1). For the induction, consider the step in S=>y x Aw in which the explicitly shown A was introduced. That is, write s kyi Aw as S I? y2 Bx 7% 72 73 Ay^x |> y 2 y 3 Ayx, KTUNOTES.IN Downloaded from Ktunotes.in 252 DETERMINISTIC CONTEXT-FREE LANGUAGES where y 2 y 3 = y x and yx = w. Then by the inductive hypothesis applied to the derivation S^y 2 Bxj>y2 y3 Ay4x9 we know that B • y 3 Ay4 is in 5{qQ , y2 ). By rule (3), B y3 • y4y4 is in <5(<70 , y 2 y 3 ), and by rule (2), A • a/?
is in <%0 , y 2 y 3 ). Since y 2 y 3 = yj, we have proved the inductive hypothesis.
Definition of LR(0) grammar We are now prepared to define an LR(0) grammar. We say that G is an LR(0) grammar if 1) its start symbol does not appear on the right side of any production, and 2) for every viable prefix y of G, whenever A a • is a complete item valid for y, then no other complete item nor any item with a terminal to the right of the dot is valid for y.| There is no prohibition against several incomplete items being valid for y, as long as no complete item is valid.
Theorem 10.9 gives a method for computing the sets of valid items for any viable prefix. Just convert the NFA whose states are items to a DFA. In the DFA, the path from the start state labeled y leads to the state that is the set of valid items for y. Thus construct the DFA and inspect each state to see if a violation of the LR(0) condition occurs.
Example 10.9 The DFA constructed from the NFA of Fig.
10.2, with the dead state (empty set of items) and transitions to the dead state removed, is shown in Fig.
10.3. Of these states, all but 70 , I u / 3 , and 7 6 consist of a single complete item.
The states with more than one item have no complete items, and surely 5", the start symbol, does not appear on the right side of any production. Hence grammar (10.2) is LK(0).
10.7 LR(0) GRAMMARS AND DPDAs We now show that every LR(0) grammar generates a DCFL, and every DCFL with the prefix property has an LR(0) grammar. Since every language with an LR(0) grammar will be shown to have the prefix property, we have an exact characterization of the DCFL's; namely L is a DCFL if and only if L$ has an LR(0) grammar.
t The only items that could be valid simultaneously with A - a are productions with a nonterminal to the right of the dot, and this can occur only if a = e; otherwise another violation of the LR(0) conditions can be shown to occur.
KTUNOTES.IN Downloaded from Ktunotes.in 10.7 | jLR(O) grammars and dpda's 253 Fig. 10.3 DFA whose states are the sets of valid items.
DPDA's from LR(0) grammars The way in which we construct a DPDA from an LR(0) grammar differs from the way in which we constructed a (nondeterministic) PDA from an arbitrary CFL in Theorem 5.3. In the latter theorem we traced out a leftmost derivation of the word on the PDA's input, using the stack to hold the suffix of a left-sentential form beginning at the leftmost variable. Now we shall trace out a rightmost derivation, in reverse, using the stack to hold a viable prefix of a right-sentential form, including all variables of that right-sentential form, allowing the remainder of the form to appear on the input.
In order to clearly describe this process, it is useful to develop a new notation for ID's of a PDA. We picture the stack with its top at the right end, rather than the left. To distinguish the new notation from the old we use brackets rather than parentheses: [q, a, w] is our synonym for (q, w, olr ).
KTUNOTES.IN Downloaded from Ktunotes.in 254 DETERMINISTIC CONTEXT-FREE LANGUAGES To simulate rightmost derivations in an LR(0) grammar not only do we keep a viable prefix on the stack, but above every symbol we keep a state of the DFA recognizing viable prefixes. If viable prefix X X X 2 "' Xk is on the stack, then the complete stack contents will be s0 X 1 s 1 • • • Xk sk , where s f is 3(q0 , X x • • • X t ) and 3 is the transition function of the DFA. The top state sk provides the valid items for X x X 2 ' ' ' Xk .
If sk contains A a • , then A a • is valid for X x Xk . Thus a is a suffix of X x • -- Xk , say ol = Xi+1 Xk (note a may be £, in which case i = k). Moreover, there is some w such that X x ••• Xk w is a right-sentential form, and there is a derivation rm 1 1 rm K Thus to obtain the right-sentential form previous to X x • • Xk w in a right deriva-tion we reduce a to A, replacing X i+ , • • • Xk on top of the stack by A. That is, by a sequence of pop moves (using distinct states so the DPDA can remember what it is doing) followed by a move that pushes A and the correct covering state onto the stack, our DPDA will enter a sequence of ID's [q, s0 X x sk . x Xk sk , w] \±- [q, s0 X x ••• A> f /ls, w], (10.4) where s = <5(s„ /I). Note that if the grammar is LR(0)9 sk contains only A a • , unless a = c, in which case sk may contain some incomplete items. However, by the LR(0) definition, none of these items have a terminal to the right of the dot, or are complete. Thus for any y such that X x • -• Xk y is a right-sentential form, X x • • • Xi Ay must be the previous right-sentential form, so reduction of a to A is correct regardless of the current input.
Now consider the case where sk contains only incomplete items. Then the right-sentential form previous to X x - - Xk w could not be formed by reducing a suffix of X x -- Xk to some variable, else there would be a complete item valid for X x X k . There must be a handle ending to the right of Xk in X x • • Xk w, as X x -Xk is a viable prefix. Thus the only appropriate action for the DPDA is to shift the next input symbol onto the stack. That is, [q, s0X x ••• sk . x Xk sk , ay]\—[q, s0 X x • s^^^at, y], (10.5) where t = 5(sk , a). If t is not the empty set of items, X x • • • Xk a is a viable prefix. If f is empty, we shall prove there is no possible previous right-sentential form for X x Xk ay, so the original input is not in the grammar's language, and the DPDA "dies" instead of making the move (10.5). We summarize the above obser-vations in the next theorem.
Theorem 10.10 If L is L(G) for an LR(0) grammar G, then L is N(M) for a DPDA M.
Proof Construct from G the DFA D, with transition function 5, that recognizes G's viable prefixes. Let the stack symbols ofM be the grammar symbols of G and KTUNOTES.IN Downloaded from Ktunotes.in 10.7 | LR(0) GRAMMARS AND DPDA'S 255 the states of D. M has state q, which is its start state, along with the additional states used to perform reductions by sequences of moves such as (10.4) above. We assume the reader can specify the set of states for each reduction and the e-transitions needed to effect a reduction. We also leave to the reader the specification of the transition function of M needed to implement the moves indicated by (10.4) and (10.5).
We have previously indicated why, if G is LR(0), reductions are the only possible way to get the previous right-sentential form when the state of the DFA on the top of Af's stack contains a complete item. We claim that when M starts with w in L(G) on its input and only s0 on its stack, it will construct a rightmost derivation for w in reverse order. The only point still requiring proof is that when a shift is called for, as in (10.5), because the top DFA state on M's stack has only incomplete items, then there could not be a handle among the grammar symbols X x Xk found on the stack at that time. If there were such a handle, then some DFA state on the stack, below the top, would have a complete item.
Suppose there were such a state containing A a . Note that each state, when it is first put on the stack either by (10.4) or (10.5), is on top of the stack.
Therefore it will immediately call for reduction of a to A. If a £, then {A - a • } is removed from the stack and cannot be buried. If a = £, then reduction of e to A occurs by (10.4), causing A to be put on the stack above X t ••• Xk . In this case, there will always be a variable above Xk on the stack as long as X t • • • Xk occupies the bottom positions on the stack. But A e at position k could not be the handle of any right-sentential form X t ••• Xk f$, where ft contains a variable.
One last point concerns acceptance by G.
If the top state on the stack is {S -> a • }, where S is G's start symbol, then G pops its stack, accepting. In this case we have completed the reverse of a rightmost derivation of the original input.
Note that as S does not appear on the right of any production, it is impossible that there is an item of the form A - Sa valid for viable prefix S. Thus there is never a need to shift additional input symbols when S alone appears on the stack. Put another way, L[G) always has the prefix property if G is LR(0).
We have thus proved that if w is in L(G), M finds a rightmost derivation of w, reduces w to S, and accepts. Conversely, if M accepts w, the sequence of right-sentential forms represented by the ID's of M provides a derivation of w from S.
Thus N(M) = L(G).
Corollary Every LR(0) grammar is unambiguous.
Proof The above argument shows that the rightmost derivation of w is unique.
Example 10.10 Consider the DFA of Fig. 10.3. Let 0, 1, .
.
., 8 be the names of the states corresponding to the sets of items /0 , I u / 8 , respectively. Let the input be aababbc. The DPDA M constructed as in Theorem 10.10 makes the sequence of moves listed in Fig.
10.4.
KTUNOTES.IN Downloaded from Ktunotes.in 256 DETERMINISTIC CONTEXT-FREE LANGUAGES Stack Remaining input Comments 1) 0 aababbc Initial ID 2) 0a3 ababbc Shift 3) 0a3a3 babbc Shift 4) Oa3a3bl abbe Shift 5) 0a3A2 abbe Reduce by A -> ab 6) 0a3S6 abbe Reduce by S -> A 7) 0a3S6a3 bbc Shift 8) 0a3S6a3b7 be Shift 9) 0a3S6A5 be Reduce by A -> ab 10) 0a3S6 be Reduce by S - SA 11) 0a3S6b% c Shift 12) 0A2 c Reduce by A -> aSb 13) 0S\ c Reduce by S -> A 14) 0S\c4 Shift 15) Accept Fig. 10.4 Sequence of moves of DPDA M.
For example, in line (1), state 0 is on top of the stack. There is no complete item in set / 0 , so we shift. The first input symbol is a, and there is a transition from / 0 to / 3 labeled a. Thus in line (2) the stack is 0a3. In line (9), 5 is the top state. / 5 consists of complete item S -> SA. We pop SA off the stack, leaving 0a3. We then push S onto the stack. There is a transition from / 3 to / 6 labeled S, so we cover S by 6, yielding the stack 0a3S6 in line (10).
LR(0) grammars from DPDA's We now begin our study of the converse result—if L is N(M) for a DPDA M, then L has an LR(0) grammar. In fact, the grammar of Theorem 5.4 is LR(0) whenever M is deterministic, but it is easier to prove that a modification of that grammar is LR(0). The change we make is to put at the beginning of the right side of each production a symbol telling which PDA move gave rise to that production.
Formally, let M = (g, Z, T, <5, qQj Z0 , 0) be a DPDA. We define grammar GM = (V, Z, P, S) such that L(GM ) = N(M). V consists of the symbol 5, the symbols [qXp] for q and p in Q and X in T, and the symbols A qaY for q in Q, a in Z u {e} and Y in T. S and the [qXp] y s play the same role as in Theorem 5.3.
Symbol A qaY indicates that the production is obtained from the move of M in 3(q, a, Y). The productions of GM are as follows (with useless symbols and pro-ductions removed).
1) S->[q0 Z0 p] for all p in Q.
2) If 3(q, a, Y) = (p, e), then there is a production [qYp] -> A qaY .
KTUNOTES.IN Downloaded from Ktunotes.in 10.7 | LR(0) GRAMMARS AND DPDA'S 257 3) If 5(q, a, Y) = (pu X X X 2 '" Xk ) for k > 1, then for each sequence of states p 2 , P3» •••» Pfc+i there is a production [qYPk + i]-+A qaY[p 1 X l p 2] ••• [phXkPk+ll 4) For all q, a, and Y, A qaY -> a.
Consider a rightmost derivation in GM . It starts with S=>[q0 Z0 p] for some state p. Suppose for the sake of argument that d(q0 , a, Z0 ) = (r, XYZ). Then the only productions for [q0 Z0 p] that derive strings beginning with a (a may be e) have right sides AqoaZo[rXs][sYt][tZp] for some states s and t. Suppose that the rightmost derivation eventually derives some string w from [tZp]. Then, if S(s, b, Y) — (u, VW), we might continue the rightmost derivation as S g> AqoaZo[rXs][sYt]w ^,oflZo[^^]^y["^][^r]w.
(10.6) Now consider the moves made by M before reading input w. The input corresponding to derivation (10.6) is of the form ax t bx 2 x 3 w y where [rXs]^>x lt [uVv] ^> x 2 , and [vWt] ^> x 3 . The corresponding sequence of moves is of the form| (qQ , ax x bx2 x$w, Z0 ) |— (r, x x bx 2 x 3 w, XYZ) p-(s, bx 2 x 3 wy YZ) x 2 x3 w, VWZ) h-(r, x 3 w, VKZ) h-(^,vv,Z).
(10.7) If we compare (10.6) and (10.7) we note that stack symbols (Z in particular) which remain on the stack at the end of (10.7) are the symbols that do not appear (with two states attached in a bracketed variable) in the longest viable prefix of (10.6). The stack symbols popped from the stack in (10.7), namely X y V, and W, are the symbols that appear in the viable prefix of (10.6). This situation makes sense, since the symbols at the left end of a sentential form derive a prefix of a sentence, and that prefix is read first by the PDA.
In general, given any viable prefix a of GM , we can find a corresponding ID / of M in which the stack contains all and only the stack symbols that were in-troduced in a rightmost derivation of some aw and later replaced by a string of terminals. Moreover, / is obtained by having M read any string derived from a. In the case that M is deterministic, we can argue that the derivations of right-sentential forms with prefix a have a specific form and translate these limitations on derivations into restrictions on the set of items for a.
Lemma 10.6 If M is a DPDA and GM is the grammar constructed from M as above, then whenever [qXp] ^> w, there is a unique computation (q y w, X) |-- (p, £, t Note that we have reverted to our original notation for ID's.
KTUNOTES.IN Downloaded from Ktunotes.in 258 DETERMINISTIC CONTEXT-FREE LANGUAGES e). Moreover, the sequence of moves made by M corresponds to the reverse of the sequence in which subscripted A^'s are replaced by a, where A^y is deemed to "correspond" to a move in which the state is s, Y is on top of the stack, and input a is used.
Proof The existence of such a computation was proved in Theorem 5.3.
Its uniqueness follows from the fact that M is deterministic. To show the correspon-dence between the moves of M and the reverse of the sequence of expansions of subscripted A's, we perform an easy induction on the length of a derivation. The key portion of the inductive step is when the first expansion is by rule (3): [qXp]=>AqaX[p 1 X l p2][p2X 2 p3 ] •• [pkXk p].
Then the explicitly shown A qaX will be expanded after all subscripted A's derived from the other variables are expanded.
As the first move of M , (q, w, X)\—(pu w\X l X 2 -- Xk\ where w = aw\ corresponds to AqaX , we have part of the induction proved. The remainder of the induction follows from observing that in the moves of M, X u X 2 , . .., Xk are removed from the stack in order, by using inputs wu vv2 , . .., w fc , where WjWj ••• wk = w', while in the rightmost derivation of w from [qXp\ the derivation of w x from [p { X x p 2 ] follows the derivation of w 2 from [p 2 X 2 p 3 ], and so on. Since all these derivations are shorter than [qXp] => w, we may use the inductive hypothesis to complete the proof.
Now, for each variable [qXp] of GM , let us fix on a particular string wqXp derived from [gXpJ.t Let h be the homomorphism from the variables of GM to L defined by KA qay) = <, H[qXp]) = WqXp .
Let N(A qay ) = 1 and N([qXp]) be the number of moves in the computation corre-sponding to [qXp]%> wqXp . Extend N to V by N(B l B2 - Bk ) = JJ-i "(H Finally, let us represent a move S(q, a, Y) of M by the triple (qaY). Let m be the homomorphism from V to moves defined by 1) m(A qaY)=(qaY); 2) m([qXp]) is the reverse of the sequence of subscripts of the A's expanded in the derivation of wqXp from [qXp]. By Lemma 10.6, m([qXp]) is also the sequence of moves (g, wqXpy X) p- (p, 6, e).
We can now complete our characterization of LR(0) grammars.
Lemma 10.7 Let y be a viable prefix of GM . (Note that by the construction of GM , y is in V). Then (q0 , h(y), Z0 ) \J^L (p, e y P) for some p and 0, by the sequence of moves m(y).
t We assume GM has no useless symbols, so wqXp exists.
KTUNOTES.IN Downloaded from Ktunotes.in 10.7 | LR(0) GRAMMARS AND DPDA'S 259 Proof As y is a viable prefix, there is some y in I such that yy is a right-sentential form. Then for some state r, [q0 Z0 r] => h(y)y. By Lemma 10.6, the last N(y) expansions of A's in that derivation take place after the right-sentential form yy is reached. Also by Lemma 10.6, there is a unique sequence ofmoves (q0 , h(y)y, Z0 ) H- (r, e, e), and the first N(y) of these must be m(y).
We are now ready to show that GM is LR(0). Since the start symbol obviously does not appear in any right side, it suffices to show that each set of items with a complete item B -> /?
• contains no other complete item and no item of the form A q(l Y 0 for a in £. We prove these facts in two lemmas.
Lemma 10.8 If / is a set of items of GM , and B ft • is in /, then there is no item A qaY - ' a in /.
Proo/ Let / be the set of items for viable prefix y.
case 1 If B -> /?
is a production from rule (1), then y = /?, and y is a single variable [q0 Z0 p], since S appears on no right side. If A qaY - • a is valid for y, then there is a derivation S ^> yA qaY y j> yay. However, no right-sentential form begins with a variable [q0 Z0 p] unless it is the first step of a derivation; all subsequent right-sentential forms begin with a subscripted A, until the last, which begins with a terminal. Thus y could not be followed by A qaY in a right-sentential form.
case 2 If B -> ft is introduced by rules (2) or (3), then we can again argue that y'f$A qaY is a viable prefix, where y = y'p. However, in any rightmost derivation, when B - /? is applied, the last symbol of /? is immediately expanded by rules (2), (3), or (4), so ft could not appear intact in a right-sentential form followed by A qaY .
case 3 If B - P is Apbz -> b introduced by rule (4), and A qaY -• a is valid for y, then b must be c, else y/l 9fl y, which is a viable prefix, has a terminal in it. As APez ^ valid for y, it follows that y^4 pfZ is a viable prefix. Thus, by Lemma 10.7 applied to yA qaY and yA pcZ , the first N(y) + 1 moves made by M when given input h{y)a are both m(y)(p£Z) and m(y)(qaY), contradicting the determinism of M.
(Note that in the first of these sequences, a is not consumed.) Lemma 10.9 If / is a set of items of GM , and B ft • is in /, then there is no other item C -> a • in /.
Proo/ Again let y be a viable prefix with set of valid items /.
case 1 Neither B -> /? nor C a is a production introduced by rule (4). Then the form of productions of types (2) and (3), and the fact that productions of type (1) are applied only at the first step tell us that as a and ft are both suffixes of y, we must have p = a. If these productions are of type (1), B = C = 5, so the two items are really the same. If the productions are of type (2) or (3), it is easy to check that B = C. For example, if a = /? = A qaY , then the productions are of type (2), and B and C are each [qYp] for some p. But rule (2) requires that d(q, a, Y) = (p, f), so the determinism of M assures that p is unique.
KTUNOTES.IN Downloaded from Ktunotes.in 260 DETERMINISTIC CONTEXT-FREE LANGUAGES case 2 B - P and C -> a are type (4) productions. Then yB and yC are viable prefixes, and Lemma 10.7 provides a contradiction to the determinism of M. That is, if f$ = a = €, then the first N(y) + 1 moves ofM on input h(y) must be m(yB) and must also be m(yC). If /? = a 6 and ai — b^e, then a = and the first N(y) + 1 moves ofM on input /i(y)a must be m(yB) and m(yC). If ft = a e and a = £, then the first N(y) + 1 moves of M on input h(y)a provides a similar contradiction.
case 3 B -> /? is from rule (1), (2), or (3) and C -> a is from rule (4), or vice versa.
Then yC is a right-sentential form, and y ends in We can rule out this possibility as in cases (1) and (2) of Lemma 10.8.
Theorem 10.11 If M is a DPDA, then GM is an LR(0) grammar.
Proof Immediate from Lemmas 10.8 and 10.9.
We can now complete our characterization of LR(0) grammars.
Theorem 10.12 A language L has an LK(0) grammar if and only if L is a DCFL with the prefix property.
Proof If: Suppose L is a DCFL with the prefix property. Then L is L(M') for a DPDA Af '. We can make M' accept L by empty stack by putting a bottom-of-stack marker on M' and causing M' to enter a new state that erases the stack whenever it enters a final state. As L has the prefix property, we do not change the language accepted, and L is accepted by empty stack by the new DPDA, M. Thus L= L(GM ), and the desired conclusion follows from Theorem 10.11.
Only if: Theorem 10.10 says that L is N(M) for a DPDA, M. We may use the construction of Theorem 5.2 to show that L is L(M') for a DPDA M '. The fact that L has the prefix property follows from the fact that a DPDA "dies" when it empties its stack.
Corollary L$ has LR(0) grammar if and only if L is a DCFL, where $ is not a symbol of L's alphabet.
Proof LS> surely has the prefix property. If LS is a DCFL, then L = L$/$ is a DCFL by Theorem 10.2. Conversely, if L is a DCFL, it is easy to construct a DPDA for LS.
10.8 LR(k) GRAMMARS It is interesting to note that if we add one symbol of "lookahead," by determining the set of following terminals on which reduction by A a could possibly be performed, then we can use DPDA's to recognize the languages of a wider class of grammars. These grammars are called LR(l) grammars, for the one symbol of lookahead.
It is known that all and only the deterministic CFL's have LK(1) KTUNOTES.IN Downloaded from Ktunotes.in 10.8 | LR(k) GRAMMARS 261 grammars. This class of grammars has great importance for compiler design, since they are broad enough to include the syntax of almost all programming languages, yet restrictive enough to have efficient parsers that are essentially DPDA's.
It turns out that adding more than one symbol of lookahead to guide the choice of reductions does not add to the class of languages definable, although for any k, there are grammars, called LR(k), that may be parsed with k symbols of lookahead but not with k — 1 symbols of lookahead.
Let us briefly give the definition and an example of LR(l) grammars, without proving any of the above contentions. The key extension of LR(0) grammars is that an LR(1 ) item consists of an LR(0) item followed by a lookahead set consisting of terminals and/or the special symbol $, which serves to denote the right end of a string. The generic form of an LR(1) item is thus A-+cc 0, {au a 2 , ...,an}.
We say LR(l) item A -> a • ft, {a} is valid for viable prefix y if there is a rightmost derivation S => SAy j> Satfty, where Sol — y, and either i) a is the first symbol of y, or ii) y — e and a is $.
Also, A -> a • /?, {a u a 2 , .
.
.
, an] is valid for y if for each i, A a • ft, {a,} is valid for y.
Like the LR(0) items, the set of LR(l) items forms the states of a viable prefix recognizing NFA, and we can compute the set of valid items for each viable prefix by converting this NFA to a DFA. The transitions of this NFA are defined as follows.
1) There is a transition on X from A -> a • Xft, {a ly a 2 , . .., an} to A olX • ft, {au a 2 , an}.
2) There is a transition on e from A -> a • Bp, {a u a 2 , .
. ., an} to B - • y, T, if B -> y is a production and T is the set of terminals and/or $ such that b is in T if and only if either i) ft derives a terminal string beginning with b, or ii) ft ^> e, and b is a x for some 1 < i < n.
3) There is an initial state q0 with transitions on e to S->- a, {$} for each production S -> a.
Example 10.11 Consider the grammar S-+A A - BA | £ B^aB\b (10.8) which happens to generate a regular set, (ab). The NFA for grammar (10.8) is shown in Fig.
10.5, and the corresponding DFA is shown in Fig. 10.6. The NFA of Fig.
10.5 is unusual in that no two items differ only in the lookahead sets. In general, we may see two items with the same dotted production.
KTUNOTES.IN Downloaded from Ktunotes.in Fig. 10.6 DFA for LR(l) items.
KTUNOTES.IN Downloaded from Ktunotes.in 10.8 | LR(k) GRAMMARS 263 To see how Fig.
10.5 is constructed, consider item 5 -> • A, {$}.
It has e-transitions to items of the form A -> • AB, T, and A -> • , 7, but what should T be?
In rule (2) above, ft is e, so (2i) yields no symbols for T. Rule (2ii) tells us that $ is in T, so T = {$}. Now consider item >4 -> • 2L4, {$}. There are e-transitions to B-+ - aB, U and B-+ - b, U for some I/. Here, /? = A It is easy to check that A derives strings beginning with a and b, so a and b are in U. A also derives e, so $ is in 1/ because it is the lookahead set of A -> • &4, {$}. Thus (7 = {a, b, $}.
A grammar is said to be LR(1) if 1) the start symbol appears on no right side, and 2) whenever the set of items I valid for some viable prefix includes some com-plete item A -> a • , {au a2 , .
.
. , a„}, then i) no a t appears immediately to the right of the dot in any item of 7, and ii) if , {b u b 2 , bk} is another complete item in 7, then a { J= b} for any 1 < i < n and 1 < j < k.
Example 10.12 Consider Fig. 10.6. Sets of items I u 74 , / 5 , and I 6 consist of only one item and so satisfy (2). Set I0 has one complete item, A->-y {$}. But $ does not appear to the right of a dot in any item of 70 . A similar remark applies to J 2 , and 7 3 has no complete items. Thus grammar (10.8) is LR(). Note that this grammar is not LR(0); its language does not have the prefix property.
The automaton that accepts an LR(l) language is like a DPDA, except that it is allowed to use the next input symbol in making its decisions even if it makes a move that does not consume its input. We can simulate such an automaton by an ordinary DPDA if we append $ to the end of the input. Then the DPDA can keep the next symbol or $ in its state to indicate the symbol scanned. The stack of our automaton is like the stack of the LR(0) grammar recognizing DPDA: it has alternating grammar symbols and sets of items. The rules whereby it decides to reduce or shift an input symbol onto the stack are: 1) If the top set of items has complete item A -> a • , {a u a 2i .
where A J= 5, reduce by A -> a if the current input symbol is in {au a 2t .
. ., an}.
2) If the top set of items has an item S -> a , {$}, then reduce by S -> a and accept if the current symbol is $, that is, the end of the input is reached.
3) If the top set of items has an item A a • aB, T, and a is the current input symbol, then shift.
Note that the definition of an LK(1) grammar guarantees that at most one of the above will apply for any particular input symbol or $. We customarily sum-marize these decisions by a table whose rows correspond to the sets of items and whose columns are the terminals and $.
KTUNOTES.IN Downloaded from Ktunotes.in 264 DETERMINISTIC CONTEXT-FREE LANGUAGES Example 10.13 The table for grammar (10.8), built from Fig.
10.6, is shown in Fig.
10.7. Empty entries indicate an error; the input is not in the language. The sequence of actions taken by the parser on input aabb is shown in Fig. 10.8. The number i on the stack represents set of items /, from Fig. 10.6. The proper set of items with which to cover a given grammar symbol is determined from the DFA transitions (Fig. 10.6) exactly as for an LR(0) grammar.
a b $ h Shift Shift Reduce by A -> c h Accept h Shift Shift Reduce by A -> e li Shift Shift u Reduce by B - b Reduce by B - b Reduce by B - b h Reduce by A -> BA h Reduce by B - aB Reduce by B-+ aB Reduce by B -> aB Fig. 10.7 Decision table for grammar (10.8).
Stack Remaining input Comments 0 aabb$ Initial 0a3 abb% Shift Oa3a3 bb% Shift Qa3a3M b% Shift Oa3a3B6 b% Reduce by B - b Oa3B6 b% Reduce by B -> aB 0B2 b% Reduce by B -> aB 0B2M $ Shift OB2B2 $ Reduce by B - b OB2B2A5 $ Reduce by A -> c 0B2A5 $ Reduce by A -> BA 0A\ $ Reduce by A -> BA $ Reduce by S -+ A and accept Fig. 10.8 Action of LR() parser on input aabb.
EXERCISES 10.1 Show that the normal form of Lemma 10.2 holds for nondeterministic PDA's.
10.2 a) Show that every DCFL is accepted by a DPDA whose only e moves are pop moves.
b) Show that the DPDA of part (a) can be made to satisfy the normal form of Lemma 10.2.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 265 10.3 Give an efficient algorithm to implement rule (4) of Lemma 10.3.
S10.4 Show that the DCFL's are not closed under union, concatenation, Kleene closure, or homomorphism.
10.5 Show that the following are not DCFL's.
Sa) {wwR | wis in (0 + 1)} b) {OT | n > 1} u {0 n l 2n | n > 1} 10.6 Prove that {WVa?
| U j > 1} u {012' | U j > 1} is a DCFL, but is not accepted by any DPDA without t-moves.
10.7 Show that if L is a DCFL, then L is accepted by a DPDA which, if it accepts Oia 2 '" an , does so immediately upon consuming an (without subsequent amoves). [Hint: Use the predicting machine.] 10.8 Does Greibach's Theorem (Theorem 8.14) apply to the DCFL's?
10.9 Construct the nonempty sets of items for the following grammars. Which are LK(0)?
a) S' - S S -> aSa | bSb \ c b) 5' - S S-+aSa\bSb\c Sc) S- Ei E l -T 3 E l \T l E2 -+T 3 E2 \T 2 7\ -> a$ | (E2 $ T 2 ->a)|(E2 ) r 3 ->0+ |(e2 + 10.10 Show the sequence of stacks used by the DPDA constructed from grammar 10.9(c) when the input is a + (a + a)$.
10.11 Construct the nonempty sets of LR(l) items for the following grammars. Which are LK(1)?
a) S^A A-+AB\t B-+aB\b b) S-+E E -> E + T\T T-+a(E) 10.12 Repeat Exercise 10.10 for grammar 10.11(b).
10.13 Let G be an LR(0) grammar with A -> a • , a ± c, valid for some viable prenx y.
Prove that no other production can be valid for y.
Solutions to selected exercises 10.4 Let U = {0T2' | ij > 0} and L2 = {tfl J2J \ ij > 0}. It is easy to show that L, and L2 are DCFL's. However, L { u L2 is the CFL shown not to be a DCFL in Example 10.1. Thus the DCFL's are not closed under union.
KTUNOTES.IN Downloaded from Ktunotes.in 266 DETERMINISTIC CONTEXT-FREE LANGUAGES For concatenation, let L3 = flL t u L2 . Then L3 is a DCFL, because the presence or absence of symbol a tells us whether to look for a word in L x or a word in L2 . Surely a is a DCFL. However aL3 is not a DCFL. If it were, then L4 = aL3 n a0l2 would be a DCFL by Theorem 10.4. But L4 = aL x u aL2 . If L4 is a DCFL, accepted by DPDA M, then we could recognize L x u L2 by simulating M on (imaginary) input a and then on the real input. As Li u L2 is not a DCFL, neither is L4 , and therefore the DCFL's are not closed under concatenation.
The proof for closure is similar, if we let L5 = {a} u L3 . L5 is a DCFL, but L? is not, by a proof similar to the above.
For homomorphism, let L6 = aL x u bL2i which is a DCFL. Let L be the homomor-phism that maps b to a and maps other symbols to themselves. Then h(L6 ) = L4 , so the DCFL's are not closed under homomorphism.
10.5(a) Suppose L x = {wwR \ w is in (0 + 1) + } were a DCFL. Then by Theorem 10.4, so would be L2 = L,n (01)(10)(01)(10).
Now L2 = {(Ol)'(lOy(Oiy(lO)' > 0, i and ; not both 0}.
By Theorem 10.3, L3 = MIN(L2 ) is a DCFL. But L3 = {(oiy(ioy(oiy(io)'|o <;</}, since if j > i, a prefix is in L2 . Let L be the homomorphism h(a) = 01 and h(b) = 10. Then U = ! (L3 ) = |0 < ; < i} is a DCFL by Theorem 10.4. However, the pumping lemma with z = an+ l b na nbn+ 1 shows that L4 is not even a CFL.
10.9(c) Before tackling this project, let us describe the language of the grammar. We first describe "expression" and "term" recursively, as follows.
1) A term is a single symbol a which stands for any "argument" of an arithmetic expres-sion or a parenthesized expression.
2) An expression is one or more terms connected by plus signs.
Then the language of this grammar is the set of all expressions followed by an endmar-ker, $. Ei and T, generate expressions and terms followed by a $. E 2 and T 2 generate expressions and terms followed by a right parenthesis, and T 3 generates a term followed by a plus sign.
It turns out that LR(0) grammars to define arithmetic expressions, of which our gram-mar is a simple example, are quite contorted, in comparison with an LR( 1 ) grammar for the same language [see Exercise 10.11(b)]. For this reason, practical compiler-writing systems never require that the syntax of a language be described by an LR(0) grammar; LK(1) grammars, or a subset of these, are preferred. Nevertheless, the present grammar will serve as a useful exercise. The DFA accepting viable prefixes has 20 states, not counting the dead state. We tabulate these sets of items in Fig. 10.9. Figure 10.10 gives the transition table for the DFA; blanks indicate transitions to the dead state.
Inspection of the sets of items tells us that certain sets, namely 1, 3, 6, 7, 8, 1 1, 14, 15, 16, 17, and 19, consist of a single complete item, while the remainder have no complete items.
Thus the grammar is LR(0).
KTUNOTES.IN Downloaded from Ktunotes.in BIBLIOGRAPHIC NOTES 267 /o / 5 /10 E, -> ( ' E2 + £2 - T 3 • £ 2 T 3 - (£2 + • r,-(-E2 $ £2--7 3 £2 £i -• T, £2-' 73E2 £2 -- r 2 T3 - d -f-IL 2 —» i 2 i 3 a -f-il \^2^> T 3 -> • (E2 + r 3 -• a + T 3 ^-(£2 + T, -> • a$ T 3 -> • (E2 + Ti (E2 $ r 2 -> • a) 7i -• (£ 2 ) £2 -» T 3 £2 • T 2 -• (E 2 ) h / , , S-E, • E l -T 3 E l • £2 - T 2 r '2 r h E, -> T 3 • Ei T 3 -» a • + T 3 - (£2 • + E! ->• T 3 E!
T 2 ^a) T 2 -» (£2 • ) Ei -» ii T 3 -• fl + T 3 -> • (E2 + Tj -> • a$ Ti--(E2 $ 7, 1 3 / ' 8 ' 13 / Ei -v T, • Ti - a% • T 3 - ( • £ 2 + 7i -»(£,)• 7i - ( • £2) £2 -• T 3 £ 2 fi 2 -» '2 T 3 -» • a + r 3 -(£2 + -r 2 -• (£ 2 ) U 19 T 3 - a • + T } - (£ 2 + Tj - a • $ r, - (£ 2 • $ Fig. 10.9 Sets of items for Exercise 10.9(c).
BIBLIOGRAPHIC NOTES Deterministic pushdown automata were first studied by Fischer , Schutzenberger , Haines , and Ginsburg and Greibach [1966a]. Lemma 10.3, the fact that DPDA's can be made to consume all their input, is from Schutzenberger ; Theorem 10.1, closure under complementation, was observed independently by various people. Most of the closure and decision properties, Theorems 10.2 through 10.8, were first proved by Ginsburg and Greibach [1966a]. An exception is the fact that it is decidable whether a DCFL is regular, which was proved by Stearns . The predicting-machine construc-tion is from Hopcroft and Ullman [1969b].
KTUNOTES.IN Downloaded from Ktunotes.in 268 DETERMINISTIC CONTEXT-FREE LANGUAGES a 0 4 1 2 4 3 4 5 12 6 7 8 9 10 12 11 12 13 12 14 Id 16 17 18 19 ( ) $ Ei 14 13 13 13 15 17 16 18 2 2 11 10 11 10 11 10 14 19 Fig. 10.10 Transition table of viable prefix recognizing DFA.
LR(k) grammars and the equivalence of DPDA's to LR(l) grammars is from Knuth . The latter work generalized a sequence of papers dealing with subclasses of the CFG's having efficient parsing algorithms. The history of this development is described in Aho and Ullman [1972, 1973]. Graham shows that a number of other classes of grammars define exactly the CFL's.
Subsequent to Knuth , a series of papers examined the class of LR() grammars for a useful subclass for which parsers of reasonable size could be built. Korenjak's was the first such method, although two subclasses of LR() grammars, called SLR() (for "simple" LR) and LALR() (for "lookahead LR"\ due to DeRemer [1969, 1971] are the methods used most commonly today. By way of comparison, a typical programming language, such as ALGOL, has an LR() parser (viable prefix recognizing DFA) with several thousand states, and even more are needed for an LR(0) parser. As the transition table must be part of the parser for a language, it is not feasible to store such a large parser in the main memory of the computer, even if the table is compacted. However, the same languages have SLR() or LALR(l) parsers of a few hundred states, which fit easily with compaction. See Aho and Johnson or Aho and Ullman for a description of how LK-based parsers are designed and used.
A good deal of research has been focused on the open question of whether equivalence is decidable for DPDA's. Korenjak and Hopcroft showed that equivalence isdecid-able for a subclass of the DCFL's called "simple" languages.! These are defined by gram-t These are not related to "simple" LR grammars in any substantial way.
KTUNOTES.IN Downloaded from Ktunotes.in BIBLIOGRAPHIC NOTES 269 mars in Greibach normal form such that no two productions A-+ act and A^>ap exist. The decidability of equivalence was extended to the LL(k) grammars of Lewis and Stearns , which are a proper subset of the LR(k) grammars, by Rosenkrantz and Stearns . Valiant showed that equivalence was decidable for finite-turn DPDA's (see Exercise 6.13 for a definition), among other classes; see also Valiant , Beeri , and Taniguchi and Kasami . Friedman showed that equivalence for DPDA's is decidable if and only if it is decidable for "monadic recursion schemes," which in terms of automata can be viewed as one-state DPDA's that can base their next move on the current input symbol, without consuming that symbol.
Additionally, work was done on extending the undecidability of containment for DCFL's to small subsets of the DCFL's. The work culminated in Friedman , which proved that containment is undecidable even for the simple languages of Korenjak and Hopcroft .
A solution to Exercise 10.5(b) is found in Ginsburg and Greibach [1966a], and Exercise 10.6 is based on Cole .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 11 CLOSURE PROPERTIES OF FAMILIES OF LANGUAGES There are striking similarities among the closure properties of the regular sets, the context-free languages, the r.e.
sets, and other classes. Not only are the closure properties similar, but so are the proof techniques used to establish these proper-ties. In this chapter we take a general approach and study all families of languages having certain closure properties. This will provide new insight into the underly-ing structure of closure properties and will simplify the study of new classes of languages.
11.1 TRIOS AND FULL TRIOS Recall that a language is a set of finite-length strings over some finite alphabet. A family of languages is a collection of languages containing at least one nonempty language. A trio is a family of languages closed under intersection with a regular set, inverse homomorphism, and 6-free (forward) homomorphism. [We say a hom-omorphism h is c-free if h(a) ± c for any symbol a.] If the family of languages is closed under all homomorphisms, as well as inverse homomorphism and intersec-tion with a regular set, then it is said to be a full trio.
Example 11.1 The regular sets, the context-free languages, and the r.e. sets are full trios. The context-sensitive languages and the recursive sets are trios but not full trios, since they are not closed under arbitrary homomorphisms. In fact, closing the CSL's or the recursive sets under arbitrary homomorphisms yields the r.e. sets (see Exercise 9.14 and its solution).
270 KTUNOTES.IN Downloaded from Ktunotes.in 11.1 I TRIOS AND FULL TRIOS 271 Theorem 3.3 showed that regular sets are closed under intersection; hence they are closed under "intersection with a regular set." Theorem 3.5 showed closure of the regular sets under homomorphisms and inverse homomorphism, completing the proof that the regular sets form a full trio. The corollary to Theorems 6.2, 6.3, and 6.5 show that the CFL's are a full trio. Exercise 9.10 and its solution provide a proof that the CSL's are closed under inverse homomorphism, intersection (hence intersection with a regular set), and substitution (hence ofree homomorphism, but not all homomorphisms, since e is not permitted in a CSL).
Thus the CSL's are a trio but not a full trio.
We shall prove that the recursive sets are a trio, leaving the proof that the r.e.
sets are a full trio as an exercise. Let h be a homomorphism and L a recursive language recognized by algorithm A. Then /i _1 (L) is recognized by algorithm B, which simply applies A to h(w), where w is B y s input. Let g be an ofree homomor-phism. Then g(L) is recognized by algorithm C which, given input w of length n, enumerates all the words x of length up to n over the domain alphabet of g. For each x such that g(x) = w, algorithm A is applied to x, and if x is in L, algorithm C accepts w. Note that since g is e-free, w cannot be g(x) if | x | > \w. Finally, ifR is a regular set accepted by DFA M, we may construct algorithm D that accepts input w if and only if A accepts w and M accepts w.
We conclude this section by observing that every full trio contains all regular sets. Thus the regular sets are the smallest full trio. Also, the ofree regular sets are the smallest trio. (A language is e-free if e is not a member of the language.) Lemma 1 1.1 Every full trio contains all regular sets; every trio contains all e-free regular sets.
Proof Let % J be a trio, Z an alphabet, and R c Z an ofree regular set. Since (€ contains at least one nonempty language, let L be in ^ and w be in L. Define Z' = {a | a is in Z} and let h be the homomorphism that maps each a in Z to c and each a' in Z' to w. Then L x = h~ l (L) is in # because ^ is a trio. As w is in L, L x contains all strings in Z'Z, and others as well. Let g be the homomorphism g(a!) = g(a) = a for all a in Z. Then g being ofree, we know that L2 = g(L t ) is in ( € and is either Z or Z + , depending on whether or not L x contains e. Thus L2 n R = R is in proving our contention that every trio contains all ofree regular sets.
If (€ is a full trio, we may modify the above proof by letting g'(a') = c and g'(a) = a for all a in Z. Then L 2 = g'{L x ) = Z. If R is any regular set whatsoever, L2 n R = R is in <€.
We leave it as an exercise to show that the ofree regular sets are a trio and hence the smallest trio. Note that they do not form a full trio, because they are not closed under all homomorphisms.
KTUNOTES.IN Downloaded from Ktunotes.in 272 CLOSURE PROPERTIES OF FAMILIES OF LANGUAGES 11.2 GENERALIZED SEQUENTIAL MACHINE MAPPINGS In studying closure properties, one quickly observes that certain properties follow automatically from others. Thus to establish a set of closure properties for a class of languages one need only establish a set of properties from which the others follow. In this section we shall establish a number of closure properties that follow from the basic properties of trios and full trios.
The first operation we consider is a generalization of a homomorphism. Con-sider a Mealy machine that is permitted to emit any string, including c, in a move.
This device is called a generalized sequential machine, and the mapping it defines is called a generalized sequential machine mapping.
More formally a generalized sequential machine (GSM) is a 6-tuple M = (Q, Z, A, 3, q0 , F), where Q, Z, and A are the states, input alphabet, and output alphabet, respectively, 3 is a mapping from Q x Z to finite subsets of Q x A, q0 is the initial state, and F is the set offinal states. The interpretation of (p, w) in 3(q, a) is that M in state q with input symbol a may, as one possible choice of move, enter state p and emit the string w.
We extend the domain of 3 to Q x Z as follows.
1) 6{q, C) = {(q, £ )}.
2) For x in Z and a in Z, 3(q, xa) = {(p, w) | w = Wj w 2 and for some p ', (p', w t ) is in d(q, x) and (p, w2 ) is in 3(p', a)}.
A GSM is e-free if 3 maps Q x Z to finite subsets of Q x A + . Let M(x), where M is a GSM as defined above, denote the set {y | (P y) is i° 5(ao x ) f°r some p in F}.
If L is a language over Z, let M(L) denote {y|y is in M{x) for some x in L}.
We say that M(L) is a GSM mapping. If M is ofree, then M(L) is an t-free GSM mapping. Note that L is a parameter of the mapping, not a given language.
Also let M _1 (x) = {y|M(y) contains x}, and M~ l (L) = {y \x is in M(y) for some x in L}.
We say that M" 1 is an inverse GSM mapping.
It is not necessarily true that M~ X (M(L)) = M(M~ 1 (L)) = L, so M~ 1 is not a true inverse.
Example 11.2 Let M = (bo, 111 {0, 1}, {a, fr}, 5, o, { 0) %o, 1) 5(qu 0) %i> 1) {(?o. «4 (9u b)}, {(?o. «)}.
0, {(?.. 0}-We may draw a GSM as a finite automaton, with an edge labeled a/w from state q to state p if <5(g, a) contains (p, w). The diagram for M above is shown in Fig. 11.1.
Intuitively, as 0's are input to M, M has the choice of either emitting two a's or one b. IfM emits the b, it goes to state q x . If 1 is input to M, and M is in state g0 , M can only output an a. In state q l9 M dies on a 0-input, but can remain in state q l with no output on a 1-input.
Let L={0 n l"\n> 1}. Then M(L) = {a 2rt6|M>0}.
For as 0's are read by M, it emits two as per 0, until at some time it guesses that it should emit the symbol b and go to state q v If Fs do not follow immediately on the input, M dies. Or ifM chooses to stay in q0 when the first 1 is read, it can never reach q { if the input is of the form CI". Thus the only output made by M when given input 0T is a 2n ' 2 b.
IfL, ={a 2nb\n>0}, then M~ l (L l ) = {wOr I i > 0 and w has an even number of l's}.
Note that M~ l (M(L))^L.
The GSM mapping is a useful tool for expressing one language in terms of a second language having essentially the same structure but different external trap-pings. For example, L x = {a nb n \ n > 1} and L2 = {a"ban \ n > 1} in some sense have the same structure, but differ slightly. L x and L2 are easily expressible in terms of each other by GSM mappings. Figure 11.2(a) shows a GSM mapping L t to L2 , and Fig. 11.2(b) shows a GSM mapping L2 to Lv Fig. 11.1 A GSM.
KTUNOTES.IN Downloaded from Ktunotes.in 274 CLOSURE PROPERTIES OF FAMILIES OF LANGUAGES (b) Fig. 11.2 Two GSM's.
Closure under GSM mappings A key fact about GSM mappings is that they can be expressed in terms of homo-morphisms, inverse homomorphisms, and intersection with regular sets. Thus any class of languages closed under the latter operations is closed under GSM mappings.
Theorem 1 1.1 Every full trio is closed under GSM mappings. Every trio is closed under ofree GSM mappings.
Proof Let % be a full trio, L a member of <€, and M = (g, I, A, d y q0 , F) a GSM.
We must prove that M(L) is in ( €. Let Ai = {[q, a, x, p] \$(q, a) contains (p, x)} and let h x and h 2 be the homomorphisms from Af to X and Af to A defined by h t ([q, 0, x, /?]) = a and h 2 ([q, a, x, p]) = x. Let R be the regular set of all strings in A such that 1) the first component of the first symbol is q0 , the start state of M; 2) the last component of the last symbol is a final state of M; 3) the last component of each symbol is the same as the first component of the succeeding symbol.
It is easy to check that R is a regular set. A DFA can verify condition (3) by remembering the previous symbol in its state and comparing it with the current symbol.
If c is not in L, then M(L) = h 2 (hi l (L) n R). That is, h\ l maps M's input to a string that has encoded in it, for each symbol of the input string, a possible state transition of M on the input symbol and a corresponding output string. The regular set forces the sequence of states to be a possible sequence of state transi-tions of M.
Finally, h 2 erases the input and state transitions, leaving only the KTUNOTES.IN Downloaded from Ktunotes.in 11.2 | GENERALIZED SEQUENTIAL MACHINE MAPPINGS 275 output string. Formally, hi 1 (L) = {[Pi, au x u q x ][p2 , a2 , x2 , q2] ••• [pk , ak9 xk , qk]\a l a2 ak is in L, the p£'s are arbitrary, and (qi9 x f ) is in 3{pi9 a £ )}.
Intersecting fcf^L) with K yields £ = {[9o> a l9 x l9 qi][q» a 2i x 2 , q2] • • • lf a, x, | a x a2 • • • ak is in L, gfc is in F, and a { ) contains (qh x,) for all f}.
Hence /i 2 (£) s M(L) by definition.
If £ is in L and q0 is not a final state, then h 2 (L) is still M(L). But if £ is in L and g0 is a final state, then M(e) = £, so we must modify the construction above to make sure e is in the resulting language.
Let E = h\ l (L) n (R + e). Then E = £u {e}, since £ is in /i -1 (L). Hence /i 2 (L') = M(L). Since every full trio is closed under intersection with a regular set, homomorphism, and inverse homo-morphism, M(L) is in The proof for trios and £-free GSM mappings proceeds in a similar fashion.
Since the GSM never emits £, the x in [q, a, x, p] is never £, and consequently h 2 is an £-free homomorphism.
Limited erasing and inverse GSM mappings Trios are not necessarily closed under homomorphisms that result in arbitrary erasing. However, trios are closed under certain homomorphisms that allow eras-ing, provided the erasing is limited. A class of languages is said to be closed under k-limited erasing if for any language L of the class and any homomorphism h such that h never maps more than k consecutive symbols of any sentence x in L to £, h(L) is in the class. The class is closed under limited erasing if it is closed under /c-limited erasing for all k. Note that if h(a) is e for some a, then whether h is a limited erasing on L depends on L.
Lemma 11.2 Every trio is closed under limited erasing.
Proof Let ^ be a trio, L £ £ be a member of ( €, and h a homomorphism that is /c-limited on L. Let I 2 = {[x]|xis in If, |x| < k + 1, and h(x) + c}.
Let h l and h 2 be homomorphisms defined by h l {[a l a 2 am]) = a l a 2 - am and h 2 ([a x a 2 am]) = % 1 fl 2 am ).
Since [a x a 2 ••• am] is only in I 2 if h(a l a 2 - am ) ± c, h 2 is £-free. Then h 2(h^ 1 {L)) is in r^ since (€ is closed under £-free homomorphisms and all inverse homomor-phisms. It is easy to check that h 2 (hy l (L)) = h(L).
KTUNOTES.IN Downloaded from Ktunotes.in 276 CLOSURE PROPERTIES OF FAMILIES OF LANGUAGES Theorem 11.2 Every trio is closed under inverse GSM mappings.
Proof Let ^ be a trio, L a member of and let M = (g, X, A, <5, g0 , F) be a GSM.
Without loss of generality assume that the sets X and A are disjoint. If not, replace symbols in A by new symbols and restore them at the end of the construction by an c-free homomorphism mapping each new symbol to the corresponding old symbol. Let h x be the homomorphism mapping (X u A) to A defined by f , , \a for a in A, h x (a) = |c for a in Z.
Let L! = hi l (L). Then is the set of strings in X^ X6 2 m ~ Xfrn X, such that b x b 2 bn is in L Let K be the regular set consisting of all words of the form a l x l a 2 x 2 • • • am xm such that 1 ) the as are in X, 2) the x's are in A, 3) there exist states q0 ,q u .
. .,qm such that qm is in F and for 1 < i < m, 1? a,) contains (g f , x f ).
Note that x, may be (. The reader may easily show R to be a regular set by constructing a nondeterministic finite automaton accepting R. This NFA guesses the sequence of states q u q 2 , qm .
Now L, n K is the set of all words in R of the form a l x l a 2 x 2 ••• am xmi m > 0, where the as are in Z, the x's are in A, x x x 2 • • • xm is in L, and (5(^f0 » fl i «2 " " " am) contains (p, x,x 2 • • • xm ), for some p in F. None of the x/s is of length greater than k, where k is the length of the longest x such that (p, x) is in 6(q, a) for some p and q in Q and a in X.
Finally, let /i 2 be the homomorphism that maps a to a for each a in X, and b to ( for each b in A. Then M' l (L) = h 2 (L l n R) is in % by Lemma 1 1.2, since h 2 never causes more than k consecutive symbols to be mapped to (.
11.3 OTHER CLOSURE PROPERTIES OF TRIOS Trios and full trios are closed under many other operations. In this section we present several of these closure properties.
Theorem 11.3 Every full trio is closed under quotient with a regular set.
Proof Let (£ be a full trio, L c X a member of %, and R c I a regular set. For each a in Z { let a' be a new symbol, and let T.\ be the set of all such symbols. Let h t KTUNOTES.IN Downloaded from Ktunotes.in 11.4 | ABSTRACT FAMILIES OF LANGUAGES 277 and h 2 be the homomorphisms from (Z x u li) to Zf defined by h x (a) = h x (ct) = a 9 h 2 (a) = e and h 2 (a) = a. Then L/R = h 2 (h x l (L) n (I'i)K), and hence L/R is in That is, h x x (L) is the words in L with each symbol primed or unprimed independently. Thus h\ l {L) n (Zi)K is those words xy such that x consists only of primed symbols, y consists only of unprimed symbols, y is in K, and if z is x with the primes removed, then zy is in L It follows that Theorem 11.4 Trios are closed under substitution by (-free regular sets, and full trios are closed under substitution by regular sets.
Proof Let (€ be a trio, L c £ a member of# and s: If -> Z£ a substitution such that for each a in Z 1? s(a) is regular. For the time being assume and I 2 are disjoint, and .s(a) does not contain c.
Let x be a string in L. By an inverse homomorphism we can insert arbitrary strings from X| among symbols of x. By intersecting with a regular set we can assure that the string inserted after the symbol a is in s(a). Then by limited erasing we can erase the symbols of x, leaving a string from s(x).
More precisely let h x \ (L x u I 2 )->Zf be the homomorphism defined by h x (a) = a for a in 2^ and /^(a) = c for a in Z 2 and let h 2 : (Zj u Z 2 ) -> Zf be the homomorphism defined by h 2 (a) = c for a "ml. l and /? 2 (a) = a for a in Z 2 . Then is a regular set, since each s(a) is. Since s(a) is (-free, /7 2 erases at most every other symbol, so s(L) is in r6 by Lemma 1 1.2. The proof that full trios are closed under substitution by regular sets is identical except for the fact that s may not be (-free.
If Zi and Z 2 are not disjoint, replace each symbol of Z 2 by a new symbol, and follow the above operations by an (-free homomorphism to restore the old sym-bols.
11.4 ABSTRACT FAMILIES OF LANGUAGES Many of the families of languages we have studied have closure properties that are not implied by the trio or full trio operations. Predominant among these are h2(K l {L) n (Z',)R) is all strings z as described above, that is, L/R.
Now KTUNOTES.IN Downloaded from Ktunotes.in 278 CLOSURE PROPERTIES OF FAMILIES OF LANGUAGES union, concatenation, and Kleene closure. For this reason, two other sets of closure properties have had their consequences heavily studied, and in fact were studied long before the trio and full trio. Define a class of languages to be an abstract family of languages (AFL) if it is a trio and also closed under union, concatenation, and positive closure (recall that L+ , the positive closure of L, is {J?
=l £)• Call a class of languages a full AFL if it is a full trio and closed under union, concatenation, and Kleene closure.
For example, we proved in Chapters 3 and 6 that the regular sets and context-free languages are full AFL's. The r.e. sets are also a full AFL, and we leave the proof as an exercise. The CSL's are an AFL, but not a full AFL, since they are not closed under general homomorphism (see Exercises 9.10 and 9.14).
We saw that the regular sets are the smallest full trio. They are also a full AFL and therefore the smallest full AFL. The c-free regular sets are the smallest AFL, as well as the smallest trio.
The next theorem states that AFL's are closed under substitution into regular sets. That is, for each symbol of an alphabet, we associate a language from an AFL r 6. Then replacing each symbol in each string in some regular set by the associated language yields a language in Theorem 11.5 Let (£ be an AFL that contains some language containing 6, and let R ^ I be a regular set. Let s be a substitution defined by s(a) — La for each a in Z, where La is a member of Then s(R) is in (€.
Proof The proof is by induction on the number of operators in a regular expres-sion denoting R. If there are zero operators, then the regular expression must be one of 0, (, or a, for a in Z. If it is a, then the result of the substitution is La , which is in Hi. If the regular expression is 0, the result of substitution is 0, which is in # by Lemma 11.1. If the regular expression is c, the result of the substitution is {e}.
We claim {(} is in r6\ because some L containing c is in f 6\ and L n {(} = {c} is in # by closure under intersection with a regular set.
The induction step is easy. AFL's are closed under union and concatenation, and we can show closure under easily, given L in c 6 containing c. That is, we already showed {c} is in %'.
If L x is any language in r6, then L\ is in ( &, so L\ = L\ u {(} is in ( £. Therefore, the AFL %' is closed under u, and , from which the inductive step follows. Thus (€ is closed under substitution into a regular set.
In general, AFL's are not closed under substitution of languages in the family into other languages in the family, although most of the common AFL's such as the CFL's, the recursive sets, and the r.e. sets are. However, any AFL closed under n is closed under substitution. The proof is similar to that of Theorem 1 1.5 and is left as an exercise. We also leave as an exercise the fact that all AFL's, even those with no language containing £, are closed under substitution into e-free regular sets.
KTUNOTES.IN Downloaded from Ktunotes.in 11.6 | SUMMARY 279 11.5 INDEPENDENCE OF THE AFL OPERATIONS The definition of an AFL requires six closure properties. However, to show that a family of languages is an AFL, one need not show all six properties, since they are not independent. For example, any family of languages closed under kj, + , e-free h, h~ l and nR is necessarily closed under -.f Similarly, u follows from the other five operations and the same holds for nR. We shall only prove the dependence of-.
Theorem 11.6 Any family of languages closed under u, + , c-free h, h~ \ and nR is closed under •.
Proof Let be such a family of languages, and let L x c X and L2 X be in ( £.
We may assume without loss of generality that c is not in L x or L2 . This assump-tion is justified by the fact that L 1 L2 = (L 1 -{e})(L2 -{e})uL' 1 u L' 2 , where Lj is L { if 6 is in L2 and 0 otherwise; L2 is L2 if e is in L t and 0 otherwise.
As ^ is closed under union, if we can show that (L x — {(])(L2 — {e}) is in we shall have shown that L 2 L2 is in c €.
Let a and 6 be symbols not in £. As # is a trio, Theorem 11.1 tells us # is closed under ofree GSM mappings. Let M l be the GSM that prints a, followed by its first input symbol, then copies its input, and let M 2 be another GSM that prints b with its first input symbol, then copies its input. Then as c is not in L x or L2 , M X (L X ) = aL x and M 2 (L 2 ) = Z?L 2 ,and both are in %. By closure under u, + , and nR, [aL x u 6L2 ) + n aZ^Z = aL^^ is in ( €. Define g to be the homomorphism g(a) = g(b) = c, and ^f(c) = c for all c in X. Then g is a 2-limited erasing, since Lj and L2 are assumed c-free. By Lemma 1 1.2, g(aL { bL2 ) = L x L2 is in <€.
11.6 SUMMARY We list in Fig.
11.3 some operations under which trios, full trios, AFL's and full AFL's are closed. The properties have all been proved in this chapter or are exercises. Recall that the regular sets, CFL's, and r.e. sets are full AFL's; the CSL's and recursive sets are AFL's. The DCFL's are not even trios, however.
Some other operations do not fit into the theory of trios and AFL's. In Fig.
11.4 we summarize the closure properties of six classes of languages under these operations. The question of whether the CSL's are closed under complementation is a long-standing open problem, and is equivalent to their closure under MIN.
t We use n R for "intersection with a regular set," h for "homomorphism," and h' 1 for "inverse homomorphism." The dot stands for concatenation.
KTUNOTES.IN Downloaded from Ktunotes.in 280 CLOSURE PROPERTIES OF FAMILIES OF LANGUAGES Trio Full trio AFL Full AFL h~ l ofree h h nR ofree GSM mappings GSM mappings Inverse GSM mappings Limited erasing Quotient with regular set INIT Substitution into regular sets Substitution by ofree regular sets Substitution by regular sets Fig. 11.3 Summary of closure properties.
While this chapter has concerned itself with closure properties and not deci-sion properties, we have reached a good point to summarize these properties as well, for the six classes of languages mentioned in Fig.
11.4. We show in Fig. 11.5 whether each of 10 important properties is decidable for the six classes. D means decidable, U means undecidable, T means trivially decidable (because the answer is always "yes"), and ? means the answer is not known. The results in Fig.
1 1.5 are proved in various chapters, chiefly Chapters 3, 6, 8, and 10.
KTUNOTES.IN Downloaded from Ktunotes.in Regular Recursive r.e.
sets CFL's DCFL's CSL's sets sets Complementation y y ?
y Intersection / / y Substitution y y MIN / ?
y MAX / CYCLE y y Reversal y V Fig. 11.4 Some other closure properties.
Question Regular sets DCFL's CFL's CSL's Recursive sets r.e.
sets Is w in L?
D D D D D U Is L= 0?
D D D U U U Is L = Z?
D D U U V U Is Li = L2 ?
D ?
U U V U Is Li c L2 ?
D V U U U u Is L,nL2 = 0?
D u U U U u Is L = where K is a given regular set?
D D U U U u Is L regular?
T D U U u u Is the intersection of two languages a language of the same type?
T U u T T T Is the complement of a language also a language of the same type?
T U u ?
T u Fig.
1 1.5 Some decision properties.
281 KTUNOTES.IN Downloaded from Ktunotes.in 282 CLOSURE PROPERTIES OF FAMILIES OF LANGUAGES EXERCISES S11.1 Show that the linear languages are a full trio but not an AFL.
11.2 Show that the c-free regular sets are an AFL.
11.3 Show that a full trio is closed under INIT, SUB, and FIN, where SUB(L) = {x | wxy is in L for some w and y}, and FIN(L) = {x | wx is in L for some w}.
11.4 Show that not every AFL is closed under , h, INIT, SUB, FIN, quotient with a regular set or substitution by regular sets.
11.5 Show that not every full trio is closed under u, \ , + , or substitution into regular sets. [Hint: The linear languages suffice for all but union. (To prove that certain languages are not linear, use Exercise 6.11). To show nonclosure under union, find two full trios and (€ 2 containing languages Li and L 2 , respectively, such that L, u L 2 is in neither ^ j nor (€ 2 . Show that ^j u # 2 is also a full trio.] 11.6 Prove each of the closure and nonclosure properties in Fig.
11.4 (some have been asked for in previous exercises or proved in previous theorems).
1 1.7 The interleaving of two languages L x and L2 , denoted IL(LU L2 ), is {w 1 x,w 2 x 2 ••• wk xk \k is arbitrary, w 1 w 2 wk is in L x and x t x 2 • • xk is in L2}.t Show that if ( € is any trio, L is in ( €, and R is a regular set, then IL(L, R) is in f €.
11.8 Are the following closed under IL?
a) regular sets b) CFL's c) CSL's d) recursive sets e) r.e. sets 11.9 An A-transducer is a GSM that may move (make output and change state) on (-input. Show that every full trio is closed under /4-transductions.
11.10 Find a GSM that maps a 1 to the set {ajb k \i <j + k < 2i} for all i.
11.11 Show that any class of languages closed under h, h~ \ and nR is closed under union.
11.12 Show that any class of languages closed under h y h~ l , -, and u is closed under nR.
11.13 Give examples of classes of languages closed under a) u, •, <-free h, h~\ and nR, but not + ; b) u, •, + , c-free h, and nR, but not h~ 1 ; c) u, •, + , fi" 1 , and nR, but not c-free h.
11.14 Show that an AFL is closed under complementation if and only if it is closed under MIN.
11.15 A scattered-context grammar, G = (V, T, P, S), has productions of the form (A l9 ...» A„)-+ (a,, a„), where each a, is in (V u T) + . If (A ly A n ) -> (a^ a„) is in P, then we write P l A l p2 A 2 • • Pn A„Pn+l =>p i oc l p 2 be the reflexive, transitive closure of =>. The language generated by G is {x \ x is in T + and S U x}.
a) Prove that the scattered-context languages form an AFL.
b) What class of languages is generated by the scattered-context grammars if we allow productions with the a,'s possibly £?
11.16 An AFL # is said to be principal if there is a language L such that # is the least AFL containing L a) Do the CFL's form a principal AFL?
b) Prove that the least AFL containing {a"bn \ n > 0} is properly contained in the CFL's.
c) Let ... be an infinite sequence of AFL's such that { £ % for all i > 0.
Prove that the union of the #,'s forms an AFL that is not principal.
d) Give an example of a nonprincipal AFL.
11.17 Show that if an AFL is closed under intersection, then it is closed under substitution.
Solutions to Selected Exercises 11.1 To prove that the linear languages are closed under homomorphism, let G be a linear grammar and h a homomorphism.
If each production A -> wBx or A -> y is replaced by A -> h(w)Bh(x) or A -> h(y), respectively, then the resulting grammar generates h(L(G)). To show closure under h~ l and nR y we could use machine-based proofs analogous to the proofs for CFL's, since by Exercise 6.13(a), the linear languages are characterized by one-turn PDA's. We shall instead give grammar-based proofs.
Let G = (K, 7, P, S) be a linear CFG, and M = (Q, T, 5, q0 , F) a DFA. Construct linear grammar G' = (V\ T, P\S') generating L(G) n L(M). Let V = {[qAp] \q and p are in Q and A in V] \j {S'}. Then define P' to have productions 0 S -+[q0 Sp] for all p in F, ^) [^P] w[rBs]x whenever A -> w£x is in P, w) = r and <5(s, x) = p, and 3) [<?/4p] -> y whenever ,4 -> y is in P and <5(g, y) = p.
An easy induction on derivation length shows that [qAp] => w if and only if A => w and <5(g, w) = p.
Thus S'=>vv if and only if S=>w and d(q 0 , vv) is a final state.
Hence L(G') = L(G) n L(M).
Now, let G = (K, T, P, S) be a linear grammar and h: Z -» T a homomorphism.
Suppose is such that for all a in Z, < /c, and if A - wBx or /I -»• w is in P, then |w| < /c and |x| < fc. Let G" = (K", Z, P", ), where K" consists of all symbols [wAx] such that /I is in V, and w and x in T are each of length at most 2k — 1. Also in V" are symbols [y], where \y \ < 3k — 1. Intuitively G" simulates a derivation of G in its variable until the string of terminals either to the right or to the left of the variable of G is of length at least k. Then G" produces a terminal a on the right or left and deletes h(a) from what is stored in the variable.
The productions of P" are: 1) If A -> w y Bx x is in P, then for all w 2 and x 2 of length at most k -1, [w 2 /Lx 2 ] - [w 2 w l Bx l x 2 ] is in P". If A -> y is in P, then [w2 /lx 2 ] - [w 2 yx 2 ] is in P".
2) For a in Z, [/i^Wi/lx^ -> fl[w i^x il [w^x^fa)] - [wjAxjJa, and [M fl )v] -> fl[y].
KTUNOTES.IN Downloaded from Ktunotes.in 284 CLOSURE PROPERTIES OF FAMILIES OF LANGUAGES 3) [€]-€.
It follows by induction on derivation length that [S]^>w l [w 2 Ax 2]x l if and only if S=j>h(w l )w 2 Ax 2 h(x l ).
Thus [S] f> v if and only if S §> h(v)y and hence L(G") = h~ '(^(G)).
To show that the linear languages are not an AFL, we show they are not closed under concatenation. Surely 1 i > 1} and {cjdj \j > 1} are linear languages, but their concaten-ation is not, by Exercise 6.12.
BIBLIOGRAPHIC NOTES The study of abstract families of languages was initiated by Ginsburg and Greibach , who proved Theorems 11.1 through 11.5 and Lemma 11.1. The central importance of the trio in this theory is pointed out by Ginsburg . Theorem 1 1.6 on independence of the operators appears in Greibach and Hopcroft ; a solution to Exercise 11.13 can also be found there. The notion of limited erasing is also due to Greibach and Hopcroft .
That AFL's closed under intersection are closed under substitution was first proved by Ginsburg and Hopcroft . An enormous amount of literature concerns itself with abstract families of languages; we mention only Ginsburg and Greibach , dealing with principal AFL's (Exercise 1 1.16), and Greibach , who attempts to work substitu-tion into the theory. A summary and additional references can be found in Ginsburg .
The theory of families of languages has, from its inception, been connected with the theory of automata. Ginsburg and Greibach show that a family of languages is a full AFL if and only if it is defined by a family of nondeterministic automata with a one-way input. Of course, the notion of a "family of automata" must be suitably defined, but, roughly, each such family is characterized by a set of rules whereby it may access or update its storage. The "if" part was proved independently in Hopcroft and Ullman [1967b].
Chandler characterized families of deterministic automata with a one-way input, in terms of closure properties, and Aho and Ullman did the same for deterministic automata with a two-way input. Curiously, no characterization for two-way nondeter-ministic automata is known.
There have also been attempts to codify a theory of grammars, chiefly subfamilies of the CFG's. Gabriellian and Ginsburg and Cremers and Ginsburg wrote the basic papers in this area.
The GSM was defined by Ginsburg , and the study of GSM mappings and their properties commenced with Ginsburg and Rose [1963b]. An important unresolved issue concerns testing for equivalence of two sequential transducers. That equivalence is decid-able for Moore machines (and hence for Mealy machines, which GSM's generalize) was known since Moore . Griffiths showed that the equivalence problem for e-free GSM's was undecidable, while Bird gave a decision algorithm for the equivalence of two-tape automata, which are more general than deterministic GSM's.
Scattered-context grammars (Exercise 11.15) are discussed in Greibach and Hopcroft .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 12 COMPUTATIONAL COMPLEXITY THEORY Language theory classifies sets by their structural complexity. Thus regular sets are regarded as "simpler" than CFL's, because the finite automaton has less complex structure than a PDA. Another classification, called computational complexity, is based on the amount of time, space, or other resource needed to recognize a language on some universal computing device, such as a Turing machine.
Although computational complexity is primarily concerned with time and space, there are many other possible measures, such as the number of reversals in the direction of travel of the tape head on a single-tape TM. In fact one can define a complexity measure abstractly and prove many of the results in a more general setting. We choose to present the results for the specific examples of time and space, since this approach renders the proofs more intuitive. In Section 12.7 we briefly outline the more abstract approach.
12.1 DEFINITIONS Space complexity Consider the off-line Turing machine M of Fig. 12.1. M has a read-only input tape with endmarkers and k semi-infinite storage tapes.
If for every input word of length h, M scans at most S(n) cells on any storage tape, then M is said to be an S(n) space-bounded Turing machine, or of space complexity S(n). The language recognized by M is also said to be of space complexity S(n).
285 KTUNOTES.IN Downloaded from Ktunotes.in 286 COMPUTATIONAL COMPLEXITY THEORY i Input $ Finite control Storage tapes V Fig. 12.1 Multitape Turing machine with read-only input.
Note that the Turing machine cannot rewrite on the input and that only the length of the storage tapes used counts in computing the tape bound. This restric-tion enables us to consider tape bounds ofless than linear growth. If the TM could rewrite on the input tape, then the length of the input would have to be included in calculating the space bound. Thus no space bound could be less than linear.
Time complexity Consider the multitape TM M of Fig. 12.2. The TM has k two-way infinite tapes, one of which contains the input. All tapes, including the input tape, may be written Finite control K\ Input region v— Storage tapes Fig. 12.2 Multitape Turing machine.
KTUNOTES.IN Downloaded from Ktunotes.in 12.1 | DEFINITIONS 287 upon. If for every input word of length n, M makes at most T(n) moves before halting, then M is said to be a T(n) time-bounded Turing machine, or of time complexity T(n). The language recognized by M is said to be of time complexity T(n).
The two different models for time and space complexity were selected with an eye toward making certain proofs simple, and some variation in the models is feasible. For example, if S(n) > n, then we can use the single tape TM as our model without changing the class of languages accepted in space S(n). We cannot, however, when discussing time complexity, use the single tape TM, or TM's with any fixed number of tapes, without possibly losing some languages from the class of languages accepted in time T(n).
Example 12.1 Consider the language L={wov|win (0+1)}.
Language L is of time complexity n + 1, since there is a Turing machine Mj, with two tapes, that copies the input to the left of the c onto the second tape. Then, when a c is found, M x moves its second tape head to the left, through the string it has just copied, and simultaneously continues to move its input tape head to the right. The symbols under the two heads are compared as the heads move. If all pairs of symbols match and if, in addition, the number of symbols to the right and left of the lone c are equal, then M, accepts. It is easy to see that M, makes at most n + 1 moves if the input is of length n.
There is another Turing machine, M 2 , of space complexity log2 rc accepting L.
M 2 uses two storage tapes for binary counters. First, the input is checked to see that only one c appears, and that there are equal numbers of symbols to the right and left of the c. Next the words on the right and left are compared symbol by symbol, using the counters to find corresponding symbols. If they disagree, M 2 halts without accepting. If all symbols match, M 2 accepts.
Special assumptions about time and space complexity functions It should be obvious that every TM uses at least one cell on all inputs, so if S(n) is a space complexity measure, we may assume S(n) > 1 for all n. We make the useful assumption that when we talk of "space complexity S(n)" we really mean max (1, [5(71)1). For example, in Example 12.1, we said that TM M 2 was of "space complexity log2 rc." This makes no sense for n = 0 or 1, unless one accepts that "log2 >i" is shorthand for max (1, Hog2 /il).
Similarly, it is reasonable to assume that any time complexity function T(n) is at least n -f 1, for this is the time needed just to read the input and verify that the end has been reached by reading the first blank. t We thus make the convention t Note, however, that there are TM's that accept or reject without reading all their input. We choose to eliminate them from consideration.
KTUNOTES.IN Downloaded from Ktunotes.in 288 COMPUTATIONAL COMPLEXITY THEORY that "time complexity T{n)" means max (n + 1, \T(n)\ For example, the value of time complexity n log2 w at n = 1 is 2, not 0, and at n = 2, its value is 3.
Nondeterministic time and space complexity The concepts of time- and space-bounded Turing machines apply equally well to nondeterministic machines. A nondeterministic TM is of time complexity T(n) if no sequence of choices of move causes the machine to make more than T(n) moves. It is of space complexity S(n) if no sequence of choices enables it to scan more than S(n) cells on any storage tape.
Complexity classes The family of languages of space complexity S(n) is denoted by DSPACE(S(rc)); the languages of nondeterministic space complexity S(n) are collectively called NSPACE(S(n)). The family of languages of time complexity T(n) is denoted DTIME(T(n)) and that of nondeterministic time complexity T(n) is denoted NTIME(T(n)). All these families of languages are called complexity classes. For example, language L of Example 12.1 is in DTIME(n)t and in DSPACE(log2 «).
L is therefore also in NTIME(n) and NSPACE(log 2 n) as well as larger classes such as DTIME(h 2 ) or NSPACE^).
12.2 LINEAR SPEED-UP, TAPE COMPRESSION, AND REDUCTIONS IN THE NUMBER OF TAPES Since the number of states and the tape alphabet size of a Turing machine can be arbitrarily large, the amount of space needed to recognize a set can always be compressed by a constant factor. This is achieved by encoding several tape sym-bols into one. Similarly one can speed up a computation by a constant factor.
Thus in complexity results it is the functional rate of growth (e.g., linear, quad-ratic, exponential) that is important, and constant factors may be ignored. For example, we shall talk about complexity log n without specifying the base of logarithms, since \ogh n and logcn differ by a constant factor, namely logb c. In this section we establish the basic facts concerning linear speed up and compres-sion as well as considering the effect of the number of tapes on complexity.
Tape compression Theorem 12.1 If L is accepted by an S(n) space-bounded Turing machine with k storage tapes, then for any c > 0, L is accepted by a cS(n) space-bounded TM.J t Recall that n really means max (n + 1, n) = n + 1 for time complexity.
+ Note that by our convention, cS(n) is regarded as max (1, IcS(h)I).
KTUNOTES.IN Downloaded from Ktunotes.in 12.2 | LINEAR SPEED-UP, TAPE COMPRESSION 289 Proof Let Mi be an S(n) tape-bounded off-line Turing machine accepting L. The proof turns on constructing a new Turing machine M 2 that simulates M u where for some constant r, each storage tape cell ofM 2 holds a symbol representing the contents of r adjacent cells of the corresponding tape ofM x . The finite control of M 2 can keep track of which of the cells of Mu among those represented, is actually scanned by M v Detailed construction of the rules of M 2 from the rules of M 1 are left to the reader. Let r be such that rc > 2. M 2 can simulate M x using no more than \S(n)/r] cells on any tape. If S(n) > r, this number is no more than cS(n). If S(n) < r, then M 2 can store in one cell the contents of any tape. Thus, M 2 uses only one cell in the latter case.
Corollary If L is in NSPACE(S(n)), then L is in NSPACE(cS(n)), where c is any constant greater than zero.
Proof If M 1 above is nondeterministic, let M 2 be nondeterministic in the above construction.
Reduction in the number of tapes for space complexity classes Theorem 12.2 If a language L is accepted by an S(n) space-bounded TM with k storage tapes, it is accepted by an S(n) space-bounded TM with a single storage tape.
Proof Let M t be an S(n) space-bounded TM with k storage tapes, accepting L.
We may construct a new TM M 2 with one storage tape, which simulates the storage tapes ofM x on k tracks. The technique was used in Theorem 7.2. M 2 uses no more than S(n) cells.
From now on we assume that any S(n) space-bounded TM has but one storage tape, and if S(n) > rc, then it is a single-tape TM, rather than an off-line TM with one storage tape and one input tape.
Linear speed up Before considering time bounds, let us introduce the following notation. Let f(n) be a function of n. The expression sup,,^^ f(n) is taken to be the limit as n -> oo of the least upper bound off(n),f(n + 1),/(h + 2), .
.
. Likewise, \n[n^^f(n) is the limit as w-> oo of the greatest lower bound off(n\f(n + l),f(n 4- 2), .
.
. If/(h) converges to a limit as n - oo, then that limit is both in^^^ f(n) and sup,,.^ f(n).
Example 12.2 Let f(n) = l/n for n even, and f(n) = n for n odd. The least upper bound off(n),f(n 4-1), ...
is clearly oo for any n, because of the terms for odd n.
Hence sup^^ f (n) = oo. However, because of the terms with n even, it is also true that inf^ co /(n) = 0.
KTUNOTES.IN Downloaded from Ktunotes.in 290 COMPUTATIONAL COMPLEXITY THEORY For another example, suppose f(n) = n/(n 4-1). Then the least upper bound of n/(n 4-1), (n 4- l)/(n 4- 2), ...
is 1 for any n. Thus The greatest lower bound of n/(n 4-1), (n 4- l)/(n 4- 2), .
.
.
is n/(n + 1) and lim,,^ n/(n + 1) = 1, so inf,,.^ n/(n 4- 1) = 1 as well.
Theorem 12.3 If L is accepted by a /c-tape T(n) time-bounded Turing machine Mi, then L is accepted by a /c-tape cT(w) time-bounded TM M 2 for any c> 0, provided that /c > 1 and inf,,.^ T(n)/n = oo.
Protf/ A TM M 2 can be constructed to simulate M, in the following manner.
First M 2 copies the input onto a storage tape, encoding m symbols into one. (The value of m will be determined later.) From this point on, M 2 uses this storage tape as the input tape and uses the old input tape as a storage tape. M 2 will encode the contents of M/s storage tapes by combining m symbols into one. During the course of the simulation, M 2 simulates a large number of moves of M, in one basic step consisting of eight moves of M 2 . Call the cells currently scanned by each of M 2 's heads the home cells. The finite control ofM 2 records, for each tape, which of the m symbols of M, represented by each home cell is scanned by the correspond-ing head of M 2 .
To begin a basic step, M 2 moves each head to the left once, to the right twice, and to the left once, recording the symbols to the left and right of the home cells in its finite control. Four moves of M 2 are required, after which M 2 has returned to its home cells.
Next, M 2 determines the contents of all of M/s tape cells represented by the home cells and their left and right neighbors at the time when some tape head of M, first leaves the region represented by the home cell and its left and right neighbors. (Note that this calculation by M 2 takes no time.
It is built into the transition rules of M 2 .) If M 1 accepts before some tape head leaves the repre-sented region, M 2 accepts.
If M x halts, M 2 halts. Otherwise M 2 then visits, on each tape, the two neighbors of the home cell, changing these symbols and that of the home cell if necessary. M 2 positions each of its heads at the cell that represents the symbol that M/s corresponding head is scanning at the end of the moves simulated. At most four moves of M 2 are needed.
It takes at least m moves for M, to move a head out of the region represented by a home cell and its neighbors. Thus, in eight moves, M 2 has simulated at least m moves of M,. Choose m such that cm > 16.
If Mj makes T(n) moves, then M 2 simulates these in at most 8[T(tt)/mlmoves.
Also, M 2 must copy and encode its input (m cells to one), then return the head of the simulated input tape to the left end. This takes n 4- In/ml moves, for a total of n+ \n/m]+S\T(n)/m] (12.1) KTUNOTES.IN Downloaded from Ktunotes.in 12.2 | LINEAR SPEED-UP, TAPE COMPRESSION 291 moves. As \x] < x 4-1 for any x, (12.1) is upper bounded by n 4- n/m 4- &T(n)/m 4- 2.
(12.2) Now we have assumed that int^^ T(n)/n = oo, so for any constant d there is an nd such that for all n > nd , T(n)/n > d, or put another way, n < T(n)/d. Thus whenever n > 2 (so n + 2 < 2n) and n > nd , (12.2) is bounded above by T(n) 8 2 J m d md (12.3) We have not yet specified d. Remembering that m was chosen so that cm > 16, choose d — m/4 4- i, and substitute 16/c for m in (12.3).
Then for all n > max (2, wd ) the number of moves made by M 2 does not exceed cT(n).
To recognize the finite number of words of length less than the maximum of 2 and rcd , M 2 uses its finite control only, taking n 4-1 moves to read its input and reach the blank marking the end of the input. Thus the time complexity ofM 2 is cT(n). Recall that for time complexity, cT(n) stands for max (n 4-1, \cT(n)]).
Corollary If inf,,.^ T(n)/n = oo and c > 0, then DTIME(T(h)) = DTIME(c7(n)).
Proof Theorem 12.3 is a direct proof for any language L accepted by a DTM with 2 or more tapes in time T(n). Clearly if L is accepted by a 1-tape TM, it is accepted by a 2-tape TM of the same time complexity.
Theorem 12.3 does not apply if T(n) is a constant multiple of m, as then inf,,.^ T(n)/n is a constant, not infinity. However, the construction of Theorem 12.3, with a more careful analysis of the time bound of M 2 shows the following.
Theorem 12.4 If L is accepted by a fc-tape cn time-bounded TM, for k > 1 and for some constant c, then for every e > 0, L is accepted by a fc-tape (1 -f ()« time-bounded TM.
Proof Pick m = l/\6c in the proof of Theorem 12.3.
Corollary If T(n) = cn for some c> 1, then DTIME(7(n)) = DTIME((1 4- c)n) for any c > 0.
Corollary (of Theorems 12.3 and 12.4) a) If inf^ T(n)/n = oo, then NTIME(7(n)) = NTIME(cT(h)) for any c> 0.
b) If T(n) = cn for some constant c, then NTIME(7») = NTIME((1 4- c)n\ for any e > 0.
-Pro?/ The proofs are analogous to Theorems 12.3 and 12.4.
KTUNOTES.IN Downloaded from Ktunotes.in 292 COMPUTATIONAL COMPLEXITY THEORY Reduction in the number of tapes for time complexity classes Now let us see what happens to time complexity when we restrict ourselves to one tape. A language like L = {wcwR | w is in (a + b)} can be recognized in linear time on a two-tape machine, as we saw in Example 12.1. However, on a one-tape machine, L requires time cn 2 for some c > 0. (The exercises give hints how this may be proved.) Thus permitting only one tape can square the time necessary to recognize a language. That this is the worst that can happen is expressed in the next theorem.
Theorem 12.5 If L is in DTIME(T(w)), then L is accepted in time T2 (n) by a one-tape TM.
Proof In the construction of Theorem 7.2, going from a multitape TM to a one-tape TM, M 2 uses at most 6T2 (n) steps to simulate T(n) steps of M. By Theorem 12.3, we may speed up M, to run in time T(n)/^/6. Then M 2 is a one-tape TM accepting L in T2 (n) steps.
Corollary If L is in NTIME(T(h)), then L is accepted by a one-tape NTM of nondeterministic time complexity T2 (n).
Proof Analogous to the proof of the theorem.
If we restrict ourselves to two tapes, the time loss is considerably less than if we restrict ourselves to one tape, as the next theorem shows.
Theorem 12.6 If L is accepted by a fc-tape T(n) time-bounded Turing machine Mj, then L is accepted by a two-storage tape TM M 2 in time T(n) log T(n).
Proof The first storage tape of M 2 will have two tracks for each storage tape of M,. For convenience, we focus on two tracks corresponding to a particular tape of Mj. The other tapes ofM x are simulated in exactly the same way. The second tape of M 2 is used only for scratch, to transport blocks of data on tape 1.
One particular cell of tape 1, known as B0 , will hold the storage symbols scanned by each of the heads of M x . Rather than moving head markers, M 2 will transport data across B0 in the direction opposite to that of the motion of the head of M , being simulated. Thus M 2 can simulate each move ofM x by looking only at B0 . To the right of cell B0 will be blocks # 2 , ... of exponentially increasing length; that is, B t is of length 2'~ !
. Likewise, to the left of B0 are blocks B_ 2 , with B_i having length 2'" l . The markers between blocks are assumed to exist, although they will not actually appear until the block is used.
Let a0 denote the contents of the cell initially scanned by this tape head of Mi-The contents of the cells to the right of this cell are au a 2 , ...,and those to the left, a_ j, a. 2 , ... The values of the a^s may change when they enter B0 \ it is not their values, but their positions on the tracks of tape 1 ofM 2 , that is important. Initially the upper track ofM 2 for the tape of M, in question is assumed to be empty, while the lower track is assumed to hold .
.
.
, a _ 2 , a_ i9 a0 , a ly a 2 , ... These are placed in blocks ...,£ 2 , B0 , B u B2 , as shown in Fig.
12.3.
KTUNOTES.IN Downloaded from Ktunotes.in 12.2 | LINEAR SPEED-UP, TAPE COMPRESSION 293 a-l a-6 a-5 a-4 0-3 a-2 a 1 a 0 u ] dt «4 ^5 « 6 tf 7 5 -3 Fig. 12.3 Blocks on tape 1.
As mentioned previously, data will be shifted across B0 and perhaps changed as it passes through. After the simulation of each move ofM u the following will hold.
1) For any i > 0, either B, is full (both tracks) and B, is empty, or B, is empty and B_, is full, or the bottom tracks of both B, and B_, are full, while the upper tracks are empty.
2) The contents of any B, or B_, represent consecutive cells on the tape ofM t represented. For i > 0, the upper track represents cells to the left of those of the lower track; for i < 0, the upper track represents cells to the right of those of the lower track.
3) For i < j, Bi represents cells to the left of those of By 4) B0 always has only its lower track filled, and its upper track is specially marked.
To see how data is transferred, imagine that the tape head of M l in question moves to the left. Then M 2 must shift the corresponding data right. To do so, M 2 moves the head of tape 1 from B0 , where it rests, and goes to the right until it finds the first block, say Bh that does not have both tracks full. Then M 2 copies all the data of B0 , Bu .
.
.
, B, l onto tape 2 and stores it in the lower track of B,, B 2 , B, ! plus the lower track of B f , assuming that the lower track of B, is not already filled. If the lower track of B, is already filled, the upper track of B, is used instead.
In either case, there is just enough room to distribute the data. Also note the data can be picked up and stored in its new location in time proportional to the length ofB,.
Next, in time proportional to the length of Bh T x can find B_, (using tape 2 to measure the distance from B, to B0 makes this easy). If B_, is completely full, T x picks up the upper track of B_, and stores it on tape 2. If B_, is half full, the lower track is put on tape 2. In either case, what has been copied to tape 2 is next copied to the lower tracks of B_ (I _ n , B_ (l _ 2) , B0 . (By Rule 1, these tracks have to be empty, since B,, B 2 , .
.
.
, B,- { were full.) Again, note that there is just enough room to store the data, and all the above operations can be carried out in time propor-tional to the length of B f . Also note that the data can be distributed in a manner that satisfies rules (1), (2), and (3), above.
We call all that we have described above a Broperation. The case in which the head of M, moves to the right is analogous. The successive contents of the blocks as M, moves its tape head in question five cells to the left are shown in Fig.
12.4.
KTUNOTES.IN Downloaded from Ktunotes.in 294 COMPUTATIONAL COMPLEXITY THEORY 5 3 5_ 2 5_, B0 B { B 2 B 3 a -l 5-6 5-5 5-4 5-3 5-2 5-1 5 0 \ 5 2 5 3 5 4 5 5 5 6 5 7 5 0 5-7 5-6 5-5 5-4 5-3 5-2 5-1 5.
5 2 5 3 5 4 5 5 5 6 5 7 5 0 3 1 5-7 5-6 5-5 5-4 5-3 5-2 5-1 5 2 5 3 5 4 5 5 5 6 5 7 5-2 5 0 a \ 5-7 5-6 5-5 5-4 5-3 5-1 a 2 5 3 5 4 5 5 5 6 5 7 5 0 5l a 2 5 3 5-7 5-6 5 5 5-4 5-3 5 -2 5-1 5 4 5 5 5 6 5 7 5 4 5 0 5l a 2 5 3 5-7 5-6 5-5 5 3 5-1 5 4 5 5 5 6 5 7 Fig. 12.4 Contents of blocks of M x .
We note that for each tape of M,, M 2 must perform a 5,-operation at most once per 2 1 " 1 moves of Mj, since it takes this long for B x , £ 2 , . .., 1? which are half empty after a #roperation, to fill. Also, a ^.-operation cannot be performed for the first time until the 2'~ th move of M,. Hence, if M, operates in time T(n), M 2 will perform only /^-operations, for those i such that i < log2 T(n) + 1.
We have seen that there is a constant ra, such that M 2 uses at most mT moves to perform a 5,-operation. If M, makes T(n) moves, M 2 makes at most log 2 T(n)+l T(n\ TM)= I ™?^T (12-4) i= 1 Z moves when simulating one tape of KTUNOTES.IN Downloaded from Ktunotes.in 12.3 | HIERARCHY THEOREMS 295 From (12.4), we obtain 7i(n) = 2mT(n)\og2 T(n) + 11, (12.5) and from (12.5), T l (n)<4mT{n) \og2 T(n).
The reader should be able to see that M 2 operates in time proportional to 7(tt) even when M l makes moves using different storage tapes rather than only the one upon which we have concentrated. By Theorem 12.3, we can modify M 2 to run in no more than T(n) log2 T(n) steps.
Corollary If L is accepted by a /c-tape NTM of time complexity T(n), then L is accepted by a two-tape NTM of time complexity T(n) log T(n).
Proof Analogous to the proof of the theorem.
12.3 HIERARCHY THEOREMS Intuitively, given more time or space, we should be able to recognize more lan-guages or compute more functions. However, the linear speed-up and compression theorems tell us that we have to increase the available space or time by more than a constant factor. But what if we multiply the space or time by a slowly growing function such as log log m? Is it possible that we cannot then recognize any new languages? Is there a time or space bound f(n) such that every recursive language is in DTIME(/(«)), or perhaps in DSPACE(/(«))?
The answer to the last question is "no," as we shall prove in the next theorem.
However, the answer to the first question depends on whether or not we start with a "well-behaved" function. In this section we shall give suitable definitions of "well behaved" and show that for well-behaved functions, small amounts of extra time and space do add to our ability to compute.
In Section 12.6 we shall consider arbitrary total recursive functions and the complexity classes they define. There we shall see that strange behavior is ex-hibited. There are "gaps" in any complexity hierarchy, that is, there exists a function T(n) for which DTIME(T2 (n)) = DTIME(T(n)), and in general, for any total recursive function / there is a time complexity T f (n) for which DTIME(7}(n)) = DTIME(/(T / (m))).
Similar statements hold for space, and indeed for any reasonable measure of computational complexity. We shall also see that there are languages L for which no "best" recognizer exists; rather there is an infinite sequence of TM's recognizing L, each of which runs much faster than the previous one.
Theorem 12.7 Given any total recursive time-bound (space-bound) T(«), there is a recursive language L not in DTIME(T(«)) or DSPACE(T(n)), respectively.
Proof We shall show the result for time; the argument for space is analogous.
The argument is basically a diagonalization. Since T(n) is total recursive, there is a KTUNOTES.IN Downloaded from Ktunotes.in 296 COMPUTATIONAL COMPLEXITY THEORY halting TM M that computes it. We construct M to accept a language L c (0 + 1) that is recursive but not in DTIME(T(n)). Let x f be the ith string in the canonical ordering of (0 + 1). In Chapter 8, we ordered single-tape TM's with tape alphabet {0, 1, B}. We can similarly order multitape TM's with arbitrary tape alphabets by replacing their transition functions by binary strings. The only sub-stantial point is that the names of the tape symbols, like those of states, don't matter, so we may assume that all TM's whose input alphabet is {0, 1} have tape alphabet 0, 1, B, X4 , X 5 , ... up to some finite Xm , then encode 0, 1, and B by 0, 00, and 000 and encode X t by 0', i > 4. We also permit an arbitrary number of l's in front of the code for M to represent M as well, so M has arbitrarily long encodings.
We are thus free to talk about M„ the ith multitape TM. Now define L = {Xi | M, does not accept x { within T( | jc t -1 ) moves}. We claim L is recursive. To recognize L, execute the following algorithm, which can surely be implemented on a Turing machine. Given input w of length n, simulate M on n to compute T(n).
Then determine i such that w = x,. The integer i written in binary is the transition function of some multitape TM M, (if i in binary is of improper form for a transition function, then M, has no moves). Simulate M, on w for T(n) moves, accepting if M, either halts without accepting or runs for more than T(n) moves and does not accept.
To see that L is not in DTIME(T(m)), suppose L = L(M,), and M, is T(n) time bounded. Is x f in L? If so, M t accepts x,- within T(n) steps, where n = | x, | . Thus by definition of L, x t is not in L, a contradiction. If x f is not in L, then M { does not accept xh so by definition of L, x, is in L, again a contradiction. Both assumptions lead to contradictions, so the supposition that M,-is T(n) time bounded must be false.
If T'(n) > T(n) for all n, it follows immediately from the definition of a time complexity class that DTIME(T(n)) c DTIME(T'(«)). If T(n) is a total recursive function, Theorem 12.7 implies there exists a recursive set L not in DTIME(T(n)).
Let t(n) be the running time of some Turing machine accepting L and let T'(n) = max {7», f(«)}. Then DTIME(T(n)) 5 DTIME(T'(n)), since L is in the latter but not the former. Thus we know that there is an infinite hierarchy of deterministic time complexity classes. A similar result holds for deterministic space complexity classes, and for nondeterministic time and space classes.
Theorem 12.7 demonstrates that for any recursive time or space complexity f(n), there is an f'(n) such that some language is in the complexity class defined by f'(n) but not/(n). We now show that for a well-behaved function/ (n) only a slight increase in the growth rate of f(n) is required to yield a new complexity class.
Theorems 12.8 and 12.9 are concerned with the increase needed in order to obtain a new deterministic complexity class. These theorems are used later to establish lower bounds on the complexity of various problems. Similar results for non-deterministic classes are very much more difficult; we shall touch on a dense hierarchy for nondeterministic space in Section 12.5.
KTUNOTES.IN Downloaded from Ktunotes.in 123 | HIERARCHY THEOREMS 297 A space hierarchy We now introduce our notion of a "well-behaved" space complexity function. A function S(n) is said to be space constructible if there is some Turing machine M that is S(n) space bounded, and for each m, there is some input of length n on which M actually uses S(n) tape cells. The set of space-constructible functions includes log n, nk , 2", and nl If S^n) and S2 (n) are space constructible, then so are S 1 (m)S2 (m), 2Sl(n) , and S 1 (n) S2(n) . Thus the set of space-constructible functions is very rich.
Note that M above need not use S(n) space on all inputs of length n, just on some one input of that length. If for all h, M in fact uses exactly S(n) cells on any input of length n, then we say S(n) is fully space constructible. Any space-constructible S(n) > w is fully space constructible (exercise).
In order to simplify the next result we prove the following lemma.
Lemma 12.1 If L is accepted by an S(n) > \og2 n space-bounded TM, then L is accepted by an S(n) space-bounded TM that halts on all inputs.
Proof Let M be an S(n) space-bounded off-line Turing machine with s states and t tape symbols accepting L. If M accepts, it does so by a sequence of at most (n + 2)sS(n)t S(n) moves, since otherwise some ID repeats. That is, there are n + 2 input head positions, s states, S(n) tape head positions, and t S(n) storage tape contents. If an additional track is added as a move counter, M can shut itself off after (4stf {n) > (n + 2)sS(n)t S(n) moves. Actually, M sets up a counter of length log n, and counts in base 4s. Whenever M scans a new cell beyond the cells con-taining the counter, M increases the counter length. Thus ifM loops having used only i tape cells, then the counter will detect this when the count reaches (4sr ynax(,Mog 2 n) > whjch js ^ + 2)SS(n)t S("K Theorem 12.8 If S2 (n) is a fully space-constructible function, n^oo S 2 {n) and and S 2 (n) are each at least log2 n, then there is a language in DSPACE(S 2 (h)) not in DSPACEfS^n)).
Proof The theorem is proved by diagonalization. Consider an enumeration of off-line Turing machines with input alphabet {0, 1} and one storage tape, based on the binary encoding of Section 8.3, but with a prefix of l's permitted, so each TM has arbitrarily long encodings. We construct a TM M that uses S 2 (n) space and disagrees on at least one input with any S^n) space-bounded TM.
On input w, M begins by marking S 2 (n) cells on a tape, where n is the length of w. Since S 2 (n) is fully space constructible, this can be done by simulating a TM that uses exactly S2 (n) cells on each input of length n. In what follows, if M attempts to leave the marked cells, M halts and rejects w. This guarantees that M is S 2 (n) space bounded.
KTUNOTES.IN Downloaded from Ktunotes.in 298 COMPUTATIONAL COMPLEXITY THEORY Next M begins a simulation on input w ofTM Mw , the TM encoded by binary string w. IfMw is S t (n) space bounded and has t tape symbols, then the simulation requires space \og2 t~\S ^n). M accepts w only ifM can complete the simulation in S2 (n) space and Mw halts without accepting jc.
Since M is S 2 (n) space bounded, L(M) is in DSPACE(S2 (n)). L(M) is not in DSPACE(5 1 (m)). For suppose there were an S^n) space-bounded TM M with t tape symbols accepting L(M). By Lemma 12.1 we may assume that M halts on all inputs. Since M appears infinitely often in the enumeration, and there exists a sufficiently long w, |w| = n, such that riog2 fl5 1 (/i) < 52 (m) and Mw is M. On input w, M has sufficient space to simulate Mw and accept if and only if Mw rejects.
Thus L(MW ) L(M), a contradiction.
Thus L(M) is in DSPACE(S 2 (n)) but not in DSPACE^Jn)).
While most common functions are fully space constructible, we need only space constructibility to make Theorem 12.8 go through. We therefore state the following.
Corollary Theorem 12.8 holds even if S2 (n) is space constructible but not fully space constructible.
Proof Let M, be a TM that constructs S2 (n) on some input. Let Z be the input alphabet of M,. We design M to accept a language over alphabet Z x {0, 1}. That is, the input to M is treated as if it had two tracks: the first is used as input to M u the second as the code of a TM with input alphabet Z x {0, 1}. The only modification to the design of M is that M must lay off blocks on tapes 1 and 2 by simulating M x on M's first track. We may show that M disagrees with any S x (n) space-bounded TM M on an input whose length, n, is sufficiently large, whose first track is a string in Z" that causes M x to use S 2 (n) cells, and whose second track is an encoding of M.
We leave as an exercise a proof that the condition S2 (n) > \og2 n in Theorem 12.8 and its corollary are not really needed. The proof is not a diagonalization, but hinges on showing that {wc'w | w is in (a + b), | w | = S 2 (n) and i = n- 2S 2 (n)} is accepted in S 2 (n) space but not in S^n) space if inf 5lW 0 n-oo S 2 (n) and S 2 (n) < \og 2 n.
Note that if inf^ [5 1 (m)/5 2 (m)] = 0 and S^n) < S 2 (n) for all «, then DSPACE^Jm)) 5 DSPACE(5 2 (m)).
KTUNOTES.IN Downloaded from Ktunotes.in 12.3 | HIERARCHY THEOREMS 299 However, if we do not have S x (n) < S2 (n), then it is possible that DSPACEfS^n)) and DSPACE(S 2 (m)) each have languages not in the other.
A time hierarchy The deterministic time hierarchy is not as tight as the space hierarchy. The reason is that a TM which diagonalizes over all multitape TM's has some fixed number of tapes. To simulate a TM with a larger number of tapes we make use of the two-tape simulation of a multitape TM, thereby introducing a logarithmic slow-down.
Before giving the construction we introduce the notion of time constructibility.
A function T(n) is said to be time constructible if there exists a T(n) time-bounded multitape Turing machine M such that for each n there exists some input on which M actually makes T(n) moves. Just as for space-constructible functions there is a rich hierarchy of time-constructible functions. We say that T(n) is fully time-constructible if there is a TM that uses T(n) time on all inputs of length n.
Again, most common functions are fully time-constructible.
Theorem 12.9 If T 2 (n) is a fully time-constructible function and T 2(n) then there is a language in DTIME(T 2 (m)) but not DTIME(T 1 (m)).
Proof The proof is similar to that of Theorem 12.8, and only a brief sketch of the necessary construction is given. A T 2 (n) time-bounded TM M is constructed to operate as follows. M treats the input w as an encoding of a Turing machine M and simulates M on w. A difficulty arises because M has some fixed number of tapes, so for some w's M will have more tapes than M. Fortunately, by Theorem 12.6, only two tapes are needed to simulate any M, although the simulation costs a factor of log T x (n). Also, since M may have many tape symbols, which must be encoded into some fixed number of symbols, the simulation of T x (n) moves ofM by M requires time cT x (n) log T x (n\ where c is a constant depending on M.
In order to assure that the simulation ofM is T 2 (n) time bounded, M simul-taneously executes steps of a TM (using additional tapes) that uses exactly T 2 (n) time on all inputs of length n. This is the reason that T 2 (n) must be fully time constructible. After T 2 (n) steps, M halts. M accepts w only if the simulation ofM is completed and M rejects w. The encoding of M is designed as in the previous theorem, so each M has arbitrarily long encodings. Thus, if M is a T x (n) time-bounded Turing machine, there will be a sufficiently large w encoding M so that cT x (\w) log TidwD^T.dwl), and the simulation will carry to completion. In this case, w is in L(M) if and only if w is not in L(M). Thus L(M) ^ L(M) for any M that is T x (n) time bounded.
Therefore L(M) is in DTIME(7 2 (m)) - DTIME ( 7^ (h)).
KTUNOTES.IN Downloaded from Ktunotes.in 300 COMPUTATIONAL COMPLEXITY THEORY Example 12.3 Let T^n) = 2 n and T 2 (n) = n 22 n . Then Til") n->oo " Thus Theorem 12.9 applies, and DTIME(2 n ) + DTIME(n 22 n ). Since T^n) < T 2 (n) for all n, we may conclude that DTIME(2 n ) £ DTIME(n 22 n ).
12.4 RELATIONS AMONG COMPLEXITY MEASURES There are several straightforward relationships and one not-so-obvious relation-ship among the complexities of a given language L according to the four complex-ity measures we have defined. The straightforward relationships are stated in one theorem.
Theorem 12.10 a) If L is in DTIME(/(n)), then L is in DSPACE(/(n)).
b) If L is in DSPACE(/(m)) and f(n) > log 2 n, then there is some constant c, depending on L, such that L is in DTIME(c /(n) ).
c) If L is in NTIME(/(m)), then there is some constant c, depending on L, such that Lis in DTIME(V (n) ).
Proof a) If TM M makes no more than f(n) moves, it cannot scan more than /(«)+!
cells on any tape. By modifying M to hold two symbols per cell we can lower the storage requirements to [f(n) + l]/2\ which is at most f(n).
b) Observe that if TM M { has s states and t tape symbols, and uses at most/(n) space, then the number of different ID's ofM x with input of length n is at most s(n + 2)f(n)t f{n) . Since/(n) > log 2 n, there is some constant c such that for all n > i,cf{n) >s(n + 2)f(n)tf{n\ Construct from M t a multitape TM M 2 that uses one tape to count to c^^K and two others to simulate M 2 .
If Mj has not accepted when the count reaches cf{n \ M 2 halts without accepting. After this number of moves, Mi must have repeated an ID and so is never going to accept. Clearly M 2 is c f(n) time bounded.
c) Let M l be an /(h) time-bounded nondeterministic TM with s states, t tape symbols, and k tapes. The number of possible ID's ofM { given input of length n is at most s(f(n)+ lftkf(n\ the product of the number of states, head positions, and tape contents. Thus d = s(t + ) 3k satisfies dfin) > s(f(n) + \ ft kf{n) for all n > 1.
A deterministic multitape TM can determine if M x accepts input w of length n by constructing a list of all the ID's ofM { that are accessible from the initial ID. This process can be carried out in time bounded by the square of KTUNOTES.IN Downloaded from Ktunotes.in 12.4 | RELATIONS AMONG COMPLEXITY MEASURES 301 the length of the list. Since the list of accessible ID's has length no greater than df(n) times the length of an ID, which can be encoded in 1 + k(f(n) + 1) symbols, the time is bounded by cf(n) for some constant c.
Theorem 12.11 (SavitcKs theorem) If L is in NSPACE(S(n)), then L is in DSPACE(S 2 (n)) provided S(n) is fully space constructible and S(n) > log2 n.
Proof Let L = L(Mj), where M t is an S(n) space-bounded nondeterministic TM.
For some constant c, there are at most c S(n) ID's for an input of length n. Thus, if Mj accepts its input, it does so by some sequence of at most c S(n) moves, since no ID is repeated in the shortest computation of M x leading to acceptance.
Let I t f^- 1 2 denote that the ID I 2 can be reached from l x by a sequence of at most 2 1 moves. For i > 1, we can determine if 7 1 (-^-/ 2 DV testing each V to see if I t - V and /' | ( ' ~ l) I 2 . Thus the space needed to determine if we can get from one ID to another in 2' moves is equal to the space needed to record the ID /' currently being tested plus the space needed to determine if we can get from one ID to another in 2'~ 1 moves. Observe that the space used to test whether one ID is reachable from another in 2'~ 1 moves can be reused for each such test.
The details for testing if w is in L(M j) are given in Fig. 12.5. The algorithm of Fig. 12.5 may be implemented on a Turing machine M 2 that uses a tape as a stack of activation records! for the calls to TEST. Each call has an activation record in which the values of parameters / l5 I 2 , and i are placed, as well as the value of local variable /'. As I l9 1 2 and V are ID's with no more than S(n) cells, we can represent each of them in S(n) space. The input head position in binary uses log n < S(n) cells. Note that the input tape in all ID's is fixed and is the same as the input to begin let | w | = n and m = Hog 2 cl; let IQ be the initial ID of M, with input w; for each final ID If of length at most S(n) do if TEST (/0 , //, mS(n)) then accept; end; procedure TEST (I u 7 2 , /); if i' = 0 and (I { = I 2 or l x \— 7 2 ) then return true; if / > 1 then for each ID V of length at most S(n) do if TEST (7 l5 /', / -1) and TEST (/', 7 2 , i -1) then return true; return false end TEST Fig. 12.5 Algorithm to simulate Mj.
t An "activation record" is the area used for the data belonging to one call of one procedure.
KTUNOTES.IN Downloaded from Ktunotes.in 302 COMPUTATIONAL COMPLEXITY THEORY M 2 , so we need not copy the input in each ID. The parameter i can be coded in binary using at most mS(n) cells. Thus each activation record takes space 0(S(n)).
As the third parameter decreases by one each time TEST is called, the initial call has i = mS(n\ and no call is made when i reaches zero, the maximum number of activation records on the stack is 0(S(n)). Thus the total space used is 0(S2 (n)), and by Theorem 12.1, we may redesign M 2 to make the space be exactly S2 (n).
Example 12.4 NSPACE(log n) c DSPACE(log2 n) NSPACE(n 2 ) c DSPACE(m4 ) and NSPACE(2n ) c DSPACE(4").
Note that for S(n) > n, Savitch's theorem holds even if S(n) is space construc-tible rather than fully space constructible. M 2 begins by simulating a TM M that constructs on each input of length m, taking the largest amount of space used as S(n) and using this length to lay out the space for the activation records.
Observe, however, that if we have no way of computing S(n) in even S2 (n) space, then we cannot cycle through all possible values of If or /' without getting some that take too much space.
12.5 TRANSLATIONAL LEMMAS AND NONDETERMINISTIC HIERARCHIES In Theorems 12.8 and 12.9 we saw that the deterministic space and time hierar-chies were very dense.
It would appear that corresponding hierarchies for non-deterministic machines would require an increase of a square for space and an exponential for time, to simulate a nondeterministic machine for diagonalization purposes. However, a translational argument can be used to give a much denser hierarchy for nondeterministic machines. We illustrate the technique for space.
A translation lemma The first step is to show that containment translates upward. For example, sup-pose it happened to be true (which it is not) that NSPACE(n 3 ) c NSPACE(n 2 ).
This relation could be translated upward by replacing n by n 2 , yielding NSPACE(n 6 ) <= NSPACE(n4 ).
Lemma 12.2 Let S x (n\ S 2 (n\ and f(n) be fully space constructible, with S 2 (n) > n and f(n) > n. Then NSPACE(Si(n)) <= NSPACE(S 2 (n)) implies NSPACE(S, (/(»))) c NSPACE(S2 (/(w))).
KTUNOTES.IN Downloaded from Ktunotes.in 12.5 | TRANSLATIONAL LEMMAS AND NONDETERMINISTIC HIERARCHIES 303 Proof Let L x be accepted by M,, a nondeterministic 5 t (/ (n)) space-bounded TM. Let L2 = {x$ l |M t accepts x in space 5 1 ( | x | 4-/)}, where $ is a new symbol not in the alphabet of Then L2 is accepted by a TM M 2 as follows. On input x$ !
, M 2 marks offS^ | x \ + i) cells, which it may do, since Si is fully constructible. Then M 2 simulates M, on x, accepting if and only ifM t accepts without using more than 5 1 ( |x | + i) cells. Clearly M 2 is S^n) space bounded.
What we have done is to take a set L v in NSPACE(S 1 (/(/i))) and pad the strings with $'s so that the padded version L2 is in NSPACE(S 1 («)). Now by the hypothesis that NSPACEfSjw)) ^ NSPACE(S 2 (n)), there is a nondeterministic S2 {n) space-bounded TM M 3 accepting L2 .
Finally we construct M4 accepting the original set L t within space S 2 (f(n)).
M4 marks offf(n) cells and then S 2(f(n)) cells, which it may do since/and S 2 are fully constructible. As S2 (n) > n,f(n) < S 2(f(n)\ so M4 has not used more than S 2 (f(n)) cells.
Next M4 on input x simulates M 3 on x$' for i = 0, 1, 2, .
.
. To do thi£, M must keep track of the head location ofM 3 on x$\ If the head ofM 3 is within x, M4's head is at the corresponding point on its input. Whenever the head of M 3 moves into the $'s, M4 records the location in a counter. The length of the counter is at most log i.
If during the simulation, M 3 accepts, then M4 accepts. IfM 3 does not accept, then M4 increases i until the counter no longer fits on S2 (f ( | x | )) tape cells. Then M4 halts. Now, if x is in Lu then x$ l is in L2 for i satisfying Si(|x| -h i) = Si (/( | x | )). Since / (n) > n, this equality is satisfied by i = /( | x | ) — | x | . Thus the counter requires log (/( | x | ) -| x | ) space. Since S 2 (f( \ x | )) > /( | x | ), it follows that the counter will fit. Thus x is in L(M4 ) if and only if x$' is in L(M 3 ) for some i.
Therefore L(M4 ) = L„ and L x is in NSPACE(S2 (/(n))).
Note that we can relax the condition that S 2 (n) > n, requiring only that S2 (n) > log 2 n, provided that S 2 (f(n)) is fully space constructible. Then M4 can lay off S 2 (f(n)) cells without having to lay offf(n) cells. As S 2 (f(n)) > log/(n), there is still room for M4 's counter.
Essentially the same argument as in Lemma 12.2 shows the analogous results for DSPACE, DTI ME, and NTIME.
Example 12.5 Using the analogous translation result for deterministic time we can prove that DTIME(2") £ DTIME(w2n ). Note that this result does not follow from Theorem 12.9, as .
2" log 2" inf —— = 1.
n-oo n2 KTUNOTES.IN Downloaded from Ktunotes.in 304 COMPUTATIONAL COMPLEXITY THEORY Suppose that DTIME(w2n ) s DTIME(2n ).
Then letting S^n) = nl n , S 2 (n) = 2", and f(n) = 2n , we get DTIME(2n2 2") c DTIME(2 2").
Similarly by letting /(rc) = n + 2" we get DTIME((w + 2n )2 n2 2n ) c DTIME(2n2 2 ").
Combining (12.6) with (12.7), we obtain (12.7) (12.6) DTIME((n + 2 n )2 n2 2n ) c DTIME(2 2").
(12.8) However, inf 2 2 " log 2 2" (n + 2n)2"2 2 ' r„ = inf n + 2 n Thus Theorem 12.9 implies that (12.8) is false, so our supposition that DTIME(w2n ) c DTIME(2n ) must be false. Since DTIME(2n ) c DTIME(n2"), we conclude that DTIME(2") £ DTIME(«2n ).
Example 12.6 The translation lemma can be used to show that NSPACE(w3 ) is properly contained in NSPACE(n4 ).
Suppose to the contrary that NSPACE(h4)c=NSPACE(h 3 ). Then letting f(n) = n\ we get NSPACE(n 12 ) NSPACE(«9 ). Similarly letting/(w) = n4 , we get NSPACE(n 16 ) c NSPACE(rc 12 ), and/(n) = n 5 gives NSPACE(n 20 ) c NSPACE(n !
5 ). Putting these together yields NSPACE(n 20 ) c NSPACE(n 9 ).
However, we know by Theorem 12.11 that NSPACE(n 9 )c= DSPACE(n 18 ), and by Theorem 12.8, DSPACE(h 18 ) £ DSPACE(h 20 ). Thus combining these results, we get a contradiction.
Therefore our assumption NSPACE(w4 ) s NSPACE(h 3 ) is wrong, and we conclude NSPACE(n 3 ) £ NSPACE(n4 ).
A nondeterministic space hierarchy Example 12.6 can be generalized to show a dense hierarchy for nondeterministic space in the polynomial range.
Theorem 12.12 If e > 0 and r > 0, then NSPACE(n r ) S= NSPACE(nr+£ ).
NSPACE(n 20 ) c NSPACE(w9 ) c DSPACE(n 18 ) £ DSPACE(n20 ) c NSPACE(n 20 ), KTUNOTES.IN Downloaded from Ktunotes.in 12.5 | TRANSLATIONAL LEMMAS AND NONDETERMINISTIC HIERARCHIES 305 Proof If r is any nonnegative real number, we can find positive integers s and t such that r < s/t and r + e > (s + 1)11. Therefore it suffices to prove for all positive integers s and r, that NSPACE(n s/r ) £ NSPACE(n(s+1)/ ').
Suppose to the contrary that NSPACE(h(s+1)/ ') <= NSPACE(n s/ ').
Then by Lemma 12.2 with /(h) = n(s+i) ', we have NSPACE(h(s+1)(s+,) ) c NSPACE(ns(s+i) ) (12.9) for i = 0, 1, .
.
.
, s. As s(s + i) < (s + l)(s + iI — 1) for i > 1, we know that NSPACE(h s(s+,) ) c NSPACE(h(s+1)(s+,1) ).
(12.10) Using (12.9) and (12.10) alternately, we have NSPACE(n(s+1)(2s) ) c NSPACE(k s(2s) ) S NSPACE(n(s+ 1)(2s ~ l >) c NSPACE(n s(2s " n ) c • • • c NSPACE(n(s+ 1)s ) c NSPACE(ns2 ).
That is, NSPACE(« 2s2 + 2s ) c NSPACE(n s2 ).
However, by Savitch's theorem, NSPACE(w s2 ) c DSPACE(n 252 ), and by Theorem 12.8, DSPACE(m 2s2 ) £ DSPACE(n 2s2 + 2s ).
Clearly, DSPACE(/7 2s2+2s ) c NSPACE(n 2s2 + 2s ).
Combining these results, we get NSPACE(n 2s2 + 2s ) 5 NSPACE(/7 2s2 + 2s ), a contradiction. We conclude that our assumption NSPACE(n(s+1)/t ) c NSPACE(n s/ ') was wrong. Since containment in the opposite direction is obvious, we conclude NSPACE(n s/') £ NSPACE(h(s+ 1)/r ) for any positive integers s and t.
KTUNOTES.IN Downloaded from Ktunotes.in 306 COMPUTATIONAL COMPLEXITY THEORY Similar dense hierarchies for nondeterministic space can be proved for ranges higher than the polynomials, and we leave some of these results as exercises.
Theorem 12.12 does not immediately generalize to nondeterministic time, because of the key role of Savitch's theorem, for which no time analog is known. However, a time analog of Theorem 12.12 has been established by Cook [1973a].
12.6 PROPERTIES OF GENERAL COMPLEXITY MEASURES: THE GAP, SPEEDUP, AND UNION THEOREMS In this section we discuss some unintuitive properties of complexity measures.
While we prove them only for deterministic space complexity, they will be seen in the next section to apply to all measures of complexity.
Theorems 12.8 and 12.9 indicate that the space and time hierarchies are very dense. However, in both theorems the functions are required to be constructive.
Can this condition be discarded? The answer is no: the deterministic space and time hierarchies have arbitrarily large gaps in them.
We say that a statement with parameter n is true almost everywhere (a.e.) if it is true for all but a finite number of values of n. We say a statement is true infinitely often (i.o.) if it is true for an infinite number of n's. Note that both a statement and its negation may be true i.o.
Lemma 12.3 If L is accepted by a TM M that is S(n) space bounded a.e., then L is accepted by an S(n) space-bounded TM.
Proof Use the finite control to accept or reject strings of length n for the finite number of n where M is not S(n) bounded. Note that the construction is not effective, since in the absence of a time bound we cannot tell which of these words M accepts.
Lemma 12.4 There is an algorithm to determine, given TM M, input length n, and integer m, whether m is the maximum number of tape cells used by M on some input of length n.
Proof For each m and n there is a limit t on the number of moves M may make on input of length n without using more than m cells of any storage tape or repeating an ID. Simulate all sequences of up to t moves, beginning with each input of length n.
Theorem 12.13 (Borodin s Gap Theorem) Given any total recursive function g(n) > h, there exists a total recursive function S(n) such that DSPACE(S(n)) — DSPACE(#(S(n))). In other words, there is a "gap" between space bounds S(n) and g(S(n)) within which the minimal space complexity of no language lies.
Proof Let M x , M 2 , ... be an enumeration of TM's. Let be the maximum number of tape cells used by M, on any input of length n. If M, always halts, then Si(n) is a total function and is the space complexity of M,, but if M, does not halt KTUNOTES.IN Downloaded from Ktunotes.in 12.6 | PROPERTIES OF GENERAL COMPLEXITY MEASURES 307 on some input of length w, then S,(«) is undefined.! We construct S(n) so that for each k either 1) Sk (n) < S(n) a.e., or 2) Sk(n)>g(S(n)) i.o.
That is, no Sk (n) lies between S(n) and g(S(n)) for almost all n.
In constructing S(n) for a given value of n, we restrict our attention to the finite set of TM's M u M 2 , • • • , M„. The value for S(n) is selected so that for no i between 1 and n does S,(w) lie between S(n) and gr(S(w)). If we could compute the largest finite value of S,(w) for 1 < i < «, then we could set S(n) equal to that value.
However, since some S,(n) are undefined, we cannot compute the largest value.
Instead, we initially set ; = 1 and see if there is some M t in our finite set for which Si(n) is between j + 1 and g(j). If there is some such S,-(n), then set ; to S,(w) and repeat the process. If not, set S(n) to j and we are done. As there is but a finite number of TM's under consideration, and by Lemma 12.4 we can tell whether Si(n) = m for any fixed m, the process will eventually compute a value for j such that for 1 < i < n either < j or S,(n) > g(j). Assign S(n) this value of Suppose there were some language L in DSPACE(#(S(n)) but not in DSPACE(S(h)). Then L = L(Mk ) for some k where Sk (n) < g(S(n)) for all n. By the construction of S(n), for all n > k, Sk (n) < S(n). That is, Sk (n) < S(n) a.e., and hence by Lemma 12.3, L is in DSPACE(5(m)), a contradiction. We conclude that DSPACE(S(n)) = DSPACE(^(5(n))).
Theorem 12.13 and its analogs for the other three complexity measures have a number of highly unintuitive consequences, such as the following.
Example 12.7 There is a total recursive function f(n) such that DTIME(/(n)) = NTIME(/(n)) = DSPACE(/(m)) = NSPACE(/(n)).
Clearly DTIME(/(n)) is contained within NTIME(/(rc)) and DSPACE(/(r?)).
Similarly, both NTIME(/(«)) and DSPACE(/(«)) are contained within NSPACE(/(m)). By Theorem 12.10, for all/(n) > log 2 w, if L is in NSPACE(/(n)), then there is a constant c, depending only on L such that L is in DTIME(c/(n) ).
Therefore, L = L(M) for some TM M whose time complexity is bounded above by f(nY (n) a.e. By the DTIME analog of Lemma 12.3, L is in DTIME(/(M)/(n) ).
Finally, the DTIME analog of Theorem 12.13 with #(x) = xx establishes the exist-ence of/(n) for which DTIME(/(n)) = DTIME(f(nY {n) \ proving the result.
Similarly, if one has two universal models of computation, but one is very simple and slow, say a Turing machine that makes one move per century, and the other is very fast, say a random-access machine with powerful built-in instructions for multiplication, exponentiation, and so on, that performs a million operations t We identify an undefined value with infinity, so an undefined value is larger than any defined value.
KTUNOTES.IN Downloaded from Ktunotes.in 308 COMPUTATIONAL COMPLEXITY THEORY per second, it is easily shown that there exists a total recursive T(n) such that any function computable in time T(n) on one model is computable in time T(n) on the other.
The speed-up theorem Another curious phenomenon regarding complexity measures is that there are functions with no best programs (Turing machines). We have already seen that every TM allows a linear speed up in time and compression in space. We now show that there are languages with no "best" program. That is, recognizers for these languages can be sped up indefinitely. We shall work only with space and show that there is a language L such that for any Turing machine accepting L, there always exists another Turing machine that accepts L and uses, for example, only the square root of the space used by the former. This new recognizer can of course be replaced by an even faster recognizer and so on, ad infinitum.
The basic idea of the proof is quite simple. By diagonalization we construct L so that L cannot be recognized quickly by any "small" machine, that is, a machine with a small integer index encoding it. As machine indices increase, the diagonali-zation process allows faster and faster machines recognizing L. Given any ma-chine recognizing L, it has some fixed index and thus can recognize L only so fast.
However, machines with larger indices can recognize L arbitrarily more quickly.
Theorem 12.14 (Blums Speed-up Theorem) Let r(n) be any total recursive func-tion. There exists a recursive language L such that for any Turing machine M, accepting L, there exists a Turing machine Mj accepting L such that r(Sj(n)) < Si(n), for almost all n.
Proof Without loss of generality assume that r(n) is a monotonically nondecreas-ing fully space-constructible function with r(n) > n 2 (see Exercise 12.9). Define h(n) by /i(l) = 2, h(n) = r(h(n-1)).
Then h(n) is a fully space-constructible function, as the reader may easily show.
Let M b M 2 y • •• be an enumeration of all off-line TM's analogous to that of Section 8.3 for single-tape TM's. In particular, we assume that the code for M, has length log 2 /. We construct L so that 1) if L(M,) = L, then S,(n) > h(n -i) a.e.; 2) for each k, there exists a Turing machine Mj such that L{M}) = L and Sj(n) r(Sj(n)) a.e.
KTUNOTES.IN Downloaded from Ktunotes.in 12.6 | PROPERTIES OF GENERAL COMPLEXITY MEASURES 309 To see this, select Mj so that Sj(n) < h(n -i -1). By (2), Af, exists. Then by (1), Si(n) > h(n -i) = r(h{n -i -1)) > r(S») a.e.
Now let us construct L ^ 0 to satisfy (1) and (2). For n = 0, 1, 2, ... in turn, we specify whether 0" is in L.
In the process, certain M, are designated as "canceled." A canceled TM surely does not accept L Let a(n) be the least integer j < n such that Sj(n) < h(n — j), and Mj is not canceled by i = 0, 1, — 1.
When we consider n, if h(n — i) a.e. Let L(M.) = L. In constructing L, all TM's Mjt for j < i, that are ever canceled are canceled after considering some finite number of n\ say up to n0 . Note that n0 cannot be effectively computed, but nevertheless exists. Sup-pose Si(n) < h(n — i) for some n > max(n0 , i). When we consider n, no MjJ < i9 is canceled. Thus a(n) = i, and A/, would be canceled had it not been previously canceled. But a TM that is canceled will surely not accept L. Thus S,(n) > h(n — i) for n > max(n0 , i), that is, S^n) > h(n — i) a.e.
To prove condition (2) we show that there exists, for given k, a TM M = M} such that L(M) = L, and Sj(n) < h(n — k)for all n. To determine whether 0" is in L, M must simulate M„ (n) on 0 n . To know what o(n) is, M must determine which M,-'s have already been canceled by Of for f < n. However, constructing the list of canceled TM's directly requires seeing if M, uses more than /?(/" — i) space for 0 < f < n and 1 ^ / < n. For i < k + f — n y this requires more than h(n — k) space.
The solution is to observe that any TM Mh i < k, that is ever canceled, is canceled when we consider some / less than a particular n x . For each f < w,, incorporate into the finite control of M whether Of is in L, and also incorporate a list of all TM's M, canceled by any f < n v Thus no space at all is needed by M if n < n v If n > to compute a(n) and simulate M a(n) on 0", it will only be neces-sary to simulate TM's M, on input (/, where n x < f < n and k < i < n, to see whether M, is canceled by /.
To test whether M, is canceled by /, we need only simulate M, using /?(/ — i) of M,'s cells, which is less than h(n — /c), as f < n and i > k. As n > it must be that o(n), if it exists, is greater than k. Thus simulating M n(n) on input 0" takes h(n — o(n)) of M a{n)'s cells, which is less than h(n — k) cells.
Lastly, we must show that M can be made to operate within space h(n — k).
We need only simulate TM's M, for k < i < n on inputs (/, < / < «, to see whether they get canceled, so we need represent no more than h(n — k — 1 ) cells of M,'s tape for any simulation. Since / < n y the integer code for M, has length no more than log 2 n. Thus any tape symbol of M, can be coded using \og 2 n of M's cells. As r(x) > x 2 , we know h(x) > 2 2x . Also, by the definition of /?, h(n — k) > [h(n -k -l)] 2 > 2 2n ' k ' x h{n - k -1). As 2 2n ' k ~ l > \og 2 n a.e., /7(« - k) space is sufficient for the simulation for almost all n.
KTUNOTES.IN Downloaded from Ktunotes.in 310 COMPUTATIONAL COMPLEXITY THEORY In addition to the space required for simulating the TM's, space is needed to maintain the list of canceled TM's. This list consists of at most n TM's, each with a code of length at most log2 n. The n log n space needed to maintain the list of canceled TM's is also less than h(n — k) a.e. By Lemma 12.3, M can be modified to recognize words 0", where n \og2 n < h(n — k) or 2 2 "~ k ~ 1 < log 2 n in its finite control. The resulting TM is of space complexity h(n — k) for all h, and is the desired M.
The union theorem The last theorem in this section, called the union theorem, has to do with the-naming of complexity classes. By way of introduction, we know that each polyno-mial such as n2 or n 3 defines a space complexity class (as well as complexity classes of the other three types). However, does polynomial space form a complexity class? That is, does there exist an S(n) such that DSPACE(S(n)) contains all sets recognizable in a polynomial space bound and no other sets? Clearly, S(n) must be almost everywhere greater than any polynomial, but it also must be small enough so that one cannot fit another function that is the space used by some TM between it and the polynomials, where "fit" must be taken as a technical term whose meaning is defined precisely in the next theorem.
Theorem 12.15 Let {fj(n) \ i = 1, 2, .
. .} be a recursively enumerable collection of recursive functions. That is, there is a TM that enumerates a list of TM's, the first computing/^, the second computing^, and so on. Also assume that for each i and nyfi( n ) ])).
i> 1 Proof We construct a function S(n) satisfying the following two conditions: 1) For each i, S(n) >fi(n) a.e.
2) If Sj(n) is the exact space complexity of some TM M } and for each i, Sj(n) > fi(n) i.o., then Sj(n) > S(n) for some n (and in fact, for infinitely many n's).
The first condition assures that U DSPACE(/(n)) ci DSPACE(S(n)).
i The second condition assures that DSPACE(S(rc)) contains only those sets that are in DSPACE(/(n)) for some /. Together the conditions imply that DSPACE(S(«)) = (J DSPACE(y;(M)).
i Setting S(n) = f n (n) would assure condition (1). However, it may not satisfy condition (2). There may be a TM M} whose space complexity Sj(n) is greater than each/j(n) i.o. but less thanf n (n) for all n. Thus there may be sets in DSPACE(/ n (n)) KTUNOTES.IN Downloaded from Ktunotes.in 12.6 | PROPERTIES OF GENERAL COMPLEXITY MEASURES 311 not in [ji DSPACE( fi(n)). To overcome this problem we construct S(n) so that it dips below each Sj(n) that is i.o. greater than each / f (n), and in fact, S(n) will dip below Sj(n) for an infinity of ris. This is done by guessing for each TM Mj an ij such that fi.(n) > Sj(n) a.e. The "guess" is not nondeterministic; rather it is subject to deterministic revision as follows. If at some point we discover that the guess is not correct, we guess a larger value for i} and for some particular n define S(n) to be less than Sj(n). If it happens that Sj grows faster than any/j, S will infinitely often be less than Sj. On the other hand, if some/ is almost everywhere greater than Sp eventually we shall guess one such f { and stop assigning values of S less than Sj.
In Fig.
12.6 we give an algorithm that generates S(n). A list called LIST of "guesses" of the form "i7 - = k" for various integers j and k is maintained. For each j, there will be at most one guess k on LIST at any time. As in the previous theorem, M u M 2 , ...
is an enumeration of all off-line TM's, and Sj(n) is the maximum amount of space used by Mj on any input of length n. Recall that Sj(n) may be undefined (infinite) for some values of n.
begin 1) LIST:= empty list 2) for«= 1,2, 3, ...do 3) if for all = k" on LIST,/(«) > S» then 4) add "i„ = n" to LIST and define S(n) =f n(n) else begin 5) Among all guesses on LIST such that f k (n) < Sj(n), let = A" be the guess with the smallest k 7 and given that k, the smallest j; 6) define S(n) = f k (n)\ 7) replace = /T by = n" on LIST; 8) add "!„ = n to LIST end end Fig. 12.6 Definition of S(n).
To prove that DSPACE(S()) = (J DSPACE(y;(«)) i we first show that S(n) satisfies conditions (1) and (2). Consider condition (1). To see that for each m, S(n) > f m (n) a.e. observe that S(n) is assigned a value only at lines (4) and (6) of Fig.
12.6. Whenever S(n) is defined at line (4) for n > m 9 the value of S(n) is at least f m (n). Thus for the values of S(n) defined at line (4), S(n) > f m(n) except for the finite set of n less than m. Now consider the values of S(n) defined at line (6). When n reaches m, LIST will have some finite number of KTUNOTES.IN Downloaded from Ktunotes.in 312 COMPUTATIONAL COMPLEXITY THEORY guesses. Each of these guesses may subsequently cause one value of S(n\ for some n > m, to be less than f m (n). However, when that happens, line (7) causes that guess to be replaced by a guess "ij = p" for some p > m, and this guess, if selected at line (5), does not cause S(n) to be made less than f m (n), since f p(n) > f m (n) whenever p>m. Thus from line (6) there are only finitely many n greater than m (at most the length of LIST when n = m) for which S(n) f m(n) a.e.
Next we must show condition (2), that if there exists TM Mj such that for each i, Sj(n) > f^n) i.o., then Sj(ri) > S(ri) for infinitely many ri. At all times after n = j, LIST will have a guess for ij9 and LIST is always finite. For n = j we place "ij = j n on LIST. As Sj(n) > fj(n) i.o., there will be arbitrarily many sub-sequent values of n for which the condition of step (3) does not hold. At each of these times, either our 4 — j n is selected at line (5), or some other one of the finite number of guesses on LIST when n = j is selected. In the latter case, that guess is replaced by a guess 4Hp = q" with q > j. All guesses added to LIST are also of the form "/ = q" for q > j, so eventually our "f . = j n is selected at step (5), and for this value of n, we have Sj(n) > fj(n) = S(n). Thus condition (2) is true.
Lastly we must show that conditions (1) and (2) imply DSPACE(S(n)) = [j DSPACE(/ i («)).
i Suppose L is in (J f DSPACE(/^)). Then L is in DSPACE(/ m (^2)) for some par-ticular m. By condition (1), S(n) >f m (n) a.e. Thus by Lemma 12.3, L is in DSPACE(S(«)). Now suppose that L is in DSPACE(S(rc)). Let L = L(M} \ where Sj(n) < S(n) for all n.
If for no i, L is in DSPACE(/(m)), then by Lemma 12.3, for every i, each TM M k accepting L has Sk (n) >fi(n) i.o. Thus by condition (2) there is some n for which Sk (n) > S(n). Letting k = j produces a contradiction.
Example 12.8 Let /(h) = ri. Then we may surely enumerate a sequence of TM's M u M 2 , ... such that Mh presented with input 0", writes 0"' on its tape and halts. Thus Theorem 12.15 says that there is some S(n) such that DSPACE(S(n)) = (J DSPACE(«').
As any polynomial p(ri) is equal to or less than some ri a.e., DSPACE(5(n)) is the union over all polynomials p(ri), of DSPACE(p(/i)). This union, which in the next chapter we shall call PSPACE, and which plays a key role in the theory of intractable problems, is thus seen to be a deterministic space complexity class.
12.7 AXIOMATIC COMPLEXITY THEORY The reader may have observed that many theorems in this chapter are not depen-dent on the fact that we are measuring the amount of time or space used, but only KTUNOTES.IN Downloaded from Ktunotes.in 12.7 | AXIOMATIC COMPLEXITY THEORY 313 that we are measuring some resource that is being consumed as the computation proceeds. In fact one could postulate axioms governing resources and give a completely axiomatic development of complexity theory. In this section we briefly sketch this approach.
The Blum axioms Let Mj, M 2 , ... be an enumeration of Turing machines defining among them every partial recursive function. For technical reasons we consider the M,'s as computing partial recursive functions fa rather than as recognizing sets. The reason is that it is notationally simpler to measure complexity as a function of the input rather than of the length of the input. Let ,(rc) be the function of one variable computed by M f , and let ^(h), Q>2 (n )i .... be a set of partial recursive functions satisfying the following two axioms (Blums axioms).
Axiom 1 O f(n) is defined if and only if $,(n) is defined.
Axiom 2 The function R(i, n, m) defined to be 1 if t (rc) = m and 0 otherwise, is a total recursive function.
The function O f (n) gives the complexity of the computation of the ith Turing machine on input n. Axiom 1 requires that ,(m) is defined if and only if the ith Turing machine halts on input n. Thus one possible O, would be the number of steps of the ith Turing machine. The amount of space used is another alternative, provided we define the space used to be infinite if the TM enters a loop.
Axiom 2 requires that we can determine whether the complexity of the ith Turing machine on input n is m. For example, if our complexity measure is the number of steps in the computation, then given i, n, and m, we can simulate M, on 0" for m steps and see if it halts. Lemma 12.4 and its analogs are claims that Axiom 2 holds for the four measures with which we have been concerned.
Example 12.9 Deterministic space complexity satisfies Blum's axioms, provided we say <!>,(«) is undefined if M, does not halt on input 0", even though the amount of space used by M, on 0" may be limited. Deterministic time complexity likewise satisfies the axioms if we say O f («) is undefined whenever M, runs forever or halts without any 0 7 on its tape. To compute R(i, n, m), simply simulate M, for m steps on input 0n .
We may establish that nondeterministic time and space satisfy the axioms if we make an intelligent definition of what it means for an NTM to compute a function. For example, we might say that = j if and only if there is some sequence of choices by M, with input 0" that halts with 0 7 on the tape, and no sequence of choices that leads to halting with some 0\ k =f= j, on the tape.
If we define O.(m) = ,(«), we do not satisfy Axiom 2. Suppose R(i', n, m) were recursive. Then there is an algorithm to tell if M, with input 0" halts with 0m on its tape. Given any TM M, we may construct M to simulate M. If M halts with any KTUNOTES.IN Downloaded from Ktunotes.in 314 COMPUTATIONAL COMPLEXITY THEORY tape, M erases its own tape. If i is an index for M, then R(i, n, 0) is true if and only ifM halts on input 0". Thus if R(i, n, m) were recursive, we could tell if a given TM M halts on a given input, which is undecidable (see Exercise 8.3).
Recursive relationships among complexity measures Many of the theorems on complexity can be proved solely from the two axioms. In particular, the fact that there are arbitrarily complex functions, the speed-up theorem, the gap theorem, and the union theorem can be so proved. We prove only one theorem here to illustrate the techniques. The theorem we select is that all measures are recursively related. That is, given any two complexity measures Q> and , there is a total recursive function r such that the complexity of the TM M, in one measure, 6,(rz), is at most r(rc, ,(rc)). For example, Theorems 12.10 and 12.11 showed that for the four measures of complexity with which we have been dealing, at most an exponential function related any pair of these complexity measures. In a sense, functions that are easy in one measure are "easy" in any other measure, although the term "easy" must be taken lightly, as r could be a very rapidly growing function, such as Ackermann's function.
Theorem 12.16 Let and 6 be two complexity measures. Then there exists a recursive function r such that for all /, r(n, > 0>,(n) a.e.
Proof Let r(n, m) = max {i>i(n) \i <n and O^rc) = m).
i The function r is recursive, since <!>,(«) = m may be tested by Axiom 2. Should it be equal to m, then ,(/?) and ,(h) must be defined, by Axiom 1, and hence the maximum can be computed.
Clearly r(n, > i, since for n > i, r(n, 0. Thus the bound of Theorem 12.5 is in a sense the best possible.
12.3 The notion of crossing sequences can be adapted to off-line TM's if we replace the notion of "state" by the state, contents of storage tapes, and positions of the storage tape heads. Theorem 12.8, the space hierarchy, applied only to space complexities of log n or above. Prove that the same holds for fully space-constructible S 2 (n) below log n. [Hint: Using a generalized crossing sequence argument, show that {wc'w | w is in (a + b) and | w | = 2S2 log log n i.o. Show the same result for nondeter-ministic space. Thus for deterministic and nondeterministic space there is a "gap" between 1 and log log n.
12.5 Show that Lemma 12.2, the "translation lemma," applies to a) deterministic space b) deterministic time, and c) nondeterministic time.
12.6 Show that DTlME(2 2 " +n ) properly includes DTIME(2 2").
12.7 Show that NSPACE((c + e) n ) properly includes NSPACE(c") for any c> 1 and c>0.
12.8 What, if any, is the relationship between each of the following pairs of complexity classes?
a) DSPACE(n 2 ) and DSPACE(/(m)), where/(n) = n for odd n and n 3 for even n.
b) DTIME(2") and DTIME(3 n ) c) NSPACE(2 n ) and DSPACE(5 n ) d) DSPACE(n) and DTIME([log2 Ml n ) 12.9 Show that if r is any total recursive function, then there is a fully space-constructible monotonically nondecreasing r' such that r'(n) > r(n), and r'(x) > x 2 for all integers x.
[Hint: Consider the space complexity of any TM computing r.] 12.10 Show that there is a total recursive function S(n) such that L is in DSPACE(S(n)) if and only if L is accepted by some c n space-bounded TM, for c > 1.
12.11 Suppose we used axioms for computational complexity theory as it pertains to languages rather than functions. That is, let M x > M 2 , ... be an enumeration of Turing machines and Li the language accepted by M t . Replace Axiom 1 by: Axiom 1': O^m) is defined if and only if M, halts on all inputs of length n.
Reprove Theorem 12.16 for Axioms 1' and 2.
KTUNOTES.IN Downloaded from Ktunotes.in 316 COMPUTATIONAL COMPLEXITY THEORY 12.12 Show that the speed-up and gap theorem hold for NSPACE, DTIME and NTIME.
[Hint: Use Theorem 12.16 and the speed-up and gap theorems for DSPACE.] 12.13 Show that the following are fully time and space constructible: a) n 2 b) 2" c) n\ 12.14 Show that the following are fully space constructible: a) Jn b) log2 M Sc) Some function that is bounded above by log 2 log 2 « and that is bounded below by c log 2 log 2 n, for some c > 0, infinitely often.
12.15 Show that if T 2 (n) is time constructible, and n-oo T 2 (n) then there is a language accepted by a T 2(n) time-bounded one-tape TM, but by no TY (n) time-bounded one-tape machine. [Hint: To simulate a one-tape TM M t by a one-tape machine, move the description ofM { so that it is always near the tape head. Similarly, carry along a "counter" to tell when M, has exceeded its time limit.] 12.16 Show that if T 2 (n) is time constructible and T 2 (n) then for all k, there is a language accepted by a fc-tape 7 2 time-bounded TM but by no /c-tape Ti(n) time-bounded TM, where \og(m) is the number of times we must take logar-ithms base 2 of m to get to 1 or below. For example, log(3) = 2 and log(2 65536 ) = 5. Note that this exercise implies Exercise 12.15.
12.17 Show that for any complexity measure O satisfying the Blum axioms there can be no total recursive function /such that O.(n) < f(n, (pi(n)). That is, one cannot bound the complexity of a function in terms of its value.
12.18 The speed-up theorem implies that for arbitrarily large recursive functions r we can find a language L for which there exists a sequence of TM's M M 2 , .
.
• , each accepting L, such that the space used by M, is at least r applied to the space used by M i + ,. However, we did not give an algorithm for finding such a sequence; we merely proved that it must exist.
Prove that speed up is not effective, in that if for every TM accepting L, there is an M, on the list using less space, then the list of TM's is not recursively enumerable.
12.19 Which of the following are complexity measures?
a) <D,(n) = the number of state changes made by M, on input n.
b) 12.27 Let L c (0 + 1) be the set accepted by some T(n) time-bounded TM. Prove that for each n there exists a Boolean circuit, with inputs x,, x„, having at most T(n) log T(n) two-input gates and producing output 1 if and only if the values ofx u .
.
.
, xn correspond to a string in L. The values of x u .
.
.
, x n correspond to the string x if x, has value true whenever the z'th symbol of x is 1 and x, has value false whenever the ith symbol of x is 0. [Hint: Simulate an oblivious TM.] 12.28 Loop programs consist of variables that take on integer values and statements. A statement is of one of the forms below.
1) (variable) := (variable) 2) (variable) := (variable) + 1 3) for i := 1 to (variable) do statement; 4) begin (statement); (statement); ... (statement) end; In (3) the value of the variable is bound before the loop, as in PL/I.
a) Prove that loop programs always terminate.
b) Prove that every loop program computes a primitive recursive function.
c) Prove that every primitive recursive function is computed by some loop program.
d) Prove that a TM with a primitive recursive running time can compute only a primitive recursive function.
12.29 Let F be a formal proof system in which we can prove theorems about one-tape TM's. Define a complexity class C T{n) = {L(Mi) | there exists a proof in F that T^n) < T(n) for all n}.
Can the time hierarchy of Exercise 12.16 be strengthened for provable complexity? [Hint: Replace the clock by a proof that T^n) < T(n).] KTUNOTES.IN Downloaded from Ktunotes.in 318 COMPUTATIONAL COMPLEXITY THEORY Solutions to Selected Exercises 12.2(a) Consider any string wcwR of length n and let £wA be the length of the crossing sequence between positions /' and i + 1, for 1 < i < n/2, made by some one-tape TM M with s states. Suppose the average of / W>1 over all words w of length (n - l)/2 is p(i). Then for at least half of all w's, SwJ < 2p(i). The number of w's is 2 {n ~ 1)/2 , so there are at least 2 {n ~ 3)12 w's for which tw i < 2p(i). As the number of crossing sequences of length 2p(i) or less is 2p(i) £ ^+1 , there must be at least 2 {n ~ 3)/2/s 2p(,)+ w's with the same crossing sequence between posi-tions i and i + 1. There are 2 (n " 1)/2 "' sequences of as and b's that may appear in positions i + 1 through (n -l)/2 in these words, so if ?2 ,"-'" 2 -i (12.U) Then two words with the same crossing sequence differ somewhere among the first i positions. Then by Exercise 12.1(b), M accepts a word it should not accept.
Thus (12.11) is false, and s 2p(i)+ 1 > 2}~ l . Therefore, ^^2Tog^-2 Surely there is some word w such that when presented with wcwR , M takes at least average time. By Exercise 12.1(a), this average is at least <"-J>/ 2 (""J,)/ 2 / -1 n 1 1 In - 3 \ /n -1 \ n -1 12.14(c) We may design an off-line TM M of space complexity S(n) to test for i = 2, 3» — whether its input length n is divisible by each i, stopping as soon as we encounter a value of / that does not divide n. As the test whether i divides n needs only log2 / storage cells, S(n) is the logarithm of the largest / such that 2, 3, / all divide n. If we let n = kl we know that S(n) > \og 2 k. As k \ < k k , we know that log2 n < k \og 2 k and log 2 log 2 « < \og 2 k + log2 log 2 /c < 2 log2 /c.
Thus for those values of n that are k ! for some k, it follows that S(n) > \ log2 log 2 n.
We must show that for all n 7 S(n) < 1 + log2 log 2 /7. It suffices to show that the smallest n for which S(n) > k, which is the least common multiple (LCM) of 2, 3, .
.
.
, 2 k ~ 1 + 1, is at least 2 2k_1 . That is, we need the fact that LCM(2, 3, ...,i) > 2i_1 . A proof requires results in the theory of numbers that we are not prepared to derive, in particular that the probability that integer / is a prime is asymptotically 1/ln z, where In is the natural logarithm (see Hardy and Wright ). Since LCM(2, 3, i) is at least the product of the primes between 2 and /, a lower bound on the order of e' for LCM (2, 3, .
.
.
, i) for large i is easy to show.
KTUNOTES.IN Downloaded from Ktunotes.in BIBLIOGRAPHIC NOTES 319 BIBLIOGRAPHIC NOTES The study of time complexity can be said to begin with Hartmanis and Stearns , where Theorems 12.3, 12.4, 12.5, and 12.9 are found. The serious study of space complexity begins with Hartmanis, Lewis, and Stearns , and Lewis, Stearns, and Hartmanis ; Theorems 12.1, 12.2, and 12.8 are from the former. Seiferas [1977a,b] presents some of the most recent results on complexity hierarchies. A number of earlier papers studied similar aspects of computation. In Grzegorczyk , Axt , and Ritchie we find hierarchies of recursive functions. Yamada studies the class of real-time comput-able functions [T(n) = n]. Rabin showed that two tapes can do more than one in real time, a result that has since been generalized by Aanderaa to k versus k — 1 tapes.
Theorem 12.6, showing that logarithmic slowdown suffices when one goes from many tapes to two is from Hennie and Stearns . Theorem 12.11, the quadratic relationship between deterministic and nondeterministic time, appears in Savitch . Translational lemmas were pioneered by Ruby and Fischer , while Theorem 12.12, the nondeter-ministic space hierarchy, is by Ibarra . The nondeterministic time hierarchy alluded to in the text is from Cook [1973a]. The best nondeterministic hierarchies known are found in Sieferas, Fischer, and Meyer . Book and Greibach characterize the lan-guages in (J c>0 NTIME(cn).
The study of abstract complexity measures originates with Blum . Theorem 12.13, the gap theorem, is from Borodin and (in essence) Trakhtenbrot ; a stronger version is due to Constable . (Note that these and all the papers mentioned in this paragraph deal with Blum complexity measures, not solely with space, as we have done.) Theorem 12.14, the speed-up theorem, is from Blum , and the union theorem is from McCreight and Meyer . Theorem 12.16, on recursive relationships among complexity measures, is from Blum . The honesty theorem mentioned in Exercse 12.20 is from McCreight and Meyer . The simplified approach to abstract complexity used in this book is based on the ideas of Hartmanis and Hopcroft .
Crossing sequences, discussed in Exercises 12.1 and 12.2, are from Hennie . The generalization of crossing sequences used in Exercises 12.3 and 12.4 is developed in Hop-croft and Ullman [1969a], although Exercise 12.4 in the deterministic case is from Hart-manis, Lewis, and Stearns . Exercise 12.14(c) is from Freedman and Ladner .
Exercise 12.16, a denser time hierarchy when TM's are restricted to have exactly k tapes, is from Paul . Exercise 12.18, showing that speed up cannot be made effective, is from Blum . Exercise 12.22 is from Hartmanis and Hopcroft . Exercise 12.23 is from Hopcroft, Paul, and Valiant . See also Paul, Tarjan, and Celoni for a proof that the method of Hopcroft et ai cannot be extended. Oblivious Turing machines and Exercises 2.26 and 2.27 are due to M. Fischer and N. Pippenger. Loop programs and Exercise 2.28 are from Ritchie and Meyer and Ritchie .
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 13 INTRACTABLE PROBLEMS In Chapter 8 we discovered that one can pose problems that are not solvable on a computer. In this chapter we see that among the decidable problems, there are some so difficult that for all practical purposes, they cannot be solved in their full generality on a computer. Some of these problems, although decidable, have been proved to require exponential time for their solution. For others the implication is very strong that exponential time is required to solve them ; if there were a faster way of solving them than the exponential one, then a great number of important problems in mathematics, computer science, and other fields—problems for which good solutions have been sought in vain over a period of many years—could be solved by substantially better means than are now known.
13.1 POLYNOMIAL TIME AND SPACE The languages recognizable in deterministic polynomial time form a natural and important class, the class (J,->, DTIME(n'), which we denote by It is an intuitively appealing notion that & is the class of problems that can be solved efficiently. Although one might quibble that an n 51 step algorithm is not very efficient, in practice we find that problems in 0 usually have low-degree polyno-mial time solutions.
There are a number of important problems that do not appear to be in & but have efficient nondeterministic algorithms. These problems fall into the class U,->i NTIME(n'), which we denote by Jr&. An example is the Hamilton circuit problem: Does a graph have a cycle in which each vertex of the graph appears exactly once? There does not appear to be a deterministic polynomial time algo-320 KTUNOTES.IN Downloaded from Ktunotes.in 13.1 | POLYNOMIAL TIME AND SPACE 321 rithm to recognize those graphs with Hamilton circuits. However, there is a simple nondeterministic algorithm ; guess the edges in the cycle and verify that they do indeed form a Hamilton circuit.
The difference between & and Jf& is analagous to the difference between efficiently finding a proof of a statement (such as "this graph has a Hamilton circuit") and efficiently verifying a proof (i.e., checking that a particular circuit is Hamilton). We intuitively feel that checking a given proof is easier than finding one, but we don't know this for a fact.
Two other natural classes are PSPACE = (J DSPACE(rt') i> 1 and NSPACE = (J NSPACE(n').
i> 1 Note that by Savitch's theorem (Theorem 12.11) PSPACE = NSPACE, since NSPACE(h') c DSPACE(« 2 ').
Obviously, & c jf& c PSPACE, yet it is not known if any of these containments are proper. Moreover, as we shall see, it is unlikely that the mathematical tools needed to resolve the questions one way or the other have been developed.
Within PSPACE we have two hierarchies of complexity classes: DSPACE(log n) £ DSPACE(log2 n) £ DSPACE(log 3 n) £ • • and NSPACE(log n) £ NSPACE(log2 n) 5 NSPACE(log 3 n) £ Clearly DSPACE(log n) c NSPACE(log n) and thus by Savitch's theorem U NSPACE(log /t n) = (J DSPACE(log ,c n).
k>l k>l Although one can show that 0> + IJ DSPACE(log fc n), k>l containment of either class in the other is unknown. Nevertheless DSPACE(log n) c & c X9> c PSPACE, and at least one of the containments is proper, since DSPACE(log n) 5 PSPACE by the space hierarchy theorem.
Bounded reductibilities Recall that in Chapter 8 we showed a language L to be undecidable by taking a known undecidable language L and reducing it to L. That is, we exhibited a mapping g computed by a TM that always halts, such that for all strings x, x is in KTUNOTES.IN Downloaded from Ktunotes.in 322 INTRACTABLE PROBLEMS L if and only if g(x) is in L. Then if L were recursive, L could be recognized by computing g(x) and deciding whether g(x) is in L By restricting g to be an easily computable function, we can establish that L is or is not in some class such as JfzP, or PSPACE. We shall be interested particularly in two types of reducibility: polynomial time reducibility and log-space reducibility. We say that L is polynomial-time reducible to L if there is a polynomial-time bounded TM that for each input x produces an output y that is in L if and only if x is in L.
Lemma 13.1 Let L be polynomial-time reducible to L. Then a) L is in .
1 if L is in Ar&, b) L is in & if L is in J.
Proof The proofs of (a) and (b) are similar. We prove only (b). Assume that the reduction is Pi(n) time bounded and that L is recognizable in time p 2 (rc), where p x and p 2 are polynomials. Then L can be recognized in polynomial time as follows.
Given input x of length m, produce y using the polynomial-time reduction. As the reduction is Pi{n) time bounded, and at most one symbol can be printed per move, it follows that \y\ < P(n). Then, we can test if y is in L in time p 2 {p{n)). Thus the total time to tell whether x is in L is Pi(n) -f p2 {P{n)), which is polynomial in n.
Therefore, L is in P.
A log-space transducer is an off-line TM that always halts, having log n scratch storage and a write-only output tape on which the head never moves left.
We say that L is log-space reducible to L if there is a log-space transducer that given input x, produces an output string y that is in L if and only if x is in L.
Lemma 13.2 If L is log-space reducible to L, then a) L is in & if L is in b) L is in NSPACE(log n) if L is in NSPACE(log n), c) L is in DSPACE(log n) if L is in DSPACE(log n).
Proof a) It suffices to show that a log-space reduction cannot take more than polyno-mial time, so the result follows from Lemma 13.1(b). In proof, note that the output tape contents cannot influence the computation, so the product of the number of states, storage tape contents, and positions of the input and storage tape heads is an upper bound on the number of moves that can be made before the log-space transducer must enter a loop, which would contradict the assumption that it always halts. If the storage tape has length log /?, the bound is easily seen to be poly-nomial in n.
There is a subtlety involved in the proofs of (b) and (c). We prove only (c), the proof of (b) being essentially the same as for (c).
KTUNOTES.IN Downloaded from Ktunotes.in 13.1 | POLYNOMIAL TIME AND SPACE 323 c) Let Mj be the log-space transducer that reduces L to L, and let M 2 be a log n space bounded TM accepting L. On input x of length n, M x produces an output of length bounded by n c for some constant c. Since the output cannot be written in log n space, M x and M 2 cannot be simulated by storing the output of M x on a tape. Instead the output of M x can be fed directly to M 2 , a symbol at a time.
This works as long as M 2 moves right on its input. Should M 2 move left, M 2 must be restarted to determine the input symbol for M 2 , since the output of Mj is not saved.
We construct M 3 to accept L as follows. One storage tape of M 3 holds the input position of M 2 in base 2 C . Since the input position cannot exceed n c , this number can be stored in log n space. The other storage tapes ofM 3 simulate the storage tapes of M x and M 2 . Suppose at some time M 2 's input head is at position z, and M 2 makes a move left or right. M 3 adjusts the state and storage tapes ofM 2 accordingly. Then M 3 restarts the simulation ofM t from the beginning, and waits until Mj has produced i — 1 or i + 1 output symbols ifM 2 's input head moved left or right, respectively. The last output symbol produced is the new symbol scanned by M 2 's head, so M 3 is ready to simulate the next move ofM 2 . As special cases, if i = 1 and M 2 moves left, we assume that M 2 next scans the left endmarker, and if Mj halts before producing i + 1 output symbols (when M 2 moves right), we assume that M 2 next scans the right endmarker. M 3 accepts its own input when-ever M 2 accepts its simulated input. Thus M 3 is a log n space bounded TM accepting L.
Lemma 13.3 The composition of two log-space (resp. polynomial-time) reduc-tions is a log-space (resp. polynomial-time) reduction.
Proof An easy generalization of the constructions in Lemmas 13.1 and 13.2.
Complete problems As we have mentioned, no one knows whether ,Ar0 includes languages not in so the issue of proper containment is open. One way to find a language in j\fg> — ^ is to look for a "hardest" problem in „V'&. Intuitively, a language Lq is a hardest problem if every language in Ar& is reducible to Lq by an easily comput-able reduction. Depending on the exact kind of reducibility, we can conclude certain things about Lq. For example, if all ofJ is log-space reducible to Lq, we can conclude that if L0 were in ^, then & would equal .
\r&. Similarly, if Lq were in DSPACE(log n\ then ^V0> = DSPACE(log n). If all of were polynomial-time reducible to Lq, then we could still conclude that if Lq were in ^, then ^ would equal A but we could not conclude from the statement Lq is in DSPACE(log n) that Jf& = DSPACE(log n).
We see from the above examples that the notion of "hardest" may depend on the kind of reducibility involved. That is, there may be languages Lq such that all KTUNOTES.IN Downloaded from Ktunotes.in 324 INTRACTABLE PROBLEMS languages in Jf0> have polynomial-time reductions to Lq, but not all have log-space reductions to Lq. Moreover, log-space and polynomial-time reductions do not exhaust the kinds of reductions we might consider. With this in mind, we define the notion of hardest (complete) problems for a general class of languages with respect to a particular kind of reduction. Clearly the following generalizes to an arbitrary type of reduction.
Let ^ be a class of languages. We say language L is completefor <6 with respect to polynomial-time (resp. log-space) reductions if L is in and every language in ^ is polynomial-time (resp. log-space) reducible to L. We say L is hard for with respect to polynomial-time (resp. log-space) reductions if every language in # is polynomial-time (resp. log-space) reducible to L, but L is not necessarily in Two special cases are of primary importance, and we introduce shorthands for them. L is NP-complete (NP-hard) if L is complete (hard) for with respect to log-space reductions.! L is PSPACE-complete (PSPACE-hard) if L is complete (hard) for PSPACE with respect to polynomial time reductions.
In order to show a first language Lq to be NP-complete, we must give a log-space reduction of each language in Jf0> to Lq. Once we have an NP-complete problem Lq, we may prove another language L x in to be NP-complete by exhibiting a log-space reduction of Lq to L l9 since the composition of two log-space reductions is a log-space reduction by Lemma 13.3. This same technique will be used for establishing complete problems for other classes as well.
13.2 SOME MP-COMPLETE PROBLEMS The significance of the class of NP-complete problems is that it includes many problems that are natural and have been examined seriously for efficient solutions.
None of these problems is known to have a polynomial-time solution. The fact that if any one of these problems were in & all would be, reinforces the notion that they are unlikely to have polynomial-time solutions. Moreover, if a new problem is proved NP-complete, then we have the same degree of confidence that the new problem is hard that we have for the classical problems.
The first problem we show to be iVP-complete, which happens to be histor-ically the first such problem, is satisfiability for Boolean expressions. We begin by defining the problem precisely.
The satisfiability problem A Boolean expression is an expression composed of variables, parentheses, and the operators a (logical AND), v (logical OR) and — i (negation). The precedence of these operators is — i highest, then a , then v . Variables take on values 0 (false) and 1 (true); so do expressions. If E x and E2 are Boolean expressions, then the t Many authors use the term "./VP-complete" to mean "complete for with respect to polynomial time reductions," or in some cases, "with respect to polynomial time Turing reductions." KTUNOTES.IN Downloaded from Ktunotes.in 13.2 | SOME JVP-COMPLETE PROBLEMS 325 value of E x a E 2 is 1 if both E x and E2 have value 1, and 0 otherwise. The value of E l v E 2 is 1 if either E x or E 2 has value 1, and 0 otherwise. The value of ~i E x is 1 if E x is 0 and 0 if E x is 1. An expression is satisfiable if there is some assignment of O's and Vs to the variables that gives the expression the value 1. The satisfiability problem is to determine, given a Boolean expression, whether it is satisfiable.
We may represent the satisfiability problem as a language L^at as follows.
Let the variables of some expression be x lf x 2 , ...,xm for some m. Code x t as the symbol x followed by i written in binary. The alphabet of L sat is thus {a, v,-n,(,),x,0, 1}.
The length of the coded version of an expression of n symbols is easily seen to be no more than \n log 2 «l, since each symbol other than a variable is coded by one symbol, there are no more than \n/2] different variables in an expression of length n, and the code for a variable requires no more than 1 + riog 2 nl symbols. We shall henceforth treat the word in L^ at representing an expression of length n as if the word itself were of length n. Our results will not depend on whether we use n or n log n for the length of the word, since log(>? log n) < 2 log n, and we shall deal with log-space reductions.
A Boolean expression is said to be in conjunctive^ normalform (CNF) if it is of the form E x a E 2 a -• • a Ek , and each called a clause (or conjunct), is of the form a f , v a i2 v • • • v a fr ., where each au is a literal, that is, either x or — i x, for some variable x. We usually write x instead of ~i x.
For example, (x y vx 2 )a (x x v x 3 v x4 ) a x 3 is in CNF. The expression is said to be in 3-CNF if each clause has exactly three distinct literals. The above example is not in 3-CNF because the first and third clauses have fewer than three literals.
Satisfiability is /VP-complete We begin by giving a log-space reduction of each language in J t^ to L^,.
Theorem 13.1 The satisfiability problem is NP-complete.
Proof The easy part of the proof is that JU a , is in X0>. To determine if an expression of length n is satisfiable, nondeterministically guess values for all the variables and then evaluate the expression. Thus L^, is in To show that every language in j\r& is reducible to L^, for each NTM M that is time bounded by a polynomial p(n\ we give a log-space algorithm that takes as input a string x and produces a Boolean formula Ex that is satisfiable if and only if M accepts x. We now describe Ex .
Let #/?0 #/?i# •• #/?p (n) be a computation of M, where each ft is an ID consisting of exactly p(n) symbols. If acceptance occurs before the p(n)lh move, we allow the accepting ID to repeat, so each computation has exactly p(n) + 1 ID's.
t "Conjunctive" is an adjective referring to the logical AND operator (conjunction). The term disjunc-tive is similarly applied to logical OR.
KTUNOTES.IN Downloaded from Ktunotes.in 326 INTRACTABLE PROBLEMS In each ID we group the state with the symbol scanned to form a single composite symbol. In addition, the composite symbol in the zth ID contains an integer m indicating the move by which the (i + l)st ID follows from the ith. Numbers are assigned to moves by arbitrarily ordering the finite set of choices that M may make given a state and tape symbol.
For each symbol that can appear in a computation and for each i, 0 < i < (p(n) + l) 2 , we create a Boolean variable cix to indicate whether the zth symbol in the computation is X. (The Oth symbol in the computation is the initial #.)The expression Ex that we shall construct will be true for a given assignment to the cix s if and only if the cix s that are true correspond to a valid computation.
The expression Ex states the following: 1 ) The Cix s that are true correspond to a string of symbols, in that exactly one C iX is true for each i.
2) The ID f}0 is an initial ID of M with input x.
3) The last ID contains a final state.
4) Each ID follows from the previous one by the move of M that is indicated.
The formula Ex is the logical AND of four formulas, each enforcing one of the above conditions. The first formula, stating that for each i between 0 and (p(n) + l) 2 -1, exactly one C ix is true is For a given value of i the term \/ x CiX forces at least one CiX to be true and ~~ 1 \Jx + y (Qv a Ct) forces at most one to be true.
Let x = a l a 2 m " a n - The second formula expressing the fact that f$0 is an initial ID is in turn the AND of the following terms.
i) c0# AcplB)+1>t . The symbols in positions 0 and p(n) + 1 are #.
ii) c 1Vl v c, y , v- v c iyjk , where Y,, Y2 , Y k are all the composite symbols that represent tape symbol a x , the start state q0 , and the number of a legal move of M in state q0 reading symbol a x . This clause states that the first symbol of/?0 is correct.
iii) f\ 2 <i< n c iQt - The 2nd through /7th symbols of /i0 are correct.
iv ) A«<'<pin) cin- Tne remaining symbols of fi0 are blank.
The third formula says that the last ID has an accepting state.
It can be written , v p{n)(p(n)+ l)<i<{p(n)+ 1)2 \ X in F V V where F is the set of composite symbols that include a final state.
KTUNOTES.IN Downloaded from Ktunotes.in 13.2 | SOME JVP-COMPLETE PROBLEMS 327 To see how to write the fourth formula stating that each ID ft, i > 1, follows from ft_ j by the move appearing in the composite symbol of ft_ l5 observe that we can essentially deduce each symbol of ft from the corresponding symbol of ft _ ! and the symbols on either side (one of which may be #). That is, the symbol in ft is the same as the corresponding symbol in ft _ j unless that symbol had the state and move, or one of the adjacent symbols had the state and move, and the move caused the head position to shift to where the symbol of ft in question was.
Note that should this symbol of ft be the one representing the state, it also represents an arbitrary legal move of M, so there may be more than one legal symbol. Also note that if the previous ID has an accepting state, the current and previous ID's are equal.
We can therefore easily specify a predicate f(W, X, Y, Z) that is true if and only if symbol Z could appear in position j of some ID given that W, X, and Y are the symbols in positions j — I, j, and j + 1 of the previous ID [W is # if j = 1 and Y is # if / = p(n)]. It is convenient also to declare /(H7 , #, X, #) to be true, so we can treat the markers between ID's as we treat the symbols within ID's.
We can now express the fourth formula as It is easy, given an accepting computation of M on x to find truth values for the clV 's that make Ex true. Just make cix true if and only if the ith symbol of the computation is X. Conversely, given an assignment of truth values making Ex true, the four formulas above guarantee that there is an accepting computation of M on x. Note that even though M is nondeterministic, the fact that a move choice is incorporated into each ID guarantees that the next state, symbol printed, and direction of head motion going from one ID to the next will all be consistent with some one choice of M.
Furthermore, the formulas composing Ex are of length 0(p 2 (n)) and are sufficiently simple that a log-space TM can generate them given x on its input. The TM only needs sufficient storage to count up to (p(n) + l) 2 . Since the logarithm of a polynomial in n is some constant times log n, this can be done with 0(log n) storage. We have thus shown that every language in NP is log-space reducible to L^,, proving that L^, is NP-complete.
We have just shown that satisfiability for Boolean expressions is NP-complete. This means that a polynomial-time algorithm for accepting L,M could be used to accept any language in „V'&. Let L be the language accepted by some p(n) time-bounded nondeterministic Turing machine M, and let A be the log-space (hence polynomial-time) transducer that converts x to £ v , where Ex is satisfiable if and only if M accepts x. Then A combined with the algorithm for L^a , A ( V (p(n) + 1)2 \ W.AW.Zsueh that f(W,X,Y,Z) KTUNOTES.IN Downloaded from Ktunotes.in 328 INTRACTABLE PROBLEMS — Algorithm constructing Ex from x Algorithm for L sat Deterministic polynomial time algorithm for arbitrary language in^V.y Fig. 13.1 Algorithm for arbitrary set L in Jf& given algorithm for L sa, .
as shown in Fig.
13.1 is a deterministic polynomial-time algorithm accepting L Thus the existence of a polynomial-time algorithm for just this one problem, the satisfiability of Boolean expressions, would imply 0> = jV0>.
Restricted satisfiability problems that are /VP-complete Recall that a Boolean formula is in conjunctive normal form (CNF) if it is the logical AND of clauses, which are the logical OR of literals. We say the formula is in k-CNF if each clause has exactly k literals. For example, (x v y) a(x vz)a (y v z) is in 2-CNF.
We shall now consider two languages, Lcsat , the set of satisfiable Boolean formulas in CNF, and L3sat , the set of satisfiable Boolean formulas in 3-CNF. We give log-space reductions of L^, to Lc sat and Lcsat to L3sal , showing the latter two problems NP-complete by Lemma 13.3. In each case we map an expression to another expression that may not be equivalent, but is satisfiable if and only if the original expression is satisfiable.
Theorem 13.2 Ltsat , the satisfiability problem for CNF expressions, is NP-complete.
Proof Clearly Ltsat is in since Iat is. We reduce to as follows. Let E be an arbitrary Boolean expression of length n.| Certainly, the number of variable occurrences in E does not exceed n, nor does the number of a and v operators.
Using the identities -i(Ei aE 2 ) = -i(E 1 )v-i(E 2 ), -i(E 1 vE 2 ) = -!(£,) a^(E 2 ), (13.1) i \E^ = E u we can transform E to an equivalent expression £', in which the —i operators are applied only to variables, never to more complex expressions. The validity of Eqs.
(13.1) may be checked by considering the four assignments of values 0 and 1 to£i t Recall that the length of a Boolean expression is the number of characters, not the length of its code, and recall that this difference is of no account where log-space reduction is concerned.
KTUNOTES.IN Downloaded from Ktunotes.in 13.2 | SOME JVP-COMPLETE PROBLEMS 329 and £ 2 . Incidentally, the first two of these equations are known as DeMorgan's laws.
The transformation can be viewed as the composition of two log-space trans-formations. As a result of the first transformation, each negation symbol that immediately precedes a variable is replaced by a bar over the variable, and each closing parenthesis whose matching opening parenthesis is immediately preceded by a negation sign is replaced by )r~. The symbol r~ indicates the end of the scope of a negation. This first transformation is easily accomplished in log-space using a counter to locate the matching parentheses.
The second transformation is accomplished by a finite automaton that scans the input from left to right, keeping track of the parity (modulo 2 sum) of the active negations, those whose immediately following opening parenthesis but not closing parenthesis has been seen. When the parity of negations is odd, x is replaced by x, x by x, v by a , and a by v . The symbols —i and r— are deleted.
That this transformation is correct may be proved using (13.1) by an easy induc-tion on the length of an expression. We now have an expression E in which all negations are applied directly to variables.
Next we create £", an expression in CNF that is satisfiable if and only if E' is satisfiable. Let V x and V 2 be sets of variables, with Vx c V 2 . We say an assignment of values to V 2 is an extension of an assignment of values to V x if the assignments agree on the variables of Vx . We shall prove by induction on r, the number of a 's and v 's in an expression £', all of whose negations are applied to variables, that if \E' | = n, then there is a list of at most n clauses, F l9 F 2 , FkJ over a set of variables that includes the variables of E' and at most n other variables, such that £' is given value 1 by an assignment to its variables if and only if there is an extension of that assignment that satisfies F x a F 2 a • • • a Fk .
Basis r = 0. Then E! is a literal, and we may take that literal in a clause by itself to satisfy the conditions.
Induction If E = E x aE 2 , let Fu F 2 , Fk and G X ,G 2 , G z be the clauses for E x and E 2 that exist by the inductive hypothesis. Assume without loss of genera-lity that no variable that is not present in E appears both among the F's and among the G's. Then F u F 2 , .
.
.
, FkJ G x , G 2 , .
.
.
, G t satisfies the conditions for E.
If E = E x v £ 2 , let the F's and G's be as above, and let y be a new variable.
Then yv F i9 yv F 2 , yw Fk , j/v G l9 yvG 2 , . .., j/vG z satisfies the conditions.
In proof, suppose an assignment of values satisfies E. Then it must satisfy E x or £ 2 . If the assignment satisfies E x , then some extension of the assignment satisfies F x , F 2 , Fk . Any further extension of this assignment that assigns y = 0 will satisfy all the clauses for E.
If the assignment satisfies £ 2 , a similar argument suffices. Conversely, suppose all the clauses for E are satisfied by some assign-ment. If that assignment sets y = 1, then all of G u G 2 , .
.
.
, G z must be satisfied, so £ 2 is satisfied. A similar argument applies ify = 0. The desired expression E" is all the clauses for E connected by a 's.
KTUNOTES.IN Downloaded from Ktunotes.in 330 INTRACTABLE PROBLEMS To see that the above transformation can be accomplished in log-space, con-sider the parse tree for E. Let yk be the variable introduced by the ith v . The final expression is the logical AND of clauses, where each clause contains a literal of the original expression. In addition, if the literal is in the left subtree of the ith v , then the clause also contains y t . If the literal is in the right subtree of the ith v , then the clause contains y. The input is scanned from left to right. Each time a literal is encountered, a clause is emitted. To determine which y^s and y ;'s to include in the clause, we use a counter of length log n to remember our place on the input. We then scan the entire input, and for each v symbol, say the ith from the left, we determine its left and right operands, using another counter of length log n to count parentheses. If the current literal is in the left operand, generate y t ; if it is in the right operand, generate yh and if in neither operand, generate neither y t nor y f .
We have thus reduced each Boolean expression £ to a CNF expression E" that is in L^, if and only if E is in L^at . Since the reduction is accomplished in log-space, the AfP-completeness of Z^at implies the ATP-completeness of Le Sal .
Example 13.1 Let E = —1(—l(x x V X 2 ) A (—IX! V X 2 )).
Applying DeMorgan's laws yields E = (Xi vx 2)v(x t AX3 ).
The transformation to CNF introduces variables y Y and y 2 to give E" = (xj vy,v y 2 ) a (x 2 v y x v y 2 ) a (x x v y2 ) a (x 3 v y2 ) Theorem 13.3 L3sat , the satisfiability problem for 3-CNF expressions, is NP-complete.
Proof Clearly, L3sal is in jV'sP, since L^at is. Let E = P x a F 2 a • • • a Fk be a CNF expression. Suppose some clause F, has more than three literals, say Pi = a i v a2 v " v fyy £ >7>.
Introduce new variables y u y2 , y^ 3 , and replace F t by (a, v a 2 v y , ) a (a 3 v y l v y 2 ) a (a4 v y 2 v j/3 ) a • • • a (a,- 2 v jv_ 4 v y( _ 3 ) a (a, _ { v v yf _ 3 ).
(13.2) Then F, is satisfied by an assignment if and only if an extension of that assignment satisfies (13.2). An assignment satisfying F- must have a7 - = 1 for some Thus assume that the assignment gives literals o^, a 2 , a7 ! the value 0 and gcj the value 1. Then ym = 1 for m < j - 2 and ym = 0 for m >j -1 is an extension of the assignment satisfying (13.2).
KTUNOTES.IN Downloaded from Ktunotes.in 13.2 | SOME WP-COMPLETE PROBLEMS 331 Conversely, we must show that any assignment satisfying (13.2) must have (Xj = 1 for some j and thus satisfies Ff . Assume to the contrary that the assignment gives all the am's the value 0. Then since the first clause has value 1, it follows that y x = 1. Since the second clause has value 1, y2 must be 1, and by induction, ym = 1 for all m. But then the last clause would have the value 0, contradicting the assumption that (13.2) is satisfied. Thus any assignment that satisfies (13.2) also satisfies F,.
The only other alterations necessary are when F£ consists of one or two literals. In the latter case replace ol x v a 2 by (o^ v a 2 v y) a (a t v a 2 v y), where y is a new variable, and in the former case an introduction of two new variables suffices.
Thus E can be converted to a 3-CNF expression that is satisfiable if and only if E is satisfiable. The transformation is easily accomplished in log-space. We have thus a log-space reduction of L^, to L3sat and conclude that L3sat is NP-complete.
The vertex cover problem It turns out that 3-CNF satisfiability is a convenient problem to reduce to other problems in order to show them NP-complete, just as Post's correspondence problem is useful for showing other problems undecidable. Another NP-complete problem that is often easy to reduce to other problems is the vertex cover problem.
Let G = (V, E) be an (undirected) graph with set of vertices V and edges E. A subset A ^ V is said to be a vertex cover of G if for every edge (v, w) in £, at least one of v or w is in A. The vertex cover problem is: Given a graph G and integer /c, does G have a vertex cover of size k or less?
We may represent this problem as a language L^, consisting of strings of the form: k in binary, followed by a marker, followed by the list of vertices, where v t is represented by v followed by i in binary, and a list of edges, where (vh Vj) is represented by the codes for v { and Vj surrounded by parentheses.
consists of all such strings representing k and G, such that G has a vertex cover of size k or less.
Theorem 13.4 L^, the vertex cover problem, is NP-complete.
Proof To show L^ c in Jf&, guess a subset of k vertices and check that it covers all edges. This may be done in time proportional to the square of the length of the problem representation.
is shown to be NP-complete by reducing 3-CNF satisfiability to U, c .
Let F = Fj aF 2 a--- AFq be an expression in 3-CNF, where each F, is a clause of the form (an val2 va l3 ), each being a literal. We construct an un-directed graph G = (K, E) whose vertices are pairs of integers (Uj), 1 < i < q, 1 < j < 3. The vertex (i, j) represents the jth literal of the ith clause. The edges of the graph are !) [('">./)> (\ )] provided; + k, and 2) [(UM. 01 if «i; = -»«• KTUNOTES.IN Downloaded from Ktunotes.in 332 INTRACTABLE PROBLEMS Each pair of vertices corresponding to the same clause are connected by an edge in (1) . Each pair of vertices corresponding to a literal and its complement are con-nected by an edge in (2).
G has been constructed so that it has a vertex cover of size 2q if and only ifF is satisfiable. To see this, assume F is satisfiable and fix an assignment satisfying F.
Each clause must have a literal whose value is 1. Select one such literal for each clause. Delete the q vertices corresponding to these literals from V. The remaining vertices form a vertex cover of size 2q. Clearly for each i, only one vertex of the form (i, j) is missing from the cover, and hence each edge in (1) is incident upont at least one vertex in the cover. Since edges in (2) are incident upon two vertices corresponding to some literal and its complement, and since we could not have deleted both a literal and its complement, one or the other of these vertices is in the cover. Thus we indeed have a cover of size 2q.
Conversely, assume we have a vertex cover of size 2q. For each i the cover must contain all but one vertex of the form (Uj), for if two such vertices were missing, an edge [(i,j), (i, k)] would not be incident upon any vertex in the cover.
For each i assign value 1 to the literal a {j corresponding to the vertex not in the cover. There can be no conflict, because two vertices not in the cover cannot correspond to a literal and its complement, else there would be an edge in group (2) not incident upon any vertex of the cover. For this assignment F has value 1.
Thus F is satisfiable. The reduction is easily accomplished in log-space. We can essentially use the variable names in the formula F as the vertices of G, appending two bits for the j-component in vertex Edges of type (1) are generated directly from the clauses, while those of type (2) require two counters to consider all pairs of literals. Thus we conclude that is NP-complete.
Example 13.2 Consider the expression F = (Xj V x 2 v X 3 ) A (Xj v x 2 v x4 ) a (x 2 v x 3 v x 5 ) a (x 3 V x4 V x 5 ).
The construction of Theorem 13.4 yields the graph of Fig.
13.2. x { = 1, x 2 = 1, x 3 = 1, x4 = 0 satisfies F and corresponds to the vertex cover [1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 3], [4, 1], and [4, 3].
The Hamilton circuit problem The Hamilton circuit problem is: Given a graph G, does G have a path that visits each vertex exactly once and returns to its starting point? The directed Hamilton circuit problem is the analogous problem for directed graphs. We represent these problems as languages and Iy h by encoding graphs as in the vertex cover problem.
t An edge (i\ w) is incident upon v and w and no other vertices.
KTUNOTES.IN Downloaded from Ktunotes.in 13.2 | SOME NP-COMPLETE PROBLEMS 333 Fig. 13.2 Graph constructed by Theorem 13.4.
Double circles indicate vertices in set cover.
Theorem 13.5 L^, the directed Hamilton circuit problem, is NP-complete.
Proof To show L dh in jV'Pf, guess a list of arcs and verify that the arcs form a simple cyclef through all the vertices. To show Iy h is iVP-complete, we reduce 3-CNF satisfiability to Ldh .
Let F = F x aF2 a-- aFq be an expression in 3-CNF, where each F, is a clause of the form (an va i2 val3 ), each a l7 being a literal. Let x u x„ be the variables of F. We construct a directed graph G that is composed of two types of subgraphs. For each variable x, there is a subgraph //, of the form shown in Fig.
13.3(a), where m, is the larger of the number of occurrences of x, and 5c f in F. The H-s are connected in a cycle, as shown in Fig. 13.3(b). That is, there are arcs from d t to ai+ !, for 1 < i < n and an arc from dn to a x .
Suppose we had a Hamilton circuit for the graph of Fig. 13.3(b). We may as well suppose it starts at a x . If it goes next to b XOy we claim it must then go to c 10 , else c 10 could never appear on the cycle. In proof, note that both predecessors of c 10 are already on the cycle, and for the cycle to later reach c 10 it would have to repeat a vertex. (This argument about Hamilton circuits occurs frequently in the proof. We shall simply say that a vertex like c 10 "would become inaccessible") Similarly, we may argue that a Hamilton circuit that begins a x ,b x0 must continue c ioy ^1 1' c \ i' ^i2> C! 2 , ... If the circuit begins a x , c 10 , then it descends the ladder of Fig. 13.3(a) in the opposite way, continuing b 10 , c x ,, b x x , c X2 , ... Likewise we may argue that when the circuit enters each H ( in turn it may go from a, to either b i0 or cl0 , but then its path through H t is fixed; in the former case it descends by the arcs cu i> an^ in the latter case by the arcs b i} ciJ+ x .ln what follows, it helps to think of the choice to go from a i to b i0 as making x, true, while the opposite choice makes x f false. With this in mind, observe that the graph of Fig. 13.3(b) has t A simple cycle has no repeated vertex.
KTUNOTES.IN Downloaded from Ktunotes.in 334 INTRACTABLE PROBLEMS (O Fig. 13.3 Graphs concerned with directed Hamilton circuits.
exactly 2n Hamilton circuits that correspond in a natural way to the 2n assign-ments to the variables of F.
For each clause F} we introduce a subgraph lp shown in Fig.
13.3(c). /,-has the properties that if a Hamilton circuit enters it at rp it must leave at u} \ if it enters at sp it must leave at vy, and if it enters at tj9 it must leave at vv,-. In proof, suppose by symmetry that the circuit enters Ij at ry case 1 The next two vertices on the circuit are Sj and t} . Then the circuit must continue with vv,, and if it leaves at vv,. or vj9 Uj is inaccessible. Thus in this case it leaves at u} .
KTUNOTES.IN Downloaded from Ktunotes.in 13.2 | SOME NP-COMPLETE PROBLEMS 335 case 2 The next two vertices on the circuit are Sj and Vj. If the circuit does not next go to ttj, then Uj will be inaccessible. If after u} it goes to wp vertex t} cannot appear on the circuit because its successors are already on the circuit. Thus in this case the circuit also leaves by ur case 3 The circuit goes directly to u} . If it next goes to vv,, the circuit cannot include t jt because its successors are already used. So again it must leave by Uj.
Observe that the above argument holds even though the circuit may enter Ij more than once. Finally, the graph /, has the additional property that entering I} at rj9 sp or tjy it can traverse all six vertices before exiting.
To complete the construction of the graph, connect the 7/s to the 77,'s as follows. Suppose the first term in F, is x,. Then pick some cip that has not yet been connected to any Ik and introduce an arc from cip to r} and from u} to b ifP + j. If the first term is xh pick an unused bip and introduce arcs b ip -+rj and Uj-citP+1 .
Make analogous connections with Sj and Vj for the second term of F j9 and analo-gous connections with t} and Wj for the third term. Each 77, was chosen sufficiently long that enough pairs of 6,/s and cf/s are available to make all the connections.
If the expression F is satisfiable, we can find a Hamilton circuit for the graph as follows. Let the circuit go from a t to bi0 if x,- is true in the satisfying assignment, and from a { to ci0 otherwise. Then, ignoring the 7y's, we have a unique Hamilton circuit for the subgraph of Fig.
13.3(b). Now, whenever the constructed circuit uses an arc bik -» c i k+ or cik -> b itk + u and b ik or c, fc , respectively, has an arc to an Ij subgraph that has not yet been visited, visit all six vertices of Ip emerging at Ci,k+i or biM+lJ respectively. The fact that F is satisfiable implies that we can traverse l} for all j.
Conversely, we must show that the existence of a Hamilton circuit implies F is satisfiable. Recall that in any Hamilton circuit an l } entered at rj9 sj9 or ti must be left at Uj, Vj, or wj9 respectively. Thus as far as paths through the 77,'s are con-cerned, connections to an Ij look like arcs in parallel with an arc bik -> cLk+ j or cik~~ bi.k+i- N excursions to the 7/s are ignored, it follows that the circuit must traverse the 7/,'s in one of the 2 n ways which are possible without the 7/s; that is, it may follow the arc a x -> b i0 or a x -> ci0 for 1 < i < n. Each set of choices determines a truth assignment for the x^s.
If one set of choices yields a Hamilton circuit, including the 7/s, then the assignment must satisfy all the clauses. For example, if we reach Ij from b ik in the circuit, then x, is a term in F; -, and it must be that the circuit goes from a { to cl0 , which corresponds to the choice x, = 0. Note that if the circuit goes from a t to b i0 , then it must traverse b i k+l before c i k + x and we could not traverse Ij between b ik and cifk+l , as bik+l could never be included in the circuit.
As a last remark, we must prove we have a log-space reduction. Given F, we can list the vertices and arcs of 77, simply by counting occurrences of x, and x f . We can list the connections between the 77,'s and 7/s easily as well. Given a term like x, in Fj, we can find a free pair of vertices in 77, to connect to Ij by counting KTUNOTES.IN Downloaded from Ktunotes.in 336 INTRACTABLE PROBLEMS occurrences of x f in F l9 F 2 , Fj- V As no count gets above the number of variables or clauses, log n space is sufficient, where n is the length of F.
Example 13.3 Let F be (Xj V X 2 V X 3 ) A (Xj V x2 v x 3 ).
The graph constructed from F by Theorem 13.5 is shown in Fig. 13.4. A Hamilton circuit corresponding to the assignment Xj = 1, x 2 = 0, x 3 = 0 is drawn in heavy lines.
Finally we show that the Hamilton circuit problem is ATP-complete by reduc-ing the directed Hamilton circuit problem to it.
Theorem 13.6 Lh , the Hamilton circuit problem for undirected graphs, is NP-complete.
Proof To show that Lh , is in J/'&, guess a list of the edges and verify that they form a Hamilton circuit. To show Lh iVT-complete we reduce Ldh to it. Let G = (V, E) be a directed graph. Construct an undirected graph G' with vertices t?0 , v u and v 2 for each v in V, and edges 1) (yo> v i) f°r eacn y n K 2) (f i, ^2) f°r eacn y ^n K 3) (^2» wo) if an^ only if v -> w is an arc in E.
Each vertex in K has been expanded into three vertices. Vertices with subscript 1 have only two edges, and since a Hamilton circuit must visit all vertices, the subscript of the vertices in any Hamilton circuit of G' must be in the order 0, 1, 2, 0, 1, ... or its reverse. Assume the order is 0, 1, 2, ... Then the edges whose subscript goes from 2 to 0 correspond to a Hamilton circuit in G. Conversely, a Hamilton circuit in G may be converted to a Hamilton circuit in G' by replacing an arc v -> w by the path from v0 to to v 2 to w0 . Thus G' has a Hamilton circuit if and only if G has a Hamilton circuit. The reduction of G to G' is easily accom-plished in log-space. Thus we conclude that Lh is NP-complete.
Integer linear programming Most known NP-complete problems are easily shown to be in jV'efi, and only the reduction from a known NP-complete problem is difficult. We shall now give an example of a problem where the opposite is the case.
It is easy to prove that integer linear programming is NF-hard but difficult to show it is in The integer linear programming problem is: Given an m x n matrix of integers A and a column vector b of n integers, does there exist a column vector of integers x such KTUNOTES.IN Downloaded from Ktunotes.in 13.2 | SOME WP-COMPLETE PROBLEMS 337 Fig. 13.4 Graph constructed for Example 13.3.
KTUNOTES.IN Downloaded from Ktunotes.in 338 INTRACTABLE PROBLEMS that Ax > b? The reader may formalize this problem as a language in an obvious way, where the words of the language are the elements of A and b written in binary.
Lemma 13.4 Integer linear programming is iVP-hard.
Proof We reduce 3-CNF satisfiability to integer linear programming.
Let E = Fj a F 2 a • • • a Fq be an expression in 3-CNF, and let x u x 2 , .
.
• , x„ be the variables of E. The matrix A will have a column for each literal x, or x f , 1 < i < n.
We may thus view the inequality Ax > b as a set of linear inequalities among the literals. For each /, 1 < i < n, we have the inequalities x, + Xi > 1, x, > 0, — x, — x, > — 1, x, > 0, which has the effect of saying that one of x, and x, is 0, the other is 1. For each clause a 1 v a 2 v a 3 , we have the inequality ct l -f a 2 -f a 3 > 1, which says that at least one literal in each clause has value 1. It is obvious that A and b can be constructed in log-space and the inequalities are all satisfied if and only if E is satisfiable. Thus linear integer programming is iVP-hard.
To show integer linear programming is in . 4 we may guess a vector x and check that Ax > b. However, if the smallest solution has elements that are too large, we may not be able to write x in polynomial time. The difficulty is to show that the elements of x need not be too large, and for this we need some concepts from linear algebra, specifically determinants of square matrices, the rank of a matrix, linear independence of vectors, and Cramer's rule for solving simultaneous linear equations, with all of which we expect the reader to be familiar.
In what follows, we assume matrix A and vector b form an instance of the integer linear programming problem and that A has m rows and n columns. Let a be the magnitude of the largest element of A or b. Note that the number of bits needed to write out A and b is at least mn + log 2 a, and we shall use this quantity as a lower bound on the input size; our nondeterministic solution finder will work in NTIME(p(mrt + log2 a)) for some polynomial p.
Further, we define a„ for 1 < i < m, to be the vector of length n consisting of the zth row of A. We let b { be the zth element of b and we let x = (x„ x 2 , .
. ., x„) be a vector of unknowns. We use | / 1 for the magnitude of integer i and det B for the determinant of matrix B. A series of technical lemmas is needed.
Lemma 13.5 If £ is a square submatrix of A, then |det£| < (aq)9 , where q = max (m, n).
Proof Recall that the determinant of a k x k matrix is the sum or difference of k\ terms, each of which is the product of k elements. Therefore, if B is a k x k KTUNOTES.IN Downloaded from Ktunotes.in 13.2 | SOME 7VP-C0MPLETE PROBLEMS 339 submatrix, k l(xk is an upper bound on | det B \ . As k ! < kk and k < g, we have our lemma.
.
" " Lemma 13.6 Let A have rank r.t If r < rc, then there is an integer vector z = (zu z 2 , .
.
z„), z not identically zero, such that Az = 0 (0 is a vector of all O's) and no Zj exceeds ( b, x > 0, then there is a solution in which for some i, b t < a 4x < b t 4- a, where a is the magnitude of the largest element of A.
Proof Let x0 be a solution to Ax > b. Suppose a,x 0 > • + a for all i. Adding or subtracting 1 from some component of x 0 must reduce some product a,x 0 .
Furthermore, no product can decrease by more than a. Thus the new x is also a solution. The process cannot be repeated indefinitely without obtaining a solution x for which there is an i such that b t < a,x < b-t + a.
Theorem 13.7 Integer linear programming is iVP-complete.
Proof By Lemma 13.4 we have to show only that the problem is in .
\r&. We begin by guessing the signs of the x f 's in some hypothetical solution and adding n constraints x i < (>)0 depending on the sign guessed. Then guess a row i and a constant c x in the range b x < c-, < b i { + a such that in some solution x 0 , we have a,x0 = ct . Now suppose that after reordering rows if necessary, we have correctly t Recall that the rank of r is equivalently defined as the maximum number of linearly independent rows, the maximum number of linearly independent columns, or the size of the largest square subma-trix with a nonzero determinant.
KTUNOTES.IN Downloaded from Ktunotes.in 340 INTRACTABLE PROBLEMS guessed cv c 2 , ck such that 1) bi < Ci < bi + (aq)2q+ \ and 2) Ax > b has a nonnegative integer solution if and only if a,x = ch 1 < i < k, a,x > b^ k < i < n% has such a solution.
Let A k be the first k rows of A, and let c be the vector (c ly c2 , ck ).
case 1 The rank of A k is less than n. By Lemma 13.6 there is an integer vector z, z 0, none of whose components has a magnitude greater than (ocq) 2q , such that A kz = 0. Therefore, if A k\ Q = c, it follows that A k (\0 + dz) = c for any integer d. If it is also true that a,x0 > b t + (ctq)2q+ 1 for all i > k, then we may repeatedly add or subtract 1 from d until for some j > k, Zj(x0 + dz) drops below b} + (<xq) 2q 1 .
Since z has some nonzero component, the row a, [corresponding to a constraint x, < (> )0] that is all zero except for a one in that component, must have f > k.
Thus some a^fxo 4- dz) for j > k must eventually drop below bj + (a^) 2?+ • Since each component of z is bounded in magnitude by (aq) 2q > changing d by 1 cannot change any a 7 (x0 + dz) by more than ctn(aq) 2q , which is no more than (ocq) 2q 1 .
Therefore a ;(x 0 + dz) > b} . By reordering rows, we may assume j = k + 1 and repeat the above process for k + 1 in place of k.
case 2 The rank of is n. In this case, there is a unique x satisfying A kx = c. By Cramer's rule, the components of x are ratios of two determinants whose magni-tudes do not exceed bj for j > k.
The nondeterministic process of guessing c/s repeats at most n times, and for any sequence of choices requires a number of arithmetic steps that is polynomial in q [since Cramer's rule can be applied in 0(r4 ) arithmetic steps tor xr matrices] applied to integers whose length in binary is polynomial in aq. The arithmetic steps that are multiplication or division of integers can be performed in time proportional to the square of the length of the integers in binaryf and addition and subtraction can be performed in linear time. Thus the entire process takes time that is polynomial in the input length, since that length is at least mn + log 2 a.
Other /VP-complete problems There is a wide variety of other known iVP-complete problems. We shall list some of them here.
t Actually in considerably less time (see Aho, Hopcroft, and Ullman ), although this is of no importance here.
KTUNOTES.IN Downloaded from Ktunotes.in 13.3 | THE CLASS co-„V& 341 1) The Chromatic Number Problem. Given a graph G and an integer k, can G be colored with k colors so that no two adjacent vertices are the same color?
2) The Traveling Salesman Problem. Given a complete graph with weights on the edges, what is the Hamilton circuit of minimum weight? To express this problem as a language, we require the weights to be integers and ask whether there is a Hamilton circuit of weight k or less. This problem is iVP-complete even if we restrict the weights to 0 and 1, when it becomes exactly the Hamil-ton circuit problem.
3) The Exact Cover Problem. Given a collection of sets S l9 S 2 , . .., Sk , all being subsets of some set U, is there a subcollection whose union is U such that each pair of sets in the subcollection is disjoint?
4) The Partition Problem. Given a list of integers i u i 2 , .
. ., ik , does there exist a subset whose sum is exactly (i x + i2 + •• + ik ). Note that this problem appears to be in & until we remember that the length of an instance is not i + i2 + + l k-> DUt tne sum of the lengths of the i-s written in binary or some other fixed base.
Among the NP-complete problems are many, including the ones mentioned in this section, for which serious effort has been expended on finding polynomial-time algorithms. Since either all or none of the iVP-complete problems are in tP, and so far none have been found to be in it is natural to conjecture that none are in More importantly, if one is faced with an iVP-complete problem to solve, it is questionable whether one should even bother to look for a polynomial-time algorithm. We believe one is much better off looking for heuristics that work well on the particular kinds of instances that one is likely to encounter.
Extended significance of /VP-completeness We have inadvertently implied that the only issue regarding NP-complete prob-lems was whether they required polynomial or exponential time. In fact, the true answer could be between these extremes; for example, they could require n,ogn time. If all languages in are log-space or even polynomial-time reducible to L, and L is in, say DTIME(M log "), then every language in .
\ is in DTIME(rcclogn ) for some constant c. In general, if L were log-space or polynomial-time complete for .
S yj>, and L were in DTIME(T(n)), then .i'^c (J DTIME(T(/i c )).
c>0 13.3 THE CLASS co-I P It is unknown whether the class .
t 'i? is closed under complementation. Should it turn out that jV& is not closed under complementation, then clearly & ± J'eP, since & is closed under complementation. There is no NP-complete problem KTUNOTES.IN Downloaded from Ktunotes.in 342 INTRACTABLE PROBLEMS whose complement is known to be in For example, to determine non-satisfiability for a Boolean formula with n variables, it appears necessary to test every one of the 2" possible assignments, even if the algorithm is nondeterministic.
In fact if any NP-complete problem is discovered to have its complement in Jf0>, then Jf& would be closed under complementation, as we show in the next theorem.
Theorem 13.8 Jf& is closed under complementation if and only if the comple-ment of some NP-complete problem is in Jf&.
Proof The "only if part is obvious. For the "if" part let 5 be an NP-complete problem, and suppose S were in Jf&. Since each L in Jf& is log-space reducible to S, each L is log-space reducible to S. Thus L is in Jf&.
We shall define the class co-J^g? to be the set of complements of the lan-guages in Jf0>. The relationship between ^\ Jf0>, zo>-Jf0> and PSPACE is shown in Fig.
13.5, although it is not known for certain that any of the regions except the one labeled e/> are nonempty.
Fig. 13.5 Relations among some language classes.
The problem of primality It is interesting to consider a problem in A'i? such as "nonprimeness" for which there is no known polynomial time algorithmf and furthermore which is not known to be NP-complete.J To test an integer to see if it is not a prime, one simply guesses a divisor and checks. The interesting observation is that the com-plementary problem is in Jf&, which suggests that there may be sets in the intersection of J/'& and co-yK that are not in We now consider a nondeterministic polynomial-time algorithm for testing whether an integer is prime.
Lemma 13.8 Let x and y be integers, with 0 < x, y < p. Then 1) x + y (mod p) can be computed in time 0(log p); t Although Miller presents strong evidence that one exists.
% This is another problem that appears to be in & until one remembers that the size of input p is log 2 P> not p itself.
KTUNOTES.IN Downloaded from Ktunotes.in 13.4 | PSPACE-COMPLETE PROBLEMS 343 2) xy (mod p) can be computed in time 0(log 2 p); 3) xy (mod p) can be computed in time 0(log 3 p).
Proof (1) and (2) are obvious since an integer mod p requires only log p bits. For (3) compute xy by repeated squaring to get x 2 , x4 , x 8 , x 2 ' mod p, where i = [log 2 y J, then multiply the appropriate powers of x to get xy .
We shall, in what follows, make use of Fermat's theorem : p > 2 is a prime if and only if there exists an x of order p — 1, that is, for some x, 1 < x < p, 1) xp ~ 1 = 1 mod p, and 2) x l 1 mod p, for 1 < i < p -1.
Theorem 13.9 The set of primes is in JfgP.
Proof If x = 2, then x is prime. If x = 1 or x is an even integer greater than 2, then x is not prime. To determine if p is prime for odd p greater than 2, guess an x, 0 < x < p, and verify that 1) xp ~ 1 = 1 mod p, and 2) x l 1 mod p for all z, 1 < i < p -1.
Condition (1) is easily checked in 0(log 3 p) steps. We cannot check condition (2) for each i directly since there are too many fs. Instead, guess the prime factoriza-tion of p — 1. Let the factorization be p — 1 = p x p2 pk . Recursively verify that each pj is a prime. Verify that p — 1 is the product of the p/s. Finally verify x(p-i)/pj | mo(j p observe that if x 17 ' 1 = 1 mod p, then the least i satisfying x 1 = 1 mod p must divide p — 1. Furthermore, any multiple of this i, say ai, must also satisfy x ai = 1 mod p. Thus, if there is an / such that x 1 = 1 mod p, then for some p^ xip ~ 1)1pj = 1 mod p.
Assume that the nondeterministic time to recognize that p is prime is bounded by c log4 p. Then we need only observe that k k X c log4py + X c i log 3A + c i IosV ^ c lQg 4p for some sufficiently large constant c.
13.4 PSPACE-COMPLETE PROBLEMS We now show several problems to be complete for PSPACE with respect to polynomial time.
Quantified Boolean formulas Quantified Boolean formulas (QBF) are built from variables, the operators a , v , and —i, parentheses, and the quantifiers 3 ("there exists") and V ("for all"). When KTUNOTES.IN Downloaded from Ktunotes.in 344 INTRACTABLE PROBLEMS defining the QBFs recursively, we find it useful simultaneously to define free occurrences of variables (occurrences to which no quantifier applies), bound occurrences of variables (occurrences to which a quantifier applies), and the scope of a quantifier (those occurrences to which the quantifier applies).
1) If x is a variable, then it is a QBF. The occurrence of x is free.
2) If E x and £ 2 are QBFs, so are -i(£i), (E t ) a (£2 ), and (£ x ) v (E2 ). An occur-rence of x is free or bound, depending on whether the occurrence is free or bound in E x or E 2 . Redundant parentheses can be omitted.
3) If £ is a QBF, then 3x(£) and Vx(£) are QBFs. The scopes of 3x and Vx are all free occurrences of x in £. (Note that there may also be bound occurrences of x in £; these are not part of the scope.) Free occurrences of x in £ are bound in 3x(£) and Vx(£). All other occurrences of variables in £ are free or bound, depending on whether they are free or bound in £.
A QBF with no free variable has a value of either true or false, which we denote by the Boolean constants 1 and 0. The value of such a QBF is determined by replacing each subexpression of the form 3x(£) by £0 v E x and each subexpres-sion of the form Vx(£) by £0 a Eu where £0 and £ t are £ with all occurrences of x in the scope of the quantifier replaced by 0 and 1, respectively. The QBF problem is to determine whether a QBF with no free variables has value true.
Example 13.4 Vx [Vx[3y(x v y)] a —ix] is a QBF. The scope of the inner Vx is the first occurrence of x; the scope of the outer Vx is the second occurrence. To test the truth of the above formula, we must check that Vx[3y(x v yj] a~ix is true when free occurrences of x (that is, the second occurrence only) are set to 0 and also when set to 1. The first clause Vx(3y(x v y)) is seen to be true, as when this x is 0 or 1 we may choose y = 1 to make x v y true. However, ~ix is not made true when x = 1, so the entire expression is false.
Note a Boolean expression £ with variables x x , x 2 , .
.
.
, xk is satisfiable if and only if the QBF 3x x 3x 2 " m 3xk (Ek ) is true. Thus the satisfiability problem is a special case of the problem of whether a QBF is true, which immediately tells us that the QBF problem is NP-hard.
It does not appear that QBF is in .
\ r9 however.
PSPACE-completeness of the QBF problem Lemma 13.9 QBF is in PSPACE.
Proof A simple recursive procedure EVAL can be used to compute the value of a QBF with no free variables. In fact, EVAL will handle a slightly more general problem, where the Boolean constants 0 and 1 have been substituted for some variables. If the QBF consists of a Boolean constant, EVAL returns that constant.
KTUNOTES.IN Downloaded from Ktunotes.in 13.4 | PSPACE-COMPLETE PROBLEMS 345 If the QBF consists of a Boolean operator applied to subformula(s), then EVAL evaluates the subformulas recursively and then applies the operator to the result(s).
If the QBF is of the form 3x(E) or Vx(£), then EVAL replaces all occurrences of x in E that are in the scope of the quantifier by 0 to obtain E0 and then evaluates E0 recursively. Next EVAL replaces the occurrences of x by 1 to obtain Eu and evaluates E x recursively. In the case of 3x(£), EVAL returns the OR of the two results. In the case of Vx(£), EVAL returns the AND.
Since the number of operators plus quantifiers is at most n for a QBF of length n, the depth of recursion is at most n. Using a Turing tape for the stack of activation records (as in Theorem 12.11), we see that the tape need never grow longer than the square of the length of the original QBF. Thus the QBF problem is in PSPACE.
Theorem 13.10 The problem of deciding whether a QBF is true is PSPACE complete.
Proof By Lemma 13.9, we need show only that the language of coded true QBF's is PSPACE-hard. That is, we must show that every language in PSPACE is polynomial-time reducible to L^.
Let M be a one-tape polynomial space-bounded DTM accepting a language L. Then for some constant c and polynomial p, M makes no more than c p(n) moves on inputs of length n. We can code ID's of M as in Theorem 13.1, using the Boolean variables clX , 1 < i < p(n), and X a tape symbol or a composite symbol representing a symbol and the state of M. Since M is deterministic, there is no need to code a choice of moves in the composite symbol. Our goal is to construct for each 0 <j < p(/i)log c, a QBF Fj(I l9 l 2 \ where 1) /j and 1 2 are each distinct sets of variables, one for each i, 1 < i < p(n), and each tape symbol or composite symbol X, analogous to the cix s of Theorem 13.1. Say = {cix \ 1 ^ < p(n) and X is such a symbol}, and 1 2 = Kt 1 1 < i < p(n) and Y is such a symbol}.
2) Fj(I u I2 ) is true if and only if I x and I 2 represent ID's /? x and p 2 °f M y that is, for each i, exactly one cix and dlX is true, and f$ x ^- fi 2 by a sequence of at most 2j moves, where p x = X x X 2 -• Xp{n) , /?2 = Y x Y 2 • • Y p{n)y and X { and Y { are the symbols such that cix . and diYt are true.
Then given x of length n we may write a QBF Qx = 3I0 3If[Fmc(Io, //) a INITIAL(/0 ) a FINAL(// )], where 3/ 0 and 31f stand for a collection of existentially quantified variables, one for each symbol X and integer i, 1 < i < p(n\ as above. INITIAL(/ 0 ) is a proposi-tional formula that says the variables in the set / 0 represent the initial ID of M KTUNOTES.IN Downloaded from Ktunotes.in 346 INTRACTABLE PROBLEMS with input x, and FINAL(// ) expresses the fact that If represents an accepting ID of M. Then Qx is true if and only if x is in L(M). INITIAL and FINAL can be written in time that is polynomial in n using the techniques of Theorem 13.1.
We now show how to construct, for each j, the formula Fj(I l9 I 2 ). The basis, j = 0, is easy. Using the technique of Theorem 13.1, we have only to express as a Boolean formula the facts that 1) l x and 1 2 represent ID's, say fi t and /? 2 ; that is, exactly one variable for each position in /?j and /?2 is true.
2) Either fi x = fi2 or p 1 |— p 2 .
For the induction step, we are tempted to write Fj{I u / 2 ) = (3I)[Fj ,(/„/) a F,,(/, / 2 )].
However, if we do so, F, has roughly double the length of Fy _ l9 and the length of Fp(„)log2C will be at least c p(n \ and therefore cannot be written in polynomial time.
Instead we use a trick that enables us to make two uses of an expression like Fj j in only a small amount (polynomial in n) more space than is required for one use. The trick is to express that there exist J and K such that if J = I x and K = I or J = I and K = / 2 , then F} _ ^J, K) must be true. The QBF for this is Fj(I !, / 2 ) = 3I[3J[3K[(^(J = I 1 aK = I) a-i (J = / aK = I 2 ))vFj .
l (J y K)]]].
(13.3) We use expressions like J = I t to mean that for each pair of corresponding variables in the sets J and l x (those representing the same position and symbol), both are true or both are false. Equation (13.3) states that whenever the pair (J, K) is either (/„ /) or (I l9 / 2 ), then F} _ t (J, K) must be true. This allows us to assert that both Fj.^Ii, I) and Fj_ x (l, l 2 ) are true using only one copy of Fj- t .
Intuitively, F} _ l is used as a "subroutine" that is "called" twice.
The number of symbols in Fj9 counting any variable as one, is 0(p(n)) plus the number of symbols in Fy_ v Since (13.3) introduces 0(p(n)) variables (in the sets /, J, and K), the number of variables in F, is 0(jp(n)). Thus we can code a variable with 0(logy" 4- log p(n)) bits. It follows by induction on j that Fj can be written in time 0(jp(n) (log j + log p(n))). If we let ; = p(n) log c and observe that for any polynomial p(n), log p(n) = 0(log n), we see that Qx can be written in 0(p 2 (n) log n) time. Thus there is a polynomial time reduction of L(M) to Lq bf . Since M is an arbitrary polynomial space-bounded TM, we have shown that Z^ bf is PSPACE-complete.
Context-sensitive recognition Another PSPACE-complete problem worth noting is: Given a CSG G and a string w, is w in L(G)? This result is surprising, since the CSL's occupy the "bottom" of PSPACE, being exactly NSPACE(n) and contained in DSPACE(n 2 ).
KTUNOTES.IN Downloaded from Ktunotes.in 13.5 | COMPLETE PROBLEMS FOR 0> AND NSPACE(log tl) 347 However, the "padding" technique used in the translation lemma (Lemma 12.2) makes a proof possible.
To begin, pick a straightforward binary code for grammars as we have done for Turing machines. Let Lcs be the language consisting of all strings x#w, where x is the code for a CSG Gx and w is a coded string from the input alphabet of Gx .
Assume that for a given grammar, all grammar symbols are coded by strings of the same length. It is easy to design an LBA that, given input x#w, guesses a deriva-tion in Gx such that no sentential form exceeds the length of the string coded by w.
The coded sentential form can be stored on a second track under the cells holding w. Moves are determined by consulting the x portion of the input (to see how this may be done it helps to assume the existence of a second tape). We see that Lcs is in NSPACE(n) and thus in PSPACE.
Theorem 13.11 Lcs , the CSL recognition problem, is PSPACE-complete.
Proof We already know Lcs to be in PSPACE. Let L be an arbitrary member of PSPACE; say L is accepted by M, a DTM of space complexity p(n). Define L to be {y$ p(iy|) |y is in L}, where $ is a new symbol.
It is easy to check that L is in DSPACE(n) and therefore is a CSL. Let G be a CSG for £, and let x be the binary encoding of G. Then the polynomial-time mapping that takes y to x#w, where w is the encoding of y$ pi^\ is a reduction of L to Lcs , showing Lcs is PSPACE-complete.
13.5 COMPLETE PROBLEMS FOR & AND NSPACE(LOG n) It is obvious that DSPACE(log n) c & by Theorem 12.10. Could it be that & = DSPACE(Iog n), or perhaps & c DSPACE(log k n) for some k'l Similarly, it is obvious that DSPACE(log n) c NSPACE(log n). Could these two classes be equal?
If so, then by a translation analogous to Lemma 12.2, it follows that NSPACE(h) c DSPACE(h), that is, deterministic and nondeterministic CSL's are the same.
We shall exhibit a language L x in & such that every language in \J f is log-space reducible to Lv Should this language be in DSPACE(log k n) for some /c, then & is contained in DSPACE(log n). Similarly we exhibit an L 2 in NSPACE(log n) such that every language in NSPACE(log n) is log-space reducible to L2 . Should L2 be in DSPACE(log n), then DSPACE(log n) would equal NSPACE(Iog n).
There is, of course, no known way to recognize L x in log n space and no known way to recognize L2 deterministically in log n space.
Languages complete for NSPACE(log n) or for & are not necessarily hard to recognize, and in fact, the languages L x and L2 are relatively easy. The results of this section serve merely to reinforce the idea that many complexity classes have complete problems. They do not suggest intractability the way MP-completeness or PSPACE-completeness results do.
KTUNOTES.IN Downloaded from Ktunotes.in 348 INTRACTABLE PROBLEMS Context-free emptiness Define L cfe to be the language of coded CFG's whose languages are empty. L cfe is the language L i alluded to above. We shall show that & is log-space reducible to L cfe .
Theorem 13.12 Lcfe , the emptiness problem for CFG's, is complete for & with respect to log-space reductions.
Proof We shall reduce an arbitrary language L in ^ to Ltfe using only log n space. Specifically we shall design a log-space transducer M x . Given input x of length n y M t writes a grammar Gx such that L(GX ) = 0 if and only if x is in L. Let M be a p(n) time-bounded TM accepting the complement of L. Since <P is effec-tively closed under complementation, we can find M. Intuitively, a derivation of Gx corresponds to a valid computation ofM on x. The nonterminals of Gx are all symbols of the form A xin where 1 ) X is a tape symbol of M, a pair [qY\ where q is a state and Y a tape symbol, or the marker symbol # used to denote the ends of ID's; 2) 0<i <p(n)+ 1; 3) 0 < t < p(n).
The intention is that A Xit ^> w for some string w if and only ifX is the ith symbol of the ID of A? at time t. The symbol S is also a nonterminal of Gx ; it is the start symbol.
The productions of Gx are: 1) S -> A [qfY]it for all /, r, and 7, where qf is a final state.
2) Let f(X, 7, Z) be the symbol in position i of the tih ID whenever XYZ occupies positions i — 1, /, and i 4-1 of the (f — l)th ID. Since M is deter-ministic,/^, V, Z) is a unique symbol and is independent oft and t. Thus for each i and r, 1 < i, r < and for each triple AT, y, Z with jy y, Z), we have the production A Wit ~ A Xj -i mt -i ^Y,i,t-1 ^Z.i+ l.f- 1-3) A 0t -> ^ and Uf -» £ for all r.
4) ^xio ^ for 1 < i < p(n) if and only if the ith symbol of the initial ID with input x is X.
Any easy induction on t shows that for 1 < i < p(n), A Wit ^> e if and only if W is the ith symbol of the ID at time f. Of course, no terminal string but e is ever derived from any nonterminal.
Basis The basis, t = 0, is immediate from rule (4).
Induction If A wit ^> £, then by rule (2) it must be that for some X, Y, and Z, W is f(X, y, Z) and each of A xi . l ^ i , A Yi4 - u and AZJ+Ut -i derive e. By the KTUNOTES.IN Downloaded from Ktunotes.in 13.5 | COMPLETE PROBLEMS FOR & AND NSPACE(log n) 349 inductive hypothesis the symbols in the ID at time t -1 in positions i — 1, i, and i + 1 are X, 7, and Z, so W is the symbol at position i and time t by the definition of/ Conversely, if W is the symbol at position i and time t > 1, then W =f(X, Y, Z), where X, 7, and Z are the symbols at time t -1 in positions i — 1, i, and i + 1.
By the inductive hypothesis, or by rule (3) if i = 0 or i = 4-1, Ax,i- l,t-l ^Y,i,t- 1 ^Z,i+ l.t-i ^ £ -Thus by rule (2), 4^ ^> e.
Then by rule (1), S ^> £ if and only if M accepts x.
Finally we need show that the productions of Gx can be produced by M x with input x of length n. First of all, recall that log 2 p(n) < c log 2 n for some constant c, since p(n) is a polynomial. Therefore M x can count from i = 0 to in log n scratch storage. Similarly M i can count from t = 0 to in log h space. The productions of Gx are easily generated by a double loop on i and Now Gx is in Lcfe if and only ifM does not accept x and hence if and only if x is in L. Thus Lcfe is complete for ^ with respect to log-space reductions.
The reachability problem Now we shall give a problem that is complete for NSPACE(log n) with respect to log-space reductions. The graph reachability problem is, given a directed graph with vertices {1, 2, n} determine if there is a path from 1 to n.
Theorem 13.13 The graph reachability problem is log-space complete for NSPACE(log n) with respect to log-space reductions.
Proof The formalization of this problem as a language is left to the reader. First we show that the graph reachability problem is in NSPACE(log n). A nondeter-ministic TM M can guess the path vertex by vertex. M does not store the path, but instead verifies the path, storing only the vertex currently reached.
Now, given a language L in NSPACE(log n) we reduce it in log n space deterministically to the language of encoded digraphs for which a path from the first vertex to the last exists. Let M be a log n space-bounded nondeterministic offline TM accepting L. An ID of M can be represented by the storage tape contents, which takes log n space to represent, the storage tape head position and state, which may be coded with the storage contents via a composite symbol [qX\ and the input head position, which requires log n bits.
We construct a log-space transducer M x that takes input x and produces a digraph Gx with a path from the first to the last vertex if and only ifM accepts x.
The vertices of Gx are the ID's ofM with input x (but with the input head position, rather than with x itself) plus a special vertex, the last one, which represents acceptance. The first vertex is the initial ID with input x.M, uses its log n storage to cycle through all the ID's of M. For each ID /, M x positions its input head at KTUNOTES.IN Downloaded from Ktunotes.in 350 INTRACTABLE PROBLEMS the correct input position, so it can see the input symbol scanned by M. M x then generates arcs / J for all the finite number of J's such that / can become J by one move of M. Since M t has / available on its storage tape, and J can be easily constructed from /, this generation requires no more than log n space. If / is an accepting ID, M x generates the arc / u, where v is the special vertex.
It is straightforward to check that there is a path in Gx from the initial ID to v if and only if M accepts x. Thus each language in NSPACE(log n) is log-space reducible to the reachability problem. We conclude that the reachability problem is complete for NSPACE(log n) with respect to log-space reductions.
13.6 SOME PROVABLY INTRACTABLE PROBLEMS Up to now we have strongly implied that certain problems require exponential time by proving them NP-complete or PSPACE-complete. We shall now prove that two problems actually require exponential time. In one case, we reduce to our problem a language which, by the space hierarchy theorem, is known to require exponential space and hence exponential time. In the second case, we show how to reduce to our problem all languages in nondeterministic exponential time and then argue by a nondeterministic time hierarchy theorem [Cook 1973a] that among them there must be one that really requires, say, 2" space.
We shall now consider a problem about regular expressions that is somewhat contrived so that (a) at least 2 c"/logrt space is required to solve it and (b) this requirement can be readily proved. After that, we consider a problem in logic that is not contrived in that it had been considered long before its complexity was analyzed, and where proof of exponentiality is far from straightforward.
Regular expressions with exponentiation Let us consider regular expressions over an alphabet assumed for convenience not to contain the symbols |, 0, or 1. Let r f i stand for the regular expression rr • • r(i times), where i is written in binary. The expression r may include the ] (exponen-tiation) operator. For example, (a | 1 1 + b | 1 1) | 10 stands for {aaaaaa, aaabbb, bbbaaa, bbbbbb}.
We assume ] has higher precedence than the other operators. The problem we shall show requiring essentially exponential space, that is, 2 p(n) space for some polynomial p(n), is whether a regular expression with exponentiation denotes all strings over its alphabet (remember |, 0, and 1 are used as operators and are not part of the alphabet).
First we give an exponential-space algorithm for the problem.
Theorem 13.14 The problem whether a regular expression with exponentiation denotes all strings over its alphabet can be solved in exponential space.
KTUNOTES.IN Downloaded from Ktunotes.in 13.6 | SOME PROVABLY INTRACTABLE PROBLEMS 351 Proof Given a regular expression of length n, we shall expand the f's to obtain an ordinary regular expression and show that it has length at most n2 n . Then we shall convert this expression to an NFA of at most n2n+2 states and test whether that NFA accepts £. (Note that this latter step must be done without conversion to a DFA, since the DFA might have 2n2n+2 states). To eliminate the fs we work from inside out. We prove by induction on j that an expression with ]\ having length m, with j O's and l's, has an equivalent ordinary regular expression of length at most m2j .
Basis j = 0. The result is immediate.
Induction Scan the expression r of length m from the left until the first | is encountered. Then scan back until the left argument r t of that | is found. We assume t has highest precedence, so its argument must be a single symbol or be surrounded by parentheses; hence this extraction is easy. Let the expression be r = r2 r t ] ir 3 . Replace r by r' = r2 r 1 r x • • • r x r3 , where r x is written i times. By the inductive hypothesis, r' has an equivalent ordinary regular expression of length at most (m + (i - l)\r 1 )2 j ~ log2i symbols. Since 2y_,082 i = 2j/i, and since \r l \ < m, we see that (m + (i -1)1^1 )2>- ,og2 ' = m + (~ HhJ x 2j < m2K If r is of length n, then surely m = n and j < n, so the equivalent ordinary regular expression has length at most ri2 n .
Now, using the algorithm of Theorem 2.3, we can produce an equivalent NFA of at most 4n2 n = n2n+2 states. Nondeterministically guess symbol by symbol an input a t a 2 that the NFA does not accept. Using n2n+2 cells we can, after each guess, compute the set of states entered after the NFA reads the sequence of symbols guessed so far. The input need not be written down, since we can compute the set of states entered from this set on any input symbol. If we ever guess an input sequence on which no accepting state of the NFA is entered, we accept; the original expression does not denote £. By Savitch's theorem we may perform this process deterministically using space n 24n . It is easy to devise an encoding of the NFA that can be stored in 0(n32n ) cells, since about n bits suffice to code a state, and the input alphabet is no larger than n. As n 24" > n 3 2", it follows that n 2 4" is an upper bound on the required space.
We shall now provide a lower bound of 2 cn/log " for some constant c on the space required for the above problem. Observe that proving a certain amount of space is required also proves that the same amount of time is required (although the opposite is not true).
Theorem 13.15 There is a constant c> 0 such that every TM accepting the language 2^.ex of regular expressions with exponentiation that denote E takes more than 2cn/logn space (and therefore 2cn,logn time) infinitely often.
KTUNOTES.IN Downloaded from Ktunotes.in 352 INTRACTABLE PROBLEMS Proof Consider an arbitrary 2" space bounded single-tape deterministic TM M.
For each input x of length n, we construct a regular expression with exponentia-tion Ex that denotes Z, where Z is the alphabet of Ex , if and only ifM does not accept x. We do so by making Ex denote all invalid computations ofM on x. Let Z consist of all tape symbols of M, the composite symbols [qX\ where q is a state and X a tape symbol, and the marker symbol #. Assume that |, 0, and 1 are none of these symbols.
A string y in Z is not an accepting computation ofM on x if and only if one or more of the following are true.
1) The initial ID is wrong.
2) There is no accepting state.
3) One ID does not follow from the previous by a move of M.
In what follows, we use sets of symbols to represent the regular expression that is the sum of those symbols. Thus, if Z = {au a2 , .
.
• , an} y then we use Z as a shorthand for the regular expression a t + a 2 + "' + an . Similarly we also use Z — a to stand for the regular expression that is the sum of all the symbols in Z except a.
A regular expression denoting all strings that do not begin with the initial ID is given by START = c + (Z - #)Z 4- A t + A 2 + •• 4- A„ + Z T (n + 1)(Z + c) T (2" - n - 1)(Z - £)Z + Z T (2"+ 1)(Z- #)Z, where /J,=ET 1(E - [9o a,])X, and for 2 < i <n, A, = E t <(£ - a,)! The next-to-last term denotes Z" + followed by up to 2" — n — 1 symbols followed by anything but a blank, and denotes strings such that some position between n + 1 and 2" of the first ID does not contain a blank. Since n and 2" — n — 1 are written in binary, the length of this term is proportional to n. The last term denotes strings in which the (2" 4- l)th symbol is not #.
It is also of length proportional to n. The remaining terms are proportional to log n in length, and there are n + 3 such terms. Thus the length of the expression is proportional to n log n. Curiously, the length of the expression denoting false initial IDs domin-ates the length of the other terms in Ex .
A regular expression enforcing the condition that there is no accepting state is given by FINISH = (Z - {[qX] \q isa final state}).
This expression is of constant length depending only on M.
KTUNOTES.IN Downloaded from Ktunotes.in 13.6 | SOME PROVABLY INTRACTABLE PROBLEMS 353 Finally, let f(X, Y, Z) denote the symbol Z such that if W, X, and Y are at positions i — 1, j, and i + 1 of one ID, then Z will be at position i of the next ID.
Then let MOVE = + Z WXYZT (2 n - 1)(Z -f(W, X, Y)Z.
That is, MOVE is the sum, over the finite number of triples (W, X, Y) of symbols in Z, of those strings with W, X, and Y occupying consecutive positions in an ID that has a wrong next symbol 2" positions to the right. As the length of each term is linear in n, the length of MOVE is linear in n.
The desired expression is Ex = START + FINISH + MOVE. IfM accepts x, then the accepting computation is not in Ex . If some string y is not in Ex , then it must begin #[^0 fl i] fl 2 a„£ 2"" n #, each ID must follow the previous by one move ofM, and acceptance must occur somewhere along the way. Thus M accepts x. Therefore Ex = Z if and only if M does not accept x.
Now, let M be a Turing machine accepting language L that can be accepted in 2" space but not in 2"/n space. The hierarchy theorem for space assures us that such an M exists. Suppose there were an S(n) space-bounded TM accepting the set L rex of regular expressions with exponentiation denoting Z, suitably coded so L rex has a finite alphabet. Then we could recognize L as follows.
1 ) From x of length «, construct Ex > whose length is proportional to n log n. We can construct Ex in space proportional to n log n in an obvious way.
2) Code Ex into the alphabet of L rex . As M has a finite number of symbols, the length of the coded Ex is cn log n for some constant c.
3) In S(cn log n) space, determine whether Ex is in L rex . If so, reject x; if not, accept x.
The total amount of space is the maximum of n log n and S(cn log n). As no TM using less than 2 n/n space and accepting L exists, it must be that n log n + S(cn log n) > 2 n/n i.o., (13.4) else L could be accepted in 2 n/n space by Lemma 12.3. There exists a constant d > 0 such that if S(m) were less than 2dm/,ogm for all but a finite set of m, then (13.4) would be false. It follows that S(m) > 2dm/Iogm for some constant d and an infinite number of m's.
Corollary Z^ex is complete for exponential space with respect to polynomial-time reduction.
Proof In Theorem 13.15, we gave a polynomial time reduction to L,cx that works for every language L in DSPACE(2 n ). We could easily have generalized it to reduce any language in DSPACE(2 p(n) ), for polynomial p, to L^.
We should observe that the n log n bound on the length of Ex is critical for Theorem 13.15, although for its corollary we could have allowed the length to be KTUNOTES.IN Downloaded from Ktunotes.in 354 INTRACTABLE PROBLEMS any polynomial in |x|.
If, for example, we could only prove that \EX \ < |x| 2 , then our lower bound on the space required by L,ex would have been 2d% ""instead.
Complexity of first-order theories Now we shall consider a problem that requires at least 2 cn time, nondeter-ministically, and is known to be solvable in exponential space and doubly expon-ential time. As the problem can also be shown nondeterministic exponential time-hard with respect to polynomial time reductions, proving a better lower bound regarding the amount of nondeterministic time would improve on Theorem 12.10, which is most unlikely.
A first-order language consists of a domain (for example, the nonnegative integers), a set of operations (for example, -f, ) a set of predicates (for example, =, < ), a set of constants chosen from the domain, and a set of axioms defining the meaning of the operators and predicates. For each theory we can define the language of true expressions over the constants, operators, predicates, variables, the logical connectives, a , v , and —1, and the quantifiers 3 and V.
Example 13.5 (N, +,,=,<, 0, 1), where N stands for the nonnegative integers, is known as number theory. GodeFs famous incompleteness theorem states that the language of true statements in number theory is undecidable. While Godel's result predated Turing machines, it is not hard to show his result. If a TM M accepts when started on blank tape, it does so by a computation in which no ID is longer than some constant m. We may treat an integer i, in binary, as a computa-tion of M with ID's of length m.
The statement that M accepts c, which is known to be undecidable, can be expressed as 3/3m(£m (i)), where Em is a predicate that is true if and only if i is the binary encoding of a computation leading to acceptance of c with no ID longer than m. (Some of the details are provided in Exercise 13.37.) Thus, number theory is an undecidable theory.
There are a number of decidable theories known. For example, (R, +, =, <, 0, 1), the theory of reals with addition, is decidable, and we shall show that it inherently requires nondeterministic exponential time. If the reals are replaced by the rationals, we get the same true statements, since without multiplication, it is impossible to find a statement like 3x(x x = 2) that is true for the reals but not the rationals. The theory of integers with addition (Z, +, =, <, 0, 1), called Presburger arithmetic, is decidable, and is known to require doubly exponential nondeterministic time. That is, 2 2cn is a lower bound on the nondeterministic time complexity of Presburger arithmetic.
Example 13.6 Before proceeding, let us consider a number of examples in the theory of reals with addition. Vx3y(y = x + 1) is true: it says that x + 1 is a real KTUNOTES.IN Downloaded from Ktunotes.in 13.6 | SOME PROVABLY INTRACTABLE PROBLEMS 355 whenever x is.
VxVy[x = yv 3z(x < z a z < y) v 3z(y < z az < x)] is also true: it states that between two different reals we can find a third real; that is, the reals are dense. The statement 3yVx(x < y v x = y) is false, since for every real number y there is a greater real. Note that we have not told how to decide whether a statement is true; the decision depends on knowing the properties of real numbers, with which we assume the reader is familiar.
A decision procedure for the reals with addition We shall begin our study of the reals with addition by giving a decision procedure that requires exponential space and doubly exponential time. To begin, let us put our given statement in prenex normal form, where all quantifiers apply to the whole expression. It is easy to obtain an expression in this form if we first rename quantified variables so they are unique, and then apply the identities -|(Vx(£)) = 3x(-i£) Vx(E l )vE 2 = Vx^ v E2 ) and four similar rules obtained from these by interchanging V and 3 and/or replacing v by a . This process does not more than double the length of the expression; the only symbols that might be added to the expression are a pair of parentheses per quantifier.! Now we have a formula QllQ 2 2 Gmm^(l, 2> m)> (13.5) where the Q/s are quantifiers, and the formula F has no quantifiers. F is therefore a Boolean expression whose operands are atoms, an atom being a Boolean con-stant or an expression of the form E 1 op E 2 , where op is = or < and E y and E 2 are sums of variables and the constants 0 and 1. We know F is of this form because no other combination of operators make sense. That is, + can be applied only to variables and constants, < and = relate only arithmetic expressions, and the Boolean operators can be applied sensibly only to expressions that have true/false as possible values.
To determine the truth or falsehood of (13.5) we repeatedly substitute for the innermost quantifier a bounded quantification, which is the logical "or" (in place of 3) or "and" (for V) of a large but finite number of terms. Suppose in (13.5) we fix the values of x u x 2 , xm _i. Every atom involving xm can be put in the form t Technically the renaming of variables may increase the length of the formula by a log n factor when we encode in a fixed alphabet. However, the complexity depends on the original number of symbols and not the length of the encoded string.
KTUNOTES.IN Downloaded from Ktunotes.in 356 INTRACTABLE PROBLEMS xm op f, where op is <, =, or > and t is of the form m- 1 Co + Z C i X i> i= 1 where the c,'s are rationals. Suppose all these atoms are xm op th 1 <i <k, where f i < t2 < ' " < tk for the given values of x 1?
.
.
. , xm _ ,. For any value of xm in the range t t < xm < ti+ u each atom has the same truth value. Thus the truth of (13.5) is independent of the actual value of xm in this range. This leads us to the observa-tion that the f f's partition the real line into a finite number of segments, and the truth of (13.5) depends only on the segment in which xm lies, and not on the actual value of xm . Thus we can test (13.5) by trying one value of xm from each of a finite number of regions as suggested in Fig.
13.6.
Fig. 13.6 Representative values of x, As the values of x { , .
.
.
, xm _ j will vary, we do not really know the order of the r,'s.
However, trying xm = t t for each /, xm = ^(f f + t} ) for each i ± and xm = ± oo,t we know that no matter what the order of the f f 's, we are sure to have a representative xm in each interval of Fig.
13.6 and also at the r,'s themselves, where atoms with the = operator may become true.
It follows that if Qm = 3, then 3xm F(x ly . .., xm ) may be replaced by F(x 1 ,...,xm _ 1 )= V F(x lf ...
f xJ, (13.6) xm = t, or xm = (l/2)(t, + tj) or xm = ±oo that is, by the logical "or" of k(k 4-l)/2 + 2 terms, each of which is F with a substitution for xm . If Qm = V, a similar replacement, with a substituting for v, may be made.
If F has k atoms, F has k[k(k + l)/2 + 2] atoms, which is at most k 3 atoms for k > 3. Also, if the coefficients in the atoms of F are each the ratio of integers of at most r bits each, then after grouping terms, solving for xm , and computing the average of two f,'s, we find that the coefficients in the atoms of F will be ratios of integers with no more than 4r + 1 bits. This follows since if a, b, c, and d are r-bit integers, a c ac bd = b~d t If .xm = +oo, then x m = t and .xm < t are false, and .x m > t is true independently of t. If xm = - 00 analogous simplifications occur.
KTUNOTES.IN Downloaded from Ktunotes.in 13.6 | SOME PROVABLY INTRACTABLE PROBLEMS 357 is the ratio of integers with at most 2r bits, and a c ad ±bc b ± d = bd is the ratio of a (2r 4- l)-bit integer and a 2r-bit integer. For r > 1, then, the coefficients in F are no more than five times the length of the coefficients in F.
If we repeat the above process to eliminate all the quantifiers and variables, we eventually produce a formula with only logical operators, =, <, and con-stants. The constants are ratios of integers with at most 5mr bits. The number of atoms is at most (...((fc 3 j 3 )...) 3 = fc 3 -.
m times As each atom is a relation between constants of 5mr bits, and k, m, and r are less than m, the length of the expression is at most 2 2Cn for some constant c (note that n 3 " < 2 22 "). We may evaluate an atom of the form a/b < c/d by computing ad — be and comparing it with 0. Thus the entire final expression may be evaluated in the square of its length. Hence our decision procedure takes 2 2dn time for some constant d.
The procedure as we have given it also takes doubly exponential space.
However, we can reduce the space to a single exponential by evaluating F recursively. We have already seen that we need consider only a finite set of values for each x f . The values for x, are given by a formula of the form a0 + Jj=\ ajXj, where the a/s are rationals that are ratios of 5m ~ j+ 1 r-bit integers, where r is the number of bits in the largest constant of the original formula, F; note r < log n.
Thus values for x Y are rationals that are at most ratios of 5mr-bit integers, the values for x 2 are ratios of at most 5m+ r-bit integers, etc. Thus we need only cycle through values for each x, that are at most 5 2mr bits. We use a recursive procedure EVAL(G) that determines whether G is true when the variables take on the values ±oo and any ratio of 5 2mr-bit integers.
If G has no quantifiers, then it consists only of arithmetic and logical relations among rationals, so its truth can be determined directly. If G = Vx(G'), EVAL(G) calls EVAL(G") for all G" formed from G' by replacing x by ±oo or a ratio of 5 2mr-bit integers. EVAL(G) is true if EVAL(G") returns true for all these expres-sions G". If G = 3x(G'), we do the same, but EVAL(G) returns true whenever some EVAL(G") is true.
It is easy to check that no more than m copies of EVAL are active simultan-eously. The arguments for the active calls to EVAL can be put on a stack, and this stack takes 0(m5 2mr) space. Thus, if F is an expression of length n, we may evaluate F in space 2 cn and time 2 2dn for some constants c and d.
A lower bound We now show that the theory of reals with addition requires essentially non-deterministic exponential time. A series of lemmas are needed showing that multi-KTUNOTES.IN Downloaded from Ktunotes.in 358 INTRACTABLE PROBLEMS plication and exponentiation by limited size integers can be expressed by short formulas.
Lemma 13.10 There exists c> 0 such that for each n there is a formula Mn (x, y, z) that is true if and only if x is a nonnegative integer strictly less than 2 2 ", and xy = z. Furthermore, |Mn (x, y, z) \ < c(n -hi), and Mn (x, y, z) can be constructed from n in time polynomial in n.
Proof For n = 0, 2 20 = 2. Thus M 0 (x, y, z) can be expressed as (x = 0 a z = 0) v (x = 1 a z = y).
Inductive step: (Construction of Mk+i from Mk ).
Let x be an integer less than 2 2k+1 . There exist integers x 1? x 2 , 3 , x 4 < 2 2 " such that x = x l x 2 + x 3 + x4 . In proof, let x t = x 2 = [y/x. Now z = xy can be expressed by z = x 1 (x 2 y)-f x 3 }? + X4 y. Thus Mk+ t (x, y, z) = 3wj ••• 3u 5 3x { •• 3x4[M fc(x 1 , x 2 , ax = uj + x 3 + x4 aM&(x 2 , y, u 2)AMk (x u u 2 , u 3 ) AMk (x 3 , y, u4 ) a M k (x4 , y, m 5 )az = u 3 + m4 + w 5 ] (13.7) That is, w 1 =x 1 x 2 , x = X! x 2 4-3 + x4 , u 2 = x 2 y, w 3 = x 1 x 2 y, w4 = x 3 y, w 5 = x4 y, and z = XjX 2 y + x 3 y + x4 y.
The condition that each x, is an integer less than 2 2k is enforced by each X; being the first argument of some Mk .
Formula (13.7) has five copies of M k , so it appears that M k + x must be at least five times as long as Mk . This would make the length of Mn exponential in n, not linear as we asserted. However, we can use the "trick" of Theorem 13. 10 to replace several copies of one predicate by a single copy. That is, we may write y, z) = 3i 3w 5 3xj 3x4 [x = W x + X 3 + X 4 A Z = W 3 + W4 + U 5 a Vr Vs Vf[—ir = x l As = x 2 At = u l ) a —i(r = x 2 a 5 = y a t = u 2 ) a —i(r = x x AS = u 2 At = u 3 ) a —(r = x 3 a 5 = y A t = w4 ) a —i(r = X4 A 5 = y A t = u 5 ) vMk(r, s, t)]], which has a constant number of symbols more than Mk does.
KTUNOTES.IN Downloaded from Ktunotes.in 13.6 | SOME PROVABLY INTRACTABLE PROBLEMS 359 One minor point is that if we introduce new variable names for each Mk , we shall eventually introduce a log n factor into the length of Mn , since variable names must be coded in a fixed alphabet in the language of true formulas.
However, the scope rules for quantified formulas allow us to reuse variables subject to the restriction that the twelve new variables introduced in Mk don't conflict with the free variables x, y, and z. Thus M n requires only 15 different variables, and its coded length is proportional to the number of symbols.
Observe that Mn (x, 0, 0) states that x is an integer less than 2 2 ". Thus we can make statements about small integers in the theory of reals with addition by using very short formulas.
Lemma 13.11 There exists a constant c> 0 such that for every n there is a formula Pn (x, y, z) that is true if and only if x and z are integers in the range 0 < x, z < 2 2 " and y x = z. Furthermore \Pn \ < c(n + 1) and Pn can be constructed from n in time polynomial in n.
Proof We construct by induction on/ca sequence of formulas £k (x, y, z, u, v, w) such that Ek has both exponentiation and multiplication built into it. The reason for doing this is that we wish to express Ek in terms of several copies of £k _ y and then use universal quantification to express Ek in terms of one copy of Ek _ j. We could not do this with Pk , since a formula for Pk involves both Pk _ l and M k _ v The formula £k (x, y, z, u, v, w) will be true if and only if x, z, and u are integers, 0 < x, z < 2 2\ z = y, 0 < u < 2 2\ and uv = w.
Basis For k = 0, E0 = (x = 0az= 1) v (x = 1 a>> = 0 az = 0) v (x = 1 a y = 1 a z = 1)a M 0 (w, v, w).
Induction To construct £fc+1 (x, y, z, w, v, w) we can use the fact that £k (0, 0, 0, w, w) = M„(u, v, w) to express the conditions on w, u, and was in Lemma 13.10. Using several copies of £ fc , we may assert that there exist integers x u x 2 , x 3 , x 4 in the range 0 < x, < 2 2k such that x = x 1 x 2 + x 3 + x4 and y x = (y xl ) X2y x yx.
Finally, we use the "trick" of Theorem 13.10 to express Ek+1 in terms of one copy of Ek and a constant number of additional symbols. Last, we may write Fn(x, y, z) = £„(x, y, z, 0, 0, 0).
This asserts that z = y x , and x and z are integers in the range 0 < x, z < 2 2 ".
To improve readability of what follows, we use the abbreviations 2 for 1 + 1, 2x for x + x, x < y for x < y v x = y, and x < y < z for (x = y v x < y) a y < z.
KTUNOTES.IN Downloaded from Ktunotes.in 360 INTRACTABLE PROBLEMS Expanding an abbreviated formula results in at most multiplying the length by a constant factor. In addition to the above abbreviations, we shall use constants like 2" and multiplications like ab in formulas. Technically these must be replaced by introducing an existentially quantified variable, say x, and asserting x = 2" or x = ab by Pn (n, 2, x) or Mn (a, b, x). This can also increase the length of the formula by a constant factor.
We intend to encode Turing machine computations as integers. Let M be a 2" time-bounded NTM. If the total number of tape symbols, composite symbols, and the marker # is b, then a computation of M is an integer x in the range 0 < x < b(2 " + 1)2+ . Asserting that an integer is a computation is facilitated by a predicate that interrogates the ith digit in the fr-ary representation of x.
Lemma 13.12 For each n and b there exists a constant c, depending only on b, such that there is a formula Dn b (x, ij) that is true if and only if x and i are integers, 0 < x < b {2n+ l)2+ \ 0 < i < 2", and xh the (i + l)th digit of x counting from the low-order end of the fr-ary representation of x, is Furthermore \Dnb \ < c(n + 1), and Dn b can be constructed from n and b in time polynomial in n and b.
Proof For each b there exists a constant s such that b i2n+ 1)2+ 1 < 2 2sn for all n.
Thus that x is an integer in the correct range can be expressed by 3m[Psn ((2" + l) 2 , b, m) a0 < x < m\ (Recall our previous remarks concerning constants like 2" and their expansions.) That i is an integer in the range 0 < i < 2" can be expressed by M n(U 0, 0) a (0 < i < 2"). Now x in base b has zeros in positions 1, 2, i + 1 if and only if it is divisible by bi+l . Thus x f =j if and only if there exist integers q and r such that x = qbi+ 1 + r and y7>' < r < (j + )b\ This fact is easily expressed using Psn and M sn .
Theorem 13.16 Any nondeterministic algorithm to decide whether a formula in the first-order theory of reals with addition is true must, for some constant c> 0, take 2 cn steps for an infinite number of n's.
Proof The proof is quite similar in spirit to that of Theorem 13.1. Let M be an arbitrary 2 n-time bounded NTM. Here ID's in a computation of M consist of 2" symbols rather than p(n) as in Theorem 13.1. Let the total number of tape sym-bols, composite symbols, and #'s be b. Then a computation of M on input of length n consists of [(2 n + l) 2 + 1] fr-ary digits. We may consider this computation to be an integer i in the range 0 < / < 2 2,n for some constant s. For convenience, we take the low-order digits of i to be at the left end of the computation.
Let x be an input of length n to M. We construct a formula Fx that is true if and only if M accepts x. Fx is of the form 3i(.
. .), where the formula within the parentheses asserts i is an accepting computation of x. This formula is analogous to that in Theorem 13.1. The first n + 1 symbols of the computation are #[g0 , a l9 m]a 2 '~an , KTUNOTES.IN Downloaded from Ktunotes.in 13.6 | SOME PROVABLY INTRACTABLE PROBLEMS 361 assuming that x = a l a 2 "-an , q0 is the initial state, and m is any choice of first move. To say that the first n + 1 symbols of the computation are correct, we say that there exist u and j such that the value of u represents #[q0 , a x , m]a 2 ' • an for some w, and i = bn+ lj + u for some integer j.
We must write this formula in 0(n)-space in time that is polynomial in n. By induction on k = 2, 3, .
.
. , n + 1 we can write a formula Ck (v) with free variable v, which asserts that the value of v is the numerical value of the first k symbols of the computation. For the basis, k = 2, we simply write a formula C 2 (v) = (v = p x vv = p 2 v-vv = p„), where the p/s are the integers represented by #[g0 > m] f°r tne finite set of values of m. For the induction, Ck (v) = 3w(Ck _ i (w) a v = 6w + ak _ x ), where t is taken to be the numerical value of tape symbol ak _ ^ To avoid using n variables to express Cn+1 , which would make its length 0(n log ri), we alternate between two variables, such as v and w, as we construct C 2 , C 3 , Cn+1 .
The desired formula asserts Cn+ t (u) and i = bn+ lj + w for integer The latter assertion is similar to what was done in Lemma 13.12, and the technique will not be repeated here.
To express that the initial ID was correct in Theorem 13.1 required asserting that "approximately" p(n) cells contained the blank symbol. This was accom-plished by the logical v of p(n) items. We must now assert that about 2" cells contain the blank symbol, and thus we cannot use a logical v of 2" formulas; this would be too long a formula. Instead we use the quantifier V/ and assert that either j is not an integer in the range n + 2 < j < 2" + 1 or the ;th symbol is the blank, which we denote by 0. Thus we write Vj[nMs„(;, 0, 0) v + 2 <j < 2" + 1) v Dn>b (i, j, 0)].
The formulas that force the last ID to contain a final state and force each ID to follow from the previous ID because of the choice of move embedded in the previous ID are similarly translated from the techniques of Theorem 13.1. Having done this, we have a formula Ex , whose length is proportional to n, that is true if and only if M accepts x.
Suppose M accepts a language L in time 2" not accepted by any 2"12 time-bounded NTM. (The existence of such a language follows from the NTIME hierarchy of Cook [1973a], which we have not proved.) We can recognize L as follows. Given x of length n, produce the formula Ex that is true if and only ifx is in L. Now, if T(n) nondeterministic time suffices to accept the set of true formulas in the first-order theory of reals with addition, we may determine whether x is in L in time p(n) + T(cn). Then p(n) + T(cn) > 2"12 for an infinity of h's, else by Lemma 12.3 we could recognize L in time at most 2 n/2 , for all n. It follows that T(n) > 2 dn i.o. for some d > 0.
KTUNOTES.IN Downloaded from Ktunotes.in 362 INTRACTABLE PROBLEMS Corollary The theory of reals with addition is nondeterministic exponential time-hard with respect to polynomial time reductions.
Proof The proof is an easy generalization of the foregoing reduction of a 2" nondeterministic time TM.
13.7 THE & = .
i W> QUESTION FOR TURING MACHINES WITH ORACLES: LIMITS ON OUR ABILITY TO TELL WHETHER 0> = . 10> The reader should recall from Section 8.9 our discussion of Turing machines with oracles. These TM's had associated languages, called oracles, and had special states in which the membership of the string written to the left of their head could be tested in one step for membership in the oracle. Any oracle TM can have any oracle "plugged in," although its behavior will naturally vary depending on the oracle chosen. If A is an oracle, we use MA for M with oracle A. The time taken by an oracle TM is one step for each query to the oracle and one step for each ordinary move of the TM.
We define &A to be the set of languages accepted in polynomial time by DTM's with oracle A. Also define Jf&A to be the set of languages accepted by NTM's with oracle A in polynomial time. We shall prove that there are oracles A and B for which 0>A = ./f&A and 0>B JT0>B . This result has implications regard-ing our ability to solve the ?f = Jf0> question for TM's without oracles. Intui-tively all known methods to resolve the question one way or the other will work when arbitrary oracles are attached. But the existence of A and B tells us that no such method can work for arbitrary oracles. Thus existing methods are probably insufficient to settle whether = Jf&. We shall provide details along these lines after we see the constructions of A and B.
An oracle for which 0 = .
\ 0> Theorem 13.17 &A = jV'&a , where A = Lqbf , the set of all true quantified Bool-ean formulas (or any other PSPACE-complete problem).
Proof Let M A be nondeterministic polynomial time bounded, and let L = L(M A ). Then M A queries its oracle a polynomial number of times on strings whose lengths are bounded by a polynomial of the length of the input to MA .
Thus we may simulate the oracle computation in polynomial space.
It follows that J 9>A c PSPACE. However, any language L in PSPACE is accepted by some DTM M A that reduces L to A in polynomial time and then queries its oracle.
Thus PSPACE c &>A . But clearly &>A c mypA , so 0>A = . V'PA .
An oracle for which & f .
I W We now show how to construct an oracle B c (0 -f 1) for which SPB ± ~\r@B > & will have at most one word of any length; exactly which words will be discussed KTUNOTES.IN Downloaded from Ktunotes.in 13.7 | THE & = Jf& QUESTION FOR TURING MACHINES WITH ORACLES 363 later. We shall be interested in the language L = {0' | B has a word of length i}.
We may easily construct an NTM with oracle B that, given input 0', guesses a string of length i in (0 + 1) and queries its oracle about the guessed string, accepting if the oracle says "yes." Thus L is in Jf&B . However, we can construct B so that the string of each length, if any, is so cleverly hidden that a DTM with oracle B cannot find it in polynomial time.
Theorem 13.18 There is an oracle B for which 0>B + Jr0>3 .
Proof We shall give a procedure to enumerate the set B. Set B will have at most one word of any length. As we generate S, we keep a list of forbidden words; these words are ruled out of consideration for possible membership in B. Assume an enumeration of DTM's with oracle and input alphabet {0, 1}, in which each TM appears infinitely often. We consider each M i9 i — 1, 2, in turn. When M, is considered we shall have generated some forbidden words and a set B { of words so far in B. There will be at most one word in B t of length 0, 1, .
.
. , i — 1, and no longer words. Furthermore, no other words of length less than i will subsequently be put in B. We simulate Mf' on input 0\ If M, queries a word of length less than i, we consult Bh which is all words in B so far, to see if the oracle responds "yes" or "no." If M, queries a word y of length i or more, we assume that y is not in B (i.e., answer "no") and to make sure y is not later placed in B, add y to the list of forbidden words.
The simulation of Mf •' on 0' continues for i log 1 steps. Afterwards, whether or not M t has halted, we make a decision about a word to put in B. If within / ,ogI steps, Mf { halts and rejects 0\ then we put a word of length i that is not on the forbidden list in 5, provided there is such a word. The word may be picked arbitrarily, say the lexicographically first word that is not forbidden. If Mf ' does not reject 0' within z ,og ' steps, then no word of length i is placed in B.
There is also no word of length i in B if all words of length i are forbidden by the time we finish simulating Mf\ However, the number of steps simulated for M Bj is j logj , so the total number of words of all lengths forbidden by M„ M 2 , . .., M, is at most £ f°J < |(/ ,0«'") < + As there are 2' words of length i, we know that not all words of length i are forbidden if 2' > i 1 +logi , that is, if i > (1 + log i) log i. But the latter relation holds for i > 32, so it is only for a finite number of small fs that all words of length i could be forbidden.
Having finished the simulation of Mf « on 0' for / ,og ' steps, we generate the selected word, if there is one, obtaining us a new set Bi+ 1 of generated words. We are now ready to repeat the process for MfVi 1 on 0,+ 1 .
KTUNOTES.IN Downloaded from Ktunotes.in 364 INTRACTABLE PROBLEMS Next we define a language L that is in Jf@B — 0>B . Let L = {0' | B has a word of length i}.
We may easily construct a linear time NTM with oracle B that, given input 0', nondeterministically guesses string w of length i in (0 + 1) and queries its oracle about w, accepting if the oracle says "yes." Thus L is in jV&b .
Suppose L is in 0>B . Let Ml accept L, where M B is a deterministic polynomial p(n) time-bounded TM with oracle B. As each TM has arbitrarily long codes, we may pick k such that k > 32 and k]ogk > p(k). IfMB accepts 0\ then 0 is in L, so £ has a word of length k. That means k rejects 0. But Mf and Mf k must behave identically on input 0\ since B and £ fc agree on words shorter than k, and B has no word of length k or more that is queried by M Bk on 0. Thus Mf rejects 0, a contradiction.
IfM B rejects 0\ so 0 is not in L, then M Bk cannot reject 0 within k logk steps.
This follows since k > 32, and had Mf fc rejected 0 within /c Iog steps, there would still be a word of length k not on the forbidden list, and that word would be in B.
Thus 0 would be in L. Hence M B does not reject 0 k within k]ogk steps. But as k]ogk > p(/c), MB does not reject 0 k at all, another contradiction. We conclude that L is in J - ePB .
Significance of oracle results Let us consider the ways used in this book to show two language classes to be the same or different, and see why Theorems 13.17 and 13.18 suggest that these methods will fail to resolve the ,J f = .YiP question. We showed certain classes to be the same by simulation. For example, Chapter 7 contains many simulations of one type of TM by another. Chapter 5 contained simulations of PDA's by CFG's and conversely.
Suppose we could simulate arbitrary polynomial time-bounded NTM's by polynomial time-bounded DTIvTs. (Note that giving a polynomial-time algorithm for any one NP-complete problem is in effect a polynomial-time simulation of all NTM's.) It is likely that the simulation would still be valid if we attached the same oracle to each TM. For example, all the simulations of Chapter 7 are still valid if we use oracle TM's. But then we would have tfB = J d?B , which was just shown to be false.
Other classes of languages were shown unequal by diagonalization. The hier-archy theorems, Theorems 12.8 and 12.9, and the proof that 1^ is an r.e. set but not a recursive set are prime examples. Diagonalizations also tend to work when oracles are attached, at least in the three examples cited above.
If we could c|jagonalize over & to show a language to be in Jf& — &>, then the same proof might well work to show Jr#A — &A ± 0. This would violate Theorem 13.17.
We also used translation lemmas to refine time and space hierarchies in Chapter 12. Could these help show & + oY&l Probably not, because the transla-tion lemmas also hold when oracles are attached.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 365 Lastly, we can use closure properties to show a difference between two lan-guage classes. For example, the DCFL's are contained in the CFL's, but the DCFL's are closed under complementation and the CFL's are not. This proves that there is a CFL that is not a DCFL. Could we find a closure property of& that is not shared by Jf&l This at first appears the most promising approach. While proofs that 0> is closed under an operation are likely to show also that &A is closed under that operation, a nonclosure result for Jf@> might not carry over to Jf&A . On the other hand, showing Jf0> not closed under an operation involves showing a particular language not to be in JT0>. This proof might be accomplished by diagonalization, but then it would likely carry over to Jf£PA . It might be done by developing a pumping lemma for JfgP, but this seems well beyond present capabi-lity. Finally, we might develop some ad hoc argument, but again, no such argu-ments have been found, and they appear very difficult.
EXERCISES 13.1 Suppose there is a 2" time-bounded reduction of Li to L 2 , and L 2 is in DTIME(2 n ).
What can we conclude about L x ?
13.2 Which of the following Boolean formulas are satisfiable.
a) X! ax 3 a(x 2 vx 3 ) h) A (xi.vXjjVxjA A vx l2 vx i3 ) i|.l2.«J i|.«2.»J where (i'i, i 2 , 13) ranges over all triples of three distinct integers between 1 and 5.
13.3 A clique in a graph G is a subgraph of G that is complete; i.e., each pair of vertices is connected by an edge. The clique problem is to determine if a given graph G contains a clique of given size k.
a) Formulate the clique problem as a language recognition problem.
b) Prove that the clique problem is AfP-complete by reducing the vertex cover problem to the clique problem.
[Hint: Consider a graph G and its complement graph G, where G has an edge if and only if G does not have that edge.] 13.4 Given a graph G and integer k, the clique cover problem is to determine if there exist k cliques in G such that each vertex of G is in at least one of the k cliques. Prove that the clique cover problem is /VP-complete by reducing the vertex cover problem to the vertex cover problem for graphs without triangles, thence to the clique cover problem. [Him: Consider graphs G = (V, E) and G' = (£, {(e u e 2 )\e x , e 2 are incident upon the same vertex in G})].
13.5 Does the graph of Fig.
13.7 a) have a Hamilton circuit?
b) a vertex cover of size 10?
c) a vertex coloring with 2 colors such that no two adjacent vertices are the same color?
13.6 Prove that the chromatic number problem is /VP-complete by reducing the 3-CNF satisfiability problem to the chromatic number problem. [Hint: The graph in Fig. 13.8 can KTUNOTES.IN Downloaded from Ktunotes.in 366 INTRACTABLE PROBLEMS Fig. 13.7 An undirected graph.
(:0 (b) V = \vr xr v ; I !</<// /•- = {(>',.
I / ^/} U {( Vf .
v ; ) | 1 < /< //} u <r ; .
v ; ), (r r v y ) I / +r Fig. 13.8 Graph used to show chromatic number problem /VP-complete, (a) Complete graph on n vertices; (b) x, and x, are connected to all v} for which i ^ j.
be used as a subgraph in your construction. Note that each v, must be colored with a distinct color, say color /. The entire graph can be colored with n + 1 colors if and only if for each /, 1 < i < n, one of x £ and ic, is colored with color /' and the other is colored with color n + 1.] 13.7 Show that the following problems are MP-complete.
a) Given a graph G, with integer distances on the edges, and two integers/and d, is there a way to select / vertices of G on which to locate "firehouses," so that no vertex is at distance more than d from a firehouse?
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 367 b) The one-register code generation problem. Suppose we have a computer with one reg-ister and instructions LOAD m bring the value in memory location m to the register STORE m store the value of the register in memory location m OP m apply OP, which may be any binary operator, with the register as left argument and location m as right argument; leave the result in the register.
Given an arithmetic expression, each of whose operands denotes a memory location and given a constant /c, is there a program that evaluates the expression in k or fewer instructions?
c) The unit execution time scheduling problem. Given a set of tasks Tu ...,T ky a number of processors p, a time limit r, and a set of constraints of the form 7] < 7}, meaning that task Ti must be processed before 7}, does there exist a schedule, that is, an assignment of at most one task to any processor at any time unit, so that if 7] < 7} is a constraint, then 7j is assigned an earlier time unit than 7}, and within t time units each task has been assigned a processor for one time unit?
d) The exact cover problem. Given a set S and a set of subsets S U S 2 , .
.
, Sk of 5, is there a subset T ^ {Si, S 2 , Sk } so that each x in S is in exactly one 5,- in 7?
13.8 The spanning tree problem. Determine whether a tree T is isomorphic to some spanning tree of G.
a) Give a log-space reduction of the Hamilton circuit problem to the spanning tree problem.
b) Give a direct log-space reduction of 3-CNF satisfiability to the spanning tree problem.
13.9 a) An n-dimensional grid is a graph G = (V, E) where ^ = {(«i.
«2, •••,'„)!
1 <ij<mj, 1 <j 1 can be replaced by an equation of the form y + x, + x 2 + x 3 = 4, provided y is constrained to be 1, 2, or 3.
The system of equations y + z Y + z 2 = 3, y = z 3 + z4 , and z t + z,= 1, 1 < i < 4, has no solution with more than four variables zero and has a solution with exactly four variables zero if and only if y = 1, 2, or 3.] 13.13 A kernel of a directed graph is a set of vertices such that 1) there is no arc from a vertex in the kernel to another vertex in the kernel, and 2) every vertex is either in the kernel or has an arc into it from the kernel.
Prove that determining whether a directed graph has a kernel is /VP-complete. [Hint: Observe that a cycle of length two or three may have only one vertex in a kernel.] 13.14 Prove that the traveling salesman problem is /VP-complete.
13.15 Consider approximations to the traveling salesman problem. Show that the exist-ence of a polynomial-time algorithm that produces a tour within twice the cost of the optimal tour would imply & = .A y/.
S13.16 Consider the traveling salesman problem where the distances satisfy the triangle inequality, that is d(v u i; 3 ) < d(v u v 2 ) + d(v 2 , v^).
Give a polynomial-time algorithm to find a tour that is within twice the cost of the optimal tour.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 369 13.17 Suppose there exists a polynomial-time algorithm for finding a clique in a graph that is of size at least one-half the size of the maximal clique.
a) Prove that there would exist a polynomial-time algorithm for finding a clique which is of size at least l/y/2 times the size of the maximal clique. [Hint: Consider replacing each vertex of a graph by a copy of the graph.] b) Prove that for any k < 1 there would exist a polynomial-time algorithm for finding a clique which is of size at least k times the size of the maximal clique.
13.18 Prove that it is NP-complete to determine whether the chromatic number of a graph is less than or equal to 3. [Hint: The graph of Fig. 13.1 1 can be used as a weak form of an OR gate when only three colors are available, in the sense that the output can be colored "true" if and only if at least one input is colored "true."] 13.19 For n > 6, let G„ = (K„, £„) be the graph where K = {('\ h k) 1 1, j, k are distinct elements of {1, 2, .
. ., n}}, En = {(«, v)\u and v are disjoint triples}.
a) Let Xm(G) be the minimum number of colors needed to assign m distinct colors to each vertex of G so that no two adjacent vertices have a color in common. Prove for n > 6 that X 3 (G n ) = n and X A(G n ) = 2/7-4.
b) Suppose there were a polynomial-time algorithm to color a graph G with at most twice the minimum number of colors needed. Then prove that & = .
\ [Hint: Combine part (a) with Exercise 13.18.] 13.20 Construct an algorithm for finding a Hamilton circuit in a graph that under the assumption that // = .
\ will find a Hamilton circuit in polynomial time whenever such a circuit exists. If no Hamilton circuit exists, the algorithm need not run in polynomial time.
Note it is not sufficient to design a nondeterministic algorithm and then use the hypothesis — .
I to claim that there is a deterministic polynomial-time algorithm. You must actually exhibit the potentially deterministic polynomial-time algorithm.
13.21 If =/= .
I prove it is undecidable for L in .
S j> whether L is in //.
13.22 Prove that the existence of an /VP-complete subset of 0 implies & = .
t ' 13.23 An integer n is composite if and only if there exists an a, 1 < a < n, such that either 1) cT~ 1 =/= 1 mod «, or 2) there exist integers b and i where n — 1 = 2'b and a b and n have a common divisor.
If n is composite, at least one-half of the integers between 1 and n satisfy (1) or (2). Give a randomized algorithm that with high probability will determine whether a number is prime in polynomial time.
Fig. 13.11 Graph used in Exercise 13.18.
KTUNOTES.IN Downloaded from Ktunotes.in 370 INTRACTABLE PROBLEMS 13.24 Suppose there exists a function / mapping integers of length k onto integers of length k such that 1) / is computable in polynomial time; 2) f~ l is not computable in polynomial time.
Prove that this would imply A = {(x, y>| /"'() < y} is in n Co-.V ^) - # 13.25 Show that the following problems are PSPACE-complete.
a) Does a given regular expression (with only the usual operators \ + , and ) define all strings over its alphabet? [Hint: The proof parallels Theorem 13.14.] Sb) The Shannon switching game. Given a graph G with two distinguished vertices s and r, suppose there are two players SHORT and CUT. Alternately, with SHORT first, the players select vertices of G other than s and t. SHORT wins by selecting vertices that, with 5 and r, form a path from s to t. CUT wins if SHORT cannot make such a path.
Can SHORT force a win on G no matter what CUT does?
13.26 Show that if PSPACE + >.J f, then there exists a proof by diagonalization. That is, there is an enumeration L u L 2 , ... of ^, and a computable function /from integers to strings and a set L in PSPACE such that for each i,/(i) is in L if and only if/(i) is not in Lh 13.27 Give a polynomial-time algorithm for converting a quantified Boolean formula to prenex normal form Q, X { Q 2 X 2 •• Q k X k (E), where £ is a Boolean expression in 3-CNF.
13.28 Can any QBF be converted in polynomial time to an equivalent formula with at most ten distinct variables?
13.29 Show that the following problems are complete for . J f with respect to log-space reductions.
a) Is x in L(G) for string x and CFG G?
b) The circuit value problem. Encode a circuit as a sequence C,, C 2y .
• .
, C„, where each C, is a variable x l7 x 2 , .
.
. or a (_/, k) or with j and k less than i. Given an encoding of a circuit and an assignment of true and false to the variables, is the output of the circuit true?
13.30 Show that the following problems are complete for NSPACE(log n) with respect to log-space reductions.
a) Is a Boolean expression in 2-CNF not satisfiable?
b) Is a directed graph strongly connected?
c) Is L(G) infinite for CFG G without (-productions or useless nonterminals?
13.31 Given CFG's G x and G 2 and integer k, show that the problem of determining whether there are words w, in L(G t ) and w 2 in L(G 2 ) that agree on the first k symbols, is complete for nondeterministic exponential time with respect to polynomial-time reductions.
13.32 Show that the problem of determining whether a regular expression with the inter-section operator permitted denotes all strings in its alphabet requires time 2 C% " i.o., for some c > 0 and can be solved in time 2dn .
13.33 a) Write a formula in the theory of integers under addition expressing that every integer greater than 5 is the sum of three distinct positive integers.
KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 371 b) Write a formula in number theory expressing that d is the greatest common divisor of a and b.
c) Write a formula in number theory expressing that z = xy .
13.34 Apply the decision procedure of Section 13.6 for the theory of reals to decide whether the formula 3y 3x[(x + y = 14) a (3x + y = 5)] is true.
13.35 a) Show that the theory of Presburger arithmetic (the integers with +, =, and <) requires nondeterministic time 2 2c" i.o., for some c > 0. [Hint: Develop the following formulas of size proportional to n: 1) R n (x, y, z): 0 < y < 2 2", and z is the residue of x mod y.
2) P„(x): 0 < x < 2 2 ", and x is a prime.
3) G„(x): x is the smallest integer divisible by all primes less than 2 2 ".
4) M„(x, y, z): x, y, and z are integers in the range 0 to 2 22 " — 1, and xy = z.] b) Show that Presburger arithmetic can be decided in 2 2" space and 2 22 time.
c) Use the algorithm of part (b) to decide 3y 3x[(x+y = 14) a (3x + y = 5)].
13.36 Extend Presburger arithmetic to allow quantification over arrays of integers. Thus we could write formulas such as Vi4 Vn 3B Vi[—1( 1 < i < n) v [3/( 1 < j < n) a /4(f) = £(;)]].
Prove that the theory of Presburger arithmetic with arrays is undecidable.
13.37 To show that number theory is undecidable, it is convenient to encode a sequence of length n + 1, x0 , x l5 xn , into an integer x such that each x, can be obtained from x by a formula.
a) Let m = max {«, x 0 , x l5 .
.
.
, x„J. Prove that the set of u ( = 1 + (i + l)w !, 0 < i < n, are pairwise relatively prime and that u, > x,. This implies that there exists an integer b < u0 u x • • u„ such that b = x, mod u iy 0 < i < n.
b) Express Godel's P function P(b, c, 0 = b mod [1 + (i + l)c] as a predicate.
c) Prove that number theory is undecidable.
13.38 Show that there are oracles C, D, and £, for which a) S/c , .
i '^S and co- \ are all different.
b) //D + .
1 y/D but .4 = Co-.
I c) ^£ = .
I is independent of the axioms of number theory.
13.39 Show that & = Jf't?
if and only if & is an AFL.
Solutions to Selected Exercises 13.16 Construct a minimum cost spanning tree by sorting the edges by increasing cost, selecting edges starting with the lowest cost edge, and discarding any edge that forms a cycle. Let T opt be the minimum cost of a Hamilton circuit and let 7^ be the cost of the KTUNOTES.IN Downloaded from Ktunotes.in 372 INTRACTABLE PROBLEMS minimum cost spanning tree. Clearly 7^ < T opt , since a spanning tree can be obtained from a Hamilton circuit by deleting an edge. Construct a path through all vertices of the graph by traversing the spanning tree. The path is not a Hamilton circuit, since each edge of the spanning tree is traversed twice. The cost of this path is at most 2T S <2T opt . Traverse the path until encountering some edge e x leading to a vertex for the second time. Let e 2 be the edge immediately following e t on the path. Replace the portion of the path consisting of ex and e 2 by a single direct edge. By the triangle inequality this cannot increase the cost.
Repeat the process of replacing pairs of edges by a single edge until a Hamilton circuit is obtained.
13.25b First we show that the Shannon switching game is in PSPACE. Consider a game tree. The root indicates the initial game position. Assume SHORT moves first. The sons of the root correspond to each possible game position after a move of SHORT. In general, a vertex in the tree corresponds to the moves so far (which determine a game position) and the sons of the vertex correspond to the board position after each possible additional move.
A position is a winning position if SHORT has a forced win from the position. Thus a leaf is a winning position only if SHORT has a path from s to t. We can recursively define winning positions as follows. If vertex v is not a leaf and corresponds to a position in which it is SHORT's move, then v is a winning position if there exists a son that is a winning position.
If it is CUTs move, then v is a winning position only if every son is a winning position.
Since the tree has depth at most n, the number of vertices of G, a recursive algorithm to determine if the root is a winning position requires space at most n. Thus the problem is in PSPACE.
To show that the Shannon switching game is PSPACE-complete, we reduce the quantified Boolean formula problem to it. Consider a quantified Boolean formula and without loss of generality assume that the quantifiers alternate (otherwise add dummy quantifiers and variables) 3xi Vx 2 3x 3 ••• Vx„_i 3x n F(xi • x„).
Consider the graph, called a ladder, shown in Fig.
13.12, where n = 3. There will be additional edges (see dashed lines) but they are unimportant for the first observation.
SHORT plays first. He must at some time select either x,(l) or x,(l). This corresponds to SHORT selecting a value for the existentially quantified variable x,. The next four moves are forced, ending up with SHORT having selected x,(l), x,(2), and 3xj and CUT having selected x,(l)and x 2 (2), or SHORT having selected x,(l), .^(2), and 3x x and CUT having selected x,(l) and x 2 (2). If SHORT does not select one of x,(l), x^l), x 1 (2),x 1 (2), or 3x u then CUT wins. If SHORT selects 3x,, then CUT is given the advantage in selecting Xi(l) or xj(l). The purpose of the vertex 3x t is to consume an additional move of SHORT, thereby allowing CUT the first selection from the set {x 2 (l), x 2 (l), x 2 (2), x 2 (2)}. This means that CUT selects the value for the universally quantified variable x 2 , and so on.
Once the values for x l5 x 2 , .
. ., x„ have been selected, the dashed portion of the graph, which corresponds to the quantifier-free portion of the formulas, comes into play. Without loss of generality we can assume that F(x u x„) is in conjunctive normal form. Let F = F,aF 2 a ,,, a Fn , where each F, is a clause. Construct the tree of Fig.
13.13. Identify the root 1 with vertex 3x„ in Fig. 13.12. From vertex F, add an edge to vertex xj( 1 ) or x,( 1 ) if Xj or Xj, respectively, appears in F,. Now observe that SHORT selects vertex 1. CUT can select either F Y or 2, and SHORT selects the other. Clearly SHORT can build a path to at KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 373 KTUNOTES.IN Downloaded from Ktunotes.in 374 INTRACTABLE PROBLEMS least one Fn and CUT can force SHORT to reach only one F, and can determine which F f .
Now SHORT has a path from s to f if F, is connected to some or x,(l) which "has value one"; that is, SHORT has selected Xj(l) or x,(l).
Observe that if the quantified formula is true, then SHORT can specify the existentially quantified variables, so that regardless of CUTs choices for the universally quantified variables, F is true. Thus regardless of which F, is forced on SHORT, that F, is true and hence connected to a selected x} or xj. Hence SHORT can win.
On the other hand, if the quantified Boolean formula is false, CUT can select the universally quantified variables so that for the assignment to the x's, F is false. Then CUT forces SHORT to reach only one F„ and in particular an F, that is false for the assignment.
Thus SHORT does not complete a path, and CUT wins. Thus SHORT is guaranteed a win if and only if the quantified Boolean formula is true, and hence the Shannon switching game on vertices is complete for PSPACE.
BIBLIOGRAPHIC NOTES Cobham was the first to devote attention to the class The first NP-complete problems, including the versions of the satisfiability problems in Theorems 13.1, 13.2, and 13.3, were introduced by Cook [1971b]. Karp gave a wide variety of NP-complete problems, and clearly demonstrated the importance of the idea. Some of these problems include the vertex cover problem (Theorem 13.4), the clique cover problem (Exercise 13.4), the exact cover problem (Exercise 13.7d), the chromatic number problem (Exercise 13.6), the Hamilton circuit (Theorem 13.6), and the traveling salesman and partition problems mentioned in Section 13.2. The clique problem (Exercise 13.3) is from Cook .
Theorem 13.7, the /VP-completeness of integer linear programming, is independently due to Gathen and Sieveking and Borosh and Treybig . The proof given is from Kannan and Monma .
An enormous number of problems have since been shown /VP-complete, and those problems come from a wide variety of areas. Garey and Johnson attempt to catalog such problems, and we shall here mention only a sample of the work that has been done and the areas that have been covered. Sethi , and Bruno and Sethi cover code generation problems (Exercise 13.7b appears in the latter). Scheduling problems are con-sidered in Coffman and Ullman ; the solution to Exercise 13.7(c) can be found in both. Garey, Johnson, and Stockmeyer , and Garey, Graham, and Johnson provide a variety of powerful results, principally for graph problems. Papadimitriou and Papadimitriou and Steiglitz study path problems in graphs. Exercise 13.18 is taken from Stockmeyer , Exercise 13.10 from Garey, Johnson, and Tarjan , and Exercise 13.12 is by J. E. Hopcroft.
A number of results showing large classes of NP-complete problems appear in Hunt and Szymanski , Hunt and Rosenkrantz , Kirkpatrick and Hell , Lewis , Schaefer , and Yannakakis .
Among the promising approaches to dealing with NP-complete problems is the idea of considering approximate algorithms to the optimization versions of problems. These algo-rithms run in polynomial time but are guaranteed to come only within some specified range of the optimum. Johnson considered approximation algorithms for some of the KTUNOTES.IN Downloaded from Ktunotes.in BIBLIOGRAPHIC NOTES 375 NP-complete problems appearing in Karp . Sahni and Gonzalez were the first to prove the approximation to an NP-complete problem to be NP-complete itself (Exercise 13.15), while Garey and Johnson showed that coming within less than a factor of two of the chromatic number of a graph (number of "colors" needed to ensure that each vertex be colored differently from adjacent vertices) is /VP-complete (Exercise 13.19). Exercise 13.17 on improving an approximation to a maximal clique is also from Garey and Johnson . Rosenkrantz, Stearns, and Lewis studied approximations to the traveling salesman problem (Exercise 13.16). Christofides has improved on their results.
A number of papers have attempted to explore the structure of. \"iJ f on the hypothesis that & , Ladner [1975a] shows, for example, that if & ± A y/, then there are prob-lems that are neither in & nor TVP-complete. Adleman and Manders show that certain problems have the property that they are in & if and only if. V& = co-t Book [1974, 1976] shows inequality among certain complexity classes, such as DTIMEf/i) or DSPACEflog n\ Exercise 13.39, relating sJ> = .
\ J> to AFL theory, is from Book .
Berman and Hartmanis look at density-preserving reductions of one problem to another. Exercise 13.22 is from Berman , and Exercise 13.20 is from Levin .
Particular attention has been given to the complexity of recognizing primes. It is easy to show that the nonprimes (written in binary) are in .Y'f? t but it was not known that the primes are in . V'd? until Pratt . Thus, if the recognition of primes is NP-complete, then by Theorem 13.8, co-.
\ '// = .
\ //. Miller gives strong evidence that the recog-nition of primes written in binary is in //>. Exercise 13.23, which shows an efficient test determining primality with high probability, is from Rabin . A similar result is found in Solovay and Strassen . Exercise 13.24 is from Brassard, Fortune, and Hopcroft .
The first PSPACE-complete problems were introduced by Karp , including CSL recognition (Theorem 13.11) and kk = I" for regular expressions (Exercise 13.25a).
PSPACE-completeness of quantified Boolean formulas was shown by Stockmeyer .
Exercise 13.25(b), PSPACE completeness of the Shannon switching game, is by Even and Tarjan , Stockmeyer gives a hierarchy of problems between .
\ \ J f and PSPACE, on the assumption that .
\ /J> + PSPACE.
Problems complete for .J / with respect to logarithmic space reductions have been con-sidered by Cook [1973b], Cook and Sethi , Jones , Jones and Laaser (including Theorem 13.12) and Ladner [1975b] (Exercise 13.29b). Problems complete for NSPACE(log n) with respect to log space reductions are considered in Savitch , including Theorem 13.13 (on reachability), Sudborough [1975a,b], Springsteel , and Jones, Lien, and Laaser . Exercise 13.30 is from Jones, Lien, and Laaser .
The first problem proved to require exponential time (in fact, exponential space) was presented by Meyer and Stockmeyer . The problem is similar in spirit to that of Theorem 13.15. The lower bounds on the complexity of the theory of reals with addition (Theorem 13.16) and of Presburger arithmetic (Exercise 13.35) are from Fischer and Rabin . The upper bounds for these problems are from Cooper , Ferrante, and Rack-off , and Oppen . Berman ; and Bruss and Meyer put what are, in a sense, more precise bounds (outside the usual time-space hierarchies) on these problems.
The undecidability of Presburger arithmetic with arrays is from Suzuki and Jefferson .
The literature contains a number of papers that deal with the complexity of a variety of problems and their special cases, dividing problems into groups, principally: polynomial, KTUNOTES.IN Downloaded from Ktunotes.in 376 INTRACTABLE PROBLEMS /VP-complete, PSPACE-complete, and provably exponential. A sample of the areas covered include Diophantine equations in Adleman and Manders , asynchronous computation in Cardoza, Lipton, and Meyer , problems about regular expressions in Hunt (including Exercise 13.32), Hunt, Rosenkrantz, and Szymanski , and Stockmeyer and Meyer , problems about context-free grammars in Hunt and Rosenkrantz [1974, 1977], Hunt and Szymanski [1975, 1976], and Hunt, Szymanski, and Ullman (including Exercise 13.31), and game theory in Schaefer .
The results of Section 13.7 and Exercise 13.38, on the & = jV0> question in the presence of oracles, are from Baker, Gill, and Solovay . However, Kozen presents another viewpoint on the issue. Exercise 13.26 is from there. Ladner, Lynch, and Selman studied the different kinds of bounded reducibility, such as many-one, Turing, and truth tables. Another attack on the & = JfgP question has been the develop-ment of models whose deterministic and nondeterministic time-bounded versions are equivalent. The vector machines (Pratt and Stockmeyer ) are the first, and other models have been proposed by Chandra and Stockmeyer and Kozen . The reader should also note the equivalence for space-bounded versions of the "auxiliary PDA's" discussed in Section 14.1.
KTUNOTES.IN Downloaded from Ktunotes.in CHAPTER 14 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES Numerous models and classes oflanguages have been introduced in the literature.
This chapter presents a few of those that appear to be of greatest interest.
Section 14.1 discusses auxiliary pushdown automata, which are PDA's with two-way input and additional general purpose storage in the form of a space-bounded Turing tape. The interesting property of auxiliary PDA's is that for a fixed amount of extra storage, the deterministic and nondeterministic versions are equivalent in language-recognizing power, and the class of languages accepted by auxiliary PDA's with a given space bound is equivalent to the class of languages accepted by Turing machines of time complexity exponential in that space bound.
Section 14.2 is concerned with stack automata, which are PDA's with the privilege of scanning the stack below the top symbol, but only in a read-only mode. Languages accepted by variants of the two-way stack automaton turn out to be time- or space-complexity classes.
Section 14.3 is devoted to indexed languages, since they arise in a number of contexts and appear to be a natural generalization of the CFL's. Finally, Section 14.4 introduces developmental systems, which attempt to model certain biological patterns of growth.
14.1 AUXILIARY PUSHDOWN AUTOMATA An S(n) auxiliary pushdown automaton (APDA) is pictured in Fig.
14.1. It consists of 1) a read-only input tape, surrounded by the endmarkers, $ and $, 2) a finite state control, 377 KTUNOTES.IN Downloaded from Ktunotes.in 378 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES Read -only input tape Finite control 1 1 Stack Fig. 14.1 Auxiliary PDA.
3) a read-write storage tape of length S(n\ where n is the length of the input string w, and 4) a stack.
A move of the APDA is determined by the state of the finite control, along with the symbols scanned by the input, storage, and stack heads. In one move, the APDA may do any or all of the following: 1) change state, 2) move its input head one position left or right, but not off the input, 3) print a symbol on the cell scanned by the storage head and move that head one position left or right, 4) push a symbol onto the stack or pop the top symbol off the stack.
If the device is nondeterministic, it has a finite number of choices of the above type. Initially the tape heads are at the left end of the input and storage tapes, with the finite control in a designated initial state and the stack consisting of a designated start symbol. Acceptance is by empty stack.
Equivalence of deterministic and nondeterministic A PDA's The interest in APDA's originates from the discovery that deterministic and non-deterministic APDA's with the same space bound are equivalent, and that S(n) space on an APDA is equivalent to c S{n) time on a Turing machine. That is, the following three statements are equivalent.
1) L is accepted by a deterministic S(n)-APDA.
2) L is accepted by a nondeterministic S(n)-APDA.
3) L is in DTIME(r S( " ) ) for some constant c.
These facts are established in the following series of lemmas.
KTUNOTES.IN Downloaded from Ktunotes.in 14.1 | AUXILIARY PUSHDOWN AUTOMATA 379 Lemma 14.1 If L is accepted by a nondeterministic 5(«)-APDA A with S(n) > log n, then L is in DTIME(cS(n) ) for some constant c.
Proof Let A have 5 states, t storage symbols, and p stack symbols. Given an input of length n, there are n + 2 possible input head positions, 5 possible states, S(n) possible storage head positions, and t S(n) possible storage tape contents, for a total of s(n) = (n + 2)sS(n)t S{n) possible configurations^ As S(n) > log n, there is a constant d such that s(n) < dSin) for all n > 1.
Construct a TM M that performs the following operations on input w of length n.
1) M constructs a PDA Pw that on £-input simulates all moves of A on input w.
2) M converts Pw to a CFG Gw by the algorithm of Theorem 5.4.
For fixed A, Pw is a different PDA for each w, with the state and contents of input and storage tapes of A encoded in the state of Pw . N(PW ) is {e} or 4> depend-ing on whether or not A accepts w.
Pw has at most s(n) < d S(n) states and p stack symbols. Therefore Gw has at most pd 2S(n) + 1 variables. As A can push only one symbol, no right side of a production of Gw has more than two variables, so there are at most rd S{n) produc-tions for any nonterminal of Gw , where r is the maximum number of choices that A has in any situation. Thus the test of Theorem 6.6, to tell whether L(GW ) is empty, takes time proportional to rp 2d5S{n\ at most. Since r, p, and d are constants, there is a constant c such that M can determine in time at most r S(n) whether L(G W ) is nonempty, i.e., whether w is accepted by A.
Lemma 14.2 If L is in DTIME(T(rc)), then L is accepted in time T(n) by a one-tape TM M that traverses its tape, making a complete scan in one direction, reaching the first cell it has never before scanned, reversing direction and repeat-ing the process, as shown in Fig.
14.2.
Fig.
14.2 Traversal pattern of TM M.
t Note that a "configuration" in the sense used here does not include the stack contents.
KTUNOTES.IN Downloaded from Ktunotes.in 380 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES Proof By Theorems 12.3 and 12.5, L is accepted by a \T2 (n) time-bounded one-tape TM M X.M simulates M l9 marking on a second track M^s head position and the cells which M l has already scanned. As long as the head ofM x travels in the same direction as M's head, M can simulate a move ofM t with each of its own moves. When M x moves in the opposite direction, M leaves the head marker, completes its scan and simulates that move on the return pass. Thus M simulates at least one move ofM x per pass, taking at most £}L / o )r2(" ) n + i < ^(n) moves to complete the simulation of M v Lemma 14.3 If L is in DTIME(c S(n) ) for any constant c, then L is accepted by a deterministic S(«)-APDA.
Proof By Lemma 14.2, L is accepted by a c4S(n) time-bounded one-tape TM M with the traversal pattern of Fig.
14.2.
Define d — c4 so that M is d S{n) time bounded. Let the triple (q, Z, t) stand for the statement that at time r,i M is in state q scanning symbol Z, where t < dS(n\ Note that since the head motion of M is independent of the data, the cell scanned at time t is easily calculated from t.
The heart of the construction of a deterministic S(n)-APDA A that accepts L is the recursive procedure TEST of Fig.
14.3, which assigns value true to the triple (q, Z, t) if and only if 1) t = 0, q is the start state, and Z is the symbol in the first tape cell of M, or 2) M scans some cell for the first time at time r, Z is the original symbol in that tape cell, and there is a triple (p, X y t — 1) that is true and implies that M enters state q after one move, or 3) M previously scanned the cell visited at time t and there are true triples (p l9 X l9 t -1) and (p 2 , X 2 , ?) such that the first triple implies that state q is entered after one move, and the second implies that Z was left on the tape cell the last time the tape cell was scanned. Recall that the head motion of M is uniform, and thus the time t' at which the cell was last visited is easily calculated from t.
As TEST only calls itself with smaller third arguments, it eventually termin-ates. The S(n)-APDA A evaluates TEST by keeping the arguments on the storage tape. When TEST calls itself, A pushes the old arguments onto the stack, and when TEST returns, A pops them off the stack and puts them on the storage tape.
The complete algorithm that A executes is for each triple (q, Z, f) such that q is an accepting state and 0 < t < d S(n) do if TEST(g, Z, t) then accept Theorem 14.1 The following are equivalent for s(n) > log n.
1) L is accepted by a deterministic S(m)-APDA.
t "At time f means "after t moves have elapsed," so initially, / = 0.
KTUNOTES.IN Downloaded from Ktunotes.in 14.2 | STACK AUTOMATA 381 procedure TEST(g, Z, r); begin if t = 0, q is the initial state of M and Z is the first input symbol then return true; if 1 < t < n and Z is the fth input symbol, or r = in + i(i — l)/2 for some integer i > 1 and Z = B then for each state p and symbol X do if M enters state q when scanning X in state p, and TEST(p, X, t — 1) then return true; / the times in + i(i — l)/2 are exactly the times when M scans a new cell / if t > n and t j= in + i(i — l)/2 for any integer i > 1 then begin let t' be the previous time M scanned the same cell as at time f; for all states p x and p 2 and symbols Xi and X 2 do if M enters state q when scanning X x in state p, and M writes Z when scanning X 2 in state p 2 and TEST(p!, X u t — 1) and TEST(p 2 , X 2 , t') then return true end; return false end Fig. 14.3 The procedure TEST.
2) L is accepted by a nondeterministic 5(n)-APDA.
3) L is in DTIME(c lS( " ) ) for some constant c.
Proof That (1) implies (2) is obvious. Lemma 14.1 established that (2) implies (3) and Lemma 14.3 established that (3) implies (1).
Corollary L is in & if and only if L is accepted by a log n-APDA.
14.2 STACK AUTOMATA The stack automaton (SA) is a PDA with the following two additional features.
1) The input is two-way, read-only with endmarkers.
2) The stack head, in addition to push and pop moves at the top of the stack can enter the stack in read-only mode, traveling up and down the stack without rewriting any symbol.
A stack automaton is shown in Fig.
14.4, in read-only mode.
A move of an SA is determined by the state, the input, and stack symbols scanned, and whether or not the top of the stack is being scanned by the stack head. In either case, in one move the state may change and the input head may move one position left or right. If the stack head is not at the top of the stack, a move may also include a stack head motion, one position up or down the stack. If the stack head is at the top, the permissible stack actions are: 1) push a symbol onto the stack, KTUNOTES.IN Downloaded from Ktunotes.in 382 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES \v Ai Finite control Top of stack Fig. 14.4 A stack automaton.
2) pop the top symbol off the stack, or 3) move one position down the stack without pushing or popping.
In actions (1) and (2) the stack head stays at the top; in action (3) it leaves the top of stack and enters the read-only mode, which it may leave only by returning to the top of the stack.
Initially, the input head is at the left end, the finite control is in a designated initial state, and the stack consists of a single designated start symbol. Acceptance is by final state.
If there is never more than one move in any situation, the device is deter-ministic (a DSA); if there is a finite number of choices of moves in any situation, the automaton is nondeterministic (an NSA). If the device never pops a symbol it is nonerasing (an NEDSA or NENSA). If the input head never moves left, the stack automaton is one-way (a 1DSA, 1 NENSA, and so on). In the absence of any statement to the contrary, we shall assume an SA is two-way, deterministic, and permits erasing.
Example 14.1 Let L = {0"\ n2"\n > 1}. We design an SA to accept L as follows.
The input head moves right at each move. While O's are encountered, they are pushed onto the stack above the bottom marker (start symbol) Z0 . The stack head remains at the top of stack in read-write mode. Fig.
14.5(a) shows the situation after reading O's. On seeing the first 1, the stack head moves down, entering the read-only mode. As successive l's are read, the stack head moves one position down for each 1 (but if the first 2 is not seen at the same time the stack head reaches the bottom marker, there is no next move, and the SA does not accept).
The situation in which the SA then finds itself is shown in Fig. 14.5(b). As 2's are scanned on the input, the stack head moves up one position for each 2. A move to an accepting state is permissible only when the stack head is at the top and $ is scanned on the input, as in Fig.
14.5(c). Of course, the state from which this move can be made is only entered after we have seen 2's, so we cannot accept inputs like or $00$.
Note that the SA we have described is one-way, deterministic, and noneras-ing.
KTUNOTES.IN Downloaded from Ktunotes.in 14.2 | STACK AUTOMATA 383 . With input positions encoded in binary, a transition table requires only cn log n bits for some constant c that depends on the number of states of the given SA.
If the SA is nondeterministic, then for each q and i, the transition table must give the set of (p, j) pairs such that started in state q, with input position /, and the stack head at the next-to-top stack symbol, the top of stack can be reached in state p and input position j. The number of possible transition tables for an s state NSA with input of length n is [2 s(M+2)+ ] s(n+2) < 2 C" 2 , so such a transition table can be encoded in cn 2 bits, where c depends only on the number of states of the NSA.
Characterization of stack languages by time and space complexity classes We shall show that a deterministic SA can be simulated by an n log n-APDA and conversely that an n log m-APDA can be simulated by a DSA establishing the equivalence of DSA and n log rc-APDA. In a similar manner we establish the KTUNOTES.IN Downloaded from Ktunotes.in 384 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES equivalence of NSA and « 2-APDA. For nonerasing SA we establish the equiv-alence of NEDSA and DSPACE(n log n) and the equivalence of NENSA and NSPACE(« 2 ). A series of lemmas is used.
Lemma 14.4 Each type of stack automaton is equivalent to one of the same type that accepts only at the top of stack.
Proof We modify a given SA so that in any accepting state it moves its stack head up the stack until the top is reached.
Lemma 14.5 If L is accepted by an NEDSA, then L is in DSPACE(« log n).
Proof Given an NEDSA A that accepts only at the top of the stack, we construct a Turing machine M that simulates A by keeping track of ,4's state, input head position, top stack symbol, and the transition table for the portion of the stack below the top symbol. The initial transition table is the table associated with the empty string ("undefined" for all q and i). We need only explain how to construct the transition table T' associated with the stack string X 1X 1 ••• Xm given the table 7 for X X X 2 • Xm . x .
For each state q and input position U execute the algorithm of Fig.
14.6. The algorithm keeps track of the sequence of state-input-position pairs (p, j) in which Xm is scanned. Each time the stack head moves to Xm „ u T is consulted to determine the next state-position pair in which Xm will be scanned if any. The variable COUNT checks that the length of the sequence of (/?, y')'s does not exceed the product of s, the number of states, and n + 2, the number of input positions. If so, A is surely in a loop, so that value of T(q y i) is "undefined." begin COUNT:- 0; (M -to); while COUNT < s(n + 2) do begin COUNT := COUNT 4-1 suppose A in state p, scanning stack symbol X m , at input position ; enters state r, moves the input head to position k and the stack head in direc-tion D; if D — "up" then return (r, k); if D — "stationary" then (p, j) : = (r, k); if D = "down" then if T(r, k) = "undefined" then return "undefined" else (p, ./):= 7(r, k) end return "undefined" end Fig. 14.6 Algorithm to compute transition table for NEDSA.
KTUNOTES.IN Downloaded from Ktunotes.in 14.2 | STACK AUTOMATA 385 Note that for given (q, i) the algorithm of Fig.
14.6 requires only 0(log n) space to hold the value of COUNT. Thus T'can be computed from Tand Xm in the space it takes to store T and T\ which is n log n. The TM M has only to simulate A directly when the stack head is at the top of the stack, consult the current transition table when the stack head leaves the top, and compute a new transition table (throwing away the old) when a stack symbol is pushed. As stack symbols are never erased, we need not preserve the stack.
Lemma 14.6 If L is accepted by a NENSA, then L is in NSPACE(« 2 ).
Proof The proof is similar to that of the previous lemma, save that n 2 space is needed to store the transition matrix, and the simulation must be nondeter-ministic.
Lemma 14.7 If L is accepted by DSA, then L is accepted by a n log n-APDA.
Proof The proof is again similar to that of Lemma 14.5. The APDA uses its stack (which it may not enter in read-only mode, of course) to hold the stack of the DSA. Between each DSA stack symbol the APDA stores a transition table. The transition table above a particular stack symbol corresponds to the entire stack, up to and including that symbol. The topmost stack symbol and the table for the stack below it are placed on the storage tape. When the DSA pushes a symbol, the APDA pushes the table that is on its storage tape along with the old top stack symbol onto its own stack, and computes the new table as in Lemma 14.5. When the DSA pops a symbol, the APDA discards the top stack symbol and then moves the top table to its storage tape.
Lemma 14.8 If L is accepted by an NSA, then L is accepted by an /7 2-APDA.
Proof The proof is a combination of the ideas introduced in Lemmas 14.6 and 14.7. Note that by Theorem 14.1 the APDA may be made deterministic.
We now turn to the simulation of space-bounded devices by stack automata.
The key idea here is that the SA can use its input of length n to count n symbols or "blocks" of symbols down its stack. A sequence of ID's representing a computa-tion of a space-bounded device is constructed on the stack by successively copying the top ID onto the stack, making changes represented by one move of the space-bounded device. The ability to count down n symbols or "blocks" of sym-bols allows the SA to copy the current ID onto the top, symbol by symbol.
As a simple introduction consider the simulation of a deterministic linear bounded automaton M by an NEDSA A. Given input w = a x •• a n , A pushes [q0 ai]a2 "' an^ onto i ts stack, where q0 is the start state and # is a special symbol separating ID's. The state is combined with the symbol scanned, so an ID is always exactly n symbols long. Suppose A has constructed a stack that is a sequence of ID's, including the first i symbols of the next ID: frX j X 2 "^n"^"^ 1 ^ 2 ^ i KTUNOTES.IN Downloaded from Ktunotes.in 386 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES (Actually one or two of the X's in the ID being constructed may differ from the corresponding symbols in the ID below, due to the move made by M). Starting at the left end of the input, A repeatedly moves one position right on the input and one position down the stack, until the right endmarker is reached on the input. At this point yfs stack head will be n + 1 symbols from the top of the stack, scanning X i+ j of the last complete ID. A looks one symbol above and below X i+ j to see if X i+ j changes in the next ID because of the move made by M. A then moves to the top of the stack and pushes either X i+ x or the symbol replacing X i+ x in the next ID due to the move of M. A accepts if and only if M enters an accepting state.
Actually a stack automaton can simulate devices with ID's of length greater than n by more clever use of the input. In particular, a DSA can manipulate ID's of length n log n, and an NSA can manipulate ID's of length n 2 . The nondeter-ministic case is easier, so we present it first.
Lemma 14.9 If L is in NSPACE(n 2 ), then L is accepted by a NENSA.
Proof Since n 2 is greater than n we may assume L is accepted by a one-tape TM rather than an off-line TM. An ID of length n 2 is represented by listing the tape symbols, combining the state with the symbol scanned by the tape head. A marker is inserted after every n symbols. The n symbols between 's make up a block.
Successive ID's are placed on the stack as in the description of the LBA above.
Suppose j blocks and i symbols of the (j + l)rcth block have been copied. The input tape is used to measure n 's down the stack to the (j + l)nth block of the previous ID. A position k y 1 < k < n, in the (j + l)nth block is guessed. Checking one symbol above and below determines if the symbol is affected by a move of the TM. If so, a move is guessed, provided a move for this ID has not been guessed previously; otherwise the symbol is recorded in the state of the SA. The input tape is then used to record k by alternately moving the input head one symbol left (starting at the right end) and the stack head one symbol down until a is encountered. Next the stack head moves to the top of the stack and compares k with i, the number of symbols of the jth block already copied. If A' j= i + 1, this sequence of choices "dies." If k = i + 1, then the next symbol of the new ID is placed on top of the stack. The input is then used to determine if i + 1 = n. If so a is printed, and then it is checked whether j + 1 = n. In the case j + 1 = n, a # is placed on the stack marking the end of an ID. Acceptance occurs if the symbol copied includes a final state. Otherwise the next symbol is copied.
A small but important point is that once a move is guessed in copying an ID, the guess cannot be changed on copying a subsequent symbol in that ID. Other-wise an invalid successor ID may be constructed.
Theorem 14.2 The family of languages accepted by nondeterministic, nonerasing stack automata is exactly NSPACE(n 2 ).
Proof Immediate from Lemmas 14.6 and 14.9.
KTUNOTES.IN Downloaded from Ktunotes.in 14.2 | STACK AUTOMATA 387 Theorem 14.3 The family of languages accepted by nondeterministic stack auto-mata is exactly (J c>0 DTIME(c" 2 ).
Proof By Theorem 14.1, L is in (Jc>0 DTIME(c" 2 ) if and only if L is accepted by an rc 2-APDA. By Lemma 14.8, if L is accepted by an NSA then L is accepted by a deterministic rc 2-APDA. Thus it suffices to show that a deterministic n 2-APDA A can be simulated by an NSA S.
We assume that the input of A is kept on the storage tape of A rather than on a read-only input, since n 2 exceeds n. The stack of S will hold the stack of A as well as a sequence of ID's representing the storage tape of A. Suppose S has the current contents of A's storage tape on top of its stack, and A pushes a symbol. S guesses the tape contents of A when that symbol is popped and places its guess on top of the stack. Then S pushes the symbol pushed by A and creates the new current tape contents of A from the old, as in Lemma 14.9. The guessed ID intervening is ignored while running up and down the stack; its symbols can be chosen from a separate alphabet, so S can skip over it.
If A pops a symbol, S checks that the guessed ID below that symbol is correct; that is, the guessed ID is the storage tape of A after the pop move. The current ID of A held on top of S's stack is popped one symbol at a time, and each symbol popped is compared with the corresponding symbol of the guessed ID by a method similar to that of Lemma 14.9.
If the guess is correct, the guessed ID becomes the current storage tape content of A, and the simulation of A proceeds; if not, this sequence of choices by S "dies." S accepts if and only if A empties its stack.
Corollary L is accepted by an NSA if and only if L is accepted by an h 2-APDA.
Proof The "only if" portion was established in Lemma 14.8. The "if" follows immediately from Theorems 14.1 and 14.3.
In the deterministic case the function n 2 is replaced by n log n in the analogs of Theorems 14.2 and 14.3. The reason for this is that in the construction of Lemma 14.9 the NSA made an essential use of its nondeterminism in copying ID's of length n 1 . A DSA is able only to copy ID's of length n log n.
Lemma 14.10 If L is in DSPACE(rc log n), then L is accepted by an NEDSA.
Proof Let L be accepted by some one-tape TM M that uses exactly n log n cells.
Let t be the number of symbols of the form X or [qX\ where X is a tape symbol and q a state. These symbols are identified with the digits 1,2, .
.
.
, fin base t + 1.
Strings of [\ogt+1 (n)\ such symbols ofM are encoded as blocks of between 0 and (n — 1) 0's. There is an integer c, depending only on M, such that an ID ofM may be represented by cn blocks of 0's, each block coding [\ogt+l (n)\ symbols, provided n > t.
Design a stack automaton S to construct a sequence of ID's of M, each ID KTUNOTES.IN Downloaded from Ktunotes.in 388 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES being a sequence of cn blocks of between 0 and (n — 1) O's separated by markers, .
Blocks are copied to the top of the stack by using the input to count cn 's down the stack, measuring the length of the block to be copied on the input, moving to the top of stack and pushing an equal number of O's onto the stack.
Before a new block is placed on the stack, it is necessary to determine which, if any, symbols change. To do so, decode 0 by repeatedly dividing by t -h 1, the successive remainders being the successive symbols of the ID. The division is accomplished by measuring k on the input and then moving the input back to the endmarker, placing an X on the stack for every t 4-1 positions the input head moves. The X's are not part of an ID. The finite control computes k mod (t H-1), and the resulting digit is placed above the X y s. The block of X's is then measured on the input and the process repeated until the block of X's has length zero. The digits written on the stack between blocks of X's are the desired block of the ID.
S checks whether the head is scanning a symbol in the block and also notes if the head moves into an adjacent block. The blocks are re-encoded into strings ofO to (n — 1) O's, making the necessary changes to reflect the move of M. The process of re-encoding is the reverse of that just described. Note that since S is nonerasing, it never gets rid of the X y s or digits on its stack; they are simply ignored in subsequent computation. Also, before copying a block, S must decode the block above, to see whether the head of M moves left into the present block.
S initializes its stack by coding its own input as an ID of M. The details of this process are omitted. S accepts if it discovers that M enters an accepting state.
Theorem 14.4 L is accepted by a deterministic nonerasing stack automaton if and only if L is in DSPACE(n log n).
Proof From Lemmas 14.5 and 14.10.
Theorem 14.5 L is accepted by a deterministic stack automaton if and only if Lis m Uoo DTIME(h c").
Proof Note that n cn = 2CM,og By Theorem 14.1, L is in |J C>0 DTIME(rc c") if and only if L is accepted by an n log h-APDA. By Lemma 14.7, if L is accepted by a DSA, then L is accepted by a deterministic n log rc-APDA. Thus it suffices to show that if L is accepted by a deterministic n log n-APDA A, then L is accepted by a DSA 5. Again we assume that /Ts input tape is combined with its storage tape.
The proof parallels Theorem 14.3, using the techniques of Lemma 14.10 to repre-sent storage tapes of A and simulate moves of A. However, when A pushes a symbol X onto its stack, S, being deterministic, cannot guess the storage tape contents of A when A eventually pops that X. Instead S cycles through all possible ID's systematically. If it has made the wrong choice, it generates the next possible ID and restarts the simulation of A from the time X was pushed by A. The fact that A empties its stack to accept assures that if A accepts, 5 will eventually get a chance to generate the correct choice.
KTUNOTES.IN Downloaded from Ktunotes.in 14.3 | INDEXED LANGUAGES 389 Corollary L is accepted by a DSA if and only if L is accepted by an n log n-APDA.
Proof The "only if" portion was established in Lemma 14.7. The "if" follows immediately from Theorems 14.1 and 14.5.
One-way stack automata are not powerful enough to simulate tape-bounded devices. However, there is one important containment relation, which we state without proof.
Theorem 14.6 If L is accepted by a 1NSA, then L is in DSPACE(rc).
14.3 INDEXED LANGUAGES Of the many generalizations of context-free grammars that have been proposed, a class called "indexed" appears the most natural, in that it arises in a wide variety of contexts. We give a grammar definition here. Other definitions of the indexed languages are cited in the bibliographic notes.
An indexed grammar is a 5-tuple (F, T, /, P, S), where V is the set of variables, T the set of terminals, / the set of indices, Sin Vis the start symbol, and P is a finite set of productions of the forms 1) A-xx, 2) A-^Bf or 3) Af- a, where A and B are in K,/is in /, and a is in (V u 7).
Derivations in an indexed grammar are similar to those in a CFG except that variables may be followed by strings of indices. (Terminals may not be followed by indices.) When a production such as A BC is applied, the string of indices for A is attached to both B and C. This feature enables many parts of a sentential form to be related to each other by sharing a common index string.
Formally, we define the relation => on sentential forms, which are strings in (VI u T), as follows. Let /? and y be in (VI u 7), 5 be in /, and X i in V u T.
1) If A -> X x X 2 ••• Xk is a production of type (1) then PASy^pX.S.X^'-' Xk Sk y y where = 5 ifX { is in Kand 3^ = c ifX { is in 7. When a production of type (1) is applied, the string of indices S distributes over all the variables on the right side.
2) If A -> B/ is a production of type (2), then fiA5y=> fiBfSy. Here /becomes the first index on the string following variable B, which replaces A.
3) If Af-+ X x X 2 •" X k is a production of type (3), then PAfSy^pX.S.X^---k S k y, where = 5 if X,- is in V and <>, = c if X, is in T. The first index on the list for A is consumed, and the remaining indices distribute over variables as in (1).
KTUNOTES.IN Downloaded from Ktunotes.in 390 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES We let ^> be the reflexive and transitive closure of => as usual, and define L(G) to be {w | S ^> w and w is in T}.
Example 14.2 Let G = ({5, T, A, B, C}, {a, by c}, {/, P, S), where P consists of S - Tg, Af-+ aA, Ag - a, T-+Tfi Bf-+bB, Bg^b, T -> ABC, Cf-+cC, Cg->c.
An example derivation in this indexed grammar is S^Tg^Tfg^AfgBfgCfg => aAgBfgCfg => aaBfgCfg => aabBgCfg => aabbCfg => aabbcCg => aabbcc.
In general, S !> Tf'g => AfgBfgCfg - ai+ >bi+l ci+ \ As the only freedom in derivations of G consists of trivial variations in order of replacement and the choice of how many times to apply T -> Tf, it should be clear that L(G) = {a nb nc"\n> 1}.
This language is not context free, of course.
We state without proof two major results about indexed languages.
Theorem 14.7 (a) If L is accepted by a one-way nondeterministic stack automa-ton, then L is an indexed language, (b) If L is an indexed language, then L is a context-sensitive language.
In fact, (a) can be strengthened by defining a generalization of an SA, called a "nested stack automaton," whose one-way nondeterministic variety exactly char-acterizes the indexed languages. The nested SA has the capability, when the stack head is inside its stack in read-only mode, to create a new stack. However, this stack must be destroyed before the stack head can move up in the original stack.
The process of creating new stacks is recursive and allows the creation of new stacks to an arbitrary depth.
14.4 DEVELOPMENTAL SYSTEMS The application of grammars to the study of growth in cellular organisms in-troduced new grammar families called /^systems. These grammar families differ from the Chomsky grammars in that 1) no distinction between terminals and nonterminals is made, and KTUNOTES.IN Downloaded from Ktunotes.in 14.4 | DEVELOPMENTAL SYSTEMS 391 2) at each step in a derivation, a production is applied to each symbol of a sentential form, rather than to just one symbol or a short substring.
The modeling of organisms by {^systems allows the testing of hypotheses concern-ing the mechanisms behind certain observable biological phenomena. Here we content ourselves with defining only the most basic family of these grammars, called OZy-systems. (The 0 stands for zero symbols of context; the L acknowledges Arvid Lindenmeyer, who first used these grammars to study growth in organisms.) A OZ^grammar is a triple G = (£, P, a), where E is a finite alphabet called the vocabulary, a is a string in Z + called the start string, and P is a set of productions of the form a -> /?, where a is in £ and ($ is in I. The relation => is defined by a l a 2 "' an =>0t l <X 2 an if a { -» a, is in P for 1 < i < n. Note that a { -> a t might be a production, permitting us to avoid substituting for a { . Otherwise, a substitution must be made for each symbol. The substitution for different occurrences of the same symbol need not be the same. The relation is the reflexive, transitive closure of =>, and L(G) is defined to be {P\ol^>P}.
Example 14.3 Let G = ({a, b}, P, a), where P consists of a -+ b and b -> ab. In this case, there is only one production for each symbol, so there is really only one (infinite length) derivation, and every word in the language appears in that deriva-tion. The derivation is «=>/?=>«/?=> bab => abbab => bababbab =>.
Note that the length of words in L(G) are exactly the Fibonacci numbers defined by/, = f 2 = 1 and / = , +/_ 2 for i > 3. One can prove by induction on / > 3 that the /th word in the derivation has , ^s and 2 as, a total of)\ symbols.
Example 14.4 The language {a, aa) is not a OL-language. Suppose L(G) = {a, aa}, where G = ({a}, F, a). Then a must be a or aa. Now all productions are of the form a -> a for some i > 0. Suppose a = a. Surely there cannot be a production a -> a\ for i > 3. Then there must be a production a -> aa, else aa could never be gen-erated. But then a=>aa=> aaaa, a contradiction. Suppose next that a = aa. There must be a production a-> c, else all strings in L(G) are of length two or more. But then aa => c, so L(G) =f= {a, aa} again.
A basic result about (^languages is the following.
Theorem 14.8 If L is a OZ^language, then L is an indexed language.
Proof Let G x = (Z, P u cc) be a C\L-grammar. Define indexed grammar G 2 = (K Z, {/ ^}, P2 , 5), where K = {5, T} u {/l fl |fl is in £}, KTUNOTES.IN Downloaded from Ktunotes.in 392 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES and P2 contains S^Tg, T-+Tf, T-+Aai Aa2 ~-Aak if ct = a 1 a2 '-aky AJ-+ A bl Ab2 • -- A bj for each production a-+b 1 b 2 b} in Pu Aa g -> a for each a in £.
Informally the string of/'s counts the number of steps in a derivation of G l9 and index g marks the end of an index string, allowing a variable to be replaced by the terminal it represents. An easy induction on the length of a derivation shows that S ^> Tfg ^> Abl f l - jgAb2 f^g • • Abk f^g in G 2 if and only if a ^> b x b 2 ' ' ' bk by a derivation of / steps in G x . Thus S^>/l bl^ 2 ^ A bk g>b l b 2 ••• fck if and only if Gt^>b { b 2 ••• bk .
14.5 SUMMARY Figure 14.7 shows the various equivalences and containments proved or stated in this chapter, plus some others that are immediate from definitions. Containments are indicated by upward edges.
EXERCISES 14.1 a) Design a one-way DSA to recognize the language {OT 2 \n > 1}.
b) Design a one-way NSA to recognize the language {ww|w is in (0 + 1)}.
14.2 Design a two-way DSA to accept the set of binary strings whose value, treated as an integer, is a power of 3.
14.3 Since every CFL can be recognized in polynomial time by the CYK algorithm, the corollary to Theorem 14.1 implies that every CFL is recognized by some deterministic log h-APDA. Give a direct construction of such an APDA from a CFG.
14.4 Show that the family of 1NSA languages and the family of 1NENSA languages form full AFL's.
14.5 Show that the families of 1DSA languages and 1NEDSA languages are closed under: a) intersection with a regular set, b) inverse GSM mappings, c) complementation, d) quotient with a regular set.
14.6 Give indexed grammars generating the following languages.
Sa) {0" | n is a perfect square} b) {0" | n is a power of 2} c) {0"|n is not a prime} d) {ww|w is in (0 + 1)} KTUNOTES.IN Downloaded from Ktunotes.in EXERCISES 393 Fig. 14.7 Containments among classes of languages.
14.7 Give OL-grammars generating the following languages.
a) {a n | n is a power of 2} b) {wcwR \ w is in (0 4- 1)} S14.8 Give a OJ^grammar with the property that every string generated is of length a perfect square and furthermore for every perfect square there is at least one string of that length generated.
14.9 Of the eight subsets of {c, a> aa}> how many are OZ^languages?
14.10 Show that the family of OJ^languages is not closed under any of the AFL operations.
14.1 1 Show that it is decidable whether the language generated by an indexed grammar is empty.
14.12 Show that Greibach's theorem (Theorem 8.14) applies to the 1NEDSA languages, and that "= £" is undecidable for this class.
14.13 Show that it is undecidable whether two OL-languages are equivalent.
Solutions to Selected Exercises 14.6 a) We make use of the fact that the nth perfect square is the sum of the first n odd integers. The indexed grammar with productions S-Ag A-+ Af A-+B KTUNOTES.IN Downloaded from Ktunotes.in 394 HIGHLIGHTS OF OTHER IMPORTANT LANGUAGE CLASSES B-+CD Df-+ B Dg-e C/-+OOC Cg-0 generates {0 n | n is a perfect square}. The derivations are trivial variations of the following derivation.
S=>Ag±>Af n - lg=>Bf n -l g => Cf-l gDf»-l g => Cf n ~ l gBf n - 2 g ^Cf"-l gCf n - 2gDf»-2 g ^Cf n -x gCf n - 2gBf n -"g=>-> ~Cf n -l gCf n - 2g-CfgCgDg =>Cf n -lgCf n -lg-CfgCg >V»-l Cf-xg-Cf 9 C9 >-|> Q2„-l 0 2„-3 ... Q3Q1 = Qn2 14.8 We again make use of the fact that the nth perfect square is the sum of the first n odd integers. Consider the OL-grammar ({a, by c}, {a - abbc\ b -> be, c - c}, a). A simple induc-tion shows that the nth string generated has one a, 2(n — I) b y s y and (n — I) 2 c's. Thus the length of the >ith string is 1 4- 2(n -1) + (n -l) 2 or n 2 .
BIBLIOGRAPHIC NOTES The auxiliary pushdown automaton and Theorem 14.1 are from Cook [1971a].
Earlier, Mager had considered "writing pushdown acceptors," which are n-APDA's. Stack automata were first considered by Ginsburg, Greibach, and Harrison [1967a, b]. Theorems 14.2 and 14.4, relating nonerasing stack automata to space complexity classes, are from Hopcroft and Ullman [1967a], although the fact that the CSL's are contained in the NEDSA languages was known from Ginsburg, Greibach, and Harrison [1967a]. Theorems 14.3 and 14.5, relating stack languages to APDA's and time complexity classes, are by Cook [1971a].
The basic closure and decision properties of one-way stack languages were treated in Ginsburg, Greibach, and Harrison [1967b]. Exercise 14.5(d), the closure of 1DSA languages under quotient with a regular set, is by Hopcroft and Ullman [1968b]. Theorem 14.6, containment of the 1NSA languages in DSPACE(n) is by Hopcroft and Ullman [1968c].
Ogden gives a "pumping lemma" for one-way stack languages. Beeri shows that two-way SA's are equivalent to two-way nested stack automata.
Indexed grammars were first studied by Aho . Theorem 14.7(b), the containment within the CSL's, is from there, as in Exercise 14.1 1, decidability of emptiness. A variety of other characterizations of the indexed languages are known. Aho discusses one-way nested stack automata, an automaton characterization. Fischer discusses macro KTUNOTES.IN Downloaded from Ktunotes.in BIBLIOGRAPHIC NOTES 395 grammars, Greibach provides another automaton characterization—a device with a stack of stacks, and Maibaum presents an algebraic characterization. These alterna-tive formulations lend credence to the idea that the indexed languages are a "natural" class.
Hayashi gives a "pumping lemma" for indexed languages.
/-systems originated with Lindenmayer , and the Ol^systems, on which we have concentrated, were considered by Lindenmayer .
Exercise 14.10, on nonclosure properties of these languages, is from Herman . Exercise 14.13, the undecidability of equivalence of 01/-languages, is implied by a stronger result of Blattner , that it is undecidable whether the sets of sentential forms generated by two CFG's are the same.
Much has been written on the subject, and the interested reader is referred to Salomaa and Herman and Rozenberg .
We have but touched on some of the multitude of species of automata and grammars that have been studied. Rosenkrantz is representative of another early step in this direction, and Salomaa covers a variety of classes not touched upon here.
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KTUNOTES.IN Downloaded from Ktunotes.in INDEX Aanderaa, S. O., 319 Abstract family of languages, 277-280 Accepting state (See Final state) Ackermann's function, 175 Ackley, S.
I., 46, 54 Adelman, L., 375-376 AFL (See Abstract family of languages) Aho, A. V., 54, 75, 106, 124, 227, 268, 284, 394 ALGOL, 268 Algorithm, 146-147 Alphabet, 2 Ambiguous grammar, 87, 200, 255 (See also Inherent ambiguity) Ancestor, 4 Arbib, M. A., 54 Arc, 2 Arden, D. N., 54 Asymmetry, 7 Atom, 355 A-transducer, 282 Auxiliary pushdown automaton, 377-381, 385, 393 Axiomatic complexity, 312-314 Axt, P., 319 Backus, J. W., 106 Backus-Naur form, 78 Baker, B. S., 216 Baker, T., 376 Bar-Hillel, Y., 76, 145, 216 Beeri, C, 269, 394 Berman, L., 375 Bird, M., 76, 284 Blank, 148 Blattner, M., 395 Blum, M., 319 Blum's axioms, 313 Blum's speed-up theorem, 308-310 Boasson, L., 145 Book, R. V., 216, 319, 375 Boolean expression, 324-325 Borodin, A., 319 Borodin's gap theorem, 306-307 Borosh, I., 374 Brainerd, W. S., 176 Bruno, J. L., 374 Bruss, A. R., 375 Brzozowski, J. A., 54 Bullen, R. H., Jr., 54 411 KTUNOTES.IN Downloaded from Ktunotes.in 412 INDEX Canonical order, 168-169 Cantor, D. C, 106, 216 Cardoza, E., 376 Cartesian product, 5 Celoni, J. R., 319 CFG/CFL (See Context free grammar/language) Chandler, W.
J., 284 Chandra, A. K., 376 Checking off symbols, 155-156 Chomsky, N., 106, 123, 145, 216, 232 Chomsky hierarchy, 217-232 Chomsky normal form, 92-94 Christofides, N., 375 Chromatic number problem, 341, 369 Church, A., 176 Church's hypothesis, 147, 166 Circuit value problem, 370 Clique cover problem, 365 Clique problem 365, 369 Closure (See Kleene closure) Closure, of a relation, 8 Closure properties, 59-63, 130-136, 230, 233, 235-247, 270-284, 365, 392 Cobham, A., 374 Cocke, J., 145 {See also CYK algorithm) Code generation problem, 367 Coffman, E. G., 374 Cole, S. N., 269 Compiler, 122-123 Complementation, 59, 135, 179-180, 204, 235-239, 279, 281, 342, 392 Complete problem, 323-354 Complexity class, 288 Computable function, 9-10 Computation, of a Turing machine, 201-202, 211-212 Computational complexity, 285-376 Concatenation, 1-2, 28, 59, 131, 230-231, 246, 266, 278, 280 Configuration, 379 (See also Instantaneous description) Congruence relation, 74 Conjunctive normal form, 325, 328 CovV^, 341-342 Constable, R.
L., 319 Containment, of sets, 5 Containment problem, 203 Containment property, 189 Context-free grammar/language, 9, 77-145, 115-120, 189, 203-206, 228, 246-247, 270, 281, 395 Context-sensitive grammar/language, 223-228, 270, 346-347, 390, 393 Conway, J. H., 54 Cook, S. A., 124, 176, 306, 319, 350, 374-375, 394 Corasick, M. J., 54 Countable set, 6 Counter machine, 123, 171-173 Cremers, A., 284 Crossing sequence, 38, 314-315, 318 CSG/CSL {See Context-sensitive grammar/language) Cudia, D.
F., 216 CYCLE 72, 142-144, 281 CYK algorithm, 139-141 Davis. M., 176 Dead state, 236 Decidable problem, 178 {See also UndeciJable problem) Decision properties, 63-65, 137-141, 230, 247, 281, 393, 395 DeRemer, F.
L., 268 Derivation, 80, 84-87, 220, 389 Derivation tree, 82-87 Descendant, 4 Deterministic CSL, 229 Deterministic finite automaton, 19 {See also Finite automaton) Deterministic language (See Deterministic pushdown automaton) Deterministic pushdown automaton, 1 12-113, 121, 233-269, 281 DFA {See Deterministic finite automaton) Diagonalization, 182-183,364 Digraph (See Directed graph) Directed graph, 2 (See also Transition diagram) Distinguishable states, 68-69 KTUNOTES.IN Downloaded from Ktunotes.in INDEX 413 Domain, 6 DSPACE (See Space complexity) DTIME (See Time complexity) Dyck language, 142 Earley, J., 145 Edge, 2 Effective closure, 59, 205 Effective procedure [See Algorithm) Emptiness problem, 63-64, 137, 189, 281, 348-349 Empty set, 28 Empty stack, acceptance by.
1 12, 1 14-115, 254 Empty string, 1-2, 28 Encoding, of a Turing machine, 181-182 Endmarker, 51, 166, 377, 381 Enumeration, 167-170, 189. 228 Epsilon-CLOSURE, 25 Epsilon-free GSM, 272 Epsilon-free GSM mapping. 272, 274, 280 Epsilon-free homomorphism. 270, 278.
280 Epsilon-free regular set, 271 Epsilon-move, 24-27, 239. 264 Epsilon-production, 90-92 Equivalence class, 7.
11-12. 66-67 Equivalence problem, 64-65. 281 Equivalence relation, 7, 65-66 Equivalent states. 68 Euclid, 8 Even, S., 375 Evey. J., 123-124 Exact cover problem, 341. 367 FA (See Finite automaton) Family of languages. 270 {See also Abstract family of languages.
Trio) Father, 4 Fermat's conjecture, 179 Ferrante, J., 375 FIN, 282 Final state.
17.
110.
148. 272 Finite automaton, 9, 13-54, 67-71, 250-253 Finite control, 17 {See also State) Finiteness problem, 63-64, 137-138, 189, 370 Finite state system, 13-14 (See also Finite automaton) Finite-turn PDA, 143 First-order theory, 354 Fischer, M. J., 232, 319, 375, 394 Fischer, P. C, 124, 176, 267, 319 Floyd, R. W., 106, 145, 216 Formal language (See Language) Freedman, A. R., 319 Friedman, A. D., 54 Friedman, E.
P., 269 Full AFL, 280 (See also Abstract family of languages) Full time/space constructibility, 297-299 Full trio, 270-277, 280 Gabriellian, A., 284 Gap theorem (See Borodin's gap theorem) Garey, M. R., 374-375 Gathen, J., 374 Generalized sequential machine, 272-276 (See also GSM mapping. Inverse GSM mapping) Gill, J., 376 Ginsburg, S., 76, 106, 124, 145, 216. 232.
267, 269, 284, 394 Godel. K..
147. 354, 371 Gonzalez. T.. 375 Graham, R.
L., 374 Graham. S.
L., 145 Grammar (See Context-free grammar.
Context-sensitive grammar. Regular grammar. Type 0 grammar) Graph, 2 Gray, J. N..
124 Greibach. S. A..
106.
124, 216, 232, 267.
269, 284. 319, 394-395 Greibach normal form, 94-99 Greibach s theorem, 205, 393 Griffiths, T. V., 284 KTUNOTES.IN Downloaded from Ktunotes.in 414 INDEX Gross, M., 106, 216 Grzegorczyk, A., 319 GSM (See Generalized sequential machine) GSM mapping, 272-274, 280 Haines, L., 267 Halting Turing machine, 149, 215 Hamilton circuit problem, 332-336 Handle, 249 Hardy, G. H., 57 Harrison, M. A., 124, 145, 394 Hartmanis, J., 76, 124, 145, 176, 216, 319. 375 Hayashi. T., 395 Hell.
P.. 374 Hennie.
F. C, 76, 176, 319 Herman. G. T.. 395 Hibbard. T. N., 232 Hilbert. D., 147 Hogben. L., 8 Homomorphism, 60-61, 132, 230-231, 246. 266, 270, 278, 280, 283 Honesty theorem, 316 Hopcroft, J. E., 75-76, 124, 216, 227, 267-269, 284, 319, 374, 394 Huffman, D. A., 54, 76 Hunt. H.
B., Ill, 216, 374, 376 Ibarra. O. H..
124. 319 ID {See Instantaneous description) Independence of operators, 279, 282 Indexed language, 389-391. 393 Inductive hypothesis, 4 Infinite set. 6 Inherent ambiguity, 99-103 INIT, 72.
142, 280, 282 Initial state, 17, 110, 272 (See also Start state) Input alphabet, 17, 110, 272 Input symbol, 148 Input tape, 377, 389 Instance, of a problem, 177 Instantaneous description, 36, III, 148-149 Integer linear programming, 336-340 Interior vertex, 4 Interleaving, 282 Intersection, 5, 59, 134, 204, 281, 283 Intersection with a regular set, 135-136.
246, 270, 278, 280, 283, 392 Intractable problems, 320-376 Invalid computation (See Computation) Inverse GSM mapping, 272, 276, 280, 392 Inverse homomorphism, 61 , 132-133, 230-231, 246, 270, 278, 280, 283 Inverse substitution, 142 Irreflexivity, 7 Item, 248-252, 261 Jefferson, D., 375 Johnson, D.
S., 374-375 Johnson, S. C, 268 Johnson, W.
L., 46, 54 Jones. N. D., 176, 375 Kannan. R., 374 Karp, R. M.. 374-375 Kasami, T..
145, 269 (See also CYK algorithm) Kernel, 368 Kirkpatrick. D. G., 374 Kleene, S. C, 54, 176, 216 Kleene closure, 28, 59.
131, 246. 266, 278. 280 A-limited erasing (See Limited erasing) Knuth, D.
E., 54, 268 Kohavi, Z., 54 Korenjak, A.
J., 268-269 Kosaraju. S.
R., 76 Kozen, D.. 376 Kuroda, S. Y., 232 Laaser, W. T., 375 Ladner, R.
E., 319, 375-376 Landweber, L. H., 176, 232 Language.
2, 18, 80-81, 112.
149.
168 LBA {See Linear bounded automaton) Left-linear grammar (See Regular grammar) Left-matching, 39 Leftmost derivation. 87 Length, of a string.
I KTUNOTES.IN Downloaded from Ktunotes.in INDEX 415 Leong, B., 176 Lesk, M. E., 46, 54 Levin, L. A., 375 Lewis, J. M., 374 Lewis, P. M., II, 106, 124, 176, 269, 319, 375 Lexical analyzer, 9, 54 Lien, E., 375 Limited erasing, 275, 280 Lindenmayer, A., 395 Linear bounded automaton, 225-226 Linear grammar/language, 105, 143,214, 283-284 Linear programming (See Integer linear programming) Linear speed-up, 289-291 Lipton, R. J., 376 Literal, 325 Logspace-complete problem, 347, 349-350 Logspace reduction, 322-323 Logspace transducer, 322 Lookahead set, 261 Loop program, 317 LR (0) grammar, 248-260, 267-268 LR item (See Item) LR(k) grammar, 260-264 L-system, 390-393 Lynch, N., 376 Machtey, M., 176 Mager, G., 394 Maibaum, T. S.
E., 395 Manders, K., 375-376 Many-one reduction, 212 MAX, 72, 142, 244-245, 281 McCarthy, J., 54 McCreight, E. M., 319 McCulloch. W. S., 54 McNaughton, R., 54, 76 Mealy, G. H., 54 Mealy machine, 42-45 Membership problem, 139-141, 281 Metalinear language, 143 Meyer, A. R., 124, 176, 319, 375-376 Millen, J. K., 54 Miller, G.
L., 232, 375 MIN. 72, 142, 244-245, 279, 281 Minimization of finite automata, 67-71 Minsky, M. L., 54, 76, 176, 216 Modified PCP, 195-196 Monma, C. L., 374 Moore, E.
F., 54, 76, 284 Moore machine, 42-45 Morris. J. H.. Jr., 54 Move.
149 Multidimensional Turing machine, 1 64-165 Multihead Turing machine, 165 Multitape Turing machine, 161-163 Myhill. J., 76. 232 Myhill-Nerode theorem, 65-67 Naur. P., 106 Nerode, A., 76 {See also Myhill-Nerode theorem) Neuron net, 47 Next move function, 148 NFA {See Nondeterministic finite automaton) Node (See Vertex) Nondeterministic finite automaton, 19-33 Nondeterministic space complexity, 288, 300-305 Nondeterministic space hierarchy, 304-305 Nondeterministic time complexity, 288, 300, 354-362 Nondeterministic time hierarchy, 306 Nondeterministic Turing machine, 163-164, 288 Nonerasing stack automaton, 382, 385-387, 393 Nonterminal {See Variable) Normal form PDA, 234 XP. 320-324, 362-365 NP-complete problem, 324-343 NP-hard problem, 324 NSPACE (See Nondeterministic space complexity) NTIME (See Nondeterministic time complexity) Number theory, 354, 371 KTUNOTES.IN Downloaded from Ktunotes.in 416 INDEX Oettinger, A. G., 123 Off-line Turing machine, 166, 174, 285-286 Ogden, W., 145, 394 Ogden's lemma, 129-130 One-way stack automaton, 382, 389, 393 Operator grammar, 105 Oppen. D. C, 375 Oracle, 209-213, 362-365, 371 Output alphabet, 42-43, 272 Output tape, 167 0>, 320-324, 362-365 Pair generator, 169 Palindrome, 2, 11-12, 105 Papadimitriou, C. H., 374 Papert, S., 76 Parikh. R. J., 145 Parser, 9, 268 Parse tree (See Derivation tree) Partial recursive function, 151 Partial solution to PCP, 197 Partition problem, 341 Paul, W.
,!., 319 Paul!, M. C 106 P-complete problem 347-349 PCP (See Post's correspondence problem) PDA (See Pushdown automaton) Perfect squares, 57 Perles. M., 76, 145, 216 Phrase structure grammar (See Type 0 grammar) Pippenger, N., 319 Pitts, W., 54 Polynomial time reduction, 322 Pop, 235, 382 Porter. J. H., 46, 54 Positive closure, 28, 230-231, 278, 280 (See also Kleene closure) Post, E., 176, 216 Post's correspondence problem, 193-201 Post tag system, 213 Power set, 5 Pratt. V.
R., 54, 375-376 Predecessor, 2 Predicting machine, 240-243 Prefix, 1 Prefix property, 121, 260 Prenex normal form, 355 Presberger arithmetic, 354, 371 Primes, 57-58, 342-343 Primitive recursive function, 175 Principal AFL, 283 Problem, 177 Production, 77-79 Proper prefix/suffix, I Proper subtraction, 151 Property, of languages.
188 PSPACE, 321 PSPACE-complete problem, 324, 343-347 PSPACE-hard problem. 324 Pumping lemma, 55-58. 72, 125-128, 143, 394-395 Push, 235, 381 Pushdown automaton. 9.
107-124. 264 (See also Deterministic pushdown automaton) Pushdown transducer, 124 Quantified Boolean formula. 343-346 Quotient, 62-63.
142, 244, 276-277, 280, 392 Rabin, M.
()., 54, 319. 375 Rackoff, C, 375 RAM (See Random access machine) Random access machine.
166-167 Range.
6 Reachability problem. 349-350 Reals with addition, 354-362 Reckhow, R. A., 176 Recursion theorem, 208 Recursive function, 175, 207-209 (See also Partial recursive function) Recursive language, 151.
169-170, 179-181, 210, 227-228, 270-271 Recursively enumerable language.
150, 168-169, 180-192, 210, 228. 230, 270 Reduction, 321-324 (See also 212-213) Reedy, A., 216 Refinement, of an equivalence relation, 65 Reflexive and transitive closure. 8 KTUNOTES.IN Downloaded from Ktunotes.in INDEX 417 Reflexivity, 7 Regular expression, 9, 28-35, 51, 350-353, 370 Regular grammar, 217-220 Regular set, 18, 55-76, 142, 189, 203, 205, 218, 228, 246, 270-271, 277-278, 280-281 Relation, 6-8 Reversal, 71-72, 142, 281 Rice, H. G., 106, 145, 216 Rice's theorem, 185-192 Right-linear grammar (See Regular grammar) Right-matching, 39 Rightmost derivation, 87 Right sentential form, 249 Ritchie, R. W., 319 Rogers, H., Jr., 176 Root, 3 Rose, G.
F., 76, 124, 145, 216, 284 Rosenberg, A.
L., 124, 176 Rosenkrantz, D. J., 106, 216, 269, 374-376, 395 Ross, D. T., 46, 54 Rozenberg, G., 395 Ruby, S., 319 Ruzzo, W.
L., 145 SA (See Stack automaton) Sahni, S., 375 Salomaa, A., 106, 395 Satisfiability problem, 325-331, 370 Savitch, W. J., 216, 319, 375 Savitch's theorem, 301-302 Scattered-context grammar/language, 282-283 Schaefer, T. J., 374, 376 Scheduling problem, 367 Scheinberg, S., 145 Schutzenberger, M.
P., 106, 123, 216, 267 Scott, D., 54 Seiferas, J. L. 76, 176, 319 Self-embedding grammar, 229 Selman, A., 376 Semilinear set, 72 Semi-Thue system {See Type 0 grammar) Sentential form, 81, 143, 389, 395 Set, 5-6 Set former, 5 Sethi, R., 374-375 Shamir, E., 76, 145, 216 Shank, H., 145 Shannon, C. E., 54, 176 Shannon switching game, 370, 372-374 Shepherdson, J. C, 54 Shifting over symbols, 156-157 Shuffle, 142 Sieveking, M., 374 Simple grammar, 229 Simulation, 364 Singletary, W. E., 216 Singleton, 192 S mn theorem. 207 Solovay, R., 375-376 Son, 4 Space-bounded Turing machine, 285 Space complexity, 285-289, 295-298, 300-319, 343-353, 384-385, 387-388, 393 Space constructibility, 297 Space hierarchy, 297-298 Spanier. E. H., 76, 145 Spanning-tree problem, 367 Speed-up (See Blum's speed-up theorem, Linear speed-up) Springsteel, F. N., 375 Stack.
107, 378, 389 Stack alphabet, 1 10 Stack automaton, 381-389, 393 Stanley. R.
J., 145 Start state, 148 (See also Initial state) Start symbol, 1 10 State, 17, 110, 148, 272, 377 Stearns, R.
E., 76, 106, 124, 176, 247, 267. 269. 319, 375 Steiglitz, K., 374 Stockmeyer, L.
J., 375-376 Storage in finite control, 153-154 Strassen, V., 375 String, 1 Strong connectivity problem, 370 SUB. 282 Subroutine, 157-158 KTUNOTES.IN Downloaded from Ktunotes.in 418 INDEX Substitution, 60-61, 131-132, 230-231, 277-278, 280-281, 283 Successor, 2 Sudborough, I. H., 375 Suffix, 1 Suzuki, N., 375 Switching circuit, 13, 47 Symbol, 1 Symmetry, 7 Syntactic category (See Variable) Szymanski, T. G., 374, 376 Taniguchi, K.. 269 Tape, 17. 36.
148, 378 Tape alphabet, 173 Tape compression, 288-289 Tape head, 36 Tape reduction, 289. 292-295 Tape symbol.
148 Tarjan, R.
E., 319, 374-375 Terminal, of a grammar, 77, 79, 389 Text editor, 46 Thompson, K., 46, 54 Three-satisfiability, 330-331 Time-bounded Turing machine, 286 Time complexity, 286-295, 299-300, 307, 313. 320-343, 378-381, 393 Time constructibility, 299 Time hierarchy, 295. 299, 303 Torii, K..
145 Total recursive function, 151 Track, of a tape, 154 Trakhtenbrot. B. A., 319 Transition diagram, 16 Transition function, 17, 48 Transition table, 383 Transitive closure (See Reflexive and transitive closure) Transitivity, 7 Translation lemma, 302-305 Traveling salesman problem. 341, 368 Tree, 3-4 (See also Derivation tree) Treybig, L. B., 374 Trio, 270-277, 280 Truth-table reduction, 214 Turing, A. M., 176, 216 Turing machine, 9, 146-176, 179, 181-183, 193, 201-204, 221-223, 285-319 Turing reduction, 212-213 Two-stack machine, 171 Two-tape finite automaton, 74 Two-tape Turing machine, 292-295 Two-way finite automaton, 36-42 Two-way infinite tape, 159-161 Two-way nondeterministic finite automaton, 51 Two-way pushdown automaton, 121, 124 Type 0 grammar, 220-223 Ullian, J. S., 106, 216 Ullman, J. D., 54. 75, 106, 124, 216, 227, 267-268, 284, 319, 376, 394 Uncountable set, 6 Undecidable problem, 178-216 Union, 5, 28, 59, 131, 180, 230, 246, 265, 278, 280 Union theorem, 310-312 Unit production, 91 Universal Turing machine, 181-185 Unrestricted grammar {See Type 0 grammar) Useless symbol, 88-89 Valiant, L. G., 145, 247, 269, 319 Valid computation (See Computation) Valid item, 249 Variable, of a grammar, 77, 79, 389 Vertex, 2 Vertex cover problem, 331-332 Viable prefix, 249-252 Wang, H., 176 Wegbreit, B., 232 Wise, D. S., 145 Word, 1 Wright, E. M., 57 Yamada, H., 54, 319 Yannakakis, M., 374 Yasuhara, A., 176 Young, P. R., 176 Younger, D. H., 145 (See also CYK algorithm) KTUNOTES.IN Downloaded from Ktunotes.in |
5340 | https://www.youtube.com/watch?v=42PjQRNObTE | Honors Geometry: 5.4 and 5.5: Four Sided Polygons and Properties of Quadrilaterals
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Honors Geometry: 5.4 and 5.5: Four Sided Polygons and Properties of Quadrilaterals lesson video
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good morning afternoon or evening how y'all doing today welcome back to Mr Morrow's honors jry class today we're going to start learning about four-sided polygons and before we can learn about four-sided polygons we have to understand what polygons are and polygons are plain figures such as the ones you see here these are plain figures guys what makes polygons what what makes a plain figure polygons if you C are made up entirely of segments all of these figures were made by segments that's a segment that's a segment that's a segment that's a segment they're made by segments where consecutive sides are noncollinear I'll explain what that means in a second okay but every single figure here is made by segments all of them so any shape any figure that is made from segments is a polygon now what am I talking about as far as polygons are made up of entirely segments we know that but we're um not where consecutive sides are noncolinear if I tell you that bcda is a polygon that works out great bcda is a polygon because you have SE M BC you have segment CD you have segment a d and you have segment AB however if I tell you that segment b c d e a is a polygon that would be erroneous that is wrong why because in this case a e and D are all collinear you you can't have that okay so this polygon exists as long as I call it bcda can I have different points along the polygon sure man I can have five million I'm not being sarcastic at all I could have a bunch of these points here but I can't name all of those points in a polygon polygon has to be made up of the segments that created only yes sir well what makes a different is the the fact that polygons are made up entirely of segments where consecutive sides are noncolinear so in this particular case BC is a side we got that CD is a side we got that now EA EA is that a side is no it's a it's it's it's cinear with d as well you can't call that a separate side now if I elevated that if I had a b c d let's go e and then a ah that we could do a b c d e over here then a then we could do it it has to be its own each segment that creates the polygon has to be its own side that's exactly correct yes sir correct to name the original polygon which we had in green we'll go back to it you have to leave out e e cannot be part of that that's why this is the polygon not this I cannot have a side that isn't its own segment yes sir can you say of course that's what I just finished saying you could have a bunch of points there but you can only call a polygon or name a polygon by the segments that make up the polygon you with me there thank you sir thank you gentlemen great job now what are not polygons these are all not polygons why because they overlap here in this case in this case they cross you can't have a cross in this case you have a curve a curve is not a segment a circle it's a figure but it's not a polygon here you have a gap you can't have a gap and again here you can't have a Crossing okay those are not polygons let's look at some more examples of not polygons and why ABC D is not a polygon because polygons consist in entirely of segments these are the definitions that I'm telling you are very important this here CD is an arc it's a curve that is not a polygon the next one AB c d is not a polygon because in polygons consecutive sides intersect only at the end points if I have this here sure I have an intersection but only at the vert at the vertex or the vertices only at the end points do I have an intersection I can't have a cross here like this see this here you can't have that that is not a polygon it's a figure but it is not a polygon now if I separated them sure I could have this a b c and then I can have another one here separated d e F sure those are two different polygons but once I have this intersection here and they're Crossing can't do it those are not polygons and last but not least this is a very good one hi J K which I will highlight in blue h i j k that's a polygon that's that's golden all right h i j KL I'll highlight it in red h j KL that's a polygon awesome hkl which I will highlight in green that's a polygon that's all good but but h i j k h j l well I messed it up h j k HL K is not what am I talking about about I'm going to I'm going to use a highlighter here okay so you could see if I say h i j k h l k that is not a polygon why because each vertex must belong to exactly two sides and in this case for this guy here H belongs to three sides H would belong to side HL to side hi and to side HK so that is not a polygon yes sir you can have a polygon with if you can draw it small enough with a million side sure as long as each side connects only at the end points of each each segment yes sir very good good question sir does that make sense promise all right let's move on yes sir um no there is not a specific symbol for a polygon you call it a polygon there'll be specific symbols for uh parallelograms squares rectangles triangles stuff like that but not for a polygon that's a good question sir okay in order to name a polygon start at any vertex continue n and continue naming each vertex by going either clockwise or counter clockwise it's pretty simple if I have uh let's do a pentagon here real quick okay if I have this Pentagon and I go A B C D E I can name this two ways I can start at one vertex let's say I choose to start at B just to change things up I could call this B C D E A or I could call this b a e d c so I can go clockwise or I can go counterclockwise what I cannot do is mix and match I cannot go b e a DC no that will not be accurate because that is implying then that I'm having a a lot of crosses within the polygon there are three types of polygons a convex polygon everything that we've well almost everything that we've seen here is a convex polygon a convex polygon is a polygon in which each interior angle has a measure of less than 180 okay so if I draw we already did a pentagon let's do a nonagon if I do a nonagon gone believe it or not I promise you each angle here is less than 180 degrees has to be less than 180 okay you have a concave polygon and a concave polygon is a polygon that can have one interior angle measure greater than 180 such as this bad boy right here I have here I have one 2 three four five sides they're all made up of segments it's it's it's a go for polygon it is a polygon however it is a concave now I'm going to go back in time Elementary they would have they would have done a little dash line here and you see how this is kind of like a cave here this part it caves in that's how you know it's a it's a a concave another way to know this angle right here would measure 300° so it's greater than 180 deg now how do you know the total number of degrees inside any polygon this is not in your book I'm doing this because I want to help you this is huge on the SAT we're going to see this later on in more detail but I always figured why do they wait so I wanted to add this in order to find the total degree sum of the interior angles of any polygon doesn't matter if it's a three-sided polygon which by the way you need at least three sides for a polygon right you can't have a two-sided polygon because then there's nothing connecting it right so yes sir no the circle's not a polygon because it's not made of segments right not because it has two sides but because it's not made of segments okay so the total degree sum of of the interior angles of any poly pogon any polygon at all is nus 2 180 where n is the is the number no is the number of sides okay the number of sides so we all know that a triangle has 180 degrees right we we've learned that already right well let's prove it 3 - 2 180 is 180 we all know that a rectangle and A square has 360 deg let's prove it 4 - 2 is 2 180 which equal 360 so in this particular case here I had five sides five sides well 5 - 2 180 is 3 180 which is 540° these are 90° angles so that already takes away 180 then another 30 and a 30 that's 60 180 and 60 is 240 so what am I left with this angle of 300° so concave concave forms a cave that caves in and there's one angle that is greater than 180 and this little formula here is hugely powerful like I mean it is it is beautiful it's it's really really important okay a regular polygon a regular polygon is not like a regular car or anything no a regular polygon is a polygon that has all equal or congruent s side lengths and equal angle measures so a square is a regular polygon because all sides are equal and all angles are equal remember how we learned about equilateral triangles right okay I don't need to draw it right equilateral triangle if it's equilateral that means that all sides are equal and that's why all angles are equal remember equal angular and equilateral okay so that's what a regular polygon is yes sir all polyx poly all regular polygons are convex polygons very good very well done good job my man yes sir they are last thing about polygons and then we get to definitions diagonals of polygons any segment that connects two non consecutive nonadjacent vertices of a polygon is a diagonal so for example I'll they're already drawn out here but you know me I like to really uh show you the stuff in action okay if I go from here all the way across that's a diagonal and that's a diagonal there are no other diagonals that I can do someone might say wait Mr MO how about this does that go from one vertex to another so it is not a diagonal here I have a hexagon got a lot of diagonals I can come down this way this way I can go from here to two vertices I can go from here to two vertices and I can come here okay so from vertex to vertex but not going along remember it's non jent non-consecutive so why do I say that this is not a diagonal that's a segment that created the polygon you feel me yes sir okay this one will not be either because okay straight down awesome that's not a a diagonal either go ahead show me from here to here oh you know what good catch good catch there is another one good good job son I'm very proud of you yeah this is also a diagonal and this is also a diagonal absolutely well done straight L um they're all straight lines but they but they come but it comes from a Vertex to a Vertex doesn't it is it okay it does it fit the description a segment that connects two non-consecutive non-adjacent vertices doesn't this and this yeah who cares if it's vertical it's not going along the segment that created it is it oh I I understand that but that is a diagonal all right exactly all right now we've got a pentagon done we've got diagonal going here and here a diagonal going here and here and this diagonal here and I think I caught all of them yeah okay does that makes sense what a diagonal is and you understand what a convex concave and regular polygon is all right now the fun part guys this has to be committed to memory but I mean like committed to memory you have to write these out a 100 times if you have to it's not me it's geometry it's the way it goes so let's get through it and remember just reading it or watching it is not going to make you learn it I promise you that you're going to have to write these out you're going to have to actually make flash cards you're going to have to study a quadrilateral is just any four-sided figure or I'm sorry that was wrong any four foursided polygon if it's got four sides you can call it a quadrilateral now there's a lot of different special types of quadrilaterals a parallelogram is a quadrilateral in which both pairs of opposite sides are parallel there's a lot more to this which we're going to learn very soon today a lot more than that but this is the basics yes sir okay that's why I said there's a lot more to prove than that a parallelogram is not does not only have opposite sides that are parallel if you just have patience we're got to walk before we run if you really want to know what a parallelogram is then it's opposite sides are also congruent the diagonals BCT each other and that's a true parallelogram all of that makes up a parallelogram but I don't want to stress you out yet let's walk before we run because just about every quadrilateral is a parallelogram manipulated in some way for example a rectangle is a parallelogram so what does that activate for you that immediately should tell you oh okay if it's a parallelogram I got opposite sides of parallel we know that for sure but a rectangle is also a parallelogram with at least one angle that is right cool cool yes sir comeing yeah absolutely that would mean that all of them are right now that's where I love it when you guys use your head if this is parallel to this using same side interior angle theorem if that's 90 doesn't this have to be 90 because it has to be supplementary and then in that case if this is 90 isn't that have to be 90 and then if that's 90 doesn't this have to be 90 great job son those are connections being made proud of you come again right yeah but in in theory the definition it's a parallelogram with one right with at least one right angle because if you have at least one right angle because of what you just finished saying you know that all of them will be right as well so you could say a parallelogram with all right angles sure I could accept that thank you my brother excellent job a rbus is a parallelogram with at least two consecutive sides that are congruent well if it's a parallelogram that means that opposite sides are parallel and it means that two consecutive sides are congruent at least so this guy has to be congruent to this guy or this guy has to be congruent to this guy and when you put all that together a rbus is a rectangle with equal sides yes sir sides two consecutive sides that are congruent correct sir well I'm sorry if the figure doesn't look like they're equal let's go ahead and assume that they are I am only a human being I did the best that I could I'm sorry yes sir yes sir OPP sesus opposite sides of a rbus will be congruent think about this if two consecutive sides are congruent okay you with me here if I go like this doesn't that mean that that's this is going to be congruent to that because has a side side side correct so then all angles will be congruent so the definition is a rumus is a parallelogram opposite sides are parallel but at least two consecutive sides are congruent but if two consecutive sides are congruent that means that all sides will be congruent so a rbus in essence is a parallelogram with all equal sides yes sir sides correct rectang no a rectangle does not say that it has all equal sides okay thank you my brother now a kite a kite is a quadrilateral it's not a rectangle it's a quadrilateral in which two disjoint pairs of consecutive sides are congruent what does disjoint mean okay see how there's a vertex here and a Vertex here right we're not talking about this and this vertex and I'm going to show you why disjoint sides mean this guy is congruent to this guy this guy here is congruent to this guy but they are not congruent to each other so bless you son so a kite is a quadrilateral which in which two disjoint pairs of consecutive sides are congruent okay what's this disjoint mean well doesn't this visually look different than the green it will always look that way in fact by Design if you've ever flown a kite you're going to have two two legs that are longer than the other legs okay those two long legs are congruent those two short legs are congruent that's what is meant by disjoint pairs of consecutive sides yes sir okay now if you're going to talk about diagonals which again this will be coming soon a kite a kite's diagonals okay are very particular they form perpendicular angles and the longer diagonal cuts the shorter diagonal it bcts it but we're not there yet okay so let's walk before we run trust me guys yes sir we're gonna have to prove a lot of things with these yes a lot of things okay a square a square is a parallelogram that is both a rectangle and a rbus so what does that mean that means that opposite sides are parallel you have four right angles and all sides are congruent that's what a square is it's a parallelogram that is both a rectangle and a rbus a trapezoid a trapezoid is a quadrilateral with exactly one pair of parallel sides so you're only going to have one pair of parallel sides and I should have said opposite because that's really opposite and the parallel sides are called the bases of the trapezoid doesn't matter how you call them you can call this base one and this base two or vice versa and then an isoceles trapezoid is a is a trapez oid with congruent legs remember what an isoceles triangle was there was a base a vertex and two congruent legs so an isoceles trapezoid has one pair of opposite sides that are parallel but then the legs are congruent okay in this case in isoceles trapezoid you have the lower bases and the upper bases a angle A and B are upper bases and just letting you know they're going to be congruent and C and D are the lower bases and just so you know they're going to be congruent a lot of similarity to isoceles triangles if you remember if the sides if the legs are congruent then the base angles are congruent well here in isoceles trapazoid since you have the upper bases and lower bases not only will the legs be congruent but the upper bases will be congruent to themselves and the lower bases will be congruent to themselves does that make sense gentlemen yes sir you will be using all of these definitions all of them thank you sir okay hope you learned a lot guys we're going to be back into about two seconds all right guys no good morning good afternoon or evening because we're still continuing now we're going to be learning about 55 today and 54 hope you just remember because we just finished it we we talked about it is recording we talked about polygons and convex concave regular we talked about the basic definitions of quadrilaterals of parallelograms squares rectangles ruses trapezoids and isos trapezoids today we're going to learn about all of their properties I promise you that all of these properties are going to come back and either Aid you in proofs or hunt you okay these proofs are very difficult I'm not going to lie because there's a lot to remember okay I'm sorry didn't invent geometry I just teach it but I know that you guys can do it I believe 100% in you so let's just look at this with an open mind this is more these are more tools for your sat Bank okay parallelograms guys we kind of saw a little bit of this I I talked about this already opposite sides are parallel by definition we know that wait a second there we go opposite sides are parallel by definition awesome we know that at least we should definitely know that opposite sides are also congruent guys okay now opposite angles are also congruent okay and if you are wondering why there's a long proof for it but I don't think you want to do it so just take my word for it opposite angles are congruent I promise you okay I promise you the diagonals bict each other now please be careful here gentlemen that does not mean that they are congruent that just means that they bict each other that's all that means that does not mean that they're congruent it means that this is cut in half by this one and vice versa and any pair of consecutive angles are supplementary you hopefully still remember that because of the same side angle theorem if you have opposite sides are parallel then the angles in between them must be supplementary the good news that's everything there is to know about a parallelogram okay opposite sides are parallel opposite sides are are congruent opposite angles are congruent the diagonals bisect each other and consecutive side angles I'm sorry consecutive angles are supplementary that's a parallelogram you got it definitely make the markings okay rectangles okay here's where it gets awesome and crazy at the same time all the properties of parallelograms apply to rectangles by definition so what does that mean well that means that opposite sides are parallel that means that opposite sides are congruent that means that the a opposite angles are congruent but remember what a rectangle is a rectangle has 90° so since they're all congruent because it's a it's a parallelogram that's why they're all 90° we proved why it was in 5 54 because my boy said wait a second what about same side interior angle theorem so that was great but now we know for sure it's a definition now here's the kicker it's a parallelogram right so that does mean let me just make this a little thinner okay that does mean that the diagonals BCT each other however however the diagonals of a rectangle are congruent so actually let make them three so you can see that it's different so not only do they bisect each other but they're congruent so every piece of that rect of that of those diagonals are congruent so opposite sides are parallel and congruent all angles are 90 okay and the diagonals BCT each other but the diagonals are congruent they're congruent okay does that make sense and what's going to happen here is if you think about this proof wise this angle is congruent to this angle because of vertical angles so that means that this triangle is congruent to this triangle and what kind of triangles are those they're isoceles and then this angle is congruent to this angle so these two guys are also two triangles that are congruent and they're also isoceles yes sir in parallelograms okay let's see can all four angles be congruent in parallelograms no no only opposite angles are congruent in parallelograms where are they all congruent in rectangles because a rectangle was a parallelogram with all 90° angles okay guys so that is a rectangle a kite a kite we already know about a kite that its two disjoint pairs of consecutive sides are congruent by definition so again CB is congruent to B b d and a is congruent to AC now the longer diagonal is the perpendicular bis sector of the smaller diagonal let's think about this the longer diagonal is the perpendicular bis sector of the shorter diagonal so what does that mean that means that if I have a point here let's call it e just for argument sake this segment C gets bisected and you create four 90° angles there okay the longer diagonal is the perpendicular bis sector of the shorter diagonal and the pair of opposite angles attached to the shorter diagonal only are congruent so that means this guy angle C is congruent to this guy angle D not the other way around it doesn't matter how hard you want it to be that way but it's not okay they definitely are not all right does that make sense gentlemen okay so that's that should be uh pretty simple you you with me here all right rbus all properties of the parallelogram apply by definition so what does that mean that means that opposite sides are parallel okay opposite sides are congruent but in this case all sides are congruent so in other words it's an equilateral quadrilateral all sides are congruent on a rum we know that the diagonals BCT each other by definition of um of uh the of the parallelogram sorry but they they BCT each other in a very particular way they are perpendicular bis sectors of each other okay that but they're not like a rectangle they're not congruent the diagonals are not congruent they are just perpendicular bisc of each other okay the diagonals also BCT the angles okay we know that opposite angles are congruent by definition of the parallelogram so this guy is congruent C is congruent to a but both C and A are uh bisected by the diagonal and angle B is congruent to angle D over here the D is hidden sorry but the diagonals BCT the angles so the diagonals are perpendicular bis sectors of each other and the diagonals are angle bis sectors as well and the diagonals divide the rbus into four congruent right triangles if you can see by the markings here yes sir okay why wouldn't CDA be congruent to b a d because they're not because a is congruent to C but a is not congruent to B they're supplementary by definition of the parallelogram but opposite angles are congruent and they're also bisected by the diagonal that's for sure but A and B angle A and B are not congruent sir at all at all at all at all okay thank you good question okay a square all the properties of a rectangle and rbus apply by definition and of course a parallelogram and the diagonals form four isos right triangles so let's check a square out since it's a parallelogram opposite sides are parallel diagonals by each other however it's also a rectangle so the angles are 90 and it's also a rectangle so that means that not only do the diagonals bisect each other but the diagonals are congruent but it's also a rumus so that means the diagonals BCT each angle so in this case this would be 45 this would be 45 and since it's a rbus the diagonals are also perpendicular bis sectors of each other and since it's AR rbus all sides are congruent let's make this two okay does that makes sense to you you promise this is what I'm going to do for you guys let me just go through the properties of the last two and then I will do the proof on my own I will do the the next example on my own and it's going to be up to you if you want to stay you may if you want to go home and watch the video you may if you want to go home and not watch the video you may even though I do not suggest that so once I finish the examples you may leave please I implore you I beg you to watch the video to its entirety I beg you to follow the the homework instructions I beg you please okay but you may you may be dismissed I'm going to keep going though I just ask that you exit through the back quietly okay thank you yes sir I got one more figure left yeah to the trapezoid right the Ice House L trapezoid okay still then the Isis trapezoid okay we have already saw it in 54 the legs are congruent by definition so CD is congruent to ba by definition okay the bases are parallel so base one and base two are parallel the lower base angles are congruent the upper base angles are congruent and the diagonals are congruent they do not BCT each other gentlemen no but they are congruent so how am I going to label that let's just say let's just go like this just on one side okay they're both congruent they do not bict each other they do not bisect each other but they are congruent and any lower base angle is supplementary to the upper base angle why because the bases are parallel so you get that U or that c shape and you have same side interior angles and that's it a lot to remember but it's not that bad and think about it it's not that bad if you know the parallelogram if you know the parallelogram the square is based on the parallelogram the rectangle based on the parallelogram the rumus is based on the parallelogram so the only two different ones that you really have are the kite and the trapezoid okay very intricate stuff again I'm a man of my word God bless you have a wonderful weekend if you don't want to stay if you do you're welcome to no big deal but I'm going to keep teaching just try to be quiet when you leave and uh have a good safe weekend guys God bless you all okay so how does this work well in example one we got a proof and proofs are pretty awesome okay I know you guys love your proofs in this proof they're telling me that AB c d is a parallelogram and this little symbol right here it won't be for rbus or for a trapezoid it's just going to be for a parallelogram because if you've noticed the parallelogram is the base of most things okay so they tell you that GHA is congruent to FEC and that HB is congruent to De and I messed up I should have put a segment there I apologize and they want to prove that GH is congruent to EF now that's just like what are you talking about okay this is a C proof all right but we're going to knock it out and we're going to do a step by step I definitely added the figure here because you're going to need it I'm going to go ahead and highlight what we're looking for so we remember and I'll leave that in red and that's the only thing I'm going to do in Red okay so number one like always we have our Givens ABC is a parallelogram that's given okay so now we're going to have to start thinking about what does this ABCD mean how is this going to help me to understand that GH and EF are congruent well I'm going to help you out if AB c d is a parallelogram then we know that by definition DC is parallel to AB right for sure for sure 100% definition of parallelogram we also know that angle ECF hold on I don't like that it's jumping like that hold on give me one second let me just extend this page so then I can hopefully move down as much as I want okay okay um we also know that ECF is congruent to to angle h a g why because of the alternate interior angle theorem remember that if we said that da is parallel to CB look at that Z use those letters that I taught you that Z that a uh that F and that U since we know that we have alternate interior angles I can easily say now um and wait a second look at what I did did look at what I did to you guys I totally messed you up man and I apologize I labeled the wrong ones parallel monkey H I'm a Fool DC is parallel to AB I apologize there's no excuse yeah I'm sure you were confused because I'm I'm a fool I didn't do it the right way sorry about that brother so ECF is congruent to h a g now because of alternate interior angles okay da is also parallel to CB but that's not what we needed right now okay now number four AB is congruent to DC why definition of parallelogram opposite sides are not only parallel but they are congruent does that make sense okay this is a good proof this is a tough proof I'm not flying because I know you guys are writing and I don't want to stress you out okay you ready next HB is congruent to D because it was given so HB is congruent to D because it was given okay now from here ha is congruent to EC because of subtraction property here's where things come together we had said that DC was congruent to AB they gave us that HB is congruent to De so by subtraction property it would make sense right guys that EC is congruent to ha you with me there all right may I continue so that's subtraction property then from there we can say GHA GHA is congruent to FEC because it was given so this angle right here is congruent to this angle right here because it was given and now if I look at that hopefully you say Wella wait a second Mr moral you got two congruent triangles there you got an angle side angle angle right here side angle angle side angle so from there I can say triangle GA a H which I will highlight in Black now should Stand Out is congruent to fce because of angle side angle and then we're done with our proof if they're congruent GH is congruent to EF because of cpctc now it takes time to be able to flow with this guys it does okay but I know you will be able to do it and last but not least not a proof but just an application VR Za is a parallelogram they tell you that AV is 2x - 4 they tell you that VR is 3 y + 5 they say that RZ is 12x x + 8 and that Za is y + 12 they want to find the perimeter what what are you talking about if you guys remember parallelograms opposite sides are congruent right so let's first solve for the X's this guy has to be congruent to this guy so 2x - 4 must equal 12x + 8 I'm going to subtract 12 x to both sides actually you know what I'm going to do nobody work with fractions let's multiply everything by two real quick this is 4x - 8 = x + 16 makes life easier subtract X to both sides add 8 to both sides so I've got 3x = 24 so x = 8 all right so if x = 8 16 - 4 that's a 12 and this has to be 12 as well and it is because 1 12 of 8 is 4 4 + 8 is 12 okay we got two opposite sides now let's do the Y's y + 12 must equal 3 y + 5 by definition because opposite sides are congruent subtract the Y and subtract the five 7 equal 2 y so y = 3.5 and y = 3.5 3.5 + 12 is 15.5 and uh 3.5 3 is 10.5 plus 10 that has to be 15.5 as well you don't have to plug it in that's the beauty I'm just plugging it in to prove to you that opposite opposite sides are congruent but once you get one of them and it told you it's a parallelogram you better believe that it is now it's good to also check just to make sure you didn't make an error in your math but it is so now the the uh perimeter 12 + 12 is 24 and uh 31 so 24 and 31 55 units is the perimeter thank you so much for hanging out with me God bless you guys hope you learned a lot have a wonderful weekend |
5341 | https://www.asha.org/practice-portal/clinical-topics/selective-mutism/?srsltid=AfmBOoomY1wRnOGG_YV1JivXitomwZeI6RzjomYXiNFi4tGnGojH3gVh | Selective Mutism
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ASHA/Practice Portal/Clinical Topics/
Selective Mutism
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Overview
The scope of this page includes information about selective mutism occurring during preschool age through adolescence. Considerations for selective mutism as it extends into adulthood are briefly discussed.
Selective mutism is a complex anxiety disorder that affects pragmatic language. Despite the term “selective,” individuals with selective mutism do not elect where to speak but are more comfortable speaking in select situations. According to the Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition, Text Revision (American Psychiatric Association, 2022, p. 222), selective mutism is an anxiety disorder, and the diagnostic criteria for selective mutism are as follows:
The child shows consistent failure to speak in specific social situations in which there is an expectation for speaking (e.g., at school), despite speaking in other situations.
The lack of verbal communication interferes with educational or occupational achievement or with social communication.
The duration of the mutism is at least 1 month (not limited to the first month of school).
The failure to speak is not attributable to a lack of knowledge of, or comfort with, the spoken language required in the social situation.
The mutism is not better explained by a communication disorder (e.g., childhood-onset fluency disorder) or exclusively due to the presence of autism spectrum disorder, schizophrenia, or another psychotic disorder.
The onset of selective mutism typically occurs between 3 and 6 years of age, with diagnosis often occurring when the child enters school (Sharp et al., 2007). Different characteristics of the three primary factors (i.e., person, place, activity) can trigger a child’s mutism and influence the child’s ability to socially engage and communicate (Schwenck et al., 2022). Some examples are as follows:
The child is generally able to speak to familiar people who they are comfortable with in familiar settings.
With the same familiar person, the child may be verbal in one setting but mute in another setting.
Within the same setting, the child may be verbal with some people but mute with others or may be mute during specific anxiety-producing activities (e.g., reading out loud, music class).
Performance is most difficult when there is an expectation for speaking (mostly at school).
Patterns of selective mutism can vary greatly and can interfere with academic, educational, and/or social performance. Speech-language pathologists are integral members of an interprofessional team and often collaborate with school-based teams (e.g., teachers, guidance counselor, school staff) and behavioral health professionals (e.g., school or clinical psychologist, psychiatrist, school social worker). Collaboration between the speech-language pathologist and assigned team members is particularly important for appropriate assessment and treatment because selective mutism is an anxiety-based disorder that can significantly impact the ability to access speech and language skills.
Incidence and Prevalence
The incidence of selective mutism refers to the number of new cases identified in a specified time period. Prevalence is the number of individuals who are living with selective mutism in a given time period.
Accurate population estimates of selective mutism are difficult to ascertain due to the relative rarity of the condition, differences in sampled populations, variations in diagnostic procedures (e.g., chart review, standardized assessment), and the use of different diagnostic criteria (Busse & Downey, 2011; Sharkey & McNicholas, 2008; Viana et al., 2009).
Most prevalence estimates for selective mutism range between 0.2% and 1.6% (Bergman et al., 2002; Chavira et al., 2004; Elizur & Perednik, 2003; Sharkey & McNicholas, 2012). Prevalence can be somewhat higher among immigrant children, language-minority children, and children with speech and language delays (Elizur & Perednik, 2003; Kristensen, 2000; Manassis et al., 2003; Steinhausen & Juzi, 1996). However, it is important to note that selective mutism must exist in all languages to confirm an accurate diagnosis in these populations (Toppelberg et al., 2005).
There is currently a lack of consensus regarding the incidence and prevalence of selective mutism and gender assigned at birth. While most studies report that selective mutism affects more females than males by a ratio of about 1.5–2.5:1.0 (Cohan et al., 2008; Cunningham et al., 2004; Dummit et al., 1997; Kumpulainen et al., 1998), some studies report that it affects more males than females with a ratio of about 1.3:1.0 (Karakaya et al., 2008) or that there is no difference between genders (Bergman et al., 2002; Elizur & Perednik, 2003).
Signs and Symptoms
As with many anxiety disorders, children with selective mutism attempt to protect themselves from the discomfort they experience by avoiding the unpleasant activity (i.e., speaking and/or communicating). Varied characteristics and behaviors associated with selective mutism are a method of self-protection but may be interpreted as deliberately oppositional (e.g., “difficult” or “rude”; Kotrba, 2015). Children with selective mutism are often anxious about communication demands. This anxiety may impair the child’s ability to attend to class instruction and participate fully in school or social expectations (Klein et al., 2019). Misunderstanding such behaviors may complicate the identification of selective mutism.
Individuals with selective mutism may demonstrate the following characteristics and behaviors in specific environments; however, they are not required for a diagnosis (Beidel et al., 1999; Doll, 2022; Kearney, 2010).
Home
Able to speak to one or more immediate family members.
Exhibit difficulty speaking to extended family members or close family friends.
May not be able to speak to immediate family members when visitors are present.
May refuse to leave home to avoid social communication demands (e.g., school, birthday parties).
May have an emotional–behavioral response (e.g., tantrum, withdrawal) when the child has an awareness of social and expressive communication expectations.
School and Community
Exhibit physical manifestations of anxiety: Fight, flight, or freeze response; rigid or restricted body movement; or minimal to no facial expression or eye contact.
May display emotional–behavioral responses (e.g., clinging to the parent, behavioral meltdowns, school refusal).
May be perceived as withdrawn, inattentive, or aloof.
May have difficulty with language processing in specific situations due to a heightened level of anxiety.
Unable to speak with adults or children in social or educational settings.
Unable to respond nonverbally or verbally when spoken to; unable to initiate speech to provide information or comment.
May use nonverbal methods of communication (e.g., body posture, eye gaze, facial expression, gesture) to respond to or initiate with people in settings where they are more comfortable and less anxious.
Unable to initiate using any mode of communication to request help.
Unable to speak at school, which impacts both educational performance and social development.
Unable to speak to immediate family outside the home or when other people are present.
Unable to speak with unfamiliar communication partners; may be able to use nonverbal modes of expression (e.g., eye gaze, head nod, pointing) over time as they become more comfortable in the social environment.
Additional conditions that may be associated with selective mutism are as follows (Capozzi et al., 2018):
enuresis (i.e., urine accidents) and encopresis (i.e., bowel accidents)
eating challenges (e.g., eating with or in front of others, food selectivity)
sleep disturbance
Causes
No single cause of selective mutism has been identified, and its causes may be multifactorial (Cohan, Price, & Stein, 2006). The following factors may coexist and play a role in selective mutism:
Psychological factors, such as social phobia, separation anxiety, and obsessive-compulsive disorder (Beidel & Turner, 2007; Black & Uhde, 1995; Manassis et al., 2003).
Hereditary or genetic predisposition of selective mutism and social anxiety disorder (Black & Uhde, 1995; Cohan, Price, & Stein, 2006; Viana et al., 2009).
Family and environmental factors, such as reduced opportunities for social contact, parenting style, or reinforced avoidance behaviors (Viana et al., 2009).
Neurological/neurodevelopmental vulnerabilities, such as delays in achieving speech, language, or fine and gross motor milestones (Viana et al., 2009).
Overactive autonomic nervous system response that impacts physiological, sensory, and emotional–behavioral responses (e.g., Melfsen et al., 2021).
Other factors, such as shy or timid temperament (American Psychiatric Association, 2022; Steinhausen & Juzi, 1996).
Roles and Responsibilities
Speech-language pathologists (SLPs) play an integral role in the screening, assessment, diagnosis, and treatment of individuals with selective mutism. The professional roles and activities in speech-language pathology include clinical services (diagnosis, assessment, planning, and treatment); prevention and advocacy; and education, administration, and research. See ASHA’s Scope of Practice in Speech-Language Pathology.
The following roles are appropriate for SLPs:
Educate other professionals on the needs of individuals with selective mutism and the role of the SLP in diagnosing and managing selective mutism.
Screen individuals who present with language and communication difficulties to determine the need for further assessment and/or referral for other services.
Conduct a comprehensive, culturally and linguistically appropriate assessment of speech, language, and communication.
Aid in diagnosing the presence or absence of selective mutism with an interdisciplinary team.
Refer to other professionals to rule out other conditions, determine etiology, and facilitate access to comprehensive services.
Make decisions about the management of selective mutism.
Develop treatment plans, provide treatment, document progress, and determine appropriate dismissal criteria.
Counsel individuals with selective mutism and their care partners regarding communication-related issues and provide education aimed at preventing further complications relating to selective mutism.
Consult and collaborate with other professionals, family members, care partners, and others to facilitate program development and to provide supervision, evaluation, and/or expert testimony, as appropriate.
Remain informed of research in selective mutism and help advance the knowledge base related to the nature and treatment of selective mutism.
Advocate for individuals with selective mutism and their families/care partners at the local, state, and national levels.
Serve as an integral member of an interdisciplinary team working with individuals with selective mutism and their families/care partners.
As indicated in the Code of Ethics (ASHA, 2023), clinicians who serve this population should be specifically educated and appropriately trained to do so. SLPs take part in the aspects of the profession that are within the scope of their professional practice and competence. If an SLP has advanced training in and knowledge of selective mutism, diagnosis is possible in accordance with existing state credentialing laws. However, a diagnosis made by an interdisciplinary team ensures that a full differential diagnosis was completed.
Assessment
Screening
Screening for selective mutism is conducted whenever selective mutism is suspected or as part of a comprehensive speech and language evaluation for a child with communication concerns. If a parent or care partner reports that a child is communicating successfully at home but not in one or more settings, the speech-language pathologist (SLP) may want to consider screening for selective mutism.
Screening typically includes
norm-referenced parent/care partner and teacher report measures,
competency-based tools such as interviews and observations, and
hearing screening to rule out hearing loss as a possible contributing factor.
See ASHA’s Practice Portal pages on Hearing Loss in Children and Hearing Loss in Adults for more information.
Comprehensive Assessment
See theSelective Mutism Evidence Map for pertinent scientific evidence, expert opinion, and client/ care partner perspective.
Please see ASHA’s resource, Assessment Tools, Techniques, and Data Sources, for information on the elements of a comprehensive assessment, considerations, and best practices. Information specific to these practices in the comprehensive assessment of individuals with selective mutism is discussed below.
Assessment of children with selective mutism involves a collaborative approach with an interdisciplinary team, which may consist of a pediatrician, a psychologist or psychiatrist, an SLP, a teacher, a school social worker or guidance counselor, and family/care partners. During the evaluation, parents/care partners may need to help elicit verbal output. The SLP can also involve parents/care partners by requesting a video recording of the child’s communicative behavior at home and then compare the child’s behavior in a clinical or school setting. Video recordings may also be used for subsequent language sample analysis.
Several techniques can be used throughout assessment to reduce stress on the child, increase participation, and improve the quality of assessment findings. See “Meeting the Child” section below for more details.
Case History
A diagnostic interview with parents/care partners and teachers is conducted without the child present to help gather information about the following:
Any suspected co-occurring disorders (e.g., schizophrenia, autism spectrum disorder).
Environmental factors (e.g., amount of language stimulation).
Circumstances of communication (Kotrba, 2015):
With whom does the child communicate?
In what circumstances is the child most likely to communicate?
Where and in what settings is the child able to communicate?
How does the child communicate—gestures? writing? sounds? whispering? short responses?
The child’s symptom history (e.g., onset and behavior).
Family history (e.g., psychiatric, personality, and/or physical problems).
Speech and language development (e.g., how well does the child express themself and understand others?).
Educational history, such as information on academic reports, parent/care partner and teacher comments, previous testing (e.g., psychological), and standardized testing.
If the child is multilingual, the SLP will need to obtain the following information (Mayworm et al., 2015; Toppelberg et al., 2005):
What languages does the child use now, and with whom?
How well does the child understand the different languages to which they are exposed?
Does the child use their primary language successfully outside the home environment? If so, in what settings and with whom?
Speech and Language Evaluation
During the speech and language evaluation, the SLP gathers information on the child’s
language comprehension;
expressive language ability;
nonverbal communication (e.g., pretend play, drawing);
pragmatic language, including situations, speakers, and contexts that encourage or discourage speech (Hungerford, 2017);
functional communication ability across various circumstances and settings (Kotrba, 2015; Selective Mutism Anxiety Research and Treatment Center & Shipon-Blum, 2012); and
oral–motor functioning, including strength, coordination, and range of motion of the lips, jaw, and tongue.
A child with selective mutism might not be able to participate in formal evaluation activities, and they may lack verbal responses and use nonverbal responses (e.g., pointing or gesturing). These behaviors provide diagnostic information regarding the child’s response to social communication. The SLP can also use audio or video recordings from home to supplement parent/care partner descriptions. Any discrepancy between the child’s communication at home and their communication in public may suggest an overarching problem of difficulty with social language.
In some situations, it may be feasible to train parents/care partners, or other familiar adults with whom the child is able to speak, to administer standardized tests (Klein et al., 2013). In these cases, parents can administer test items with the SLP in or out of the room to promote verbal responses from the child; however, the SLP is still responsible for scoring and interpreting test performance. It is the responsibility of the SLP to review the examiner manual to see if parents are listed as a potential assessor based on the prescribed educational and expertise requirements. Using standardized tools in a nonstandardized way may invalidate standardized scores; however, information gleaned from the assessment can still be reported.
Speech Sound Production
Speech sound disorders may occur in children with selective mutism and may magnify the child’s anxiety of interacting with others (Anstendig, 1999). These children may benefit from direct assessment and treatment with parental involvement and support.
See ASHA’s Practice Portal page on Speech Sound Disorders – Articulation and Phonology for more information related to speech assessment and treatment.
Voice
Some children with selective mutism have reported that they do not like their voice, they don’t want their voice to be heard, or their voice “sounds funny” (Henkin & Bar-Haim, 2015; Vogel et al., 2019). Voicing requires control and coordination of airflow and the vocal mechanism that may be disrupted by their level of anxiety and may present a challenge (e.g., increased laryngeal tension) for an individual with selective mutism (Ruiz & Klein, 2018). Even in cases where a child verbalizes in front of the clinician, this speech may be produced in a whisper, at a decreased vocal intensity, or in an altered vocal quality. The SLP documents vocal quality at the time of the initial evaluation and then reassesses during intervention. The altered vocal quality can lessen as anxiety decreases. Clinicians may also want to evaluate the level of vocal tension during the assessment.
Language Ability
Receptive and expressive language skills may vary in children with selective mutism. For example, expressive–receptive and receptive language disorders may coexist with selective mutism (e.g., Viana et al., 2009). Some children with selective mutism with average receptive language abilities may demonstrate shorter, less detailed, and more linguistically simplistic narratives (McInnes et al., 2004). These subtle deficits in expressive language are theorized to be a compilation of anxiety, mild language deficits, and lack of experience with high-level language skills.
It may be beneficial to use low-stress tasks, such as a picture-pointing task when assessing language ability. If the child is unable to speak, SLPs acknowledge and respond to the child’s gestures or written/typed responses, assess the effectiveness of the child’s attempts at nonverbal communication, and assess the child’s behaviors when engaged in communication. There may be cultural differences within nonverbal communication that the SLP needs to consider when assessing communication.
See ASHA’s Practice Portal pages on Cultural Responsiveness and Social Communication Disorder for further information as well as ASHA’s Practice Portal page on Spoken Language Disorders for more information related to language assessment and treatment.
Cognitive Abilities
While children with selective mutism may demonstrate average cognitive and academic abilities (Manassis et al., 2003; McInnes et al., 2004), some children with selective mutism may have impaired visual memory or auditory–verbal memory (Kristensen & Oerbeck, 2006; Manassis et al., 2007). Difficulty responding using verbal and nonverbal responses, avoidance of interacting with unfamiliar adults, and slowness to respond can lead to lower test scores and misinterpretation of the child’s ability (Kotrba, 2015).
Social Communication Skills
Social communication skills for children with selective mutism typically appear limited outside the home and other familiar environments and, at times, may appear limited in the home as well. Research is not clear as to whether children with selective mutism have pragmatic language challenges beyond avoiding communicating in certain circumstances outside the home setting (McInnes et al., 2004). Social immaturity is not uncommon because the child with selective mutism has fewer social interactions and may lack social awareness (Kotrba, 2015). Children with selective mutism can display decreased nonverbal and verbal indicators of social engagement, such as proxemics, facial expressions, gestures, eye contact, turn-taking, participation in joint activity routines, and joint attention (Hungerford et al., 2003). Home video samples may be helpful in assessing social communication variations across settings.
Please see ASHA’s Practice Portal page on Social Communication Disorder for more information related to assessment and treatment.
Assessment Considerations
Meeting the Child
Prior to initiating speech and language services, the SLP can provide parental or teacher questionnaires regarding selective mutism or conduct a diagnostic interview with parents, care partners, and teachers to prepare for the initial meeting. Clinicians may consider meeting the child one-on-one or with the parent/care partner present prior to formal assessment. Conditions of meeting the child with selective mutism may vary based on the school, home, or private practice setting. The clinician can reassure the parents/care partners and child of the expectations for the first meeting, such as the child will not be pressured to speak, there will be no interruptions, and no one else will be present in the meeting setting (Doll, 2022).
First sessions may be informal and flexible. The SLP may develop a relationship with the child prior to the evaluation by scheduling two to three sessions for age-appropriate recreational or play-based interactions without the expectation for speech. Clinicians may play at the child’s level and follow their lead with open-ended, creative play involving arts and crafts, building blocks, and/or board games (Kotrba, 2015).
The child and parent/care partner may benefit from playing in the assessment room for 5–10 minutes without the SLP in the room to increase comfort and familiarity with the setting. During this time, parents are encouraged to actively engage with their child or ask their child questions to promote verbal output. The SLP can observe if an observation room or video is available. This allows for comparison of the child’s communication with and without an unfamiliar person in the area. Then, the SLP can enter the room, allow the child and parent/care partner to continue playing for several minutes, and then enter the child’s circle of play (Middendorf & Buringrud, 2009).
The following defocused communication strategies can help build a positive rapport and establish trust (Oerbeck et al., 2014):
Minimize eye contact. Sustaining eye contact from unfamiliar people can make children with selective mutism uncomfortable.
Maintain a calm demeanor.
Make environmental modifications. For example, some children may prefer that the SLP sit by their side rather than face-to-face, whereas this may be too close for others.
Create opportunities for joint attention using an activity that the child enjoys.
Think aloud by providing behavioral descriptions of what the child is doing, rather than by asking direct questions (e.g., “I see that you’re playing with the truck!” instead of “What are you doing?”).
Use phrases and terms that encourage the child to communicate, including using the terms “words” or “voice” rather than “talk” or “speak .” The latter two words may have negative connotations for the child (Kotrba, 2015). Also, encourage the child to show, gesture, write, or draw if they are not able to speak (Schum, 2006).
Reflect back language that the child shares.
Offer choices or options to respond instead of open-ended or yes/no questions.
Allow plenty of time for the child to process and respond rather than talking for the child.
Continue the conversation, even when the child does not respond verbally.
Receive the child’s responses in a neutral way.
Within an evaluation process, it is also important to be mindful of the communication demands required for specific tasks completed in the evaluation. An SLP may need to modify the order in which they present materials, starting with tasks with no verbal communication demands and moving to verbal communication tasks based on the child’s presentation and responsiveness.
Visit the Selective Mutism Association’s Educator Toolkit for more information.
Interprofessional Collaboration and Referrals
During evaluation and treatment, the SLP may collaborate with and refer to the following professionals:
audiologist
behavior analyst/behavioral specialist
extended family and/or care partners
family
guidance counselor
pediatrician
psychiatrist
school or clinical psychologist
social worker
teacher
The SLP’s role on the evaluation team is to identify and describe (a) the child’s communication skills and coexisting communication disorders and (b) the impact of those skills on the child’s ability to consistently participate in various settings (Kotrba, 2015). If the SLP is the first professional that a family encounters, the SLP can initiate the collaborative process and provide referrals to behavioral health professionals with training and experience in working with children with anxiety disorders (e.g., behavioral therapists, cognitive therapists). A collaborative interprofessional team that develops a treatment plan and communicates regularly can optimize treatment outcomes and promote generalization of effective communication skills across people, settings, and situations.
See ASHA’s webpage on Interprofessional Education/Interprofessional Practice (IPE/IPP).
Differential Diagnosis
The major difference between selective mutism and other disorders is that the child with selective mutism can talk in certain situations but not others due to anxiety (Kotrba, 2015). SLPs consider whether a child’s absence of speech may be due to a communication disorder, a developmental disorder, or other psychiatric disorders (Kearney, 2010). Diagnosis by an interdisciplinary team, including behavioral health care professionals, ensures a complete differential diagnosis process.
Although selective mutism is not better explained by a communication disorder or psychological disorder, selective mutism may occur simultaneously with the following (Driessen et al., 2020; Steffenburg et al., 2018):
social anxiety
generalized anxiety
separation anxiety
autism
specific phobia
obsessive-compulsive disorder
speech and/or language disorder (Viana et al., 2009)
SLPs also consider if the child is immersed in a new language environment because acquiring another language is a complex process. When children are exposed to a new language, they may experience a brief silent period in which they are quiet and speak little. Although children may not speak in situations in which the new language is used, children with typical second-language acquisition demonstrate appropriate social communication skills in settings and with people who speak the child’s primary language (Doll, 2022). When working with a multilingual child, diagnosing selective mutism depends on understanding typical multilingual child development. Multilingual children with true selective mutism present with mutism in both languages, in several unfamiliar settings, and for significant periods of time (Toppelberg et al., 2005). Interviewing parents/care partners to determine if the child speaks in their primary language successfully outside of the home environment is important information for the SLP to gather to inform differential diagnosis (Mayworm et al., 2015). Please see ASHA’s Practice Portal page on Multilingual Service Delivery in Audiology and Speech-Language Pathologyfor further information.
It is necessary to collaborate with an interpreter or a translator if the SLP does not speak the language(s) of the child. The SLP should be mindful of the number of people in the room and consider how the introduction of an additional person may impact performance. The SLP may need to consider asking a family member to act as an interpreter in this circumstance so as not to create additional anxiety or stress for the child. See ASHA’s Practice Portal page on Collaborating With Interpreters, Transliterators, and Translators for more information.
Some children will not speak after a traumatic event or ongoing social–emotional difficulties, such as parental divorce. Children who do not speak following trauma are mute in all settings (Manassis et al., 2003). If the child spoke well prior to these events, then a diagnosis of selective mutism may not be appropriate. Instead, the child may require assistance in adjusting to the trauma or other life challenges (Kearney, 2010); in which case, referral to a behavioral health professional is appropriate. See ASHA’s resource on trauma-informed care.
Determining Educational Eligibility
Interprofessional practice and family involvement are essential in assessing and diagnosing selective mutism; the SLP is a key member of a multidisciplinary team. The multidisciplinary team reaches a consensus that assessment results are consistent with the diagnostic characteristics of selective mutism.
Within school settings, children can be supported through informal services, Section 504 plans, or individualized education programs. There is no single, preferred, consistent diagnostic category for children and youth with selective mutism in the school setting. Eligibility for special education services under the Individuals with Disabilities Education Improvement Act of 2004 could be determined to fall within the disability categories of Other Health Impairment, Speech-Language Impairment, or Emotional Disturbance/Disability.
The level of accommodations will depend on the functional impact of selective mutism in the school setting. For example, a newly identified student may need regular access to a “buddy,” someone who the child can speak to throughout the day, versus a student farther in the treatment process may need opportunities to work in small groups with less familiar peers (Doll, 2022). Some children with selective mutism may benefit from the classroom accommodations offered through a Section 504 plan, whereas others may need more direct services within special education to address the communication concerns.
Treatment
See the Selective Mutism Evidence Map for pertinent scientific evidence, expert opinion, and client/ care partner perspective.
Early intervention for selective mutism is key to remediation. Communication partners sometimes speak for the child with selective mutism when the child demonstrates distress. This “rescuing” behavior may discourage the child’s future speech attempts and results in negatively reinforcing the child’s avoidance of speaking. Treatment works to break the cycle of negative reinforcement. Consistency in the intervention and expectations, at home and in school, of everyone on the team is important. Speech-language pathologists (SLPs) work to provide predictability and control for children with selective mutism, which may decrease anxiety and improve self-image based on mastery of skills in a variety of settings (Kotrba, 2015). Pharmacological treatment may be prescribed by the individual’s treating pediatrician or psychiatrist (Manassis et al., 2016). Clinicians consider the behavioral influences and side effects of medications (e.g., selective serotonin reuptake inhibitors) on speech and language interventions and collaborate with behavioral health professionals, as appropriate.
Monitoring the individual’s success at each level of the treatment plan through ongoing assessment will determine the overall success for consistent communication with a variety of people in different settings. Anxiety and avoidance behaviors will indicate the need to break down communication steps, locations, or audience size into more manageable steps of facing a fear (Kotrba, 2015).
Behavioral and Cognitive-Behavioral Strategies and Definitions
The behavioral perspective views selective mutism as a learned behavior that the individual has developed as a coping mechanism for anxiety. The purpose of treatment is to decrease anxiety and increase verbal communication in a variety of settings, incorporating practice and reinforcement for speaking in subtle, nonthreatening ways (Camposano, 2011; Cohan, Chavira, & Stein, 2006; Kotrba, 2015). Reinforcements may be
verbal (e.g., praise);
tangible (e.g., toys, special outings, belongings); and/or
privileges (e.g., staying up later, having additional time to play a video game, choosing a movie or board game to enjoy with a parent/care partner).
The trained behavioral health professional, SLP, and school staff collaborate to incorporate behavioral and cognitive-behavioral strategies into interventions across settings for children with selective mutism. These strategies include the following.
Exposure-based practice involves the child saying words in gradually but increasingly difficult or anxiety-provoking situations. Exposure-based practice aims to (a) replace anxious feelings/behaviors with more relaxed feelings and (b) increase the child’s feelings of independence by gradually improving their ability to speak in different situations (Kearney, 2010; Middendorf & Buringrud, 2009).
Systematic desensitization involves the use of relaxation techniques along with gradual exposure to subsequently more anxiety-provoking situations (Cohan, Chavira, & Stein, 2006; Kearney, 2010).
Stimulus fading involves gradually increasing exposure to a fear-evoking stimulus (e.g., the number of people present or the presence of an unfamiliar person in the room while the child is speaking). For example, if a child does not speak in school, then a child’s parent would be brought into the child’s classroom. When the child speaks to the parent, the clinician slowly brings a new person into the room (e.g., a teacher). This process usually includes rewarding the child when they are speaking in the presence of someone to whom they do not typically speak (Middendorf & Buringrud, 2009; Viana et al., 2009).
Contingency management, positive reinforcement, and shaping includes (a) providing positive reinforcement contingent upon verbalization and (b) reinforcing attempts and approximations to communicate (i.e., shaping) until such attempts are shaped into verbalizations, with the goal of making verbalizing more rewarding than not responding. Shaping is commonly used in combination with contingency management and positive reinforcement.
Treatment Options and Techniques
The treatment options below include approaches that are within the scope of an SLP, may involve an SLP in an interprofessional team, or may require additional training.
Augmentative and Alternative Communication
Augmentative and alternative communication (AAC) involves supplementing or replacing natural speech with aided symbols (e.g., pictures, line drawings, tangible objects, and writing) and/or unaided symbols (e.g., gestures). Some children who have been diagnosed with selective mutism may adapt an AAC system to facilitate classroom communication. Some individuals may use AAC only in the initial stages of intervention. Some individuals may use AAC only in the initial stages of intervention, with AAC faded over time as an individual with selective mutism finds more success with verbal communication. Other clients and their care partners may have long-term preferences for AAC as their primary communication method. In such cases, language and communication treatment goals incorporating the client’s preferred communication modality may be appropriate.
Please see ASHA’s Practice Portal page on Augmentative and Alternative Communication for further information.
Augmented Self-Modeling
In augmented self-modeling, the individual with selective mutism watches a video segment or listens to an audio segment in which they are engaging in a positive verbal interaction in a comfortable setting (typically, at home). This approach may be paired with additional behavioral and cognitive-behavioral strategies, such as positive reinforcement and stimulus fading (Kehle et al., 2011). This process may also involve making a video recording of the child and editing it so that the video shows the child speaking in settings where the child does not speak, such as the classroom. The child watches and listens to themself speaking to learn to think positively about speaking in front of others.
DIR Floortime®
DIR (Developmental, Individual Differences, Relationship-Based) Floortime is a developmental and interdisciplinary framework based on functional emotional developmental capacities (FEDCs). It utilizes the concepts of self-regulation, attention, engagement, intentional communication, and purposeful problem-solving communication. Goals are based on evaluating the child’s FEDC (i.e., moving from nonresponsive to using gestures, to making sounds, and then to being verbal) and supporting individual differences (sensory processing, praxis, speech and language challenges, visual–spatial processing, postural stability) to move the child up the FEDC ladder. It incorporates sensorimotor and play-based activities (often having co-treatments with an occupational therapist) and instruction regarding antianxiety strategies from a social worker or other behavioral and mental health professionals (Fernald, 2011).
ECHO Program
ECHO: A Vocal Language Program for Easing Anxiety in Conversation (Ruiz et al., 2022) aims to support individuals, who are of late elementary age through adolescence, who may experience social anxiety related to speaking in certain situations or with certain individuals. This program, which can be implemented by SLPs, bridges the gap from vocalization to conversation. The following three modules include both face-to-face and computer-based interactive activities:
Module 1—Voice Control: The individual learns how to initiate voice, modulate intonation and volume, and produce speech sounds in words and sentences.
Module 2—Social Pragmatic Language: The individual learns to use language for different purposes, change language for the listeners or situation, and follow rules for conversation and storytelling.
Module 3—Role Play: The individual uses the skills learned in the previous two modules to participate in conversational role plays that simulate real life (e.g., school, home, social, public). A cognitive behavioral therapy framework is used to help reduce cognitive distortions (e.g., “Everyone will laugh at me if I talk because my voice sounds funny”).
EXPRESS Program
EXPanding Receptive and Expressive Skills through Stories (EXPRESS): Language Formulation in Children With Selective Mutism and Other Communication Needs (Klein et al., 2018) aims to expand receptive and expressive language skills with five levels of communication (i.e., nonvocal communication through spontaneous vocalization). The EXPRESS approach, which supports the Common Core State Standards for English Language Arts, uses classic children’s stories to correspond with each module to help expand vocabulary and grammar, engage in question–answer routines, improve sentence formulation, and generate narrative language.
Integrated Behavioral Therapy for Selective Mutism
Integrated behavioral therapy for selective mutism, originally developed for children ages 4–8 years, aims to increase successful speaking behaviors in anxiety-provoking situations, habituate speaking-related anxiety, and positively reinforce speaking (Bergman, 2013). Using a combination of behavioral techniques (e.g., stimulus fading, shaping, desensitization) and exposure-based interventions, the clinician systematically and gradually exposes the child to increasingly difficult speaking situations. This program takes place over 24 weeks during the school year.
Intensive Group Behavioral Treatment
Intensive Group Behavioral Treatment focuses on providing a full course of intervention for selective mutism in a condensed period, such as a 1-week summer camp program (Cornacchio et al., 2019). In a 1:1 child–staff ratio, trained counselors and at least one clinical psychologist incorporate aspects of the parent–child interaction therapy and cognitive behavioral therapy in a group setting. Components of the Intensive Group Behavioral Treatment may also include parent training and coaching.
Parent–Child Interaction Therapy for Children With Selective Mutism
Parent–child interaction therapy for children with selective mutism aims to increase verbal interactions in social settings and decrease avoidance behaviors (Cotter et al., 2018). Intervention includes the following two phases that involve specific techniques, procedures, and tasks to promote verbalization:
Child-directed interaction (CDI)—This phase focuses on building the child’s comfort with the communication partner and environment (Doll, 2022). The communication partner uses strategies (e.g., labeled praise, reflection, enthusiasm) to provide verbal models of communication and to take away the pressure of the child speaking.
Verbal-directed interaction (VDI)—Once rapport is established, VDI is introduced to prompt the child’s speech. Exposure tasks are used to begin generalizing speech to new environments and people (Cotter et al., 2018). Clinicians continue to address CDI skills during the VDI phase.
Social Communication Anxiety Treatment®
Social Communication Anxiety Treatment (S-CAT) uses a multimodal approach to increase the social engagement, verbal communication, and confidence of the person with selective mutism (SMart Center, n.d.). S-CAT focuses on reducing the child’s anxiety about speaking and the parent/care partner’s rescuing behaviors that enable the child’s avoidance behaviors (Klein et al., 2016). Using behavioral and cognitive-behavioral strategies, the clinician helps the individual move through the four stages of communication (i.e., noncommunicative, nonverbal, transition to verbal, and verbal).
The clinician can incorporate the Ritual Sound Approach® into the S-CAT program to systematically increase the child’s comfort with making sounds and words (Shipon-Blum, n.d.). In the Ritual Sound Approach, the clinician teaches and models how sounds are made through a mechanical perspective. Once the child with selective mutism is comfortable with making nonspeech sounds, the clinician can gradually introduce different phonemes. Eventually, the clinician can help the child blend the phonemes into simple words. Involvement of the child, parent/care partner, and school staff is integral to establishing skills across all environments and communication partners.
Social–Pragmatic Approach
This integrated approach emphasizes participation in social engagement (nonverbal and verbal) at increasingly difficult levels. Shaping and reinforcement, in the context of interactive routines, are used to move the child with selective mutism through
building acceptance of joining social activities (e.g., games, art, social play);
using nonverbal communication during social activities (reaching, pointing, gesturing “yes” or “no,” facial expression); and
using a hierarchy of sound production (i.e., from nonspeech sounds to speech sounds to using words).
The clinician considers the hierarchy of language functions at the word level and beyond. For example, the child may begin with answering noninvasive questions (e.g., “What color is your shirt?”) and progress to answering increasingly more personal questions (likes/dislikes, family and friends) before eventually being able to ask noninvasive personal questions and participate in conversation over multiple turns.
Tasks may need to be simplified when the child changes communication partners or contexts. The approach considers different variables of the communication context, as follows:
who the child is communicating with (familiar vs. unfamiliar)
where the child is communicating (e.g., treatment room, school library, classroom before school starts, in small group inside the classroom)
the purpose of communication (e.g., regulating another’s behavior, social interaction, joint attention)
the ability to manage conversation (i.e., multiple turns, repair conversation, select/maintain/terminate conversation, take another’s perspective; Hungerford et al., 2003)
Generalization/Carryover
Several of the treatment programs described above incorporate ways to generalize speaking in new environments and with new communication partners. Overall, generalizing spontaneous speech to different settings and communication partners may involve (Kotrba, 2015; Middendorf & Buringrud, 2009)
having the individual with selective mutism rate situations and people from “most difficult” to “least difficult” in a hierarchy;
preparing and reassuring the individual with selective mutism of their abilities by thoroughly explaining the plan to generalize their skills;
establishing a keyworker, an adult trained in behavioral and cognitive-behavioral strategies, in the school setting;
changing only one variable at a time (e.g., either the location or the people present);
moving from structured and carefully planned occurrences to spontaneous and unplanned situations; and/or
practicing frequently and repetitively.
Interprofessional Collaboration
Continued collaboration between the SLP and behavioral health professionals, classroom teachers, and the family is necessary for treatment continuity, clear delineation of roles and responsibilities, and appropriate hierarchical goal setting. Having the SLP on the team helps the child with selective mutism gain confidence in what they may perceive as decreased communication skills (Dow et al., 1995).
The SLP can work with the child’s teacher and school staff to use the following strategies: Form small, cooperative learning groups that include the child’s preferred peers.
Help the child communicate with peers in a group by first using nonverbal methods (e.g., signals, gestures, pictures, writing) and then gradually working toward verbal participation.
Watch for opportunities to reinforce small improvements.
Reassure others that the child is still comprehending even if they are not talking.
Try to minimize symptoms—the child may not want to talk, but they can point, show, gesture, or draw.
Avoid speaking forthe child, justifying the child’s silences, or pressuring the child to speak, all of which may reinforce mutism.
Support peer acceptance of nonverbal participation in classroom and recreational activities.
Find nonverbal jobs that the child with selective mutism can perform to build confidence.
Maintain the classroom routine and try making the same request of the child at the same point in the schedule to decrease anxiety.
Strategize speaking assignments that the individual agrees to complete with the teacher prompting or reminding the student as necessary (Bergman, 2013).
Arrange one-on-one time with the teacher and student so that they can seek assistance quietly rather than in front of peers (Richard, 2011; Schum, 2002, 2006).
See also ASHA’s webpage on Interprofessional Education/Interprofessional Practice (IPE/IPP).
Special Considerations
Structuring Treatment
Initially, children may require individual treatment sessions to establish rapport and practice relaxation techniques and pragmatic skills in a comfortable setting. Typically, treatment progresses from CDI to VDI. During CDI, the adult observes the child performing an activity that the child chooses, and then, the adult joins in by imitating, describing, and demonstrating enjoyment without asking questions, giving commands, or using negative talk. VDI allows adults and peers to ask questions, direct play, and give instructions (Kurtz, 2015; Mac, 2015). A trained keyworker could also provide behavioral interventions, CDI, and VDI throughout the school environment (Kotrba, 2015).
English Language Learners
The SLP and the interprofessional team incorporate the following considerations when an English language learner is suspected or confirmed of having selective mutism (Mayworm et al., 2015):
early identification and intervention
the language(s) that providers use to implement intervention and the role of language development in the type of treatment
willingness and training of team members to implement interventions in multiple contexts
When treating an English language learner with selective mutism, the SLP is aware of possible stressors within the child’s school setting that will need to be addressed through staff development, interventions, and accommodations (Toppelberg et al., 2005). These may include
lack of class support for learning another language,
the potential for negative views of the child’s culture or language used at home, and/or
limited communication between the parent/care partner and the school.
See Multilingual Service Delivery in Audiology and Speech-Language Pathologyand Cultural Responsivenessfor more information related to providing culturally and linguistically appropriate services.
Treating Concomitant Speech and Language Problems
Children with selective mutism can also have a concomitant communication delay, disorder, or weakness (Richard, 2011). Children with selective mutism may avoid speaking out of fear of being teased regarding speech sound production or vocal quality (Anstendig, 1999). Evidence of a concomitant communication disorder is not restricted to specific settings or social situations, even when co-occurring with selective mutism.
Before addressing specific speech and language deficits, the child may benefit from addressing only selective mutism goals to increase their confidence in communicating and to establish rapport with the SLP.
Adolescents and Adults
Selective mutism may be resolved in childhood; however, selective mutism in childhood may persist into adolescence and adulthood, or it may develop into another anxiety disorder or phobia (Steinhausen et al., 2006). Adolescents and adults with selective mutism may report not wanting to talk because they do not see the benefits of speaking. At times, young adults may have the desire to speak but are unable to speak because of significant anxiety or lack of strategies (Walker & Tobbell, 2015). The inability to speak may bring about feelings of shame, isolation, frustration, and hopelessness because they have difficulty fulfilling expected social roles.
Older individuals often develop strategies to avoid talking and may have defined themselves as being primarily nonverbal. Motivational interviewing is a client-centered counseling technique that helps the adolescent or adult explore and resolve ambivalence through discussion and aims to increase internal motivation for behavioral change. A motivational interview for someone with selective mutism could include asking about the positive and negative aspects of selective mutism, exploring life goals and values, and then determining goals (Kotrba, 2015; Rollnick & Miller, 1995). The client may be more comfortable with sharing their experiences and concerns through online interview methods (Walker & Tobbell, 2015).
Coding and Reimbursement
Payment and coverage of services related to the evaluation and treatment of selective mutism varies based on factors such as the patient’s diagnosis(es), the payer (e.g., Medicare, Medicaid, or commercial insurance), and the patient’s specific health insurance plan. It is important for clinicians to understand coverage policies for the payers they commonly bill, to verify coverage for each patient prior to initiating services, and to be familiar with correct diagnosis and procedure coding for accurate claims submission.
Clinicians use International Classification of Diseases and Related Health Problems, 10th Revision, Clinical Modification (ICD-10-CM) codes to describe the patient’s diagnosis and Current Procedural Terminology codes to describe related evaluation and treatment services.
The term “selective mutism” is used to classify this diagnosis within the ICD-10-CM family of codes for behavioral and emotional disorders with onset usually occurring in childhood and adolescence. Clinicians may also report specific diagnosis codes for speech, language, and communication disorders, as needed. Payer policies may outline specific guidelines based on this diagnosis, such as who may assign a diagnosis for behavioral and emotional disorders and what types of services are covered. For example, some payers may only cover services related to this diagnosis under a mental health benefit; this could require an initial evaluation by a physician or mental health professional and limit coverage of evaluation and treatment by an SLP.
For more information about coding, see the following ASHA resources:
ICD-10-CM Diagnosis Codes Related to Speech, Language, and Swallowing Disorders [PDF]
Current Procedural Terminology (CPT) Codes for Speech-Language Pathology Services
Service Delivery
Format
Format refers to the structure of the treatment session (e.g., group vs. individual) provided.
Children may require individual treatment sessions initially, depending on the strategies and techniques being applied, to establish rapport and to practice relaxation techniques and pragmatic skills in a safe, comfortable setting.
Small-group treatment can facilitate communication with peers, beginning with nonverbal play using scripted interactions involving single words and phrases and moving toward the goal of speaking spontaneously (Klein & Armstrong, 2013). Another format involves forming groups of individuals with selective mutism who are of similar age, cognitive functioning, and speech-language skills. Groups may need to be adjusted based on each individual’s progress (Kearney, 2010).
Some individuals may prefer telepractice to receive treatment. Many of the treatment strategies noted above can be implemented through virtual means (Busman et al., 2020; Hong et al., 2022). The clinician can use technology (e.g., mobile device) to coach the child or care partner through in vivo exposure activities in school or in the community. Telepractice provides increased access to services for children who may not otherwise have access to trained professionals with experience treating selective mutism. It can also allow for increased collaboration between professionals and family members in different settings.
See ASHA’s Practice Portal page on Telepractice for more information.
Provider
Provider refers to the person offering the treatment (e.g., SLP, speech-language pathology assistant, care partner). In treating selective mutism in a school setting, an established and trained keyworker may be the provider of interventions (Kotrba, 2015). A keyworker is an adult in the school setting who is trained to provide consistent behavioral interventions to the student. The keyworker can help the student generalize skills throughout the school environment and communicate with the treatment team.
Dosage
Dosagerefers to the frequency, intensity, and duration of service.
Intensive treatment sessions for selective mutism may be helpful for some individuals and can take place in a variety of settings. For example, in the school setting, using stimulus fading and/or shaping can take place over the course of a week; however, an intensive treatment can disrupt the child’s schedule for the duration of a week. With this type of treatment schedule, school staff receive training in the intervention approach to continue with appropriate treatment and provide accommodations after the intensive treatment ends (Kotrba, 2015).
Intensive group treatment in a summer camp simulates a school setting, and the child with selective mutism can receive intensive practice in a safe setting without interruption to their school schedule. Families also receive the benefit of meeting other families who have a child with selective mutism (Cornacchio et al., 2019; Kotrba, 2015).
Setting
Setting refers to the location of treatment (e.g., home, school, community-based). Generalization of skills to new environments is an important aspect to selective mutism treatment. Treatment may occur within the clinical office, school, and community to reinforce the individual’s speaking skills.
Resources
ASHA Resources
Assessment Tools, Techniques, and Data Sources
Consumer Information: Selective Mutism
Conquering Challenges of Interprofessional Treatment for Selective Mutism
Current Procedural Terminology (CPT) Codes for Speech Language Pathology Services
ICD-10-CM Diagnosis Codes Related to Speech, Language, and Swallowing Disorders [PDF]
Interprofessional Education/Interprofessional Practice (IPE/IPP)
Selective Mutism: An Integrated Approach
Selective Mutism in Elementary School: Multidisciplinary Interventions
Teaching the Language of Feelings to Students With Severe Emotional and Behavioral Handicaps
Trauma-Informed Care
Other Resources
This list of resources is not exhaustive, and the inclusion of any specific resource does not imply endorsement from ASHA.
Bergman, R. L., Keller, M. L., Piacentini, J., & Bergman, A. J. (2008). The development and psychometric properties of the Selective Mutism Questionnaire. Journal of Clinical Child & Adolescent Psychology, 37(2), 456–464.
Child Mind Institute: Complete Guide to Selective Mutism
Kurtz Psychology: Selective Mutism Learning University
RCSLT: New Long COVID Guidance and Patient Handbook
Striving to Speak: Children’s Books About Selective Mutism
Selective Mutism Association
Selective Mutism Association: Educator Toolkit [PDF]
Selective Mutism Association: Speech Language Therapy and Selective Mutism
References
American Psychiatric Association. (2022). Anxiety disorders. In Diagnostic and statistical manual of mental disorders(5th ed., text rev.).
American Speech-Language-Hearing Association. (2023). Code of ethics [Ethics]. www.asha.org/policy/
Anstendig, K. D. (1999). Is selective mutism an anxiety disorder? Rethinking its DSM-IV classification. Journal of Anxiety Disorders, 13(4), 417–434.
Beidel, D. C., & Turner, S. M. (2007). Shy children, phobic adults: Nature and treatment of social anxiety disorders (2nd ed.). American Psychological Association.
Beidel, D. C., Turner, S. M., & Morris, T. L. (1999). Psychopathology of childhood social phobia. Journal of the American Academy of Child & Adolescent Psychiatry, 38(6), 643–650.
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About This Content
Acknowledgments
Content for ASHA’s Practice Portal is developed and updated through a comprehensive process that includes multiple rounds of subject matter expert input and review. ASHA extends its gratitude to the following subject matter experts who were involved in the development of the Selective Mutism page:
Sharon Lee Armstrong, PhD
Brittany Bice-Urbach, PhD
Rachel Cortese, MS, CCC-SLP
Emily R. Doll, MA, MS, CCC-SLP
Joleen Fernald, PhD, CCC-SLP
Suzanne Hungerford, PhD, CCC-SLP
Evelyn Klein, PhD, CCC-SLP
Janet Middendorf, MA, CCC-SLP
Gail Richard, PhD, CCC-SLP
Robert Schum, PhD
Donna Spillman-Kennedy, MS, CCC-SLP
Robert Thompson, PhD, CCC-SLP
ASHA seeks input from subject matter experts representing differing perspectives and backgrounds. At times a subject matter expert may request to have their name removed from our acknowledgment. We continue to appreciate their work.
Citing Practice Portal Pages
The recommended citation for this Practice Portal page is:
American Speech-Language-Hearing Association. (n.d.). Selective mutism [Practice Portal].
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Home / Dictionary / Tree Diagram
Tree Diagram
Tree diagrams are essential visual tools for probability calculations and decision-making, breaking down complex scenarios into structured branches. They help visualize possible outcomes, calculate probabilities, and are widely used in educational and professional contexts for solving intricate problems.
Updated 25 Oct, 2024
9 min read
Simplifying the Power of Tree Diagrams for Complex Probability Problems
Tree diagrams are a visual tool that maps out different outcomes in probability calculations and decision-making processes. They provide a structured way to visualize and solve problems that involve multiple events or decisions. By breaking down complex scenarios into branches, tree diagrams make it easier to see all potential outcomes and their probabilities, ensuring a clearer understanding of various possibilities. Tree diagrams have become essential to educational curriculums and professional applications, from basic event calculations to complex decision-making scenarios.
What are Tree Diagrams?
A tree diagram is a visual depiction illustrating the various possibilities resulting from a series of occurrences. It starts with a single point, known as the root, and branches into multiple lines representing possible choices or outcomes. These branches further split into additional branches, depicting all potential results of subsequent decisions or events. Tree diagrams are widely used in probability, statistics, and decision-making because of their simplicity and effectiveness in visualizing all possible paths and outcomes.
Importance of Tree Diagrams in Probability Calculations
Tree diagrams play a crucial role in probability because they allow individuals to systematically explore all possible outcomes and their associated probabilities. Whether calculating the likelihood of independent or dependent events, tree diagrams provide a visual roadmap that ensures accuracy and comprehension. They help break down complex probability problems into manageable parts, making it easier to calculate probabilities, even for scenarios with multiple events or conditions.
Basic Concepts of Tree Diagrams
Definition of a Tree Diagram
A tree diagram is a powerful visual tool to map out all potential outcomes of a series of interconnected events. By branching out from a single starting point, the diagram visually represents each possible result at every stage, making it easier to calculate probabilities and understand complex scenarios. This structure allows users to follow the progression of decisions or random events, making it ideal for visualizing probability calculations, particularly in statistics and decision-making processes. Tree diagrams help identify dependent and independent events, making them valuable in evaluating outcomes. This systematic approach simplifies the analysis of multi-step processes and ensures clarity when determining the probability of a specific outcome.
Components of a Tree Diagram
Each tree diagram comprises nodes and branches, the building blocks for visualizing potential outcomes. The nodes, represented as points where branches split, signify decision points or events. Branches, depicted as lines extending from these nodes, illustrate the various choices or potential outcomes that stem from a specific decision. The diagram begins with a single starting point, the root node. From the root, subsequent branches extend, representing different paths that could unfold. As each branch leads to another node, more branches may appear, creating a cascading structure of decisions and outcomes. This hierarchical visualization helps systematically break down complex scenarios, making it easier to analyze dependencies and outcomes. Tree diagrams are particularly useful in decision analysis, probability calculations, and problem-solving, as they offer a straightforward, visual method of displaying the progression of choices and their consequences.
Steps to Create a Tree Diagram
How to Start with Tree Diagrams
Identify the initial event or decision and draw a single point as the root. From this point, draw branches for each possible outcome of the first event.
Drawing Branches Correctly
Each branch in a tree diagram should represent an event’s distinct and unique outcome. To construct a complete tree diagram, list all the potential outcomes for the first event. Then, continue to add branches for each subsequent event, ensuring that all possible outcomes are accounted for. This process allows for a visual representation of the events’ sequence and probabilities. Each branch should be clearly labeled to reflect the event and associated probability. This helps understand the likelihood of different scenarios occurring and is helpful in fields like probability theory and decision analysis for breaking down complex events into more superficial, manageable visual structures.
Calculating Probabilities Using Tree Diagrams
Tree diagrams are handy for calculating probabilities because they visually represent all potential outcomes:
Using a Tree Diagram to Calculate Single-Event Probabilities
To calculate the probability of a single event, multiply the probabilities along the branches leading to that event. This product gives the likelihood of reaching that particular outcome.
Combining Probabilities in Tree Diagrams
Combine probabilities for scenarios with multiple events by multiplying them along each path. Add up the probabilities of all paths leading to a desired outcome to find the total probability.
Tree Diagrams for Independent Events
Independent events are those where the outcome of one event does not affect the outcome of another:
Concept of Independent Events in Tree Diagrams
In tree diagrams, independent events are represented by branches that do not influence one another. Each event’s probability remains the same, regardless of previous outcomes.
Examples of Independent Events
An example of independent events is flipping a coin twice. The outcome of the first flip does not impact the second flip, making them independent events.
Tree Diagrams for Dependent Events
Dependent events occur when the outcome of one event affects the outcome of another:
Understanding Dependent Events
In a tree diagram, dependent events are represented by branches that change based on the outcomes of preceding events. The probabilities of subsequent branches depend on the previous outcomes.
How to Represent Dependent Events on a Tree Diagram
To illustrate dependent events, adjust the probabilities along branches to reflect how earlier outcomes impact the likelihood of future events.
Solving Probability Problems with Tree Diagrams
Tree diagrams simplify solving probability problems by providing a clear, step-by-step visual representation.
Step-by-Step Approach Using Tree Diagrams
Identify all events: Start by determining all events involved in the problem. For example, if you are drawing cards, consider the number of cards and possible outcomes (e.g., drawing a king or queen).
Draw branches for each possible outcome: Create a tree diagram by drawing branches for each event. Each branch represents a possible outcome. For instance, if you are drawing two cards, your first set of branches will show the outcomes of the first draw (king, queen, etc.), and subsequent branches will represent the outcomes of the second draw based on the first.
Multiply probabilities along the branches: Calculate the probability for each branch and multiply the probabilities along each path. For example, if you drew a king first, the probability of drawing a queen following changes since you’re not replacing the cards. Multiply the probabilities along each step.
Add probabilities of relevant paths: Once you’ve calculated the probabilities along the branches, sum up the probabilities of the paths that correspond to the outcome you’re interested in. This will give you the total probability.
Example Problems and Solutions
Consider a problem where you draw two cards from a deck without replacement. Use a tree diagram to represent all possible outcomes and calculate the probability of drawing a king and a queen. Draw branches for each draw, calculate the changing probabilities without replacement, and sum the relevant paths to find the total probability.
Advanced Applications of Tree Diagrams
Tree diagrams can be used for more than just fundamental probability problems. They are also valuable in complex scenarios:
Using Tree Diagrams for Complex Probability Scenarios
Tree diagrams offer a straightforward way to track outcomes and calculate probabilities accurately for scenarios involving multiple stages or conditional probabilities.
Case Studies Involving Multiple Events
For example, tree diagrams can be used in medical decision-making to evaluate the probability of different diagnostic outcomes based on a series of tests.
Tree Diagrams in Educational Curriculum
Tree diagrams are a vital part of the educational curriculum, especially in mathematics and statistics:
How Tree Diagrams Are Taught in Schools
Students are introduced to tree diagrams in schools to solve probability problems and visualize outcomes.
Importance of Tree Diagrams in GCSE Maths
Tree diagrams are included in the GCSE Maths curriculum to help students understand probability concepts and solve related problems efficiently.
Interactive Tree Diagram Exercises
Interactive exercises can enrich the learning process and augment the understanding of tree diagrams:
Engaging Students with Interactive Tree Diagram Problems
Use software tools and online resources to create interactive exercises that allow students to manipulate tree diagrams and explore different scenarios.
Resources for Teachers and Students
Numerous online platforms and worksheets provide practice problems and interactive exercises for mastering tree diagrams.
Tree Diagrams in Statistics Versus Pure Maths
The application of tree diagrams varies between statistics and pure mathematics:
Differences in the Application of Tree Diagrams in Different Fields
In statistics, tree diagrams are often used to represent real-world scenarios with data-driven probabilities, while in pure maths, they illustrate theoretical concepts.
Importance of Contextual Understanding
Understanding the context in which tree diagrams are applied is crucial for accurately interpreting results and making informed decisions.
Tree Diagram Software Tools
Software tools can simplify the creation of tree diagrams and make it easier to handle complex scenarios:
Software That Can Be Used to Create Tree Diagrams
Several software options, such as Microsoft Visio, Lucidchart, and specialized probability software, can be used to create detailed tree diagrams.
Benefits of Using Software over Manual Drawing
Using software allows for easier adjustments, accurate probability calculations, and more complex scenarios with multiple events.
Common Mistakes in Drawing Tree Diagrams
There are several common mistakes that people make when creating tree diagrams:
Identifying and Correcting Common Errors
One standard error represents only some possible outcomes, leading to inaccurate probabilities. Another mistake is miscalculating probabilities along branches.
Tips for Accurate Tree Diagram Creation
Double-check that all branches are accounted for and that probabilities along each path add up to 1 to ensure the diagram is accurate.
Tree Diagram Worksheets and Practice Questions
Worksheets provide structured practice for mastering tree diagrams:
Resources for Practice
Look for worksheets that cover a variety of problems, from basic probability to advanced scenarios involving multiple events.
How Worksheets Can Improve Understanding of Tree Diagrams
Regular practice with worksheets helps reinforce the concepts and improve problem-solving skills related to tree diagrams.
Assessment Techniques for Tree Diagram Problems
Assessing students’ understanding of tree diagrams requires a combination of problem-solving and conceptual questions:
How Educators Can Assess Understanding of Tree Diagrams
Teachers can use problem-based assessments, quizzes, and interactive exercises to gauge students’ proficiency with tree diagrams.
Example Assessment Questions
Ask students to create a tree diagram for a specific scenario and calculate the probabilities of certain outcomes based on their diagram.
Comparing Tree Diagrams with Other Probability Tools
Tree diagrams are not the only tool for probability calculations, but they have unique advantages:
How Tree Diagrams Stack Against Other Probability Tools
Compared to tools like Venn diagrams or probability tables, tree diagrams provide a more explicit visual representation of sequential events and conditional probabilities.
When to Use Tree Diagrams over Other Methods
Tree diagrams are ideal when dealing with multiple stages or conditional probabilities that require a step-by-step breakdown.
The Role of Tree Diagrams in Decision Making
Tree diagrams are not limited to probability problems—they also have applications in decision-making processes:
Real-World Applications of Tree Diagrams
Businesses use tree diagrams for decision analysis, evaluating risks, and determining the best course of action in complex scenarios.
Case Examples from Business and Science
In healthcare, tree diagrams help visualize treatment options and their outcomes, aiding doctors and patients in making informed decisions.
Teaching Tips for Tree Diagrams
Effective teaching strategies can make learning tree diagrams more engaging and impactful:
Effective Methods to Teach Tree Diagrams in Classrooms
Incorporate real-life scenarios and interactive exercises to make learning more relatable and engaging for students.
Engaging Students with Real-Life Problems
Use examples from sports, finance, and everyday decisions to show how tree diagrams apply to real-life situations.
Visual Learning Aids for Tree Diagrams
Visual aids enhance the comprehension of tree diagrams, especially for students who learn best through visual representation:
Importance of Visual Aids in Teaching Tree Diagrams
Visual aids make abstract concepts more concrete and help students grasp the relationships between different outcomes.
Examples of Effective Visual Aids
Use colour-coded branches, icons, and interactive diagrams to illustrate complex scenarios effectively.
Feedback and Revision Techniques Using Tree Diagrams
Feedback and revision are crucial for mastering tree diagrams:
How to Use Feedback Effectively in Tree Diagram Problems
Provide detailed feedback on errors and suggest strategies for improvement, such as rechecking branch probabilities.
Techniques for Revising and Improving Understanding
Encourage students to revisit problematic areas and practice with additional problems to solidify their knowledge.
Digital Resources and Online Platforms for Tree Diagrams
Online platforms and digital resources provide a wealth of information and practice opportunities for learning tree diagrams:
Online Resources Available for Learning Tree Diagrams
Websites like Khan Academy, Math is Fun, and BBC Bitesize offer tutorials, exercises, and videos that explain the use of tree diagrams in probability and decision-making.
Recommended Websites and Tools
Tools such as Lucidchart and Creately are excellent choices for interactive tree diagram creation. These platforms allow users to build and manipulate tree diagrams, helping them understand complex scenarios visually.
Tree Diagram Challenges and Competitions
Challenges and competitions can motivate students to master tree diagrams and apply their knowledge creatively:
Encouraging Mastery of Tree Diagrams Through Competitions
Organize classroom competitions or participate in national math contests with tree diagram problems to boost students’ skills and confidence.
Examples of National and International Contests
Competitions like the American Mathematics Competitions (AMC) and the International Mathematical Olympiad (IMO) often include problems that require a solid understanding of tree diagrams.
Future Trends in Teaching Probability with Tree Diagrams
As education evolves, new trends are emerging in the teaching of probability concepts using tree diagrams:
Innovations in Teaching Tree Diagrams
Innovations include using AI-powered educational platforms, virtual reality simulations, and gamified learning experiences that make mastering tree diagrams more engaging and interactive.
Predictions for Future Curriculum Developments
Future curriculums may emphasize digital tools and interactive learning platforms to teach tree diagrams, making it easier for students to visualize and solve complex probability problems.
Expert Opinions on Tree Diagrams in Education
Educators and mathematicians recognize the importance of tree diagrams in teaching probability and decision-making:
Insights from Educators and Mathematicians
Experts suggest tree diagrams are essential for understanding probability and developing critical thinking and decision-making skills.
Importance of Tree Diagrams in Modern Education
In modern education, tree diagrams bridge theoretical concepts and real-world applications, making them a valuable tool for academic and practical learning.
FAQs
What is a tree diagram used for in probability? Tree diagrams represent all possible outcomes of a series of events in probability, making calculating the likelihood of different outcomes easier.
How do you create a tree diagram? Start with a single point representing the initial event and draw branches for each possible outcome. Continue adding branches for subsequent events until all possible outcomes are defined.
What are the differences between independent and dependent events in a tree diagram? Independent events do not affect each other’s outcomes, while dependent events have outcomes that are influenced by previous events, represented by varying probabilities along the branches.
Can tree diagrams be used for decision-making? In the realm of decision-making, tree diagrams serve as a valuable tool. They enable the evaluation of diverse options and their corresponding outcomes, aiding strategic planning and risk assessment.
What tools can be used to create tree diagrams? Tools like Lucidchart, Creately, and Microsoft Visio can create and manipulate tree diagrams, offering features that simplify the visualization and calculation of probabilities.
What Tools Can Be Used to Create Tree Diagrams?
Tools like Lucidchart, Creately, and Microsoft Visio can create and manipulate tree diagrams, offering features that simplify the visualization and calculation of probabilities.
Awais Jawad
Content Writer at OneMoneyWay
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A Big Step on the CMMC Rollout Timeline
September 10, 2025
A Big Step on the CMMC Rollout Timeline
Written by
Chris Camejo
CMMC
Information Security Compliance
Table of contents
CMMC Rollout Start Date Estimate
CMMC Rollout Phases
Next Steps
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A major step on the CMMC rollout timeline was completed recently as the regulatory change that will create the CMMC contract clause made its way to the Office of Information and Regulatory Affairs (OIRA). This post covers what that means for contractors that want to know when to expect CMMC clauses in their contracts.
September 10, 2025 update: This post has been updated to reflect the publication of the finalized CMMC contract clause with a start date of November 10, 2025.
There are three pieces of information that a contractor needs in order to determine when they can expect to see CMMC requirements in their contracts:
The CMMC level and assessment requirements the contractor expects to be subject to
The start date of the CMMC rollout
The phases of the CMMC rollout
Contractors that are unsure of their future CMMC level and assessment requirements should take a detour to our previous blog post on CMMC Level and Assessment Requirements for Defense Contractors, as the phased rollout information below will be more useful with this knowledge.
A specific start date for the CMMC rollout to begin has not been released. However, based on the submission to OIRA we can estimate that it will be no earlier than October 2025 and likely no later than January 2026. This post will be updated once we have a definitive start date. Additional information on how this range was estimated is provided below.
Some contractors may encounter supposed CMMC clauses even though the CMMC rollout had not yet begun or, if you’re reading this in the future, CMMC clauses that don’t align with the phased rollout schedule. These clauses are suspicious, and contractors should refer to our previous blog post on The Proliferation of “Fake” CMMC Contract Clauses for more information on this topic.
CMMC Rollout Start Date Estimate
The regulation that defines the CMMC program and its requirements, known as 32 CFR 170, was finalized on October 15, 2024. This regulation lays out a phased rollout for CMMC that will take place over three years (explained below); however, it does not set the start date for that rollout, nor does it define the contracting procedures that would impose CMMC obligations on contractors.
Another regulation that creates the CMMC contracting procedures, including the CMMC contract clause and start date for the CMMC phased rollout, was finalized and published on September 10, 2025. The start date for the CMMC rollout will be November 10, 2025.
The publication of the start date doesn’t mean defense contractors will suddenly need to be CMMC compliant on November 10 for a few reasons:
The CMMC rules do not alter any prior existing contracts so contractors will continue to protect information they handle today under the terms of the specific contract that information was received or created under. CMMC obligations will only apply to information received or created under new contracts containing CMMC clauses that the contractor receives after the rollout commences.
The phased CMMC rollout defined in 32 CFR 170 (and explained below) means that contracts that should eventually contain CMMC clauses will not contain these clauses or will contain clauses that require a lower CMMC level than would be expected based on the information that is being handled during the first three years after the start of the rollout.
The final version of the CMMC contract clause will be inserted at 48 CFR 252.204-7021 (better known as DFARS 7021) to replace the old clause from CMMC 1.0. Organizations interested in what to expect can review the final rule published in the Federal Register. The most interesting parts are the DFARS 7021 clause itself and the DFARS 7025 clause that will be inserted into solicitations to warn of CMMC requirements.
CMMC Rollout Phases
The phases of the CMMC rollout are based on which CMMC level and assessment requirements should apply to a specific contract. The phased rollout of CMMC is defined in 32 CFR 170.3 and contains four phases that dictate when certain types of CMMC requirements will start appearing in new DoD contracts:
Phase 1
| | |
--- |
| Starts: | Effective date of the Title 48 CMMC Acquisition Final Rule |
| Start Date: | November 10, 2025 |
| Level 1 and 2 Self-Assessments: | Included as a condition of contract award May be required to exercise an option period on a contract that was awarded prior to the CMMC rollout start date at DoD discretion |
| Level 2 C3PAO Certification Assessments: | May be included in some contracts at DoD discretion |
| Level 3 DIBCAC Certification Assessments: | Not included in contracts |
Phase 2
| | |
--- |
| Starts: | One calendar year following the start date of Phase 1 |
| Start Date: | November 10, 2026 |
| Level 1 and 2 Self-Assessments: | Included as a condition of contract award May be required to exercise an option period on a contract that was awarded prior to the CMMC rollout start date at DoD discretion |
| Level 2 C3PAO Certification Assessments: | Included as a condition of contract award May be delayed to an option period instead of as a condition of contract award at DoD discretion |
| Level 3 DIBCAC Certification Assessments: | May be included in some contracts at DoD discretion |
Phase 3
| | |
--- |
| Starts: | One calendar year following the start date of Phase 2 |
| Start Date: | November 10, 2027 |
| Level 1 and 2 Self-Assessments: | Included as a condition of contract award May be required to exercise an option period on a contract that was awarded prior to the CMMC rollout start date at DoD discretion |
| Level 2 C3PAO Certification Assessments: | Included as a condition of contract award and option exercise |
| Level 3 DIBCAC Certification Assessments: | Included as a condition of contract award May be delayed to an option period instead of as a condition of contract award at DoD discretion |
Phase 4 (Full Implementation)
| | |
--- |
| Starts: | One calendar year following the start date of Phase 3 |
| Start Date: | November 10, 2028 |
| Level 1 and 2 Self-Assessments: | Included as a condition of contract award and option exercise |
| Level 2 C3PAO Certification Assessments: | Included as a condition of contract award and option exercise |
| Level 3 DIBCAC Certification Assessments: | Included as a condition of contract award and option exercise |
Next Steps
Based on the information above, a contractor should now have a reasonable understanding of which CMMC level and assessment requirements they will eventually be subject to and when to expect these requirements. Contractors should begin work on achieving compliance at the appropriate level prior to receiving CMMC contract clauses. Contractors that will require Level 2 or Level 3 certification assessments may also want to schedule the necessary third-party assessments with a CMMC Third-Party Assessor Organization (C3PAO) and/or DIBCAC so they are prepared for phase 2 and 3 contract requirements.
Unlike other compliance frameworks, the CMMC program differentiates between organizations that help contractors implement compliance (CMMC Registered Practitioner Organizations or RPOs) and organizations that assess contractors for certification or C3PAOs. TrustedSec is a CMMC RPO and is available to help understand and implement CMMC requirements. Get in touch with us if you need assistance!
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Mathematics Interventions for Students with Learning Disabilities (LD) in Secondary School: A Review of the Literature
2015, Learning Disabilities: A Contemporary Journal
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Abstract
The purpose of our literature review was to extend and update Maccini, Mulcahy, and Wilson's (2007) review of the literature on mathematics interventions for secondary students with learning disabilities (LD). An extensive search of the research literature netted 15 research studies that focused on mathematics interventions for secondary students with LD. The findings are presented in terms of three main instructional approaches for improving the mathematics achievement of these students: (a) cognitive and metacognitive strategies for solving word problems; (b) use of representations to increase conceptual knowledge and problem-solving skills; and (c) Enhanced Anchored Instruction. Results include evidence of the efficacy of Enhanced Anchored Instruction and Solve It! We discuss additional results, implications for improving classroom practice and professional development, and recommendations for future research.
Related papers
2006
Rarely is there a single cause for a learning disability. Often there is a constellation of causes operating together that result in a child’s learning difficulties. Mathematics is a complex process requiring visual and cognitive perception abilities, comprehension ability, and adequate prior knowledge. This article addresses students’ learning difficulties in mathematics and provides some suggested activities that have been proven to work with students with learning disabilities.
Review of …, 2009
The purpose of this meta-analysis was to synthesize findings from 42 interventions (randomized control trials and quasi-experimental studies) on instructional approaches that enhance the mathematics proficiency of students with learning disabilities. We examined the impact of four categories of instructional components: (a) approaches to instruction and/or curriculum design, (b) formative assessment data and feedback to teachers on students' mathematics performance, (c) formative data and feedback to students with LD on their performance, and (d) peer-assisted mathematics instruction. All instructional components except for student feedback with goal-setting and peer-assisted learning within a class resulted in significant mean effects ranging from 0.21 to 1.56. We also examined the effectiveness of these components conditionally, using hierarchical multiple regressions. Two instructional components provided practically and statistically important increases in effect size–teaching students to use heuristics and explicit instruction. Limitations of the study, suggestions for future research, and applications for improvement of current practice are discussed.
2015
Aim: The current study tested the comparative efficacy of various strategies on basic mathematical skills of learning disabled children. Methods: Learning disabled children were randomly assigned to multimedia, cognitive, eclectic and control conditions. Assessment included the use of IQ, Diagnostic Test of Learning Disability, and Pre and post-test administration of the Children with Specific Learning Disabilities in Arithmetic scale. Results: All the tested strategies significantly enhanced basic mathematical skills of learning disabled children. Conclusions: Multimedia, cognitive strategy and eclectic approach can be used for enhancing the mathematical skills of learning disabled children.
Iranian Evolutionary Educational Psychology Journal, 2021
The aim of this study was to evaluate the effectiveness of the educational package of cognitive-metacognitive strategies and cognitive intervention in solving the mathematical verbal problem and cognitive functions in students with special math learning disabilities. The design of the present study was a pretest-posttest with a control group. The statistical population of the study consisted of all fourth grade elementary students with special learning disabilities in Bahmaei city, Iran in 2020. From this population, 16 people were selected as the research sample by purposive sampling. The training package was presented in the form of 9 sessions of 45 minutes. Content validation was used to validate the training package. The research instruments were a researcher-made questionnaire for solving mathematical verbal problem, Wechsler scale index version 5 and Key Math scale. Research hypotheses were tested by repeated measures analysis of variance. The results indicated that the educational package and cognitive intervention could significantly increase cognitive functions and mathematical verbal problem solving in students with special math learning disabilities (p < .01). Comparison of means of post-test and follow-up in the variables of mathematical verbal problem solving (MD = 0.31, p < .52,), planning (MD = 0.13, p = .99), concurrent processing (MD = 0.38, p <0.49) and sequence processing (MD = 0.25, p <0.12) showed that there was no significant change in dependent variables after two months. Explaining this result, it can be said that the training package has been effective in the long time. On the other hand, teaching cognitive-metacognitive strategies compared to cognitive intervention had a greater effect on the variables of mathematical verbal problem solving (t = 3.12, p < .01) and simultaneous processing (t = 3.22, p <.01). Accordingly, the teaching of cognitive-metacognitive strategies can be used to improve cognitive functions and verbal problem solving in students with special math learning disabilities.
Journal of Indian Association For Child and Adolescent Mental Health, 2008
The current study tested the comparative efficacy of various strategies on basic mathematical skills of learning disabled children. Methods: Learning disabled children were randomly assigned to multimedia, cognitive, eclectic and control conditions. Assessment included the use of IQ, Diagnostic Test of Learning Disability, and Pre and post-test administration of the Children with Specific Learning Disabilities in Arithmetic scale. Results: All the tested strategies significantly enhanced basic mathematical skills of learning disabled children. Conclusions: Multimedia, cognitive strategy and eclectic approach can be used for enhancing the mathematical skills of learning disabled children.
To improve student success in mathematics, the use of research-based interventions is necessary to help secondary students with learning disabilities (LD) access the algebra curriculum. The authors provide an overview of the following research-based approaches: explicit instruction, graduated instructional sequence, technology, and graphic organizers. For each approach, a summary of the research is provided along with suggestions for how the approach can be used by classroom teachers to support algebra instruction.
The Educator Journal
Mathematics, commonly taken as a complicated subject, has been further challenging for students with learning difficulties. Learning difficulties are the difficulties of learners acquiring the knowledge and skills to the average level expected of those of the same age, primarily because of mental disability or cognitive disorder. Effective teaching approaches are required to meet the needs of diverse students in enhancing mathematics learning. This paper explores the approaches for enhancing mathematics learning for students with learning difficulties. The armchair research adopts document study or library study to explore the techniques to enhance students’ mathematics learning with learning difficulties. The findings showed students’ perception of the learning of mathematics greatly affects students’ motivation to engage in learning mathematics and achievement. The positive perception of students encourages them for effective learning. Some of the approaches to enhance mathematic...
Learning Disability Quarterly, 2008
Increased attention is being paid to students who demonstrate difficulty in learning and applying mathematics concepts. The purpose of this special series was to address issues related to students and mathematics learning disabilities (LD). We identify Response to Intervention (RtI) as it relates to early mathematics instruction and a multi-tiered service delivery system. Further, because RtI has focused primarily on young children and the prevention of LD, we present information about older students who have been identified as having mathematics LD and provide strategies for helping them access the general education curriculum. Six papers on various mathematics topics, grade levels, and service delivery will be provided in this special series. Authors report findings on research efforts and offer implications for practice.
ADDRESSING DIFFICULTIES IN TEACHING MATHEMATICS TO STUDENTS WITH MATHEMATICAL LEARNING DISABILITIES, 2022
The essence and reason for the inability to master mathematics are described as a lack of working memory. This paper describes two main approaches to solving the problem of teaching mathematics to students with learning difficulties in mathematics (MLD): (1) training working memory and (2) reducing the load on working memory in the instructional process. It was found that the results of the first approach are ambiguous: Training working memory leads to its improvement, which is confirmed by the test results but may not lead to improvement of the mathematical learning process associated with the student's working memory. This justifies the primacy of the second approach. Both previously known methods for reducing the load on working memory in mathematics instruction are presented. A computer-based mathematics learning system developed by the author aims at automating basic computational skills (arithmetics, trigonometry, geometry). It is explained how to work with the developed computer-assisted learning system, which is based on the method of interval repetitions, and empirical data on the results of the system implementation are given.
1990
Based on suggestions of participants at the second annual Instructional Methods Forum held in June 1989, the monograph considers cognitive-based approaches to teaching mathematics to students with learning problems. It addresses identification of characteristics of successful cognitive approaches and the role of media and materials. An introductory chapter looks at possible reasons for United States students' difficulties with mathematics and proposes that cognitive-based approaches be used. The second chapter looks at mathematical learning amdg students with disabilities. Cognitive-based principles for teaching mathematics are presented in chapter 3, including findings from research and a taxonomy of word problem types. Chapter 4 offers two examples of such approaches including the Cognitively Guided Instruction program ()Mich improves teacher understanding of how children learn mathematics) and the Verbal Pmblem Solving Among the Mildly Handicapped Project (which uses specially designed materials in its problem-solving focus). The fifth chapter con3iders instructional components of cognitive-based mathematics concerned with both content and teaching methods. The sixth chapter considers the role of the teacher in cognitive-oriented programs, and a summary chapter suggests areas for further research. Appended are a list of the forum participants, sample records from the instructional materials database of ae Information Center for Special Education Media and Materials, and a bibliography of approximately 170 items. (DB)
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References (30)
Related papers
Learning Disabilities Research and Practice, 2007
AbstractThis article extends a previous review of the literature (Maccini & Hughes, 1997) on mathematics interventions for secondary school students with learning disabilities (LD). A systematic review of the literature from 1995 to 2006 yielded 23 articles that met the criteria for inclusion. It was determined that a number of practices demonstrated significant gains for secondary school students with LD in math, including mnemonic strategy instruction, graduated instructional approach, cognitive strategy instruction involving planning, schema-based instruction, and contextualized videodisc instruction. We also discuss the nature and focus of math interventions and implications for both research and practice based on the findings.
Ldv Forum
Mathematical literacy is the ability to apply skills and concepts to reason through, communicate about, and solve mathematical problems. Mathematics instruction must involve methodology and curricular materials that assist students in mastering instructional objectives that are relevant to the development of mathematical literacy and appear on Individualized Education Programs (IEPs) of students with learning disabilities (LD).
Journal of Learning Disabilities, 1997
Secondary students with learning disabilities generally make inadequate progress in mathematics. Their achievement is often limited by a variety of factors, including prior low achievement, low expectations for success, and inadequate instruction. This article will discuss techniques that have been demonstrated to be effective with secondary students who have learning disabilities in mathematics.
Center on Instruction, 2008
Focus on Autism and Other Developmental Disabilities, 2019
Educational programs for students with moderate and severe disabilities (MSD) have undergone drastic changes since the mandate for access to the general curriculum was provided by Individuals With Disabilities Education Act. Since then, educators have struggled to find methods to use to promote optimal learning, including in the area of mathematics. The purpose of this systematic literature review was to provide an update on research related to teaching mathematics to students with MSD published from 2005 to 2017. Results from the included studies indicated that mathematics research has started to diversify in the skills that are being taught to this population. In addition to skills taught, current research has continued to inform the field on some promising methods that can be used to teach a broader range of mathematics skills. Emerging strategies that were identified included the use of concrete representations, anchored instruction, and instructional technology. Suggestions for...
I like it instead of maths': how pupils with moderate learning difficulties in Scottish primary special schools intuitively solved mathematical word problemsb jsp_461 130..138
2003
Abstract This article presents the results of a meta-analysis of 58 studies of mathematics interventions for elementary students with special needs. Interventions in three different domains were selected: preparatory mathematics, basic skills, and problem-solving strategies. The majority of the included studies described interventions in the domain of basic skills. In general, these interventions were also the most effective. Furthermore, a few specific characteristics were found to influence the outcomes of the studies.
Learning Disabilities: A Contemporary Journal, 2020
As today's classrooms become more and more diverse, there is a growing need to explore the intersection between English Learners (ELs) and students with learning disabilities (LD) in the content-specific instruction of mathematics problem solving. The aim of this study was to determine which types of instructional scaffolds may be used by math teachers to effectively support ELs with LD learning multiplicative reasoning. To this end, we employed an exploratory case study based on a frequency count analysis of four scaffold types used by the students and the teacher in their sessions. The results showed that kinesthetic and linguistic scaffolds were the most beneficial for helping ELs with LD to cultivate mathematical thinking with both concrete and abstract units, while also helping to increase the sophistication of their mathematical content-language usage. In combination with small-group interactions, these scaffolds provide an effective instructional method for improving multiplicative reasoning among ELs with LD.
Intervention in School and Clinic, 2021
State exams frequently use word problems to measure mathematics performance, making difficulties with word problem solving a barrier for many students with learning disabilities (LD) in mathematics. Based on meta-analytic data for students with LD, five empirically validated word-problem strategies are presented, with components of model-based problem solving (MBPS) highlighted.
2017
Explicit instruction is commonly used for helping students with mathematics learning difficulties. However, some research findings indicate that students’ mastery of conceptual understanding and mathematical process skills was often ignored if this approach was used solely. This research was aimed to investigate the teaching and learning processes during a remedial intervention The intervention was carried out using a mixed instructional approach which combined the perspective of the behaviorists and constructivists. It was expected to enhance the mathematical knowledge and process skills of students. A case study research design was employed. Data was collected and analyzed using a qualitative approach. Results showed that the students were able to improve their mathematics conceptual and procedural knowledge, and mathematical process skills through active interaction and mind-on activities. During the intervention, the teacher used explicit instruction for introduction and followe...
Related topics |
5345 | https://www.datascienceblog.net/post/basic-statistics/distributions/ | Understand. Implement. Succeed.
Using probability distributions in R: dnorm, pnorm, qnorm, and rnorm
Basic Statistical Concepts for Data Science
Distribution functions in R
Every distribution has four associated functions whose prefix indicates the type of function and the suffix indicates the distribution. To exemplify the use of these functions, I will limit myself to the normal (Gaussian) distribution. The four normal distribution functions are:
dnorm: density function of the normal distribution
pnorm: cumulative density function of the normal distribution
qnorm: quantile function of the normal distribution
rnorm: random sampling from the normal distribution
The probability density function: dnorm
The probability density function (PDF, in short: density) indicates the probability of observing a measurement with a specific value and thus the integral over the density is always 1. For a value (x), the normal density is defined as [{\displaystyle f(x\mid \mu ,\sigma ^{2})={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\text{exp}\left(-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}\right)}] where (\mu) is the mean, (\sigma) is the standard deviation, and (\sigma^2) is the variance.
Using the density, it is possible to determine the probabilities of events. For example, you may wonder: What is the likelihood that a person has an IQ of exactly 140?. In this case, you would need to retrieve the density of the IQ distribution at value 140. The IQ distribution can be modeled with a mean of 100 and a standard deviation of 15. The corresponding density is:
sample.range <- 50:150 iq.mean <- 100 iq.sd <- 15 iq.dist <- dnorm(sample.range, mean = iq.mean, sd = iq.sd) iq.df <- data.frame("IQ" = sample.range, "Density" = iq.dist) library(ggplot2) ggplot(iq.df, aes(x = IQ, y = Density)) + geom_point()
From these data, we can now answer the initial question as well as additional questions:
pp <- function(x) { print(paste0(round(x 100, 3), "%")) } # likelihood of IQ == 140? pp(iq.df$Density[iq.df$IQ == 140])
```
"0.076%"
```
```
likelihood of IQ >= 140? pp(sum(iq.df$Density[iq.df$IQ >= 140]))
```
```
"0.384%"
```
```
likelihood of 50 < IQ <= 90? pp(sum(iq.df$Density[iq.df$IQ <= 90]))
```
```
"26.284%"
```
The cumulative density function: pnorm
The cumulative density (CDF) function is a monotonically increasing function as it integrates over densities via [f(x | \mu, \sigma) = \displaystyle {\frac {1}{2}}\left[1+\operatorname {erf} \left({\frac {x-\mu }{\sigma {\sqrt {2}}}}\right)\right]] where (\text{erf}(x) = \frac {1}{\sqrt {\pi }}\int _{-x}^{x}e^{-t^{2}}\,dt) is the error function.
To get an intuition of the CDF, let’s create a plot for the IQ data:
cdf <- pnorm(sample.range, iq.mean, iq.sd) iq.df <- cbind(iq.df, "CDF_LowerTail" = cdf) ggplot(iq.df, aes(x = IQ, y = CDF_LowerTail)) + geom_point()
As we can see, the depicted CDF shows the probability of having an IQ less or equal to a given value. This is because pnorm computes the lower tail by default, i.e. (P[X <= x]). Using this knowledge, we can obtain answers to some of our previous questions in a slightly different manner:
```
likelihood of 50 < IQ <= 90? pp(iq.df$CDF_LowerTail[iq.df$IQ == 90])
```
```
"25.249%"
```
```
set lower.tail to FALSE to obtain P[X >= x] cdf <- pnorm(sample.range, iq.mean, iq.sd, lower.tail = FALSE) iq.df <- cbind(iq.df, "CDF_UpperTail" = cdf) # Probability for IQ >= 140? same value as before using dnorm! pp(iq.df$CDF_UpperTail[iq.df$IQ == 140])
```
```
"0.383%"
```
Note that the results from pnorm are the same as those obtained from manually summing up the probabilities obtained via dnorm. Moreover, by setting lower.tail = FALSE, dnorm can be used to directly compute p-values, which measure how the likelihood of an observation that is at least as extreme as the obtained one.
To remember that pnorm does not provide the PDF but the CDF, just imagine that the function carries a p in its name such that pnorm is lexicographically close to qnorm, which provides the inverse of the CDF.
The quantile function: qnorm
The quantile function is simply the inverse of the cumulative density function (iCDF). Thus, the quantile function maps from probabilities to values. Let’s take a look at the quantile function for (P[X <= x]):
```
input to qnorm is a vector of probabilities prob.range <- seq(0, 1, 0.001) icdf.df <- data.frame("Probability" = prob.range, "IQ" = qnorm(prob.range, iq.mean, iq.sd)) ggplot(icdf.df, aes(x = Probability, y = IQ)) + geom_point()
```
Using the quantile function, we can answer quantile-related questions:
```
what is the 25th IQ percentile? print(icdf.df$IQ[icdf.df$Probability == 0.25])
```
```
89.88265
```
```
what is the 75 IQ percentile? print(icdf.df$IQ[icdf.df$Probability == 0.75])
```
```
110.1173
```
```
note: this is the same results as from the quantile function quantile(icdf.df$IQ)
```
```
0% 25% 50% 75% 100% ## -Inf 89.88265 100.00000 110.11735 Inf
```
The random sampling function: rnorm
When you want to draw random samples from the normal distribution, you can use rnorm. For example, we could use rnorm to simulate random samples from the IQ distribution.
```
fix random seed for reproducibility set.seed(1) # law of large numbers: mean will approach expected value for large N n.samples <- c(100, 1000, 10000) my.df <- do.call(rbind, lapply(n.samples, function(x) data.frame("SampleSize" = x, "IQ" = rnorm(x, iq.mean, iq.sd)))) # show one facet per random sample of a given size ggplot() + geom_histogram(data = my.df, aes(x = IQ)) + facet_wrap(.~SampleSize, scales = "free_y")
```
```
note: we can also implement our own sampler using the densities my.sample <- sample(iq.df$IQ, 100, prob = iq.df$Density, replace = TRUE) my.sample.df <- data.frame("IQ" = my.sample) ggplot(my.sample.df, aes(x = IQ)) + geom_histogram()
```
Note that we called set.seed in order to ensure that the random number generator always generates the same sequence of numbers for reproducibility.
Summary
Of the four functions dealing with distributions, dnorm is the most important one. This is because the values from pnorm, qnorm, and rnorm are based on dnorm. Still, pnorm, qnorm, and rnorm are very useful convenience functions when dealing with the normal distribution. If you would like to learn about the corresponding functions for the other distributions, you can simply call ?distribtuion to obtain more information.
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About Matthias Döring
Matthias Döring is a data scientist and AI architect. He is currently driving the digitization of the German railway system at DB Systel. Previously, he completed a PhD at the Max Planck Institute for Informatics in which he researched computational methods for improving treatment and prevention of viral infections.
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Comments
Sam07 Oct 20 07:00 UTC
Perhaps I’m wrong, but it seems to me that there is a mistake in the interpretation of the density function. With continuous random variables, the probability of having an IQ of 140 is not the value of the density function at 140. Technically, the probability of having a specific value with a continuous r.v. is always zero. If one wants to makes an approximation, then he would write
pnorm(140.1, mean=100,sd=15)-pnorm(139.9,mean=100,sd=15) 0.0001519534
Which is not identical to dnorm(140,mean=100,sd=15) 0.0007597324
The small interval around 140 which we use to make our calculations with, will depend on the level of precision with which we make our measurements, I believe.
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5348 | https://flexbooks.ck12.org/cbook/ck-12-math-analysis-concepts/section/2.10/primary/lesson/polynomial-and-rational-inequalities-mat-aly/ | Polynomial and Rational Inequalities | CK-12 Foundation
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2.10
Polynomial and Rational Inequalities
FlexBooks 2.0>
CK-12 Math Analysis Concepts>
Polynomial and Rational Inequalities
Written by:Raja Almukkahal
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 02, 2025
Often it is easier to use and remember new terms when you have a 'hook' or comparison to a term you know already.
"Polynomial inequality" is a term generally used to refer to inequalities where the x variable has a degree of 3 or greater.
The prefix "poly" means 'multiple' or 'many', and the root word "nomial" means 'name' or 'term'.
Therefore a "polynomial" is literally: "many terms".
The prefix "in" means 'not', and the root word "equal" of course means 'the same'.
Therefore and "inequality" refers to things which are "not same" or "not equal".
Can you use this logic to identify the origin of some of the other terms used in this lesson?
Polynomial and Rational Inequalities
Polynomial Inequalities
Solving polynomial inequalities is very similar to solving quadratic inequalities. The basic steps are the same:
Set up the inequality in the form p(x)>0 (or p(x)<0,p(x)≤0,p(x)≥0)
Find the solutions to the equation p(x)=0.
Divide the number line into intervals based on the solutions to p(x)=0.
Use test points to find solution sets to the equation.
Rational Inequalities
There is one step added to the process of solving rational inequalities because a rational function can also change signs at its vertical asymptotes or at a break in the graph. For instance, look at the graph of the function r(x)=x x 2−9 below.
If we want to solve the inequality x x 2−9>0, then we need to use the following critical points: x=0,x=3, and x=−3. x=0 is the solution of setting the numerator equal to 0, and this gives us the only root of the function. x=±3 are the vertical asymptotes, the x−coordinates that make the function undefined because putting in 3 or -3 for x will cause a division by zero.
Testing the intervals between each critical point to see if the values in that interval satisfy the function gives us:
| Interval | Test Point | Positive/Negative? | Part of Solution Set? |
--- --- |
| (−∞,−3) | -4 no |
| (-3, 0) | -2 | + | yes |
| (0, 3) | 2 no |
| (3,+∞) | 4 | + | yes |
Thus, the solutions to x x 2−9>0 are x∈(−3,0)∪(3,+∞).
Examples
Example 1
Earlier, you were given a question about identifying the origins of other terms in this lesson.
How many of the terms we have used recently were you able to track down the origins of? A few are listed below, did you find others?
Bi-nomial: "two-named" or "two-terms" - from "bi", meaning 'two' and "nomial", meaning 'name' or 'term'. Quad-ratic: "related to a square" - from "quadratus", meaning 'square'. Ratio-nal: "related to a ratio" - from "ratio", meaning 'reason' (as in "to reason" or "calculate") and "-al", meaning "related to".
Example 2
Solve x 3−3 x 2+2 x≥0.
The polynomial is already in the correct form p(x)≥0 so we solve the equation
x 3−3 x 2+2 x=0 x(x 2−3 x+2)=0 x(x−2)(x−1)=0
The zeros are at x=0,x=1, and x=2.
| Interval | Test Point | Positive/Negative? | Part of Solution Set? |
--- --- |
| (−∞,0) | -5 no |
| (0, 1) | 1 2 | + | yes |
| (1, 2) | 3 2 no |
| (2,+∞) | 3 | + | yes |
Notice that this inequality is greater than or equal to zero, so we include the zeros in the solution set. Therefore the solutions are x∈[0,1]∪[2,+∞].
Example 3
Solve 6 x 4+5 x 2<25.
First we will change the inequality to 6 x 4+5 x 2−25<0. Now, solve the equation 6 x 4+5 x 2−25=0.
6 x 4+5 x 2−25=0(3 x 2−5)(2 x 2+5)=0
The first term yields the solutions x=±5 3 and there are no real solutions for the second term.
| Interval | Test Point | Positive/Negative? | Part of Solution Set? |
--- --- |
| (−∞,−5 3) | -3 | + | no |
| (−5 3,5 3) | 0 yes |
| (5 3,+∞) | 3 | + | no |
Finally, the solution set is x∈(−5 3,5 3).
Example 4
Find the solution set of the inequality
4 x−12 3 x−2<0
From the numerator we solve 4 x−12=0 or x=3. In the denominator, solve 3 x−2=0 and we find the critical point x=2 3.
Making the table
| Interval | Test Point | Positive/Negative? | Part of Solution Set? |
--- --- |
| (−∞,2 3) | 0 | + | no |
| (2 3,3) | 1 yes |
| (3,+∞) | 5 | + | no |
Therefore, the solution set includes the numbers in the interval (2 3,3). Or in set-builder notation, the solution is {x|2 3<x<3}.
Example 5
(using technology) The McNeil Surf Company makes wetsuits. For a given number of wetsuits x, McNeil's profit, in dollars, is given by the function P(x)=−0.01 x 2+25 x−3000.
If the manager of McNeil wants the profit to stay above $9,000, what is the minimum and maximum number of wetsuits they can manufacture to maintain that level of profit?
Set up the inequality
−0.01 x 2+25 x−3000>9000−0.01 x 2+25 x−12000>0
With a calculator you can graph the function Y 1=−0.01 x 2+25 x−12000.
On a TI-83: use the window [−1000,4000]×[−5000,15000].
The settings are:
X m i n=−1000,X m a x=4000,X s c l=500
Y m i n=−5000,Y m a x=5000,Y s c l=1000 x r e s=1
On a software grapher, the image looks like this with the window x:0→2400 and y:0→3500
Using the CALC menu (2ND TRACE), and selecting the option ZEROS, we can see that the zeros of Y 1=−0.01 x 2+25 x−12000 are at x=647.920 and x=1852.080.
By inspecting the graph, we can see that the solution set to the inequality −0.01 x 2+25 x−12000>0 is x∈(648,1852).
Visually that looks like:
What is the maximum profit McNeil can make?
Keeping the same graph open, use CALC MAXIMUM to solve for the maximum profit. The maximum is at (1250, 3625), indicating that the maximum profit is $3625 above the minimum we set: of $9000.
So the actual maximum profit is $12,625 when 1250 wetsuits are produced.
Can you explain why this shape might make sense for a profit function?
One possible reason the profit function might take this shape is labor costs. If McNeil wants to make a very large number of wetsuits in a short period of time, then that may require paying overtime for workers, and this could reduce the profit margin.
Example 6
For the following rational function, determine limitations on the domain and the asymptotes, and then sketch the graph.
f(x)≥2 x+5 x−1
To identify the graph of the inequality f(x)≥2 x+5 x−1, first treat it as if it were the equality f(x)≥2 x+5 x−1
For f(x)≥2 x+5 x−1:
To find the critical points, identify the value(s) which make the denominator = 0: x≠1
That gives us a vertical asymptote of x=1
The horizontal asymptote becomes apparent as x becomes truly huge and the "+5" and "-1" no longer matter. At that point, we have f(x)=2 x x→f(x)=2 So the horizontal asymptote is y=2
Now that you know the shape of the graph, simply shade the area above the lines, since the original function was f(x) is greater-than function, and leave the lines solid since it was a greater-than or equal to.
The final graph should look like:
Review
Find the solution set of the following inequalities without using a calculator. Display the solution set on a number line.
x 2+2 x−3≤0
3 x 2−7 x+2>0
−6 x 2−13 x+5≥0
5 x−1 x−2>0
1−x x<1
Solve: 4 x 3−4 x 2−3 x>0
Solve: x 4 4−x 2<0
Solve: 4 x 3−8 x 2−x+2≥0
n 3−2 n 2−n+2 n 3+3 n 2+4 n+12<0
n 3+3 n 2−4 n−12 n 3−5 n 2+4 n−20≤0
2 n 3+5 n 2−18 n−45 3 n 3−n 2+27 n−9≥0
12 n 3+16 n 2−3 n−4 8 n 3+12 n 2+10 n+15>0
Use a calculator to solve the following inequalities. Round your answer to three places after the decimal.
−9.8 t 2+357.6 t≥0
x 3−5 x+7≤−4 x 2+18
x 2−2 x x−5>x 2−25
Solve and graph: f(x)>9 x 2−4 3 x+2
The total resistance of two electronics components wired in parallel is given by R 1 R 2 R 1+R 2 where R 1 and R 2 are the individual resistances (in Ohms) of the two components. a) If the resistance of R 1 is 20 Ohms, what is the maximum resistance of R 2 if the total resistance must be less than 15 Ohms? b) What is the maximum theoretical resistance of this circuit? How do you know?
A rectangular lot of land has a length that is 7 meters more than twice its width. If the area of the lot is greater than 60 square meters, what are the possible values of the widths of the lot?
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A discounted cash flow analysis considers the (time-adjusted) present value of future cash flows to determine the value of an investment and choose business projects under consideration, M&A buyout candidates, or securities investments.
What is Discounted Cash Flow?
Discounted cash flow is a financial analysis computing future years’ forecasted cash flows at today’s lower value. The DCF formula considers a time period, the time value of money, and risk with a selected discount rate. Businesses and investors use DCF to assess potential projects, the value of a company for M&A, and the expected return from securities investments.
Discounted Cash Flow Formula
The discounted cash flow (DCF) formula is:
DCF = CF1 + CF2+ … + CFn
(1+r)1 (1+r)2 (1+r)n
The discounted cash flow formula uses a cash flow forecast for future years, discounted back to the equivalent value if received in today’s dollars, then sums the discounted value for every year projected. CF1 is cash flows for year 1, CF2 is cash flows for year 2. Cash flows for the final year being considered in the DCF financial analysis are denoted as n.
The discount factor (discount rate) in the denominator is 1+ a risk rate adjusted above the risk-free cost of capital, the rate of return expected from other competing investments, or the company’s weighted average cost of capital (WACC).
Cash flow is different than business profit, which is revenue minus expenses.
You can use a Google Sheet, Excel spreadsheet, or an online discounted cash flow calculator to compute discounted cash flow.
Try using this online calculator (discounted payback) to calculate discounted cash flow. Set the Initial Investment to $0 (because discounted cash flow doesn’t consider it) and provide Cash Flow per year (year 1), Increase in cash flow, Number of Years, and Discount Rate.
If you do this correctly, you’ll get the same result as the discounted cash flow example below that was calculated using a spreadsheet. The example assumes year 1 cash flow of $225,000, a 10% increase in cash flow per year, 5 as the number of years, and an 8% discount rate.
Purpose of Discounted Cash Flow
Discounted cash flow analysis is applied in different areas, including business investment project selection, M&A valuation, and investors determining the market value of stock and other investments.
Business Investment Project Selection
Businesses often use their weighted average cost of capital (WACC) rate as the risk rate to justify investment projects.
The investment project evaluated should produce positive cash flows at their present value and yield a return of at least the company’s hurdle rate for the investment. The terminal value (residual value) of a project in the year of project completion is also included in the DCF financial model.
A considered project should be ranked compared to other potential investment projects competing for funding. The terminal value (residual value) of a project in the year of project completion is also included in the DCF financial model.
Another discounted cash flow methodology that businesses and commercial real estate companies use to select a project for investment is the internal rate of return (IRR).
M&A DCF Valuation
In investment banking and corporate finance, M&A analysts use discounted cash flow as one method to determine the value of a business.
M&A teams also set the enterprise value using company comparables for other recent deals in the industry. Public companies have a higher per-share valuation price than private companies. Privately-held companies are discounted to a lower fair value due to a lack of share marketability and higher risk profiles.
Equity Value Assessment by Investors
Investors use discounted cash flow as one method of analyzing potential stock investments and their estimated market value.
HBO released a documentary in 2022 titled Icahn: The Restless Billionaire that mentions the use of discounted cash flow. Carl Icahn, an activist investor and chairman of Icahn Enterprises LP (IEP), uses a DCF model and other analyses as valuation methods to identify underperforming companies with low equity value that could be improved. Icahn or a team member takes a Board of Directors position after buying a substantial stock position in an undervalued publicly-traded company or acquiring the company.
Investors can also determine the intrinsic value of a stock or a company valuation based on the present value of future dividends with a perpetual growth rate by using the Gordon Growth Model (GGM). The Gordon Growth Model is a dividend discount model to determine stock price or enterprise valuation with a simpler formula. GGM has been proven using a discounted cash flow calculation.
Discounted Cash Flow vs. Net Present Value
The main difference between discounted cash flow vs. net present value is that net present value subtracts upfront year 0 costs (in actual dollars estimated) from the sum of the present value of the cash flows. The discounted cash flow method doesn’t subtract these initial costs that include capital expenditures.
Example of DCF Discounted Cash Flow vs. Net Present Value
Cash flow projections by year for the forecast period and the risk rate selected as an appropriate discount rate to determine the present value of these cash flows are used for both NPV and DCF calculations in the examples below.
The expected cash flow growth rate is forecast at 10% per year. The company’s weighted average cost of capital of 8% is used in this calculation as the discount rate in the present value table. The present value discount factors are from a Present Value of 1 table.
Only the net present value example assumes and deducts the initial costs in year 0 of $150,000. The table below shows the initial investment cost ($150,000) as a negative cash flow.
In the financial modeling examples for the DCF method and NPV below, the terminal value or cash salvage value from selling used investment project equipment is assumed to be zero. (Any salvage value or residual value received at the project termination would be discounted back to year 0.)
| Year | Cash Flow Forecast | Present Value Discount Factor @ 8% rate | Discounted Cash Flow by Year |
--- --- |
| 0 | ($150,000) | Not applicable to year 0 | ($150,000) |
| 1 | $225,000 | 0.9259 | $208,328 |
| 2 | $247,500 | 0.8573 | $212,182 |
| 3 | $272,250 | 0.7938 | $216,112 |
| 4 | $299,475 | 0.7350 | $220,114 |
| 5 | $329,423 | 0.6806 | $224,205 |
Discounted Cash Flow Calculation Example
DCF = ($208,328 + $212,182 + $216,112 + $220,114 + $224,205)
DCF = $1,080,941
Net Present Value (NPV) Calculation Example
Net Present Value = – $150,000 + ($208,328 + $212,182 + $216,112 + $220,114 + $224,205)
Net Present Value = – $150,000 + $1,080,941
Net Present Value = $ 930,941
Difference Between DCF and NPV Calculations
The difference between discounted cash flow and net present value equals the initial cost as an estimated upfront cash investment in year 0.
Reconciliation of discounted cash flow and net present value calculations:
Discounted cash flow $1,080,941
Net present value $ 930,941
Difference = $ – 150,000
When is DCF Analysis Most Useful?
DCF analysis is most useful when future cash flows are predictable, an appropriate discount rate for risk is used, and objectivity is needed to select the best potential investment.
Constraints of DCF
Constraints of DCF (discounted cash flow) include the use of estimates and the variability in forecasting future cash flows. Both external and internal factors can impact cash flows.
Possible cash flow impacts affecting forecast accuracy include:
Economic recessions or improved business cycles
Inflation rate increases
Unanticipated effects of interest rate hikes
Geopolitical events
Sales forecast, cost estimates, and cash conversion cycle inaccuracy
New competitors entering the market or taking market share
The corporate finance team often prepares cash forecasts in businesses with scenarios for best case, expected case, and worst case. Then the analyst and management select one scenario. Is it the right case for evaluating an investment?
The discounted cash flow calculation can also be affected by using the wrong risk rate, also known as the discount rate.
If the estimated cash flows used in the calculation of DCF are too high vs. realistic estimates that can be achieved, an investment may be selected that will underperform expectations. Conversely, if the forecast cash flows are erroneously too low, the company may miss an investment opportunity that should have been selected.
Takeaways
Discounted cash flow (DCF) financial models are used as cash flow valuations to value and select investments.
Discounted cash flow analysis uses projected future cash flows from an investment for a selected time period. It discounts them to the present value by incorporating a risk rate then sums the present values of cash flows. Projected cash flows are the net inflows and outflows of cash for a year.
Investments may have a salvage value of equipment sold or residual value that produces cash flow when the project ends. The terminal value is also discounted to its present value.
A risk rate often used by businesses for calculating the discounted cash flow of investment projects is the weighted average cost of capital (WACC). Other risk rates, including alternative project returns, may be used in evaluating investments with the DCF metric.
The difference between discounted cash flow and net present value is that net present value (NPV) subtracts the initial cash investment, but DCF doesn’t.
Discounted cash flow models may produce incorrect valuation results if forecast cash flows or the risk rate are inaccurate. The result is missed projects for underestimated cash flows or acceptance of underperforming projects if cash flow estimates are too high.
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5350 | https://getsyeducated.substack.com/p/a-brief-note-on-intellectual-sloth | Get Syeducated
A Brief Note on Intellectual Sloth; or, Are 70% of U.S. Faculty Really Adjuncts?
moin syed
Sep 13, 2024
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Call me old-fashioned, but it makes me sad when people rely on shitty data to support their arguments. It makes me even sadder when they minimize, dismiss, or get angry when confronted with the shittiness of the data. These experiences highlight that, for many people, what the data actually are doesn’t really matter. All that matters is that there are some data, somewhere, that they can point to in support of their position.
I use the rate of adjunct instructors in U.S. higher education as an example, but this same phenomenon applies to many different topics, so I encourage you to read on even if you are none too interested in the intricacies of U.S. higher education.
Before going any further, I want to be clear about something, which some of you will nevertheless ignore or even downright misrepresent: The reliance on adjunct instructors is bad. It’s bad for students, it’s bad for institutions, and most importantly, it’s bad for the people in those roles. Adjunct instructors are generally not paid well, not provided benefits, not guaranteed employment from term to term, and not accorded much respect by their colleagues or supervisors. This is bad. Other bad things about the U.S. higher education scene include the growing legions of administrators who draw large salaries and the major declines in state funding for public institutions. All very bad1.
The way we know about these trends is not just intuition or through observing what is happening in our local environments. We know about this stuff because we collect, analyze, and disseminate data on the topics. We do that because we think the data matter. And if we think the data matter, we should also think it matters that they are as accurate as possible.
Yesterday I came across this post on Bluesky:
70% of all U.S. faculty are adjuncts! 70% of all faculty are making close to minimum wage!
Best I could tell, these claims went totally unchecked and were readily accepted. C’mon. This is the kind of number that, when you see it, you should immediately question its accuracy. If you come across a statistic that you consider “mind-blowing,” you may want to at least look into whether or not it is true before sharing it with others.
The source for this figure seems to be from the American Association of University Professors (AAUP). On their website, its states the following:
“Contingent positions that are ineligible for tenure now account for nearly 70 percent of all instructional staff appointments in American higher education, including 49 percent part time and 19 percent full-time non-tenure-track.”
Wait, the statement references “contingent positions that are ineligible for tenure.” It does not say “adjuncts.” What exactly are contingent positions not eligible for tenure?
Well, it says right in the quote that it includes the 19% who are full-time non-tenure track positions. These positions are certainly sometimes low-paid positions with no benefits or guarantees of employment, but not always. Many institutions have “teaching professor” tracks that are well-paid, include benefits, and come with multi-year contracts, so this is a noisy category that is difficult to understand.
You may say I am nitpicking with that one, but the 70% also includes the 20% that are graduate student instructors. Sorry, graduate students are not faculty! We definitely can—and should—discuss how graduate student instructors are underpaid and poorly treated, but we should not include them in faculty statistics2.
The more detailed breakdown is readily available on the AAUP website. This is the default figure that is shown on the landing page (other than I switched from headcount to percentage):
The figure clearly shows that in 2022 the actual number of part-time faculty, what people generally think of when referring to “adjuncts” is about 38%, not 70%. The 70% figure seems to come from adding together part-time, full-time non-tenure track, and graduate students. (Not really the point of the post, but I am struck by how stable these numbers are across 20 years.)
I had wrongly assumed that these data only pertained to instructional staff, but as Dave Vanness pointed out, the graduate student data also include those who have research appointments and likely are not teaching3. Removing them brings the part-time rate up to 42% from 38%. Still not even close to 70%.
These are the kinds of details you can get by clicking on the “drilldown” tab on the AAUP page for the data, and I encourage interested folks to poke around. Importantly, there is a great deal of variability among institutional types, and among schools within those types. For example, the situation is much more dire at community colleges, which tend to be ignored in discussions of higher education, relative to the R1 universities that dominate our attention.
Is 42% still bad? Yes! Is 42% the same as 70%? No! It is simply false to say that 70% of U.S. faculty are adjuncts making nearly minimum wage, and if we are going to share such statistics—especially in such a dramatic way—we should ensure they are correct. Several folks did not take too kindly to me pointing this out on Bluesky, and that brings me back to what is really the main point of this post: there are people out there who want to have data to support their views, but don’t really care if the data are accurate or not. What’s more, questioning the accuracy of the data is equivalent to minimizing the issue.
For me, this is some of the worst behavior of the supposedly scientific, empirically-minded person. What seems to matter is that there are some data out there, somewhere, that they can point to that support their position and allow them to argue that their view is based in evidence. This is the kind of thing that we see quite frequently in research on the negative implications of social media and tech use for well-being. A lot of the data are just bad and don’t provide even modest evidence for the claims, but pointing to some data—any data—gives the arguments extra force and credibility.
If you want to argue for some policy based on moral, ethical, and rational grounds, or because you just think it’s a good idea, then go on ahead. But don’t say that the policy is backed by scienceTM just because you have a crapheap of data you can point to.
This whole issue reminds me of Geoffrey Pullum’s classic takedown of the myth of the “Eskimo words for snow4.” If you have never read it, do so now. It is a quick, entertaining, insightful read that is well worth your time. I assign it for the first day of my undergraduate research methods class because, as Pullum describes in the Appendix, the essay is fundamentally about intellectual sloth:
The tragedy is not that so many people got the facts wildly wrong; it is that in the mentally lazy and anti-intellectual world we live in today, hardly anyone cares enough to think about trying to determine what the facts are. (Pullum, 1991, p. 171)
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1
I’m sure there are many other things that are bad, and I apologize if I left out the issue that you are particularly passionate about.
2
A lot of folks will say that 20% graduate instructors is too high. That may be so, I don’t really know what an optimal number is. But a lot of (the same) folks will also say we don’t provide enough training to prepare graduate students to teach. Doing so does requires that they actually teach, not just take classes in pedagogy.
3
Of course, some have both types of appointments at the same time.
4
I include the following note when referring people to this essay: The term “Eskimo” is still widely used but is generally considered offensive, or at the very least outdated, as it is a label that was developed and applied by people who are outside of the relevant communities. It is preferable to use the labels that are used by the specific communities in question, generally the Inuit and Yup’ik peoples. The term is sometimes still used by Inuit and Yup’ik peoples to describe themselves, but should not be used by outsiders. This is similar to the use of “Indian” in the U.S., which is often used by those are Indigenous or Native American, but should not be used by those who are not. The great Eskimo vocabulary hoax takes on an issue that is related to the cultural homogenization and exotification that is wrapped up in the label “Eskimo.” Thus, despite the outdated terminology, I continue to find it an appropriate reading to assign and ask that you do so while being mindful and critical about the labels we use to describe people, especially those from marginalized backgrounds.
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Siva
Sep 13, 2024Edited
Interesting note - but I am confused. Are you expecting a social media post to be accurate or nuanced? Aren’t we supposed to assume they are mostly inaccurate or imprecise?
Or is your point you see this in media that is supposed to be rigorous and trust worthy?
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Spartacus
Sep 13, 2024
Your point about 70% adjuncts is a similar point to this article about the transformation of news media. And there are other "true facts" that I don't want to get into.
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What is the probability of rolling a sum of 7 with two dice?
There are 6 ways to roll a sum of 7, out of a total of 36 possible outcomes, so the probability is 6/36 = 1/6.
Mohamed C.
2 years ago
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To find the probability of rolling a sum of 7 with two dice, we need to first determine the total number of possible outcomes when rolling two dice. Each die has six sides, so the total number of possible outcomes is 6 x 6 = 36.
Next, we need to determine the number of outcomes that result in a sum of 7. We can do this by listing all the possible combinations of two dice that add up to 7, which are: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1).
So, there are 6 possible outcomes that result in a sum of 7. Therefore, the probability of rolling a sum of 7 with two dice is:
P(sum of 7) = number of outcomes resulting in a sum of 7 / total number of possible outcomes
= 6 / 36
= 1/6
Therefore, the probability of rolling a sum of 7 with two dice is 1/6 or approximately 0.1667 (or 16
.67%).
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Understanding the Discriminant of a Quadratic Equation
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Example SolutionsPractice Questions
Using the Discriminant to Determine the Nature of Roots of a Quadratic Equation
Step 1: Identify a, b, and c in the quadratic equation a x 2+b x+c=0.
Step 2: Substitute the values found in step 1 into the formula for the discriminant, Δ=b 2−4 a c
Step 3: Analyze the results. If Δ>0, there are two real-valued roots. If Δ<0, there are two imaginary roots. If Δ=0, there is one real-valued root.
Using the Discriminant to Determine the Nature of Roots of a Quadratic Equation - Vocabulary and Equations
Quadratic Equation: A quadratic equation is an equation of the form a x 2+b x+c=0 where a, b, and c are real numbers.
Discriminant: The discriminant, often denoted by Δ, is the part of the quadratic formula that was underneath the square root. That is, Δ=b 2−4 a c. The discriminant can tell you the types of roots a quadratic equation has but does not tell you the exact values of the solutions.
We will use these steps, definitions, and equations to use the discriminant to determine the nature of roots of a quadratic equation in the following three examples.
Example Problem 1: Using the Discriminant to Determine the Nature of Roots of a Quadratic Equation - Two Real Solutions
Use the discriminant to determine the nature of the roots of the equation 3 x 2+7 x−2=0.
Step 1: Identify a, b, and c in the quadratic equation a x 2+b x+c=0.
For the equation 3 x 2+7 x−2=0, we have a = 3, b = 7, and c = -2.
Step 2: Substitute the values found in step 1 into the formula for the discriminant, Δ=b 2−4 a c
Substituting the values into the discriminant, we have:
Δ=7 2−4(3)(−2)=49+24=73
Step 3: Analyze the results. If Δ>0, there are two real-valued roots. If Δ<0, there are two imaginary roots. If Δ=0, there is one real-valued root.
Since we found that Δ=73, which is positive, there are two real-valued roots to the equation 3 x 2+7 x−2=0.
Example Problem 2: Using the Discriminant to Determine the Nature of Roots of a Quadratic Equation - Two Imaginary Solutions
Use the discriminant to determine the nature of the roots of the equation 4 x 2−x+3=0.
Step 1: Identify a, b, and c in the quadratic equation a x 2+b x+c=0.
We have a = 4, b = -1, and c = 3.
Step 2: Substitute the values found in step 1 into the formula for the discriminant, Δ=b 2−4 a c
We have:
Δ=(−1)2−4(4)(3)=1−48=−47
Step 3: Analyze the results. If Δ>0, there are two real-valued roots. If Δ<0, there are two imaginary roots. If Δ=0, there is one real-valued root.
Since Δ=−47, which is negative, there are two imaginary roots to the equation 4 x 2−x+3=0.
Example Problem 3: Using the Discriminant to Determine the Nature of Roots of a Quadratic Equation - One Real Solution
Use the discriminant to determine the nature of the roots of the equation −x 2+2 x−1=0.
Step 1: Identify a, b, and c in the quadratic equation a x 2+b x+c=0.
We have a = -1, b = 2, and c = -1.
Step 2: Substitute the values found in step 1 into the formula for the discriminant, Δ=b 2−4 a c
Δ=2 2−4(−1)(−1)=0
Step 3: Analyze the results. If Δ>0, there are two real-valued roots. If Δ<0, there are two imaginary roots. If Δ=0, there is one real-valued root.
Since Δ=0, there is one real-valued root to the equation −x 2+2 x−1=0.
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Table of Contents
Using the Discriminant to Determine the Nature of Roots of a Quadratic Equation
Using the Discriminant to Determine the Nature of Roots of a Quadratic Equation - Vocabulary and Equations
Example Problem 1
Example Problem 2
Example Problem 3
Test your current knowledge
Practice
Understanding the Discriminant of a Quadratic Equation
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5353 | https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf?srsltid=AfmBOorIpkHxAJTFJiVP5OQnsQzSgdxG9dP_y3tHSNLowC1QA8MS_ZFP | Olympiad Inequalities
Thomas J. Mildorf December 22, 2005
It is the purpose of this document to familiarize the reader with a wide range of theorems and techniques that can be used to solve inequalities of the variety typically appearing on mathematical olympiads or other elementary proof contests. The Standard Dozen is an exhibition of twelve famous inequalities which can be cited and applied without proof in a solution. It is expected that most problems will fall entirely within the span of these inequalities. The Examples section provides numerous complete solutions as well as remarks on inequality-solving intuition, all intended to increase the reader’s aptitude for the material covered here. It is organized in rough order of difficulty. Finally, the Problems section contains exercises without solutions, ranging from straightforward to quite difficult, for the purpose of practicing techniques contained in this document. I have compiled much of this from posts by my peers in a number of mathematical communities, particularly the Mathlinks-Art of Problem Solving forums, 1 as well as from various MOP lectures, 2 Kiran Kedlaya’s inequalities packet, 3 and John Scholes’ site. 4 I have tried to take note of original sources where possible. This work in progress is distributed for personal educational use only. In particular, any publication of all or part of this manuscript without prior consent of the author, as well as any original sources noted herein, is strictly prohibited. Please send comments - suggestions, corrections, missing information, 5 or other interesting problems - to the author at tmildorf@mit.edu .Without further delay...
1
and respectively, though they have merged into a single, very large and robust group. The forums there are also host to a considerable wealth of additional material outside of inequalities.
2
Math Olympiad Program. Although some people would try to convince me it is the Math Olympiad Summer Program and therefore is due the acronym MOSP, those who know acknowledge that the traditional “MOP” is the preferred appellation.
3
The particularly diligent student of inequalities would be interested in this document, which is available online at Further ma-terial is also available in the books Andreescu-Cartoaje-Dospinescu-Lascu, Old and New Inequalities , GIL Publishing House, and Hardy-Littlewood-P´ olya, Inequalities , Cambridge University Press. (The former is elementary and geared towards contests, the latter is more technical.)
4
where a seemingly inexhaustible supply of Olympiads is available.
5
Such as the source of the last problem in this document.
11 The Standard Dozen
Throughout this lecture, we refer to convex and concave functions. Write I and I′ for the intervals [ a, b ] and ( a, b ) respectively. A function f is said to be convex on I if and only if
λf (x) + (1 − λ)f (y) ≥ f (λx + (1 − λ)y) for all x, y ∈ I and 0 ≤ λ ≤ 1. Conversely, if the inequality always holds in the opposite direction, the function is said to be concave on the interval. A function f that is continuous on I and twice differentiable on I′ is convex on I
if and only if f ′′ (x) ≥ 0 for all x ∈ I (Concave if the inequality is flipped.) Let x1 ≥ x2 ≥ · · · ≥ xn; y1 ≥ y2 ≥ · · · ≥ yn be two sequences of real numbers. If
x1 + · · · + xk ≥ y1 + · · · + yk for k = 1 , 2, . . . , n with equality where k = n, then the sequence
{xi} is said to majorize the sequence {yi}. An equivalent criterion is that for all real numbers
t,
|t − x1| + |t − x2| + · · · + |t − xn| ≥ | t − y1| + |t − y2| + · · · + |t − yn|
We use these definitions to introduce some famous inequalities.
Theorem 1 (Jensen) Let f : I → R be a convex function. Then for any x1, . . . , x n ∈ I
and any nonnegative reals ω1, . . . , ω n,
ω1f (x1) + · · · + ωnf (xn) ≥ (ω1 + · · · + ωn) f
(ω1x1 + · · · + ωnxn
ω1 + · · · + ωn
)
If f is concave, then the inequality is flipped.
Theorem 2 (Weighted Power Mean) If x1, . . . , x n are nonnegative reals and ω1, . . . , ω n
are nonnegative reals with a postive sum, then
f (r) :=
(ω1xr
1
· · · + ωnxrn
ω1 + · · · + ωn
)1
r
is a non-decreasing function of r, with the convention that r = 0 is the weighted geometric mean. f is strictly increasing unless all the xi are equal except possibly for r ∈ (−∞ , 0] ,where if some xi is zero f is identically 0. In particular, f (1) ≥ f (0) ≥ f (−1) gives the AM-GM-HM inequality.
Theorem 3 (H¨ older) Let a1, . . . , a n; b1, . . . , b n; · · · ; z1, . . . , z n be sequences of nonnegative real numbers, and let λa, λ b, . . . , λ z positive reals which sum to 1. Then
(a1 + · · · + an)λa (b1 + · · · + bn)λb · · · (z1 + · · · + zn)λz ≥ aλa
1
bλb
1
· · · zλz
1
· · · + aλz
n
bλb
n
· · · zλz
n
This theorem is customarily identified as Cauchy when there are just two sequences.
Theorem 4 (Rearrangement) Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be two nondecreasing sequences of real numbers. Then, for any permutation π of {1, 2, . . . , n }, we have
a1b1 + a2b2 + · · · + anbn ≥ a1bπ(1) + a2bπ(2) + · · · + anbπ(n) ≥ a1bn + a2bn−1 + · · · + anb1
with equality on the left and right holding if and only if the sequence π(1) , . . . , π (n) is de-creasing and increasing respectively.
2Theorem 5 (Chebyshev) Let a1 ≤ a2 ≤ · · · ≤ an; b1 ≤ b2 ≤ · · · ≤ bn be two nondecreas-ing sequences of real numbers. Then
a1b1 + a2b2 + · · · + anbn
n ≥ a1 + a2 + · · · + an
n ·b1 + b2 + · · · + bn
n ≥ a1bn + a2bn−1 + · · · + anb1
n
Theorem 6 (Schur) Let a, b, c be nonnegative reals and r > 0. Then
ar(a − b)( a − c) + br(b − c)( b − a) + cr(c − a)( c − b) ≥ 0
with equality if and only if a = b = c or some two of a, b, c are equal and the other is 0.
Remark - This can be improved considerably. (See the problems section.) However, they are not as well known (as of now) as this form of Schur, and so should be proven whenever used on a contest.
Theorem 7 (Newton) Let x1, . . . , x n be nonnegative real numbers. Define the symmetric polynomials s0, s 1, . . . , s n by (x + x1)( x + x2) · · · (x + xn) = snxn + · · · + s1x + s0, and define the symmetric averages by di = si/(ni
). Then
d2
i
≥ di+1 di−1
Theorem 8 (Maclaurin) Let di be defined as above. Then
d1 ≥ √d2 ≥ 3
√d3 ≥ · · · ≥ n
√dn
Theorem 9 (Majorization) Let f : I → R be a convex on I and suppose that the sequence
x1, . . . , x n majorizes the sequence y1, . . . , y n, where xi, y i ∈ I. Then
f (x1) + · · · + f (xn) ≥ f (y1) + · · · + f (yn)
Theorem 10 (Popoviciu) Let f : I → R be convex on I, and let x, y, z ∈ I. Then for any positive reals p, q, r ,
pf (x) + qf (y) + rf (z) + (p + q + r)f
(px + qy + rz p + q + r
)
≥ (p + q)f
(px + qy p + q
)
( q + r)f
(qy + rz q + r
)
( r + p)f
(rz + px r + p
)
Theorem 11 (Bernoulli) For all r ≥ 1 and x ≥ − 1,
(1 + x)r ≥ 1 + xr
3Theorem 12 (Muirhead) Suppose the sequence a1, . . . , a n majorizes the sequence b1, . . . , b n.Then for any positive reals x1, . . . , x n,
∑
sym
xa1
1
xa2
2
· · · xan
n
≥ ∑
sym
xb1
1
xb2
2
· · · xbn
n
where the sums are taken over all permutations of n variables.
Remark - Although Muirhead’s theorem is a named theorem, it is generally not favor-ably regarded as part of a formal olympiad solution. Essentially, the majorization criterion guarantees that Muirhead’s inequality can be deduced from a suitable application of AM-GM. Hence, whenever possible, you should use Muirhead’s inequality only to deduce the correct relationship and then explicitly write all of the necessary applications of AM-GM. For a particular case this is a simple matter.
We now present an array of problems and solutions based primarily on these inequalities and ideas.
2 Examples
When solving any kind of problem, we should always look for a comparatively easy solu-tion first, and only later try medium or hard approaches. Although what constitutes this notoriously indeterminate “difficulty” varies widely from person to person, I usually con-sider “Dumbassing,” AM-GM (Power Mean), Cauchy, Chebyshev (Rearrangement), Jensen, H¨ older, in that order before moving to more clever techniques. (The first technique is de-scribed in remarks after example 1.) Weak inequalities will fall to AM-GM, which blatantly pins a sum to its smallest term. Weighted Jensen and H¨ older are “smarter” in that the effect of widely unequal terms does not cost a large degree of sharpness 6 - observe what happens when a weight of 0 appears. Especially sharp inequalities may be assailable only through clever algebra. Anyway, I have arranged the following with that in mind. 1. Show that for positive reals a, b, c
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥ 9a2b2c2
Solution 1. Simply use AM-GM on the terms within each factor, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥
(
3 3
√a3b3c3
) (
3 3
√a3b3c3
)
= 9 a2b2c2
6The sharpness of an inequality generally refers to the extent to which the two sides mimic each other, particularly near equality cases.
4Solution 2. Rearrange the terms of each factor and apply Cauchy,
(a2b + b2c + c2a) ( bc 2 + ca 2 + ab 2) ≥
(√a3b3c3 + √a3b3c3 + √a3b3c3
)2
= 9 a2b2c2
Solution 3. Expand the left hand side, then apply AM-GM, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) = a3b3 + a2b2c2 + a4bc
ab 4c + b3c3 + a2b2c2
a2b2c2 + abc 4 + a3c3
≥ 9 9
√a18 b18 c18 = 9 a2b2c2
We knew this solution existed by Muirhead, since (4 , 1, 1) , (3 , 3, 0), and (2 , 2, 2) all majorize (2 , 2, 2). The strategy of multiplying out all polynomial expressions and ap-plying AM-GM in conjunction with Schur is generally knowing as dumbassing because it requires only the calculational fortitude to compute polynomial products and no real ingenuity. As we shall see, dumbassing is a valuable technique. We also remark that the AM-GM combining all of the terms together was a particularly weak inequality, but the desired was a multiple of a2b2c2’s, the smallest 6th degree symmetric polynomial of three variables; such a singular AM-GM may not always suffice. 2. Let a, b, c be positive reals such that abc = 1. Prove that
a + b + c ≤ a2 + b2 + c2
Solution. First, we homogenize the inequality. that is, apply the constraint so as to make all terms of the same degree. Once an inequality is homogenous in degree
d, we may scale all of the variables by an arbitrary factor of k, causing both sides of the inequality to scale by the factor kd. This is valid in that it does not change the correctness of an inequality for any positive k, and if d is even, for any nonzero
k. Hence, we need consider a nonhomogenous constraint no futher. In this case, we multiply the left hand side by 3
√abc , obtaining
a43 b13 c13 + a13 b43 c13 + a13 b13 c43 ≤ a2 + b2 + c2
As abc = 1 is not homogenous, the above inequality must be true for all nonnegative
a, b, c . As (2 , 0, 0) majorizes (4 /3, 1/3, 1/3), we know it is true, and the necessary AM-GM is 2a2
3 + b2
6 + c2
6 = a2 + a2 + a2 + a2 + b2 + c2
6 ≥ 6
√a8b2c2 = a43 b13 c13
Let P (x) be a polynomial with positive coefficients. Prove that if
P
( 1
x
)
≥ 1
P (x)5holds for x = 1, then it holds for all x > 0.
Solution. Let P (x) = anxn + an−1xn−1 + · · · + a1x + a0. The first thing we notice is that the given is P (1) ≥ 1. Hence, the natural strategy is to combine P (x) and P ( 1
x
)
into P (1) in some fashion. The best way to accomplish this is Cauchy, which gives
P (x)P
( 1
x
)
= (anxn + · · · + a1x + a0)
(
an
1
xn + · · · + a1
1
x + a0
)
≥ (an + · · · + a1 + a0)2 = P (1) 2 ≥ 1as desired. This illustrates a useful means of eliminating denominators - by introducing similar factors weighted by reciprocals and applying Cauchy / H¨ older. 4. (USAMO 78/1) a, b, c, d, e are real numbers such that
a + b + c + d + e = 8
a2 + b2 + c2 + d2 + e2 = 16 What is the largest possible value of e?
Solution. Observe that the givens can be effectively combined by considering squares: (a − r)2 + ( b − r)2 + ( c − r)2 + ( d − r)2 + ( e − r)2 = (a2 + b2 + c2 + d2 + e2)
− 2r(a + b + c + d + e) + 5 r2
= 16 − 16 r + 5 r2
Since these squares are nonnegative, e ≤ √5r2 − 16 r + 16 + r = f (r) for all r. Since equality e = f (r) can be achieved when a = b = c = d = r, we need only compute the smallest value f (r). Since f grows large at either infinity, the minimum occurs when
f ′(r) = 1 + 10 r−16 2√5r2−16 r+16 = 0. The resultant quadratic is easily solved for r = 65 and
r = 2, with the latter being an extraneous root introduced by squaring. The largest possible e and greatest lower bound of f (r) is then f (6 /5) = 16 /5, which occurs when
a = b = c = d = 6 /5 and e = 16 /5. Alternatively, proceed as before except write
a = b = c = d = 8−e
4
since the maximum e must occur when the other four variables are equal. The second condition becomes a quadratic, and the largest solution is seen to be e = 16 5 .The notion of equating a, b, c, d is closely related to the idea of smoothing and Jensen’s inequality. If we are working with S1 = f (x1) + · · · + f (xn) under the constraint of a fixed sum x1 + · · · + xn, we can decrease S1 by moving several xi in the same interval
I together (that is, replacing xi1 < x i2 with x′
i1
= xi1 + ≤ < x i2 − ≤ = x′
i2
for any sufficiently small ≤) for any I where f is convex. S1 can also be decreased by spreading
xi in the same interval where f is concave. When seeking the maximum of S1, we proceed in the opposite fashion, pushing xi on the concave intervals of f together and moving xi on the convex intervals apart. 65. Show that for all positive reals a, b, c, d ,1
a + 1
b + 4
c + 16
d ≥ 64
a + b + c + d
Solution. Upon noticing that the numerators are all squares with √1 + √1 + √4 + √16 = √64, Cauchy should seem a natural choice. Indeed, multiplying through by
a + b + c + d and applying Cauchy, we have (a + b + c + d)
(12
a + 12
b + 22
c + 42
d
)
≥ (1 + 1 + 2 + 4) 2 = 64 as desired. 6. (USAMO 80/5) Show that for all non-negative reals a, b, c ≤ 1,
ab + c + 1 + bc + a + 1 + ca + b + 1 + (1 − a)(1 − b)(1 − c) ≤ 1
Solution. Let f (a, b, c ) denote the left hand side of the inequality. Since ∂2
∂a 2
f =
2b
(c+a+1) 3
2c
(a+b+1) 3
≥ 0, we have that f is convex in each of the three variables; hence, the maximum must occur where a, b, c ∈ { 0, 1}. Since f is 1 at each of these 8 points, the inequality follows. Second derivative testing for convexity/concavity is one of the few places where the use of Calculus is not seriously loathed by olympiad graders. It is one of the standard techniques in inequalities and deserves to be in any mental checklist of inequality solving. In this instance, it led to an easy solution. 7. (USAMO 77/5) If a, b, c, d, e are positive reals bounded by p and q with 0 < p ≤ q,prove that (a + b + c + d + e)
( 1
a + 1
b + 1
c + 1
d + 1
e
)
≤ 25 + 6
(√ pq −
√ qp
)2
and determine when equality holds.
Solution. As a function f of five variables, the left hand side is convex in each of
a, b, c, d, e ; hence, its maximum must occur when a, b, c, d, e ∈ { p, q }. When all five variables are p or all five are q, f is 25. If one is p and the other four are q, or vice versa, f becomes 17 + 4( pq + qp ), and when three are of one value and two of the other,
f = 13 + 6( pq + qp ). pq + qp ≥ 2, with equality if and only if p = q. Clearly, equality holds where p = q. Otherwise, the largest value assumed by f is 13 + 6( pq + qp ), which is obtained only when two of a, b, c, d, e are p and the other three are q, or vice versa. In such instances, f is identically the right hand side. This is a particular case of the Schweitzer inequality, which, in its weighted form, is sometimes known as the Kantorovich inequality. 78. a, b, c, are non-negative reals such that a + b + c = 1. Prove that
a3 + b3 + c3 + 6 abc ≥ 14
Solution. Multiplying by 4 and homogenizing, we seek 4a3 + 4 b3 + 4 c3 + 24 abc ≥ (a + b + c)3
= a3 + b3 + c3 + 3 (a2(b + c) + b2(c + a) + c2(a + b)) + 6 abc
⇐⇒ a3 + b3 + c3 + 6 abc ≥ a2(b + c) + b2(c + a) + c2(a + b)Recalling that Schur’s inequality gives a3 +b3 +c3 +3 abc ≥ a2(b+c)+ b2(c+a)+ c2(a+b), the inequality follows. In particular, equality necessitates that the extra 3 abc on the left is 0. Combined with the equality condition of Schur, we have equality where two of a, b, c are 12 and the third is 0. This is a typical dumbass solution.
Solution 2. Without loss of generality, take a ≥ b ≥ c. As a+b+c = 1, we have c ≤ 13
or 1 −3c ≥ 0. Write the left hand side as ( a+b)3 −3ab (a+b−2c) = ( a+b)3 −3ab (1 −3c). This is minimized for a fixed sum a + b where ab is made as large as possible. As by AM-GM ( a + b)2 ≥ 4ab , this minimum occurs if and only if a = b. Hence, we need only consider the one variable inequality 2 (1−c
2
)3 + c3 + 6 (1−c
2
)2 c = 14 · (9 c3 − 9c2 + 3 c + 1). Since c ≤ 13 , 3 c ≥ 9c2. Dropping this term and 9 c3, the inequality follows. Particularly, 9c3 = 0 if and only if c = 0, and the equality cases are when two variables are 12 and the third is 0. 9. (IMO 74/5) If a, b, c, d are positive reals, then determine the possible values of
aa + b + d + bb + c + a + cb + c + d + da + c + d
Solution. We can obtain any real value in (1 , 2). The lower bound is approached by
a → ∞ , b = d = √a, and c = 1. The upper bound is approached by a = c → ∞ ,
b = d = 1. As the expression is a continuous function of the variables, we can obtain all of the values in between these bounds. Finally, these bounds are strict because
aa + b + d + bb + c + a + cb + c + d + da + c + d >aa + b + c + d + ba + b + c + d + ca + b + c + d + da + b + c + d = 1 and
aa + b + d + bb + c + a + cb + c + d + da + c + d <aa + b + ba + b + cc + d + dc + d = 2 Whenever extrema occur for unusual parameterizations, we should expect the need for non-classical inequalities such as those of this problem where terms were completely dropped. 810. (IMO 95/2) a, b, c are positive reals with abc = 1. Prove that 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) ≥ 32
Solution 1. Let x = 1
a
, y = 1
b
, and z = 1
c
. We perform this substitution to move terms out of the denominator. Since abc = xyz = 1, we have 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) = x2
y + z + y2
x + z + z2
x + y
Now, multiplying through by ( x + y)( y + z)( z + x), we seek
x4 + y4 + z4 + x3y + x3z + y3z + xy 3 + xz 3 + yz 3 + x2yz + xy 2z + xyz 2 ≥
3
√xyz ·
(
3xyz + 32 · (x2y + x2z + y2x + xy 2 + xz 2 + yz 2))
which follows immediately by AM-GM, since x2yz +xy 2z+xyz 2 ≥ 3 3
√x4y4z4, x3y+xy 3+x3z
3
≥
3
√x7y4z and 7x4+4 y4+z4
12
≥ 3
√x7y4z - as guaranteed by Muirhead’s inequality.
Solution 2. Substitute x, y, z as before. Now, consider the convex function f (x) = x−1
for x > 0. ( f (x) = xc is convex for c < 0 and c ≥ 1, and concave for 0 < c ≤ 1, verify this with the second derivative test.) Now, by Jensen,
x2
y + z + y2
z + x + z2
x + y = xf
(y + zx
)
yf
(z + xy
)
zf
(x + yz
)
≥ (x + y + z)f
((y + z) + ( z + x) + ( x + y)
x + y + z
)
= x + y + z
2But x + y + z ≥ 3 3
√xyz = 3, as desired.
Solution 3. Perform the same substitution. Now, multiplying by ( x + y + z) and applying Cauchy, we have 12 (( y + z) + ( z + x) + ( x + y))
( x2
y + z + y2
z + x + z2
x + y
)
≥ 12(x + y + z)2
Upon recalling that x+y +z ≥ 3 we are done. Incidentally, the progress of this solution with Cauchy is very similar to the weighted Jensen solution shown above. This is no coincidence, it happens for many convex f (x) = xc.
Solution 4. Apply the same substitution, and put x ≥ y ≥ z. Simultaneously,
xy+z
≥ yz+x ≥ zx+y . Hence, by Chebyshev,
x ·
( xy + z
)
y ·
( yz + x
)
z ·
( zx + y
)
≥ x + y + z
3
( xy + z + yx + z + zx + y
)
Again, x + y + z ≥ 3. But now we have Nesbitt’s inequality, xy+z + yx+z + zx+y ≥ 32 . This follows immediately from AM-HM upon adding 1 to each term. 911. Let a, b, c be positive reals such that abc = 1. Show that 2(a + 1) 2 + b2 + 1 + 2(b + 1) 2 + c2 + 1 + 2(c + 1) 2 + a2 + 1 ≤ 1
Solution. The previous problem showed the substitution offers a way to rewrite an inequality in a more convenient form. Substitution can also be used to implicity use a given. First, expand the denominators and apply AM-GM, obtaining 2(a + 1) 2 + b2 + 1 = 2
a2 + b2 + 2 a + 2 ≤ 1
ab + a + 1 Now, write a = xy , b = yz , c = zx . We have 1
ab +a+1
= 1
xz+xy+1
= yz xy +yz +zx . It is now evident that the sum of the new fractions is 1. 12. (USAMO 98/3) Let a0, . . . , a n real numbers in the interval (0 , π
2
) such that tan
(
a0 − π
4
)
tan
(
a1 − π
4
)
· · · + tan
(
an − π
4
)
≥ n − 1Prove that tan( a0) tan( a1) · · · tan( an) ≥ nn+1
Solution 1. Let yi = tan (x − π
4
). We have tan( xi) = tan ((xi − π
4
) + π
4
) = yi+1 1−yi .Hence, given s = y0 + · · · + yn ≥ n − 1 we seek to prove ∏ni=0 1+ yi
1−yi
≥ nn+1 . Observe that for any a > b and fixed sum a + b, the expression
( 1 + a
1 − a
)
·
( 1 + b
1 − b
)
= 1 + 2( a + b)(1 − a)(1 − b)can be decreased by moving a and b any amount closer together. Hence, for any sequence y0, . . . , y n, we can replace any yi > sn+1 and yj < sn+1 with y′
i
= sn+1 and
y′
j
= yi + yj − sn+1 , decreasing the product. Now we have
n
∏
i=0
1 + yi
1 − yi
≥
( 1 + sn+1
1 − sn+1
)n+1
≥
( 2nn+1 2
n+1
)n+1
= nn+1
Where the last inequality follows from the fact that 1+ x
1−x
is an increasing function of x.
Solution 2. Perform the same substitution. The given can be written as 1 + yi ≥
∑
j6=i
(1 − yj ), which by AM-GM gives 1+ yn
n
≥ ∏
j6=i
(1 − yj ) 1
n
. Now we have
n
∏
i=0
1 + yi
n ≥
n
∏
i=0
∏
j6=i
(1 − yj ) 1
n
=
n
∏
i=0
(1 − yi)as desired. 10 13. Let a, b, c be positive reals. Prove that 1
a(1 + b) + 1
b(1 + c) + 1
c(1 + a) ≥ 31 + abc
with equality if and only if a = b = c = 1.
Solution. Multiply through by 1+ abc and add three to each side, on the left obtaining 1 + a + ab + abc a(1 + b) + 1 + b + bc + abc b(1 + c) + 1 + c + ac + abc c(1 + a)= (1 + a) + ab (1 + c)
a(1 + b) + (1 + b) + bc (1 + a)
b(1 + c) + (1 + c) + ac (1 + b)
c(1 + a)which is at least 6 by AM-GM, as desired. In particular, this AM-GM asserts the equivalence of (1+ a)
a(1+ b)
and a(1+ b)1+ a , or that they are both one. Likewise, all of the other terms must be 1. Now, (1 + a)2 = a2(1 + b)2 = a2b2(1 + c)2 = a2b2c2(1 + a)2, so the product abc = 1. Hence, 1+ aa(1+ b) = bc (1+ a)1+ b = bc (1+ a)
b(1+ c)
so that 1 + b = b + bc = b + 1
a
. It is now easy to see that equality holds if and only if a = b = c = 1. 14. (Romanian TST) Let a, b, x, y, z be positive reals. Show that
xay + bz + yaz + bx + zax + by ≥ 3
a + b
Solution. Note that ( a + b)( xy + yz + xz ) = ( x(ay + bz ) + y(az + bx ) + z(ax + by )). We introduce this factor in the inequality, obtaining (x(ay + bz ) + y(az + bx ) + z(ax + by ))
( xay + bz + yaz + bx + zax + by
)
≥
(x + y + z)2 ≥ 3( xy + yz + xz )Where the last inequality is simple AM-GM. The desired follows by simple algebra. Again we have used the idea of introducing a convenient factor to clear denominators with Cauchy. 15. The numbers x1, x 2, . . . , x n obey −1 ≤ x1, x 2, . . . , x n ≤ 1 and x 31 + x 32 + · · · + x 3
n
= 0. Prove that
x1 + x2 + · · · + xn ≤ n
3
Solution 1. Substitute yi = x3
i
so that y1 + · · · + yn = 0. In maximizing 3
√y1 +
· · · + 3
√yn, we note that f (y) = y 13 is concave over [0 , 1] and convex over [ −1, 0], with
|f ′(y1)| ≥ | f ′(y2)| ⇐⇒ 0 < |y1| ≤ | y2|. Hence, we may put y1 = · · · = yk = −1;
−1 ≤ yk+1 < 0, and yk+2 = · · · = yn = k−yk+1
n−k−1
. We first show that yk+1 leads to a maximal sum of 3
√yi if it is -1 or can be made positive. If |yk+1 | < |yk+2 |, we set 11 y′
k+1
= y′
k+2
= yk+1 +yk+2
2
, increasing the sum while making yk+1 positive. Otherwise, set y′
k+1
= −1 and y′
k+2
= 1 − yk+1 − yk+2 , again increasing the sum of the 3
√yi. Now we may apply Jensen to equate all positive variables, so that we need only show
k 3
√−1 + ( n − k) 3
√ kn − k ≤ n
3But we have (n + 3 k)3 − 27( n − k)2k = n3 − 18 n2k + 81 nk 2 = n(n − 9k)2 ≥ 0as desired. Particularly, as k is an integer, equality can hold only if 9 |n and then if and only if one ninth of the variables yi are -1 and the rest are 1/8.
Solution 2. Let xi = sin( αi), and write 0 = x31 + · · · + x3
n
= sin 3(α1) + · · · + sin 3(αn) =
14
((3 sin( α1) − sin(3 α1)) + · · · + (3 sin( αn) − sin(3 αn))). It follows that x1 + · · · + xn =sin( α1) + · · · + sin( αn) = sin(3 α1)+ ··· +sin(3 αn)3 ≤ n
3
. The only values of sin( α) which lead to sin(3 α) = 1 are 12 and -1. The condition for equality follows. 16. (Turkey) Let n ≥ 2 be an integer, and x1, x 2, . . . , x n positive reals such that x21 + x22 +
· · · + x2
n
= 1. Determine the smallest possible value of
x51
x2 + x3 + · · · + xn
x52
x3 + · · · + xn + x1
· · · + x5
n
x1 + · · · + xn−1
Solution. Observe that ∑ni=1 xi
∑
j6=i
xj ≤ n − 1, so that
( n∑
i=1
xi
(∑
j6=i
xj
)) ( n∑
i=1
x5
i
∑
j6=i
xi
)
≥ (x31 + · · · + x3
n
)2
= n2
(x31 + · · · + x3
n
n
)2
≥ n2
(x21 + · · · + x2
n
n
)3
= 1
n
Leads to n∑
i=1
x5
i
∑
j6=i
xi
≥ 1
n(n − 1) with equality if and only if x1 = · · · = xn = 1√n .17. (Poland 95) Let n be a positive integer. Compute the minimum value of the sum
x1 + x22
2 + x33
3 + · · · + xnn
n
12 where x1, x 2, . . . , x n are positive reals such that 1
x1
1
x2
· · · + 1
xn
= n
Solution. The given is that the harmonic mean of x1, . . . , x n is 1, which implies that the product x1x2 · · · xn is at least 1. Now, we apply weighted AM-GM
x1 + x22
2 + x33
3 + · · · + xnn
n ≥
(
1 + 12 + 13 + · · · + 1
n
)
1+ 12 +··· + 1
n
√x1x2 · · · xn
= 1 + 12 + 13 + · · · + 1
n
Prove that for all positive reals a, b, c, d ,
a4b + b4c + c4d + d4a ≥ abcd (a + b + c + d)
Solution. By AM-GM, 23 a4b + 7 b4c + 11 c4d + 10 ad 4
51 ≥ 51
√a102 b51 c51 d51 = a2bcd
from which the desired follows easily. Indeed, the most difficult part of this problem is determining suitable weights for the AM-GM. One way is to suppose arbitrary weights
x1, x 2, x 3, x 4 for a4b, b 4c, c 4d, ad 4 respectively, and solve the system
x1 + x2 + x3 + x4 = 14x1 + x2 = 24x2 + x3 = 14x3 + x4 = 119. (USAMO 01/3) Let a, b, c be nonnegative reals such that
a2 + b2 + c2 + abc = 4 Prove that 0 ≤ ab + bc + ca − abc ≤ 2
Solution [by Tony Zhang.] For the left hand side, note that we cannot have a, b, c >
Suppose WLOG that c ≤ 1. Then ab +bc +ca −abc ≥ ab +bc +ca −ab = c(a+b) ≥ 0. For the right, 4 = a2 + b2 + c2 + abc ≥ 4( abc )43 =⇒ abc ≤ 1. Since by the pigeon hole principle, among three numbers either two exceed 1 or two are at most 1. Hence, we assume WLOG that ( a − 1)( b − 1) ≥ 0, which gives ab + 1 ≥ a + b ⇐⇒ abc + c ≥ ac +
bc ⇐⇒ c ≥ ac + bc − abc . Now, we have ab + bc + ca − abc ≤ ab + c. Either we are done or ab +c > 2. But in the latter case, 4 = ( a2 +b2)+ c(c+2 ab ) > 2ab +2 c = 2( ab +c) > 4, a contradiction. 13 20. (Vietnam 98) Let x1, . . . , x n be positive reals such that 1
x1 + 1998 + 1
x2 + 1998 + · · · + 1
xn + 1998 = 11998 Prove that n
√x1x2 · · · xn
n − 1 ≥ 1998
Solution. Let yi = 1
xi+1998
so that y1 + · · · + yn = 11998 and xi = 1
yi
− 1998. Now
n
∏
i=1
xi =
n
∏
i=1
( 1
yi
− 1998
)
= e
Pni=1 ln
“1
yi−1998
”
Hence, to minimize the product of the xi, we equivalently minimize the sum of ln
( 1
yi
− 1998
)
.In particular,
ddy
(
ln
( 1
y − 1998
))
= 1
( 1
y
− 1998
)2 · −1
y2
= −1
y − 1998 y2
d2
dy 2
(
ln
( 1
y − 1998
))
= 1 − 3996 y
(y − 1998 y2)2
So ln
( 1
y
− 1998
)
is convex on [0 , 1/3996]. If we had 0 < y i ≤ 1/3996 for all i we could apply Jensen. Since yi + yj ≤ 1/1998 for all i, j , we consider
( 1
a − 1998
) ( 1
b − 1998
)
≥
( 2
a + b − 1998
)2
⇐⇒ 1
ab − 1998
( 1
a + 1
b
)
≥ 4(a + b)2 − 4 · 1998
a + b
⇐⇒ (a + b)2 − 1998( a + b)3 ≥ 4ab − 4ab (a + b) · 1998
⇐⇒ (a − b)2 ≥ 1998( a + b)( a − b)2
which incidentally holds for any a + b ≤ 11998 . Hence, any two yi and yj may be set to their average while decreasing the sum in question; hence, we may assume yi ∈ (0 , 13996 ]. Now Jensen’s inequality shows that the minimum occurs when yi = 11998 n for all i, or when xi = 1998( n − 1) for all i. It is easy to see that this yields equality. 21. (Romania 99) Show that for all positive reals x1, . . . , x n with x1x2 · · · xn = 1, we have 1
n − 1 + x1
· · · + 1
n − 1 + xn
≤ 114 Solution. First, we prove a lemma: the maximum of the sum occurs when n − 1 of the xi are equal. Consider f (y) = 1
k+ey
for an arbitrary nonnegative constant k. We have f ′(y) = −ey
(k+ey)2
and f ′′ (y) = ey (ey −k)(k+ey )3 . Evidently f ′′ (y) ≥ 0 ⇐⇒ ey ≥ k. Hence,
f (y) has a single inflexion point where y = ln( k), where f (y) is convex over the interval ((ln( k), ∞). Now, we employ the substitution yi = ln( xi) so that y1 + · · · + yn = 0 and
n
∑
i=1
1
n − 1 + xi
=
n
∑
i=1
f (yi)We take k = n − 1 and write k0 = ln( n − 1). Suppose that y1 ≥ · · · ≥ ym ≥ k0 ≥
ym+1 ≥ · · · xn for some positive m. Then by, Majorization,
f (y1) + · · · + f (ym) ≤ (m − 1) f (k0) + f (y1 + · · · + ym − (m − 1) k0)But then, also by Majorization, (m − 1) f (k0) + f (ym+1 ) + · · · + f (yn) ≤ (n − 1) f
((m − 1) k0 + ym+1 + · · · + yn
n − 1
)
Otherwise, all of the yi are less than k0. In that case we may directly apply Majorization to equate n − 1 of the yi whilst increasing the sum in question. Hence, the lemma is valid. 7 N
Applying the lemma, it would suffice to show
kk + x + 1
k + 1
xk
≤ 1Clearing the denominators,
(
k2 + kxk
)
( k + x) ≤ k2 + k
(
x + 1
xk
)
x1−k
−xk + x + k ≤ x1−k
But now this is evident. We have Bernoulli’s inequality, since x1−k = (1 + ( x − 1)) 1−k ≥
1 + ( x − 1)(1 − k) = x + k − xk . Equality holds only where x = 1 or n = 2. 22. (Darij Grinberg) Show that for all positive reals a, b, c ,
√b + ca +
√c + ab +
√a + bc ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
7This n−1 equal value principle is particularly useful. If a differentiable function has a single inflexion point and is evaluated at narbitrary reals with a fixed sum, any minimum or maximum must occur where some n−1 variables are equal.
15 Solution 1. By Cauchy, we have √(a + b)( a + c) ≥ a + √bc . Now,
∑
cyc
√b + ca ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
⇐⇒ ∑
cyc
b + ca
√(a + b)( a + c) ≥ 4( a + b + c)Substituting our result from Cauchy, it would suffice to show
∑
cyc
(b + c)
√bc a ≥ 2( a + b + c)WLOG a ≥ b ≥ c, implying b + c ≤ c + a ≤ a + b and √bc a ≤ √ca b ≤ √ab c . Hence, by Chebyshev and AM-GM,
∑
cyc
(b + c)
√bc a ≥
(2( a + b + c))
(√bc a + √ca b + √ab c
)
3
≥ 2( a + b + c)as desired.
Solution 2. Let x = √b + c, y = √c + a, z = √a + b. Then x, y, z are the sides of acute triangle XY Z (in the typical manner), since x2 + y2 = a + b + 2 c > a + b = z2.The inequality is equivalent to
∑
cyc
xy2 + z2 − x2 ≥ x2 + y2 + z2
xyz
Recalling that y2 + z2 − x2 = 2 yz cos( X), we reduce this to the equivalent
∑
cyc
x2
cos( X) ≥ 2( x2 + y2 + z2)WLOG, we have x ≥ y ≥ z, implying 1cos( X) ≥ 1cos( Y ) ≥ 1cos( Z) , so that applying Chebyshev to the left reduces the desired to proving that the sum of the reciprocals of the cosines is at least 6. By AM-HM, 1cos( X) + 1cos( Y ) + 1cos( Z) ≥ 9cos( X) + cos( Y ) + cos( Z)But recall from triangle geometry that cos( X) + cos( Y ) + cos( Z) = 1 + rR and R ≥ 2r.The desired is now evident. 16 23. Show that for all positive numbers x1, . . . , x n,
x31
x21 + x1x2 + x22
x32
x22 + x2x3 + x23
· · · + x3
n
x2
n
xnx1 + x21
≥ x1 + · · · + xn
3
Solution. Observe that 0 = ( x1 −x2)+( x2 −x3)+ · · · +( xn −x1) = ∑ni=1
x3
i−x3
i+1
x2
i+xixi+1 +x2
i+1
.Hence, (where xn+1 = x1)
n
∑
i=1
x3
i
x2
i
xixi+1 x2
i+1
= 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
But now a3 + b3 ≥ 13 a3 + 23 a2b + 23 ab 2 + 13 b3 = 13 (a + b)( a2 + ab + b2). Hence, 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
≥ 12
n
∑
i=1
xi + xi+3
3 = 13
n
∑
i=1
xi
as desired. 24. Let a, b, c be positive reals such that a + b ≥ c; b + c ≥ a; and c + a ≥ b, we have 2a2(b + c) + 2 b2(c + a) + 2 c2(a + b) ≥ a3 + b3 + c3 + 9 abc
Solution. After checking that equality holds for ( a, b, c ) = ( t, t, t ) and (2 t, t, t ), it is apparent that more than straight AM-GM will be required. To handle the condition, put a = y + z, b = z + x, c = x + y with x, y, z ≥ 0. Now, the left hand side becomes 4x3 + 4 y3 + 4 z3 + 10 x2(y + z) + 10 y2(z + x) + 10 z2(x + y) + 24 xyz while the right hand side becomes 2 x3 + 2 y3 + 2 z3 + 12 x2(y + z) + 12 y2(z + x) + 12 z2(x + y) + 18 xyz .The desired is seen to be equivalent to x3 + y3 + z3 + 3 xyz ≥ x2(y + z) + y2(z +
x) + z2(x + y), which is Schur’s inequality. Equality holds where x = y = z, which gives ( a, b, c ) = ( t, t, t ), or when two of x, y, z are equal and the third is 0, which gives (a, b, c ) ∈ { (2 t, t, t ), (t, 2t, t ), (t, t, 2t)}.25. Let a, b, c be the lengths of the sides of a triangle. Prove that
a
√2b2 + 2 c2 − a2 + b
√2c2 + 2 a2 − b2 + c
√2a2 + 2 b2 − c2 ≥ √3
Solution 1. Again write a = y + z, b = z + x, and c = x + y, noting that x, y, z are positive. (Triangles are generally taken to be non-degenerate when used in inequalities.) We have
∑
cyc
a
√2b2 + 2 c2 − a2 = ∑
cyc
y + z
√4x2 + 4 xy + 4 xz + y2 + z2 − 2yz
17 Consider the convex function f (x) = 1√x . (As we shall see, Jensen almost always provides a tractable means of eliminating radicals from inequalities.) Put x+y +z = 1. We have
∑
cyc
(y + z)f (4x2 + 4 xy + 4 xz + y2 + z2 − 2yz ) ≥
(( y + z) + ( z + x) + ( x + y)) f
(∑
cyc
(y + z) (4 x2 + 4 xy + 4 xz + y2 + z2 − 2yz )(y + z) + ( z + x) + ( x + y)
)
= 2√2
√∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2
Noting that ∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2 =
∑
cyc
2x3 + 7 x2(y + z) + 8 xyz ,8( x + y + z)3 ≥ 3 ∑
cyc
2x3 + 7 x2(y + z) + 8 xyz
⇐⇒ ∑
sym
4x3 + 24 x2y + 8 xyz ≥ ∑
sym
3x3 + 21 x2y + 12 xyz
⇐⇒ 2x3 + 2 y3 + 2 z3 + 3 (x2(y + z) + y2(z + x) + z2(x + y)) ≥ 24 xyz
which follows by AM-GM. As a follow up on an earlier mentioned connection, oberserve the similarity of the above application of Jensen and the following inequality (which follows by H¨ older’s inequality)
(∑
i
αiβi
) ( ∑
i
αi
1
√βi
)2
≥
(∑
i
αi
)3
Solution 2 [by Darij Grinberg.] Let ABC be a triangle of side lengths a, b, c
in the usual order. Denote by ma, m b, m c the lengths of the medians from A, B, C
respectively. Recall from triangle goemetry that ma = 12
√2b2 + 2 c2 − a2, so that we need only show ama + bmb + cmc ≥ 2√3. But a triangle with side lengths ma, m b, m c, in turn, has medians of length 3a
4
, 3b
4
, and 3c
4
. The desired inequality is therefore equivalent to 43 ma
a
43 mb
b
43 mc
c
≥ 2√3 where we refer to the new triangle ABC . Recalling that
23
ma = AG , where G is the centroid, the desired is seen to be equivalent to the geometric inequality AG a + BG b + CG c ≥ √3. But we are done as we recall from triangle geometry that AM a + BM b + CM c ≥ √3 holds for any point inside triangle ABC .8
8For a complete proof of this last inequality, see post #14.
18 26. (IMO 99/2) For n ≥ 2 a fixed positive integer, find the smallest constant C such that for all nonnegative reals x1, . . . , x n,
∑
1≤i<j ≤n
xixj (x2
i
x2
j
) ≤ C
( n∑
i=1
xi
)4
Solution. The answer is C = 18 , which is obtained when any two xi are non-zero and equal and the rest are 0. Observe that by AM-GM, (x1 + · · · + xn)4 =
( n∑
i=1
x2
i
2 ∑
1≤i<j ≤n
xixj
)2
≥ 4
( n∑
i=1
x2
i
) (
2 ∑
1≤i<j ≤n
xixj
)
= 8 ∑
1≤i<j ≤n
xixjn∑
k=1
x2
k
But x21 + · · · + x2
n
≥ x2
i
x2
j
with equality iff xk = 0 for all k 6 = i, j . It follows that (x1 + · · · + xn)4 ≥ 8 ∑
1≤i<j ≤n
xixj
(x2
i
x2
j
)
as desired. 27. Show that for nonnegative reals a, b, c ,2a6 + 2 b6 + 2 c6 + 16 a3b3 + 16 b3c3 + 16 c3a3 ≥ 9a4(b2 + c2) + 9 b4(c2 + a2) + 9 c4(a2 + b2)
Solution 1. Consider
∑
cyc
(a − b)6 = ∑
cyc
a6 − 6a5b + 15 a4b2 − 20 a3b3 + 15 a2b4 − 6ab 5 + b6 ≥ 0and ∑
cyc
ab (a − b)4 = ∑
cyc
a5b − 4a4b2 + 6 a3b3 − 4a2b4 + ab 5 ≥ 0Adding six times the latter to the former yields the desired result.
Solution 2. We shall prove a6 − 9a4b2 + 16 a3b3 − 9a2b4 + b6 ≥ 0. We have
a6 − 2a3b3 + b6 = (a3 − b3)2
= ((a − b)( a2 + ab + b2))2
≥ (a − b)2(3 ab )2 = 9 a4b2 − 18 a3b3 + 9 a2b4
19 As desired. The result now follows from adding this lemma cyclicly. The main difficulty with this problem is the absence of a5b terms on the right and also the presence of
a4b2 terms on the right - contrary to where Schur’s inequality would generate them. Evidently AM-GM is too weak to be applied directly, since a6 + 2 a3b3 ≥ 3a4b2 cannot be added symmetrically to deduce the problem. By introducing the factor ( a − b)2,however, we weight the AM-GM by a factor which we “know” will be zero at equality, thereby increasing its sharpness. 28. Let 0 ≤ a, b, c ≤ 12 be real numbers with a + b + c = 1. Show that
a3 + b3 + c3 + 4 abc ≤ 932
Solution. Let f (a, b, c ) = a3 + b3 + c3 + 4 abc and g(a, b, c ) = a + b + c = 1. Because
f and g are polynomials, they have continuous first partial derivatives. Moreover, the gradient of g is never zero. Hence, by the theorem of Lagrange Multipliers ,any extrema occur on the boundary or where ∇f = λ∇g for suitable scalars λ. As
∇f =< 3a2 + 4 bc, 3b2 + 4 ca, 3c2 + 4 ab > and ∇g =< 1, 1, 1 >, we have
λ = 3a2 + 4 bc
= 3b2 + 4 ca
= 3c2 + 4 ab g(a, b, c ) = a + b + c = 1 We have 3 a2 + 4 bc = 3 b2 + 4 ca or ( a − b)(3 a + 3 b − 4c) = ( a − b)(3 − 7c) = 0 for any permutation of a, b, c . Hence, without loss of generality, a = b. Now, 3 a2 + 4 ac =3c2 + 4 a2 and a2 − 4ac + 3 c2 = ( a − c)( a − 3c) = 0. The interior local extrema therefore occur when a = b = c or when two of {a, b, c } are three times as large as the third. Checking, we have f (13 , 13 , 13 ) = 7 /27 < 13 /49 = f (17 , 37 , 37 ). Recalling that f (a, b, c ) is symmetric in a, b, c , the only boundary check we need is f (12 , t, 12 −t) ≤ 932 for 0 ≤ t ≤ 12 .We solve
h(t) = f
(12, t, 12 − t
)
= 18 + t3 +
(12 − t
)3
2 t
(12 − t
)
= 14 + t
4 − t2
2
h(t) is 14 at either endpoint. Its derivative h′(t) = 14 − t is zero only at t = 14 . Checking,
h(14 ) = f (12 , 14 , 14 ) = 932 . Since h(t) has a continuous derivative, we are done. (As a further check, we could observe that h′′ (t) = −1 < 0, which guarantees that h(14 ) is a local minimum.) 20 Usage Note. The use of Lagrange Multipliers in any solution will almost certainly draw hostile review, in the sense that the tiniest of errors will be grounds for null marks. If you consider multipliers on Olympiads, be diligent and provide explicit, kosher remarks about the continuous first partial derivatives of both f (x1, . . . , x n) and the constraint g(x1, . . . , x n) = k, as well as ∇g 6 = 0, before proceeding to solve the system ∇f = λ∇g. The main reason this approach is so severely detested is that, given sufficient computational fortitude (if you are able to sort through the relevant algebra and Calculus), it can and will produce a complete solution. The example provided here is included for completeness of instruction; typical multipliers solutions will not be as clean or painless. 9
(Vascile Cartoaje) Let p ≥ 2 be a real number. Show that for all nonnegative reals
a, b, c ,
3
√
a3 + pabc
1 + p + 3
√
b3 + pabc
1 + p + 3
√
c3 + pabc
1 + p ≤ a + b + c
Solution. By H¨ older,
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤
(∑
cyc
11 + p
) ( ∑
cyc
a
) ( ∑
cyc
a2 + pbc
)
But a2 + b2 + c2 ≥ ab + bc + ca (proven by AM-GM, factoring, or a number of other methods) implies that
∑
cyc
a2 + pbc ≤ (p + 1) ∑
cyc
a2 + 2 bc
3 = p + 1 3 (a + b + c)2
From which we conclude
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤ (a + b + c)3
as desired. 30. Let a, b, c be real numbers such that abc = −1. Show that
a4 + b4 + c4 + 3( a + b + c) ≥ a2
b + a2
c + b2
c + b2
a + c2
a + c2
b
Solution. First we homogenize, obtaining a4 + b4 + c4 + a3(b + c) + b3(c + a) + c3(a +
b) − 3abc (a + b + c) ≥ 0. As this is homogenous in the fourth degree, we can scale a, b, c
9Just how painful can the calculations get? Most multipliers solutions will tend to look more like than this solution.
21 by any real k and hence may now ignore abc = −1. Equality holds at a = b = c = 1, but also at a = b = 1 , c = −2, a = 1 , b = 0 , c = −1, and a number of unusual locations with the commonality that a + b + c = 0. Indeed, c = −a − b is a parametric solution, and we discover the factorization ( a + b + c)2(a2 + b2 + c2 − ab − bc − ca ) ≥ 0. (We are motivated to work with factorizations because there are essentially no other inequalities with a + b + c = 0 as an equality condition.) 31. (MOP 2003) Show that for all nonnegative reals a, b, c ,
a4(b2 + c2) + b4(c2 + a2) + c4(a2 + b2) +2abc (a2b + a2c + b2c + b2a + c2a + c2b − a3 − b3 − c3 − 3abc )
≥ 2a3b3 + 2 b3c3 + 2 c3a3
Solution. As was suggested by the previous problem, checking for equality cases is important when deciding how to solve a problem. We see that setting a = b produces equality. As the expression is symmetric, this certainly implies that b = c and c = a are equality cases. Hence, if P (a, b, c ) is the difference LHS - RHS, then ( a − b)( b − c)( c −
a)|P (a, b, c ). Obviously, if the problem is going to be true, ( a−b) must be a double root of P , and accordingly we discover the factorization P (a, b, c ) = ( a − b)2(b − c)2(c − a)2.The result illustrated above was no accident. If ( x−y) divides a symmetric polynomial
P (x, y, z ), then ( x − y)2 divides the same polynomial. If we write P (x, y, z ) = ( x −
y)Q(x, y, z ), then ( x − y)Q(x, y, z ) = P (x, y, z ) = P (y, x, z ) = ( y − x)Q(y, x, z ), which gives Q(x, y, z ) = −Q(y, x, z ). Hence Q(x, x, z ) = 0, and ( x − y) also divides Q(x, y, z ). 32. (Cezar Lupu) Let a, b, c be positive reals such that a + b + c + abc = 4. Prove that
a
√b + c + b
√c + a + c
√a + b ≥√22 · (a + b + c)
Solution. By Cauchy
(∑
cyc
a√b + c
) ( ∑
cyc
a
√b + c
)
≥ (a + b + c)2
But, also by Cauchy,
√(a + b + c) ( a(b + c) + b(c + a) + c(a + b)) ≥ ∑
cyc
a√b + c
Hence, ∑
cyc
a
√b + c ≥√22 · (a + b + c) ·
√ a + b + cab + bc + ca
22 And we need only show a + b + c ≥ ab + bc + ca . Schur’s inequality for r = 1 can be expressed as 9abc a+b+c ≥ 4( ab + bc + ca ) − (a + b + c)2. Now, we suppose that
ab + bc + ca > a + b + c, and have 9abc a + b + c ≥ 4( ab + bc + ca ) − (a + b + c)2
(a + b + c) (4 − (a + b + c)) = abc (a + b + c)Hence, a + b + c < 3. But then abc < 1, which implies 4 = a + b + c + abc < 4. Contradiction, as desired. 33. (Iran 1996) Show that for all positive real numbers a, b, c ,(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
≥ 94
Solution. Fearless courage is the foundation of all success. 10 When everything else fails, return to the sure-fire strategy of clearing all denominators. In this case, we obtain 4( a + b)2(b + c)2(c + a)2(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
=
∑
sym
4a5b + 8 a4b2 + 10 a4bc + 6 a3b3 + 52 a3b2c + 16 a2b2c2
on the left, and on the right, 9( a + b)2(b + c)2(c + a)2 =
∑
sym
9a4b2 + 9 a4bc + 9 a3b3 + 54 a3b2c + 15 a2b2c2
Canceling like terms, we seek
∑
sym
4a5b − a4b2 + a4bc − 3a3b3 − 2a3b2c + a2b2c2
Sure enough, this is true, since 3a5b+ab 5
4
≥ a4b2 and a4b2+a2b4
2
≥ a3b3 by AM-GM, and
abc (a3 + b3 + c3 − a2(b + c) + b2(c + a) + c2(a + b) + 3 abc ) ≥ 0 by Schur. 34. (Japan 1997) Show that for all positive reals a, b, c ,(a + b − c)2
(a + b)2 + c2 + (b + c − a)2
(b + c)2 + a2 + (c + a − b)2
(c + a)2 + b2 ≥ 35
10 Found on a fortune cookie by Po-Ru Loh while grading an inequality on 2005 Mock IMO Day 2 that was solved by brutal force.
23 Solution. Put a + b + c = 3 so that equality will hold at a = b = c = 1 and suppose that there exists some k for which (b + c − a)2
(b + c)2 + a2 = (3 − 2a)2
(3 − a)2 + a2 ≥ 15 + ka − k
for all positive a, b, c ; such an inequality would allow us to add cyclicly to deduce the desired inequality. As the inequality is parametrically contrived to yield equality where
a = 1, we need to find k such that a = 1 is a double root. At a = 1, the derivative on the left is (2(3 −2a)·− 2)((3 −a)2+a2)−((3 −2a)2)(2(3 −a)·− 1+2 a)((3 −a)2+a2)2 = −18 25 . The derivative on the right is k, so we set k = −18 25 . But for this k we find (3 − 2a)2 −
(15 + ka − k
) ((3 − a)2 + a2) = 18 25 − 54 a2
25 + 36 a3
25 = 18 25 (a − 1) 2(2 a + 1) ≥ 0as desired. Alternatively, we could have used AM-GM to show a3 + a3 + 1 ≥ 3a2. As hinted at by a previous problem, inequalities are closely linked to polynomials with roots of even multiplicity. The isolated manipulation idea used in this solution offers a completely different approach to the inequalities which work with every term. 35. (MOP 02) Let a, b, c be positive reals. Prove that
( 2ab + c
)23
+
( 2bc + a
)23
+
( 2ca + b
)23
≥ 3
Solution. Suppose that there exists some r such that
( 2ab + c
)23
≥ 3ar
ar + br + cr
We could sum the inequality cyclicly to deduce what we want. Since equality holds at
a = b = c = 1, we use derivatives to find a suitable r. At the said equality case, on the left, the partial derivative with respect to a is 23 , while the same derivative on the right is 23 r. Equating the two we have r = 1. (This is necessary since otherwise the inequality will not hold for either a = 1 + ≤ or a = 1 − ≤.) 11 Now, 3aa + b + c ≤ 3a
3 3
√
a · (b+c
2
)2
11 Actually, even this is a special case of the general sense that the convexity of one side must exceed the convexity of the other. More precisely, we have the following result: Let fand gfunctions over the domain Dwith continuous partial derivatives. If f(ν)≥g(ν) for all ν∈D, then at every equality case ν0,
∇(f−g)( ν0) = 0and every component of ∇2(f−g) ( ν0) is nonnegative.
24 = a23
(b+c
2
)23
=
( 2ab + c
)23
by AM-GM, as desired. 36. (Mildorf) Let n ≥ 2 be an integer. Prove that for all reals a1, a 2, . . . , a n > 0 and reals
p, k ≥ 1, ( a1 + a2 + · · · + an
ap
1
ap
2
· · · + apn
)k
≥ ak
1
ak
2
· · · + akn
apk
1
apk
2
· · · + apk n
where inequality holds iff p = 1 or k = 1 or a1 = a2 = · · · = an, flips if instead 0 < p < 1, and flips (possibly again) if instead 0 < k < 1.
Solution. Taking the kth root of both sides, we see that the inequality is equivalent to n∑
i=1
k
√
aki
ak
1
ak
2
· · · + akn
≥
n
∑
i=1
k
√
apk i
apk
1
apk
2
· · · apk n
WLOG, suppose that a1 ≥ a2 ≥ · · · ≥ an. We prove a lemma. Let Si = api
ap
1+··· +apn
and
Ti = aqi
aq
1+··· +aqn
for i = 1 , 2, . . . , n where 0 < q < p . Then the sequence S1, S 2, . . . , S n
majorizes the sequence T1, T 2, . . . , T n.To prove the claim, we note that S1 ≥ · · · ≥ Sn and T1 ≥ · · · ≥ Tn and have, for
m ≤ n,
m
∑
i=1
Si ≥
m
∑
i=1
Ti
⇐⇒ (ap
1
· · · + apm) ( aq
1
· · · + aqn) ≥ (aq
1
· · · + aqm) ( ap
1
· · · + apn)
⇐⇒ (ap
1
· · · + apm) (aqm+1 + · · · + aqn
) ≥ (aq
1
· · · + aqm) (apm+1 + · · · + apn
)
⇐⇒ ∑
(i,j )| { 1≤i≤m<j ≤n}
api aqj − aqi apj ≥ 0Which is obvious. In particular, m = n is the equality case, and the claim is established. But now the desired is a direct consequence of the Majorization inequality applied to the sequences in question and the function f (x) = k
√x.37. (Vascile Cartoaje) Show that for all real numbers a, b, c,(a2 + b2 + c2)2 ≥ 3 (a3b + b3c + c3a)
25 Solution. We will be content to give the identity (a2 + b2 + c2)2 − 3( a3b + b3c + c3a) = 12
∑
cyc
(a2 − 2ab + bc − c2 + ca )2
Any Olympiad partipant should be comfortable constructing various inequalities through well-chosen squares. Here, we could certainly have figured we were summing the square of a quadratic that is 0 when a = b = c such that no term a2bc is left uncancelled. A good exercise is to show that equality actually holds iff a = b = c or, for some cyclic permutation, a : b : c ≡ sin 2 (4π
7
) : sin 2 (2π
7
) : sin 2 (π
7
).38. (Anh-Cuong) Show that for all nonnegative reals a, b, c ,
a3 + b3 + c3 + 3 abc ≥ ab √2a2 + 2 b2 + bc √2b2 + 2 c2 + ca √2c2 + 2 a2
Solution. Upon observing that this inequality is stronger than Schur’s inequality for
r = 1, we are inspired to prove a sharp lemma to eliminate the radical. Knowing that √2x2 + 2 y2 ≥ x + y ≥ 2xy x+y , we seek a combination of the latter two that exceeds the former. We find 3x2 + 2 xy + 3 y2
2( x + y) ≥ √2x2 + 2 y2
This follows from algebra, since (3 x2 + 2 xy + 3 y2)2 = 9 x4 + 12 x3y + 22 x2y2 + 12 xy 3 +9y4 ≥ 8x4 + 16 x3y + 16 x2y2 + 16 xy 3 + 8 y4 = 4( x + y)2(2 x2 + 2 y2), so that (3 x2 + 2 xy +3y2)2 − 4( x + y)2(2 x2 + 2 y2) = x4 − 4x3y + 6 x2y2 − 4xy 3 + y4 = ( x − y)4 ≥ 0. Now,
∑
cyc
ab √2a2 + 2 b2 ≤ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b)So it would suffice to show
∑
cyc
a(a − b)( a − c) = ∑
cyc
(a3 + abc − ab (a + b))
≥ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b) − ab (a + b)= ∑
cyc
3a3b + 2 a2b2 + 3 ab 3 − 2a3b − 4a2c2 − 2ab 3
2( a + b)= ∑
cyc
ab (a − b)2
2( a + b)But ∑
cyc
(b + c − a)( b − c)2 = 2 ∑
cyc
a(a − b)( a − c)26 so that the desired is
∑
cyc
(
b + c − a − bc b + c
)
(b − c)2 ≥ 0which is evident, since without loss of generality we may assume a ≥ b ≥ c and find
(
a + b − c − ab a + b
)
(a − b)2 ≥ 0
(
c + a − b − ac a + c
) ((a − c)2 − (b − c)2) ≥ 0
(
b + c − a − bc b + c
)
(b − c)2 +
(
c + a − b − ac a + c
)
(b − c)2 ≥ 0The key to this solution was the sharp upper bound on the root-mean-square. At first glance our lemma seems rather arbitrary and contrived. Actually, it is a special case of a very sharp bound on the two variable power mean that I have conjectured and proved.
Mildorf’s Lemma 1 Let k ≥ − 1 be an integer. Then for all positive reals a and b,
(1 + k)( a − b)2 + 8 ab
4( a + b) ≥ k
√ak + bk
2
with equality if and only if a = b or k = ±1, where the power mean k = 0 is interpreted to be the geometric mean √ab . Moreover, if k < −1, then the inequality holds in the reverse direction, with equality if and only if a = b.
Usage Note. As of early November 2005, I have proven an extension of this lemma to additional values of k.12 Thus, you may rest assured that the result stated above is true. I was unable to get this result published, so I have instead posted the proof here as “ASharpBound.pdf.” However, the proof is rather difficult (or at least so I think, being as though it took me nearly half a year) and the lemma is far from mainstream. Thus, should you require it on an Olympiad, you should prove it for whatever particular value of k you are invoking. This is not terribly difficult if k is a small integer. One simply takes the kth power of both sides and factors the difference of the two sides as
(a − b)4 · P (a, b ), etc.
For x ≥ y ≥ 1, prove that
x
√x + y + y
√y + 1 + 1
√x + 1 ≥ y
√x + y + x
√x + 1 + 1
√y + 1
12 In particular, the inequality holds for all kin ( −∞ ,−1) ,{− 1,0,1},(1 ,3/2] ,[2 ,∞) with the signs ≤,≥,≤
,≥respectively, with equality iff a=bor k=±1.
27 Solution. By observation, equality holds when y = 1 and when x = y. Combining this with the restriction, it makes sense to write x = y + a and y = 1 + b where a, b ≥ 0. Now we can write
x − y
√x + y + y − 1
√y + 1 + 1 − x
√1 + x ≥ 0
⇐⇒ a
√2 + a + 2 b + b
√2 + b ≥ a + b
√2 + a + b
But this is evident by Jensen’s inequality applied to the convex function f (x) = 1√x ,since
af (2 + a + 2 b) + bf (2 + b) ≥ (a + b)f
(a(2 + a + 2 b) + b(2 + b)
a + b
)
= (a + b)f
((a + b)2 + 2( a + b)
a + b
)
= a + b
√2 + a + b
as desired. 40. (MOP) For n ≥ 2 a fixed positive integer, let x1, . . . , x n be positive reals such that
x1 + x2 + · · · + xn = 1
x1
1
x2
· · · + 1
xn
Prove that 1
n − 1 + x1
1
n − 1 + x2
· · · + 1
n − 1 + xn
≤ 1
Solution. We will prove the contrapositive. (We are motivated to do this for two good reasons: 1) it is usually difficult the show that the sum of some reciprocals is bounded above, and 2) the given relation in its current form is an abomination.) Take
yi = 1
n−1+ xi
, and for the sake of contradiction assume y1 + · · · + yn > 1. Since the yi
are too large, the xi are too small and we shall prove 1
x1
· · · + 1
xn
x 1 + · · · + xn.Since xiyi = 1 − (n − 1) yi, we have (n − 1) yi > (n − 1)
(
yi + 1 −
n
∑
j=1
yj
)
= (n − 1) yi − 1 +
n
∑
j=1
(1 − (n − 1) yj )= −xiyi +
n
∑
j=1
xj yj (∗)=⇒ n − 1
xi
−1 +
n
∑
j=1
xj yj
xiyi
(∗∗ )28 Summing () over i,(n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(( n∑
j=1
1
xj yj
)
− 1
xiyi
)
But by Cauchy and (), we have
( n∑
j=1
1
xj yj
)
− 1
xiyi
≥ (n − 1) 2
(∑nj=1 xj yj
)
− xiyi
(n − 1) 2
(n − 1) yi
= n − 1
yi
Hence, (n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(n − 1
yi
)
= ( n − 1)( x1 + · · · + xn)as desired. 41. (Vascile Cartoaje) Show that for positive reals a, b, c ,14a2 − ab + 4 b2 + 14b2 − bc + 4 c2 + 14c2 − ca + 4 a2 ≥ 97( a2 + b2 + c2)
Solution. Upon expansion, we see that it is equivalent to
∑
sym
56 a6 − 28 a5b + 128 a4b2 + 44 a3b3 + 95 2 a4bc + 31 a3b2c − 45 2 a2b2c2 ≥ 0We conjure up the following inequalities:
∑
sym
a6 − 2a5b + a4bc ≥ 0 (1)
∑
sym
a5b − 4a4b2 + 3 a3b3 ≥ 0 (2)
∑
sym
a4b2 − a4bc − a3b3 + 2 a3b2c − a2b2c2 ≥ 0 (3)
∑
sym
a4bc − 2a3b2c + a2b2c2 ≥ 0 (4) (1) and (4) follow from Schur’s inequality for r = 4 and r = 1 (multiplied by abc )respectively. (2) is the result of expanding ∑
cyc
ab (a − b)4 ≥ 0, and (3) is the expanded form of the famous ( a − b)2(b − c)2(c − a)2 ≥ 0. The desired now follows by subtracting 56 times (1), 84 times (2), 208 times (3), 399 2 times (4), and then simple AM-GM to clear the remaining a2b2c2.29 This is about as difficult as a dumbass solution can get. A good general strategy is to work with the sharpest inequalities you can find until you reduce a problem to something obvious, starting with the most powerful (most bunched, in this case ∑
sym
a6) term and work your way down to the weak terms while keeping the most powerful term’s coefficient positive. My solution to this problem starts with (1), Schur with r = 4 (Schur is stronger for larger r), which is almost certainly sharper than the inequality in question. Next, inequality (2) is a sharp cyclic sum to use the a5b terms. In particular, it relates terms involving only two of the three variables. Most of the time, the only inequality that can “pull up” symmetric sums involving three variables to stronger ones involving just two is Schur, although it does so at the expense of a very strong term with only one variable. Hence, we made a logical choice. Inequality (3) is extremely sharp, and allowed us to obtain more a4bc and a3b3 terms simultaneously. In particular, it was necessary to cancel the a3b3 terms. I’ll note that this inequality is peculiar to sixth degree symmetry in three variables - it does not belong to a family of similar, nice inequalities. Finally, inequality (4), which is a handy corollary to (3), is another Schur. Every inequality we have used so far is quite sharp, and so it is no surprise that the leftovers are the comparatively weak AM-GM. 42. (Reid Barton, IMO Shortlist 03/A6.) Let n ≥ 2 be a positive integer and x1, x 2, . . . , x n,y1, y 2, . . . , y n a sequence of 2 n positive reals. Suppose z2, z 3, . . . , z 2n is such that z2
i+j
≥
xiyj for all i, j ∈ { 1, . . . , n }. Let M = max {z2, z 3, . . . , z 2n}. Prove that
(M + z2 + z3 + · · · + z2n
2n
)2
≥
(x1 + · · · + xn
n
) ( y1 + · · · + yn
n
)
Reid’s official solution. Let max( x1, . . . , x n) = max( y1, . . . , y n) = 1. (We can do this by factoring X from every xi, Y from every yj , and √XY from every zi+j
without changing the sign of the inequality.) We will prove M + z2 + · · · + z2n ≥
x1 + x2 + · · · + xn + y1 + y2 + · · · + yn, after which the desired follows by AM-GM. We will show that the number of terms on the left which are greater than r is at least as large as the number of terms on the right which are greater than r, for all r ≥ 0. For r ≥ 1, the claim is obvious, since all terms on the right are at most 1. Now take
r < 1. Let A and B denote the set of i for which xi > r and the set of j for which
yj > r respectively, and write a = |A|, b = |B|. Evidently, from our scaling, a, b ≥ 1. Now, xi > r and yj > r implies zi+j ≥ √xiyj ≥ r. Hence, if C is the set of k for which zk > r , we have |C| ≥ | A + B|, where the set addition is defined by the set of possible values if we take an element of A and add it to an element of B. How-ever, |A + B| ≥ | A| + |B| − 1, since if A and B consist of the values p1 < · · · < p a and
q1 < · · · < q b respectively we have all of the values p1 +q1 < . . . < p a +q1 < · · · < p a +qb
in A + B. Hence, |C| ≥ a + b − 1. Since |C| ≥ 1, there is some zk > r , and hence,
M > r . Therefore, the left side of the inequality in question has at least a + b terms which exceed r, as desired. •
30 The preponderance of difficulty here stemmed from dealing with the superabundance of givens, especially the mysterious M . Scaling allowed us to introduce some degree of control and, with marked audacity, a profoundly clever idea. As it turned out, the in-equality was no sharper than simple AM-GM! It is my opinion that it is highly unlikely that a problem as staggeringly pernicious as this one will appear on an Olympiad - at least in the foreseeable future. Nevertheless, I have included it here for the purpose of illustrating just how unusual and creative a solution can be.
3 Problems
(MOP 04) Show that for all positive reals a, b, c ,
( a + 2 ba + 2 c
)3
+
( b + 2 cb + 2 a
)3
+
(c + 2 ac + 2 b
)3
≥ 32. (MOP) Show that if k is a positive integer and a1, a 2, . . . , a n are positive reals which sum to 1, then n∏
i=1
1 − aki
aki
≥ (nk − 1)n
Let a1, a 2, . . . , a n be nonnegative reals with a sum of 1. Prove that
a1a2 + a2a3 + · · · + an−1an ≤ 144. (Ukraine 01) Let a, b, c, x, y, z be nonnegative reals such that x + y + z = 1. Show that
ax + by + cz + 2 √(ab + bc + ca )( xy + yz + zx ) ≤ a + b + c
Let n > 1 be a positive integer and a1, a 2, . . . , a n positive reals such that a1a2 . . . a n = 1. Show that 11 + a1
· · · + 11 + an
≤ a1 + · · · + an + n
(Aaron Pixton) Let a, b, c be positive reals with product 1. Show that 5 + ab + bc + ca ≥ (1 + a)(1 + b)(1 + c)7. (Valentin Vornicu 13 ) Let a, b, c, x, y, z be arbitrary reals such that a ≥ b ≥ c and either
x ≥ y ≥ z or x ≤ y ≤ z. Let f : R → R+0 be either monotonic or convex, and let k be a positive integer. Prove that
f (x)( a − b)k(a − c)k + f (y)( b − c)k(b − a)k + f (z)( c − a)k(c − b)k ≥ 0
13 This improvement is more widely known than the other one in this packet, and is published in his book,
Olimpiada de Matematica... de la provocare la experienta , GIL Publishing House, Zalau, Romania. (In English, “The Math Olympiad... from challenge to experience.”)
31 8. (IMO 01/2) Let a, b, c be positive reals. Prove that
a
√a2 + 8 bc + b
√b2 + 8 ca + c
√c2 + 8 ab ≥ 19. (USAMO 04/5) Let a, b, c be positive reals. Prove that
(a5 − a2 + 3 ) ( b5 − b2 + 3 ) ( c5 − c2 + 3 ) ≥ (a + b + c)3
(Titu Andreescu) Show that for all nonzero reals a, b, c ,
a2
b2 + b2
c2 + c2
a2 ≥ ac + cb + ba
(IMO 96 Shortlist) Let a, b, c be positive reals with abc = 1. Show that
ab a5 + b5 + ab + bc b5 + c5 + bc + ca c5 + a5 + ca ≤ 112. Let a, b, c be positive reals such that a + b + c = 1. Prove that
√ab + c + √bc + a + √ca + b ≥ 1 + √ab + √bc + √ca
(APMO 2005/2) Let a, b, c be positive reals with abc = 8. Prove that
a2
√(a3 + 1) ( b3 + 1) + b2
√(b3 + 1) ( c3 + 1) + c2
√(c3 + 1) ( a3 + 1) ≥ 4314. Show that for all positive reals a, b, c ,
a3
b2 − bc + c2 + b3
c2 − ca + a2 + c3
a2 − ab + b2 ≥ a + b + c
(USAMO 97/5) Prove that for all positive reals a, b, c ,1
a3 + b3 + abc + 1
b3 + c3 + abc + 1
c3 + a3 + abc ≤ 1
abc
(Mathlinks Lore) Show that for all positive reals a, b, c, d with abcd = 1, and k ≥ 2, 1(1 + a)k + 1(1 + b)k + 1(1 + c)k + 1(1 + d)k ≥ 22−k
(IMO 05/3) Prove that for all positive a, b, c with product at least 1,
a5 − a2
a5 + b2 + c2 + b5 − b2
b5 + c2 + a2 + c5 − c2
c5 + a2 + b2 ≥ 032 18. (Mildorf) Let a, b, c, k be positive reals. Determine a simple, necessary and sufficient condition for the following inequality to hold: (a + b + c)k (akbk + bkck + ckak) ≤ (ab + bc + ca )k(ak + bk + ck)19. Let a, b, c be reals with a + b + c = 1 and a, b, c ≥ − 34 . Prove that
aa2 + 1 + bb2 + 1 + cc2 + 1 ≤ 910 20. (Mildorf) Show that for all positive reals a, b, c ,
3
√4a3 + 4 b3 + 3
√4b3 + 4 c3 + 3
√4c3 + 4 a3 ≤ 4a2
a + b + 4b2
b + c + 4c2
c + a
Let a, b, c, x, y, z be real numbers such that (a + b + c)( x + y + z) = 3 , (a2 + b2 + c2)( x2 + y2 + z2) = 4 Prove that
ax + by + cz ≥ 022. (Po-Ru Loh) Let a, b, c be reals with a, b, c > 1 such that 1
a2 − 1 + 1
b2 − 1 + 1
c2 − 1 = 1 Prove that 1
a + 1 + 1
b + 1 + 1
c + 1 ≤ 123. (Weighao Wu) Prove that (sin x)sin x < (cos x)cos x
for all real numbers 0 < x < π
4
.24. (Mock IMO 05/2) Let a, b, c be positive reals. Show that 1 < a
√a2 + b2 + b
√b2 + c2 + c
√c2 + a2 ≤ 3√2225. (Gabriel Dospinescu) Let n ≥ 2 be a positive integer. Show that for all positive reals
a1, a 2, . . . , a n with a1a2 . . . a n = 1,
√a21 + 1 2 + · · · +
√a2
n
1 2 ≤ a1 + · · · + an
33 26. Let n ≥ 2 be a positive integer, and let k ≥ n−1
n
be a real number. Show that for all positive reals a1, a 2, . . . , a n,
( (n − 1) a1
a2 + · · · + an
)k
+
( (n − 1) a2
a3 + · · · + an + a1
)k
· · · +
( (n − 1) an
a1 + · · · + an−1
)k
≥ n
(Mildorf) Let a, b, c be arbitrary reals such that a ≥ b ≥ c, and let x, y, z be nonnegative reals with x + z ≥ y. Prove that
x2(a − b)( a − c) + y2(b − c)( b − a) + z2(c − a)( c − b) ≥ 0and determine where equality holds. 28. (USAMO 00/6) Let n ≥ 2 be an integer and S = {1, 2, . . . , n }. Show that for all nonnegative reals a1, a 2, . . . , a n, b 1, b 2, . . . , b n,
∑
i,j ∈S
min {aiaj , b ibj } ≤ ∑
i,j ∈S
min {aibj , a j bi}
(Kiran Kedlaya) Show that for all nonnegative a1, a 2, . . . , a n,
a1 + √a1a2 + · · · + n
√a1 · · · an
n ≤ n
√
a1 · a1 + a2
2 · · · a1 + · · · + an
n
(Vascile Cartoaje) Prove that for all positive reals a, b, c such that a + b + c = 3,
aab + 1 + bbc + 1 + cca + 1 ≥ 3231. (Gabriel Dospinescu) Prove that ∀a, b, c, x, y, z ∈ R+| xy + yz + zx = 3,
a(y + z)
b + c + b(z + x)
c + a + c(x + y)
a + b ≥ 332. (Mildorf) Let a, b, c be non-negative reals. Show that for all real k,
∑
cyc
max( ak, b k)( a − b)2
2 ≥ ∑
cyc
ak(a − b)( a − c) ≥ ∑
cyc
min( ak, b k)( a − b)2
2(where a, b, c 6 = 0 if k ≤ 0) and determine where equality holds for k > 0, k = 0, and
k < 0 respectively. 33. (Vascile Cartoaje) Let a, b, c, k be positive reals. Prove that
ab + ( k − 3) bc + ca
(b − c)2 + kbc + bc + ( k − 3) ca + ab
(c − a)2 + kca + ca + ( k − 3) ab + bc
(a − b)2 + kab ≥ 3( k − 1)
k
(Taiwan? 02) Show that for all positive a, b, c, d ≤ k, we have
abcd
(2 k − a)(2 k − b)(2 k − c)(2 k − d) ≤ a4 + b4 + c4 + d4
(2 k − a)4 + (2 k − b)4 + (2 k − c)4 + (2 k − d)4
34 |
5354 | https://wayground.com/admin/quiz/5f903584c23e43001f51f715/converting-between-pounds-and-kilograms | Converting between pounds and kilograms 1st - 12th Grade Quiz | Wayground (formerly Quizizz)
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Converting between pounds and kilograms
1st - 12th Grade
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20 Qs
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Converting between pounds and kilograms
Quiz
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Mathematics
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1st - 12th Grade
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Easy
•
CCSS
6.RP.A.3D, 8.EE.A.4, 6.NS.B.3
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Katherine Furgeson
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20 questions
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[x]
1.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Preview
Edit
A patient comes into a medical office and says he weighs 173 lbs. How many kg does he weigh?
62.3 kg
78.6 kg
380.6 kg
247.1 kg
Answer explanation
173 divided by 2.2 gives you 78.6. The actual number is 78.63636363.....but we only want to take our answer to the tenths place so he weighs 78.6 kg.
Tags
CCSS.6.RP.A.3D
2.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
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Edit
117 lbs = _____ kg?
49.1 kg
257.4 kg
59.2 kg
53.1 kg
Tags
CCSS.6.RP.A.3D
3.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Preview
Edit
475 lbs = ______ kg?
215.9 kg
200.0 kg
375.4 kg
198.5 kg
Tags
CCSS.6.RP.A.3D
4.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Preview
Edit
What do I do to convert pounds to kilograms?
Divide by 2.2
Multiply by 2.2
Tags
CCSS.6.RP.A.3D
5.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Preview
Edit
You are traveling in Europe and become ill. You go to a medical clinic and you are told you weigh 65 kg. How many lbs is that?
178 lbs
29.5 lbs
143 lbs
150.1 lbs
Answer explanation
To convert from kg to lbs you would multiply by 2.2. Therefore, 65 X 2.2 gives us 143. The weight would equal 143 pounds.
Tags
CCSS.6.RP.A.3D
6.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Preview
Edit
184 kg = ______ lbs?
404.8 lbs
360.2 lbs
480.1 lbs
424.1 lbs
Tags
CCSS.6.RP.A.3D
7.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Preview
Edit
536 kg = ______ lbs?
1,000 lbs
673.2 lbs
1,854.3 lbs
1,179.2 lbs
Tags
CCSS.6.RP.A.3D
8.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Preview
Edit
How many lbs are in 45 kg?
99 lbs
108 lbs
67 lbs
450 lbs
Tags
CCSS.6.RP.A.3D
9.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Preview
Edit
If I'm trying to end up with pounds do I multiply or divide?
Multiply
Divide
Answer explanation
If you are trying to express a value in pounds you will multiply the number of kg.
Tags
CCSS.6.RP.A.3D
10.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Preview
Edit
If you are trying to express a value in kg that began as pounds do I multiply or divide?
Multiply
Divide
Answer explanation
If you are trying to express a value in kg you will divide the number of pounds.
Tags
CCSS.6.RP.A.3D
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5355 | http://cecs.wright.edu/~sthomas/htchapter03.pdf | STEADY HEAT CONDUCTION I n heat transfer analysis, we are often interested in the rate of heat transfer through a medium under steady conditions and surface temperatures. Such problems can be solved easily without involving any differential equations by the introduction of the thermal resistance concept in an analogous manner to electrical circuit problems. In this case, the thermal resistance corresponds to electrical resistance, temperature difference corresponds to voltage, and the heat transfer rate corresponds to electric current.
We start this chapter with one-dimensional steady heat conduction in a plane wall, a cylinder, and a sphere, and develop relations for thermal resis-tances in these geometries. We also develop thermal resistance relations for convection and radiation conditions at the boundaries. We apply this concept to heat conduction problems in multilayer plane walls, cylinders, and spheres and generalize it to systems that involve heat transfer in two or three dimen-sions. We also discuss the thermal contact resistance and the overall heat transfer coefficient and develop relations for the critical radius of insulation for a cylinder and a sphere. Finally, we discuss steady heat transfer from finned surfaces and some complex geometrics commonly encountered in practice through the use of conduction shape factors.
135 CHAPTER 3 OBJECTIVES When you finish studying this chapter, you should be able to: ■ Understand the concept of thermal resistance and its limitations, and develop thermal resistance networks for practical heat conduction problems, ■ Solve steady conduction problems that involve multilayer rectangular, cylindrical, or spherical geome-tries, ■ Develop an intuitive understand-ing of thermal contact resistance, and circumstances under which it may be significant, ■ Identify applications in which insulation may actually increase heat transfer, ■ Analyze finned surfaces, and as-sess how efficiently and effectively fins enhance heat transfer, and ■ Solve multidimensional practical heat conduction problems using conduction shape factors.
136 STEADY HEAT CONDUCTION 3–1 STEADY HEAT CONDUCTION IN PLANE WALLS Consider steady heat conduction through the walls of a house during a winter day. We know that heat is continuously lost to the outdoors through the wall.
We intuitively feel that heat transfer through the wall is in the normal direc-tion to the wall surface, and no significant heat transfer takes place in the wall in other directions (Fig. 3–1).
Recall that heat transfer in a certain direction is driven by the temperature gradient in that direction. There is no heat transfer in a direction in which there is no change in temperature. Temperature measurements at several loca-tions on the inner or outer wall surface will confirm that a wall surface is nearly isothermal. That is, the temperatures at the top and bottom of a wall surface as well as at the right and left ends are almost the same. Therefore, there is no heat transfer through the wall from the top to the bottom, or from left to right, but there is considerable temperature difference between the in-ner and the outer surfaces of the wall, and thus significant heat transfer in the direction from the inner surface to the outer one.
The small thickness of the wall causes the temperature gradient in that di-rection to be large. Further, if the air temperatures in and outside the house re-main constant, then heat transfer through the wall of a house can be modeled as steady and one-dimensional. The temperature of the wall in this case de-pends on one direction only (say the x-direction) and can be expressed as T(x).
Noting that heat transfer is the only energy interaction involved in this case and there is no heat generation, the energy balance for the wall can be ex-pressed as or in out (3–1) But dEwall/dt = 0 for steady operation, since there is no change in the temper-ature of the wall with time at any point. Therefore, the rate of heat transfer into the wall must be equal to the rate of heat transfer out of it. In other words, the rate of heat transfer through the wall must be constant, cond, wall constant.
Consider a plane wall of thickness L and average thermal conductivity k.
The two surfaces of the wall are maintained at constant temperatures of T1 and T2. For one-dimensional steady heat conduction through the wall, we have T(x). Then Fourier’s law of heat conduction for the wall can be expressed as cond, wall kA (W) (3–2) where the rate of conduction heat transfer cond, wall and the wall area A are Q dT dx Q Q dEwall dt Q Q £ Rate of heat transfer into the wall ≥ £ Rate of heat transfer out of the wall ≥ £ Rate of change of the energy of the wall ≥ ■ Q · 20°C 20°C 20°C 20°C 20°C 20°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 3°C 20°C 20°C 11°C 11°C 11°C 11°C 11°C 11°C 11°C 11°C A x z y T(x) FIGURE 3–1 Heat transfer through a wall is one-dimensional when the temperature of the wall varies in one direction only.
FIGURE 3–2 Under steady conditions, the temperature distribution in a plane wall is a straight line.
x 0 L dx dT T2 T1 T(x) Qcond · A constant. Thus dT/dx constant, which means that the temperature through the wall varies linearly with x. That is, the temperature distribution in the wall under steady conditions is a straight line (Fig. 3–2).
137 CHAPTER 3 Separating the variables in the preceding equation and integrating from x 0, where T(0) T1, to x L, where T(L) T2, we get cond, wall dx kA dT Performing the integrations and rearranging gives cond, wall kA (W) (3–3) which is identical to Eq. 1–21. Again, the rate of heat conduction through a plane wall is proportional to the average thermal conductivity, the wall area, and the temperature difference, but is inversely proportional to the wall thickness. Also, once the rate of heat conduction is available, the tem-perature T(x) at any location x can be determined by replacing T2 in Eq. 3–3 by T, and L by x.
Thermal Resistance Concept Equation 3–3 for heat conduction through a plane wall can be rearranged as cond, wall (W) (3–4) where Rwall (K/W) (3–5) is the thermal resistance of the wall against heat conduction or simply the conduction resistance of the wall. Note that the thermal resistance of a medium depends on the geometry and the thermal properties of the medium.
Note that thermal resistance can also be expressed as Rwall T/ cond, wall, Q L kA T1 T2 Rwall Q T1 T2 L Q T2 TT1 Q L x0 T2 (a) Heat flow (b) Electric current flow R Re V2 T1 V1 T1 – T2 ——— R Q = · V1 – V2 ——— Re I = FIGURE 3–3 Analogy between thermal and electrical resistance concepts.
which is the ratio of the driving potential T to the corresponding transfer rate cond, wall.
This equation for heat transfer is analogous to the relation for electric current flow I, expressed as I (3–6) where Re L/se A is the electric resistance and V1 V2 is the voltage dif-ference across the resistance (se is the electrical conductivity). Thus, the rate of heat transfer through a layer corresponds to the electric current, the thermal resistance corresponds to electrical resistance, and the temperature difference corresponds to voltage difference across the layer (Fig. 3–3).
Consider convection heat transfer from a solid surface of area As and tem-perature Ts to a fluid whose temperature sufficiently far from the surface is T, with a convection heat transfer coefficient h. Newton’s law of cooling for con-vection heat transfer rate conv hAs(Ts T) can be rearranged as conv (W) (3–7) Ts T Rconv Q Q V1 V2 Re Q where Rconv (K/W) (3–8) is the thermal resistance of the surface against heat convection, or simply the convection resistance of the surface (Fig. 3–4). Note that when the convec-tion heat transfer coefficient is very large (h →), the convection resistance becomes zero and Ts T. That is, the surface offers no resistance to convec-tion, and thus it does not slow down the heat transfer process. This situation is approached in practice at surfaces where boiling and condensation occur. Also note that the surface does not have to be a plane surface. Equation 3–8 for con-vection resistance is valid for surfaces of any shape, provided that the as-sumption of h constant and uniform is reasonable.
When the wall is surrounded by a gas, the radiation effects, which we have ignored so far, can be significant and may need to be considered. The rate of radiation heat transfer between a surface of emissivity e and area As at tem-perature Ts and the surrounding surfaces at some average temperature Tsurr can be expressed as rad es As( ) hrad As(Ts Tsurr) (W) (3–9) where Rrad (K/W) (3–10) is the thermal resistance of a surface against radiation, or the radiation resistance, and hrad es( (Ts Tsurr) (W/m2K) (3–11) is the radiation heat transfer coefficient. Note that both Ts and Tsurr must be in K in the evaluation of hrad. The definition of the radiation heat transfer co-efficient enables us to express radiation conveniently in an analogous manner to convection in terms of a temperature difference. But hrad depends strongly on temperature while hconv usually does not.
A surface exposed to the surrounding air involves convection and radiation simultaneously, and the total heat transfer at the surface is determined by adding (or subtracting, if in the opposite direction) the radiation and convec-tion components. The convection and radiation resistances are parallel to each other, as shown in Fig. 3–5, and may cause some complication in the thermal resistance network. When Tsurr T, the radiation effect can properly be ac-counted for by replacing h in the convection resistance relation by hcombined hconv hrad (W/m2K) (3–12) where hcombined is the combined heat transfer coefficient discussed in Chapter 1. This way all complications associated with radiation are avoided.
T 2 surr) T 2 s Qrad As(Ts Tsurr) 1 hrad As Ts Tsurr Rrad T 4 surr T 4 s Q 1 hAs 138 STEADY HEAT CONDUCTION Solid Q · T h As T Ts Rconv = 1 — hAs Ts FIGURE 3–4 Schematic for convection resistance at a surface.
Solid Q · As Ts Qconv · Rconv T Qrad · Q = Qconv + Qrad · · · Rrad Tsurr FIGURE 3–5 Schematic for convection and radiation resistances at a surface.
Thermal Resistance Network Now consider steady one-dimensional heat transfer through a plane wall of thickness L, area A, and thermal conductivity k that is exposed to convection on both sides to fluids at temperatures T1 and T2 with heat transfer coeffi-cients h1 and h2, respectively, as shown in Fig. 3–6. Assuming T2 T1, the variation of temperature will be as shown in the figure. Note that the temper-ature varies linearly in the wall, and asymptotically approaches T1 and T2 in the fluids as we move away from the wall.
Under steady conditions we have or h1 A(T1 T1) kA h2 A(T2 T2) (3–13) which can be rearranged as (3–14) T1 T1 Rconv, 1 T1 T2 Rwall T2 T2 Rconv, 2 T1 T1 1/h1A T1 T2 L/kA T2 T2 1/h2A Q T1 T2 L Q £ Rate of heat convection into the wall ≥ £ Rate of heat conduction through the wall ≥ £ Rate of heat convection from the wall ≥ 139 CHAPTER 3 Thermal network Electrical analogy Q · T1 T2 Rconv,2 Rconv,1 Rwall Re,3 Re,1 I Re,2 T2 V2 T1 V1 T1 – T2 —————————— Rconv,1 + Rwall + Rconv,2 Q = · V1 – V2 ———————— Re,1 + Re,2 + Re,3 I = Wall T2 T1 T2 T1 FIGURE 3–6 The thermal resistance network for heat transfer through a plane wall subjected to convection on both sides, and the electrical analogy.
Once the rate of heat transfer is calculated, Eq. 3–14 can also be used to determine the intermediate temperatures T1 or T2. Adding the numerators and denominators yields (Fig. 3–7) (W) (3–15) where Rtotal Rconv, 1 Rwall Rconv, 2 (K/W) (3–16) Note that the heat transfer area A is constant for a plane wall, and the rate of heat transfer through a wall separating two media is equal to the overall tem-perature difference (T1 T2) divided by the total thermal resistance between the media. Also note that the thermal resistances are in series, and the equiva-lent thermal resistance is determined by simply adding the individual resis-tances, just like the electrical resistances connected in series. Thus, the electrical analogy still applies. We summarize this as the rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.
Another observation that can be made from Eq. 3–15 is that the ratio of the temperature drop to the thermal resistance across any layer is constant, and thus the temperature drop across any layer is proportional to the thermal re-sistance of the layer. The larger the resistance, the larger the temperature drop.
In fact, the equation T/R can be rearranged as T R (C) (3–17) which indicates that the temperature drop across any layer is equal to the rate of heat transfer times the thermal resistance across that layer (Fig. 3–8). You may recall that this is also true for voltage drop across an electrical resistance when the electric current is constant.
It is sometimes convenient to express heat transfer through a medium in an analogous manner to Newton’s law of cooling as UA T (W) (3–18) where U is the overall heat transfer coefficient with the unit W/m2·K. The overall heat transfer coefficient is usually used in heat transfer calculations as-sociated with heat exchangers (Chapter 11). It is also used in heat transfer cal-culations through windows (Chapter 9), commonly referred to as U-factor. A comparison of Eqs. 3–15 and 3–18 reveals that UA (W/K) (3–19) Therefore, for a unit area, the overall heat transfer coefficient is equal to the inverse of the total thermal resistance.
Note that we do not need to know the surface temperatures of the wall in or-der to evaluate the rate of steady heat transfer through it. All we need to know is the convection heat transfer coefficients and the fluid temperatures on both 1 Rtotal Q Q Q 1 h1 A L kA 1 h2 A T1 T2 Rtotal Q 140 STEADY HEAT CONDUCTION If then For example, and a1 — b1 a1 + a2 + . . . + an ——————— b1 + b2 + . . . + bn a2 — b2 = = . . .
= c = c = an — bn = 1 — 4 2 — 8 = 5 — 20 = 0.25 1 + 2 + 5 ———— 4 + 8 + 20 = 0.25 FIGURE 3–7 A useful mathematical identity.
Rwall Rconv,2 Rconv,1 T1 T2 T2 T1 T1 T2 2°C/W 20°C 150°C 30°C 15°C/W 3°C/W ΔT = QR T2 T1 · Q = 10 W · FIGURE 3–8 The temperature drop across a layer is proportional to its thermal resistance.
sides of the wall. The surface temperature of the wall can be determined as described above using the thermal resistance concept, but by taking the sur-face at which the temperature is to be determined as one of the terminal sur-faces. For example, once is evaluated, the surface temperature T1 can be determined from (3–20) Multilayer Plane Walls In practice we often encounter plane walls that consist of several layers of dif-ferent materials. The thermal resistance concept can still be used to determine the rate of steady heat transfer through such composite walls. As you may have already guessed, this is done by simply noting that the conduction resis-tance of each wall is L/kA connected in series, and using the electrical analogy.
That is, by dividing the temperature difference between two surfaces at known temperatures by the total thermal resistance between them.
Consider a plane wall that consists of two layers (such as a brick wall with a layer of insulation). The rate of steady heat transfer through this two-layer composite wall can be expressed as (Fig. 3–9) (3–21) where Rtotal is the total thermal resistance, expressed as Rtotal Rconv, 1 Rwall, 1 Rwall, 2 Rconv, 2 (3–22) 1 h1 A L1 k1 A L2 k2 A 1 h2 A T1 T2 Rtotal Q T1 T1 Rconv, 1 T1 T1 1/h1 A Q Q 141 CHAPTER 3 Wall 1 Wall 2 T1 T2 T2 T1 T1 T2 L1 k2 k1 T3 Rwall,1 L1 k1A T1 A T2 T3 h2 h1 = —–– Rwall,2 L2 k2A = —–– Rconv,2 1 h2A = —–– Rconv,1 1 h1A = —–– Q · L2 FIGURE 3–9 The thermal resistance network for heat transfer through a two-layer plane wall subjected to convection on both sides.
The subscripts 1 and 2 in the Rwall relations above indicate the first and the second layers, respectively. We could also obtain this result by following the approach already used for the single-layer case by noting that the rate of steady heat transfer through a multilayer medium is constant, and thus it must be the same through each layer. Note from the thermal resistance net-work that the resistances are in series, and thus the total thermal resistance is simply the arithmetic sum of the individual thermal resistances in the path of heat transfer.
This result for the two-layer case is analogous to the single-layer case, ex-cept that an additional resistance is added for the additional layer. This result can be extended to plane walls that consist of three or more layers by adding an additional resistance for each additional layer.
Once is known, an unknown surface temperature Tj at any surface or interface j can be determined from (3–23) where Ti is a known temperature at location i and Rtotal, i j is the total thermal resistance between locations i and j. For example, when the fluid temperatures T1 and T2 for the two-layer case shown in Fig. 3–9 are available and is calculated from Eq. 3–21, the interface temperature T2 between the two walls can be determined from (Fig. 3–10) (3–24) The temperature drop across a layer is easily determined from Eq. 3–17 by multiplying by the thermal resistance of that layer.
The thermal resistance concept is widely used in practice because it is intu-itively easy to understand and it has proven to be a powerful tool in the solu-tion of a wide range of heat transfer problems. But its use is limited to systems through which the rate of heat transfer remains constant; that is, to systems involving steady heat transfer with no heat generation (such as resistance heating or chemical reactions) within the medium.
Q Q T1 T2 Rconv, 1 Rwall, 1 T1 T2 1 h1 A L1 k1 A Q Q Ti Tj Rtotal, ij Q Q Q 142 STEADY HEAT CONDUCTION EXAMPLE 3–1 Heat Loss through a Wall Consider a 3-m-high, 5-m-wide, and 0.3-m-thick wall whose thermal conduc-tivity is k 0.9 W/m·K (Fig. 3–11). On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be 16°C and 2°C, respectively. Determine the rate of heat loss through the wall on that day.
T1 – T1 Rconv,1 To find T1: = ———— T1 – T2 Rconv,1 + Rwall,1 To find T2: = —————— T3 – T2 Rconv,2 To find T3: = ———— Q · Q · Q · Q · Wall 1 Wall 2 T1 T2 T2 T1 T2 Rwall,1 Rconv,1 Rconv,2 Rwall,2 T3 T1 FIGURE 3–10 The evaluation of the surface and interface temperatures when T1 and T2 are given and is calculated.
Q Wall 2°C 16°C L = 0.3 m A 5 m 3 m Q · FIGURE 3–11 Schematic for Example 3–1.
143 CHAPTER 3 SOLUTION The two surfaces of a wall are maintained at specified tempera-tures. The rate of heat loss through the wall is to be determined.
Assumptions 1 Heat transfer through the wall is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients exist in the direction from the indoors to the outdoors. 3 Thermal conductivity is constant.
Properties The thermal conductivity is given to be k 0.9 W/m·K.
Analysis Noting that heat transfer through the wall is by conduction and the area of the wall is A 3 m 5 m 15 m2, the steady rate of heat transfer through the wall can be determined from Eq. 3–3 to be kA (0.9 W/mC)(15 m2) 630 W We could also determine the steady rate of heat transfer through the wall by making use of the thermal resistance concept from where Rwall 0.02222C/W Substituting, we get 630 W Discussion This is the same result obtained earlier. Note that heat conduction through a plane wall with specified surface temperatures can be determined directly and easily without utilizing the thermal resistance concept. However, the thermal resistance concept serves as a valuable tool in more complex heat transfer problems, as you will see in the following examples. Also, the units W/m·°C and W/m·K for thermal conductivity are equivalent, and thus inter-changeable. This is also the case for °C and K for temperature differences.
(16 2)°C 0.02222°C/ W Q L kA 0.3 m (0.9 W/m.C)(15 m2) Twall Rwall Q (16 2)C 0.3 m T1 T2 L Q EXAMPLE 3–2 Heat Loss through a Single-Pane Window Consider a 0.8-m-high and 1.5-m-wide glass window with a thickness of 8 mm and a thermal conductivity of k 0.78 W/m·K. Determine the steady rate of heat transfer through this glass window and the temperature of its inner sur-face for a day during which the room is maintained at 20°C while the temper-ature of the outdoors is 10°C. Take the heat transfer coefficients on the inner and outer surfaces of the window to be h1 10 W/m2·K and h2 40 W/m2·K, which includes the effects of radiation.
SOLUTION Heat loss through a window glass is considered. The rate of heat transfer through the window and the inner surface temperature are to be determined.
144 STEADY HEAT CONDUCTION Assumptions 1 Heat transfer through the window is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer through the wall is one-dimensional since any significant temperature gradients exist in the direction from the indoors to the outdoors. 3 Thermal conductivity is constant.
Properties The thermal conductivity is given to be k 0.78 W/m·K.
Analysis This problem involves conduction through the glass window and con-vection at its surfaces, and can best be handled by making use of the thermal resistance concept and drawing the thermal resistance network, as shown in Fig. 3–12. Noting that the area of the window is A 0.8 m 1.5 m 1.2 m2, the individual resistances are evaluated from their definitions to be Ri Rconv, 1 0.08333C/W Rglass 0.00855C/W Ro Rconv, 2 0.02083C/W Noting that all three resistances are in series, the total resistance is Rtotal Rconv, 1 Rglass Rconv, 2 0.08333 0.00855 0.02083 0.1127C/W Then the steady rate of heat transfer through the window becomes 266 W Knowing the rate of heat transfer, the inner surface temperature of the window glass can be determined from ⎯→ T1 T1 Rconv, 1 20C (266 W)(0.08333C/W) 2.2C Discussion Note that the inner surface temperature of the window glass is 2.2°C even though the temperature of the air in the room is maintained at 20°C. Such low surface temperatures are highly undesirable since they cause the formation of fog or even frost on the inner surfaces of the glass when the humidity in the room is high.
Q T1 T1 Rconv, 1 Q T1 T2 Rtotal [20 (10)]C 0.1127C/ W Q 1 h2 A 1 (40 W/m2K)(1.2 m2) L kA 0.008 m (0.78 W/mK)(1.2 m2) 1 h1 A 1 (10 W/m2K)(1.2 m2) T2 Rglass Ro L = 8 mm Ri T1 T2 T1 h1 = 10 W/m2·K h2 = 40 W/m2·K Glass 20°C –10°C T1 T2 FIGURE 3–12 Schematic for Example 3–2.
EXAMPLE 3–3 Heat Loss through Double-Pane Windows Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 4-mm-thick layers of glass (k 0.78 W/m·K) separated by a 10-mm-wide stagnant air space (k 0.026 W/m·K). Determine the steady rate of heat 145 CHAPTER 3 transfer through this double-pane window and the temperature of its inner sur-face for a day during which the room is maintained at 20°C while the temper-ature of the outdoors is 10°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1 10 W/m2·K and h2 40 W/m2·K, which includes the effects of radiation.
SOLUTION A double-pane window is considered. The rate of heat transfer through the window and the inner surface temperature are to be determined.
Analysis This example problem is identical to the previous one except that the single 8-mm-thick window glass is replaced by two 4-mm-thick glasses that enclose a 10-mm-wide stagnant air space. Therefore, the thermal resis-tance network of this problem involves two additional conduction resistances corresponding to the two additional layers, as shown in Fig. 3–13. Noting that the area of the window is again A 0.8 m 1.5 m 1.2 m2, the individual resistances are evaluated from their definitions to be Ri Rconv, 1 0.08333C/W R1 R3 Rglass 0.00427C/W R2 Rair 0.3205C/W Ro Rconv, 2 0.02083C/W Noting that all three resistances are in series, the total resistance is Rtotal Rconv, 1 Rglass, 1 Rair Rglass, 2 Rconv, 2 0.08333 0.00427 0.3205 0.00427 0.02083 0.4332C/W Then the steady rate of heat transfer through the window becomes 69.2 W which is about one-fourth of the result obtained in the previous example. This explains the popularity of the double- and even triple-pane windows in cold climates. The drastic reduction in the heat transfer rate in this case is due to the large thermal resistance of the air layer between the glasses.
The inner surface temperature of the window in this case will be T1 T1 R conv, 1 20C (69.2 W)(0.08333C/W) 14.2C which is considerably higher than the 2.2°C obtained in the previous exam-ple. Therefore, a double-pane window will rarely get fogged. A double-pane window will also reduce the heat gain in summer, and thus reduce the air-conditioning costs.
Q T1 T2 Rtotal [20 (10)]C 0.4332C/ W Q 1 h2 A 1 (40 W/m2K)(1.2 m2) L2 k2 A 0.01 m (0.026 W/mK)(1.2 m2) L1 k1 A 0.004 m (0.78 W/mK)(1.2 m2) 1 h1 A 1 (10 W/m2K)(1.2 m2) T1 T2 T3 R1 Ri R3 R2 T4 10 mm 20°C Air Glass Glass –10°C 4 mm 4 mm Ro T2 T1 FIGURE 3–13 Schematic for Example 3–3.
3–2 THERMAL CONTACT RESISTANCE In the analysis of heat conduction through multilayer solids, we assumed “perfect contact” at the interface of two layers, and thus no temperature drop at the in-terface. This would be the case when the surfaces are perfectly smooth and they produce a perfect contact at each point. In reality, however, even flat surfaces that appear smooth to the eye turn out to be rather rough when examined under a microscope, as shown in Fig. 3–14, with numerous peaks and valleys. That is, a surface is microscopically rough no matter how smooth it appears to be.
When two such surfaces are pressed against each other, the peaks form good material contact but the valleys form voids filled with air in most cases. As a result, an interface contains numerous air gaps of varying sizes that act as insulation because of the low thermal conductivity of air. Thus, an interface offers some resistance to heat transfer, and this resistance for a unit interface area is called the thermal contact resistance, Rc. The value of Rc is deter-mined experimentally using a setup like the one shown in Fig. 3–15, and as expected, there is considerable scatter of data because of the difficulty in char-acterizing the surfaces.
Consider heat transfer through two metal rods of cross-sectional area A that are pressed against each other. Heat transfer through the interface of these two rods is the sum of the heat transfers through the solid contact spots (solid-to-solid conduction) and the gaps (conduction and/or radiation across the gaps) in the noncontact areas (which is a major contributor to heat transfer) and can be expressed as contact gap (3–25) It can also be expressed in an analogous manner to Newton’s law of cooling as hc A Tinterface (3–26) Q Q Q Q ■ 146 STEADY HEAT CONDUCTION FIGURE 3–14 Temperature distribution and heat flow lines along two solid plates pressed against each other for the case of perfect and imperfect contact.
Layer 1 (a) Ideal (perfect) thermal contact (b) Actual (imperfect) thermal contact Layer 2 Layer 1 Layer 2 Temperature drop ΔT No temperature drop Temperature distribution Interface T1 = T2 T1 T2 Interface Cold fluid Applied load Loading shaft Alignment collar Top plate Steel ball Pencil heaters Heaters block Bell jar base plate Bottom plate Steel ball Lower heat flux meter Lower test specimen Upper test specimen Load cell Cold plate Thermocouples Interface FIGURE 3–15 A typical experimental setup for the determination of thermal contact resistance (from Song et al.).
where A is the apparent interface area (which is the same as the cross-sectional area of the rods) and Tinterface is the effective temperature difference at the interface. The quantity hc, which corresponds to the convection heat transfer coefficient, is called the thermal contact conductance and is expressed as hc (W/m2 ) (3–27) It is related to thermal contact resistance by Rc (m2K/W) (3–28) That is, thermal contact resistance is the inverse of thermal contact conduc-tance. Usually, thermal contact conductance is reported in the literature, but the concept of thermal contact resistance serves as a better vehicle for ex-plaining the effect of interface on heat transfer. Note that Rc represents thermal contact resistance for a unit area. The thermal resistance for the entire inter-face is obtained by dividing Rc by the apparent interface area A.
The thermal contact resistance can be determined from Eq. 3–28 by mea-suring the temperature drop at the interface and dividing it by the heat flux un-der steady conditions. The value of thermal contact resistance depends on the surface roughness and the material properties as well as the temperature and pressure at the interface and the type of fluid trapped at the interface. The sit-uation becomes more complex when plates are fastened by bolts, screws, or rivets since the interface pressure in this case is nonuniform. The thermal con-tact resistance in that case also depends on the plate thickness, the bolt radius, and the size of the contact zone. Thermal contact resistance is observed to decrease with decreasing surface roughness and increasing interface pres-sure, as expected. Most experimentally determined values of the thermal con-tact resistance fall between 0.000005 and 0.0005 m2·K/W (the corresponding range of thermal contact conductance is 2000 to 200,000 W/m2·K).
When we analyze heat transfer in a medium consisting of two or more lay-ers, the first thing we need to know is whether the thermal contact resistance is significant or not. We can answer this question by comparing the magni-tudes of the thermal resistances of the layers with typical values of thermal contact resistance. For example, the thermal resistance of a 1-cm-thick layer of an insulating material for a unit surface area is Rc, insulation 0.25 m2K/W whereas for a 1-cm-thick layer of copper, it is Rc, copper 0.000026 m2K/W Comparing the values above with typical values of thermal contact resistance, we conclude that thermal contact resistance is significant and can even domi-nate the heat transfer for good heat conductors such as metals, but can be L k 0.01 m 386 W/mK L k 0.01 m 0.04 W/mK 1 hc Tinterface Q/A Q/A Tinterface 147 CHAPTER 3 disregarded for poor heat conductors such as insulations. This is not surpris-ing since insulating materials consist mostly of air space just like the inter-face itself.
The thermal contact resistance can be minimized by applying a thermally conducting liquid called a thermal grease such as silicon oil on the surfaces before they are pressed against each other. This is commonly done when at-taching electronic components such as power transistors to heat sinks. The thermal contact resistance can also be reduced by replacing the air at the in-terface by a better conducting gas such as helium or hydrogen, as shown in Table 3–1.
Another way to minimize the contact resistance is to insert a soft metallic foil such as tin, silver, copper, nickel, or aluminum between the two surfaces.
Experimental studies show that the thermal contact resistance can be reduced by a factor of up to 7 by a metallic foil at the interface. For maximum effec-tiveness, the foils must be very thin. The effect of metallic coatings on thermal contact conductance is shown in Fig. 3–16 for various metal surfaces.
There is considerable uncertainty in the contact conductance data reported in the literature, and care should be exercised when using them. In Table 3–2 some experimental results are given for the contact conductance between sim-ilar and dissimilar metal surfaces for use in preliminary design calculations.
Note that the thermal contact conductance is highest (and thus the contact re-sistance is lowest) for soft metals with smooth surfaces at high pressure.
148 STEADY HEAT CONDUCTION EXAMPLE 3–4 Equivalent Thickness for Contact Resistance The thermal contact conductance at the interface of two 1-cm-thick aluminum plates is measured to be 11,000 W/m2·K. Determine the thickness of the alu-minum plate whose thermal resistance is equal to the thermal resistance of the interface between the plates (Fig. 3–17).
SOLUTION The thickness of the aluminum plate whose thermal resistance is equal to the thermal contact resistance is to be determined.
Properties The thermal conductivity of aluminum at room temperature is k 237 W/m·K (Table A–3).
Analysis Noting that thermal contact resistance is the inverse of thermal con-tact conductance, the thermal contact resistance is Rc 0.909 104 m2K/W For a unit surface area, the thermal resistance of a flat plate is defined as R where L is the thickness of the plate and k is the thermal conductivity. Setting R Rc, the equivalent thickness is determined from the relation above to be L kRc (237 W/mK)(0.909 104 m2K/W) 0.0215 m 2.15 cm L k 1 hc 1 11,000 W/m2K 105 102 103 Bronze Uncoated Coated with nickel alloy Coated with aluminum alloy Nickel Stainless Steel Coated with tin/nickel alloy 104 103 Thermal contact conductance (W/m2·K) Thermal contact conductance Btu ——–— h·ft2·°F ( ) 104 103 102 102 103 Contact pressure (kN/m2) Contact pressure (psi) 104 Coated FIGURE 3–16 Effect of metallic coatings on thermal contact conductance (from Peterson, 1987).
TABLE 3–1 Thermal contact conductance for aluminum plates with different fluids at the interface for a surface roughness of 10 μm and interface pressure of 1 atm (from Fried, 1969).
Contact Fluid at the conductance, hc, interface W/m2·K Air 3640 Helium 9520 Hydrogen 13,900 Silicone oil 19,000 Glycerin 37,700 149 CHAPTER 3 Discussion Note that the interface between the two plates offers as much re-sistance to heat transfer as a 2.15-cm-thick aluminum plate. It is interesting that the thermal contact resistance in this case is greater than the sum of the thermal resistances of both plates.
EXAMPLE 3–5 Contact Resistance of Transistors Four identical power transistors with aluminum casing are attached on one side of a 1-cm-thick 20-cm 20-cm square copper plate (k 386 W/m·K) by screws that exert an average pressure of 6 MPa (Fig. 3–18). The base area of each transistor is 8 cm2, and each transistor is placed at the center of a 10-cm 10-cm quarter section of the plate. The interface roughness is estimated to be about 1.5 μm. All transistors are covered by a thick Plexiglas layer, which is a poor conductor of heat, and thus all the heat generated at the junction of the transistor must be dissipated to the ambient at 20°C through the back surface of the copper plate. The combined convection/radiation heat transfer coefficient at the back surface can be taken to be 25 W/m2·K. If the case temperature of Plate 1 1 cm Plate 2 1 cm Plate 1 Equivalent aluminum layer 1 cm Plate 2 2.15 cm 1 cm Interface FIGURE 3–17 Schematic for Example 3–4.
TABLE 3–2 Thermal contact conductance of some metal surfaces in air (from various sources) Surface Pressure, hc, Material condition Roughness, μm Temperature, C MPa W/m2·K Identical Metal Pairs 416 Stainless steel Ground 2.54 90–200 0.17–2.5 3800 304 Stainless steel Ground 1.14 20 4–7 1900 Aluminum Ground 2.54 150 1.2–2.5 11,400 Copper Ground 1.27 20 1.2–20 143,000 Copper Milled 3.81 20 1–5 55,500 Copper (vacuum) Milled 0.25 30 0.17–7 11,400 Dissimilar Metal Pairs Stainless steel– 10 2900 Aluminum 20–30 20 20 3600 Stainless steel– 10 16,400 Aluminum 1.0–2.0 20 20 20,800 Steel Ct-30– 10 50,000 Aluminum Ground 1.4–2.0 20 15–35 59,000 Steel Ct-30– 10 4800 Aluminum Milled 4.5–7.2 20 30 8300 5 42,000 Aluminum-Copper Ground 1.17–1.4 20 15 56,000 10 12,000 Aluminum-Copper Milled 4.4–4.5 20 20–35 22,000 Divide the given values by 5.678 to convert to Btu/h·ft2·F.
150 STEADY HEAT CONDUCTION the transistor is not to exceed 70°C, determine the maximum power each transistor can dissipate safely, and the temperature jump at the case-plate interface.
SOLUTION Four identical power transistors are attached on a copper plate.
For a maximum case temperature of 70°C, the maximum power dissipation and the temperature jump at the interface are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer can be ap-proximated as being one-dimensional, although it is recognized that heat con-duction in some parts of the plate will be two-dimensional since the plate area is much larger than the base area of the transistor. But the large thermal con-ductivity of copper will minimize this effect. 3 All the heat generated at the junction is dissipated through the back surface of the plate since the transis-tors are covered by a thick Plexiglas layer. 4 Thermal conductivities are constant.
Properties The thermal conductivity of copper is given to be k 386 W/m·K.
The contact conductance is obtained from Table 3–2 to be hc 42,000 W/m2·K, which corresponds to copper-aluminum interface for the case of 1.17–1.4 μm roughness and 5 MPa pressure, which is sufficiently close to what we have.
Analysis The contact area between the case and the plate is given to be 8 cm2, and the plate area for each transistor is 100 cm2. The thermal resis-tance network of this problem consists of three resistances in series (interface, plate, and convection), which are determined to be Rinterface 0.030C/W Rplate 0.0026C/W Rconv 4.0C/W The total thermal resistance is then Rtotal Rinterface Rplate Rambient 0.030 0.0026 4.0 4.0326C/W Note that the thermal resistance of a copper plate is very small and can be ignored altogether. Then the rate of heat transfer is determined to be 12.4 W Therefore, the power transistor should not be operated at power levels greater than 12.4 W if the case temperature is not to exceed 70°C.
The temperature jump at the interface is determined from Tinterface R interface (12.4 W)(0.030C/W) 0.37C which is not very large. Therefore, even if we eliminate the thermal contact re-sistance at the interface completely, we lower the operating temperature of the transistor in this case by less than 0.4°C.
Q T Rtotal (70 20)C 4.0326C/ W Q 1 ho A 1 (25 W/m2K)(0.01 m2) L kA 0.01 m (386 W/mK)(0.01 m2) 1 hc Ac 1 (42,000 W/m2K)(8 104 m2) 20°C Copper plate 20 cm 1 cm Plexiglas cover 70°C FIGURE 3–18 Schematic for Example 3–5.
3–3 GENERALIZED THERMAL RESISTANCE NETWORKS The thermal resistance concept or the electrical analogy can also be used to solve steady heat transfer problems that involve parallel layers or combined series-parallel arrangements. Although such problems are often two- or even three-dimensional, approximate solutions can be obtained by assuming one-dimensional heat transfer and using the thermal resistance network.
Consider the composite wall shown in Fig. 3–19, which consists of two par-allel layers. The thermal resistance network, which consists of two parallel re-sistances, can be represented as shown in the figure. Noting that the total heat transfer is the sum of the heat transfers through each layer, we have 1 2 (T1 T2) (3–29) Utilizing electrical analogy, we get (3–30) where ⎯→ Rtotal (3–31) since the resistances are in parallel.
Now consider the combined series-parallel arrangement shown in Fig. 3–20.
The total rate of heat transfer through this composite system can again be ex-pressed as (3–32) where Rtotal R12 R3 Rconv R3 Rconv (3–33) and R1 R2 R3 Rconv (3–34) Once the individual thermal resistances are evaluated, the total resistance and the total rate of heat transfer can easily be determined from the relations above.
The result obtained is somewhat approximate, since the surfaces of the third layer are probably not isothermal, and heat transfer between the first two lay-ers is likely to occur.
Two assumptions commonly used in solving complex multidimensional heat transfer problems by treating them as one-dimensional (say, in the 1 hA3 L 3 k3 A3 L 2 k2 A2 L1 k1 A1 R1R2 R1 R2 T1 T Rtotal Q R1R2 R1 R2 1 Rtotal 1 R1 1 R2 T1 T2 Rtotal Q a 1 R1 1 R2b T1 T2 R1 T1 T2 R2 Q Q Q ■ 151 CHAPTER 3 T1 T2 R1 R2 k1 A1 A2 T1 T2 k2 Insulation L 1 2 Q2 · Q = Q1 + Q2 · · · Q1 · Q · Q · FIGURE 3–19 Thermal resistance network for two parallel layers.
T1 R1 R2 R3 Rconv k1 A1 A2 T1 A3 k2 Insulation 1 2 3 k3 T Q2 · Q1 · Q · Q · h, T L1 = L2 L3 FIGURE 3–20 Thermal resistance network for combined series-parallel arrangement.
x-direction) using the thermal resistance network are (1) any plane wall nor-mal to the x-axis is isothermal (i.e., to assume the temperature to vary in the x-direction only) and (2) any plane parallel to the x-axis is adiabatic (i.e., to assume heat transfer to occur in the x-direction only). These two assumptions result in different resistance networks, and thus different (but usually close) values for the total thermal resistance and thus heat transfer. The actual result lies between these two values. In geometries in which heat transfer occurs pre-dominantly in one direction, either approach gives satisfactory results.
152 STEADY HEAT CONDUCTION EXAMPLE 3–6 Heat Loss through a Composite Wall A 3-m-high and 5-m-wide wall consists of long 16-cm 22-cm cross section horizontal bricks (k 0.72 W/m·K) separated by 3-cm-thick plaster layers (k 0.22 W/m·K). There are also 2-cm-thick plaster layers on each side of the brick and a 3-cm-thick rigid foam (k 0.026 W/m·K) on the inner side of the wall, as shown in Fig. 3–21. The indoor and the outdoor temperatures are 20°C and 10°C, respectively, and the convection heat transfer coefficients on the inner and the outer sides are h1 10 W/m2·K and h2 25 W/m2·K, respec-tively. Assuming one-dimensional heat transfer and disregarding radiation, de-termine the rate of heat transfer through the wall.
SOLUTION The composition of a composite wall is given. The rate of heat transfer through the wall is to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer can be approximated as being one-dimensional since it is predominantly in the x-direction. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is negligible.
Properties The thermal conductivities are given to be k 0.72 W/m·K for bricks, k 0.22 W/m·K for plaster layers, and k 0.026 W/m·K for the rigid foam.
Analysis There is a pattern in the construction of this wall that repeats itself every 25-cm distance in the vertical direction. There is no variation in the hor-izontal direction. Therefore, we consider a 1-m-deep and 0.25-m-high portion of the wall, since it is representative of the entire wall.
Assuming any cross section of the wall normal to the x-direction to be isothermal, the thermal resistance network for the representative section of the wall becomes as shown in Fig. 3–21. The individual resistances are evalu-ated as: Ri Rconv, 1 0.40C/W R1 Rfoam 4.62C/W R2 R6 Rplaster, side 0.36C/W R3 R5 Rplaster, center 48.48C/W L kA 0.16 m (0.22 W/mK)(0.015 1 m2) L kA 0.02 m (0.22 W/mK)(0.25 1 m2) L kA 0.03 m (0.026 W/mK)(0.25 1 m2) 1 h1 A 1 (10 W/m2K)(0.25 1 m2) h1 T1 h2 T2 22 cm 1.5 cm 1.5 cm 3 x 2 16 cm 2 Plaster R5 R4 R3 R6 Ri R1 R2 Ro T2 T1 Brick Foam FIGURE 3–21 Schematic for Example 3–6.
153 CHAPTER 3 R4 Rbrick 1.01C/W Ro Rconv, 2 0.16C/W The three resistances R3, R4, and R5 in the middle are parallel, and their equiv-alent resistance is determined from 1.03 W/C which gives Rmid 0.97C/W Now all the resistances are in series, and the total resistance is Rtotal Ri R1 R2 Rmid R6 Ro 0.40 4.62 0.36 0.97 0.36 0.16 6.87C/W Then the steady rate of heat transfer through the wall becomes 4.37 W (per 0.25 m2 surface area) or 4.37/0.25 17.5 W per m2 area. The total area of the wall is A 3 m 5 m 15 m2. Then the rate of heat transfer through the entire wall becomes total (17.5 W/m2)(15 m2) 263 W Of course, this result is approximate, since we assumed the temperature within the wall to vary in one direction only and ignored any temperature change (and thus heat transfer) in the other two directions.
Discussion In the above solution, we assumed the temperature at any cross section of the wall normal to the x-direction to be isothermal. We could also solve this problem by going to the other extreme and assuming the surfaces par-allel to the x-direction to be adiabatic. The thermal resistance network in this case will be as shown in Fig. 3–22. By following the approach outlined above, the total thermal resistance in this case is determined to be Rtotal 6.97°C/W, which is very close to the value 6.85°C/W obtained before. Thus either approach gives roughly the same result in this case. This example demon-strates that either approach can be used in practice to obtain satisfactory results.
Q T1 T2 Rtotal [20 (10)]C 6.87C/ W Q 1 Rmid 1 R3 1 R4 1 R5 1 48.48 1 1.01 1 48.48 1 h2 A 1 (25 W/m2K)(0.25 1 m2) L kA 0.16 m (0.72 W/mK)(0.22 1 m2) x Ro T2 Ri T1 Adiabatic lines FIGURE 3–22 Alternative thermal resistance network for Example 3–6 for the case of surfaces parallel to the primary direction of heat transfer being adiabatic.
3–4 HEAT CONDUCTION IN CYLINDERS AND SPHERES Consider steady heat conduction through a hot-water pipe. Heat is continu-ously lost to the outdoors through the wall of the pipe, and we intuitively feel that heat transfer through the pipe is in the normal direction to the pipe surface and no significant heat transfer takes place in the pipe in other directions (Fig. 3–23). The wall of the pipe, whose thickness is rather small, separates two fluids at different temperatures, and thus the temperature gradient in the radial direction is relatively large. Further, if the fluid temperatures inside and outside the pipe remain constant, then heat transfer through the pipe is steady. Thus heat transfer through the pipe can be modeled as steady and one-dimensional. The temperature of the pipe in this case depends on one direction only (the radial r-direction) and can be expressed as T T(r). The tempera-ture is independent of the azimuthal angle or the axial distance. This situation is approximated in practice in long cylindrical pipes and spherical containers.
In steady operation, there is no change in the temperature of the pipe with time at any point. Therefore, the rate of heat transfer into the pipe must be equal to the rate of heat transfer out of it. In other words, heat transfer through the pipe must be constant, cond, cyl constant.
Consider a long cylindrical layer (such as a circular pipe) of inner radius r1, outer radius r2, length L, and average thermal conductivity k (Fig. 3–24). The two surfaces of the cylindrical layer are maintained at constant temperatures T1 and T2. There is no heat generation in the layer and the thermal conductiv-ity is constant. For one-dimensional heat conduction through the cylindrical layer, we have T(r). Then Fourier’s law of heat conduction for heat transfer through the cylindrical layer can be expressed as cond, cyl kA (W) (3–35) where A 2prL is the heat transfer area at location r. Note that A depends on r, and thus it varies in the direction of heat transfer. Separating the variables in the above equation and integrating from r r1, where T(r1) T1, to r r2, where T(r2) T2, gives dr k dT (3–36) Substituting A 2prL and performing the integrations give cond, cyl 2pLk (W) (3–37) since cond, cyl constant. This equation can be rearranged as cond, cyl (W) (3–38) where Rcyl (3–39) ln(Outer radius/Inner radius) 2p Length Thermal conductivity ln(r2/r1) 2pLk T1 T2 Rcyl Q Q T1 T2 ln(r2/r1) Q T2 TT1 r2 rr1 Q cond, cyl A dT dr Q Q ■ 154 STEADY HEAT CONDUCTION h T r Q · FIGURE 3–23 Heat is lost from a hot-water pipe to the air outside in the radial direction, and thus heat transfer from a long pipe is one-dimensional.
r1 k r2 T1 T2 FIGURE 3–24 A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T1 and T2.
is the thermal resistance of the cylindrical layer against heat conduction, or simply the conduction resistance of the cylinder layer. Note that Eq. 3–37 is identical to Eq. 2–59 which was obtained by using the “standard” approach by first solving the heat conduction equation in cylindrical coordinates, Eq. 2–29, to obtain the temperature distribution, Eq. 2–58, and then using the Fourier’s law to obtain the heat transfer rate. The method used in obtaining Eq. 3–37 can be considered an “alternative” approach. However, it is restricted to one-dimensional steady heat conduction with no heat generation.
We can repeat the analysis for a spherical layer by taking A 4pr2 and per-forming the integrations in Eq. 3–36. The result can be expressed as cond, sph (3–40) where Rsph (3–41) is the thermal resistance of the spherical layer against heat conduction, or sim-ply the conduction resistance of the spherical layer. Note also that Eq. 3–40 is identical to Eq. 2–61 which was obtained by solving the heat conduction equation in spherical coordinates.
Now consider steady one-dimensional heat transfer through a cylindrical or spherical layer that is exposed to convection on both sides to fluids at tempera-tures T1 and T2 with heat transfer coefficients h1 and h2, respectively, as shown in Fig. 3–25. The thermal resistance network in this case consists of one con-duction and two convection resistances in series, just like the one for the plane wall, and the rate of heat transfer under steady conditions can be expressed as (3–42) where Rtotal Rconv, 1 Rcyl Rconv, 2 (3–43) for a cylindrical layer, and Rtotal Rconv, 1 Rsph Rconv, 2 (3–44) for a spherical layer. Note that A in the convection resistance relation Rconv 1/hA is the surface area at which convection occurs. It is equal to A 2prL for a cylindrical surface and A 4pr2 for a spherical surface of radius r. Also note that the thermal resistances are in series, and thus the total thermal resistance is determined by simply adding the individual resistances, just like the electrical resistances connected in series.
1 (4pr 2 1)h1 r2 r1 4pr1r2k 1 (4pr 2 2)h2 1 (2pr1L)h1 ln(r2/r1) 2pLk 1 (2pr2L)h2 T1 T2 Rtotal Q Outer radius Inner radius 4p(Outer radius)(Inner radius)(Thermal conductivity) r2 r1 4pr1r2k T1 T2 Rsph Q 155 CHAPTER 3 FIGURE 3–25 The thermal resistance network for a cylindrical (or spherical) shell subjected to convection from both the inner and the outer sides.
r1 r2 T1 Rconv,2 Rconv,1 Rtotal = Rconv,1 + Rcyl + Rconv,2 Rcyl T2 T2 T1 h2 h1 Q · Multilayered Cylinders and Spheres Steady heat transfer through multilayered cylindrical or spherical shells can be handled just like multilayered plane walls discussed earlier by simply adding an additional resistance in series for each additional layer. For example, the steady heat transfer rate through the three-layered composite cylinder of length L shown in Fig. 3–26 with convection on both sides can be expressed as (3–45) where Rtotal is the total thermal resistance, expressed as Rtotal Rconv, 1 Rcyl, 1 Rcyl, 2 Rcyl, 3 Rconv, 2 (3–46) where A1 2pr1L and A4 2pr4L. Equation 3–46 can also be used for a three-layered spherical shell by replacing the thermal resistances of cylindri-cal layers by the corresponding spherical ones. Again, note from the thermal resistance network that the resistances are in series, and thus the total thermal resistance is simply the arithmetic sum of the individual thermal resistances in the path of heat flow.
Once Q . is known, we can determine any intermediate temperature Tj by applying the relation Q . (Ti Tj)/Rtotal, i j across any layer or layers such that Ti is a known temperature at location i and Rtotal, i j is the total thermal re-sistance between locations i and j (Fig. 3–27). For example, once Q . has been calculated, the interface temperature T2 between the first and second cylindri-cal layers can be determined from 1 h1 A1 ln(r2/r1) 2pLk1 ln(r3/r2) 2pLk2 ln(r4/r3) 2pLk3 1 h2 A4 T1 T2 Rtotal Q 156 STEADY HEAT CONDUCTION FIGURE 3–26 The thermal resistance network for heat transfer through a three-layered composite cylinder subjected to convection on both sides.
T2 T4 T3 T2 T1 h1 r1 r2 r3 r4 T1 T1 h2 T2 Rconv,1 Rcyl,1 Rcyl,2 Rcyl,3 Rconv,2 k1 k2 k3 1 2 3 (3–47) We could also calculate T2 from (3–48) Although both relations give the same result, we prefer the first one since it in-volves fewer terms and thus less work.
The thermal resistance concept can also be used for other geometries, pro-vided that the proper conduction resistances and the proper surface areas in convection resistances are used.
T2 T2 R2 R3 Rconv, 2 T2 T2 ln(r3/r2) 2pLk2 ln(r4/r3) 2pLk3 1 ho(2pr4L) Q T1 T2 Rconv, 1 Rcyl, 1 T1 T2 1 h1(2pr1L) ln(r2/r1) 2pLk1 Q 157 CHAPTER 3 EXAMPLE 3–7 Heat Transfer to a Spherical Container A 3-m internal diameter spherical tank made of 2-cm-thick stainless steel (k 15 W/m·K) is used to store iced water at T1 0°C. The tank is located in a room whose temperature is T2 22°C. The walls of the room are also at 22°C.
The outer surface of the tank is black and heat transfer between the outer sur-face of the tank and the surroundings is by natural convection and radiation.
The convection heat transfer coefficients at the inner and the outer surfaces of the tank are h1 80 W/m2·K and h2 10 W/m2·K, respectively. Determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0°C that melts during a 24-h period.
SOLUTION A spherical container filled with iced water is subjected to con-vection and radiation heat transfer at its outer surface. The rate of heat trans-fer and the amount of ice that melts per day are to be determined.
Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the midpoint. 3 Thermal conductivity is constant.
Properties The thermal conductivity of steel is given to be k 15 W/m·K. The heat of fusion of water at atmospheric pressure is hif 333.7 kJ/kg. The outer surface of the tank is black and thus its emissivity is e 1.
Analysis (a) The thermal resistance network for this problem is given in Fig. 3–28. Noting that the inner diameter of the tank is D1 3 m and the outer diameter is D2 3.04 m, the inner and the outer surface areas of the tank are A1 p p(3 m)2 28.3 m2 A2 p p(3.04 m)2 29.0 m2 Also, the radiation heat transfer coefficient is given by hrad es( )(T2 T2) But we do not know the outer surface temperature T2 of the tank, and thus we cannot calculate hrad. Therefore, we need to assume a T2 value now and check T 2 2 T 2 2 D2 2 D2 1 T1 – T1 Rconv,1 = ———— T1 – T2 Rconv,1 + R1 = ———— T1 – T3 R1 + R2 = ———— T2 – T3 R2 = ———— T2 – T2 R2 + Rconv,2 = ———— = . . .
T2 T1 R1 T1 T2 T3 Rconv,1 Rconv,2 R2 Q · FIGURE 3–27 The ratio T/R across any layer is equal to , which remains constant in one-dimensional steady conduction.
Q T1 T1 T2 Ri R1 Rrad Ro T2 h2 1.5 m 2 cm 0°C Iced water h1 FIGURE 3–28 Schematic for Example 3–7.
158 STEADY HEAT CONDUCTION the accuracy of this assumption later. We will repeat the calculations if neces-sary using a revised value for T2.
We note that T2 must be between 0°C and 22°C, but it must be closer to 0°C, since the heat transfer coefficient inside the tank is much larger. Taking T2 5°C 278 K, the radiation heat transfer coefficient is determined to be hrad (1)(5.67 108 W/m2K4)[(295 K)2 (278 K)2][(295 278) K] 5.34 W/m2K 5.34 W/m2C Then the individual thermal resistances become Ri Rconv, 1 0.000442C/W R1 Rsphere 0.000047C/W Ro Rconv, 2 0.00345C/W Rrad 0.00646C/W The two parallel resistances Ro and Rrad can be replaced by an equivalent resistance Requiv determined from 444.7 W/C which gives Requiv 0.00225°C/W Now all the resistances are in series, and the total resistance is Rtotal Ri R1 Requiv 0.000442 0.000047 0.00225 0.00274°C/W Then the steady rate of heat transfer to the iced water becomes 8029 W (or 8.029 kJ/s) To check the validity of our original assumption, we now determine the outer surface temperature from ⎯→ T2 T2 R equiv 22C (8029 W)(0.00225C/W) 4C which is sufficiently close to the 5°C assumed in the determination of the ra-diation heat transfer coefficient. Therefore, there is no need to repeat the cal-culations using 4°C for T2.
Q T2 T2 Requiv Q Q T2 T1 Rtotal (22 0)C 0.00274C/ W Q 1 Requiv 1 Ro 1 Rrad 1 0.00345 1 0.00646 1 hrad A2 1 (5.34 W/m2K)(29.0 m2) 1 h2 A2 1 (10 W/m2K)(29.0 m2) r2 r1 4pkr1r2 (1.52 1.50) m 4p (15 W/mK)(1.52 m)(1.50 m) 1 h1 A1 1 (80 W/m2K)(28.3 m2) 159 CHAPTER 3 (b) The total amount of heat transfer during a 24-h period is Q t (8.029 kJ/s)(24 3600 s) 693,700 kJ Noting that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the amount of ice that will melt during a 24-h period is mice 2079 kg Therefore, about 2 metric tons of ice will melt in the tank every day.
Discussion An easier way to deal with combined convection and radiation at a surface when the surrounding medium and surfaces are at the same tempera-ture is to add the radiation and convection heat transfer coefficients and to treat the result as the convection heat transfer coefficient. That is, to take h 10 5.34 15.34 W/m2·K in this case. This way, we can ignore radiation since its contribution is accounted for in the convection heat transfer coeffi-cient. The convection resistance of the outer surface in this case would be Rcombined 0.00225C/W which is identical to the value obtained for equivalent resistance for the paral-lel convection and the radiation resistances.
1 hcombined A2 1 (15.34 W/m2K)(29.0 m2) Q hif 693,700 kJ 333.7 kJ/kg Q EXAMPLE 3–8 Heat Loss through an Insulated Steam Pipe Steam at T1 320°C flows in a cast iron pipe (k 80 W/m·K) whose inner and outer diameters are D1 5 cm and D2 5.5 cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation with k 0.05 W/m·K. Heat is lost to the surroundings at T2 5°C by natural convection and radiation, with a combined heat transfer coefficient of h2 18 W/m2·K. Taking the heat trans-fer coefficient inside the pipe to be h1 60 W/m2·K, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the tem-perature drops across the pipe shell and the insulation.
SOLUTION A steam pipe covered with glass wool insulation is subjected to convection on its surfaces. The rate of heat transfer per unit length and the temperature drops across the pipe and the insulation are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Ther-mal conductivities are constant. 4 The thermal contact resistance at the inter-face is negligible.
Properties The thermal conductivities are given to be k 80 W/m·K for cast iron and k 0.05 W/m·K for glass wool insulation.
3–5 CRITICAL RADIUS OF INSULATION We know that adding more insulation to a wall or to the attic always decreases heat transfer. The thicker the insulation, the lower the heat transfer rate. This is expected, since the heat transfer area A is constant, and adding insulation al-ways increases the thermal resistance of the wall without increasing the con-vection resistance.
Adding insulation to a cylindrical pipe or a spherical shell, however, is a dif-ferent matter. The additional insulation increases the conduction resistance of ■ 160 STEADY HEAT CONDUCTION FIGURE 3–29 Schematic for Example 3–8.
Analysis The thermal resistance network for this problem involves four resis-tances in series and is given in Fig. 3–29. Taking L 1 m, the areas of the surfaces exposed to convection are determined to be A1 2pr1L 2p(0.025 m)(1 m) 0.157 m2 A3 2pr3L 2p(0.0575 m)(1 m) 0.361 m2 Then the individual thermal resistances become Ri Rconv, 1 0.106C/W R1 Rpipe 0.0002C/W R2 Rinsulation 2.35C/W Ro Rconv, 2 0.154C/W Noting that all resistances are in series, the total resistance is determined to be Rtotal Ri R1 R2 Ro 0.106 0.0002 2.35 0.154 2.61°C/W Then the steady rate of heat loss from the steam becomes 121 W (per m pipe length) The heat loss for a given pipe length can be determined by multiplying the above quantity by the pipe length L.
The temperature drops across the pipe and the insulation are determined from Eq. 3–17 to be Tpipe R pipe (121 W)(0.0002C/W) 0.02C Tinsulation R insulation (121 W)(2.35C/W) 284C That is, the temperatures between the inner and the outer surfaces of the pipe differ by 0.02°C, whereas the temperatures between the inner and the outer surfaces of the insulation differ by 284°C.
Discussion Note that the thermal resistance of the pipe is too small relative to the other resistances and can be neglected without causing any significant error. Also note that the temperature drop across the pipe is practically zero, and thus the pipe can be assumed to be isothermal. The resistance to heat flow in insulated pipes is primarily due to insulation.
Q Q T1 T2 Rtotal (320 5)C 2.61C/W Q 1 h2 A3 1 (18 W/m2K)(0.361 m2) ln(r3 /r2) 2pk2L ln(5.75/2.75) 2p(0.05 W/mK)(1 m) ln(r2 /r1) 2pk1L ln(2.75/2.5) 2p(80 W/mK)(1 m) 1 h1 A1 1 (60 W/m2K)(0.157 m2) the insulation layer but decreases the convection resistance of the surface be-cause of the increase in the outer surface area for convection. The heat trans-fer from the pipe may increase or decrease, depending on which effect dominates.
Consider a cylindrical pipe of outer radius r1 whose outer surface tempera-ture T1 is maintained constant (Fig. 3–30). The pipe is now insulated with a material whose thermal conductivity is k and outer radius is r2. Heat is lost from the pipe to the surrounding medium at temperature T, with a convection heat transfer coefficient h. The rate of heat transfer from the insulated pipe to the surrounding air can be expressed as (Fig. 3–31) (3–49) The variation of with the outer radius of the insulation r2 is plotted in Fig. 3–31. The value of r2 at which reaches a maximum is determined from Q Q T1 T Rins Rconv T1 T ln(r2/r1) 2pLk 1 h(2pr2L) Q 161 CHAPTER 3 Rconv r2 Insulation T T h T1 Rins r1 k FIGURE 3–30 An insulated cylindrical pipe exposed to convection from the outer surface and the thermal resistance network associated with it.
r2 k h r1 r1 0 r2 rcr = k/h Q · Q · Qmax · Qbare · FIGURE 3–31 The variation of heat transfer rate with the outer radius of the insulation r2 when r1 rcr.
the requirement that d /dr2 0 (zero slope). Performing the differentiation and solving for r2 yields the critical radius of insulation for a cylindrical body to be rcr, cylinder (m) (3–50) Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h. The rate of heat transfer from the cylinder increases with the addition of insulation for r2 rcr, reaches a maximum when r2 rcr, and starts to decrease for r2 rcr. Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r2 rcr.
The important question to answer at this point is whether we need to be con-cerned about the critical radius of insulation when insulating hot-water pipes or even hot-water tanks. Should we always check and make sure that the outer radius of insulation sufficiently exceeds the critical radius before we install any insulation? Probably not, as explained here.
The value of the critical radius rcr is the largest when k is large and h is small. Noting that the lowest value of h encountered in practice is about 5 W/m2·K for the case of natural convection of gases, and that the thermal conductivity of common insulating materials is about 0.05 W/m2·K, the largest value of the critical radius we are likely to encounter is rcr, max 0.01 m 1 cm This value would be even smaller when the radiation effects are considered.
The critical radius would be much less in forced convection, often less than 1 mm, because of much larger h values associated with forced convection.
Therefore, we can insulate hot-water or steam pipes freely without worrying about the possibility of increasing the heat transfer by insulating the pipes.
The radius of electric wires may be smaller than the critical radius. There-fore, the plastic electrical insulation may actually enhance the heat transfer 0.05 W/mK 5 W/m2 K kmax, insulation hmin k h Q from electric wires and thus keep their steady operating temperatures at lower and thus safer levels.
The discussions above can be repeated for a sphere, and it can be shown in a similar manner that the critical radius of insulation for a spherical shell is rcr, sphere (3–51) where k is the thermal conductivity of the insulation and h is the convection heat transfer coefficient on the outer surface.
2k h 162 STEADY HEAT CONDUCTION EXAMPLE 3–9 Heat Loss from an Insulated Electric Wire A 3-mm-diameter and 5-m-long electric wire is tightly wrapped with a 2-mm-thick plastic cover whose thermal conductivity is k 0.15 W/m·K. Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a medium at T 30°C with a heat transfer coefficient of h 12 W/m2·K, de-termine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature.
SOLUTION An electric wire is tightly wrapped with a plastic cover. The inter-face temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Ther-mal conductivities are constant. 4 The thermal contact resistance at the inter-face is negligible. 5 Heat transfer coefficient incorporates the radiation effects, if any.
Properties The thermal conductivity of plastic is given to be k 0.15 W/m·K.
Analysis Heat is generated in the wire and its temperature rises as a result of resistance heating. We assume heat is generated uniformly throughout the wire and is transferred to the surrounding medium in the radial direction. In steady operation, the rate of heat transfer becomes equal to the heat generated within the wire, which is determined to be VI (8 V)(10 A) 80 W The thermal resistance network for this problem involves a conduction resis-tance for the plastic cover and a convection resistance for the outer surface in series, as shown in Fig. 3–32. The values of these two resistances are A2 (2pr2)L 2p(0.0035 m)(5 m) 0.110 m2 Rconv 0.76C/W Rplastic 0.18C/W ln(r2 /r1) 2pkL ln(3.5/1.5) 2p(0.15 W/mK)(5 m) 1 hA2 1 (12 W/m2K)(0.110 m2) W e Q h T T1 T2 T Rplastic Rconv r2 Q · Q · k T1 r1 T2 FIGURE 3–32 Schematic for Example 3–9.
3–6 HEAT TRANSFER FROM FINNED SURFACES The rate of heat transfer from a surface at a temperature Ts to the surrounding medium at T is given by Newton’s law of cooling as conv hAs(Ts T) where As is the heat transfer surface area and h is the convection heat transfer coefficient. When the temperatures Ts and T are fixed by design considera-tions, as is often the case, there are two ways to increase the rate of heat trans-fer: to increase the convection heat transfer coefficient h or to increase the surface area As. Increasing h may require the installation of a pump or fan, or replacing the existing one with a larger one, but this approach may or may not be practical. Besides, it may not be adequate. The alternative is to increase the surface area by attaching to the surface extended surfaces called fins made of highly conductive materials such as aluminum. Finned surfaces are manufac-tured by extruding, welding, or wrapping a thin metal sheet on a surface. Fins enhance heat transfer from a surface by exposing a larger surface area to con-vection and radiation.
An interesting application of fins from about 150 million years ago, the Jurassic era, is shown in Fig. 3–33. The dinosaur stegosaurus lived during this Q ■ 163 CHAPTER 3 FIGURE 3–33 Presumed cooling fins on dinosaur stegosaurus. (© Alamy RF.) and therefore Rtotal Rplastic Rconv 0.76 0.18 0.94°C/W Then the interface temperature can be determined from ⎯→ T1 T R total 30C (80 W)(0.94C/W) 105C Note that we did not involve the electrical wire directly in the thermal resis-tance network, since the wire involves heat generation.
To answer the second part of the question, we need to know the critical radius of insulation of the plastic cover. It is determined from Eq. 3–50 to be rcr 0.0125 m 12.5 mm which is larger than the radius of the plastic cover. Therefore, increasing the thickness of the plastic cover will enhance heat transfer until the outer radius of the cover reaches 12.5 mm. As a result, the rate of heat transfer will increase when the interface temperature T1 is held constant, or T1 will decrease when is held constant, which is the case here.
Discussion It can be shown by repeating the calculations above for a 4-mm-thick plastic cover that the interface temperature drops to 90.6°C when the thickness of the plastic cover is doubled. It can also be shown in a similar manner that the interface reaches a minimum temperature of 83°C when the outer radius of the plastic cover equals the critical radius.
Q Q k h 0.15 W/mK 12 W/m2K Q T1 T Rtotal Q era and it had two rows of big (and bizarre) bony plates down its back. For a long time, scientists thought that the plates were some kind of armor to pro-tect the vegetarian from predators. We now know that a lot of blood flowed through the plates, and they may have acted like a car radiator. The heart pumped blood through the plates, and the plates acted like cooling fins to cool the blood down.
Finned surfaces are commonly used in practice to enhance heat transfer, and they often increase the rate of heat transfer from a surface severalfold. The car radiator shown in Fig. 3–34 is an example of a finned surface. The closely packed thin metal sheets attached to the hot-water tubes increase the surface area for convection and thus the rate of convection heat transfer from the tubes to the air many times. There are a variety of innovative fin designs available in the market, and they seem to be limited only by imagination (Fig. 3–35).
In the analysis of fins, we consider steady operation with no heat generation in the fin, and we assume the thermal conductivity k of the material to remain constant. We also assume the convection heat transfer coefficient h to be constant and uniform over the entire surface of the fin for convenience in the analysis. We recognize that the convection heat transfer coefficient h, in gen-eral, varies along the fin as well as its circumference, and its value at a point is a strong function of the fluid motion at that point. The value of h is usually much lower at the fin base than it is at the fin tip because the fluid is sur-rounded by solid surfaces near the base, which seriously disrupt its motion to the point of “suffocating” it, while the fluid near the fin tip has little contact with a solid surface and thus encounters little resistance to flow. Therefore, adding too many fins on a surface may actually decrease the overall heat trans-fer when the decrease in h offsets any gain resulting from the increase in the surface area.
Fin Equation Consider a volume element of a fin at location x having a length of x, cross-sectional area of Ac, and a perimeter of p, as shown in Fig. 3–36. Under steady conditions, the energy balance on this volume element can be expressed as ° Rate of heat conduction into the element at x ¢ ° Rate of heat conduction from the element at x x ¢ ° Rate of heat convection from¢ the element 164 STEADY HEAT CONDUCTION FIGURE 3–35 Some innovative fin designs.
FIGURE 3–36 Volume element of a fin at location x having a length of x, cross-sectional area of Ac, and perimeter of p.
FIGURE 3–34 The thin plate fins of a car radiator greatly increase the rate of heat transfer to the air. (left: ©Yunus Çengel, photo by James Kleiser, right: © The McGraw-Hill Companies, Inc./Christopher Kerrigan, Photographer.) or cond, x cond, x x conv where conv h(p x)(T T) Substituting and dividing by x, we obtain hp(T T) 0 (3–52) Taking the limit as x →0 gives hp(T T) 0 (3–53) From Fourier’s law of heat conduction we have cond kAc (3–54) where Ac is the cross-sectional area of the fin at location x. Substitution of this re-lation into Eq. 3–53 gives the differential equation governing heat transfer in fins, hp(T T) 0 (3–55) In general, the cross-sectional area Ac and the perimeter p of a fin vary with x, which makes this differential equation difficult to solve. In the special case of constant cross section and constant thermal conductivity, the differential equa-tion 3–55 reduces to (T T) 0 or m2u 0 (3–56) where m2 (3–57) and u T T is the temperature excess. At the fin base we have ub Tb T.
Equation 3–56 is a linear, homogeneous, second-order differential equation with constant coefficients. A fundamental theory of differential equations states that such an equation has two linearly independent solution functions, and its general solution is the linear combination of those two solution func-tions. A careful examination of the differential equation reveals that subtract-ing a constant multiple of the solution function u from its second derivative yields zero. Thus we conclude that the function u and its second derivative must be constant multiples of each other. The only functions whose deriva-tives are constant multiples of the functions themselves are the exponential functions (or a linear combination of exponential functions such as sine and cosine hyperbolic functions). Therefore, the solution functions of the differ-ential equation above are the exponential functions emx or emx or constant multiples of them. This can be verified by direct substitution. For example, the second derivative of emx is m2emx, and its substitution into Eq. 3–56 hp kAc d 2u dx 2 hp kAc d2T dx2 d dx akAc dT dx b dT dx Q d Q cond dx Q cond, x x Q cond, x x Q Q Q Q 165 CHAPTER 3 yields zero. Therefore, the general solution of the differential equation Eq. 3–56 is u(x) C1emx C2emx (3–58) where C1 and C2 are arbitrary constants whose values are to be determined from the boundary conditions at the base and at the tip of the fin. Note that we need only two conditions to determine C1 and C2 uniquely.
The temperature of the plate to which the fins are attached is normally known in advance. Therefore, at the fin base we have a specified temperature boundary condition, expressed as Boundary condition at fin base: u(0) ub Tb T (3–59) At the fin tip we have several possibilities, including infinitely long fins, neg-ligible heat loss (idealized as an adiabatic tip), specified temperature, and con-vection (Fig. 3–37). Next, we consider each case separately.
1 Infinitely Long Fin (Tfin tip T) For a sufficiently long fin of uniform cross section (Ac constant), the tem-perature of the fin at the fin tip approaches the environment temperature T and thus u approaches zero. That is, Boundary condition at fin tip: u(L) T(L) T 0 as L → This condition is satisfied by the function emx, but not by the other prospec-tive solution function emx since it tends to infinity as x gets larger. Therefore, the general solution in this case will consist of a constant multiple of emx. The value of the constant multiple is determined from the requirement that at the fin base where x 0 the value of u is ub. Noting that emx e0 1, the proper value of the constant is ub, and the solution function we are looking for is u(x) ubemx. This function satisfies the differential equation as well as the requirements that the solution reduce to ub at the fin base and approach zero at the fin tip for large x. Noting that u T T and m , the varia-tion of temperature along the fin in this case can be expressed as Very long fin: emx (3–60) Note that the temperature along the fin in this case decreases exponentially from Tb to T, as shown in Fig. 3–38. The steady rate of heat transfer from the entire fin can be determined from Fourier’s law of heat conduction Very long fin: long fin kAc (Tb T) (3–61) where p is the perimeter, Ac is the cross-sectional area of the fin, and x is the distance from the fin base. Alternatively, the rate of heat transfer from the fin could also be determined by considering heat transfer from a differential vol-ume element of the fin and integrating it over the entire surface of the fin: fin h[T(x) T] dAfin hu(x) dAfin (3–62) Afin Afin Q 2hpkAc dT dx x0 Q ex2hp/kAc T(x) T Tb T 2hp/kAc 166 STEADY HEAT CONDUCTION Tb 0 1. Infinitely long fin 2. Negligible heat loss (adiabatic tip) 3. Specified temperature 4. Convection T L x FIGURE 3–37 Boundary conditions at the fin base and the fin tip.
Tb T Tb T(x) = T + (Tb – T)e T L 0 x Ab = Ac k h, T hp —– kAc –x D (p = pD, Ac = pD2/4 for a cylindrical fin) FIGURE 3–38 A long circular fin of uniform cross section and the variation of temperature along it.
The two approaches described are equivalent and give the same result since, under steady conditions, the heat transfer from the exposed surfaces of the fin is equal to the heat transfer to the fin at the base (Fig. 3–39).
2 Negligible Heat Loss from the Fin Tip (Adiabatic fin tip, fin tip 0) Fins are not likely to be so long that their temperature approaches the sur-rounding temperature at the tip. A more realistic situation is for heat transfer from the fin tip to be negligible since the heat transfer from the fin is propor-tional to its surface area, and the surface area of the fin tip is usually a negli-gible fraction of the total fin area. Then the fin tip can be assumed to be adiabatic, and the condition at the fin tip can be expressed as Boundary condition at fin tip: 0 (3–63) The condition at the fin base remains the same as expressed in Eq. 3–59. The application of the boundary conditions given by Eqs. (3–59) and (3–63) on the general solution (Eq. 3–58) requires that u(0) ub C1C2 and mC1emL mC2emL 0, respectively. Solving these two equations simultaneously for C1 and C2 yields C1 ub /(1e2mL) and C2 ub /(1e2mL). Substituting the rela-tions for C1 and C2 into Eq. 3–58 and using the definition of the hyperbolic co-sine function cosh x (ex ex)/2 gives the desired relation for the temperature distribution: Adiabatic fin tip: (3–64) The rate of heat transfer from the fin can be determined again from Fourier’s law of heat conduction: Adiabatic fin tip: adiabatic tip kAc (Tb T) tanh mL (3–65) where the equation for the hyperbolic tangent function is tanh x sinh x/cosh x (ex ex)/(ex ex).
Note that the heat transfer relations for the very long fin and the fin with neg-ligible heat loss at the tip differ by the factor tanh mL, which approaches 1 as L becomes very large.
3 Specified Temperature (Tfin,tip TL) In this case the temperature at the end of the fin (the fin tip) is fixed at a spec-ified temperature TL. This case could be considered as a generalization of the case of Infinitely Long Fin where the fin tip temperature was fixed at The condition at the fin tip for this case is Boundary condition at fin tip: (3–66) The fin base boundary condition remains the same as given in Eq. 3–59. Ap-plying the boundary conditions given by Eqs. 3–59 and 3–66 on the general solution (Eq. 3–58) gives, after some lengthy algebra and using the definition u(L) uL TL T T.
2hpkAc dT dx x 0 Q T(x) T Tb T cosh m(L x) cosh mL du dx x L Q 167 CHAPTER 3 Qbase · Qbase = Qfin · · Qfin · FIGURE 3–39 Under steady conditions, heat transfer from the exposed surfaces of the fin is equal to heat conduction to the fin at the base.
of the hyperbolic sine function, the desired temperature distribution: Specified fin tip temperature: (3–67) Using the Fourier’s law of heat conduction, the rate of heat transfer from the fin is Specified fin tip temperature: (3–68) Note that Eqs. 3–67 and 3–68 reduce to Eqs. 3–60 and 3–61 for the case of in-finitely long fin (L →).
4 Convection from Fin Tip The fin tips, in practice, are exposed to the surroundings, and thus the proper boundary condition for the fin tip is convection that may also include the ef-fects of radiation. Consider the case of convection only at the tip. The condi-tion at the fin tip can be obtained from an energy balance at the fin tip That is, Boundary condition at fin tip: (3–69) The boundary condition at the fin base is Eq. 3–59, which is the same as the three previous cases. Substituting the two boundary conditions given by Eqs. 3–59 and 3–69 in the general solution (Eq. 3–58), it may be shown, after some lengthy manipulation that the temperature distribution is Convection from fin tip: (3–70) The rate of heat transfer from the fin can be found by substituting the tem-perature gradient at the base of the fin, obtained from Eq. 3–70, into the Fourier’s law of heat conduction. The result is Convection from fin tip: (3–71) The solution to the general fin equation for the case of convection from fin tip is rather complex. An approximate, yet practical and accurate, way of ac-counting for the loss from the fin tip is to replace the fin length L in the rela-tion for the insulated tip case by a corrected fin length (defined as (Fig. 3–40) Corrected fin length: Lc L (3–72) Ac p 2h p k Ac (Tb T) sinh mL (h
mk) cosh mL cosh mL (h
mk) sinh mL Q .
convection kAc dT dx x0 T(x) T Tb T cosh m(L x) (h
mk) sinh m(L x) cosh mL (h
mk) sinh mL kAc dT dx xL hAc[T(L) T] (Q .
cond Q .
conv). 2h p k Ac(Tb T)cosh mL [(TL T)
(Tb T)] sinh mL Q .
specified temp. kAc dT dx x0 T(x) T Tb T [(TL T)/(Tb T)]sinh mx sinh m(Lx) sinh mL sinh x (exex)/2, 168 STEADY HEAT CONDUCTION Lc Insulated (b) Equivalent fin with insulated tip Qfin · L Convection (a) Actual fin with convection at the tip Qfin · Ac –— p FIGURE 3–40 Corrected fin length Lc is defined such that heat transfer from a fin of length Lc with insulated tip is equal to heat transfer from the actual fin of length L with convection at the fin tip.
where Ac is the cross-sectional area and p is the perimeter of the fin at the tip.
Multiplying the relation above by the perimeter gives Acorrected Afin (lateral) Atip, which indicates that the fin area determined using the corrected length is equivalent to the sum of the lateral fin area plus the fin tip area.
The corrected length approximation gives very good results when the vari-ation of temperature near the fin tip is small (which is the case when mL 1) and the heat transfer coefficient at the fin tip is about the same as that at the lateral surface of the fin. Therefore, fins subjected to convection at their tips can be treated as fins with insulated tips by replacing the actual fin length by the corrected length in Eqs. 3–64 and 3–65.
Using the proper relations for Ac and p, the corrected lengths for rectangu-lar and cylindrical fins are easily determined to be Lc, rectangular fin L and Lc, cylindrical fin L where t is the thickness of the rectangular fins and D is the diameter of the cylindrical fins.
Fin Efficiency Consider the surface of a plane wall at temperature Tb exposed to a medium at temperature T. Heat is lost from the surface to the surrounding medium by convection with a heat transfer coefficient of h. Disregarding radiation or ac-counting for its contribution in the convection coefficient h, heat transfer from a surface area As is expressed as Q . hAs(Ts T).
Now let us consider a fin of constant cross-sectional area Ac Ab and length L that is attached to the surface with a perfect contact (Fig. 3–41). This time heat is transferrd from the surface to the fin by conduction and from the fin to the surrounding medium by convection with the same heat transfer coefficient h. The temperature of the fin is Tb at the fin base and gradually decreases to-ward the fin tip. Convection from the fin surface causes the temperature at any cross section to drop somewhat from the midsection toward the outer surfaces.
However, the cross-sectional area of the fins is usually very small, and thus the temperature at any cross section can be considered to be uniform. Also, the fin tip can be assumed for convenience and simplicity to be adiabatic by using the corrected length for the fin instead of the actual length.
In the limiting case of zero thermal resistance or infinite thermal conductiv-ity (k →), the temperature of the fin is uniform at the base value of Tb. The heat transfer from the fin is maximum in this case and can be expressed as fin, max hAfin (Tb T) (3–73) In reality, however, the temperature of the fin drops along the fin, and thus the heat transfer from the fin is less because of the decreasing temperature dif-ference T(x) T toward the fin tip, as shown in Fig. 3–42. To account for the effect of this decrease in temperature on heat transfer, we define a fin effi-ciency as hfin (3–74) Qfin Qfin, max Actual heat transfer rate from the fin Ideal heat transfer rate from the fin if the entire fin were at base temperature Q D 4 t 2 169 CHAPTER 3 (a) Surface without fins Tb (b) Surface with a fin Afin = 2 × w × L + w × t ≅ 2 × w × L Ab = w × t L w t Ab Afin FIGURE 3–41 Fins enhance heat transfer from a surface by enhancing surface area.
80°C 70 65 61 58 56°C (b) Actual 80°C 80 80 80 80 80°C (a) Ideal FIGURE 3–42 Ideal and actual temperature distribution along a fin.
170 STEADY HEAT CONDUCTION Straight rectangular fins Lc L t/2 Afin 2wLc Straight triangular fins Straight parabolic fins Circular fins of rectangular profile r2c r2 t/2 Pin fins of rectangular profile Lc L D/4 Afin pDLc Pin fins of triangular profile Pin fins of parabolic profile C3 1 2(D/L)2 Pin fins of parabolic profile (blunt tip) Afin pD4 96L2e[16(L/D)2 1]3/2 1f m 24h/kD C4 21 (D/L)2 Afin pL3 8D 3C3C4 L 2D ln(2DC4/L C3)4 m 24h/kD Afin pD 2 2L2 (D/2)2 m 24h/kD m 24h/kD Afin 2p(r2c 2 r1 2) m 22h/kt C1 21 (t/L)2 Afin wL3C1 (L/t) ln(t/L C1)4 m 22h/kt Afin 2w2L2 (t/2)2 m 22h/kt m 22h/kt I2(x) I0 (x) (2/x)I1 (x) where x 2mL hfin 3 2mL l1(4mL/3) l0(4mL/3) hfin 2 1 2(2mL/3)2 1 hfin 2 mL l2(2mL) l1(2mL) hfin tanh mLc mLc C2 2r1/m r2c 2 r1 2 hfin C2 K1(mr1)I1(mr2c) I1(mr1)K1(mr2c) I0(mr1)K1(mr2c) K0(mr1)I1(mr2c) hfin 2 1 2(2mL)2 1 hfin 1 mL l1(2mL) l0(2mL) hfin tanh mLc mLc TABLE 3–3 Efficiency and surface areas of common fin configurations w L x t y = (t/2) (1 – x/L) t w L t L w y = (t/2) (1 x/L)2 r2 r1 t L D L y = (D/2) (1 – x/L) D L y = (D/2) (1 – x/L)2 D L y = (D/2) (1 – x/L)1/2 D L or fin hfin fin, max hfin hAfin (Tb T) (3–75) where Afin is the total surface area of the fin. This relation enables us to deter-mine the heat transfer from a fin when its efficiency is known. For the cases of constant cross section of very long fins and fins with adiabatic tips, the fin ef-ficiency can be expressed as hlong fin (3–76) and hadiabatic tip (3–77) since Afin pL for fins with constant cross section. Equation 3–77 can also be used for fins subjected to convection provided that the fin length L is replaced by the corrected length Lc.
Table 3–3 provides fin efficiency relations for fins with uniform and non-uniform cross section. For fins with non-uniform profile, Eq. 3–56 is no longer valid and the general form of the differential equation governing heat transfer in fins of arbitrary shape, Eq. 3–55, must be used. For these cases the solution is no longer in the form of simple exponential or hyperbolic functions.
The mathematical functions I and K that appear in some of these relations are the modified Bessel functions, and their values are given in Table 3–4. Efficiencies are plotted in Fig. 3–43 for fins on a plain surface and in Fig. 3–44 for circular fins of constant thickness. For most fins of constant thickness encountered in practice, the fin thickness t is too small relative to the fin length L, and thus the fin tip area is negligible.
Note that fins with triangular and parabolic profiles contain less material and are more efficient than the ones with rectangular profiles, and thus are more suitable for applications requiring minimum weight such as space applications.
An important consideration in the design of finned surfaces is the selection of the proper fin length L. Normally the longer the fin, the larger the heat transfer area and thus the higher the rate of heat transfer from the fin. But also the larger the fin, the bigger the mass, the higher the price, and the larger the fluid friction. Therefore, increasing the length of the fin beyond a certain value cannot be justified unless the added benefits outweigh the added cost.
Also, the fin efficiency decreases with increasing fin length because of the decrease in fin temperature with length. Fin lengths that cause the fin effi-ciency to drop below 60 percent usually cannot be justified economically and should be avoided. The efficiency of most fins used in practice is above 90 percent.
Fin Effectiveness Fins are used to enhance heat transfer, and the use of fins on a surface cannot be recommended unless the enhancement in heat transfer justifies the added cost and complexity associated with the fins. In fact, there is no assurance that tanh mL mL Q # fin Q # fin, max 2hpkAc (Tb T) tanh mL hAfin (Tb T) Q fin Q fin, max 2hpkAc (Tb T) hAfin (Tb T) 1 L B kAc hp 1 mL Q Q 171 CHAPTER 3 TABLE 3–4 Modified Bessel functions of the first and second kinds x exI0(x) exI1(x) exK0(x) exK1(x) 0.0 1.0000 0.0000 — — 0.2 0.8269 0.0823 2.1408 5.8334 0.4 0.6974 0.1368 1.6627 3.2587 0.6 0.5993 0.1722 1.4167 2.3739 0.8 0.5241 0.1945 1.2582 1.9179 1.0 0.4658 0.2079 1.1445 1.6362 1.2 0.4198 0.2153 1.0575 1.4429 1.4 0.3831 0.2185 0.9881 1.3011 1.6 0.3533 0.2190 0.9309 1.1919 1.8 0.3289 0.2177 0.8828 1.1048 2.0 0.3085 0.2153 0.8416 1.0335 2.2 0.2913 0.2121 0.8057 0.9738 2.4 0.2766 0.2085 0.7740 0.9229 2.6 0.2639 0.2047 0.7459 0.8790 2.8 0.2528 0.2007 0.7206 0.8405 3.0 0.2430 0.1968 0.6978 0.8066 3.2 0.2343 0.1930 0.6770 0.7763 3.4 0.2264 0.1892 0.6580 0.7491 3.6 0.2193 0.1856 0.6405 0.7245 3.8 0.2129 0.1821 0.6243 0.7021 4.0 0.2070 0.1788 0.6093 0.6816 4.2 0.2016 0.1755 0.5953 0.6627 4.4 0.1966 0.1725 0.5823 0.6454 4.6 0.1919 0.1695 0.5701 0.6292 4.8 0.1876 0.1667 0.5586 0.6143 5.0 0.1835 0.1640 0.5478 0.6003 5.2 0.1797 0.1614 0.5376 0.5872 5.4 0.1762 0.1589 0.5280 0.5749 5.6 0.1728 0.1565 0.5188 0.5634 5.8 0.1697 0.1542 0.5101 0.5525 6.0 0.1667 0.1521 0.5019 0.5422 6.5 0.1598 0.1469 0.4828 0.5187 7.0 0.1537 0.1423 0.4658 0.4981 7.5 0.1483 0.1380 0.4505 0.4797 8.0 0.1434 0.1341 0.4366 0.4631 8.5 0.1390 0.1305 0.4239 0.4482 9.0 0.1350 0.1272 0.4123 0.4346 9.5 0.1313 0.1241 0.4016 0.4222 10.0 0.1278 0.1213 0.3916 0.4108 Evaluated from EES using the mathematical functions Bessel_I(x) and Bessel_K(x) adding fins on a surface will enhance heat transfer. The performance of the fins is judged on the basis of the enhancement in heat transfer relative to the no-fin case. The performance of fins is expressed in terms of the fin effective-ness efin defined as (Fig. 3–45) efin (3–78) Qfin Qno fin Qfin hAb (Tb T) Heat transfer rate from the fin of base area Ab Heat transfer rate from the surface of area Ab 172 STEADY HEAT CONDUCTION FIGURE 3–44 Efficiency of annular fins of constant thickness t.
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 r2c = r2 + t/2 Lc = L + t/2 Ap = Lct 0 3 1 Fin efficiency, ηfin 0 2 3 5 1 = r2c/r1 4 r2 r1 t L ξ = Lc 3/2(h/kAp)1/2 FIGURE 3–43 Efficiency of straight fins of rectangular, triangular, and parabolic profiles.
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 0 3 ξ = Lc 3/2(h/kAp)1/2 1 Fin efficiency, ηfin w L t t w L t L w Lc = L Ap = Lct/3 y = (t/2) (1 – x/L)2 Lc = L Ap = Lct/2 x Lc = L + t/2 Ap = Lct Here, Ab is the cross-sectional area of the fin at the base and no fin represents the rate of heat transfer from this area if no fins are attached to the surface. An effectiveness of efin 1 indicates that the addition of fins to the surface does not affect heat transfer at all. That is, heat conducted to the fin through the base area Ab is equal to the heat transferred from the same area Ab to the sur-rounding medium. An effectiveness of efin 1 indicates that the fin actually acts as insulation, slowing down the heat transfer from the surface. This situ-ation can occur when fins made of low thermal conductivity materials are used. An effectiveness of efin 1 indicates that fins are enhancing heat trans-fer from the surface, as they should. However, the use of fins cannot be justi-fied unless efin is sufficiently larger than 1. Finned surfaces are designed on the basis of maximizing effectiveness for a specified cost or minimizing cost for a desired effectiveness.
Note that both the fin efficiency and fin effectiveness are related to the per-formance of the fin, but they are different quantities. However, they are related to each other by efin hfin (3–79) Therefore, the fin effectiveness can be determined easily when the fin effi-ciency is known, or vice versa.
The rate of heat transfer from a sufficiently long fin of uniform cross section under steady conditions is given by Eq. 3–61. Substituting this relation into Eq. 3–78, the effectiveness of such a long fin is determined to be elong fin (3–80) since Ac Ab in this case. We can draw several important conclusions from the fin effectiveness relation above for consideration in the design and selec-tion of the fins: • The thermal conductivity k of the fin material should be as high as possible. Thus it is no coincidence that fins are made from metals, with copper, aluminum, and iron being the most common ones. Perhaps the most widely used fins are made of aluminum because of its low cost and weight and its resistance to corrosion.
• The ratio of the perimeter to the cross-sectional area of the fin p/Ac should be as high as possible. This criterion is satisfied by thin plate fins and slender pin fins.
• The use of fins is most effective in applications involving a low convection heat transfer coefficient. Thus, the use of fins is more easily justified when the medium is a gas instead of a liquid and the heat transfer is by natural convection instead of by forced convection. Therefore, it is no coincidence that in liquid-to-gas heat exchangers such as the car radiator, fins are placed on the gas side.
When determining the rate of heat transfer from a finned surface, we must consider the unfinned portion of the surface as well as the fins. Therefore, the rate of heat transfer for a surface containing n fins can be expressed as Q fin Q no fin 2hpkAc (Tb T) hAb (Tb T) B kp hAc Qfin Qno fin Qfin hAb (Tb T) hfin hAfin (Tb T) hAb (Tb T) Afin Ab Q 173 CHAPTER 3 Ab Ab = —— — Tb Tb Qfin · Qno fin · Qfin · Qno fin · efin FIGURE 3–45 The effectiveness of a fin.
total, fin unfin fin hAunfin (Tb T) hfin hAfin (Tb T) h(Aunfin hfin Afin)(Tb T) (3–81) We can also define an overall effectiveness for a finned surface as the ratio of the total heat transfer from the finned surface to the heat transfer from the same surface if there were no fins, efin, overall (3–82) where Ano fin is the area of the surface when there are no fins, Afin is the total sur-face area of all the fins on the surface, and Aunfin is the area of the unfinned por-tion of the surface (Fig. 3–46). Note that the overall fin effectiveness depends on the fin density (number of fins per unit length) as well as the effectiveness of the individual fins. The overall effectiveness is a better measure of the per-formance of a finned surface than the effectiveness of the individual fins.
Proper Length of a Fin An important step in the design of a fin is the determination of the appropriate length of the fin once the fin material and the fin cross section are specified.
You may be tempted to think that the longer the fin, the larger the surface area and thus the higher the rate of heat transfer. Therefore, for maximum heat transfer, the fin should be infinitely long. However, the temperature drops along the fin exponentially and reaches the environment temperature at some length. The part of the fin beyond this length does not contribute to heat trans-fer since it is at the temperature of the environment, as shown in Fig. 3–47.
Therefore, designing such an “extra long” fin is out of the question since it re-sults in material waste, excessive weight, and increased size and thus in-creased cost with no benefit in return (in fact, such a long fin will hurt performance since it will suppress fluid motion and thus reduce the convection heat transfer coefficient). Fins that are so long that the temperature approaches the environment temperature cannot be recommended either since the little in-crease in heat transfer at the tip region cannot justify the disproportionate in-crease in the weight and cost.
To get a sense of the proper length of a fin, we compare heat transfer from a fin of finite length to heat transfer from an infinitely long fin under the same conditions. The ratio of these two heat transfers is tanh mL (3–83) Using a hand calculator, the values of tanh mL are evaluated for some values of mL and the results are given in Table 3–5. We observe from the table that heat transfer from a fin increases with mL almost linearly at first, but the curve reaches a plateau later and reaches a value for the infinitely long fin at about mL 5. Therefore, a fin whose length is L L m can be considered to be an infinitely long fin. We also observe that reducing the fin length by half in that case (from mL 5 to mL 2.5) causes a drop of just 1 percent in 1 5 Qfin Qlong fin 2hpkAc (Tb T) tanh mL 2hpkAc (Tb T) Heat transfer ratio: Aunfin hfin Afin Ano fin Qtotal, fin Qtotal, no fin h(Aunfin hfin Afin)(Tb T) hAno fin (Tb T) Q Q Q 174 STEADY HEAT CONDUCTION t w L Afin Ano fin = w × H Aunfin = w × H – 3 × (t × w) Afin = 2 × L × w + t × w ≅ 2 × L × w (one fin) Aunfin H FIGURE 3–46 Various surface areas associated with a rectangular surface with three fins.
T Tb T h, T 0 x T(x) High heat transfer Low heat transfer ΔT = low ΔT = high ΔT = 0 No heat transfer L Tb ΔT FIGURE 3–47 Because of the gradual temperature drop along the fin, the region near the fin tip makes little or no contribution to heat transfer.
heat transfer. We certainly would not hesitate sacrificing 1 percent in heat transfer performance in return for 50 percent reduction in the size and possi-bly the cost of the fin. In practice, a fin length that corresponds to about mL 1 will transfer 76.2 percent of the heat that can be transferred by an infinitely long fin, and thus it should offer a good compromise between heat transfer performance and the fin size.
A common approximation used in the analysis of fins is to assume the fin temperature to vary in one direction only (along the fin length) and the tem-perature variation along other directions is negligible. Perhaps you are won-dering if this one-dimensional approximation is a reasonable one. This is certainly the case for fins made of thin metal sheets such as the fins on a car radiator, but we wouldn’t be so sure for fins made of thick materials. Studies have shown that the error involved in one-dimensional fin analysis is negligi-ble (less than about 1 percent) when 0.2 where d is the characteristic thickness of the fin, which is taken to be the plate thickness t for rectangular fins and the diameter D for cylindrical ones.
Specially designed finned surfaces called heat sinks, which are commonly used in the cooling of electronic equipment, involve one-of-a-kind complex geometries, as shown in Table 3–6. The heat transfer performance of heat sinks is usually expressed in terms of their thermal resistances R in °C/W, which is defined as fin hAfin hfin (Tb T) (3–84) A small value of thermal resistance indicates a small temperature drop across the heat sink, and thus a high fin efficiency.
Tb T R Q hd k 175 CHAPTER 3 TABLE 3–5 The variation of heat transfer from a fin relative to that from an infinitely long fin mL tanh mL 0.1 0.100 0.2 0.197 0.5 0.462 1.0 0.762 1.5 0.905 2.0 0.964 2.5 0.987 3.0 0.995 4.0 0.999 5.0 1.000 Q .
fin Q .
long fin EXAMPLE 3–10 Maximum Power Dissipation of a Transistor Power transistors that are commonly used in electronic devices consume large amounts of electric power. The failure rate of electronic components increases almost exponentially with operating temperature. As a rule of thumb, the fail-ure rate of electronic components is halved for each 10°C reduction in the junction operating temperature. Therefore, the operating temperature of elec-tronic components is kept below a safe level to minimize the risk of failure.
The sensitive electronic circuitry of a power transistor at the junction is pro-tected by its case, which is a rigid metal enclosure. Heat transfer characteris-tics of a power transistor are usually specified by the manufacturer in terms of the case-to-ambient thermal resistance, which accounts for both the natural convection and radiation heat transfers.
The case-to-ambient thermal resistance of a power transistor that has a max-imum power rating of 10 W is given to be 20°C/W. If the case temperature of (Continued on page 177) 176 STEADY HEAT CONDUCTION TABLE 3–6 Combined natural convection and radiation thermal resistance of various heat sinks used in the cooling of electronic devices between the heat sink and the surroundings. All fins are made of aluminum 6063T-5, are black anodized, and are 76 mm (3 in) long.
R 0.9C/W (vertical) R 1.2C/W (horizontal) Dimensions: 76 mm 105 mm 44 mm Surface area: 677 cm2 R 5C/W Dimensions: 76 mm 38 mm 24 mm Surface area: 387 cm2 R 1.4C/W (vertical) R 1.8C/W (horizontal) Dimensions: 76 mm 92 mm 26 mm Surface area: 968 cm2 R 1.8C/W (vertical) R 2.1C/W (horizontal) Dimensions: 76 mm 127 mm 91 mm Surface area: 677 cm2 R 1.1C/W (vertical) R 1.3C/W (horizontal) Dimensions: 76 mm 102 mm 25 mm Surface area: 929 cm2 R 2.9C/W (vertical) R 3.1C/W (horizontal) Dimensions: 76 mm 97 mm 19 mm Surface area: 290 cm2 HS 5030 HS 6065 HS 6071 HS 6105 HS 6115 HS 7030 177 CHAPTER 3 the transistor is not to exceed 85°C, determine the power at which this tran-sistor can be operated safely in an environment at 25°C.
SOLUTION The maximum power rating of a transistor whose case temperature is not to exceed 85°C is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The transistor case is isothermal at 85°C.
Properties The case-to-ambient thermal resistance is given to be 20°C/W.
Analysis The power transistor and the thermal resistance network associated with it are shown in Fig. 3–48. We notice from the thermal resistance network that there is a single resistance of 20°C/W between the case at Tc 85°C and the ambient at T 25°C, and thus the rate of heat transfer is 3 W Therefore, this power transistor should not be operated at power levels above 3 W if its case temperature is not to exceed 85°C.
Discussion This transistor can be used at higher power levels by attaching it to a heat sink (which lowers the thermal resistance by increasing the heat transfer surface area, as discussed in the next example) or by using a fan (which lowers the thermal resistance by increasing the convection heat trans-fer coefficient).
(85 25)°C 20°C/ W Tc T Rcase-ambient aT R b case-ambient Q EXAMPLE 3–11 Selecting a Heat Sink for a Transistor A 60-W power transistor is to be cooled by attaching it to one of the commer-cially available heat sinks shown in Table 3–6. Select a heat sink that will al-low the case temperature of the transistor not to exceed 90°C in the ambient air at 30°C.
SOLUTION A commercially available heat sink from Table 3–6 is to be se-lected to keep the case temperature of a transistor below 90°C.
Assumptions 1 Steady operating conditions exist. 2 The transistor case is isothermal at 90°C. 3 The contact resistance between the transistor and the heat sink is negligible.
Analysis The rate of heat transfer from a 60-W transistor at full power is 60 W. The thermal resistance between the transistor attached to the heat sink and the ambient air for the specified temperature difference is determined to be ⎯→ R 1.0C/W Therefore, the thermal resistance of the heat sink should be below 1.0°C/W. An examination of Table 3–6 reveals that the HS 5030, whose thermal resis-tance is 0.9°C/W in the vertical position, is the only heat sink that will meet this requirement.
T Q (90 30)C 60 W T R Q Q Q · T Tc R FIGURE 3–48 Schematic for Example 3–10.
178 STEADY HEAT CONDUCTION EXAMPLE 3–12 Heat Transfer from Fins of Variable Cross-Section Aluminum pin fins of parabolic profile with blunt tips are attached on a plane wall with surface temperature of 200°C (Fig. 3–49). Each fin has a length of 20 mm and a base diameter of 5 mm. The fins are exposed to an ambient air condition of 25°C and the convection heat transfer coefficient is 50 W/m2·K. If the thermal conductivity of the fins is 240 W/m2·K, determine the efficiency, heat transfer rate, and effectiveness of each fin.
SOLUTION The efficiency, heat transfer rate, and effectiveness of a pin fin of parabolic profile with blunt tips are to be determined.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible.
Properties The thermal conductivity of the fin is given as 240 W/m2·K.
Analysis From Table 3–3, for pin fins of parabolic profile (blunt tip), we have (0.020m) 0.2582 hfin 3 2mL I1(4mL3) I0(4mL3) 3 2(0.2582) I1[4(0.2582)3] I0[4(0.2582)3] 5.8095I1[0.3443] I0[0.3443] 2.106 104 m2 Afin pD4 96L2ec16aL Db 2 1d 32 1f p (0.005 m)4 96(0.020 m)2 ec16a0.020 m 0.005 mb 2 1d 32 1f mL B 4h kDL B 4(50 W/m2K) (240 W/mK)(0.005 m) L = 20 mm D = 5 mm Air, 25°C h = 50 W/m2·K k = 240 W/m·K Tb = 200°C FIGURE 3–49 Schematic for Example 3–12.
The values of the Bessel functions corresponding to x 0.3443 are deter-mined from Table 3–4 to be I0 1.0350 and I1 0.1716. Substituting, the fin efficiency is determined to be The heat transfer rate for a single fin is The fin effectiveness is That is, over a 10-fold increase in heat transfer is achieved by using a pin fin in this case.
10.3 1.77 W (50W/m2K) [p(0 .005 m)24] (200 25)C efin Q # fin hAb(Tb T ) Q # fin h(pD24) (Tb T ) (0.9632) (50 W/m2 K) (2 .106 104 m2)(20025)C 1.77 W Q # fin hfin hAfin (Tb T ) hfin 5.8095 0.1716 1.0350 0.9632 3–7 HEAT TRANSFER IN COMMON CONFIGURATIONS So far, we have considered heat transfer in simple geometries such as large plane walls, long cylinders, and spheres. This is because heat transfer in such geometries can be approximated as one-dimensional, and simple analytical solutions can be obtained easily. But many problems encountered in practice are two- or three-dimensional and involve rather complicated geometries for which no simple solutions are available.
An important class of heat transfer problems for which simple solutions are obtained encompasses those involving two surfaces maintained at constant temperatures T1 and T2. The steady rate of heat transfer between these two sur-faces is expressed as Q Sk(T1 T2) (3–85) where S is the conduction shape factor, which has the dimension of length, and k is the thermal conductivity of the medium between the surfaces. The conduction shape factor depends on the geometry of the system only.
Conduction shape factors have been determined for a number of configura-tions encountered in practice and are given in Table 3–7 for some common cases. More comprehensive tables are available in the literature. Once the value of the shape factor is known for a specific geometry, the total steady heat transfer rate can be determined from the equation above using the specified two constant temperatures of the two surfaces and the thermal conductivity of the medium between them. Note that conduction shape factors are applicable only when heat transfer between the two surfaces is by conduction. Therefore, they cannot be used when the medium between the surfaces is a liquid or gas, which involves natural or forced convection currents.
A comparison of Eqs. 3–4 and 3–85 reveals that the conduction shape fac-tor S is related to the thermal resistance R by R 1/kS or S 1/kR. Thus, these two quantities are the inverse of each other when the thermal conductiv-ity of the medium is unity. The use of the conduction shape factors is illus-trated with Examples 3–13 and 3–14.
■ 179 CHAPTER 3 Discussion The fin efficiency can be determined more accurately by avoiding the interpolation error by using an equation solver with built-in mathematical functions such as EES. Copying the line on a blank EES screen and hitting the ‘solve’ button gives the fin efficiency to be which is about 2 percent higher than the result obtained above using the tables.
hfin 0.9855, eta_fin 3
(20.2582)Bessel_| 1(40.2582/3)
Bessel_|0 (40.2582/3) 180 STEADY HEAT CONDUCTION TABLE 3–7 Conduction shape factors S for several configurations for use in kS(T1 T2) to determine the steady rate of heat transfer through a medium of thermal conductivity k between the surfaces at temperatures T1 and T2 Q (1) Isothermal cylinder of length L buried in a semi-infinite medium (L >> D and z > 1.5D) (2) Vertical isothermal cylinder of length L buried in a semi-infinite medium (L >> D) (3) Two parallel isothermal cylinders placed in an infinite medium (L >> D1, D2, z) (4) A row of equally spaced parallel isothermal cylinders buried in a semi-infinite medium (L >> D, z, and w > 1.5D) (5) Circular isothermal cylinder of length L in the midplane of an infinite wall (z > 0.5D) (6) Circular isothermal cylinder of length L at the center of a square solid bar of the same length (7) Eccentric circular isothermal cylinder of length L in a cylinder of the same length (L > D2) (8) Large plane wall S = ———– 2πL ln (4z/D) S = ———– 2πL ln(4L/D) S = —–————–———— 2πL D2 1 + D2 2 – 4z2 2D1D2 cosh–1 –––——————— S = —————–—— 2πL 2w 2πz w ln — — sinh — — S = ———–– 2πL ln(8z/πD) S = —————– 2πL ln (1.08w/D) S = — A L T1 z T2 T1 T1 T2 T2 T1 T2 T2 T1 A L L w D L z L z L z z D (per cylinder) T2 πD T2 D1 D2 L T1 z D2 D1 L L D T2 T1 D w w w T2 T1 D S = —–————–———— 2πL 4z2 – D2 1 – D2 2 2D1D2 cosh–1 –––——————— 181 CHAPTER 3 TABLE 3–7 (Continued) (9) A long cylindrical layer (10) A square flow passage (a) For a/b > 1.4, (11) A spherical layer (12) Disk buried parallel to the surface in a semi-infinite medium (z >> D) (13) The edge of two adjoining walls of equal thickness (14) Corner of three walls of equal thickness L (inside) L L T1 (outside) T2 (15) Isothermal sphere buried in a semi-infinite medium (16) Isothermal sphere buried in a semi-infinite medium at T2 whose surface is insulated S = ———–— 2πL ln (D2/D1) S = ———— 2πD1D2 D2 – D1 S = ———–———– 2πL 0.93 ln (0.948a/b) (b) For a/b < 1.41, S = ——–———– 2πL 0.785 ln (a/b) S = ————— 2πD 1 – 0.25D/z S = ————— 2πD 1 + 0.25D/z S = 4 D (S = 2D when z = 0) S = 0.54 w S = 0.15L D1 D2 T1 T2 L L T2 T2 T1 D T1 T2 Insulated w z T2 (medium) D T2 T2 T1 T1 z b a L D T1 z T2 T1 D1 L D2 182 STEADY HEAT CONDUCTION EXAMPLE 3–13 Heat Loss from Buried Steam Pipes A 30-m-long, 10-cm-diameter hot-water pipe of a district heating system is buried in the soil 50 cm below the ground surface, as shown in Fig. 3–50. The outer surface temperature of the pipe is 80°C. Taking the surface temperature of the earth to be 10°C and the thermal conductivity of the soil at that location to be 0.9 W/m·K, determine the rate of heat loss from the pipe.
SOLUTION The hot-water pipe of a district heating system is buried in the soil. The rate of heat loss from the pipe is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the soil is constant.
Properties The thermal conductivity of the soil is given to be k 0.9 W/m·K.
Analysis The shape factor for this configuration is given in Table 3–7 to be S since z 1.5D, where z is the distance of the pipe from the ground surface, and D is the diameter of the pipe. Substituting, S 62.9 m Then the steady rate of heat transfer from the pipe becomes Sk(T1 T2) (62.9 m)(0.9 W/m·K)(80 10)C 3963 W Discussion Note that this heat is conducted from the pipe surface to the sur-face of the earth through the soil and then transferred to the atmosphere by convection and radiation.
Q 2p (30 m) ln(4 0.5/0.1) 2pL ln(4z/D) EXAMPLE 3–14 Heat Transfer between Hot- and Cold-Water Pipes A 5-m-long section of hot- and cold-water pipes run parallel to each other in a thick concrete layer, as shown in Fig. 3–51. The diameters of both pipes are 5 cm, and the distance between the centerline of the pipes is 30 cm. The sur-face temperatures of the hot and cold pipes are 70°C and 15°C, respectively.
Taking the thermal conductivity of the concrete to be k 0.75 W/m·K, deter-mine the rate of heat transfer between the pipes.
SOLUTION Hot- and cold-water pipes run parallel to each other in a thick concrete layer. The rate of heat transfer between the pipes is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the concrete is constant.
Properties The thermal conductivity of concrete is given to be k 0.75 W/m·K.
D2 = 5 cm T1 = 70°C L = 5 m z = 30 cm T2 = 15°C D1 = 5 cm FIGURE 3–51 Schematic for Example 3–14.
L = 30 m D = 10 cm T1 = 80°C z = 0.5 m T2 = 10°C FIGURE 3–50 Schematic for Example 3–13.
It is well known that insulation reduces heat transfer and saves energy and money. Decisions on the right amount of insulation are based on a heat trans-fer analysis, followed by an economic analysis to determine the “monetary value” of energy loss. This is illustrated with Example 3–15.
183 CHAPTER 3 Analysis The shape factor for this configuration is given in Table 3–7 to be S where z is the distance between the centerlines of the pipes and L is their length. Substituting, S 6.34 m Then the steady rate of heat transfer between the pipes becomes Sk(T1 T2) (6.34 m)(0.75 W/m· )(70 15)C 262 W Discussion We can reduce this heat loss by placing the hot- and cold-water pipes farther away from each other.
Q 2p (5 m) cosh1a4 0.32 0.052 0.052 2 0.05 0.05 b 2pL cosh1a4z2 D2 1 D2 2 2D1D2 b EXAMPLE 3–15 Cost of Heat Loss through Walls in Winter Consider an electrically heated house whose walls are 9 ft high and have an R-value of insulation of 13 (i.e., a thickness-to-thermal conductivity ratio of L/k 13 h·ft2°F/Btu). Two of the walls of the house are 40 ft long and the others are 30 ft long. The house is maintained at 75°F at all times, while the temperature of the outdoors varies. Determine the amount of heat lost through the walls of the house on a certain day during which the average tem-perature of the outdoors is 45°F. Also, determine the cost of this heat loss to the home owner if the unit cost of electricity is $0.075/kWh. For combined convection and radiation heat transfer coefficients, use the ASHRAE (American Society of Heating, Refrigeration, and Air Conditioning Engineers) recom-mended values of hi 1.46 Btu/h·ft2·°F for the inner surface of the walls and ho 6.0 Btu/h·ft2°F for the outer surface of the walls under 15 mph wind con-ditions in winter.
SOLUTION An electrically heated house with R-13 insulation is considered.
The amount of heat lost through the walls and its cost are to be determined.
Assumptions 1 The indoor and outdoor air temperatures have remained at the given values for the entire day so that heat transfer through the walls is steady.
2 Heat transfer through the walls is one-dimensional since any significant temperature gradients in this case exist in the direction from the indoors to the outdoors. 3 The radiation effects are accounted for in the heat transfer coefficients.
184 STEADY HEAT CONDUCTION Analysis This problem involves conduction through the wall and convection at its surfaces and can best be handled by making use of the thermal resistance concept and drawing the thermal resistance network, as shown in Fig. 3–52.
The heat transfer area of the walls is A Circumference Height (2 30 ft 2 40 ft)(9 ft) 1260 ft2 Then the individual resistances are evaluated from their definitions to be Ri Rconv, i 0.00054 h·F/Btu Rwall 0.01032 h·F/Btu Ro Rconv, o 0.00013 h·F/Btu Noting that all three resistances are in series, the total resistance is Rtotal Ri Rwall Ro 0.00054 0.01032 0.00013 0.01099 h·F/Btu Then the steady rate of heat transfer through the walls of the house becomes 2730 Btu/h Finally, the total amount of heat lost through the walls during a 24-h period and its cost to the home owner are t (2730 Btu/h)(24-h/day) 65,514 Btu/day 19.2 kWh/day since 1 kWh 3412 Btu, and Heating cost (Energy lost)(Cost of energy) (19.2 kWh/day)($0.075/kWh) $1.44/day Discussion The heat losses through the walls of the house that day cost the home owner $1.44 worth of electricity. Most of this loss can be saved by insulation.
Q Q T1 T2 Rtotal (75 45)F 0.01099 hF/Btu Q 1 ho A 1 (6.0 Btu/hft2F)(1260 ft2) L kA R value A 13 hft2F/Btu 1260 ft2 1 hi A 1 (1.46 Btu/hft2F)(1260 ft2) Rwall Ro Ri T1 T2 Wall, R-13 75°F 45°F T1 T2 FIGURE 3–52 Schematic for Example 3–15.
TOPIC OF SPECIAL INTEREST Heat Transfer through Walls and Roofs Under steady conditions, the rate of heat transfer through any section of a building wall or roof can be determined from Q · UA(Ti To) (3–86) where Ti and To are the indoor and outdoor air temperatures, A is the heat transfer area, U is the overall heat transfer coefficient (the U-factor), and A(Ti To) R This section can be skipped without a loss of continuity.
CHAPTER 3 185 R 1/U is the overall unit thermal resistance (the R-value). Walls and roofs of buildings consist of various layers of materials, and the structure and op-erating conditions of the walls and the roofs may differ significantly from one building to another. Therefore, it is not practical to list the R-values (or U-factors) of different kinds of walls or roofs under different conditions.
Instead, the overall R-value is determined from the thermal resistances of the individual components using the thermal resistance network. The over-all thermal resistance of a structure can be determined most accurately in a lab by actually assembling the unit and testing it as a whole, but this ap-proach is usually very time consuming and expensive. The analytical ap-proach described here is fast and straightforward, and the results are usually in good agreement with the experimental values.
The unit thermal resistance of a plane layer of thickness L and thermal conductivity k can be determined from R L/k. The thermal conductivity and other properties of common building materials are given in the appen-dix. The unit thermal resistances of various components used in building structures are listed in Table 3–8 for convenience.
Heat transfer through a wall or roof section is also affected by the con-vection and radiation heat transfer coefficients at the exposed surfaces. The effects of convection and radiation on the inner and outer surfaces of walls and roofs are usually combined into the combined convection and radiation heat transfer coefficients (also called surface conductances) hi and ho, TABLE 3–8 Unit thermal resistance (the R-value) of common components used in buildings R-value Component m2·°C/W ft2·h·°F/Btu Outside surface (winter) 0.030 0.17 Outside surface (summer) 0.044 0.25 Inside surface, still air 0.12 0.68 Plane air space, vertical, ordinary surfaces (eff 0.82): 13 mm ( in) 0.16 0.90 20 mm ( in) 0.17 0.94 40 mm (1.5 in) 0.16 0.90 90 mm (3.5 in) 0.16 0.91 Insulation, 25 mm (1 in): Glass fiber 0.70 4.00 Mineral fiber batt 0.66 3.73 Urethane rigid foam 0.98 5.56 Stucco, 25 mm (1 in) 0.037 0.21 Face brick, 100 mm (4 in) 0.075 0.43 Common brick, 100 mm (4 in) 0.12 0.79 Steel siding 0.00 0.00 Slag, 13 mm ( in) 0.067 0.38 Wood, 25 mm (1 in) 0.22 1.25 Wood stud, nominal 2 in 4 in (3.5 in or 90 mm wide) 0.63 3.58 1 2 3 4 1 2 R-value Component m2·°C/W ft2·h·°F/Btu Wood stud, nominal 2 in 6 in (5.5 in or 140 mm wide) 0.98 5.56 Clay tile, 100 mm (4 in) 0.18 1.01 Acoustic tile 0.32 1.79 Asphalt shingle roofing 0.077 0.44 Building paper 0.011 0.06 Concrete block, 100 mm (4 in): Lightweight 0.27 1.51 Heavyweight 0.13 0.71 Plaster or gypsum board, 13 mm ( in) 0.079 0.45 Wood fiberboard, 13 mm ( in) 0.23 1.31 Plywood, 13 mm ( in) 0.11 0.62 Concrete, 200 mm (8 in): Lightweight 1.17 6.67 Heavyweight 0.12 0.67 Cement mortar, 13 mm ( in) 0.018 0.10 Wood bevel lapped siding, 13 mm 200 mm ( in 8 in) 0.14 0.81 1 2 1 2 1 2 1 2 1 2 186 STEADY HEAT CONDUCTION respectively, whose values are given in Table 3–9 for ordinary surfaces (e 0.9) and reflective surfaces (e 0.2 or 0.05). Note that surfaces hav-ing a low emittance also have a low surface conductance due to the reduc-tion in radiation heat transfer. The values in the table are based on a surface temperature of 21°C (72°F) and a surface–air temperature difference of 5.5°C (10°F). Also, the equivalent surface temperature of the environment is assumed to be equal to the ambient air temperature. Despite the conve-nience it offers, this assumption is not quite accurate because of the addi-tional radiation heat loss from the surface to the clear sky. The effect of sky radiation can be accounted for approximately by taking the outside tem-perature to be the average of the outdoor air and sky temperatures.
The inner surface heat transfer coefficient hi remains fairly constant throughout the year, but the value of ho varies considerably because of its dependence on the orientation and wind speed, which can vary from less than 1 km/h in calm weather to over 40 km/h during storms. The com-monly used values of hi and ho for peak load calculations are hi 8.29 W/m2·K 1.46 Btu/h·ft2·°F (winter and summer) ho which correspond to design wind conditions of 24 km/h (15 mph) for win-ter and 12 km/h (7.5 mph) for summer. The corresponding surface thermal resistances (R-values) are determined from Ri 1/hi and Ro 1/ho. The surface conductance values under still air conditions can be used for inte-rior surfaces as well as exterior surfaces in calm weather.
Building components often involve trapped air spaces between various layers. Thermal resistances of such air spaces depend on the thickness of the layer, the temperature difference across the layer, the mean air temper-ature, the emissivity of each surface, the orientation of the air layer, and the direction of heat transfer. The emissivities of surfaces commonly encoun-tered in buildings are given in Table 3–10. The effective emissivity of a plane-parallel air space is given by 1 (3–87) where e1 and e2 are the emissivities of the surfaces of the air space. Table 3–10 also lists the effective emissivities of air spaces for the cases where (1) the emissivity of one surface of the air space is e while the emis-sivity of the other surface is 0.9 (a building material) and (2) the emissivity of both surfaces is e. Note that the effective emissivity of an air space between building materials is 0.82/0.03 27 times that of an air space be-tween surfaces covered with aluminum foil. For specified surface tempera-tures, radiation heat transfer through an air space is proportional to effective emissivity, and thus the rate of radiation heat transfer in the ordi-nary surface case is 27 times that of the reflective surface case.
Table 3–11 lists the thermal resistances of 20-mm-, 40-mm-, and 90-mm-(0.75-in, 1.5-in, and 3.5-in) thick air spaces under various conditions. The 1 effective 1 1 1 2 e34.0 W/m2.K 6.0 Btu/h . ft2 . °F (winter) 22.7 W/m2 .K 4.0 Btu/h . ft2 . °F (summer) TABLE 3–9 Combined convection and radiation heat transfer coefficients at window, wall, or roof surfaces (from ASHRAE Handbook of Fundamentals, Chap. 22, Table 1).
Direc-tion of Posi-Heat tion Flow 0.90 0.20 0.05 Still air (both indoors and outdoors) Horiz.
Up ↑ 9.26 5.17 4.32 Horiz.
Down ↓ 6.13 2.10 1.25 45° slope Up ↑ 9.09 5.00 4.15 45° slope Down ↓ 7.50 3.41 2.56 Vertical Horiz. →8.29 4.20 3.35 Moving air (any position, any direction) Winter condition (winds at 15 mph or 24 km/h) 34.0 — — Summer condition (winds at 7.5 mph or 12 km/h) 22.7 — — Multiply by 0.176 to convert to Btu/h·ft2·°F.
Surface resistance can be obtained from R 1/h.
h, W/m2·K Surface Emittance, CHAPTER 3 187 thermal resistance values in the table are applicable to air spaces of uniform thickness bounded by plane, smooth, parallel surfaces with no air leakage.
Thermal resistances for other temperatures, emissivities, and air spaces can be obtained by interpolation and moderate extrapolation. Note that the presence of a low-emissivity surface reduces radiation heat transfer across an air space and thus significantly increases the thermal resistance. The thermal effectiveness of a low-emissivity surface will decline, however, if the condition of the surface changes as a result of some effects such as con-densation, surface oxidation, and dust accumulation.
The R-value of a wall or roof structure that involves layers of uniform thickness is determined easily by simply adding up the unit thermal resis-tances of the layers that are in series. But when a structure involves com-ponents such as wood studs and metal connectors, then the thermal resistance network involves parallel connections and possible two-dimensional effects. The overall R-value in this case can be determined by assuming (1) parallel heat flow paths through areas of different construc-tion or (2) isothermal planes normal to the direction of heat transfer. The first approach usually overpredicts the overall thermal resistance, whereas the second approach usually underpredicts it. The parallel heat flow path approach is more suitable for wood frame walls and roofs, whereas the isothermal planes approach is more suitable for masonry or metal frame walls.
The thermal contact resistance between different components of building structures ranges between 0.01 and 0.1 m2°C/W, which is negligible in most cases. However, it may be significant for metal building components such as steel framing members.
The construction of wood frame flat ceilings typically involve 2-in 6-in joists on 400-mm (16-in) or 600-mm (24-in) centers. The fraction of framing is usually taken to be 0.10 for joists on 400-mm centers and 0.07 for joists on 600-mm centers.
Most buildings have a combination of a ceiling and a roof with an attic space in between, and the determination of the R-value of the roof–attic– ceiling combination depends on whether the attic is vented or not. For ade-quately ventilated attics, the attic air temperature is practically the same as the outdoor air temperature, and thus heat transfer through the roof is gov-erned by the R-value of the ceiling only. However, heat is also transferred between the roof and the ceiling by radiation, and it needs to be considered (Fig. 3–53). The major function of the roof in this case is to serve as a ra-diation shield by blocking off solar radiation. Effectively ventilating the at-tic in summer should not lead one to believe that heat gain to the building through the attic is greatly reduced. This is because most of the heat trans-fer through the attic is by radiation.
Radiation heat transfer between the ceiling and the roof can be mini-mized by covering at least one side of the attic (the roof or the ceiling side) by a reflective material, called radiant barrier, such as aluminum foil or aluminum-coated paper. Tests on houses with R-19 attic floor insulation have shown that radiant barriers can reduce summer ceiling heat gains by 16 to 42 percent compared to an attic with the same insulation level and no Air intake Air intake 3 in 3 in Radiant barrier Air exhaust 6 in FIGURE 3–53 Ventilation paths for a naturally ventilated attic and the appropriate size of the flow areas around the radiant barrier for proper air circulation (from DOE/CE-0335P, U.S. Dept. of Energy).
TABLE 3–10 Emissivities of various surfaces and the effective emissivity of air spaces (from ASHRAE Handbook of Fundamentals, Chap. 22, Table 3).
Effective Emissivity of Air Space 1 1 Surface 2 0.9 2 Aluminum foil, bright 0.05 0.05 0.03 Aluminum sheet 0.12 0.12 0.06 Aluminum-coated paper, polished 0.20 0.20 0.11 Steel, galvanized, bright 0.25 0.24 0.15 Aluminum paint 0.50 0.47 0.35 Building materials: Wood, paper, masonry, nonmetallic paints 0.90 0.82 0.82 Ordinary glass 0.84 0.77 0.72 Surface emissivity of aluminum foil increases to 0.30 with barely visible condensation, and to 0.70 with clearly visible condensation.
TABLE 3–11 Unit thermal resistances (R-values) of well-sealed plane air spaces (from ASHRAE Handbook of Fundamentals, Chap. 22, Table 2) (a) SI units (in m2·°C/W) 20-mm Air Space 40-mm Air Space 90-mm Air Space Effective Effective Effective Emissivity, eff Emissivity, eff Emissivity, eff Position Direction Mean Temp.
of Air of Heat Temp., Diff., Space Flow °C °C 0.03 0.05 0.5 0.82 0.03 0.05 0.5 0.82 0.03 0.05 0.5 0.82 32.2 5.6 0.41 0.39 0.18 0.13 0.45 0.42 0.19 0.14 0.50 0.47 0.20 0.14 10.0 16.7 0.30 0.29 0.17 0.14 0.33 0.32 0.18 0.14 0.27 0.35 0.19 0.15 Horizontal Up ↑ 10.0 5.6 0.40 0.39 0.20 0.15 0.44 0.42 0.21 0.16 0.49 0.47 0.23 0.16 17.8 11.1 0.32 0.32 0.20 0.16 0.35 0.34 0.22 0.17 0.40 0.38 0.23 0.18 32.2 5.6 0.52 0.49 0.20 0.14 0.51 0.48 0.20 0.14 0.56 0.52 0.21 0.14 10.0 16.7 0.35 0.34 0.19 0.14 0.38 0.36 0.20 0.15 0.40 0.38 0.20 0.15 45° slope Up ↑ 10.0 5.6 0.51 0.48 0.23 0.17 0.51 0.48 0.23 0.17 0.55 0.52 0.24 0.17 17.8 11.1 0.37 0.36 0.23 0.18 0.40 0.39 0.24 0.18 0.43 0.41 0.24 0.19 32.2 5.6 0.62 0.57 0.21 0.15 0.70 0.64 0.22 0.15 0.65 0.60 0.22 0.15 10.0 16.7 0.51 0.49 0.23 0.17 0.45 0.43 0.22 0.16 0.47 0.45 0.22 0.16 Vertical Horizontal → 10.0 5.6 0.65 0.61 0.25 0.18 0.67 0.62 0.26 0.18 0.64 0.60 0.25 0.18 17.8 11.1 0.55 0.53 0.28 0.21 0.49 0.47 0.26 0.20 0.51 0.49 0.27 0.20 32.2 5.6 0.62 0.58 0.21 0.15 0.89 0.80 0.24 0.16 0.85 0.76 0.24 0.16 10.0 16.7 0.60 0.57 0.24 0.17 0.63 0.59 0.25 0.18 0.62 0.58 0.25 0.18 45° slope Down ↓ 10.0 5.6 0.67 0.63 0.26 0.18 0.90 0.82 0.28 0.19 0.83 0.77 0.28 0.19 17.8 11.1 0.66 0.63 0.30 0.22 0.68 0.64 0.31 0.22 0.67 0.64 0.31 0.22 32.2 5.6 0.62 0.58 0.21 0.15 1.07 0.94 0.25 0.17 1.77 1.44 0.28 0.18 10.0 16.7 0.66 0.62 0.25 0.18 1.10 0.99 0.30 0.20 1.69 1.44 0.33 0.21 Horizontal Down ↓ 10.0 5.6 0.68 0.63 0.26 0.18 1.16 1.04 0.30 0.20 1.96 1.63 0.34 0.22 17.8 11.1 0.74 0.70 0.32 0.23 1.24 1.13 0.39 0.26 1.92 1.68 0.43 0.29 (b) English units (in h·ft2·°F/Btu) 0.75-in Air Space 1.5-in Air Space 3.5-in Air Space Effective Effective Effective Emissivity, eff Emissivity, eff Emissivity, eff Position Direction Mean Temp.
of Air of Heat Temp., Diff., Space Flow °F °F 0.03 0.05 0.5 0.82 0.03 0.05 0.5 0.82 0.03 0.05 0.5 0.82 90 10 2.34 2.22 1.04 0.75 2.55 2.41 1.08 0.77 2.84 2.66 1.13 0.80 50 30 1.71 1.66 0.99 0.77 1.87 1.81 1.04 0.80 2.09 2.01 1.10 0.84 Horizontal Up ↑ 50 10 2.30 2.21 1.16 0.87 2.50 2.40 1.21 0.89 2.80 2.66 1.28 0.93 0 20 1.83 1.79 1.16 0.93 2.01 1.95 1.23 0.97 2.25 2.18 1.32 1.03 90 10 2.96 2.78 1.15 0.81 2.92 2.73 1.14 0.80 3.18 2.96 1.18 0.82 50 30 1.99 1.92 1.08 0.82 2.14 2.06 1.12 0.84 2.26 2.17 1.15 0.86 45° slope Up ↑ 50 10 2.90 2.75 1.29 0.94 2.88 2.74 1.29 0.94 3.12 2.95 1.34 0.96 0 20 2.13 2.07 1.28 1.00 2.30 2.23 1.34 1.04 2.42 2.35 1.38 1.06 90 10 3.50 3.24 1.22 0.84 3.99 3.66 1.27 0.87 3.69 3.40 1.24 0.85 50 30 2.91 2.77 1.30 0.94 2.58 2.46 1.23 0.90 2.67 2.55 1.25 0.91 Vertical Horizontal → 50 10 3.70 3.46 1.43 1.01 3.79 3.55 1.45 1.02 3.63 3.40 1.42 1.01 0 20 3.14 3.02 1.58 1.18 2.76 2.66 1.48 1.12 2.88 2.78 1.51 1.14 90 10 3.53 3.27 1.22 0.84 5.07 4.55 1.36 0.91 4.81 4.33 1.34 0.90 50 30 3.43 3.23 1.39 0.99 3.58 3.36 1.42 1.00 3.51 3.30 1.40 1.00 45° slope Down ↓ 50 10 3.81 3.57 1.45 1.02 5.10 4.66 1.60 1.09 4.74 4.36 1.57 1.08 0 20 3.75 3.57 1.72 1.26 3.85 3.66 1.74 1.27 3.81 3.63 1.74 1.27 90 10 3.55 3.29 1.22 0.85 6.09 5.35 1.43 0.94 10.07 8.19 1.57 1.00 50 30 3.77 3.52 1.44 1.02 6.27 5.63 1.70 1.14 9.60 8.17 1.88 1.22 Horizontal Down ↓ 50 10 3.84 3.59 1.45 1.02 6.61 5.90 1.73 1.15 11.15 9.27 1.93 1.24 0 20 4.18 3.96 1.81 1.30 7.03 6.43 2.19 1.49 10.90 9.52 2.47 1.62 188 CHAPTER 3 189 radiant barrier. Considering that the ceiling heat gain represents about 15 to 25 percent of the total cooling load of a house, radiant barriers will reduce the air conditioning costs by 2 to 10 percent. Radiant barriers also reduce the heat loss in winter through the ceiling, but tests have shown that the percentage reduction in heat losses is less. As a result, the percentage reduction in heating costs will be less than the reduction in the air-conditioning costs. Also, the values given are for new and undusted radiant barrier installations, and percentages will be lower for aged or dusty radi-ant barriers.
Some possible locations for attic radiant barriers are given in Figure 3–54.
In whole house tests on houses with R-19 attic floor insulation, radiant bar-riers have reduced the ceiling heat gain by an average of 35 percent when the radiant barrier is installed on the attic floor, and by 24 percent when it is attached to the bottom of roof rafters. Test cell tests also demonstrated that the best location for radiant barriers is the attic floor, provided that the attic is not used as a storage area and is kept clean.
For unvented attics, any heat transfer must occur through (1) the ceiling, (2) the attic space, and (3) the roof (Fig. 3–55). Therefore, the overall R-value of the roof–ceiling combination with an unvented attic depends on the combined effects of the R-value of the ceiling and the R-value of the roof as well as the thermal resistance of the attic space. The attic space can be treated as an air layer in the analysis. But a more practical way of ac-counting for its effect is to consider surface resistances on the roof and ceil-ing surfaces facing each other. In this case, the R-values of the ceiling and the roof are first determined separately (by using convection resistances for the still-air case for the attic surfaces). Then it can be shown that the over-all R-value of the ceiling–roof combination per unit area of the ceiling can be expressed as R Rceiling Rroof (3–88) Aceiling Aroof Rroof Rceiling To Ti Deck Shingles Rafter Attic Ceiling joist Aroof Tattic Aceiling FIGURE 3–55 Thermal resistance network for a pitched roof–attic–ceiling combination for the case of an unvented attic.
FIGURE 3–54 Three possible locations for an attic radiant barrier (from DOE/CE-0335P, U.S. Dept. of Energy).
(a) Under the roof deck (b) At the bottom of rafters (c) On top of attic floor insulation Air space Rafter Rafter Roof decking Radiant barrier Roof decking Roof decking Joist Joist Joist Insulation Insulation Insulation Radiant barrier Radiant barrier Rafter 190 STEADY HEAT CONDUCTION where Aceiling and Aroof are the ceiling and roof areas, respectively. The area ratio is equal to 1 for flat roofs and is less than 1 for pitched roofs. For a 45° pitched roof, the area ratio is Aceiling/Aroof 1/ 0.707. Note that the pitched roof has a greater area for heat transfer than the flat ceiling, and the area ratio accounts for the reduction in the unit R-value of the roof when expressed per unit area of the ceiling. Also, the direction of heat flow is up in winter (heat loss through the roof) and down in summer (heat gain through the roof).
The R-value of a structure determined by analysis assumes that the materials used and the quality of workmanship meet the standards. Poor workmanship and substandard materials used during construction may result in R-values that deviate from predicted values. Therefore, some en-gineers use a safety factor in their designs based on experience in critical applications.
22 EXAMPLE 3–16 The R-Value of a Wood Frame Wall Determine the overall unit thermal resistance (the R-value) and the overall heat transfer coefficient (the U-factor) of a wood frame wall that is built around 38-mm 90-mm (2 4 nominal) wood studs with a center-to-center distance of 400 mm. The 90-mm-wide cavity between the studs is filled with glass fiber insulation. The inside is finished with 13-mm gypsum wallboard and the out-side with 13-mm wood fiberboard and 13-mm 200-mm wood bevel lapped siding. The insulated cavity constitutes 75 percent of the heat transmission area while the studs, plates, and sills constitute 21 percent. The headers con-stitute 4 percent of the area, and they can be treated as studs.
Also, determine the rate of heat loss through the walls of a house whose perimeter is 50 m and wall height is 2.5 m in Las Vegas, Nevada, whose win-ter design temperature is 2°C. Take the indoor design temperature to be 22°C and assume 20 percent of the wall area is occupied by glazing.
SOLUTION The R-value and the U-factor of a wood frame wall as well as the rate of heat loss through such a wall in Las Vegas are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant.
Properties The R-values of different materials are given in Table 3–8.
Analysis The schematic of the wall as well as the different elements used in its construction are shown here. Heat transfer through the insulation and through the studs meets different resistances, and thus we need to analyze the thermal resistance for each path separately. Once the unit thermal resistances and the U-factors for the insulation and stud sections are available, the overall average thermal resistance for the entire wall can be determined from Roverall 1/Uoverall where Uoverall (U farea)insulation (U farea)stud 191 CHAPTER 3 and the value of the area fraction farea is 0.75 for the insulation section and 0.25 for the stud section since the headers that constitute a small part of the wall are to be treated as studs. Using the available R-values from Table 3–8 and calculating others, the total R-values for each section can be determined in a systematic manner in the table below.
Schematic R-value, m2·°C/W Between At Construction Studs Studs 1.
Outside surface, 24 km/h wind 0.030 0.030 2.
Wood bevel lapped siding 0.14 0.14 3.
Wood fiberboard sheeting, 13 mm 0.23 0.23 4a. Glass fiber insulation, 90 mm 2.45 — 4b. Wood stud, 38 mm 90 mm — 0.63 5.
Gypsum wallboard, 13 mm 0.079 0.079 6.
Inside surface, still air 0.12 0.12 Total unit thermal resistance of each section, R (in m2·°C/W) 3.05 1.23 The U-factor of each section, U 1/R, in W/m2·K 0.328 0.813 Area fraction of each section, farea 0.75 0.25 Overall U-factor: U farea, i Ui 0.75 0.328 0.25 0.813 0.449 W/m2·K Overall unit thermal resistance: R 1/U 2.23 m2·°C/W We conclude that the overall unit thermal resistance of the wall is 2.23 m2°C/W, and this value accounts for the effects of the studs and head-ers. It corresponds to an R-value of 2.23 5.68 12.7 (or nearly R-13) in English units. Note that if there were no wood studs and headers in the wall, the overall thermal resistance would be 3.05 m2°C/W, which is 37 percent greater than 2.23 m2°C/W. Therefore, the wood studs and headers in this case serve as thermal bridges in wood frame walls, and their effect must be consid-ered in the thermal analysis of buildings.
The perimeter of the building is 50 m and the height of the walls is 2.5 m.
Noting that glazing constitutes 20 percent of the walls, the total wall area is Awall 0.80(Perimeter)(Height) 0.80(50 m)(2.5 m) 100 m2 Then the rate of heat loss through the walls under design conditions becomes Q · wall (UA)wall (Ti To) (0.449 W/m2·K)(100 m2)[22 (2)°C] 1078 W Discussion Note that a 1-kW resistance heater in this house will make up al-most all the heat lost through the walls, except through the doors and windows, when the outdoor air temperature drops to 2°C.
4a 1 3 2 4b 5 6 192 STEADY HEAT CONDUCTION EXAMPLE 3–17 The R-Value of a Wall with Rigid Foam The 13-mm-thick wood fiberboard sheathing of the wood stud wall discussed in the previous example is replaced by a 25-mm-thick rigid foam insulation.
Determine the percent increase in the R-value of the wall as a result.
SOLUTION The overall R-value of the existing wall was determined in Exam-ple 3–16 to be 2.23 m2°C/W. Noting that the R-values of the fiberboard and the foam insulation are 0.23 m2°C/W and 0.98 m2°C/W, respectively, and the added and removed thermal resistances are in series, the overall R-value of the wall after modification becomes Rnew Rold Rremoved Radded 2.23 0.23 0.98 2.98 m2·°C/W This represents an increase of (2.98 2.23)/2.23 0.34 or 34 percent in the R-value of the wall. This example demonstrated how to evaluate the new R-value of a structure when some structural members are added or removed.
EXAMPLE 3–18 The R-Value of a Masonry Wall Determine the overall unit thermal resistance (the R-value) and the overall heat transfer coefficient (the U-factor) of a masonry cavity wall that is built around 6-in-thick concrete blocks made of lightweight aggregate with 3 cores filled with perlite (R 4.2 h·ft2·°F/Btu). The outside is finished with 4-in face brick with -in cement mortar between the bricks and concrete blocks. The inside finish consists of -in gypsum wallboard separated from the concrete block by -in-3 4 1 2 1 2 thick (1-in 3-in nominal) vertical furring (R 4.2 h·ft2·°F/Btu) whose cen-ter-to-center distance is 16 in. Both sides of the -in-thick air space between the concrete block and the gypsum board are coated with reflective aluminum foil (e 0.05) so that the effective emissivity of the air space is 0.03. For a mean temperature of 50°F and a temperature difference of 30°F, the R-value of the air space is 2.91 hft2°F/Btu. The reflective air space constitutes 80 percent of the heat transmission area, while the vertical furring constitutes 20 percent.
SOLUTION The R-value and the U-factor of a masonry cavity wall are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant.
Properties The R-values of different materials are given in Table 3–8.
Analysis The schematic of the wall as well as the different elements used in its construction are shown below. Following the approach described here and using the available R-values from Table 3–8, the overall R-value of the wall is determined in the following table.
3 4 193 CHAPTER 3 Schematic R-value, h·ft2·°F/Btu Between At Construction Furring Furring 1.
Outside surface, 15 mph wind 0.17 0.17 2.
Face brick, 4 in 0.43 0.43 3.
Cement mortar, 0.5 in 0.10 0.10 4.
Concrete block, 6 in 4.20 4.20 5a. Reflective air space, in 2.91 — 5b. Nominal 1 3 vertical furring — 0.94 6.
Gypsum wallboard, 0.5 in 0.45 0.45 7.
Inside surface, still air 0.68 0.68 Total unit thermal resistance of each section, R 8.94 6.97 The U-factor of each section, U 1/R, in Btu/h·ft2·°F 0.112 0.143 Area fraction of each section, farea 0.80 0.20 Overall U-factor: U farea, i Ui 0.80 0.112 0.20 0.143 0.118 Btu/h·ft2·°F Overall unit thermal resistance: R 1/U 8.46 h·ft2·°F/Btu Therefore, the overall unit thermal resistance of the wall is 8.46 hft2°F/Btu and the overall U-factor is 0.118 Btu/hft2°F. These values account for the ef-fects of the vertical furring.
3 4 7 6 5a 4 3 2 1 5b EXAMPLE 3–19 The R-Value of a Pitched Roof Determine the overall unit thermal resistance (the R-value) and the overall heat transfer coefficient (the U-factor) of a 45° pitched roof built around nominal 2-in 4-in wood studs with a center-to-center distance of 16 in. The 3.5-in-wide air space between the studs does not have any reflective surface and thus its effective emissivity is 0.84. For a mean temperature of 90°F and a temper-ature difference of 30°F, the R-value of the air space is 0.86 hft2°F/Btu. The lower part of the roof is finished with -in gypsum wallboard and the upper part with -in plywood, building paper, and asphalt shingle roofing. The air space constitutes 75 percent of the heat transmission area, while the studs and headers constitute 25 percent.
SOLUTION The R-value and the U-factor of a 45° pitched roof are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the roof is one-dimensional. 3 Thermal properties of the roof and the heat transfer coefficients are constant.
5 8 1 2 194 STEADY HEAT CONDUCTION Properties The R-values of different materials are given in Table 3–8.
Analysis The schematic of the pitched roof as well as the different elements used in its construction are shown below. Following the approach described above and using the available R-values from Table 3–8, the overall R-value of the roof can be determined in the table here.
Schematic R-value, h·ft2·°F/Btu Between At Construction Studs Studs 1.
Outside surface, 15 mph wind 0.17 0.17 2.
Asphalt shingle roofing 0.44 0.44 3.
Building paper 0.06 0.06 4.
Plywood deck, in 0.78 0.78 5a. Nonreflective air space, 3.5 in 0.86 — 5b. Wood stud, 2 in 4 in — 3.58 6.
Gypsum wallboard, 0.5 in 0.45 0.45 7.
Inside surface, 45° slope, still air 0.63 0.63 Total unit thermal resistance of each section, R 3.39 6.11 The U-factor of each section, U 1/R, in Btu/h·ft2·°F 0.292 0.163 Area fraction of each section, farea 0.75 0.25 Overall U-factor: U farea, i Ui 0.75 0.292 0.25 0.163 0.260 Btu/h·ft2·°F Overall unit thermal resistance: R 1/U 3.85 h·ft2·°F/Btu Therefore, the overall unit thermal resistance of this pitched roof is 3.85 hft2°F/Btu and the overall U-factor is 0.260 Btu/hft2°F. Note that the wood studs offer much larger thermal resistance to heat flow than the air space between the studs.
5 8 45° 1 2 3 4 5a 5b 6 7 SUMMARY One-dimensional heat transfer through a simple or composite body exposed to convection from both sides to mediums at temperatures T1 and T2 can be expressed as where Rtotal is the total thermal resistance between the two mediums. For a plane wall exposed to convection on both sides, the total resistance is expressed as Rtotal Rconv, 1 Rwall Rconv, 2 This relation can be extended to plane walls that consist of two or more layers by adding an additional resistance for each additional layer. The elementary thermal resistance relations can be expressed as follows: Conduction resistance (plane wall): Rwall Conduction resistance (cylinder): Rcyl Conduction resistance (sphere): Rsph Convection resistance: Rconv Interface resistance: Rinterface Radiation resistance: Rrad 1 hrad A 1 hc A Rc A 1 hA r2 r1 4pr1r2k ln(r2/r1) 2pLk L kA 1 h1 A L kA 1 h2 A T1 T2 Rtotal Q CHAPTER 3 195 where hc is the thermal contact conductance, Rc is the thermal contact resistance, and the radiation heat transfer coefficient is defined as hrad es (Ts Tsurr) Once the rate of heat transfer is available, the temperature drop across any layer can be determined from T R The thermal resistance concept can also be used to solve steady heat transfer problems involving parallel layers or combined series-parallel arrangements.
Adding insulation to a cylindrical pipe or a spherical shell increases the rate of heat transfer if the outer radius of the in-sulation is less than the critical radius of insulation, defined as rcr, cylinder rcr, sphere The effectiveness of an insulation is often given in terms of its R-value, the thermal resistance of the material for a unit sur-face area, expressed as R-value (flat insulation) where L is the thickness and k is the thermal conductivity of the material.
Finned surfaces are commonly used in practice to enhance heat transfer. Fins enhance heat transfer from a surface by ex-posing a larger surface area to convection. The temperature dis-tribution along the fin are given by Very long fin: Adiabatic fin tip: Specified temperature at fin tip: Convection from fin tip: where m , p is the perimeter, and Ac is the cross-sectional area of the fin. The rates of heat transfer for these cases are given to be Very long fin: long fin kAc dT dx x 0 2hpkAc (Tb T) Q 2hp/kAc T(x) T Tb T cosh m(L x) (h
mk) sinh m(L x) cosh mL (h
mk) sinh mL T(x) T Tb T [(TL T)
(Tb T)] sinh mx sinh m(L x) sinh mL T(x) T Tb T cosh m(L x) cosh mL T(x) T Tb T ex2hp/kAc L k 2kins h kins h Q (T 2 s T 2 surr) Adiabatic fin tip: adiabatic tip kAc Specified temperature at fin tip: Convection from the fin tip: Fins exposed to convection at their tips can be treated as fins with adiabatic tips by using the corrected length Lc L Ac/p instead of the actual fin length.
The temperature of a fin drops along the fin, and thus the heat transfer from the fin is less because of the decreasing temperature difference toward the fin tip. To account for the effect of this de-crease in temperature on heat transfer, we define fin efficiency as hfin When the fin efficiency is available, the rate of heat transfer from a fin can be determined from fin hfin fin, max hfinhAfin (Tb T) The performance of the fins is judged on the basis of the en-hancement in heat transfer relative to the no-fin case and is ex-pressed in terms of the fin effectiveness efin, defined as efin Here, Ab is the cross-sectional area of the fin at the base and no fin represents the rate of heat transfer from this area if no fins are attached to the surface. The overall effectiveness for a finned surface is defined as the ratio of the total heat transfer from the finned surface to the heat transfer from the same sur-face if there were no fins, efin, overall Fin efficiency and fin effectiveness are related to each other by efin hfin Certain multidimensional heat transfer problems involve two surfaces maintained at constant temperatures T1 and T2. The steady rate of heat transfer between these two surfaces is expressed as Sk(T1 T2) where S is the conduction shape factor that has the dimension of length and k is the thermal conductivity of the medium be-tween the surfaces.
Q Afin Ab Qtotal, fin Qtotal, no fin h(Aunfin hfin Afin)(Tb T) hAno fin (Tb T) Q Qfin Qno fin Qfin hAb (Tb T) Heat transfer rate from the fin of base area Ab Heat transfer rate from the surface of area Ab Q Q Qfin Qfin, max Actual heat transfer rate from the fin Ideal heat transfer rate from the fin if the entire fin were at base temperature Q # convection 2h p k Ac (Tb T) sinh mL (h
mk) cosh mL cosh mL (h
mk) sinh mL Q # specified temp. 2h p k Ac (Tb T) cosh mL [(TL T)
(Tb T)] sinh mL dT dx x 0 2hpkAc (Tb T) tanh mL Q REFERENCES AND SUGGESTED READING 1. American Society of Heating, Refrigeration, and Air Conditioning Engineers. Handbook of Fundamentals.
Atlanta: ASHRAE, 1993.
2. R. V. Andrews. “Solving Conductive Heat Transfer Problems with Electrical-Analogue Shape Factors.” Chemical Engineering Progress 5 (1955), p. 67.
3. R. Barron. Cryogenic Systems. New York: McGraw-Hill, 1967.
4. L. S. Fletcher. “Recent Developments in Contact Conductance Heat Transfer.” Journal of Heat Transfer 110, no. 4B (1988), pp. 1059–79.
5. E. Fried. “Thermal Conduction Contribution to Heat Transfer at Contacts.” Thermal Conductivity, vol. 2, ed. R. P. Tye. London: Academic Press, 1969.
6. K. A. Gardner. “Efficiency of Extended Surfaces.” Trans.
ASME 67 (1945), pp. 621–31. Reprinted by permission of ASME International.
7. D. Q. Kern and A. D. Kraus. Extended Surface Heat Transfer. New York: McGraw-Hill, 1972.
8. G. P. Peterson. “Thermal Contact Resistance in Waste Heat Recovery Systems.” Proceedings of the 18th ASME/ETCE Hydrocarbon Processing Symposium.
Dallas, TX, 1987, pp. 45–51. Reprinted by permission of ASME International.
9. S. Song, M. M. Yovanovich, and F. O. Goodman.
“Thermal Gap Conductance of Conforming Surfaces in Contact.” Journal of Heat Transfer 115 (1993), p. 533.
10. J. E. Sunderland and K. R. Johnson. “Shape Factors for Heat Conduction through Bodies with Isothermal or Convective Boundary Conditions.” Trans. ASME 10 (1964), pp. 2317–41.
11. W. M. Edmunds. “Residential Insulation.” ASTM Standardization News (Jan. 1989), pp. 36–39.
196 STEADY HEAT CONDUCTION PROBLEMS Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them.
Problems with the icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with the icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text.
2 mm 3 mm 2 mm Aluminum Copper FIGURE P3–4C Steady Heat Conduction in Plane Walls 3–1C Consider heat conduction through a wall of thickness L and area A. Under what conditions will the temperature dis-tributions in the wall be a straight line?
3–2C Consider heat conduction through a plane wall. Does the energy content of the wall change during steady heat con-duction? How about during transient conduction? Explain.
3–3C Consider steady heat transfer through the wall of a room in winter. The convection heat transfer coefficient at the outer surface of the wall is three times that of the inner surface as a result of the winds. On which surface of the wall do you think the temperature will be closer to the surrounding air tem-perature? Explain.
3–4C The bottom of a pan is made of a 4-mm-thick alu-minum layer. In order to increase the rate of heat transfer through the bottom of the pan, someone proposes a design for the bottom that consists of a 3-mm-thick copper layer sandwiched between two 2-mm-thick aluminum layers. Will the new design conduct heat better? Explain. Assume perfect contact between the layers.
3–5C Consider one-dimensional heat conduction through a cylindrical rod of diameter D and length L. What is the heat transfer area of the rod if (a) the lateral surfaces of the rod are insulated and (b) the top and bottom surfaces of the rod are insulated?
3–6C What does the thermal resistance of a medium represent?
3–7C How is the combined heat transfer coefficient defined?
What convenience does it offer in heat transfer calculations?
CHAPTER 3 197 3–8C Can we define the convection resistance for a unit surface area as the inverse of the convection heat transfer coefficient?
3–9C Why are the convection and the radiation resistances at a surface in parallel instead of being in series?
3–10C Consider a surface of area A at which the convection and radiation heat transfer coefficients are hconv and hrad, re-spectively. Explain how you would determine (a) the single equivalent heat transfer coefficient, and (b) the equivalent ther-mal resistance. Assume the medium and the surrounding sur-faces are at the same temperature.
3–11C How does the thermal resistance network associated with a single-layer plane wall differ from the one associated with a five-layer composite wall?
3–12C Consider steady one-dimensional heat transfer through a multilayer medium. If the rate of heat transfer is known, explain how you would determine the temperature drop across each layer.
3–13C Consider steady one-dimensional heat transfer through a plane wall exposed to convection from both sides to environments at known temperatures T1 and T2 with known heat transfer coefficients h1 and h2. Once the rate of heat trans-fer has been evaluated, explain how you would determine the temperature of each surface.
3–14C Someone comments that a microwave oven can be viewed as a conventional oven with zero convection resistance at the surface of the food. Is this an accurate statement?
3–15C Consider a window glass consisting of two 4-mm-thick glass sheets pressed tightly against each other. Compare the heat transfer rate through this window with that of one consisting of a single 8-mm-thick glass sheet under identical conditions.
3–16C Consider two cold canned drinks, one wrapped in a blanket and the other placed on a table in the same room.
Which drink will warm up faster?
3–17 Consider a 3-m-high, 6-m-wide, and 0.25-m-thick brick wall whose thermal conductivity is k 0.8 W/mK. On a cer-tain day, the temperatures of the inner and the outer surfaces of the wall are measured to be 14°C and 5°C, respectively. Deter-mine the rate of heat loss through the wall on that day.
3–18 Water is boiling in a 25-cm-diameter aluminum pan (k 237 W/mK) at 95°C. Heat is transferred steadily to the boiling water in the pan through its 0.5-cm-thick flat bottom at a rate of 800 W. If the inner surface temperature of the bottom of the pan is 108°C, determine (a) the boiling heat transfer co-efficient on the inner surface of the pan and (b) the outer sur-face temperature of the bottom of the pan.
3–19 Consider a 1.5-m-high and 2.4-m-wide glass window whose thickness is 6 mm and thermal conductivity is k 0.78 W/mK. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 24°C while the tem-perature of the outdoors is 5°C. Take the convection heat Q # Q transfer coefficients on the inner and outer surfaces of the win-dow to be h1 10 W/m2K and h2 25 W/m2K, and disre-gard any heat transfer by radiation.
3–20 Consider a 1.5-m-high and 2.4-m-wide double-pane window consisting of two 3-mm-thick layers of glass (k 0.78 W/mK) separated by a 12-mm-wide stagnant air space (k 0.026 W/mK). Determine the steady rate of heat transfer through this double-pane window and the temperature of its in-ner surface for a day during which the room is maintained at 21°C while the temperature of the outdoors is 5°C. Take the convection heat transfer coefficients on the inner and outer sur-faces of the window to be h1 10 W/m2K and h2 25 W/m2K, and disregard any heat transfer by radiation.
Answers: 153 W, 14.6°C Frame 3 12 3 mm Glass FIGURE P3–20 3–21 Repeat Prob. 3–20, assuming the space between the two glass layers is evacuated.
3–22 Reconsider Prob. 3–20. Using EES (or other) soft-ware, plot the rate of heat transfer through the window as a function of the width of air space in the range of 2 mm to 20 mm, assuming pure conduction through the air.
Discuss the results.
3–23E Consider an electrically heated brick house (k 0.40 Btu/hft°F) whose walls are 9 ft high and 1 ft thick. Two of the walls of the house are 50 ft long and the others are 35 ft long. The house is maintained at 70°F at all times while the temperature of the outdoors varies. On a certain day, the tem-perature of the inner surface of the walls is measured to be at 55°F while the average temperature of the outer surface is ob-served to remain at 45°F during the day for 10 h and at 35°F at night for 14 h. Determine the amount of heat lost from the house that day. Also determine the cost of that heat loss to the home owner for an electricity price of $0.09/kWh.
198 STEADY HEAT CONDUCTION surface temperature of the transparent film. Assume thermal contact resistance is negligible.
Answers: 127°C, 103°C 35 ft 50 ft 9 ft Tin = 70°F FIGURE P3–23E 3–24 A cylindrical resistor element on a circuit board dissi-pates 0.15 W of power in an environment at 35°C. The resistor is 1.2 cm long, and has a diameter of 0.3 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-h period; (b) the heat flux on the surface of the resistor, in W/m2; and (c) the surface temperature of the resistor for a combined con-vection and radiation heat transfer coefficient of 9 W/m2K.
3–25 To defog the rear window of an automobile, a very thin transparent heating element is attached to the inner surface of the window. A Uniform heat flux of 1300 W/m2 is provided to the heating element for defogging a rear window with thick-ness of 5 mm. The interior temperature of the automobile is 22°C and the convection heat transfer coefficient is 15 W/m2·K.
The outside ambient temperature is 5°C and the convection heat transfer coefficient is 100 W/m2·K. If the thermal conduc-tivity of the window is 1.2 W/m·K, determine the inner surface temperature of the window.
T1 Rear window k = 1.2 W/m·K Inside air, 22°C h = 15 W/m2·K Outside air, –5°C h = 100 W/m2·K L = 5 mm q .
h = 1300 W/m2 Heating element FIGURE P3–25 3–26 A transparent film is to be bonded onto the top surface of a solid plate inside a heated chamber. For the bond to cure properly, a temperature of 70°C is to be maintained at the bond, between the film and the solid plate. The transparent film has a thickness of 1 mm and thermal conductivity of 0.05 W/m·K, while the solid plate is 13 mm thick and has a thermal conductivity of 1.2 W/m·K. Inside the heated chamber, the convection heat transfer coefficient is 70 W/m2·K. If the bottom surface of the solid plate is maintained at 52°C, deter-mine the temperature inside the heated chamber and the T2 = 52°C T1 Air, h = 70 W/m2·K Transparent film kf = 0.05 W/m·K Lf = 1 mm Ls = 13 mm Tb = 70°C Solid plate ks = 1.2 W/m·K FIGURE P3–26 Power transistor 0.15 W 0.5 cm 30°C 0.4 cm FIGURE P3–27 3–27 Consider a power transistor that dissipates 0.15 W of power in an environment at 30°C. The transistor is 0.4 cm long and has a diameter of 0.5 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-h period, in kWh; (b) the heat flux on the surface of the transistor, in W/m2; and (c) the surface temperature of the resistor for a combined convection and radiation heat transfer coefficient of 18 W/m2K.
3–28 A 12-cm 18-cm circuit board houses on its surface 100 closely spaced logic chips, each dissipating 0.06 W in an environment at 40°C. The heat transfer from the back surface of the board is negligible. If the heat transfer coefficient on the surface of the board is 10 W/m2K, determine (a) the heat flux on the surface of the circuit board, in W/m2; (b) the surface temperature of the chips; and (c) the thermal resistance be-tween the surface of the circuit board and the cooling medium, in °C/W.
3–29 Consider a person standing in a room at 20°C with an exposed surface area of 1.7 m2. The deep body temperature of the human body is 37°C, and the thermal conductivity of the human tissue near the skin is about 0.3 W/mK. The body is losing heat at a rate of 150 W by natural convection and CHAPTER 3 199 radiation to the surroundings. Taking the body temperature 0.5 cm beneath the skin to be 37°C, determine the skin temperature of the person.
Answer: 35.5C 3–30 A 1.0 m 1.5 m double-pane window consists of two 4-mm-thick layers of glass (k 0.78 W/mK) that are the sep-arated by a 5-mm air gap (kair 0.025 W/mK). The heat flow through the air gap is assumed to be by conduction. The inside and outside air temperatures are 20°C and 20°C, respectively, and the inside and outside heat transfer coefficients are 40 and 20 W/m2K. Determine (a) the daily rate of heat loss through the window in steady operation and (b) the temperature differ-ence across the largest thermal resistence.
3–31E A wall is constructed of two layers of 0.6-in-thick sheetrock (k 0.10 Btu/hft°F), which is a plasterboard made of two layers of heavy paper separated by a layer of gypsum, placed 7 in apart. The space between the sheetrocks is filled with fiberglass insulation (k 0.020 Btu/hft°F). Determine (a) the thermal resistance of the wall and (b) its R-value of in-sulation in English units.
3–33 A 2-m 1.5-m section of wall of an industrial furnace burning natural gas is not insulated, and the temperature at the outer surface of this section is measured to be 110°C. The tem-perature of the furnace room is 32°C, and the combined con-vection and radiation heat transfer coefficient at the surface of the outer furnace is 10 W/m2K. It is proposed to insulate this section of the furnace wall with glass wool insulation (k 0.038 W/mK) in order to reduce the heat loss by 90 percent.
Assuming the outer surface temperature of the metal section still remains at about 110°C, determine the thickness of the in-sulation that needs to be used.
The furnace operates continuously and has an efficiency of 78 percent. The price of the natural gas is $1.10/therm (1 therm 105,500 kJ of energy content). If the installation of the insula-tion will cost $250 for materials and labor, determine how long it will take for the insulation to pay for itself from the energy it saves.
3–34 Repeat Prob. 3–33 for expanded perlite insulation as-suming conductivity is k 0.052 W/mK.
3–35 Reconsider Prob. 3–33. Using EES (or other) soft-ware, investigate the effect of thermal conductiv-ity on the required insulation thickness. Plot the thickness of insulation as a function of the thermal conductivity of the insu-lation in the range of 0.02 W/mK to 0.08 W/mK, and discuss the results.
3–36 Consider a house that has a 10-m 20-m base and a 4-m-high wall. All four walls of the house have an R-value of 2.31 m2°C/W. The two 10-m 4-m walls have no windows.
The third wall has five windows made of 0.5-cm-thick glass (k 0.78 W/mK), 1.2 m 1.8 m in size. The fourth wall has the same size and number of windows, but they are double-paned with a 1.5-cm-thick stagnant air space (k 0.026 W/mK) enclosed between two 0.5-cm-thick glass layers. The thermo-stat in the house is set at 24°C and the average temperature out-side at that location is 8°C during the seven-month-long heating season. Disregarding any direct radiation gain or loss through the windows and taking the heat transfer coefficients at the inner and outer surfaces of the house to be 7 and Sheetrock 7 in 0.6 in 0.6 in Fiberglass insulation FIGURE P3–31E 20 m 15 m Tin = 20°C Tair = 10°C Tsky = 100 K Concrete roof 15 cm FIGURE P3–32 3–32 The roof of a house consists of a 15-cm-thick concrete slab (k 2 W/mK) that is 15 m wide and 20 m long. The con-vection heat transfer coefficients on the inner and outer sur-faces of the roof are 5 and 12 W/m2K, respectively. On a clear winter night, the ambient air is reported to be at 10°C, while the night sky temperature is 100 K. The house and the interior surfaces of the wall are maintained at a constant temperature of 20°C. The emissivity of both surfaces of the concrete roof is 0.9. Considering both radiation and convection heat transfers, determine the rate of heat transfer through the roof, and the in-ner surface temperature of the roof.
If the house is heated by a furnace burning natural gas with an efficiency of 80 percent, and the price of natural gas is $1.20/therm (1 therm 105,500 kJ of energy content), deter-mine the money lost through the roof that night during a 14-h period.
18 W/m2K, respectively, determine the average rate of heat transfer through each wall.
If the house is electrically heated and the price of electricity is $0.08/kWh, determine the amount of money this household will save per heating season by converting the single-pane win-dows to double-pane windows.
3–37 The wall of a refrigerator is constructed of fiberglass in-sulation (k 0.035 W/mK) sandwiched between two layers of 1-mm-thick sheet metal (k 15.1 W/mK). The refrigerated space is maintained at 2°C, and the average heat transfer coef-ficients at the inner and outer surfaces of the wall are 4 W/m2K and 9 W/m2K, respectively. The kitchen tempera-ture averages 28°C. It is observed that condensation occurs on the outer surfaces of the refrigerator when the temperature of the outer surface drops to 20°C. Determine the minimum thick-ness of fiberglass insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces.
200 STEADY HEAT CONDUCTION Btu/hft°F) that are 7 in 9 in in size. Determine the effective thermal conductivity of the board along its 9-in-long side.
What fraction of the heat conducted along that side is con-ducted through copper?
Kitchen air 28°C L 1 mm 1 mm Insulation Refrigerated space 2°C Sheet metal FIGURE P3–37 T1 = 0°C k = 1.4 W/m·K Windshield Outside air, –10°C ho = 200 W/m2·K L = 5 mm Inside air, 25°C FIGURE P3–41 0.05 in 7 in 9 in Epoxy boards Copper plate FIGURE P3–40E 3–38 Reconsider Prob. 3–37. Using EES (or other) software, investigate the effects of the thermal conductivities of the insulation material and the sheet metal on the thickness of the insulation. Let the thermal conductivity vary from 0.02 W/mK to 0.08 W/mK for insulation and 10 W/mK to 400 W/mK for sheet metal. Plot the thickness of the insula-tion as the functions of the thermal conductivities of the insula-tion and the sheet metal, and discuss the results.
3–39 Heat is to be conducted along a circuit board that has a copper layer on one side. The circuit board is 15 cm long and 15 cm wide, and the thicknesses of the copper and epoxy lay-ers are 0.1 mm and 1.2 mm, respectively. Disregarding heat transfer from side surfaces, determine the percentages of heat conduction along the copper (k 386 W/mK) and epoxy (k 0.26 W/mK) layers. Also determine the effective thermal con-ductivity of the board.
Answers: 0.8 percent, 99.2 percent, and 29.9 W/m·K 3–40E A 0.05-in-thick copper plate (k 223 Btu/hft°F) is sandwiched between two 0.15-in-thick epoxy boards (k 0.15 3–41 To defrost ice accumulated on the outer surface of an automobile windshield, warm air is blown over the inner sur-face of the windshield. Consider an automobile windshield with thickness of 5 mm and thermal conductivity of 1.4 W/m·K. The outside ambient temperature is 10°C and the convection heat transfer coefficient is 200 W/m2·K, while the ambient temperature inside the automobile is 25°C. Determine the value of the convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield nec-essary to cause the accumulated ice to begin melting.
3–42 An aluminum plate of 25 mm thick (k 235 W/m·K) is attached on a copper plate with thickness of 10 mm. The cop-per plate is heated electrically to dissipate a uniform heat flux of 5300 W/m2. The upper surface of the aluminum plate is ex-posed to convection heat transfer in a condition such that the convection heat transfer coefficient is 67 W/m2·K and the CHAPTER 3 201 Thermal Contact Resistance 3–43C What is thermal contact resistance? How is it related to thermal contact conductance?
3–44C Will the thermal contact resistance be greater for smooth or rough plain surfaces?
3–45C A wall consists of two layers of insulation pressed against each other. Do we need to be concerned about the ther-mal contact resistance at the interface in a heat transfer analy-sis or can we just ignore it?
3–46C A plate consists of two thin metal layers pressed against each other. Do we need to be concerned about the ther-mal contact resistance at the interface in a heat transfer analy-sis or can we just ignore it?
3–47C Consider two surfaces pressed against each other.
Now the air at the interface is evacuated. Will the thermal con-tact resistance at the interface increase or decrease as a result?
3–48C Explain how the thermal contact resistance can be minimized.
3–49 The thermal contact conductance at the interface of two 1-cm-thick copper plates is measured to be 14,000 W/m2·K.
Determine the thickness of the copper plate whose thermal re-sistance is equal to the thermal resistance of the interface be-tween the plates.
3–50 A 1-mm-thick copper plate (k 386 W/mK) is sand-wiched between two 7-mm-thick epoxy boards (k 0.26 W/mK) that are 15 cm 20 cm in size. If the thermal contact conductance on both sides of the copper plate is esti-mated to be 6000 W/mK, determine the error involved in the total thermal resistance of the plate if the thermal contact con-ductances are ignored.
3–51 Two 5-cm-diameter, 15-cm-long aluminum bars (k 176 W/mK) with ground surfaces are pressed against each other with a pressure of 20 atm. The bars are enclosed in an in-sulation sleeve and, thus, heat transfer from the lateral surfaces is negligible. If the top and bottom surfaces of the two-bar sys-tem are maintained at temperatures of 150°C and 20°C, re-spectively, determine (a) the rate of heat transfer along the cylinders under steady conditions and (b) the temperature drop at the interface.
Answers: (a) 142.4 W, (b) 6.4C Generalized Thermal Resistance Networks 3–52C What are the two approaches used in the development of the thermal resistance network for two-dimensional problems?
3–53C The thermal resistance networks can also be used approximately for multidimensional problems. For what kind of multidimensional problems will the thermal resistance approach give adequate results?
3–54C When plotting the thermal resistance network associ-ated with a heat transfer problem, explain when two resistances are in series and when they are in parallel.
3–55 A typical section of a building wall is shown in Fig. P3–55. This section extends in and out of the page and is Air, 20°C h = 67 W/m2·K Aluminum plate k = 235 W/m·K L = 25 mm T1 = 100°C q · elec = 5300 W/m2 FIGURE P3–42 Q .
7 mm 7 mm Epoxy hc Copper Epoxy FIGURE P3–50 surrounding room temperature is 20°C. Other surfaces of the two attached plates are insulated such that heat only dissipates through the upper surface of the aluminum plate. If the surface of the copper plate that is attached to the aluminum plate has a temperature of 100°C, determine the thermal contact conduc-tance of the aluminum/copper interface.
Answer: 16 kW/m2·K LA LB 0 1 2 3 4 5 FIGURE P3–55 repeated in the vertical direction. The wall support members are made of steel (k 50 W/mK). The support members are 8 cm (t23) 0.5 cm (LB). The remainder of the inner wall space is filled with insulation (k 0.03 W/mK) and measures 8 cm (t23) 60 cm (LB). The inner wall is made of gypsum board (k 0.5 W/mK) that is 1 cm thick (t12) and the outer wall is made of brick (k 1.0 W/mK) that is 10 cm thick (t34). What is the average heat flux through this wall when T1 20°C and T4 35°C?
3–56 A 4-m-high and 6-m-wide wall consists of a long 15-cm 25-cm cross section of horizontal bricks (k 0.72 W/mK) separated by 3-cm-thick plaster layers (k 0.22 W/mK). There are also 2-cm-thick plaster layers on each side of the wall, and a 2-cm-thick rigid foam (k 0.026 W/mK) on the inner side of the wall. The indoor and the out-door temperatures are 22°C and 4°C, and the convection heat transfer coefficients on the inner and the outer sides are h1 10 W/m2K and h2 20 W/m2K, respectively. Assuming one-dimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall.
202 STEADY HEAT CONDUCTION thermal resistance of the wall considering a representative section of it and (b) the rate of heat transfer through the wall.
3–59 A 10-cm-thick wall is to be constructed with 2.5-m-long wood studs (k 0.11 W/mK) that have a cross section of 10 cm 10 cm. At some point the builder ran out of those studs and started using pairs of 2.5-m-long wood studs that have a cross section of 5 cm 10 cm nailed to eachother in-stead. The manganese steel nails (k 50 W/mK) are 10 cm long and have a diameter of 0.4 cm. A total of 50 nails are used to connect the two studs, which are mounted to the wall such that the nails cross the wall. The temperature difference be-tween the inner and outer surfaces of the wall is 8°C. Assum-ing the thermal contact resistance between the two layers to be negligible, determine the rate of heat transfer (a) through a solid stud and (b) through a stud pair of equal length and width nailed to each other. (c) Also determine the effective conduc-tivity of the nailed stud pair.
3–60E A 10-in-thick, 30-ft-long, and 10-ft-high wall is to be constructed using 9-in-long solid bricks (k 0.40 Btu/hft°F) of cross section 7 in 7 in, or identical size bricks with nine square air holes (k 0.015 Btu/hft°F) that are 9 in long and have a cross section of 1.5 in 1.5 in. There is a 0.5-in-thick plaster layer (k 0.10 Btu/hft°F) between two adjacent bricks on all four sides and on both sides of the wall. The house is maintained at 80°F and the ambient temperature outside is 35°F. Taking the heat transfer coefficients at the inner and outer surfaces of the wall to be 1.5 and 6 Btu/hft2°F, respectively, determine the rate of heat transfer through the wall constructed of (a) solid bricks and (b) bricks with air holes.
Plaster 0.5 in 0.5 in Brick 0.5 in 7 in 0.5 in Air channels 1.5 in × 1.5 in × 9 in 9 in FIGURE P3–60E 1.5 cm 25 cm 1.5 cm 2 15 cm 2 Plaster Foam 2 Brick FIGURE P3–56 3–57 Reconsider Prob. 3–56. Using EES (or other) software, plot the rate of heat transfer through the wall as a function of the thickness of the rigid foam in the range of 1 cm to 10 cm. Discuss the results.
3–58 A 12-m-long and 5-m-high wall is constructed of two layers of 1-cm-thick sheetrock (k 0.17 W/mK) spaced 16 cm by wood studs (k 0.11 W/mK) whose cross section is 16 cm 5 cm. The studs are placed vertically 60 cm apart, and the space between them is filled with fiberglass insulation (k 0.034 W/mK). The house is maintained at 20°C and the ambient temperature outside is 9°C. Taking the heat trans-fer coefficients at the inner and outer surfaces of the house to be 8.3 and 34 W/m2K, respectively, determine (a) the 3–61 Consider a 5-m-high, 8-m-long, and 0.22-m-thick wall whose representative cross section is as given in the figure. The thermal conductivities of various materials used, in W/mK, are kA kF 2, kB 8, kC 20, kD 15, and kE 35. The left and right surfaces of the wall are maintained at uniform tem-peratures of 300°C and 100°C, respectively. Assuming heat CHAPTER 3 203 3–62 Repeat Prob. 3–61 assuming that the thermal contact resistance at the interfaces D-F and E-F is 0.00012 m2°C/W.
3–63 Clothing made of several thin layers of fabric with trapped air in between, often called ski clothing, is commonly used in cold climates because it is light, fashionable, and a very effective thermal insulator. So it is no surprise that such cloth-ing has largely replaced thick and heavy old-fashioned coats.
Consider a jacket made of five layers of 0.15-mm-thick syn-thetic fabric (k 0.13 W/mK) with 1.5-mm-thick air space (k 0.026 W/mK) between the layers. Assuming the inner surface temperature of the jacket to be 25°C and the surface area to be 1.25 m2, determine the rate of heat loss through the jacket when the temperature of the outdoors is 0°C and the heat transfer coefficient at the outer surface is 25 W/m2K.
What would your response be if the jacket is made of a sin-gle layer of 0.75-mm-thick synthetic fabric? What should be the thickness of a wool fabric (k 0.035 W/mK) if the person is to achieve the same level of thermal comfort wearing a thick wool coat instead of a five-layer ski jacket?
3–65 In an experiment to measure convection heat transfer coefficients, a very thin metal foil of very low emissivity (e.g., highly polished copper) is attached on the surface of a slab of material with very low thermal conductivity. The other surface of the metal foil is exposed to convection heat transfer by flowing fluid over the foil surface. This setup diminishes heat conduction through the slab and radiation on the metal foil surface, while heat convection plays the prominent role. The slab on which the metal foil is attached to has a thickness of 25 mm and a thermal conductivity of 0.023 W/m·K. In a con-dition where the surrounding room temperature is 20°C, the metal foil is heated electrically with a uniform heat flux of 5000 W/m2. If the bottom surface of the slab is 20°C and the metal foil has an emissivity of 0.02, determine.
(a) The convection heat transfer coefficient if air is flowing over the metal foil and the surface temperature of the foil is 150°C.
(b) The convection heat transfer coefficient if water is flow-ing over the metal foil and the surface temperature of the foil is 30°C.
E 6 cm 300°C 100°C 6 cm 10 cm 5 cm 1 cm 6 cm 8 m 4 cm 4 cm 4 cm F D C A B C Q · FIGURE P3–61 T2 = 20°C Metal foil e = 0.02 Tsurr = 20°C Fluid, T = 20°C L = 25 mm T1 qelec = 5000 W/m2 Slab k = 0.023 W/m.K .
FIGURE P3–65 40 m 5 m 4 m 20 cm Tin = 40°C Tout = –4°C FIGURE P3–66 FIGURE P3–63 3–64 Repeat Prob. 3–63 assuming the layers of the jacket are made of cotton fabric (k 0.06 W/mK).
3–66 A 5-m-wide, 4-m-high, and 40-m-long kiln used to cure concrete pipes is made of 20-cm-thick concrete walls and ceil-ing (k 0.9 W/mK). The kiln is maintained at 40°C by in-jecting hot steam into it. The two ends of the kiln, 4 m 5 m in size, are made of a 3-mm-thick sheet metal covered with 2-cm-thick Styrofoam (k 0.033 W/mK). The convection transfer through the wall to be one-dimensional, determine (a) the rate of heat transfer through the wall; (b) the tempera-ture at the point where the sections B, D, and E meet; and (c) the temperature drop across the section F. Disregard any contact resistances at the interfaces.
heat transfer coefficients on the inner and the outer surfaces of the kiln are 3000 W/m2K and 25 W/m2K, respectively. Disre-garding any heat loss through the floor, determine the rate of heat loss from the kiln when the ambient air is at 4°C.
3–67 Reconsider Prob. 3–66. Using EES (or other) software, investigate the effects of the thickness of the wall and the convection heat transfer coefficient on the outer surface of the rate of heat loss from the kiln. Let the thickness vary from 10 cm to 30 cm and the convection heat transfer coefficient from 5 W/m2K to 50 W/m2K. Plot the rate of heat transfer as functions of wall thickness and the convec-tion heat transfer coefficient, and discuss the results.
3–68E Consider a 10-in 12-in epoxy glass laminate (k 0.10 Btu/hft°F) whose thickness is 0.05 in. In order to reduce the thermal resistance across its thickness, cylindrical copper fillings (k 223 Btu/hft°F) of 0.02 in diameter are to be planted throughout the board, with a center-to-center distance of 0.06 in. Determine the new value of the thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification.
Answer: 0.000256 h°F/Btu 204 STEADY HEAT CONDUCTION 3–72 An 8-m-internal-diameter spherical tank made of 1.5-cm-thick stainless steel (k 15 W/mK) is used to store iced water at 0°C. The tank is located in a room whose tem-perature is 25°C. The walls of the room are also at 25°C. The outer surface of the tank is black (emissivity e 1), and heat transfer between the outer surface of the tank and the sur-roundings is by natural convection and radiation. The convec-tion heat transfer coefficients at the inner and the outer surfaces of the tank are 80 W/m2K and 10 W/m2K, respec-tively. Determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0°C that melts during a 24-h period. The heat of fusion of water at atmospheric pres-sure is hif 333.7 kJ/kg.
0.02 in 0.06 in Copper filling Epoxy board FIGURE P3–68E Iced water 1.5 cm Di = 8 m Tin = 0°C Troom = 25°C FIGURE P3–72 Heat Conduction in Cylinders and Spheres 3–69C What is an infinitely long cylinder? When is it proper to treat an actual cylinder as being infinitely long, and when is it not?
3–70C Can the thermal resistance concept be used for a solid cylinder or sphere in steady operation? Explain.
3–71C Consider a short cylinder whose top and bottom sur-faces are insulated. The cylinder is initially at a uniform tem-perature Ti and is subjected to convection from its side surface to a medium at temperature T, with a heat transfer coefficient of h. Is the heat transfer in this short cylinder one- or two-dimensional? Explain.
3–73 Steam at 280°C flows in a stainless steel pipe (k 15 W/mK) whose inner and outer diameters are 5 cm and 5.5 cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation (k 0.038 W/mK). Heat is lost to the sur-roundings at 5°C by natural convection and radiation, with a combined natural convection and radiation heat transfer coeffi-cient of 22 W/m2K. Taking the heat transfer coefficient inside the pipe to be 80 W/m2K, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the tem-perature drops across the pipe shell and the insulation.
3–74 Reconsider Prob. 3–73. Using EES (or other) software, investigate the effect of the thickness of the insulation on the rate of heat loss from the steam and the temperature drop across the insulation layer. Let the insulation thickness vary from 1 cm to 10 cm. Plot the rate of heat loss and the temperature drop as a function of insulation thickness, and discuss the results.
3–75 A 50-m-long section of a steam pipe whose outer diameter is 10 cm passes through an open space at 15°C. The average temperature of the outer surface of the pipe is measured to be 150°C. If the combined heat transfer co-efficient on the outer surface of the pipe is 20 W/m2K, deter-mine (a) the rate of heat loss from the steam pipe; (b) the CHAPTER 3 205 annual cost of this energy lost if steam is generated in a natural gas furnace that has an efficiency of 75 percent and the price of natural gas is $0.52/therm (1 therm 105,500 kJ); and (c) the thickness of fiberglass insulation (k 0.035 W/mK) needed in order to save 90 percent of the heat lost. Assume the pipe tem-perature to remain constant at 150°C.
are very small and can be neglected. The price of electricity is $0.08/kWh, and the home owner pays $280 a year for water heating. Determine the fraction of the hot-water energy cost of this household that is due to the heat loss from the tank.
Hot-water tank insulation kits consisting of 3-cm-thick fiber-glass insulation (k 0.035 W/mK) large enough to wrap the entire tank are available in the market for about $30. If such an insulation is installed on this water tank by the home owner himself, how long will it take for this additional insulation to pay for itself?
Answers: 15.2 percent, 21 months 3–77 Reconsider Prob. 3–76. Using EES (or other) soft-ware, plot the fraction of energy cost of hot water due to the heat loss from the tank as a function of the hot-water temperature in the range of 40°C to 90°C. Discuss the results.
3–78 Chilled water enters a thin-shelled 4-cm-diameter, 200-m-long pipe at 7°C at a rate of 0.98 kg/s and leaves at 8°C. The pipe is exposed to ambient air at 30°C with a heat transfer co-efficient of 9 W/m2K. If the pipe is to be insulated with glass wool insulation (k 0.05 W/mK) in order to decrease the temperature rise of water to 0.25°C, determine the required thickness of the insulation.
3–79 Superheated steam at an average temperature 200°C is transported through a steel pipe (k 50 W/mK, Do 8.0 cm, Di 6.0 cm, and L 20.0 m). The pipe is insulated with a 4-cm thick layer of gypsum plaster (k 0.5 W/mK). The in-sulated pipe is placed horizontally inside a warehouse where the average air temperature is 10°C. The steam and the air heat transfer coefficients are estimated to be 800 and 200 W/m2K, respectively. Calculate (a) the daily rate of heat transfer from the superheated steam, and (b) the temperature on the outside surface of the gypsum plaster insulation.
3–80E Steam at 450°F is flowing through a steel pipe (k 8.7 Btu/hft°F) whose inner and outer diameters are 3.5 in and 4.0 in, respectively, in an environment at 55°F. The pipe is insulated with 2-in-thick fiberglass insulation (k 0.020 Btu/hft°F). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and 5 Btu/hft2°F, respec-tively, determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations?
3–81 Hot water at an average temperature of 90°C is flow-ing through a 15-m section of a cast iron pipe (k 52 W/mK) whose inner and outer diameters are 4 cm and 4.6 cm, respectively. The outer surface of the pipe, whose emissivity is 0.7, is exposed to the cold air at 10°C in the basement, with a heat transfer coefficient of 15 W/m2K. The heat transfer coefficient at the inner surface of the pipe is 120 W/m2K. Taking the walls of the basement to be at 10°C also, determine the rate of heat loss from the hot water. Also, determine the average velocity of the water in the pipe if the temperature of the water drops by 3°C as it passes through the basement.
150°C Tair = 15°C 50 m Fiberglass insulation Steam FIGURE P3–75 3 cm Foam insulation 40 cm Water heater Tw = 60°C 27°C 1.5 m FIGURE P3–76 3–76 Consider a 1.5-m-high electric hot-water heater that has a diameter of 40 cm and maintains the hot water at 60°C. The tank is located in a small room whose average temperature is 27°C, and the heat transfer coefficients on the inner and outer surfaces of the heater are 50 and 12 W/m2K, respectively. The tank is placed in another 46-cm-diameter sheet metal tank of negligible thickness, and the space between the two tanks is filled with foam insulation (k 0.03 W/mK). The thermal re-sistances of the water tank and the outer thin sheet metal shell 3–82 Repeat Prob. 3–81 for a pipe made of copper (k 386 W/mK) instead of cast iron.
3–83E Steam exiting the turbine of a steam power plant at 100°F is to be condensed in a large condenser by cooling water flowing through copper pipes (k 223 Btu/hft°F) of inner di-ameter 0.4 in and outer diameter 0.6 in at an average temperature of 70°F. The heat of vaporization of water at 100°F is 1037 Btu/lbm. The heat transfer coefficients are 2400 Btu/hft2°F on the steam side and 35 Btu/hft2°F on the water side. Deter-mine the length of the tube required to condense steam at a rate of 250 lbm/h.
Answer: 2380 ft 206 STEADY HEAT CONDUCTION pipe on the length of the tube required. Let the thermal con-ductivity vary from 10 Btu/hft°F to 400 Btu/hft°F and the outer diameter from 0.5 in to 1.0 in. Plot the length of the tube as functions of pipe conductivity and the outer pipe diameter, and discuss the results.
3–86 The boiling temperature of nitrogen at atmospheric pressure at sea level (1 atm pressure) is 196°C. Therefore, ni-trogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the at-mosphere will remain constant at 196°C until it is depleted.
Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of 198 kJ/kg and a density of 810 kg/m3 at 1 atm.
Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and 196°C. The tank is exposed to ambient air at 15°C, with a combined convection and radia-tion heat transfer coefficient of 35 W/m2K. The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is (a) not insulated, (b) insulated with 5-cm-thick fiberglass insulation (k 0.035 W/mK), and (c) insulated with 2-cm-thick superinsulation which has an effective thermal conductivity of 0.00005 W/mK.
Steel pipe Insulation Steam 450°F FIGURE P3–81 1 atm Liquid N2 –196°C N2 vapor Insulation Tair = 15°C FIGURE P3–86 Liquid water Steam, 100°F 250 lbm/h Cooling water FIGURE P3–83E 3–87 Repeat Prob. 3–86 for liquid oxygen, which has a boiling temperature of 183°C, a heat of vaporization of 213 kJ/kg, and a density of 1140 kg/m3 at 1 atm pressure.
3–88 A 2.2-mm-diameter and 14-m-long electric wire is tightly wrapped with a 1-mm-thick plastic cover whose thermal conductivity is k 0.15 W/mK. Electrical measurements indi-cate that a current of 13 A passes through the wire and there is a 3–84E Repeat Prob. 3–83E, assuming that a 0.01-in-thick layer of mineral deposit (k 0.5 Btu/hft°F) has formed on the inner surface of the pipe.
3–85E Reconsider Prob. 3–83E. Using EES (or other) software, investigate the effects of the thermal conductivity of the pipe material and the outer diameter of the CHAPTER 3 207 3–89 In a pharmaceutical plant, a copper pipe (kc 400 W/m·K) with inner diameter of 20 mm and wall thickness of 2.5 mm is used for carrying liquid oxygen to a storage tank.
The liquid oxygen flowing in the pipe has an average tempera-ture of 200°C and a convection heat transfer coefficient of 120 W/m2·K. The condition surrounding the pipe has an ambi-ent air temperature of 20°C and a combined heat transfer coef-ficient of 20 W/m2·K. If the dew point is 10°C, determine the thickness of the insulation (ki 0.05 W/m·K) around the cop-per pipe to avoid condensation on the outer surface. Assume thermal contact resistance is negligible.
3–92C A pipe is insulated to reduce the heat loss from it.
However, measurements indicate that the rate of heat loss has increased instead of decreasing. Can the measurements be right?
3–93C Consider a pipe at a constant temperature whose ra-dius is greater than the critical radius of insulation. Someone claims that the rate of heat loss from the pipe has increased when some insulation is added to the pipe. Is this claim valid?
3–94C A pipe is insulated such that the outer radius of the in-sulation is less than the critical radius. Now the insulation is taken off. Will the rate of heat transfer from the pipe increase or decrease for the same pipe surface temperature?
3–95E A 0.083-in-diameter electrical wire at 90°F is covered by 0.02-in-thick plastic insulation (k 0.075 Btu/hft°F). The wire is exposed to a medium at 50°F, with a combined convec-tion and radiation heat transfer coefficient of 2.5 Btu/hft2°F.
Determine if the plastic insulation on the wire will increase or decrease heat transfer from the wire.
Answer: It helps 3–96E Repeat Prob. 3–95E, assuming a thermal contact re-sistance of 0.001 hft2°F/Btu at the interface of the wire and the insulation.
3–97 A 4-mm-diameter spherical ball at 50°C is covered by a 1-mm-thick plastic insulation (k 0.13 W/mK). The ball is exposed to a medium at 15°C, with a combined convection and radiation heat transfer coefficient of 20 W/m2K. Determine if the plastic insulation on the ball will help or hurt heat transfer from the ball.
D1 = 20 mm Liq. O2, –200°C h = 120 W/m2·K D2 = 25 mm D3 Pipe, kc = 400 W/m·K Surrounding air, 20°C hcombined = 20 W/m2·K Dew point = 10°C Insulation ki = 0.05 W/m·K FIGURE P3–89 4 mm 1 mm Plastic insulation FIGURE P3–97 14 m Insulation Electrical wire T = 30°C FIGURE P3–88 voltage drop of 8 V along the wire. If the insulated wire is ex-posed to a medium at T 30°C with a heat transfer coefficient of h 24 W/m2K, determine the temperature at the interface of the wire and the plastic cover in steady operation. Also de-termine if doubling the thickness of the plastic cover will in-crease or decrease this interface temperature.
Critical Radius of Insulation 3–90C What is the critical radius of insulation? How is it defined for a cylindrical layer?
3–91C Consider an insulated pipe exposed to the atmos-phere. Will the critical radius of insulation be greater on calm days or on windy days? Why?
3–98 Reconsider Prob. 3–97. Using EES (or other) soft-ware, plot the rate of heat transfer from the ball as a function of the plastic insulation thickness in the range of 0.5 mm to 20 mm. Discuss the results.
Heat Transfer from Finned Surfaces 3–99C Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attach-ing the fins inside or outside the tubes? Why? When would you recommend attaching fins both inside and outside the tubes?
3–100C What is the reason for the widespread use of fins on surfaces?
3–101C What is the difference between the fin effectiveness and the fin efficiency?
3–102C The fins attached to a surface are determined to have an effectiveness of 0.9. Do you think the rate of heat transfer from the surface has increased or decreased as a result of the addition of these fins?
3–103C Explain how the fins enhance heat transfer from a surface. Also, explain how the addition of fins may actually de-crease heat transfer from a surface.
3–104C How does the overall effectiveness of a finned sur-face differ from the effectiveness of a single fin?
3–105C Hot water is to be cooled as it flows through the tubes exposed to atmospheric air. Fins are to be attached in or-der to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why?
3–106C Consider two finned surfaces that are identical ex-cept that the fins on the first surface are formed by casting or extrusion, whereas they are attached to the second surface af-terwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat trans-fer? Explain.
3–107C The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?
3–108C Does the (a) efficiency and (b) effectiveness of a fin increase or decrease as the fin length is increased?
3–109C Two pin fins are identical, except that the diameter of one of them is twice the diameter of the other. For which fin is the (a) fin effectiveness and (b) fin efficiency higher? Explain.
3–110C Two plate fins of constant rectangular cross section are identical, except that the thickness of one of them is twice the thickness of the other. For which fin is the (a) fin effective-ness and (b) fin efficiency higher? Explain.
3–111C Two finned surfaces are identical, except that the convection heat transfer coefficient of one of them is twice that of the other. For which finned surface is the (a) fin effective-ness and (b) fin efficiency higher? Explain.
3–112 Obtain a relation for the fin efficiency for a fin of constant cross-sectional area Ac, perimeter p, length L, and thermal conductivity k exposed to convection to a medium at T with a heat transfer coefficient h. Assume the fins are suffi-ciently long so that the temperature of the fin at the tip is nearly T. Take the temperature of the fin at the base to be Tb and neglect heat transfer from the fin tips. Simplify the rela-tion for (a) a circular fin of diameter D and (b) rectangular fins of thickness t.
208 STEADY HEAT CONDUCTION 3–113 The case-to-ambient thermal resistance of a power transistor that has a maximum power rating of 15 W is given to be 25°C/W. If the case temperature of the transistor is not to exceed 80°C, determine the power at which this transistor can be operated safely in an environment at 35°C.
3–114 A 4-mm-diameter and 10-cm-long aluminum fin (k 237 W/mK) is attached to a surface. If the heat transfer coeffi-cient is 12 W/m2K, determine the percent error in the rate of heat transfer from the fin when the infinitely long fin assump-tion is used instead of the adiabatic fin tip assumption.
h, T L = 10 cm Tb k D = 4 mm FIGURE P3–112 h, T L = 10 cm Tb k D = 4 mm FIGURE P3–114 Air, 20°C h = 25 W/m2·K Stainless steel shaft k = 15.1 W/m·K D = 25 mm, L = 25 cm Qs .
TL = 22°C Th DC motor Welec = 300 W .
As = 0.075 m2 Qh .
FIGURE P3–116 3–115 Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essen-tially that of the surrounding air, i.e. 20°C. Its width is 5.0 cm; thickness is 1.0 mm; thermal conductivity is 200 W/mK; and base temperature is 40°C. The heat transfer coefficient is 20 W/m2K. Estimate the fin temperature at a distance of 5.0 cm from the base and the rate of heat loss from the entire fin.
3–116 A DC motor delivers mechanical power to a rotating stainless steel shaft (k 15.1 W/m·K) with a length of 25 cm and a diameter of 25 mm. In a surrounding with ambient air tempera-ture of 20°C and convection heat transfer coefficient of 25 W/m2·K, the surface area of the motor housing that is exposed CHAPTER 3 209 to the ambient air is 0.075 m2. The motor uses 300 W of electri-cal power and delivers 55% of it as mechanical power to rotate the stainless steel shaft. If the tip of the stainless steel shaft has a temperature of 22°C, determine the surface temperature of the motor housing. Assume the base temperature of the shaft is equal to the surface temperature of the motor housing.
Answer: 87.7°C.
3–117 Two 4-m-long and 0.4-cm-thick cast iron (k 52 W/mK) steam pipes of outer diameter 10 cm are connected to each other through two 1-cm-thick flanges of outer diameter 18 cm. The steam flows inside the pipe at an average tempera-ture of 200°C with a heat transfer coefficient of 180 W/m2K.
The outer surface of the pipe is exposed to an ambient at 12°C, with a heat transfer coefficient of 25 W/m2K. (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. (b) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes?
3–121 Pipes with inner and outer diameters of 50 mm and 60 mm, respectively, are used for transporting superheated va-por in a manufacturing plant. The pipes with thermal conduc-tivity of 16 W/m·K are connected together by flanges with combined thickness of 20 mm and outer diameter of 90 mm.
Air condition surrounding the pipes has a temperature of 25°C and a convection heat transfer coefficient of 10 W/m2·K. If the inner surface temperature of the pipe is maintained at a con-stant temperature of 150°C, determine the temperature at the base of the flange and the rate of heat loss through the flange.
Answers: 148°C, 18W 90°C 40 W Tair = 20°C FIGURE P3–118 Tair = 12°C Steam 200°C 9.2 cm 1 cm 1 cm 10 cm 18 cm FIGURE P3–117 TL Turbine blade k = 17 W/m·K p = 11 cm, L = 5.3 cm Ac = 5.13 cm2 Hot gas, 973°C h = 538 W/m2·K Tb = 450°C FIGURE P3–120 Di = 50 mm Do = 60 mm t = 20 mm Tb Ti = 150°C Df = 90 mm Pipe, k = 16 W/m·K Air, 25°C h = 10 W/m2·K FIGURE P3–121 3–118 A 40-W power transistor is to be cooled by attaching it to one of the commercially available heat sinks shown in Table 3–6. Select a heat sink that will allow the case temperature of the transistor not to exceed 90°C in the ambient air at 20°C.
3–119 A 25-W power transistor is to be cooled by attaching it to one of the commercially available heat sinks shown in Table 3–6. Select a heat sink that will allow the case temperature of the transistor not to exceed 55°C in the ambient air at 18°C.
3–120 A turbine blade made of a metal alloy (k 17 W/m·K) has a length of 5.3 cm, a perimeter of 11 cm, and a cross-sectional area of 5.13 cm2. The turbine blade is exposed to hot gas from the combustion chamber at 973°C with a con-vection heat transfer coefficient of 538 W/m2·K. The base of the turbine blade maintains a constant temperature of 450°C and the tip is adiabatic. Determine the heat transfer rate to the turbine blade and temperature at the tip.
3–123 Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 130°C. Circular aluminum alloy 2024-T6 fins (k 186 W/mK) of outer diameter 6 cm and constant thick-ness 1 mm are attached to the tube. The space between the fins is 3 mm, and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding air at T 25°C, with a heat transfer coefficient of 40 W/m2K. Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.
Answer: 1788 W 210 STEADY HEAT CONDUCTION 3–125E Repeat Prob. 3–124E for a silver spoon (k 247 Btu/hft°F).
3–126E Reconsider Prob. 3–124E. Using EES (or other) software, investigate the effects of the thermal conductivity of the spoon material and the length of its extension in the air on the temperature difference across the exposed surface of the spoon handle. Let the thermal conduc-tivity vary from 5 Btu/hft°F to 225 Btu/hft°F and the length from 5 in to 12 in. Plot the temperature difference as the func-tions of thermal conductivity and length, and discuss the results.
3–127 A 0.4-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.04 W. The board is impregnated with copper fill-ings and has an effective thermal conductivity of 30 W/mK.
All the heat generated in the chips is conducted across the cir-cuit board and is dissipated from the back side of the board to a medium at 40°C, with a heat transfer coefficient of 52 W/m2K. (a) Determine the temperatures on the two sides of the circuit board. (b) Now a 0.2-cm-thick, 12-cm-high, and 18-cm-long aluminum plate (k 237 W/mK) with 864 2-cm-long aluminum pin fins of diameter 0.25 cm is attached to the back side of the circuit board with a 0.02-cm-thick epoxy ad-hesive (k 1.8 W/mK). Determine the new temperatures on the two sides of the circuit board.
3–128 Repeat Prob. 3–127 using a copper plate with copper fins (k 386 W/mK) instead of aluminum ones.
3–129 A hot surface at 100°C is to be cooled by attaching 3-cm-long, 0.25-cm-diameter aluminum pin fins (k 237 W/mK) to it, with a center-to-center distance of 0.6 cm. The temperature of the surrounding medium is 30°C, and the heat transfer coefficient on the surfaces is 35 W/m2K. Determine the rate of heat transfer from the surface for a 1-m 1-m section of the plate. Also determine the overall effectiveness of the fins.
3–122 A plane wall with surface temperature of 350°C is at-tached with straight rectangular fins (k 235 W/m·K). The fins are exposed to an ambient air condition of 25°C and the convection heat transfer coefficient is 154 W/m2·K. Each fin has a length of 50 mm, a base of 5 mm thick and a width of 100 mm. Determine the efficiency, heat transfer rate, and effec-tiveness of each fin, using (a) Table 3–3 and (b) Figure 3–43.
w = 100 mm x t = 5 mm Air, 25°C h = 154 W/m2·K k = 235 W/m·K Tb = 350°C L = 50 mm FIGURE P3–122 7 in Boiling water 200°F Tair = 75°F Spoon FIGURE P3–124E 1 mm 3 mm 130°C T = 25°C 3 cm 2.5 cm FIGURE P3–123 3–124E Consider a stainless steel spoon (k 8.7 Btu/hft°F) partially immersed in boiling water at 200°F in a kitchen at 75°F. The handle of the spoon has a cross section of 0.08 in 0.5 in, and extends 7 in in the air from the free surface of the water. If the heat transfer coefficient at the exposed surfaces of the spoon handle is 3 Btu/hft2°F, determine the temperature difference across the exposed surface of the spoon handle.
State your assumptions.
Answer: 124.6°F CHAPTER 3 211 3–130 Repeat Prob. 3–129 using copper fins (k 386 W/mK) instead of aluminum ones.
3–131 Reconsider Prob. 3–129. Using EES (or other) soft-ware, investigate the effect of the center-to-center distance of the fins on the rate of heat transfer from the surface and the overall effectiveness of the fins. Let the center-to-center dis-tance vary from 0.4 cm to 2.0 cm. Plot the rate of heat transfer and the overall effectiveness as a function of the center-to-center distance, and discuss the results.
3–132 Circular cooling fins of diameter D 1 mm and length L 30 mm, made of copper (k 380 W/mK), are used to enhance heat transfer from a surface that is maintained at temperature Ts1 132°C. Each rod has one end attached to this surface (x 0), while the opposite end (x L) is joined to a second surface, which is maintained at Ts2 0°C. The air flow-ing between the surfaces and the rods is also at T 0°C, and the convection coefficient is h 100 W/m2K.
3 cm 0.6 cm 0.25 cm FIGURE P3–129 L D x Ts2 Ts1 T, h FIGURE P3–132 8 in 15 ft 350°F 60°F 1 in 8 in 8 in 3 ft FIGURE P3–137E (a) Express the function u(x) T(x) T along a fin, and calculate the temperature at x L/2.
(b) Determine the rate of heat transferred from the hot sur-face through each fin and the fin effectiveness. Is the use of fins justified? Why?
(c) What is the total rate of heat transfer from a 10-cm by 10-cm section of the wall, which has 625 uniformly dis-tributed fins? Assume the same convection coefficient for the fin and for the unfinned wall surface.
Heat Transfer in Common Configurations 3–133C What is a conduction shape factor? How is it related to the thermal resistance?
3–134C What is the value of conduction shape factors in engineering?
3–135 Hot- and cold-water pipes 12 m long run parallel to each other in a thick concrete layer. The diameters of both pipes are 6 cm, and the distance between the centerlines of the pipes is 40 cm. The surface temperatures of the hot and cold pipes are 60°C and 15°C, respectively. Taking the thermal conductivity of the concrete to be k 0.75 W/mK, determine the rate of heat transfer between the pipes.
Answer: 555 W 3–136 Reconsider Prob. 3–135. Using EES (or other) software, plot the rate of heat transfer between the pipes as a function of the distance between the centerlines of the pipes in the range of 10 cm to 1.0 m. Discuss the results.
3–137E A row of 3-ft-long and 1-in-diameter used uranium fuel rods that are still radioactive are buried in the ground par-allel to each other with a center-to-center distance of 8 in at a depth of 15 ft from the ground surface at a location where the thermal conductivity of the soil is 0.6 Btu/hft°F. If the surface temperature of the rods and the ground are 350°F and 60°F, respectively, determine the rate of heat transfer from the fuel rods to the atmosphere through the soil.
3–138 A 12-m-long and 8-cm-diameter hot-water pipe of a district heating system is buried in the soil 80 cm below the ground surface. The outer surface temperature of the pipe is 60°C. Taking the surface temperature of the earth to be 2°C and the thermal conductivity of the soil at that location to be 0.9 W/m·K, determine the rate of heat loss from the pipe.
3–139 Reconsider Prob. 138. Using EES (or other) software, plot the rate of heat loss from the pipe as a function of the burial depth in the range of 20 cm to 2.0 m.
Discuss the results.
3–140 Hot water at an average temperature of 53°C and an average velocity of 0.4 m/s is flowing through a 5-m section of a thin-walled hot-water pipe that has an outer diameter of 2.5 cm. The pipe passes through the center of a 14-cm-thick wall filled with fiberglass insulation (k 0.035 W/mK). If the surfaces of the wall are at 18°C, determine (a) the rate of heat transfer from the pipe to the air in the rooms and (b) the tem-perature drop of the hot water as it flows through this 5-m-long section of the wall.
Answers: 19.6 W, 0.024°C 212 STEADY HEAT CONDUCTION 3–142 Consider a house with a flat roof whose outer dimen-sions are 12 m 12 m. The outer walls of the house are 6 m high. The walls and the roof of the house are made of 20-cm-thick concrete (k 0.75 W/mK). The temperatures of the inner and outer surfaces of the house are 15°C and 3°C, respectively. Accounting for the effects of the edges of adjoin-ing surfaces, determine the rate of heat loss from the house through its walls and the roof. What is the error involved in ig-noring the effects of the edges and corners and treating the roof as a 12 m 12 m surface and the walls as 6 m 12 m surfaces for simplicity?
3–143 Consider a 25-m-long thick-walled concrete duct (k 0.75 W/mK) of square cross section. The outer dimensions of the duct are 20 cm 20 cm, and the thickness of the duct wall is 2 cm. If the inner and outer surfaces of the duct are at 100°C and 30°C, respectively, determine the rate of heat transfer through the walls of the duct.
Answer: 47.1 kW 12 m D = 8 cm 60°C 80 cm 2°C FIGURE P3–138 Hot water pipe –3°C 5°C 3 m 20 m 80°C FIGURE P3–141 Hot water Wall 2.5 cm 53°C 18°C 5 m FIGURE P3–140 30°C 16 cm 20 cm 25 m 100°C FIGURE P3–143 3–141 Hot water at an average temperature of 80°C and an average velocity of 1.5 m/s is flowing through a 25-m section of a pipe that has an outer diameter of 5 cm. The pipe extends 2 m in the ambient air above the ground, dips into the ground (k 1.5 W/mK) vertically for 3 m, and continues horizontally at this depth for 20 m more before it enters the next building.
The first section of the pipe is exposed to the ambient air at 5°C, with a heat transfer coefficient of 22 W/m2K. If the sur-face of the ground is covered with snow at 3°C, determine (a) the total rate of heat loss from the hot water and (b) the tem-perature drop of the hot water as it flows through this 25-m-long section of the pipe.
3–144 A 3-m-diameter spherical tank containing some ra-dioactive material is buried in the ground (k 1.4 W/mK).
The distance between the top surface of the tank and the ground surface is 4 m. If the surface temperatures of the tank and the ground are 140°C and 15°C, respectively, determine the rate of heat transfer from the tank.
CHAPTER 3 213 3–145 Reconsider Prob. 3–144. Using EES (or other) software, plot the rate of heat transfer from the tank as a function of the tank diameter in the range of 0.5 m to 5.0 m. Discuss the results.
3–146 Hot water at an average temperature of 90°C passes through a row of eight parallel pipes that are 4 m long and have an outer diameter of 3 cm, located vertically in the middle of a concrete wall (k 0.75 W/mK) that is 4 m high, 8 m long, and 15 cm thick. If the surfaces of the concrete walls are ex-posed to a medium at 32°C, with a heat transfer coefficient of 12 W/m2K, determine the rate of heat loss from the hot water and the surface temperature of the wall.
Special Topics: Heat Transfer through the Walls and Roofs 3–147C What is the R-value of a wall? How does it differ from the unit thermal resistance of the wall? How is it related to the U-factor?
3–148C What is effective emissivity for a plane-parallel air space? How is it determined? How is radiation heat transfer through the air space determined when the effective emissivity is known?
3–149C The unit thermal resistances (R-values) of both 40-mm and 90-mm vertical air spaces are given in Table 3–9 to be 0.22 m2C/W, which implies that more than doubling the thickness of air space in a wall has no effect on heat transfer through the wall. Do you think this is a typing error?
Explain.
3–150C What is a radiant barrier? What kind of materials are suitable for use as radiant barriers? Is it worthwhile to use ra-diant barriers in the attics of homes?
3–151C Consider a house whose attic space is ventilated effectively so that the air temperature in the attic is the same as the ambient air temperature at all times. Will the roof still have any effect on heat transfer through the ceiling? Explain.
3–152 Determine the summer R-value and the U-factor of a wood frame wall that is built around 38-mm 140-mm wood studs with a center-to-center distance of 400 mm. The 140-mm-wide cavity between the studs is filled with mineral fiber batt insulation. The inside is finished with 13-mm gypsum wallboard and the outside with 13-mm wood fiberboard and 13-mm 200-mm wood bevel lapped siding. The insulated cavity constitutes 80 percent of the heat transmission area, while the studs, headers, plates, and sills constitute 20 percent. Answers: 3.213 m2K/W, 0.311 W/m2K 3–153 The 13-mm-thick wood fiberboard sheathing of the wood stud wall in Prob. 3–152 is replaced by a 25-mm-thick rigid foam insulation. Determine the percent increase in the R-value of the wall as a result.
3–154 The overall heat transfer coefficient (the U-value) of a wall under winter design conditions is U 2.25 W/m2K. Now a layer of 100-mm face brick is added to the outside, leaving a 20-mm air space between the wall and the bricks. Determine the new U-value of the wall. Also, determine the rate of heat transfer through a 3-m-high, 7-m-long section of the wall after modification when the indoor and outdoor temperatures are 22°C and 25°C, respectively.
4a 1 3 2 4b 5 6 FIGURE P3–153 Existing wall Face brick FIGURE P3–154 3–155 Consider a flat ceiling that is built around 38-mm 90-mm wood studs with a center-to-center distance of 400 mm.
The lower part of the ceiling is finished with 13-mm gypsum wallboard, while the upper part consists of a wood subfloor (R 0.166 m2°C/W), a 13-mm plywood, a layer of felt (R 0.011 m2°C/W), and linoleum (R 0.009 m2°C/W). Both sides of the ceiling are exposed to still air. The air space consti-tutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter R-value and the U-factor of the ceiling assuming the 90-mm-wide air space between the studs (a) does not have any reflec-tive surface, (b) has a reflective surface with e 0.05 on one side, and (c) has reflective surfaces with e 0.05 on both sides.
Assume a mean temperature of 10°C and a temperature differ-ence of 5.6°C for the air space.
214 STEADY HEAT CONDUCTION face bricks, 100-mm common bricks, 25-mm urethane rigid foam insulation, and 13-mm gypsum wallboard.
Answers: 1.404 m2°C/W, 0.712 W/m2K 3–159 The overall heat transfer coefficient (the U-value) of a wall under winter design conditions is U 1.40 W/m2K. De-termine the U-value of the wall under summer design conditions.
3–160E Determine the winter R-value and the U-factor of a masonry cavity wall that is built around 4-in-thick concrete blocks made of lightweight aggregate. The outside is finished with 4-in face brick with -in cement mortar between the bricks and concrete blocks. The inside finish consists of -in gypsum wallboard separated from the concrete block by -in-thick (1-in by 3-in nominal) vertical furring whose center-to-center distance is 16 in. Neither side of the -in-thick air space between the concrete block and the gypsum board is coated with any reflective film. When determining the R-value of the air space, the temperature difference across it can be taken to be 30°F with a mean air temperature of 50°F. The air space constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent.
3 4 3 4 1 2 1 2 7 6 5a 5b 4 3 2 1 FIGURE P3–160E 5 4 3 1 7 6 2 FIGURE P3–156 1 2 3 4 5 6 7 8 FIGURE P3–155 3–161 Determine the summer and winter R-values, in m2°C/W, of a masonry wall that consists of 100-mm face bricks, 13-mm of cement mortar, 100-mm lightweight concrete block, 40-mm air space, and 20-mm plasterboard.
Answers: 0.809 and 0.795 m2°C/W 3–162E The overall heat transfer coefficient of a wall is de-termined to be U 0.075 Btu/hft2F under the conditions of still air inside and winds of 7.5 mph outside. What will the U-factor be when the wind velocity outside is doubled? Answer: 0.0755 Btu/hft2°F 3–163 Two homes are identical, except that the walls of one house consist of 200-mm lightweight concrete blocks, 20-mm air space, and 20-mm plasterboard, while the walls of the other house involve the standard R-2.4 m2°C/W frame wall con-struction. Which house do you think is more energy efficient?
3–164 Determine the R-value of a ceiling that consists of a layer of 19-mm acoustical tiles whose top surface is covered 3–156 Determine the winter R-value and the U-factor of a masonry cavity wall that consists of 100-mm common bricks, a 90-mm air space, 100-mm concrete blocks made of light-weight aggregate, 20-mm air space, and 13-mm gypsum wall-board separated from the concrete block by 20-mm-thick (1-in 3-in nominal) vertical furring whose center-to-center distance is 400 mm. Neither side of the two air spaces is coated with any reflective films. When determining the R-value of the air spaces, the temperature difference across them can be taken to be 16.7°C with a mean air temperature of 10°C. The air space constitutes 84 percent of the heat transmission area, while the vertical furring and similar structures constitute 16 percent.
Answers: 1.02 m2°C/W, 0.978 W/m2K 3–157 Repeat Prob. 3–156 assuming one side of both air spaces is coated with a reflective film of e 0.05.
3–158 Determine the winter R-value and the U-factor of a masonry wall that consists of the following layers: 100-mm CHAPTER 3 215 with a highly reflective aluminum foil for winter conditions.
Assume still air below and above the tiles.
3–167E Steam is produced in the copper tubes (k 223 Btu/hft°F) of a heat exchanger at a temperature of 280°F by another fluid condensing on the outside surfaces of the tubes at 350°F. The inner and outer diameters of the tube are 1 in and 1.3 in, respectively. When the heat exchanger was new, the rate of heat transfer per foot length of the tube was 2 104 Btu/h.
Determine the rate of heat transfer per foot length of the tube when a 0.01-in-thick layer of limestone (k 1.7 Btu/hft°F) has formed on the inner surface of the tube after extended use.
3–168E Repeat Prob. 3–167E, assuming that a 0.01-in-thick limestone layer has formed on both the inner and outer surfaces of the tube.
3–169 Hot water is flowing at an average velocity of 1.5 m/s through a cast iron pipe (k 52 W/mK) whose inner and outer diameters are 3 cm and 3.5 cm, respectively. The pipe passes through a 15-m-long section of a basement whose temperature is 15°C. If the temperature of the water drops from 70°C to 67°C as it passes through the basement and the heat transfer coefficient on the inner surface of the pipe is 400 W/m2K, determine the combined convection and radia-tion heat transfer coefficient at the outer surface of the pipe.
Answer: 272.5 W/m2K 3–170 Newly formed concrete pipes are usually cured first overnight by steam in a curing kiln maintained at a temperature of 45°C before the pipes are cured for several days outside. The heat and moisture to the kiln is provided by steam flowing in a pipe whose outer diameter is 12 cm. During a plant inspection, it was noticed that the pipe passes through a 8-m section that is completely exposed to the ambient air before it reaches the kiln. The temperature measurements indicate that the average temperature of the outer surface of the steam pipe is 90°C when 19 mm Highly reflective foil Acoustical tiles FIGURE P3–164 D1 = 15 mm Fuel rod 1 MW/m3 Air gap, h = 10 W/m2·K D2 = 35 mm D3 = 110 mm T1 T3 = 30°C Ceramic k = 0.07 W/m·K FIGURE P3–165 t = 2 mm S = 3 mm T h Tb r2 = 3 cm r1 = 1.5 cm FIGURE P3–166 8 m Steam pipe Tair = 8°C 90°C Kiln Furnace Steam 12 cm FIGURE P3–170 Review Problems 3–165 A cylindrical nuclear fuel rod of 15 mm in diameter is encased in a concentric hollow ceramic cylinder with inner di-ameter of 35 mm and outer diameter of 110 mm. This created an air gap between the fuel rod and the hollow ceramic cylin-der with a convection heat transfer coefficient of 10 W/m2·K.
The hollow ceramic cylinder has a thermal conductivity of 0.07 W/m·K and its outer surface maintains a constant tempera-ture of 30 °C. If the fuel rod generates heat at a rate of 1 MW/m3, determine the temperature at the surface of the fuel rod. Answer: 1026°C 3–166 Steam in a heating system flows through tubes whose outer diameter is 3 cm and whose walls are maintained at a tem-perature of 120°C. Circular aluminum alloy fins (k 180 W/m·K) of outer diameter 6 cm and constant thickness t 2 mm are attached to the tube, as shown in Fig. P3–166. The space between the fins is 3 mm, and thus there are 200 fins per meter length of the tube. Heat is transferred to the surrounding air at 25°C, with a combined heat transfer coefficient of 60 W/m2·K. Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.
216 STEADY HEAT CONDUCTION pipe is estimated to be 40 W/m2K, and the heat of fusion of water is 333.7 kJ/kg. Assuming the pipe to contain stationary water initially at 0°C, determine if the water in that section of the pipe will completely freeze that night.
3–173 Repeat Prob. 3–172 for the case of a heat transfer co-efficient of 18 W/m2K on the outer surface as a result of putting a fence around the pipe that blocks the wind.
3–174E The surface temperature of a 3-in-diameter baked potato is observed to drop from 300°F to 200°F in 5 min in an en-vironment at 70°F. Determine the average heat transfer coefficient between the potato and its surroundings. Using this heat transfer coefficient and the same surface temperature, determine how long it will take for the potato to experience the same temperature drop if it is wrapped completely in a 0.12-in-thick towel (k 0.035 Btu/hft°F). You may use the properties of water for potato.
3–175E Repeat Prob. 3–174E assuming there is a 0.02-in-thick air space (k 0.015 Btu/hft°F) between the potato and the towel.
3–176 A 6-m-wide, 2.8-m-high wall is constructed of one layer of common brick (k 0.72 W/mK) of thickness 20 cm, one inside layer of light-weight plaster (k 0.36 W/mK) of thickness 1 cm, and one outside layer of cement based cover-ing (k 1.40 W/mK) of thickness 2 cm. The inner surface of the wall is maintained at 23°C while the outer surface is ex-posed to outdoors at 8°C with a combined convection and radi-ation heat transfer coefficient of 17 W/m2K. Determine the rate of heat transfer through the wall and temperature drops across the plaster, brick, covering, and surface-ambient air.
3–177 Reconsider Prob. 3–176. It is desired to insulate the wall in order to decrease the heat loss by 90 percent. For the same inner surface temperature, determine the thickness of in-sulation and the outer surface temperature if the wall is insu-lated with (a) polyurethane foam (k 0.025 W/mK) and (b) glass fiber (k 0.036 W/mK).
3–178 A 0.2-cm-thick, 10-cm-high, and 15-cm-long circuit board houses electronic components on one side that dissipate 18 cm 18 cm Epoxy glass Copper FIGURE P3–171 0.3 cm 0.2 cm 20 fins Fin 2 mm 2 cm 1 mm 15 cm Electronic components 10 cm FIGURE P3–178 Exposed water pipe 2.4 cm Air Tair = –5°C Soil Water FIGURE P3–172 the ambient temperature is 8°C. The combined convection and radiation heat transfer coefficient at the outer surface of the pipe is estimated to be 35 W/m2K. Determine the amount of heat lost from the steam during a 10-h curing process that night.
Steam is supplied by a gas-fired steam generator that has an efficiency of 85 percent, and the plant pays $1.20/therm of nat-ural gas (1 therm 105,500 kJ). If the pipe is insulated and 90 percent of the heat loss is saved as a result, determine the amount of money this facility will save a year as a result of in-sulating the steam pipes. Assume that the concrete pipes are cured 110 nights a year. State your assumptions.
3–171 Consider an 18-cm 18-cm multilayer circuit board dissipating 27 W of heat. The board consists of four layers of 0.2-mm-thick copper (k 386 W/mK) and three layers of 1.5-mm-thick epoxy glass (k 0.26 W/mK) sandwiched together, as shown in the figure. The circuit board is attached to a heat sink from both ends, and the temperature of the board at those ends is 35°C. Heat is considered to be uniformly generated in the epoxy layers of the board at a rate of 0.5 W per 1-cm 18-cm epoxy laminate strip (or 1.5 W per 1-cm 18-cm strip of the board). Considering only a portion of the board because of sym-metry, determine the magnitude and location of the maximum temperature that occurs in the board. Assume heat transfer from the top and bottom faces of the board to be negligible.
3–172 The plumbing system of a house involves a 0.5-m sec-tion of a plastic pipe (k 0.16 W/mK) of inner diameter 2 cm and outer diameter 2.4 cm exposed to the ambient air. During a cold and windy night, the ambient air temperature remains at about 5°C for a period of 14 h. The combined convection and radiation heat transfer coefficient on the outer surface of the 3–183 Cold conditioned air at 12°C is flowing inside a 1.5-cm-thick square aluminum (k 237 W/mK) duct of inner cross section 22 cm 22 cm at a mass flow rate of 0.8 kg/s. The duct is exposed to air at 33°C with a combined convection-radiation heat transfer coefficient of 13 W/m2K.
The convection heat transfer coefficient at the inner surface is 75 W/m2K. If the air temperature in the duct should not in-crease by more than 1°C determine the maximum length of the duct.
3–184 When analyzing heat transfer through windows, it is important to consider the frame as well as the glass area. Con-sider a 2-m-wide, 1.5-m-high wood-framed window with 85 percent of the area covered by 3-mm-thick single-pane glass (k 0.7 W/mK). The frame is 5 cm thick, and is made of pine wood (k 0.12 W/mK). The heat transfer coefficient is 7 W/m2K inside and 13 W/m2K outside. The room is main-tained at 24°C, and the outdoor temperature is 40°C. Deter-mine the percent error involved in heat transfer when the window is assumed to consist of glass only.
3–185 Steam at 260°C is flowing inside a steel pipe (k 61 W/mK) whose inner and outer diameters are 10 cm and 12 cm, respectively, in an environment at 20°C. The heat transfer coefficients inside and outside the pipe are 120 W/m2K and 14 W/m2K, respectively. Determine (a) the thickness of the insulation (k 0.038 W/mK) needed to reduce the heat loss by 95 percent and (b) the thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to 40°C for safety reasons.
3–186 When the transportation of natural gas in a pipeline is not feasible for economic or other reasons, it is first liquefied at CHAPTER 3 217 a total of 15 W of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of 12 W/mK. All the heat generated in the com-ponents is conducted across the circuit board and is dissipated from the back side of the board to a medium at 37°C, with a heat transfer coefficient of 45 W/m2K. (a) Determine the sur-face temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15-cm-long alu-minum plate (k 237 W/mK) with 20 0.2-cm-thick, 2-cm-long, and 15-cm-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a 0.03-cm-thick epoxy adhesive (k 1.8 W/mK). Determine the new temperatures on the two sides of the circuit board.
3–179 Repeat Prob. 3–178 using a copper plate with copper fins (k 386 W/mK) instead of aluminum ones.
3–180 A row of 10 parallel pipes that are 5 m long and have an outer diameter of 6 cm are used to transport steam at 145°C through the concrete floor (k 0.75 W/mK) of a 10-m 5-m room that is maintained at 24°C. The combined convection and radiation heat transfer coefficient at the floor is 12 W/m2K. If the surface temperature of the concrete floor is not to exceed 38°C, determine how deep the steam pipes should be buried below the surface of the concrete floor.
D = 6 cm 38°C Room 24°C Steam pipes Concrete floor 10 m FIGURE P3–180 Fiberglass insulation 0.8 cm 0.8 cm 22 cm Steel plates 99 cm 1 cm FIGURE P3–182 3–181 Consider two identical people each generating 60 W of metabolic heat steadily while doing sedentary work, and dissipating it by convection and perspiration. The first person is wearing clothes made of 1-mm-thick leather (k 0.159 W/mK) that covers half of the body while the second one is wearing clothes made of 1-mm-thick synthetic fabric (k 0.13 W/mK) that covers the body completely. The ambi-ent air is at 30°C, the heat transfer coefficient at the outer sur-face is 15 W/m2K, and the inner surface temperature of the clothes can be taken to be 32°C. Treating the body of each per-son as a 25-cm-diameter, 1.7-m-long cylinder, determine the fractions of heat lost from each person by perspiration.
3-182 A 4-m-high and 6-m-long wall is constructed of two large 0.8-cm-thick steel plates (k 15 W/m·K) separated by 1-cm-thick and 22-cm wide steel bars placed 99 cm apart. The remaining space between the steel plates is filled with fiber-glass insulation (k 0.035 W/m·K). If the temperature differ-ence between the inner and the outer surfaces of the walls is 22°C, determine the rate of heat transfer through the wall. Can we ignore the steel bars between the plates in heat transfer analysis since they occupy only 1 percent of the heat transfer surface area?
about 160°C, and then transported in specially insulated tanks placed in marine ships. Consider a 4-m-diameter spheri-cal tank that is filled with liquefied natural gas (LNG) at 160°C. The tank is exposed to ambient air at 24°C with a heat transfer coefficient of 22 W/m2K. The tank is thin-shelled and its temperature can be taken to be the same as the LNG temperature. The tank is insulated with 5-cm-thick super insulation that has an effective thermal conductivity of 0.00008 W/mK. Taking the density and the specific heat of LNG to be 425 kg/m3 and 3.475 kJ/kg°C, respectively, esti-mate how long it will take for the LNG temperature to rise to 150°C.
3–187 A 15-cm 20-cm hot surface at 85°C is to be cooled by attaching 4-cm-long aluminum (k 237 W/mK) fins of 2-mm 2-mm square cross section. The temperature of sur-rounding medium is 25°C and the heat transfer coefficient on the surfaces can be taken to be 20 W/m2K. If it is desired to triple the rate of heat transfer from the bare hot surface, deter-mine the number of fins that needs to be attached.
3–188 Reconsider Prob. 3–187. Using EES (or other) software, plot the number of fins as a function of the increase in the heat loss by fins relative to no fin case (i.e., overall effectiveness of the fins) in the range of 1.5 to 5. Discuss the results. Is it realistic to assume the heat transfer coefficient to remain constant?
3–189 An agitated vessel is used for heating 500 kg/min of an aqueous solution at 15°C by saturated steam condensing in the jacket outside the vessel. The vessel can hold 6200 kg of the aqueous solution. It is fabricated from 15-mm-thick sheet of 1.0 percent carbon steel (k 43 W/mK), and it provides a heat transfer area of 12.0 m2. The heat transfer coefficient due to agitation is 5.5 kW/m2K, while the steam condensation at 115°C in the jacket gives a heat transfer coefficient of 10.0 kW/m2K. All properties of the aqueous solution are com-parable to those of pure water. Calculate the temperature of the outlet stream in steady operation.
3–190 A 0.6-m-diameter, 1.9-m-long cylindrical tank con-taining liquefied natural gas (LNG) at 160°C is placed at the center of a 1.9-m-long 1.4-m 1.4-m square solid bar made of an insulating material with k 0.0002 W/mK. If the outer surface temperature of the bar is 12°C, determine the rate of heat transfer to the tank. Also, determine the LNG temperature after one month. Take the density and the specific heat of LNG to be 425 kg/m3 and 3.475 kJ/kg°C, respectively.
3–191 A typical section of a building wall is shown in Fig. P3–191. This section extends in and out of the page and is repeated in the vertical direction. The wall support members are made of steel (k 50 W/mK). The support members are 8 cm (t23) 0.5 cm (LB). The remainder of the inner wall space is filled with insulation (k 0.03 W/mK) and measures 8 cm (t23) 60 cm (LB). The inner wall is made of gypsum board (k 0.5 W/mK) that is 1 cm thick (t12) and the outer wall is 218 STEADY HEAT CONDUCTION made of brick (k 1.0 W/mK) that is 10 cm thick (t34). What is the temperature on the interior brick surface, 3, when T1 20 °C and T4 35°C?
LA LB 0 1 2 3 4 5 FIGURE P3–191 3–192 A total of 10 rectangular aluminum fins (k 203 W/mK) are placed on the outside flat surface of an elec-tronic device. Each fin is 100 mm wide, 20 mm high and 4 mm thick. The fins are located parallel to each other at a center-to-center distance of 8 mm. The temperature at the outside surface of the electronic device is 72°C. The air is at 20°C, and the heat transfer coefficient is 80 W/m2K. Determine (a) the rate of heat loss from the electronic device to the surrounding air and (b) the fin effectiveness.
3–193 One wall of a refrigerated warehouse is 10.0-m-high and 5.0-m-wide. The wall is made of three layers: 1.0-cm-thick aluminum (k 200 W/mK), 8.0-cm-thick fibreglass (k 0.038 W/mK), and 3.0-cm thick gypsum board (k 0.48 W/mK). The warehouse inside and outside temperatures are 10°C and 20°C, respectively, and the average value of both inside and outside heat transfer coefficients is 40 W/m2K.
(a) Calculate the rate of heat transfer across the warehouse wall in steady operation.
(b) Suppose that 400 metal bolts (k 43 W/mK), each 2.0 cm in diameter and 12.0 cm long, are used to fasten (i.e., hold together) the three wall layers. Calculate the rate of heat transfer for the “bolted” wall.
(c) What is the percent change in the rate of heat transfer across the wall due to metal bolts?
3–194 A 2.2-m-diameter spherical steel tank filled with iced water at 0°C is buried underground at a location where the ther-mal conductivity of the soil is k 0.55 W/mK. The distance between the tank center and the ground surface is 2.4 m. For ground surface temperature of 18°C, determine the rate of heat transfer to the iced water in the tank. What would your answer be if the soil temperature were 18°C and the ground surface were insulated?
CHAPTER 3 219 3–196 A spherical vessel, 3.0 m in diameter (and negligible wall thickness), is used for storing a fluid at a temperature of 0°C. The vessel is covered with a 5.0-cm-thick layer of an in-sulation (k 0.20 W/mK). The surrounding air is at 22°C.
The inside and outside heat transfer coefficients are 40 and 10 W/m2K, respectively. Calculate (a) all thermal resistances, in K/W, (b) the steady rate of heat transfer, and (c) the temper-ature difference across the insulation layer.
3–197 A plane wall with surface temperature of 300°C is attached with straight aluminum triangular fins (k 236 W/m·K). The fins are exposed to an ambient air condi-tion of 25°C and the convection heat transfer coefficient is 25 W/m2·K. Each fin has a length of 55 mm, a base of 4 mm thick and a width of 110 mm. Using Table 3–4, determine the efficiency, heat transfer rate, and effectiveness of each fin.
a single fin and the increase in the rate of heat transfer per m2 surface area as a result of attaching fins. Assume there are 100 fins per m2 surface area.
3–199 Circular fins of uniform cross section, with diameter of 10 mm and length of 50 mm, are attached to a wall with sur-face temperature of 350°C. The fins are made of material with thermal conductivity of 240 W/m·K, and they are exposed to an ambient air condition of 25°C and the convection heat transfer coefficient is 250 W/m2·K. Determine the heat transfer rate and plot the temperature variation of a single fin for the following boundary conditions: (a) Infinitely long fin (b) Adiabatic fin tip (c) Fin with tip temperature of 250°C (d) Convection from the fin tip 2 cm 2 cm 12 cm Copper Epoxy composite FIGURE P3–195 Tb Ab = Ac x = 0 L k h, T D FIGURE P3–199 k = 236 W/m·K Air, 25°C h = 25 W/m2·K t = 4 mm L = 55 mm Tb = 300°C w = 110 mm FIGURE P3–197 T2 L w D = 127 mm Air, –5°C h = 20 W/m2.K Concrete bar k = 1.7 W/m.K T1 = 120°C FIGURE P3–200 3–195 A 12-cm-long bar with a square cross-section, as shown in Fig. P3–195, consists of a 1-cm-thick copper layer (k 380 W/mK) and a 1-cm-thick epoxy composite layer (k 0.4 W/mK). Calculate the rate of heat transfer under a thermal driving force of 50°C, when the direction of steady one-dimensional heat transfer is (a) from front to back (i.e., along its length), (b) from left to right, and (c) from top to bottom.
3–200 In a combined heat and power (CHP) generation process, by-product heat is used for domestic or industrial heat-ing purposes. Hot steam is carried from a CHP generation plant by a tube with diameter of 127 mm centered at a square cross-section solid bar made of concrete with thermal conductivity of 1.7 W/m·K. The surface temperature of the tube is constant at 120°C, while the square concrete bar is exposed to air with temperature of –5°C and convection heat transfer coefficient of 20 W/m2·K. If the temperature difference between the outer surface of the square concrete bar and the ambient air is to be maintained at 5°C, determine the width of the square concrete bar and the rate of heat loss per meter length. Answers: 1.32 m, 530 W/m 3–198 A plane wall surface at 200°C is to be cooled with alu-minum pin fins of parabolic profile with blunt tips. Each fin has a length of 25 mm and a base diameter of 4 mm. The fins are exposed to an ambient air condition of 25°C and the heat transfer coefficient is 45 W/m2·K. If the thermal conductivity of the fins is 230 W/m·K, determine the heat transfer rate from Fundamentals of Engineering (FE) Exam Problems 3–201 Heat is lost at a rate of 275 W per m2 area of a 15-cm-thick wall with a thermal conductivity of k 1.1 W/mK. The temperature drop across the wall is (a) 37.5°C (b) 27.5°C (c) 16.0°C (d) 8.0°C (e) 4.0°C 3–202 Consider a wall that consists of two layers, A and B, with the following values: kA 1.2 W/mK, LA 8 cm, kB 0.2 W/mK, LB 5 cm. If the temperature drop across the wall is 18°C, the rate of heat transfer through the wall per unit area of the wall is (a) 56.8 W/m2 (b) 72.1 W/m2 (c) 114 W/m2 (d) 201 W/m2 (e) 270 W/m2 3–203 A plane furnace surface at 150°C covered with 1-cm-thick insulation is exposed to air at 30°C, and the combined heat transfer coefficient is 25 W/m2K. The thermal conductivity of insulation is 0.04 W/mK. The rate of heat loss from the surface per unit surface area is (a) 35 W (b) 414 W (c) 300 W (d) 480 W (e) 128 W 3–204 Heat is generated steadily in a 3-cm-diameter spheri-cal ball. The ball is exposed to ambient air at 26°C with a heat transfer coefficient of 7.5 W/m2K. The ball is to be covered with a material of thermal conductivity 0.15 W/mK. The thickness of the covering material that will maximize heat gen-eration within the ball while maintaining ball surface tempera-ture constant is (a) 0.5 cm (b) 1.0 cm (c) 1.5 cm (d) 2.0 cm (e) 2.5 cm 3–205 Consider a 1.5-m-high and 2-m-wide triple pane win-dow. The thickness of each glass layer (k 0.80 W/mK) is 0.5 cm, and the thickness of each air space (k 0.025 W/mK) is 1.2 cm. If the inner and outer surface temperatures of the window are 10°C and 0°C, respectively, the rate of heat loss through the window is (a) 3.4 W (b) 10.2 W (c) 30.7 W (d) 61.7 W (e) 86.8 W 3–206 Consider a furnace wall made of sheet metal at an av-erage temperature of 800°C exposed to air at 40°C. The com-bined heat transfer coefficient is 200 W/m2K inside the furnace, and 80 W/m2K outside. If the thermal resistance of the furnace wall is negligible, the rate of heat loss from the fur-nace per unit surface area is (a) 48.0 kW/m2 (b) 213 kW/m2 (c) 91.2 kW/m2 (d) 151 kW/m2 (e) 43.4 kW/m2 3–207 Consider a jacket made of 5 layers of 0.1-mm-thick cotton fabric (k 0.060 W/mK) with a total of 4 layers of 1-mm-thick air space (k 0.026 W/mK) in between. Assum-ing the inner surface temperature of the jacket is 25°C and the 220 STEADY HEAT CONDUCTION surface area normal to the direction of heat transfer is 1.1 m2, determine the rate of heat loss through the jacket when the tem-perature of the outdoors is 0°C and the heat transfer coefficient on the outer surface is 18 W/m2K.
(a) 6 W (b) 115 W (c) 126 W (d) 287 W (e) 170 W 3–208 Consider two metal plates pressed against each other.
Other things being equal, which of the measures below will cause the thermal contact resistance to increase?
(a) Cleaning the surfaces to make them shinier.
(b) Pressing the plates against each other with a greater force.
(c) Filling the gab with a conducting fluid.
(d) Using softer metals.
(e) Coating the contact surfaces with a thin layer of soft metal such as tin.
3–209 A 10-m-long 8-cm-outer-radius cylindrical steam pipe is covered with 3-cm thick cylindrical insulation with a thermal conductivity of 0.05 W/mK. If the rate of heat loss from the pipe is 1000 W, the temperature drop across the insulation is (a) 58°C (b) 101°C (c) 143°C (d) 282°C (e) 600°C 3–210 Steam at 200°C flows in a cast iron pipe (k 80 W/mK) whose inner and outer diameters are D1 0.20 m and D2 0.22 m, respectively. The pipe is covered with 2-cm-thick glass wool insulation (k 0.05 W/mK). The heat transfer coeffi-cient at the inner surface is 75 W/m2K. If the temperature at the interface of the iron pipe and the insulation is 194°C, the temperature at the outer surface of the insulation is (a) 32 °C (b) 45 °C (c) 51 °C (d) 75 °C (e) 100 °C 3–211 A 5-m-diameter spherical tank is filled with liquid oxygen (r 1141 kg/m3, cp 1.71 kJ/kg°C) at 184°C. It is observed that the temperature of oxygen increases to 183°C in a 144-hour period. The average rate of heat transfer to the tank is (a) 124 W (b) 185 W (c) 246 W (d) 348 W (e) 421 W 3–212 A 2.5 m-high, 4-m-wide, and 20-cm-thick wall of a house has a thermal resistance of 0.025°C/W. The thermal con-ductivity of the wall is (a) 0.8 W/m·K (b) 1.2 W/m·K (c) 3.4 W/m·K (d) 5.2 W/m·K (e) 8.0 W/m·K 3–213 Consider two walls, A and B, with the same surface ar-eas and the same temperature drops across their thicknesses.
The ratio of thermal conductivities is kA/kB 4 and the ratio of the wall thicknesses is LA/LB 2. The ratio of heat transfer rates through the walls Q .
A/Q .
B is (a) 0.5 (b) 1 (c) 2 (d) 4 (e) 8 CHAPTER 3 221 3–214 A hot plane surface at 100°C is exposed to air at 25°C with a combined heat transfer coefficient of 20 W/m2K. The heat loss from the surface is to be reduced by half by covering it with sufficient insulation with a thermal conductivity of 0.10 W/mK. Assuming the heat transfer coefficient to remain constant, the required thickness of insulation is (a) 0.1 cm (b) 0.5 cm (c) 1.0 cm (d) 2.0 cm (e) 5 cm 3–215 Consider a 4.5-m-long, 3.0-m-high, and 0.22-m-thick wall made of concrete (k 1.1 W/mK). The design tempera-tures of the indoor and outdoor air are 24°C and 3°C, respec-tively, and the heat transfer coefficients on the inner and outer surfaces are 10 and 20 W/m2K. If a polyurethane foam insula-tion (k 0.03 W/mK) is to be placed on the inner surface of the wall to increase the inner surface temperature of the wall to 22°C, the required thickness of the insulation is (a) 3.3 cm (b) 3.0 cm (c) 2.7 cm (d) 2.4 cm (e) 2.1 cm 3–216 Steam at 200°C flows in a cast iron pipe (k 80 W/m ) whose inner and outer diameters are D1 0.20 m and D2 0.22 m. The pipe is exposed to room air at 35°C. The heat transfer coefficients at the inner and outer surfaces of the pipe are 90 and 20 W/m2K, respectively. The pipe is to be cov-ered with glass wool insulation (k 0.05 W/mK) to decrease the heat loss from the stream by 90 percent. The required thick-ness of the insulation is (a) 1.2 cm (b) 2.0 cm (c) 2.8 cm (d) 3.4 cm (e) 4.0 cm 3–217 A 50-cm-diameter spherical tank is filled with iced water at 0°C. The tank is thin-shelled and its temperature can be taken to be the same as the ice temperature. The tank is ex-posed to ambient air at 20°C with a heat transfer coefficient of 12 W/m2K. The tank is to be covered with glass wool in-sulation (k 0.05 W/mK) to decrease the heat gain to the iced water by 90 percent. The required thickness of the insu-lation is (a) 4.6 cm (b) 6.7 cm (c) 8.3 cm (d) 25.0 cm (e) 29.6 cm 3–218 A room at 20°C air temperature is loosing heat to the outdoor air at 0°C at a rate of 1000 W through a 2.5-m-high and 4-m-long wall. Now the wall is insulated with 2-cm thick insu-lation with a conductivity of 0.02 W/mK. Determine the rate of heat loss from the room through this wall after insulation. As-sume the heat transfer coefficients on the inner and outer sur-face of the wall, the room air temperature, and the outdoor air temperature to remain unchanged. Also, disregard radiation.
(a) 20 W (b) 561 W (c) 388 W (d) 167 W (e) 200 W 3–219 A 1-cm-diameter, 30-cm-long fin made of aluminum (k 237 W/mK) is attached to a surface at 80°C. The surface is exposed to ambient air at 22°C with a heat transfer coeffi-cient of 18 W/m2K. If the fin can be assumed to be very long, the rate of heat transfer from the fin is (a) 2.0 W (b) 3.2 W (c) 4.4 W (d) 5.5 W (e) 6.0 W 3–220 A 1-cm-diameter, 30-cm-long fin made of aluminum (k 237 W/mK) is attached to a surface at 80°C. The surface is exposed to ambient air at 22°C with a heat transfer coeffi-cient of 11 W/m2K. If the fin can be assumed to be very long, its efficiency is (a) 0.60 (b) 0.67 (c) 0.72 (d) 0.77 (e) 0.88 3–221 A hot surface at 80°C in air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylindrical fins. The combined heat transfer coefficient is 30 W/m2K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins is (a) 325 W (b) 707 W (c) 566 W (d) 424 W (e) 754 W 3–222 A cylindrical pin fin of diameter 1 cm and length 5 cm with negligible heat loss from the tip has an effectiveness of 15.
If the fin base temperature is 280°C, the environment tempera-ture is 20°C, and the heat transfer coefficient is 65 W/m2K, the rate of heat loss from this fin is (a) 20 W (b) 48 W (c) 156 W (d) 398 W (e) 418 W 3–223 A cylindrical pin fin of diameter 0.6 cm and length of 3 cm with negligible heat loss from the tip has an efficiency of 0.7. The effectiveness of this fin is (a) 0.3 (b) 0.7 (c) 2 (d) 8 (e) 14 3–224 A 3-cm-long, 2-mm 2-mm rectangular cross-section aluminum fin (k 237 W/mK) is attached to a sur-face. If the fin efficiency is 65 percent, the effectiveness of this single fin is (a) 39 (b) 30 (c) 24 (d) 18 (e) 7 3–225 Aluminum square pin fins (k 237 W/mK) of 3-cm-long, 2 mm 2 mm cross-section with a total number of 150 are attached to an 8-cm-long, 6-cm-wide surface. If the fin efficiency is 78 percent, the overall fin effectiveness for the sur-face is (a) 3.4 (b) 4.2 (c) 5.5 (d) 6.7 (e) 8.4 3–226 Two finned surfaces with long fins are identical, ex-cept that the convection heat transfer coefficient for the first finned surface is twice that of the second one. What statement below is accurate for the efficiency and effectiveness of the first finned surface relative to the second one?
(a) Higher efficiency and higher effectiveness (b) Higher efficiency but lower effectiveness (c) Lower efficiency but higher effectiveness (d) Lower efficiency and lower effectiveness (e) Equal efficiency and equal effectiveness 3–227 A 20-cm-diameter hot sphere at 120°C is buried in the ground with a thermal conductivity of 1.2 W/mK. The dis-tance between the center of the sphere and the ground surface is 0.8 m and the ground surface temperature is 15°C. The rate of heat loss from the sphere is (a) 169 W (b) 20 W (c) 217 W (d) 312 W (e) 1.8 W 3–228 A 25-cm-diameter, 2.4-m-long vertical cylinder con-taining ice at 0°C is buried right under the ground. The cylin-der is thin-shelled and is made of a high thermal conductivity material. The surface temperature and the thermal conductivity of the ground are 18°C and 0.85 W/mK respectively. The rate of heat transfer to the cylinder is (a) 37.2 W (b) 63.2 W (c) 158 W (d) 480 W (e) 1210 W 3–229 Hot water (cp 4.179 kJ/kgK) flows through a 80-m-long PVC (k 0.092 W/mK) pipe whose inner diame-ter is 2 cm and outer diameter is 2.5 cm at a rate of 1 kg/s, en-tering at 40°C. If the entire interior surface of this pipe is maintained at 35°C and the entire exterior surface at 20°C, the outlet temperature of water is (a) 35°C (b) 36°C (c) 37°C (d) 38°C (e) 39°C 3–230 The walls of a food storage facility are made of a 2-cm-thick layer of wood (k 0.1 W/mK) in contact with a 5-cm-thick layer of polyurethane foam (k 0.03 W/mK). If the temperature of the surface of the wood is 10°C and the temperature of the surface of the polyurethane foam is 20°C, the temperature of the surface where the two layers are in contact is (a) 7°C (b) 2°C (c) 3°C (d) 8°C (e) 11°C 3–231 Heat transfer rate through the wall of a circular tube with convection acting on the outer surface is given per unit of its length by where i refers to the innertube surface and o the outer tube surface. Increasing ro will reduce the heat transfer as long as (a) ro k/h (b) ro k/h (c) ro k/h (d) ro 2k/h (e) Increasing ro will always reduce the heat transfer.
3–232 A typical section of a building wall is shown in Fig. P3–232. This section extends in and out of the page and is q # 2pL(Ti To) ln(ro/ri) k 1 roh 222 STEADY HEAT CONDUCTION R23A R23A R23B R23B R23A R23B R34 R34 R34 R45 R01 R01 R12 R12 R12 R23A R34 R01 R12 T5 T0 T5 T5 T0 T0 T5 T0 LA LB 0 1 2 3 4 5 (a) (b) (c) (d) (e) None of them FIGURE P3–232 3–233 The 700 m2 ceiling of a building has a thermal resis-tance of 0.52 m2K/W. The rate at which heat is lost through this ceiling on a cold winter day when the ambient tempera-ture is 10°C and the interior is at 20°C is (a) 23.1 kW (b) 40.4 kW (c) 55.6 kW (d) 68.1 kW (e) 88.6 kW 3–234 A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at 90 K. The tank consists of a 0.5-cm-thick aluminum (k 170 W/mK) shell whose exte-rior is covered with a 10-cm-thick layer of insulation (k 0.02 W/mK). The insulation is exposed to the ambient air at 20°C and the heat transfer coefficient on the exterior side of the insulation is 5 W/m2K. The rate at which the liquid oxygen gains heat is (a) 141 W (b) 176 W (c) 181 W (d) 201 W (e) 221 W 3–235 A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at 90 K. The tank consists of a 0.5-cm-thick aluminum (k 170 W/mK) shell whose exte-rior is covered with a 10-cm-thick layer of insulation (k 0.02 W/mK). The insulation is exposed to the ambient air at 20°C and the heat transfer coefficient on the exterior side of the insulation is 5 W/m2K. The temperature of the exterior surface of the insulation is (a) 13°C (b) 9°C (c) 2°C (d) 3°C (e) 12°C repeated in the vertical direction. The correct thermal resis-tance circuit for this wall is CHAPTER 3 223 3–236 The fin efficiency is defined as the ratio of the actual heat transfer from the fin to (a) The heat transfer from the same fin with an adiabatic tip (b) The heat transfer from an equivalent fin which is infi-nitely long (c) The heat transfer from the same fin if the temperature along the entire length of the fin is the same as the base temperature (d) The heat transfer through the base area of the same fin (e) None of the above 3–237 Computer memory chips are mounted on a finned metallic mount to protect them from overheating. A 152 MB memory chip dissipates 5 W of heat to air at 25°C. If the tem-perature of this chip is to not exceed 60°C, the overall heat transfer coefficient–area product of the finned metal mount must be at least (a) 0.14 W/°C (b) 0.20 W/°C (c) 0.32 W/°C (d) 0.48 W/°C (e) 0.76 W/°C 3–238 In the United States, building insulation is specified by the R-value (thermal resistance in hft2°F/Btu units). A home owner decides to save on the cost of heating the home by adding additional insulation in the attic. If the total R-value is increased from 15 to 25, the home owner can expect the heat loss through the ceiling to be reduced by (a) 25% (b) 40% (c) 50% (d) 60% (e) 75% 3–239 Coffee houses frequently serve coffee in a paper cup that has a corrugated paper jacket surrounding the cup like that shown here. This corrugated jacket: (a) Serves to keep the coffee hot.
(b) Increases the coffee-to-surrounding thermal resistance.
(c) Lowers the temperature where the hand clasps the cup.
(d) All of the above.
(e) None of the above.
aluminum (k 150 W/mK). This fin is exposed to air with a convective heat transfer coefficient of 30 W/m2K acting on its surfaces. The efficiency of the fin is 75 percent. If the fin base temperature is 130°C and the air temperature is 25°C, the heat transfer from this fin per unit width is (a) 32 W/m (b) 57 W/m (c) 102 W/m (d) 124 W/m (e) 142 W/m 3–241 A plane brick wall (k 0.7 W/mK) is 10 cm thick.
The thermal resistance of this wall per unit of wall area is (a) 0.143 m2K/W (b) 0.250 m2K/W (c) 0.327 m2K/W (d) 0.448 m2K/W (e) 0.524 m2K/W Design and Essay Problems 3–242 The temperature in deep space is close to absolute zero, which presents thermal challenges for the astronauts who do space walks. Propose a design for the clothing of the astro-nauts that will be most suitable for the thermal environment in space. Defend the selections in your design.
3–243 In the design of electronic components, it is very de-sirable to attach the electronic circuitry to a substrate material that is a very good thermal conductor but also a very effective electrical insulator. If the high cost is not a major concern, what material would you propose for the substrate?
3–244 Using cylindrical samples of the same material, devise an experiment to determine the thermal contact resistance.
Cylindrical samples are available at any length, and the thermal conductivity of the material is known.
3–245 Find out about the wall construction of the cabins of large commercial airplanes, the range of ambient conditions under which they operate, typical heat transfer coefficients on the inner and outer surfaces of the wall, and the heat genera-tion rates inside. Determine the size of the heating and air-conditioning system that will be able to maintain the cabin at 20°C at all times for an airplane capable of carrying 400 people.
3–246 Repeat Prob. 3–245 for a submarine with a crew of 60 people.
3–247 A house with 200-m2 floor space is to be heated with geothermal water flowing through pipes laid in the ground un-der the floor. The walls of the house are 4 m high, and there are 10 single-paned windows in the house that are 1.2 m wide and 1.8 m high. The house has R-19 (in hft2°F/Btu) insulation in the walls and R-30 on the ceiling. The floor temperature is not to exceed 40°C. Hot geothermal water is available at 90°C, and the inner and outer diameter of the pipes to be used are 2.4 cm and 3.0 cm. Design such a heating system for this house in your area.
FIGURE P3–239 3–240 A triangular shaped fin on a motorcycle engine is 0.5-cm thick at its base and 3-cm long (normal distance between the base and the tip of the triangle), and is made of 3–248 Using a timer (or watch) and a thermometer, conduct this experiment to determine the rate of heat gain of your refrigerator. First, make sure that the door of the refrigerator is not opened for at least a few hours to make sure that steady operating conditions are established. Start the timer when the refrigerator stops running and measure the time t1 it stays off before it kicks in. Then measure the time t2 it stays on.
Noting that the heat removed during t2 is equal to the heat gain of the refrigerator during t1 t2 and using the power 224 STEADY HEAT CONDUCTION consumed by the refrigerator when it is running, determine the average rate of heat gain for your refrigerator, in watts.
Take the COP (coefficient of performance) of your refrigera-tor to be 1.3 if it is not available.
Now, clean the condenser coils of the refrigerator and re-move any obstacles on the way of airflow through the coils. By replacing these measurements, determine the improvement in the COP of the refrigerator. |
5356 | https://virtualnerd.com/common-core/hsa-algebra/HSA-REI-equations-inequalities-reasoning/B/3/variables-and-fraction-both-sides-example | Real math help.
How Do You Solve an Equation with Variables on Both Sides and Fractions?
How Do You Solve an Equation with Variables on Both Sides and Fractions?
Note:
Trying to solve an equation with variables and fractions on both sides of the equation? You can bet it involves finding a common denominator! To see what it takes, watch this tutorial.
Keywords:
problem
equation
solve
fractions
linear equation
variable on both sides
1 variable
distributive property
multiplication property of equality
solve by subtraction
solve by division
Background Tutorials
Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
How Do You Find a Common Denominator and a Least Common Denominator?
This tutorial gives you some practice finding a common denominator and the least common denominator of three fractions. There's only one least common denominator, but there are many common denominators. This tutorial gives you one. Can you find another?
#### Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations).
What is a Variable?
You can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!
#### Apply the properties of operations to generate equivalent expressions.
How Do You Use the Distributive Property to Simplify an Expression?
In this tutorial you'll see how to apply the distributive property. Remember that this is important when you are trying to simplify an expression and get rid of parentheses!
#### Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers.
What's the Multiplication Property of Equality?
Solving equations can be tough, especially if you've forgotten or have trouble understanding the tools at your disposal. One of those tools is the multiplication property of equality, and it lets you multiply both sides of an equation by the same number. Watch the video to see it in action!
#### Apply properties of operations as strategies to add and subtract rational numbers.
How Do You Subtract a Whole Number from a Fraction?
Subtracting a whole number from a fraction can be tricky. Luckily, watching this tutorial can make this subtraction no big deal!
#### Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
How Do You Solve an Equation with Variables on Both Sides?
Trying to solve an equation with variables on both sides of the equal sign? Figure out how to get those variables together and find the answer with this tutorial!
Further Exploration
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
How Do You Solve a Multi-Step Equation with Fractions by Multiplying Away the Fraction?
Trying to solve an equation involving a fraction? Just multiply the fraction away and then perform the order of operations in reverse! See how in this tutorial.
+ #### How Do You Solve a Multi-Step Equation with Fractions Using Reverse Order of Operations and Reciprocals?
Trying to solve an equation involving a fraction? Just perform the order of operations in reverse! See how in this tutorial.
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5357 | https://www.youtube.com/watch?v=GT1TXcB5yFI | Substitution in Quadratic Equations
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Substitution in Quadratic Equations
I put together a book to help support students in their Algebra I journey. Click the link below to learn more.
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Transcript:
okay so there's sometimes when you look at an equation and you're like oh my goodness i'm going to have to do so much work to solve that equation so for example we're looking at an equation here and as far as what it looks like on first glance it looks like we're going to have to take this and we're going to have to foil you know so multiply this by itself and this by itself and this by this and this like this and it just looks like a whole lot going on just in this one little part not to mention we have to distribute this that that and simplify it and then solve a lot of stuff going on or not well when you see an equation that looks pretty complex and like there's just too much going on you want to try and find an opportunity to solve it a little bit in a little bit easier way so if you take notice here there's a 3y minus 2 here and there is a 3y minus 2 here so you can recognize that maybe perhaps if we just call this whole thing one little thing then we might have an easier time so perhaps we can go ahead and call 3 minus 2 x we can substitute x for 3 minus two three y minus two so let's say everywhere we see three minus two we're going to put an x so we have it there and we have it there as well so what we now end up having is x square minus five x minus six is equal to zero all right so all i did was i took x and replace it for three minus two like a substitution actually it is a substitution all right now that we have that done this looks a lot more manageable so we can factor and solve this equation here so we're looking for two numbers that multiply to get negative six and add to get negative five don't mind the sounds in the background all right those two numbers are negative six and positive one so now we've factored it and we just have one more step to solve it in this form it's not quite done but we're almost there so two numbers are being multiplied x minus six is being multiplied by x plus one and the answer is zero which means that either x minus six is zero or x plus one is zero because the only way to get an answer of 0 is to multiply by a 0. so that means x is equal to adding 6 to both sides here x could be equal to 6 or subtracting 1 from both sides here x could be equal to negative 1. but again we're not done remember initially we created a relationship that 3y minus 2 is x but the equation originally wanted us to find y there was no x we created that so in order to finish this question we have to figure out what y is but since we know that 3 y minus 2 is x and we now know that x is either 6 or negative 1 we can solve so we have here 3y minus 2 is in the first case we have that x is 6. so let's go ahead and solve that we can add 2 to both sides and we get that y is equal to eight now we can divide both sides by three and we get that y is equal to eight over three but that's just one of our answers we still need to find what happens when x is negative one so three y minus two is equal to x that's the relationship we we created so three y minus two is equal to negative one we can add two to both sides once again and we get that three y is equal to one and finally we divide both sides by three and we get that y is a third and we're done |
5358 | https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/02%3A_Vectors/2.04%3A__Coordinate_Systems_and_Components_of_a_Vector_(Part_1) | 2.4: Coordinate Systems and Components of a Vector (Part 1) - Physics LibreTexts
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2: Vectors
University Physics I - Mechanics, Sound, Oscillations, and Waves (OpenStax)
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Sun, 16 Mar 2025 21:22:14 GMT
2.4: Coordinate Systems and Components of a Vector (Part 1)
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3973
Delmar Larsen
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[ "article:topic", "authorname:openstax", "vector components", "the component form of a vector", "scalar components", "direction angle", "license:ccby", "showtoc:no", "program:openstax", "licenseversion:40", "source@ ]
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2.4: Coordinate Systems and Components of a Vector (Part 1)
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2.3: Scalars and Vectors (Part 2)
2.5: Coordinate Systems and Components of a Vector (Part 2)
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( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Example 2.4.1: Displacement of a Mouse Pointer
1. Solution
2. Significance
Exercise 2.4
Example 2.4.2: Magnitude and Direction of the Displacement Vector
Solution
Exercise 2.5
Example 2.4.3: Components of Displacement Vectors
Solution
Exercise 2.6
Learning Objectives
Describe vectors in two and three dimensions in terms of their components, using unit vectors along the axes.
Distinguish between the vector components of a vector and the scalar components of a vector.
Explain how the magnitude of a vector is defined in terms of the components of a vector.
Identify the direction angle of a vector in a plane.
Explain the connection between polar coordinates and Cartesian coordinates in a plane.
Vectors are usually described in terms of their components in a coordinate system. Even in everyday life we naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions to a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction 37° north of east.
In a rectangular (Cartesian) xy-coordinate system in a plane, a point in a plane is described by a pair of coordinates (x, y). In a similar fashion, a vector A→ in a plane is described by a pair of its vector coordinates. The x-coordinate of vector A→ is called its x-component and the y-coordinate of vector A→ is called its y-component. The vector x-component is a vector denoted by A→x. The vector y-component is a vector denoted by A→y. In the Cartesian system, the x and y vector components of a vector are the orthogonal projections of this vector onto the x- and y-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of its vector components:
(2.4.1)A→=A→x+A→y.
As illustrated in Figure 2.4.1, vector A→ is the diagonal of the rectangle where the x-component A→x is the side parallel to the x-axis and the y-component A→y is the side parallel to the y-axis. Vector component A→x is orthogonal to vector component A→y.
Figure 2.4.1: Vector A→ in a plane in the Cartesian coordinate system is the vector sum of its vector x- and y-components. The x-vector component A→x is the orthogonal projection of vector A→ onto the x-axis. The y-vector component A→y is the orthogonal projection of vector A→ onto the y-axis. The numbers A x and A y that multiply the unit vectors are the scalar components of the vector.
It is customary to denote the positive direction on the x-axis by the unit vector i^ and the positive direction on the y-axis by the unit vector j^. Unit vectors of the axes, i^ and j^, define two orthogonal directions in the plane. As shown in Figure 2.4.1, the x- and y- components of a vector can now be written in terms of the unit vectors of the axes:
(2.4.2){A→x=A xi^A→y=A yj^
The vectors A→x and A→y defined by Equation 2.11 are the vector components of vector A→. The numbers A x and A y that define the vector components in Equation 2.4.2 are the scalar components of vector A→. Combining Equation 2.4.1 with Equation 2.4.2, we obtain the component form of a vector:
(2.4.3)A→=A xi^+A yj^.
If we know the coordinates b(x b,y b) of the origin point of a vector (where b stands for “beginning”) and the coordinates e(x e, y e) of the end point of a vector (where e stands for “end”), we can obtain the scalar components of a vector simply by subtracting the origin point coordinates from the end point coordinates:
(2.4.4){A x=x e−x b A y=y e−y b.
Example 2.4.1: Displacement of a Mouse Pointer
A mouse pointer on the display monitor of a computer at its initial position is at point (6.0 cm, 1.6 cm) with respect to the lower left-side corner. If you move the pointer to an icon located at point (2.0 cm, 4.5 cm), what is the displacement vector of the pointer?
Strategy
The origin of the xy-coordinate system is the lower left-side corner of the computer monitor. Therefore, the unit vector i^ on the x-axis points horizontally to the right and the unit vector j^ on the y-axis points vertically upward. The origin of the displacement vector is located at point b(6.0, 1.6) and the end of the displacement vector is located at point e(2.0, 4.5). Substitute the coordinates of these points into Equation 2.4.4 to find the scalar components D x and D y of the displacement vector D→. Finally, substitute the coordinates into Equation 2.4.3 to write the displacement vector in the vector component form.
Solution
We identify x b = 6.0, x e = 2.0, y b = 1.6, and y e = 4.5, where the physical unit is 1 cm. The scalar x- and y-components of the displacement vector are
(2.4.5)D x=x e−x b=(2.0−6.0)cm=−4.0 cm,
(2.4.6)D y=y e−y b=(4.5−1.6)cm=+2.9 cm.
The vector component form of the displacement vector is
(2.4.7)D→=D x i^+D y j^=(−4.0 cm)i^+(2.9 cm)j^=(−4.0 i^+2.9 j^)cm.
This solution is shown in Figure 2.4.2.
Figure 2.4.2: The graph of the displacement vector. The vector points from the origin point at b to the end point at e.
Significance
Notice that the physical unit—here, 1 cm—can be placed either with each component immediately before the unit vector or globally for both components, as in Equation 2.4.7. Often, the latter way is more convenient because it is simpler.
The vector x-component D→x = −4.0 i^ = 4.0(−i^) of the displacement vector has the magnitude |D→x| = |− 4.0||i^| = 4.0 because the magnitude of the unit vector is |i^| = 1. Notice, too, that the direction of the x-component is −i^, which is antiparallel to the direction of the +x-axis; hence, the x-component vector D→x points to the left, as shown in Figure 2.4.2. The scalar x-component of vector D→ is D x = −4.0. Similarly, the vector y-component D→y = +2.9j^ of the displacement vector has magnitude |D→y| = |2.9||j^| = 2.9 because the magnitude of the unit vector is |j^| = 1. The direction of the y-component is +j^, which is parallel to the direction of the +y-axis. Therefore, the y-component vector D→y points up, as seen in Figure 2.4.2. The scalar y-component of vector D→ is D y = + 2.9. The displacement vector D→ is the resultant of its two vector components.
The vector component form of the displacement vector Equation 2.4.7 tells us that the mouse pointer has been moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.
Exercise 2.4
A blue fly lands on a sheet of graph paper at a point located 10.0 cm to the right of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the left edge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lower left-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing.
When we know the scalar components A x and A y of a vector A→, we can find its magnitude A and its direction angle θ A. The direction angle—or direction, for short—is the angle the vector forms with the positive direction on the x-axis. The angle θ A is measured in the counterclockwise direction from the +x-axis to the vector (Figure 2.4.3). Because the lengths A, A x, and A y form a right triangle, they are related by the Pythagorean theorem:
(2.4.8)A 2=A x 2+A y 2⇔A=A x 2+A y 2.
This equation works even if the scalar components of a vector are negative. The direction angle θ A of a vector is defined via the tangent function of angle θ A in the triangle shown in Figure 2.4.3:
(2.4.9)tanθ=A y A x⇒θ=tan−1(A y A x).
Figure 2.4.3: For vector A→, its magnitude A and its direction angle θ A are related to the magnitudes of its scalar components because A, A x, and A y form a right triangle.
When the vector lies either in the first quadrant or in the fourth quadrant, where component A x is positive (Figure 2.4.4), the angle θ in Equation 2.4.9) is identical to the direction angle θ A. For vectors in the fourth quadrant, angle θ is negative, which means that for these vectors, direction angle θ A is measured clockwise from the positive x-axis. Similarly, for vectors in the second quadrant, angle θ is negative. When the vector lies in either the second or third quadrant, where component A x is negative, the direction angle is θ A = θ + 180° (Figure 2.4.4).
Figure 2.4.4: Scalar components of a vector may be positive or negative. Vectors in the first quadrant (I) have both scalar components positive and vectors in the third quadrant have both scalar components negative. For vectors in quadrants II and III, the direction angle of a vector is θ A = θ + 180°.
Example 2.4.2: Magnitude and Direction of the Displacement Vector
You move a mouse pointer on the display monitor from its initial position at point (6.0 cm, 1.6 cm) to an icon located at point (2.0 cm, 4.5 cm). What is the magnitude and direction of the displacement vector of the pointer?
Strategy
In Example 2.4.1, we found the displacement vector D→ of the mouse pointer (see Equation 2.4.7). We identify its scalar components D x = −4.0 cm and D y = + 2.9 cm and substitute into Equation 2.4.8 and Equation 2.4.9 to find the magnitude D and direction θ D, respectively.
Solution
The magnitude of vector D→ is
(2.4.10)D=D x 2+D y 2=(−4.0 cm)2+(2.9 cm)2=(4.0)2+(2.9)2 cm=4.9 cm.
The direction angle is
(2.4.11)tanθ=D y D x=+2.9 cm−4.0 cm=−0.725⇒θ=tan−1(−0.725)=−35.9 o.
Vector D→ lies in the second quadrant, so its direction angle is
(2.4.12)θ D=θ+180 o=−35.9 o+180 o=144.1 o.
Exercise 2.5
If the displacement vector of a blue fly walking on a sheet of graph paper is D→=(−5.00 i^−3.00 j^) cm, find its magnitude and direction.
In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach is to find vector components when the direction and magnitude of a vector are known.
Let us return to the right triangle in Figure 2.4.3. The quotient of the adjacent side A x to the hypotenuse A is the cosine function of direction angle θ A, A x/A = cos θ A, and the quotient of the opposite side A y to the hypotenuse A is the sine function of θ A, A y/A = sin θ A. When magnitude A and direction θ A are known, we can solve these relations for the scalar components:
(2.4.13){A x=Acosθ A A y=Asinθ A.
When calculating vector components with Equation 2.4.13, care must be taken with the angle. The direction angle θ A of a vector is the angle measured counterclockwisefrom the positive direction on the x-axis to the vector. The clockwise measurement gives a negative angle.
Example 2.4.3: Components of Displacement Vectors
A rescue party for a missing child follows a search dog named Trooper. Trooper wanders a lot and makes many trial sniffs along many different paths. Trooper eventually finds the child and the story has a happy ending, but his displacements on various legs seem to be truly convoluted. On one of the legs he walks 200.0 m southeast, then he runs north some 300.0 m. On the third leg, he examines the scents carefully for 50.0 m in the direction 30° west of north. On the fourth leg, Trooper goes directly south for 80.0 m, picks up a fresh scent and turns 23° west of south for 150.0 m. Find the scalar components of Trooper’s displacement vectors and his displacement vectors in vector component form for each leg.
Strategy
Let’s adopt a rectangular coordinate system with the positive x-axis in the direction of geographic east, with the positive y-direction pointed to geographic north. Explicitly, the unit vector i^ of the x-axis points east and the unit vector j^ of the y-axis points north. Trooper makes five legs, so there are five displacement vectors. We start by identifying their magnitudes and direction angles, then we use Equation 2.4.13 to find the scalar components of the displacements and Equation 2.4.3 for the displacement vectors.
Solution
On the first leg, the displacement magnitude is L 1 = 200.0 m and the direction is southeast. For direction angle θ 1 we can take either 45° measured clockwise from the east direction or 45° + 270° measured counterclockwise from the east direction. With the first choice, θ 1 = −45°. With the second choice, θ 1 = + 315°. We can use either one of these two angles. The components are
(2.4.14)L 1x=L 1cosθ 1=(200.0 m)cos315 o=141.4 m,
(2.4.15)L 1y=L 1sinθ 1=(200.0 m)sin315 o=−141.4 m,
The displacement vector of the first leg is
(2.4.16)L→1=L 1x i^+L 1y j^=(14.4 i^−141.4 j^)m.
On the second leg of Trooper’s wanderings, the magnitude of the displacement is L 2 = 300.0 m and the direction is north. The direction angle is θ 2 = + 90°. We obtain the following results:
(2.4.17)L 2x=L 2cosθ 2=(300.0 m)cos90 o=0.0,
(2.4.18)L 2y=L 2sinθ 2=(300.0 m)sin90 o=300.0 m,
(2.4.19)L→2=L 2x i^+L 2y j^=(300.0 m)j^.
On the third leg, the displacement magnitude is L 3 = 50.0 m and the direction is 30° west of north. The direction angle measured counterclockwise from the eastern direction is θ 3 = 30° + 90° = + 120°. This gives the following answers:
(2.4.20)L 3x=L 3cosθ 3=(50.0 m)cos120 o=−25.0 m,
(2.4.21)L 3y=L 3sinθ 3=(50.0 m)sin120 o=+43.3 m,
(2.4.22)L→3=L 3x i^+L 3y j^=(−25.0 i^+43.3 j^)m.
On the fourth leg of the excursion, the displacement magnitude is L 4 = 80.0 m and the direction is south. The direction angle can be taken as either θ 4 = −90° or (\theta_{4} = + 270°. We obtain
(2.4.23)L 4x=L 4cosθ 4=(80.0 m)cos(−90 o)=0,
(2.4.24)L 4y=L 4sinθ 4=(80.0 m)sin(−90 o)=−80.0 m,
(2.4.25)L→4=L 4x i^+L 4y j^=(−80.0 m)j^.
On the last leg, the magnitude is L 5 = 150.0 m and the angle is θ 5 = −23° + 270° = + 247° (23° west of south), which gives
(2.4.26)L 5x=L 5cosθ 5=(150.0 m)cos247 o=−58.6 m,
(2.4.27)L 5y=L 5sinθ 5=(150.0 m)sin247 o=−138.1 m,
(2.4.28)L→5=L 5x i^+L 5y j^=(−58.6 i^−138.1 j^)m.
Exercise 2.6
If Trooper runs 20 m west before taking a rest, what is his displacement vector?
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5359 | https://uomustansiriyah.edu.iq/media/lectures/4/4_2020_12_27!01_45_46_PM.pdf | ONE-COMPARTMENT OPEN MODEL: INTRAVENOUS BOLUS ADMINISTRATION: INTRODUCTION The one-compartment open model offers the simplest way to describe the process of drug distribution and elimination in the body. This model assumes that the drug can enter or leave the body (ie, the model is "open"), and the body acts like a single, uniform compartment.
The simplest route of drug administration from a modelling perspective is a rapid intravenous injection (IV bolus).
The simplest kinetic model that describes drug disposition in the body is to consider that the drug is injected all at once into a box, or compartment, and that the drug distributes instantaneously and homogenously throughout the compartment.
Drug elimination also occurs from the compartment immediately after injection.
Of course, this model is a simplistic view of drug disposition in the body, which in reality is infinitely more complex than a single compartment. In the body, when a drug is given in the form of an IV bolus, 1- the entire dose of drug enters the bloodstream immediately, and the drug absorption process is considered to be instantaneous. In most cases, 2- the drug distributes via the circulatory system to potentially all the tissues in the body.
Uptake of drugs by various tissue organs will occur at varying rates, depending on the 1-blood flow to the tissue, 2-the lipophilicity of the drug, 3- the molecular weight of the drug, and the 4-binding affinity of the drug for the tissue mass. Most drugs are eliminated from the body either through the kidney and/or by being metabolized in the liver.
nX5CQOoCSjLsP2rSRwAg&q=different+body+organs&oq=different+body+organs&gs_lcp=CgN pbWcQAzICCAAyBggAEAgQHjIGCAAQCBAeMgQIABAYOgUIABCxAzoECAAQHjoGCAAQChAYUB 9Yrj1gnE5oAnAAeACAAfsBiAG-FJIBBjAuMTguMpgBAKABAaoBC2d3cy13aXotaW1n&sclient=img&ved=0ahUKEwjQ2IqiheztAh UACWMBHVpaBIgQ4dUDCAc&uact=5#imgrc=uplAMEJL2sJjIM Because of rapid drug equilibration between the blood and tissue, drug elimination occurs as if the dose is all dissolved in a tank of uniform fluid (a single compartment) from which the drug is eliminated. The volume in which the drug is distributed is termed the apparent volume of distribution, VD. The apparent volume of distribution assumes that the drug is uniformly distributed in the body. The VD is determined from the preinjected amount of the dose in the syringe and the plasma drug concentration resulting immediately after the dose is injected.
1-The apparent volume of distribution is a parameter of the one-compartment model and governs the plasma concentration of the drug after a given dose. A second pharmacokinetic parameter is 2-the elimination rate constant, k, which governs the rate at which the drug concentration in the body declines over time. The one compartment model that describes the distribution and elimination after an IV bolus dose is given.
The one-compartment open model does not predict actual drug levels in the tissues. However, the model assumes that changes in the plasma levels of a drug will result in proportional changes in tissue drug levels, since their kinetic profile is consistent with inclusion within the vascular compartment and the various drug concentrations within the compartment are in equilibrium. The drug in the body, DB, cannot be measured directly; however, accessible body fluids (such as blood) can be sampled to determine drug concentrations.
ELIMINATION RATE CONSTANT The rate of elimination for most drugs from a tissue or from the body is a first-order process, in which the rate of elimination is dependent on the amount or concentration of drug present. The elimination rate constant, k, is a first-order elimination rate constant with units of time 1 (eg, hr-1 or 1/hr). Generally, the parent or active drug is measured in the vascular compartment. Total removal or elimination of the parent drug from this compartment is effected by metabolism (biotransformation) and excretion. The elimination rate constant represents the sum of each of these processes: where k m = first-order rate process of metabolism and k e = first-order rate process of excretion. There may be several routes of elimination of drug by metabolism or excretion. In such a case, each of these processes has its own first-order rate constant.
A rate expression for is This expression shows that the rate of elimination of drug in the body is a first-order process, depending on the overall elimination rate constant, k, and the amount of drug in the body, DB, remaining at any given time, t.
Integration of Equation 3.2 gives the following expression: where DB = drug in the body at time t and D B0 = drug in the body at t = 0. When log D B is plotted against t for this equation, a straight line is obtained. In practice, instead of transforming values of DB to their corresponding logarithms, each value of DB is placed at logarithmic intervals on semilog paper.
er&oq=semilog+pa&gs_lcp=CgNpbWcQARgAMgIIADICCAAyAggAMgIIADICCAAyAggAMgYIABAIEB4yBg AgQHjIGCAAQCBAeMgYIABAIEB46BwgjEOoCECc6BAgjECc6BQgAELEDUMkrWM5LYL5iaABwAHgAgAH2 BqQuSAQYwLjEwLjGYAQCgAQGqAQtnd3Mtd2l6LWltZ7ABCg&sclient=img#imgrc=t0H1REXCeT5_-M Equation 3.3 can also be expressed as APPARENT VOLUME OF DISTRIBUTION In general, drug equilibrates rapidly in the body. When plasma or any other biologic compartment is sampled and analyzed for drug content, the results are usually reported in units of concentration instead of amount. Each individual tissue in the body may contain a different concentration of drug due to differences in drug affinity for that tissue. Therefore, the amount of drug in a given location can be related to its concentration by a proportionality constant that reflects the volume of fluid the drug is dissolved in.
The volume of distribution represents a volume that must be considered in estimating the amount of drug in the body from the concentration of drug found in the sampling compartment. The volume of distribution is also the apparent volume (VD) in which the drug is dissolved (Eq. 3.5). Because the value of the volume of distribution does not have a true physiologic meaning in terms of an anatomic space, the term apparent volume of distribution is used.
The amount of drug in the body is not determined directly. Instead, a blood sample is removed at periodic intervals and analyzed for its concentration of drug. The VD relates the concentration of drug in plasma (Cp) and the amount of drug in the body (DB), as in the following equation: By substituting Equation 3.5 into Equation 3.3, a similar expression based on drug concentration in plasma is obtained for the first-order decline of drug plasma levels: where C p = concentration of drug in plasma at time t and C p0 = concentration of drug in plasma at t = 0.
Equation 3.6 can also be expressed as The relationship between apparent volume, drug concentration, and total amount of drug may be better understood by the following example.
Example Exactly 1 g of a drug is dissolved in an unknown volume of water. Upon assay, the concentration of this solution is 1 mg/mL. What is the original volume of this solution?
The original volume of the solution may be obtained by the following proportion, remembering that 1 g = 1000 mg: Therefore, the original volume was 1000 mL or 1 L.
If, in the above example, the volume of the solution is known to be 1 L, and the concentration of the solution is 1 mg/mL, then, to calculate the total amount of drug present, Therefore, the total amount of drug in the solution is 1000 mg, or 1 g.
From the preceding example, if the volume of solution in which the drug is dissolved and the drug concentration of the solution are known, then the total amount of drug present in the solution may be calculated. This relationship between drug concentration, volume in which the drug is dissolved, and total amount of drug present is given in the following equation: where D = total amount of drug, V = total volume, and C = drug concentration. From Equation 3.8, which is similar to Equation 3.5, if any two parameters are known, then the third term may be calculated The body may be considered as a constant-volume system or compartment.
Therefore, the apparent volume of distribution for any given drug is generally a constant. If both the concentration of drug in the plasma and the apparent volume of distribution for the drug are known, then the total amount of drug in the body (at the time in which the plasma sample was obtained) may be calculated from Equation 3.5.
Calculation of Volume of Distribution In a one-compartment model (IV administration), the VD is calculated with the following equation: When Cp0 is determined by extrapolation, it represents the instantaneous drug concentration (concentration of drug at t = 0) after drug equilibration in the body. The dose of drug given by IV bolus (rapid IV injection) represents the amount of drug in the body, DB0, at t = 0. Because both DB0 and C p0 are known at t = 0, then the apparent volume of distribution, V D, may be calculated from Equation 3.9.
From Equation 3.2 (repeated here), the rate of drug elimination is By substitution of Equation 3.5, DB = VD Cp, into Equation 3.2, the following expression is obtained: Rearrangement of Equation 3.10 gives As both k and V D are constants, Equation 3.10 may be integrated as follows: Equation 3.12 shows that a small change in time (dt) results in a small change in the amount of drug in the body, DB.
The integral represents the AUC ∞0, which is the summation of the area under the curve from t = 0 to t = ∞ . Thus, the apparent V D may also be calculated from knowledge of the dose, elimination rate constant, and the area under the curve (AUC) from t = 0 to t = ∞. The AUC ∞ 0 is usually estimated by the trapezoidal rule (see chapter 12). After integration, Equation 3.12 becomes Significance of the Apparent Volume of Distribution The apparent volume of distribution is not a true physiologic volume. Most drugs have an apparent volume of distribution smaller than, or equal to, the body mass. For some drugs, the volume of distribution may be several times the body mass.
Equation 3.9 shows that the apparent V D is dependent on C p0. For a given dose, a very small Cp0 may occur in the body due to concentration of the drug in peripheral tissues and organs. For this dose, the small Cp0 will result in a large V D.
Drugs with a large apparent VD are more concentrated in extravascular tissues and less concentrated intravascularly. If a drug is highly bound to plasma proteins or remains in the vascular region, then Cp0 will be higher, resulting in a smaller apparent VD.
Consequently, binding of a drug to peripheral tissues or to plasma proteins will significantly affect VD.
The apparent VD is a volume term that can be expressed as a simple volume or in terms of percent of body weight. In expressing the apparent VD in terms of percent body weight, a 1-L volume is assumed to be equal to the weight of 1 kg. For example, if the VD is 3500 mL for a subject weighing 70 kg, the VD expressed as percent of body weight is If If VD is a very large number ie, >100% of body weight then it may be assumed that the drug is concentrated in certain tissue compartments. Thus, the apparent VD is a useful parameter in considering the relative amounts of drug in the vascular and in the extravascular tissues.
Given the apparent VD for a particular drug, the total amount of drug in the body at any time after administration of the drug may be determined by the measurement of the drug concentration in the plasma (Eq. 3.5). Because the magnitude of the apparent VD is a useful indicator for the amount of drug outside the sampling compartment (usually the blood), the larger the apparent VD, the greater the amount of drug in the extravascular tissues. For each drug, the apparent VD is a constant. In certain pathologic cases, the apparent VD for the drug may be altered if the distribution of the drug is changed. For example, in edematous conditions, the total body water and total extracellular water increase; this is reflected in a larger apparent VD value for a drug that is highly water soluble. Similarly, changes in total body weight and lean body mass (which normally occur with age) may also affect the apparent VD.
CLEARANCE Clearance is a measure of drug elimination from the body without identifying the mechanism or process.
Drug Clearance in the One-Compartment Model The body is considered as a system of organs perfused by plasma and body fluids. Drug elimination from the body is an ongoing process due to both metabolism (biotransformation) and drug excretion through the kidney and other routes. The mechanisms of drug elimination are complex, but collectively drug elimination from the body may be quantitated using the concept of drug clearance. 1- Drug clearance refers to the volume of plasma fluid that is cleared of drug per unit time. 2-Clearance may also be considered as the fraction of drug removed per unit time multiplied by the VD.
DRUG ELIMINATION EXPRESSED AS AMOUNT PER TIME UNIT The expression of drug elimination from the body in terms of mass per unit time (eg, mg/min, or mg/hr) is simple, absolute, and unambiguous. For a zero-order elimination process, expressing the rate of drug elimination as mass per unit time is convenient because the rate is constant. In contrast, the rate of drug elimination for a first-order elimination process is not constant and changes with respect to the drug concentration in the body. For a first-order elimination, drug clearance expressed as volume per unit time (eg, L/hr or mL/min) is convenient because it is a constant.
DRUG ELIMINATION EXPRESSED AS VOLUME PER TIME UNIT The concept of expressing a rate in terms of volume per unit time is common in pharmacy. For example, a patient may be dosed at the rate of 2 teaspoonsful (10 mL) of a liquid medicine (10 mg/mL) daily, or alternatively, a dose (weight) of 100 mg of the drug daily.
Clearance is a concept that expresses "the rate of drug removal" in terms of volume of drug solution removed per unit time (at whatever drug concentration in the body prevailing at that time) . In contrast to a solution in a bottle, the drug concentration in the body will gradually decline by a first-order process such that the mass of drug removed over time is not constant. The plasma volume in the healthy state is relatively constant because water lost through the kidney is rapidly replaced with fluid absorbed from the gastrointestinal tract.
Since a constant volume of plasma (about 120 mL/min in humans) is filtered through the glomeruli of the kidneys, the rate of drug removal is dependent on the plasma drug concentration at all times.
This observation is based on a first-order process governing drug elimination. For many drugs, the rate of drug elimination is dependent on the plasma drug concentration, multiplied by a constant factor (dC/dt = kC).
When the plasma drug concentration is high, the rate of drug removal is high, and vice versa.
Clearance (volume of fluid removed of drug) for a first-order process is constant regardless of the drug concentration because clearance is expressed in volume per unit time rather than drug amount per unit time.
Mathematically, the rate of drug elimination is similar to Equation 3.10: where dD B/dt is the rate of drug elimination from the body (mg/hr), Cp is the plasma drug concentration (mg/L), k is a first-order rate constant (hr- 1 or 1/hr), and VD is the apparent volume of distribution (L). Cl is clearance and has the units L/hr in this example. In the example in , Cl is in mL/min.
Clearance, Cl, is expressed as volume/time. Equation 3.15 shows that clearance is a constant because VD and k are both constants. DB is the amount of drug in the body, and dD B/dt is the rate of change (of amount) of drug in the body with respect to time. The negative sign refers to the drug exiting from the body.
One-Compartment Model Equation in Terms of Cl and V D Equation 3.20 may be rewritten in terms of clearance and volume of distribution by substituting Cl/V D for k.
The clearance concept may also be applied a biologic system in physiologic modeling without the need of a theoretical compartment.
Clearance from Drug-Eliminating Tissues Clearance may be applied to any organ that is involved in drug elimination from the body. As long as first order elimination processes are involved, clearance represents the sum of the clearances for each drug eliminating organ as shown in Equation 3.26: where Cl R is renal clearance or drug clearance through the kidney, and Cl NR is nonrenal clearance through other organs. Generally, clearance is considered as the sum of renal, Cl R, and nonrenal drug clearance, Cl NR.
Cl NR is assumed to be due primarily to hepatic clearance (Cl H) in the absence of other significant drug clearances, such as elimination through the lung or the bile, as shown in Equation 3.27: Drug clearance considers that the drug in the body is uniformly dissolved in a volume of fluid (apparent volume of distribution, VD) from which drug concentrations can be measured easily.
Typically, plasma fluid concentration is measured and drug clearance is then calculated as the fixed volume of plasma fluid (containing the drug) cleared of drug per unit of time. The units for clearance are volume/time (eg, mL/min, L/hr).
Alternatively, Cl T may be defined as the rate of drug elimination divided by the plasma drug concentration.
Thus, clearance is expressed in terms of the volume of plasma containing drug that is eliminated per unit time. This clearance definition is equivalent to the previous definition and provides a practical way to calculate clearance based on plasma drug concentration data.
Therefore Cl T is a constant for a specific drug and represents the slope of the line obtained by plotting dDE/dt versus Cp, as shown in Equation 3.30.
For drugs that follow first-order elimination, the rate of drug elimination is dependent on the amount of drug remaining in the body.
Substituting the elimination rate in Equation 3.30 for kC pV D in Equation 3.31 and solving for Cl T gives Equation 3.32: Equation 3.32 shows that clearance, Cl T, is the product of V D and k, both of which are constant. This Equation 3.32 is similar to Equation 3.19 shown earlier. As the plasma drug concentration decreases during elimination, the rate of drug elimination, dD E/dt, will decrease accordingly, but clearance will remain constant. Clearance will be constant as long as the rate of drug elimination is a first-order process.
For some drugs, the elimination rate process is more complex and a non compartment method may be used to calculate certain pharmacokinetic parameters such as clearance. In this case, clearance can be determined directly from the plasma drug concentration-versus-time curve by |
5360 | https://www.who.int/publications/i/item/9789241548373 | Skip to main content
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Pocket book of hospital care for children
guidelines for the management of common childhood illnesses | Second edition
3 June 2013
| Publication
Download (3 MB)
Overview
This is the second edition of the World Health Organization (WHO) Pocket book of hospital care for children, which was first published in 2005. It is a compilation of the updated WHO guidelines for the management of common childhood illnesses at the first-referral level in low-resource countries. It presents relevant, up-to-date, evidence-based clinical guidelines that can be used by clinicians in their daily work in hospitals with basic laboratory facilities and inexpensive medicines. The guidelines focus on inpatient management of children who are severely ill with conditions that are major causes of childhood mortality, such as neonatal illness, pneumonia, diarrhoea, fever (mainly malaria, meningitis and septicaemia), severe acute malnutrition and HIV/AIDS. It also includes guidance on common surgical problems, appropriate supportive care and monitoring of patients on the ward.
The Pocket book is part of a series of tools for improving the quality of care for severely ill children and is consistent with the Integrated Management of Childhood Illness (IMCI) guidelines for outpatient management of sick children. It is for use by doctors, senior nurses and other senior health workers who are responsible for the care of young children at the first referral level in developing countries.
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5361 | https://scispace.com/papers/the-rigidity-of-graphs-ii-4nzh197174 | (Open Access) The rigidity of graphs, II (1979) | L Asimow | 367 Citations
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The rigidity of graphs, II
Open Access Journal Article10.1016/0022-247X(79)90108-2
The rigidity of graphs, II
L Asimow,B Roth +1 moreUniversity of Wyoming
-01 Mar1979
-Journal of Mathematical Analysis and App...
-Vol. 68, Iss: 1, pp 171-190
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TL;DR: In this article, a graph G is viewed as a set 1,…,v together with a nonempty set E of two-element subsets of 1,..,v.
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About:This article is published in Journal of Mathematical Analysis and Applications. The article was published on 01 Mar 1979. and is currently open access.
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Stabilisation of infinitesimally rigid formations of multi-robot networks -------------------------------------------------------------------------
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Journal Article•10.1007/BF01534980
On graphs and rigidity of plane skeletal structures ---------------------------------------------------
G. LamanUniversity of Amsterdam
-01 Oct1970
-Journal of Engineering Mathematics
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TL;DR: In this paper, the combinatorial properties of rigid plane skeletal structures are investigated, and the properties are found to be adequately described by a class of graph-structured graphs.
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•Journal Article•10.1090/S0002-9947-1978-0511410-9
The rigidity of graphs ----------------------
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Almost all simply connected closed surfaces are rigid -----------------------------------------------------
Herman GluckUniversity of Pennsylvania
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The Many Facets of Graph Theory -------------------------------
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5362 | https://www.poison.org/articles/teething-gels-a-warning | Teething Gels: A Warning | Poison Control
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Poison Prevention for Infants| Medicines
Teething Gels: A Warning
The Bottom Line
Benzocaine is a local anesthetic (medicine that numbs skin and gums). Teething gels are among the over-the-counter preparations that contain benzocaine. However, even small amounts of benzocaine are dangerous for infants; it can prevent the bloodstream from carrying oxygen throughout the body.
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The Full Story
Teething gels and lotions with benzocaine should not be used on children under the age of two. Benzocaine is a medicine applied directly to children's gums to ease pain. Unfortunately, benzocaine also can affect red blood cells, lowering the amount of oxygen in the bloodstream. This rare condition, called methemoglobinemia, can be fatal.
The Food and Drug Administration (FDA) warns parents not to use benzocaine for children under two years old, even if there is no warning on the label. Using even the labeled amounts on infants can cause serious illness. Children with methemoglobinemia may develop blue or gray skin and lips, have seizures, have trouble breathing, or stop breathing altogether. Symptoms can occur within a few minutes or up to two hours later. This type of reaction can happen the first time the drug is used or after it's been used many times.
The American Academy of Pediatrics notes that these gels and lotions may not even be useful. They mix with saliva and are swallowed within a short time. Instead, the pediatricians recommend hard rubber teething rings (but not frozen) and massaging children's gums for comfort. If teething infants remain extremely uncomfortable, contacting the pediatrician is the best next step.
Benzocaine is also found in pain-relieving creams and lotions for adults. Following label instructions exactly is extremely important. Adults can also develop methemoglobinemia, and even die, if too much benzocaine is applied, if it is applied too often, or if the treated area is covered up too tightly.
Prevent poisonings from benzocaine teething preparations:
Do not use teething preparations containing benzocaine on children younger than two years old.
Be sure to follow label instructions for children age two and older and for adults.
ALWAYS store medicines with benzocaine (and all medicines) locked up, out of sight and reach of children.
If you mistakenly use benzocaine on an infant, or if someone may have swallowed a medicine containing benzocaine, use thewebPOISONCONTROL®online tool for guidance or call the poison specialists immediately at 1-800-222-1222. webPOISONCONTROLand the poison specialists will help you figure out if you should go to the emergency room or if the situation can be managed safely at home.
Rose Ann Gould Soloway, RN, BSN, MSEd, DABAT emerita Clinical Toxicologist
Poisoned?
Call 1-800-222-1222 or
HELP ME online
Prevention Tips
Do not use teething preparations containing benzocaine on children younger than two years old.
Be sure to follow label instructions for children age two and older and for adults.
ALWAYS store medicines with benzocaine (and all medicines) locked up, out of sight and reach of children.
This Really Happened
A mother treated her child's tooth pain with an over-the-counter remedy containing benzocaine. The next morning, the child swallowed a small amount of this medicine. Within 2 hours, his lips and nail beds had turned blue and his skin was "dusky". In the emergency room, his blood was found to be dark brown. Nearly half of his hemoglobin, which carries oxygen in the blood, had turned to methemoglobin, which cannot carry oxygen. The child was treated with an antidote and admitted to the hospital. He had recovered by the following morning.
Reference:Darracq MA, Daubert GP. A cyanotic toddler. Pediatric Emergency Care. 2007; 23: 195-199.
An 18-month-old girl played with a tube of Baby Orajel®(benzocaine 7.5%). She had the product all over her hands and her mom smelled it on her breath. Poison Control advised her mom to wash her skin, rinse her mouth and watch her closely over the next hour for drowsiness and/or pale or dusky skin color. The child was fine according to her mom during a follow-up call from Poison Control about an hour later.
Share this:
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For More Information
American Academy of Pediatrics. Ages & Stages. Teething: 4 to 7 Months.Accessed March 12, 2014.
References
U.S. Food and Drug Administration. FDA Drug Safety Communication: Reports of a rare, but serious and potentially fatal adverse effect with the use of over-the-counter (OTC) benzocaine gels and liquids applied to the gums or mouth. Accessed Mar 12, 2014.
Poisoned?
Call 1-800-222-1222 or
HELP ME online
Prevention Tips
Do not use teething preparations containing benzocaine on children younger than two years old.
Be sure to follow label instructions for children age two and older and for adults.
ALWAYS store medicines with benzocaine (and all medicines) locked up, out of sight and reach of children.
This Really Happened
A mother treated her child's tooth pain with an over-the-counter remedy containing benzocaine. The next morning, the child swallowed a small amount of this medicine. Within 2 hours, his lips and nail beds had turned blue and his skin was "dusky". In the emergency room, his blood was found to be dark brown. Nearly half of his hemoglobin, which carries oxygen in the blood, had turned to methemoglobin, which cannot carry oxygen. The child was treated with an antidote and admitted to the hospital. He had recovered by the following morning.
Reference:Darracq MA, Daubert GP. A cyanotic toddler. Pediatric Emergency Care. 2007; 23: 195-199.
An 18-month-old girl played with a tube of Baby Orajel®(benzocaine 7.5%). She had the product all over her hands and her mom smelled it on her breath. Poison Control advised her mom to wash her skin, rinse her mouth and watch her closely over the next hour for drowsiness and/or pale or dusky skin color. The child was fine according to her mom during a follow-up call from Poison Control about an hour later.
Items of interest:
Poison Statistics
Jingles
Spike Program
© 2012 - 2025 NCPC Terms of Use (Websites)Terms of Use (Triage)About UsPrivacy PolicyContact UsFeedbackMedical Toxicology Fellowship |
5363 | https://math.stackexchange.com/questions/4528125/why-does-ab-imply-a-b-2-0 | real analysis - Why does $a>b$ imply $(a-b)/2 > 0$ - Mathematics Stack Exchange
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Why does a>b imply (a−b)/2>0
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Sorry if this is an easy explanation but I'm having trouble understanding why a>b implies that (a−b)/2>0. I'm trying to understand this theorem and its proof:
a0 implies a≤b
I understand that the best way to go about it is to prove by contradiction. So you would assume a>b. From the other proofs I've seen it says that a>b implies (a−b)/2>0.
I tried reorganizing it, but don't understand where the divide by 2 comes from.
Sorry if this has been asked before or if I'm missing something simple.
real-analysis
inequality
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edited Sep 9, 2022 at 16:49
Aaron Hendrickson
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asked Sep 9, 2022 at 16:43
reclaimer2reclaimer2
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9 a>b implies a−b>0, implies 1 2(a−b)>0.Dietrich Burde –Dietrich Burde 2022-09-09 16:44:36 +00:00 Commented Sep 9, 2022 at 16:44
1 Because if a>b, then (a−b) is positive.Mauro ALLEGRANZA –Mauro ALLEGRANZA 2022-09-09 16:45:14 +00:00 Commented Sep 9, 2022 at 16:45
3 Other than MathJax this question meets quality standards for the site. Try not to confuse "low-quality" with "easy". (+1)Aaron Hendrickson –Aaron Hendrickson 2022-09-09 16:48:07 +00:00 Commented Sep 9, 2022 at 16:48
2 It can be any other r>0, but look at your title. You didn't ask for (a−b)/5>0, but rather for (a−b)/2>0.Dietrich Burde –Dietrich Burde 2022-09-09 16:54:27 +00:00 Commented Sep 9, 2022 at 16:54
1 This may not be your particular confusion, but it's noteworthy that in mathematics, "implies" does not mean "naturally leads to." It simply means that if the first thing is true, the second thing is also true. I might also say that if n is a perfect square, then n>−7. There's no particular reason I would want to say that, necessarily, but it's still true.Brian Tung –Brian Tung 2022-09-09 17:07:39 +00:00 Commented Sep 9, 2022 at 17:07
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Let consider the following chain of implications
a>b⟹a−b>b−b⟹a−b>0⟹a−b 2>0 2⟹a−b 2>0
that is
a>b⟹a−b 2>0
Note also that in the same way we can prove that the reverse is also true a−b 2>0⟹a>b that is
a>b⟺a−b 2>0
Moreover, as noticed in the comments, for any n>0 we can show in the same way that
a>b⟺a−b n>0
The proof for the particular theorem you are referring to:
a0 implies a≤b
should be as follows:
let assume by contradiction a>b
then for any n>0 we have a−b n>0 and therefore exists ε>0 such that
0<ε n≤a−b n⟹a−b≥ε⟹a≥b+ε
As an alternative, we can proceed by a direct proof by exhaustion, that is:
a=b⟹a−b=0<ε⟹a−b<ε, ∀ε>0
a<b⟹a−b<0<ε⟹a−b<ε, ∀ε>0
a>b let assume ε=a−b 2>0 then a−b>ε
therefore
a0⟹a=b∨a<b
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edited Sep 9, 2022 at 17:54
answered Sep 9, 2022 at 17:04
useruser
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For that particular theorem I posted in my question, a0 implies a≤b, is there a particular reason as to why n is set to 2 for the proofs I've seen.reclaimer2 –reclaimer2 2022-09-09 17:12:07 +00:00 Commented Sep 9, 2022 at 17:12
@reclaimer2 I've indicated the reason to justify the implication and there is no reason to take n=2 since any n>0 works fine. For the particular theorem, maybe it would be useful add some more detail on the full proof.user –user 2022-09-09 17:19:31 +00:00 Commented Sep 9, 2022 at 17:19
Anyway, since the theorem refers to any ϵ>0 there is no reason to take n=2 or any other value since we can always set ε=ε n such that 0<ε n<a−b n.user –user 2022-09-09 17:26:05 +00:00 Commented Sep 9, 2022 at 17:26
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5364 | https://people.math.harvard.edu/~engelwar/MathS305/Sequences%20An%20Introduction.pdf | Mathematical Sequences (sourced from Wikipedia) In mathematics, informally speaking, a sequence is an ordered list of objects (or events). Like a set, it contains members (also called elements, or terms). The number of ordered elements (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence. Most precisely, a sequence can be defined as a function whose domain is a countable totally ordered set, such as the natural numbers. For example, (M, A, R, Y) is a sequence of letters with the letter 'M' first and 'Y' last. This sequence differs from (A, R, M, Y). Also, the sequence (1, 1, 2, 3, 5, 8), which contains the number 1 at two different positions, is a valid sequence. Sequences can be finite, as in this example, or infinite, such as the sequence of all even positive integers (2, 4, 6,...). Finite sequences are sometimes known as strings or words and infinite sequences as streams. The empty sequence ( ) is included in most notions of sequence, but may be excluded depending on the context. Examples and notation A sequence can be thought of as a list of elements with a particular order. Sequences are useful in a number of mathematical disciplines ... In particular, sequences are the basis for series, which are important in differential equations and analysis. Sequences are also of interest in their own right and can be studied as patterns or puzzles, such as in the study of prime numbers. There are a number of ways to denote a sequence, some of which are more useful for specific types of sequences. One way to specify a sequence is to list the elements. For example, the first four odd numbers form the sequence (1,3,5,7). This notation can be used for infinite sequences as well. For instance, the infinite sequence of positive odd integers can be written (1,3,5,7,...). Listing is most useful for infinite sequences with a pattern that can be easily discerned from the first few elements. Other ways to denote a sequence are discussed after the examples. Important examples A tiling with squares whose sides are successive Fibonacci numbers in length. There are many important integer sequences. The prime numbers are numbers that have no divisors but 1 and themselves. Taking these in their natural order gives the sequence (2,3,5,7,11,13,17,...). The study of prime numbers has important applications for mathematics and specifically number theory. The Fibonacci numbers are the integer sequence whose elements are the sum of the previous two elements. The first two elements are either 0 and 1 or 1 and 1 so that the sequence is (0,1,1,2,3,5,8,13,21,34,...). For a list of important examples of integers sequences see On-line Encyclopedia of Integer Sequences. Other important examples of sequences include ones made up of rational numbers, real numbers, and complex numbers. The sequence (.9,.99,.999,.9999,...) approaches the number 1. In fact, every real number can be written as the limit of a sequence of rational numbers. For instance, for a sequence (3,3.1,3.14,3.141,3.1415,...) the limit of a sequence can be written as π. It is this fact that allows us to write any real number as the limit of a sequence of decimals. The decimal for π, however, does not have any pattern like the one for the sequence (0.9,0.99,...). Indexing Other notations can be useful for sequences whose pattern cannot be easily guessed, or for sequences that do not have a pattern such as the digits of π. The terms of a sequence are commonly denoted by a single variable, say an, where the index n indicates the nth element of the sequence. Indexing notation is used to refer to a sequence in the abstract. It is also a natural notation for sequences whose elements are related to the index n (the element's position) in a simple way. For instance, the sequence of the first 10 square numbers could be written as This represents the sequence (1,4,9,...100). This notation is often simplified further as Here the subscript {k=1} and superscript 10 together tell us that the elements of this sequence are the ak such that k = 1, 2, ..., 10. Sequences can be indexed beginning and ending from any integer. The infinity symbol ∞ is often used as the superscript to indicate the sequence including all integer k-values starting with a certain one. The sequence of all positive squares is then denoted In cases where the set of indexing numbers is understood, such as in analysis, the subscripts and superscripts are often left off. That is, one simply writes ak for an arbitrary sequence. In analysis, k would be understood to run from 1 to ∞. However, sequences are often indexed starting from zero, as in In some cases the elements of the sequence are related naturally to a sequence of integers whose pattern can be easily inferred. In these cases the index set may be implied by a listing of the first few abstract elements. For instance, the sequence of squares of odd numbers could be denoted in any of the following ways. Moreover, the subscripts and superscripts could have been left off in the third, fourth, and fifth notations if the indexing set was understood to be the natural numbers. Finally, sequences can more generally be denoted by writing a set inclusion in the subscript, such as in |
5365 | https://wikishire.co.uk/wiki/Terminology_of_the_British_Isles | Terminology of the British Isles
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Various terms are used to describe the different (and sometimes overlapping) geographical and political areas of the islands of Great Britain, Ireland, and surrounding islands. The terminology is often a source of confusion, partly owing to the similarity between some of the actual words used, but also because they are often used loosely. The purpose of this article is to explain the meanings of and relationships among those terms.
In brief, the main terms and their simple explanations are as follows.
Geographical terms:
The British Isles is an archipelago consisting of the two large islands of Great Britain and Ireland, and many smaller surrounding islands.
Great Britain is the largest island of the archipelago.
Ireland is the second largest island of the archipelago and lies directly to the west of Great Britain.
The full list of islands in the British Isles includes over 1,000 islands, of which 51 have an area larger than 20 km².
Political terms
The United Kingdom of Great Britain and Northern Ireland is the constitutional monarchy occupying the island of Great Britain, the small nearby islands (but not the Isle of Man or the Channel Islands), and the north-eastern part of the island of Ireland. Usually, it is shortened to United Kingdom, the UK, Great Britain or Britain. The abbreviation GB is generally used for the United Kingdom of Great Britain and Northern Ireland in international agreements e.g. Universal Postal Union and Road Traffic Convention, as well as in the ISO 3166 country codes (GB and GBR).
Ireland is the sovereign republic occupying the larger portion of the island of Ireland. However, to distinguish the state from the island it occupies the majority of, or to distinguish either of these from Northern Ireland, it is also called "the Republic of Ireland" or simply "the Republic". Periodically, its Irish-language name, Éire, will be used in an English-language context to distinguish it from "Northern Ireland", even though the word "Éire" directly translates as "Ireland".
England and Wales, Scotland, and Northern Ireland are legal jurisdictions within the United Kingdom.
Great Britain means the jurisdictions of England and Wales and Scotland considered as a unit.
British Islands consists of the United Kingdom, the Channel Islands and the Isle of Man. These are the states within the British Isles that have the British monarch as head of state.
Linguistic terms
The United Kingdom and (the Republic of) Ireland are referred to as countries or nations in formal documents while England, Wales, Scotland and (to a lesser extent) Northern Ireland can also be referred to as such. In everyday language the terms nation and country are used almost interchangeably.
British is an adjective pertaining to the United Kingdom; for example, a citizen of the UK is called a British citizen.
Wales is sometimes referred to as the 'Principality of Wales', or just 'the Principality', although this has no modern geographical or constitutional basis.
Northern Ireland can also be referred to, by those of a unionist persuasion, as the Province, in relation to its locality within the Province of Ulster.
Sport
The constituent parts of the United Kingdom often compete separately in international competition as nations (and can be described as "the home nations"). For example in association football, Scotland, Wales, Northern Ireland and England play as nations. An additional complication is that in some sports, such as rugby union, players from Northern Ireland and the Republic of Ireland play as one team, Ireland, in international competitions.
Rugby union players from both Ireland and Great Britain play for British and Irish Lions representing the four "Home Unions" of England, Ireland, Scotland and Wales.
Great Britain is often used to mean United Kingdom. Usually this is simply sloppy language, but it is sometimes used as an official shortening of United Kingdom of Great Britain and Northern Ireland. For example, at the Olympic Games, the team officially called "Great Britain" represents the political entity the United Kingdom, which includes Northern Ireland. The "Ireland" Olympic team represents the whole island of Ireland, a geographical entity. Athletes from Northern Ireland have the choice of participating in either the "Great Britain" team or the "Ireland" team.
In the majority of individual sports (e.g. tennis and athletics), at international level competitors are identified as GB if they are from Great Britain or Northern Ireland. A small number of sports (e.g. golf) identify participants as representing their constituent part of the UK. In the Commonwealth Games, England, Northern Ireland, Scotland and Wales each compete as separate nations, as do each of the three Crown Dependencies.
1 At a glance
2 Terminology in detail
3 Geographical distinctions
3.1 The British Isles
3.2 Great Britain
3.3 Scotland
3.4 England
3.5 Wales
3.6 Ireland
3.7 Isle of Man
3.8 Channel Islands
4 Political distinctions
4.1 The United Kingdom
4.2 Ireland
4.3 Ulster
4.4 British Islands
5 Historical aspects
5.1 Origins of terms
5.2 Romans
5.3 Mediæval period
5.4 Renaissance mapmakers
5.5 18th and 19th Centuries
5.6 Evolution of kingdoms and states
6 Adjectives
7 Further information
7.1 Isle of Man and Channel Islands
7.2 Celtic names
7.3 The terms for the British Isles in the Irish language
8 Slang
9 Europe
10 References
At a glance
Below is a visual reference guide to some of the main concepts and territories described in this article:
The British Isles
The British Islands
The Isle of Man
The Channel Islands (Jersey, Guernsey)
The Republic of Ireland
The United Kingdom
Northern Ireland
Great Britain
England
Scotland
Wales
Terminology in detail
Britain is a political and geographic term which can refer to the United Kingdom of Great Britain and Northern Ireland, or sometimes the island of Great Britain.
Great Britain is the largest island in Europe encompassing:
: : England and Wales is a political and administrative term referring to England and Wales, which share the same legal system. Between 1746 and 1967 the term "England" did legally include Wales.
: England (see also the historical Kingdom of England).
Wales (see also the historical Principality of Wales).
Scotland (see also the historical Kingdom of Scotland)
: The historical Kingdom of Great Britain is Britain for the period 1707–1801.
Britannia is the Roman province of Britain, or a poetic reference to later Britain, or a personification of Britain.
The United Kingdom of Great Britain and Northern Ireland, usually shortened to the United Kingdom (abbreviation UK), is Great Britain plus Northern Ireland since 1927. (The Partition of Ireland took place in 1922, but the consequent change in the official title of the UK was only made by Act of Parliament five years later.) The United Kingdom of Great Britain and Northern Ireland is often shortened to Britain, even on official websites.
: The historical United Kingdom of Great Britain and Ireland was Great Britain plus Ireland, for the period 1801 to 1922, although the name change after the secession/independence of most of Ireland only took place in 1927.
: N.B.: While "United Kingdom" is normally abbreviated UK, the official ISO 3166 two-letter country code is GB and the three letter code is GBR (Ukraine has the two letter code UA and the three letter code UKR). Due to a pre-existing convention originating in the UK's JANET academic computer network, the UK's Internet top-level domain is .uk, a break from the practice of following ISO 3166 (a .gb domain has also been used to a limited extent in the past but is now defunct). GB was also used on car number plates to indicate the United Kingdom between 1910 and 2021.
Ireland (in Irish, Éire) refers, geographically, to the island of Ireland, or to any of the following:
: Historically:
The Kingdom of Ireland was Ireland for the period 1541-1801. (The King of Ireland remained Head of State in the Irish Free State and Ireland/Éire until the Republic of Ireland Act 1948 abolished that status).
The Irish Republic, established by the Irish Declaration of Independence, was a 32-county republic encompassing the entire island, during the period 1919–22—though its de facto rule did not encompass all of the island. During this period, according to British law, Ireland remained part of the UK though its independence was recognised by Russia.
Southern Ireland was a proposed Home Rule 26-county state under the Government of Ireland Act 1920. It never came into practical existence, being superseded by:
The Irish Free State is Ireland excepting Northern Ireland during the period 1922-37.
: The terms Irish Republic, Southern Ireland, the Irish Free State and Éire (in English-language texts) have been used synonymously with the (Republic of) Ireland, although their use is inappropriate and potentially offensive if used today.
: However it has to be noted that as of 2010 on the BBC News website, the Republic of Ireland is still widely and almost exclusively referred to as the "Irish Republic".
: Present:
Ireland (in Irish, Éire) is the political entity consisting of the island of Ireland excepting Northern Ireland, 1937-present. This is the name of the state according to the Irish Constitution.
The Republic of Ireland a legal "description" of Ireland excepting Northern Ireland, 1949-present. This form is used where tact or disambiguity demands. It is also the name used by the international Association Football team.
Northern Ireland 1922-present. That part of the island of Ireland north of the line of partition of 1922, and which is still part of the United Kingdom. It is sometimes referred to as "the North of Ireland", "the Six Counties" or (in extremist usage) the "occupied six counties", especially by Irish Nationalists.
Ulster The name of one of Ireland's four traditional provinces. The area contains the nine northernmost counties, six of which make up Northern Ireland, and three of which are part of the Republic of Ireland. It is also often used to refer to the smaller Northern Ireland. Though Ulster has not been a political entity since the ancient Gaelic provincial Kingdoms, it remains associated with a geographical area and is used in sporting and cultural contexts.
: In sport
In Gaelic games, no distinction is recognised between the counties of the Republic and those of Northern Ireland. County teams play in their provincial championships (where the six counties of Ulster within Northern Ireland and three within the Republic all play in the Ulster championship) and the winners of these play in the All-Ireland championship (which has been recently complicated by the introduction of a back-door system). Even within Northern Ireland, a tricolour, the flag of the Republic of Ireland, is flown at all games. At bigger games, where an anthem is played, it is always the national anthem of the Republic. In the case of the International Rules series against Australia, an Irish national team is chosen from all thirty-two counties.
In Association Football, the teams correspond to political entities: Northern Ireland and the Republic of Ireland. In accordance with UEFA and FIFA's rules, each of these countries has its own football league: the Irish League and the League of Ireland respectively.
In rugby union, rugby league, field hockey , cricket, boxing, golf, athletics and others the Ireland team is drawn from the whole island (ie. both the Republic and Northern Ireland). Many sports organisations are subdivided along provincial lines e.g. Gaelic Athletic Association, golf.
The British Isles is a term used to mean the island of Great Britain plus the island of Ireland and many smaller surrounding islands, including the Isle of Man and, in some contexts, the Channel Islands (Guernsey and Jersey).
Great Britain and Ireland, or variants like "Britain and Ireland" or "The UK and Ireland" are sometimes used as alternatives to the term British Isles.
Anglo-Celtic Isles is an alternative term (in limited use) for the geographic region comprising Britain & Ireland, more commonly referred to as the 'British Isles'. 'Anglo-Celtic Islands' is a derivative of this. It is intended as a geographic term free of any political implication and uses the macro-cultural grouping term Anglo-Celtic, referring to the peoples from which the majority of the island group's population are descended—the Anglo-Saxons and the Celts (it can also be inclusive of the Anglo-Normans).
Islands of the North Atlantic is another suggested replacement term for 'British Isles', without the same political connotations. However, its convolution and impracticality due to implying inclusion of fellow North Atlantic islands such as Iceland have made it unworkable and it has not come into common use. The term was used as part of the Strand 3 level of negotiations for the Belfast agreement. (Its acronym, IONA, is also the name of the small but historically important island of Iona off the coast of Scotland.)
British Islands (a legal term not in common usage) is the UK, the Isle of Man, and the Channel Islands.
Brittany, itself a corruption of 'Britain', and sometimes formerly known as 'Little Britain' is a historical Duchy in the West of France, now a French region.
Geographical distinctions
The British Isles
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| | The British Isles is an archipelago in the Atlantic Ocean off the coast of Continental Europe. It includes Ireland, Great Britain, and the Isle of Man. In a technical geographical sense, the British Isles does not include the Channel Islands, but in practice, and despite the fact that "British Isles" is a geographical term, the British-owned Channel Islands are often included because of their political associations with and relative proximity to Britain. Also included are the thousands of small islands off the coast of both the larger islands such as Shetland and Orkney. |
Great Britain
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| | Great Britain refers to the largest of the British Isles. In present the sense of the word "Great" means "larger" in contrast to Brittany in modern-day France. A historical term for a peninsula in France that largely corresponds to the modern French province is Lesser or Little Britain. That region was settled by many British immigrants during the period of Anglo-Saxon migration into Britain, and named "Little Britain" by them. The French term "Bretagne" now refers to the French "Little Britain", not to the British "Great Britain", which in French is called Grande-Bretagne. In classical times, the Roman geographer Ptolemy in his Almagest also called the larger island Megale Brettania (Great Britain). At that time, it was in contrast to the smaller island of Ireland, which he called Mikra Brettania (Little Britain). In this later work, Geography, Ptolemy refers to Great Britain as Albion and to Ireland as Iwernia. These "new" names were likely to have been the native names for the islands at the time. The earlier names, in contrast, were likely to have been coined before direct contact with local peoples were made. |
Scotland
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| | Scotland is the second-largest part of the UK, occupying the Northern part of the island of Great Britain, and includes hundreds of small islands off the coast, such as Arran, the Hebrides, Shetland and Orkney. Scotland borders England to the south and is bounded by the North Atlantic Ocean to the west and the North Sea to the East. |
England
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| | The name England specifically refers to the largest part of the UK. It occupies the south-eastern part of the island of Great Britain, and includes small islands off the coast, such as the Isle of Wight and the Isles of Scilly. Between the Wales and Berwick Act 1746 and the Welsh Language Act 1967, the term "England" legally included Wales. |
Wales
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| | Wales is the third-largest part of the UK, occupying part of the western area of the southern half of the island of Great Britain, as well as some smaller islands. Wales has a land border with England, and adjoins the Irish Sea to the north, St George's Channel to the west, and the Bristol Channel to the south. |
Ireland
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| | The second largest island in the archipelago is Ireland. Most of the island is in the Republic of Ireland. The north-east of the island (Northern Ireland) is a part of the United Kingdom. That Ireland is a part of the geographical "British Isles" does not mean that all of the island is politically "British", as the term predates the name "Great Britain", which is historically known as "Albion". There are also numerous smaller islands off the coast of Ireland. |
Isle of Man
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| | The Isle of Man lies between Great Britain and Ireland. It is governed as a British Crown dependency, having its own parliament, but with the United Kingdom responsible for its defence and external relations. |
Channel Islands
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| | Although the Channel Islands are associated with the United Kingdom politically as Crown dependencies, they are an outcrop of the nearby French mainland, and historically they are the last remaining parts of the former Duchy of Normandy. |
Political distinctions
The United Kingdom
The United Kingdom of Great Britain and Northern Ireland is the official full title of the state. This name appears on official documentation such as British passports. For convenience, the name is usually shortened to United Kingdom, UK, Great Britain or Britain.
The United Kingdom is a sovereign state. Wales can also be described as a principality of the United Kingdom (Prince of Wales is a title usually given to the heir apparent to the British throne but has no political or other role in respect of Wales). Northern Ireland is sometimes described by United Kingdom citizens as a province of the United Kingdom, which derives from the province of Ulster, of which Northern Ireland is part. Northern Ireland also had, until 1972, a far greater degree of self-government than the other constituent parts of the UK.
Great Britain is both a geographical and a political entity. Geographically, it is one island, but politically it also contains the surrounding islands, notably Anglesey, the Isle of Wight, the Inner Hebrides, Outer Hebrides, Orkney Islands and Shetland Islands).
The abbreviation GB is sometimes officially used for the United Kingdom, for example in the Olympics, or as the vehicle registration plate country identification code for UK-registered cars between 1910 and 2021.
The internet code ".gb", although allocated to the UK, is virtually unused and UK domains use ".uk".
The four constituent parts of the UK are also known to some as Home Nations or the "Four Nations". The BBC refers to its regional broadcasting operation as Nations and Regions ("regions" referring to the Regions of England).
In sport, England, Wales, Scotland, Northern Ireland have their own separate national teams in certain sports - for example in association football. Sporting contests between them are known as "Home internationals" (an example is the British Home Championship in football).
The governing body for football in Northern Ireland is called the Irish Football Association, having been in existence since some forty years before partition. Its counterpart in the Republic (plus Derry City FC) is the Football Association of Ireland. The Northern Ireland national team retained the name "Ireland" for some fifty years after partition. Since around 1970 the two teams have been consistently referred to as "Northern Ireland" and "Republic of Ireland" respectively.
UK teams in the Olympics have competed under several different names - for instance in Beijing the athletes were presented at the Opening Ceremony under a banner which said simply Great Britain, rather than the full Great Britain and Northern Ireland. Olympic athletes from Northern Ireland may choose whether to represent the UK or Ireland.
Since the Good Friday Agreement, and the subsequent implementation legislation, sporting organisation (and several other organisations, e.g. tourism, Irish Gaelic and Ulster Scots language boards) on the island of Ireland has increasingly been cross-border.
Citizens of the UK are called British or Britons. The term Brits may also be used, sometimes pejoratively. Some older slang names for Britons are Tommy (for British soldiers), Anglo and Limey. Anglo properly refers only to England, but it is sometimes used as a broader reference as an element in compound adjectives: for example, "Anglo-French relations" may be used in newspaper articles when referring to relations between the political entities France and the United Kingdom. Anglo-Saxon may be used (particularly in Continental European languages) when referring to the whole English-speaking world, the Anglosphere, although ethnically very few of the world's one thousand million English-speakers are of Anglo-Saxon origin.
Ireland
Since the adoption of the Constitution of Ireland in 1937, Ireland has been the constitutional name of the state which covers approximately five-sixths of the island of Ireland. (Northern Ireland covers the remaining sixth of the island, in its north-east. Northern Ireland remains a constituent part of the United Kingdom.)
Since the Republic of Ireland Act 1948, the term "Republic of Ireland" is the term used as the additional description of the state. This term is useful in avoiding ambiguity between the name of the island and the name of the state. However, the term "Ireland" is always used in formal diplomatic contexts such as the European Union or the United Nations. The passport of the Republic of Ireland bears the name Éire - Ireland.
Before the introduction of the 1937 constitution and the new name, the Irish Free State occupied the same territory as the modern state of Ireland. The Irish Free State became an autonomous dominion of the British Empire in 1922 when it seceded from the United Kingdom through the Anglo-Irish Treaty. The King ceased to be its Head of State in 1936 and the state ceased to be a Dominion and left the Commonwealth in 1948.
Traditionally, the island of Ireland is divided into four provinces - Leinster, Connacht, Munster and Ulster, with each of the provinces further divided into counties. The Republic of Ireland takes up 83% of the island, and twenty-six of the thirty-two traditional counties of Ireland. Northern Ireland takes up the remaining area and six of the traditional nine counties of Ulster, although these counties no longer exist as official administrative units.
On the island of Ireland the naming of places often raises political issues. The usage of "Ireland" as the official name of the state causes offence to some Unionists in Northern Ireland, who believe it implies that the state still has a territorial claim to the whole island - the terminology of "Republic of Ireland" or "Éire" is much preferred by Northern Irish unionists when referring to that political state. Similarly, some Nationalists in Northern Ireland also prefer to reserve the usage of "Ireland" to refer to the whole island.
In Northern Ireland, Irishness is a highly contested identity, with fundamentally different perceptions between unionists who perceive themselves as being both British and Irish, and nationalists who consider both communities to be part of the Irish nation.
The Republic of Ireland is often referred to by the Nationalist and Republican communities by the term "the Twenty-six Counties", with the connotation that the state constituted as such forms only a portion of the ideal political unit of the Irish Republic, which would consist of all of the thirty-two counties into which the island is divided. Additionally, the term "the Six Counties" (in reference to Northern Ireland's six counties) is also used. Other Nationalist terms in use include: "the occupied six counties", but more popularly, "the North of Ireland" and, "the North", these are terms also used by the Irish national broadcaster RTÉ.
The Irish passport is available to Irish citizens and can also be applied for abroad through Irish Consular services and the local Irish Embassy. As per the Irish nationality law, any person born on the island of Ireland before 2005, or otherwise a first generation descendant of such a person, is allowed to apply for an Irish passport. As such, people born in Northern Ireland and their children may be Irish citizens and hold an Irish passport if they choose.
Ulster
The terminology and usage of the name Ulster in Irish and British culture varies. Many within the unionist community refer to Northern Ireland as Ulster – although this is officially wrong, as the Irish province of Ulster is a nine-county entity incorporating the three counties of Donegal, Cavan and Monaghan, which are in the Republic of Ireland, along with the six counties of Northern Ireland. The term Ulster (and "the Province") are sometimes preferred by Unionists, sometimes because it can suggest an origin of the polity of Northern Ireland that pre-dates 1922, referring back to the Act of Union 1800, the Glorious Revolution of 1689, the Plantation of Ulster in 1610, the ancient migrations between Ulster and Scotland, and even to biblical tradition. This use for the term Ulster by Unionists to mean Northern Ireland, is perceived as offensive by some members of the Nationalist community, as Ulster includes, but is not exclusive, to Northern Ireland. Certain local place names are also still disputed, Derry/Londonderry for example.
British Islands
Under the Interpretation Act 1978 of the United Kingdom, the legal term British Islands (as opposed to the geographical term British Isles) refers to the United Kingdom of Great Britain and Northern Ireland, together with the Crown dependencies: the Bailiwicks of Jersey and of Guernsey (which in turn includes the smaller islands of Alderney, Herm and Sark) in the Channel Islands; and the Isle of Man. On the front of passports issued to residents of the Crown dependencies, the words "United Kingdom of Great Britain and Northern Ireland" are replaced with "British Islands" followed by the name of the issuing state or island.
Historical aspects
Origins of terms
The earliest known names for the islands come from ancient Greek writings. Though the original texts have been lost, excerpts were quoted or paraphrased by later authors. Parts of the Massaliote Periplus, a merchants' handbook describing searoutes of the 6th century BC, were used in translation in the writings of Avienus around AD 400. Ireland was referred to as Ierne (Insula sacra, the sacred island, as the Greeks interpreted it) "inhabited by the race of Hiberni" (gens hiernorum), and Britain as insula Albionum, "island of the Albions". Several sources from around 150 BC to AD 70 include fragments of the travel writings of the ancient Greek Pytheas around 320 BC, which also used the terms "Albion and 'Ierne." and have been described as referring to the British Isles, including Ireland, as the Pretanic Islands. Greek writers used the term αι Βρεττανιαι, which has been translated as the Brittanic Isles, and the peoples of these islands of Prettanike were called the Πρεττανοι, Priteni or Pretani. These names derived from a "Celtic language" term which is likely to have reached Pytheas from the Gauls who may have used it as their term for the inhabitants of the islands.
The Romans called the inhabitants of Gaul (modern France) Galli or Celtae. The latter term came from the Greek name Κελτοι for a central European people, and 17th century antiquarians who found language connections developed the idea of a race of Celts inhabiting the area, but this term was not used by the Greeks or Romans for the inhabitants of Britain or Ireland, nor is there any record of the inhabitants of the British Isles referring to themselves as such. Nevertheless, Roman administration would later incorporate the province of Britannia into the praetorian prefecture of Gaul, in common with Hispania, which had Celtiberians. Armorica, where the Bretons would settle, was part of Gallia Celtica, so there were tertiary relations between the Britons and Gallic Celts at least. In addition, the Parisii of Gallia Celtica are thought to have founded Aldborough in Britain. Belgae and Silures also came from Gallic areas, although not strictly "Celtic", but from Gallia Belgica and Aquitainia.
Priteni is the source of the Welsh language term Prydain, Britain, and has the same source as the Goidelic term Cruithne. The latter referred to the early Brythonic-speaking inhabitants of the Scottish highlands and the north of Scotland, who are known as the Cruithne in Scottish Gaelic, and who the Romans called Picts or Caledonians.
Romans
Caesar's invasions of Britain brought descriptions of the peoples of what he called Britannia pars interior, "inland Britain", in 55 BC. Throughout Book 4 of his Geography, Strabo is consistent in spelling the island Britain (transliterated) as Prettanikee; he uses the terms Prettans or Brettans loosely to refer to the islands as a group - a common generalisation used by classical geographers. For example, in Geography 2.1.18, …οι νοτιώτατοι των Βρεττανών βορειότεροι τούτων εισίν ("…the most southern of the Brettans are further north than this"). He was writing around AD 10, although the earliest surviving copy of his work dates from the 6th century.
Pliny the Elder writing around AD 70 uses a Latin version of the same terminology in section 4.102 of his Naturalis Historia. He writes of Great Britain: Albion ipsi nomen fuit, cum Britanniae vocarentur omnes de quibus mox paulo dicemus. ("Albion was its own name, when all [the islands] were called the Britannias; I will speak of them in a moment"). In the following section, 4.103, Pliny enumerates the islands he considers to make up the Britannias, listing Great Britain, Ireland, and many smaller islands.
In his Geography written in the mid 2nd century and probably describing the position around AD 100, Ptolemy includes both Britain and Ireland – he calls it Hibernia – in the island group he calls Britannia. He entitles Book II, Chapter 1 of as Hibernia, Island of Britannia, and Chapter 2 as Albion, Island of Britannia.
The name Albion for Great Britain fell from favour, and the island was described in Greek as Πρεττανία or Βρεττανία, in Latin Britannia, an inhabitant as Βρεττανός, Britannus, with the adjective Βρεττανικός, Britannicus, equating to "British". With the Roman conquest of Britain the name Britannia was used for the province of Roman Britain. The Emperor Claudius was honoured with the agnomen Britannicus as if he were the conqueror, and coins were struck from AD 46 inscribed DE BRITAN, DE BRITANN, DE BRITANNI, or DE BRITANNIS. With the visit of Hadrian in AD 121 coins introduced a female figure with the label BRITANNIA as a personification or goddess of the place. These and later Roman coins introduced the seated figure of Britannia which would be reintroduced in the 17th century.
In the later years of Roman rule Britons who left Latin inscriptions, both at home and elsewhere in the Empire, often described themselves as Brittanus or Britto, and where describing their citizenship gave it as cives of a British tribe or of a patria (homeland) of Britannia, not Roma. From the 4th century, many Britons migrated from Roman Britain across the English Channel and founded Brittany.
Mediæval period
While Latin remained the language of learning, from the early mediæval period records begin to appear in native languages. The earliest indigenous source to use a collective term for the archipelago is the Life of Saint Columba, a hagiography recording the missionary activities of the sixth century Irish monk Saint Columba among the peoples of what is now Scotland. It was written in the late seventh century by Adomnán of Iona, an Irish monk living on the Inner Hebridean island. The collective term for the archipelago used within this work is Oceani Insulae meaning "Islands of the Ocean" (Book 2, 46 in the Sharpe edition = Book 2, 47 in Reeves edition), it is used sparingly and no Priteni-derived collective reference is made.
Another early native source to use a collective term is the Historia ecclesiastica gentis Anglorum of Bede written in the early eighth century. The collective term for the archipelago used within this work is insularum meaning "islands" (Book 1, 8) and it too is used sparingly. He stated that Britain "studies and confesses one and the same knowledge of the highest truth in the tongues of five nations, namely the Angles, the Britons, the Scots, the Picts, and the Latins", distinguishing between the Brythonic languages of the "ancient Britons" or Old Welsh speakers and other language groups.
Early Celtic, Saxon and Viking kingdoms such as Rheged, Strathclyde and Wessex amalgamated, leading to the formation of Scotland, England and Wales. In Norman Ireland, local lords gained considerable autonomy from the Lordship of Ireland until it became the Kingdom of Ireland under direct English rule.
Renaissance mapmakers
Abraham Ortelius makes clear his understanding that England, Scotland and Ireland were politically separate in 1570 by the full title of his map: "Angliae, Scotiae et Hiberniae, sive Britannicar. insularum descriptio" which translates as "A representation of England, Scotland and Ireland, or Britannica's islands." Additionally many maps from this period show Wales and Cornwall as separate nations, most notably those of Mercator.
Maps of the Mediæval, Renaissance and later periods often referred to Albion. This archaic term was originally used by Ptolemy and Pliny to mean the island of Great Britain. In later centuries its meaning changed to refer only to the area we now call Scotland (Albany, or Alba in Gaelic). Albion has survived as a poetic name for Britain but it is not in everyday use.
18th and 19th Centuries
Following the Acts of Union 1707, a fashion arose, particularly in Scotland, for referring to Scotland and England as North Britain and South Britain respectively. These terms gained in popularity during the nineteenth century. The most lasting example of this usage was in the name of the North British Railway, which became part of the London and North Eastern Railway in 1923, and in the name of the North British Hotel, built by the railway in Edinburgh in 1902, which retained the name until it reopened in 1991 as the Balmoral Hotel.
Evolution of kingdoms and states
The diagram on the right gives an indication of the further evolution of kingdoms and states. In 1603 the Scottish King James VI inherited the English throne as "James I of England". He styled himself as James I of Great Britain, although both states retained their sovereignty and independent parliaments, the Parliament of Scotland and the Parliament of England. (The term "Great Britain" itself reportedly dates from as early as 1474, and was in common usage from the mid-16th century onwards.)
The 1707 Act of Union united England and Scotland in the United Kingdom of Great Britain under the Parliament of Great Britain, then in 1800 Ireland was brought under British government control by the Act of Union 1800 creating the United Kingdom of Great Britain and Ireland. Irish unrest culminated in the Irish War of Independence and the 1922 separation of the Irish Free State, which later became a republic with the name Ireland. The majority northeast continued to be part of what became the United Kingdom of Great Britain and Northern Ireland.
British overseas territories such as Bermuda, Gibraltar, Hong Kong, the Falkland Islands, and the British Antarctic Territory have (or have had) various relationships with the UK. The Commonwealth of Nations, initially formalised in 1931 (the British Commonwealth until 1949), is an association of independent states roughly corresponding to the former British Empire. (This has no connection with the Commonwealth of England, a short-lived republic replacing the previous kingdoms during the English Interregnum (1649–1660).)
Adjectives
In the absence of a single adjective to refer to the United Kingdom, British is generally used to refer to the United Kingdom as a whole. However, in a specifically physical geographical sense, British is used to refer to the island of Great Britain. Members of the Unionist communities in Northern Ireland might describe themselves as British even though they are not on the island of Great Britain, as this reflects a political and cultural identity.
The cumbersome adjective Great British is very rarely used to refer to Great Britain, other than to contrive a pun on the word great, as in "Great British Food".
Irish, in a political sense, is used to refer to the Republic of Ireland. Northern Ireland, as a constituent part of the United Kingdom, would be included within the umbrella of the political term British, though many Unionists in Northern Ireland also consider themselves Irish. In order to be more specific, Northern Irish is therefore in common usage. Members of the Nationalist communities would not describe themselves as British and would only use the terms Irish, or specifically Northern Irish where needed.
The term Ulster can also be used as an adjective (e.g. "Royal Ulster Constabulary"), but this is more likely to be used by Unionists and has political connotations in the same fashion as its use as a proper noun (because only six of the traditional nine counties of Ulster, namely Antrim, Armagh, Down, Fermanagh, Londonderry and Tyrone, are included in Northern Ireland with the remaining three counties Cavan, Donegal and Monaghan forming part of the Republic). The term Ulsterman (or Ulsterwoman) is common and holds no such political connotation. Likewise, Irish Nationalists might describe, say, a lake in Northern Ireland as Irish. However, some Nationalists might attribute what they see as less attractive aspects of Northern Ireland to Britain or even to England.
Note that the geographical term Irish Sea thus far appears to have escaped political connotations, even though territorial control of the waters of the Irish Sea is divided between both the Republic of Ireland and the UK, and also includes a British Crown dependency, the Isle of Man—as yet there appears to be no controversy with the term’s usage to mirror that of "British Isles". The North Channel is found off the east coast of Northern Ireland and it stretches the length of roughly two-thirds of Northern Ireland's eastern coastline.
The "Northern" in "Northern Ireland" is not completely accurate. The most northerly point on the island, Malin Head, is in the Republic of Ireland — in County Donegal's Inishowen Peninsula.
Scottish, English and Welsh are self-explanatory but the term English is often incorrectly used to mean British as well. Americans and Europeans often use the term in this fashion.
Further information
Isle of Man and Channel Islands
The Isle of Man and the two bailiwicks of the Channel Islands are Crown dependencies; that is, non-sovereign nations, self-governing but whose sovereignty is held by the British Crown. They control their own internal affairs, but not their defence or foreign relations. They are not part of the United Kingdom nor part of the European Union.
The Isle of Man is part of the British Isles, situated in the Irish Sea between Great Britain and Ireland.
The Channel Islands consist politically of two self-governing bailiwicks: the Bailiwick of Guernsey and the Bailiwick of Jersey. They are the remnants of the Duchy of Normandy, which was once in personal union with the Kingdom of England. They are sometimes, despite their location next to mainland France, considered part of the British Isles. This usage is political rather than geographic.
The Isle of Man and the Channel Islands are British Islands in United Kingdom law.
Celtic names
There are five living Celtic languages in the region. Each have names for the islands and countries of the British Isles. They are divided into two branches:
Brythonic – which includes Welsh and Cornish
Goidelic – which includes Irish, Scottish Gaelic and Manx
Some of the above are:
| English | Cornwall | Wales | Ireland | Northern Ireland | Republic of Ireland | Scotland | Mann | England |
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| Cornish (Kernewek) | Kernow | Kembra | Iwerdhon | Iwerdhon Gledh | Repoblek Iwerdhon | Alban | Manow | Pow an Sawson |
| Welsh (Cymraeg) | Cernyw | Cymru | Iwerddon | Gogledd Iwerddon | Gweriniaeth Iwerddon | Yr Alban | Manaw | Lloegr |
| Irish (Gaeilge) | an Chorn | an Bhreatain Bheag | Éire | Tuaisceart Éireann | Poblacht na hÉireann | Albain | Manainn | Sasana |
| Scottish Gaelic (Gàidhlig) | a' Chòrn | a' Chuimrigh | Èirinn | Èirinn a Tuath | Poblachd na h-Èireann | Alba | Manainn | Sasainn |
| Manx (Gaelg) | yn Chorn | Bretyn Beg | Nerin | Nerin Twoaie | Pobblaght Nerin | Nalbin | Mannin | Sostyn |
Note: In Irish there are actually several terms for Northern Ireland: An Tuaisceart, meaning "the North", is usually used, but a more recent term for official use is Tuaisceart Éireann.
The English word Welsh is from a common Germanic root meaning "foreigner" (cognate with Wallonia and Wallachia, and also cognate with the word used in Mediaeval German to refer to the French and Italians). The English names Albion and Albany are related to Alba and used poetically for either England or Scotland, or the whole island of Great Britain. English Erin is a poetic name for Ireland derived from Éire (or rather, from its dative form Éirinn).
The terms for the British Isles in the Irish language
In Irish, the term Oileáin Bhriotanacha is attested as a version of the English term British Isles. In this sense, Briotanach refers to British people in the sense of the islands belonging to them. Another translation is Oileáin Bhreataineacha, which is used in a 1937 geography book translated into Irish from English. In this instance, Breataineach refers to the people of the island of Great Britain, again in the sense of the islands belonging to them. Neither of these two terms is often used in Irish.
Earlier dictionaries give Oileáin Iarthair Eorpa as the translation, literally meaning West European Isles. Today the most common term Éire agus an Bhreatain Mhór is used, meaning literally as Ireland and Great Britain, as provided by terminological dictionaries.
Slang
Blighty is a slang word for Britain derived from the Hindustani word bilāyatī ("foreign"). Depending on the user, it is meant either affectionately or archly. It was often used by British soldiers abroad in the First World War to refer to home.
Europe
The term "Europe" may be used in one of several different contexts by British and Irish people; either to refer to the whole of the European continent, to refer to only to Mainland Europe, sometimes called "continental Europe" or simply "the Continent" by some people in the archipelago—as in the apocryphal newspaper headline "Fog shrouds Channel, continent cut off."
Europe and the adjective European may also be used in reference to the European Union, particularly in a derogative context such as "The new regulations handed out by Europe".
↑ "Britain", Oxford English Dictionary: "More fully Great Britain. As a geographical and political term: (the main island and smaller offshore islands making up) England, Scotland, and Wales, sometimes with the Isle of Man"
↑ New Oxford American Dictionary: "Britain: an island that consists of England, Wales, and Scotland. The name is broadly synonymous with Great Britain, but the longer form is more usual for the political unit."
↑ "Britain", Oxford English Dictionary (Online Edition): "Britain: 1a - The proper name of the whole island containing England, Wales, and Scotland, with their dependencies; more fully called Great Britain; now also used for the British state or empire as a whole."
↑ The are no official definitions, but Scotland has over 790 offshore islands - see Haswell-Smith, Hamish (2004). The Scottish Islands. Edinburgh: Canongate. ISBN 1-84195-454-3. plus numerous freshwater islands so a complete list of the British Isles would probably have between 1,000 and 2,000 entries.
↑ Guardian Unlimited Style Guide, Guardian News and Media Limited, 2007
↑ 6.0 6.1 "Great Britain", New Oxford American Dictionary: "Great Britain: England, Wales, and Scotland considered as a unit. The name is also often used loosely to refer to the United Kingdom."
↑ "BBC SPORT | Athletics | Nine NI athletes in Irish squad". BBC News. 2007-06-14. Retrieved 2010-01-24.
↑ "the term 'Britain' is used informally to mean the United Kingdom of Great Britain and Northern Ireland" — quote from British Government website
↑ "UK Government's "Guide to Government"". Direct.gov.uk. Retrieved 2010-06-19.
↑ "Office for National Statistics". Statistics.gov.uk. Retrieved 2010-06-19.
↑ "Electronic Mail address changes" On the transition form big-endian to little-endian notation (Dept of Mechanical Engineering, The University of Leeds
↑ Philip Freeman, Ireland and the Classical World, University of Texas Press, 2001
↑ BBC Press Office. "BBC Nations & Regions".
↑ CAIN: Democratic Dialogue: With all due respect - pluralism and parity of esteem (Report No. 7) by Tom Hennessey and Robin Wilson, Democratic Dialogue (1997)
↑ "States of Guernsey passports". Guernsey Government Website.
↑ 16.0 16.1 16.2 16.3 Snyder, Christopher A. (2003). The Britons. Blackwell Publishing. ISBN 0-631-22260-X.
↑ "Entry for Albion a 1911 Encyclopedia". Historymedren.about.com. 2010-06-14. Retrieved 2010-06-19.
↑ 18.0 18.1 18.2 18.3 18.4 Foster (editor), R F; Donnchadh O Corrain, Professor of Irish History at University College Cork: (Chapter 1: Prehistoric and Early Christian Ireland) (1 November 2001). The Oxford History of Ireland. Oxford University Press. ISBN 0-19-280202-X.
↑ Encyclopedia of the Celts: Pretani
↑ "The earliest Celts in Europe". WalesPast. Retrieved 2010-06-19.
↑ Translation by Roseman, op.cit.
↑ Ptolemy's Geography
↑ Britannia on Roman Coins, Roman coins in Britain
↑ General survey of Lothian
↑ Royal Styles and Titles in England and Great Britain, heraldica.org
↑ "The majority of English people still behave as if 'English' and 'British' are synonymous", historian Norman Davies quoted in The English: Europe's lost tribe, BBC News Story, January 14, 1999
↑ For example on english-irishdictionary.com.
↑ Tír-Eóluíocht na h-Éireann ('The Geography of Ireland') by T. J. Dunne, translated by Toirdhealbhach Ó Raithbheartaigh, Government Publications Office, Dublin
> Tá Éire ar cheann de na h-oileáin a dtugar na h-Oileáin Bhreataineacha ortha agus atá ar an taobh Thiar-Thuaidh de'n Eóraip. Tá siad tuairim a's ar chúig mhíle oileán ar fad ann. (Oileánradh an t-ainm a bheirtear ar áit ar bith i n-a bhfuil a lán oileán agus iad i n-aice a chéile mar seo.) Éire agus an Bhreatain Mhór (Sasain, an Bhreatain Bheag, agus Alba) an dá oileán is mó de na h-Oileáin Bhreataineacha.
Ireland is one of the islands which are called the British Isles and which are on the North-Western side of Europe. It is thought that there are five thousand islands in total there. (Archipelago is the name which is borne by a place in which there are many islands next to each other like these.) Ireland and Great Britain (England, Wales, and Scotland) are the two largest islands of the British Isles.
29. ↑ See for example Google searches for Oileáin Bhriotanacha (less references to the British Virgin Isles) or Oileáin Bhreataineacha, which produces only mirrors of Wikipedia.
30. ↑ Patrick S. Dineen, Foclóir Gaeilge Béarla, Irish-English Dictionary, Dublin, 1927
31. ↑ focail.ie, "The British Isles", Foras na Gaeilge, 2006
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2009 AIME II Problems/Problem 15
Contents
[hide]
1 Problem
2 Solutions
2.1 Solution 1 (Quick Calculus)
2.2 Solution 2 (Projective)
2.3 Solution 3 (Calculus)
2.4 Solution 4
2.5 Solution 5 (Projective)
3 Video Solution
4 See Also
Problem
Let be a diameter of a circle with diameter 1. Let and be points on one of the semicircular arcs determined by such that is the midpoint of the semicircle and . Point lies on the other semicircular arc. Let be the length of the line segment whose endpoints are the intersections of diameter with chords and . The largest possible value of can be written in the form , where and are positive integers and is not divisible by the square of any prime. Find .
Solutions
Solution 1 (Quick Calculus)
Let and . Further more let and . Angle chasing reveals and . Additionally and by the Pythagorean Theorem.
By the Angle Bisector Formula,
As we compute and , and finally . Taking the derivative of with respect to , we arrive at Clearly the maximum occurs when . Plugging this back in, using the fact that and , we get
with
~always_correct
Solution 2 (Projective)
Since , point lies between and on the semicircular arc. We will first compute the length of . By the law of cosines, , so . Then , so .
Let and , and let , , . Note thatthat is,orHence , and we also know . Now AM-GM givesThis gives the quadratic inequality , which solves as![Image 71: [d \in \left(-\infty, 7-4\sqrt3\right] \cup \left7+4\sqrt3, \infty\right).]But , so the greatest possible value of is . The answer is .
~MSTang
Solution 3 (Calculus)
Let be the center of the circle. Define , , and let and intersect at points and , respectively. We will express the length of as a function of and maximize that function in the interval .
Let be the foot of the perpendicular from to . We compute as follows.
(a) By the Extended Law of Sines in triangle , we have
(b) Note that and . Since and are similar right triangles, we have , and hence,
(c) We have and , and hence by the Law of Sines,
(d) Multiplying (a), (b), and (c), we have
,
which is a function of (and the constant ). Differentiating this with respect to yields
,
and the numerator of this is
,
which vanishes when . Therefore, the length of is maximized when , where is the value in that satisfies .
Note that
,
so . We compute
,
so the maximum length of is , and the answer is .
Solution 4
Suppose and intersect at and , respectively, and let and . Since is the midpoint of arc , bisects , and we get To find , we note that and , so
Writing , we can substitute known values and multiply the equations to get
The value we wish to maximize is
By the AM-GM inequality, , so giving the answer of . Equality is achieved when subject to the condition , which occurs for and .
Solution 5 (Projective)
By Pythagoras in we get
Since cross ratios are preserved upon projecting, note that By definition of a cross ratio, this becomes Let such that We know that so the LHS becomes
In the RHS, we are given every value except for However, Ptolemy's Theorem on gives Substituting, we get where we use
Again using we have Then Since this is a function in we differentiate WRT to find its maximum. By quotient rule, it suffices to solve Substituting back yields so is the answer.
~Generic_Username
Video Solution
~MathProblemSolvingSkills.com
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5367 | https://arxiv.org/pdf/1307.0690 | 1
The density of states of graphene underneath a metal electrode and its correlation with the contact resistivity
Ryota Ifuku, Kosuke Nagashio, Tomonori Nishimura, and Akira Toriumi Department of Materials Engineering, The University of Tokyo, Tokyo, 113-8656 JAPAN Email: nagashio@material.t.u-tokyo.ac.jp
The density of states ( DOS ) of graphene underneath a metal is estimated through a quantum capacitance measurement of the metal/graphene/SiO 2/n +-Si contact structure fabricated by a resist-free metal deposition process. Graphene underneath Au maintains a linear DOS - energy relationship except near the Dirac point, whereas the DOS of graphene underneath Ni is broken and largely enhanced around the Dirac point, resulting in only a slight modulation of the Fermi energy. Moreover, the
DOS of graphene in the contact structure is correlated with the contact resistivity measured using devices fabricated by the resist-free process.
To make the best use of the extremely high carrier mo-bility of graphene in electric devices, contact resistivity (
c)should be seriously addressed 1,2 because it is necessary to be lowered by several orders of magnitude from the present status. 1 Generally, the current injection from metal to graphene is proportional to the transmission probability and the density of states ( DOS ) in the metal and graphene. 3,4 In terms of the transmission probability, an intrinsic problem exists for momentum matching because the Dirac cone at the K point of the Brillouin zone boundary is generally la-ger than the Fermi wavenumber for typical metals. 5 The number of empty states in graphene as the final state is much smaller than the number of occupied states in the metal. This DOS bottleneck is considered to be the limiting factor in the total performance of graphene devices. 6,7 This limitation results from the inapplicability of conventional doping techniques such as ion implantation 8 because the thermodynamically stable C-C bonding prevents the substitutional doping of B or N for C. 9 The DOS of graphene in the contact structure, i.e., the met-al/graphene/SiO 2 structure, could be different from the ideal linear relation because the electrical properties of graphene with monoatomic thickness are easily modulated by the environment. Therefore, determining the DOS of graphene in the contact structure provides information on the strength of the metal/graphene/SiO 2 interaction, which is key to set-ting guidelines to further reduce
c.Here, in the case of graphene grown on metals, the modulation of the linear dispersion is reported to depend strongly on the metal element used, according to both the-oretical calculations 10,11 and experimental results. 12,13 The modulation in the electron dispersion relation occurs on a chemisorption group (e.g., Ni, Co, and Pd) and not on an adsorption group (e.g., Au, Ag, and Pt). 10 Based on these reports, we selected Ni as a contact electrode to measure
c1because the DOS of graphene in the contact structure is expected to increase due to the strong
-d coupling. 14 However, in the graphene field effect transistors (FETs) fabricated using the conventional resist process, it has been shown that graphene underneath the Ni electrodes roughly maintains linear dispersion based on the transport meas-urements. 15 This discrepancy could be caused by the resist residue in the device fabrication process. Many researchers have reported that the resist residue remains on graphene. 16-18 The resist residue is a serious concern in light of the fact that activated carbon, whose hydrophobic surface attracts organic materials, is composed of graphene. 19 Indeed, in the case of the direct deposition of Ni on graphene, Ni(111) grew epitaxially, whereas Ni was amorphous or had a tiny crystalline structure on graphene after the conventional resist process. 15
In this report, the resist-free graphene/metal interac-tion is studied for two typical metals, Ni and Au, using the
DOS - energy relationship determined by the quantum ca-pacitance (CQ) measurement of the met-al/graphene/SiO 2/n + -Si stack system, as shown in Fig. 1(a). The CQ of graphene extracted from capacitance measure-ments provides direct information on the DOS of graphene because it is regarded as the energy cost of inducing carri-ers in graphene and is directly related to the DOS according to CQ =e 2DOS . 20,21 Although many other techniques, such as photoemission spectroscopy 12,13 and scanning tunneling spectroscopy 22 , have been used to extract the DOS of graphene, these techniques cannot be applied to the contact
SiO 2n+-Si M
CSub CSub CQVCh VTG CSub VTG
(a) (c) n+-Si MSiO 2
VTG VTG
(d) n+-Si
3 nm
~300 nm graphene
Wavelength (nm) 400 750 020 SiO 2 thickness (nm) 1.2 ×10 -2 -6.7 ×10 -3 0(b) Ni or Au Cu Au W prober SiO 2
C’
FIG 1 (a) Schematic of the metal/graphene/SiO 2/n +-Si device. (b) Color plot of the optical contrast for graphene on the thin SiO 2/Si substrate calculated as functions of wavelength and dSiO2 . The white line indicates the zero contrast. (c) Schematic of the capacitor with graphene (side view) and its equivalent circuit. (d) Schematic of the capacitor without graphene (side view) and its equivalent circuit.
2structure due to the lack of an accessible graphene surface. Moreover, in this report,
c is determined in the graphene FETs fabricated using the resist-free metal deposition tech-nique. Finally, the correlation between the DOS of graphene in the contact structure and
c is discussed.
There is a critical requirement for the SiO 2 thickness (d SiO2 ) in the metal/graphene/SiO 2/n + -Si capacitor in Fig. 1(a) in order to extract CQ. 23 Because CQ is introduced in series with the geometrical capacitance ( Cox ) in the equiva-lent circuit (1/ C = 1/ Cox + 1/ CQ), Cox should be at least comparable to CQ. Therefore, CQ has so far only been ex-tracted in top gate devices with a thin high-k oxide layer (<10 nm). 21,24-26 In this study, the thickness of the SiO 2 lay-er in metal/graphene/SiO 2/n + -Si structure is adjusted to be thinner than 5 nm. Although a thickness of 90 or 280 nm for SiO 2 is generally selected to efficiently visualize graphene, we have noticed that the optical contrast exists even for thin SiO 2 layers. Figure 1(b) shows the color plot of the contrast as functions of wavelength and d SiO2 calcu-lated using the Fresnel equation assuming a trilayer model of graphene, SiO 2, and Si. 27 Although the optical contrast (< 0.01) for graphene on thin SiO 2 is generally weaker than that on 90-nm SiO 2 (~0.1), a thinner d SiO2 provides better visibility for thicknesses less than ~10 nm. Meanwhile, the gate leakage current caused by direct tunneling increases exponentially with decreasing oxide thickness, especially for thicknesses less than ~2 nm for the SiO 2/Si substrate. Therefore, a SiO 2 layer with ~ 3-nm thickness is used in this study. Graphene was mechanically exfoliated from Kish graphite onto ~ 3-nm SiO 2/n + -Si substrates (0.01 cm). Graphene was detected after the contrast was optimized by adjusting the RGB values of the image, which revealed that the graphene contrast was indeed brighter than that of SiO 2,as suggested by the negative contrast in Fig. 1(b). The con-tours of graphene were clearly observed using an optical microscope in dark-field mode. 23 The monolayer graphene was confirmed by Raman spectroscopy. A motorized posi-tion-alignment system was used for the resist-free metal deposition, as shown in Fig. 2(a). The pre-patterned poly(methyl methacrylate) (PMMA) stencil mask, sup-ported by the Si substrate with a 200-m square window (Fig. 2(b)), was adjusted to the position of graphene on the SiO 2/n + -Si substrate under the optical microscope. 23 Next, this PMMA stencil mask was fixed to the SiO 2/n + -Si sub-strate with tape. The topgate contact metal was thermally evaporated under vacuum at a background pressure of 10 -4 Pa. A three-layered metal stack (composed of a Au or Ni; contact layer ~ 20 nm thick, a Cu protection layer ~ 200 nm thick, a Au; probing layer ~60 nm thick) was used to pre-vent mechanical damage to graphene by the direct probing to the metal contact, as shown in Figs. 1(a) and 1(c). Moreover, it is difficult to precisely deposit the metal along the contours of graphene. The devices without graphene in Fig. 1(d) were also fabricated using the same metal deposi-tion steps for comparison, as observed in Fig. 2(b). Alterna-tively, the PMMA resist (495 PMMA A11, MicroChem Corp.) was intentionally spin-coated onto graphene and removed by acetone to clarify the effect of the resist residue. After the adjustment of the mask position, the metals were deposited onto graphene. The capacitance measurements were performed at room temperature in ambient air at a frequency of 1 MHz to suppress the leakage current and to reduce the slow time-constant response. Although the re-sistance was concerned to be high near the Dirac point (DP) due to the very small DOS , it was confirmed by a simple
RC consideration that 1 MHz was low enough because the carrier density near the DP in graphene on SiO 2 is large enough (~10 11 cm -2 ) due to the charged impurities. 21,28 Moreover, backgate graphene FETs with a device structure appropriate for the transfer length method (TLM) were fabricated by the resist-free metal deposition tech-nique to determine
c. The long as-transferred graphene (~50 m) was selected for use in this study, as shown in Fig. 2(c). A SiN membrane mask patterned by a fo-cused-ion-beam apparatus, as shown in Fig. 2(c), was used for the I-V measurements instead of the PMMA mask that was used for the C-V measurements because the hardness of SiN enables a more complex pattern. The pad patterns were also included in this mask. Because the graphene channel region was exposed to air after the metal deposition unlike in the case of the process used with the C-V devices, all of the devices were annealed in vacuum at 300 °C for 1
100 μm
(b)
100 μm
Ni w. graphene Ni w.o graphene SiO 2
PMMA Si
SiN 100 m
10 mgraphene
(c)
1234Ni Si
xyz θstage Si
SiO 2/n +-Si
(a)
PMMA mask SiN mask or Graphene
FIG 2 (a) Schematic of the position ad-justment system for the PMMA mask and graphene. The position of the PMMA mask pattern can be adjusted relative to that of graphene by the xyz stage driven by the stepping motor. (b) (top) The PMMA stencil mask supported by the Si substrate with a 200-m square window. (bottom) Optical image of the C-V device with and without graphene. (c) (top) The SiN stencil mask supported by the Si substrate with a 200-m square window. The TLM pattern was fabricated by the FIB instrument. (bottom) Optical image of the I-V device with the TLM structure fabricated by the resist-free metal deposi-tion technique. The electrodes are always numbered sequentially from 1, which represents the shortest channel.
3hour and the electrical measurements were performed in vacuum at room temperature.
First, the effect of the resist residue on the met-al/graphene interaction is addressed. Figures 3(a) and 3(b) show the capacitance as a function of the topgate voltage (VTG ) for the resist-free and resist-processed Ni devices, respectively. The blue data were obtained from the device with graphene in Fig. 1(c), while the red data were obtained from the device without graphene in Fig. 1(d). The broad capacitance change observed in all of the data as a function of VTG is a result of the formation of the depletion layer in the n + -Si substrate. In addition, the dip attributed to the CQ
of graphene is clearly observed near VTG = 0 V in the re-sist-processed Ni device in Fig. 3(b), whereas there is no dip in the resist-free Ni device in Fig. 3(a). The CQ of graphene in contact with Ni was extracted. The equivalent circuits of two types of measured capacitors are modeled in Figs. 1(c) and 1(d), where CSub is the series capacitance of SiO 2 and n + -Si and VCh is the channel volt-age. To extract CQ, the Csub measured for the device with-out graphene (red data) was first subtracted from the total capacitance of the device with graphene (blue data) after careful determination of the ratio of the metal area on SiO 2to the area on graphene. Thus, the series capacitance (1/ C’ =1/ CQ+1/ CSub ) was obtained. Note that the parasitic capacitance is not shown in the equivalent circuits in Fig. 1 because the parasitic capacitance can be removed by this subtraction, which is the main advantage of this method. Finally, CSub was again subtracted from C’ to extract CQ. In this CQ analysis, the keys to obtain accurate data are the following two techniques. One is to increase the area of graphene compared with that of the metal; therefore, when the monolayer graphene was found on ~3 nm-SiO 2, the PMMA mask pattern was selected from the pre-fabricated PMMA masks with different pattern sizes to fit the graphene’s size. The other is to precisely measure the graphene area, which was achieved using the clear contours in dark-field mode. Figure 3(d) shows the CQ extracted for the Ni devices as a function of the Fermi energy ( EF). The right vertical axis indicates the DOS calculated by CQ = e 2DOS . EF is indeed the charging energy and is expressed as EF =eV Ch .When VTG ’ is defined as VTG ’ = VTG - VDP , VCh can be ex-pressed as VCh = VTG ’ - ܥ/’ܥ ௌ௨்ܸ݀ீ VTG ᇱ Ԣ. 26 VDP is the DP voltage that is determined by the minimum value in the dip. Thus, the experimentally estimated CQ can be com-pared with the theoretical CQ (= 2e 2EF /
(v Fħ)2), where vF is the Fermi velocity (1×10 8 cm/s) and ħ is Planck’s constant. In addition, the CQ estimated for the Y 2O 3 top-gate device in which graphene is sandwiched between the oxides is also added as a reference in this figure. 21 The resist-processed Ni device exhibits a slightly broken ambipolar EF modula-tion, while the DOS for the resist-free Ni device increases near the DP and the range of EF modulation is very limited. It should be noted that the gate voltage range is the same for both devices ( 1.5 V). This result suggests that the uti-lization of the organic resist considerably weakens the graphene/metal interaction and also explains the ambipolar behavior in the previous I-V measurement of graphene un-derneath the Ni electrode. 15 The intrinsic difference between the Ni and Au in the metal/graphene interaction is examined using devices fab-ricated by the resist-free process. Figure 3(c) shows the capacitance as a function of VTG for the resist-free Au de-vice. Compared with the dip for the Ni device in Fig. 3(b), the dip observed near VTG = ~0 V for the Au electrode is much larger. Therefore, the CQ extracted for the resist-free Au device in Fig. 3(d) exhibits a large ambipolar behavior similar to the Y 2O 3 topgate device, suggesting that Au has
012345678012345-0.3 -0.2 -0.1 00.1 0.2 0.3 Quantum capacitance ( Fcm -2 )DOS (10 13 eV -1 cm -2 )
Fermi energy / eV 0.6 0.7 0.8 0.9 1-1.5 0 1.5 Capacitance ( Fcm -2 )Topgate voltage (V) 0.5 0.55 0.6 0.65 0.7 -1.5 0 1.5 Capacitance ( Fcm -2 )Topgate voltage (V) (a) Ni w/o resist (b) Ni w resist (c) Au w/o resist Dip 0.75 0.8 0.85 0.9 -1.5 -1 -0.5 0 0.5 1 1.5 Capacitance ( Fcm -2 )Topgate voltage (V) Au Dip Ni w/o resist Au w/o resist Ni w resist (d) (e) Theory Y2 O3 topgate device Au Ni Strong Very weak Interaction EF modulation Weak w/o w w/o Resist Ni Ni FIG 3 Total capacitance as a function of VTG
for the (a) resist-free Ni, (b) resist-processed Ni and (c) resist-free Au devices. The blue data is obtained from the device with graphene in Fig. 1(c), while the red data is obtained from the device without graphene in Fig. 1(d). (d) The experimentally extracted CQ
of graphene with the theoretical line. Solid circles with green, red, and blue colors repre-sent the resist-processed Ni, resist-free Ni, and resist-free Au devices, respectively. The black circle represents the CQof graphene obtained from the Y 2O3topgate device as a reference. 21 The arrows indicate the highest
DOS for the three devices. (e) Summary of the metal/graphene interaction suggested from the CQmeasurements.
4little influence on the DOS in graphene. Using the re-sist-free metal deposition technique, it is evident that the interaction of graphene with Ni is much stronger than that with Au. The schematic in Fig. 3(e) summarizes the met-al/graphene interaction. Next, the electrical transport is discussed for the TLM devices fabricated by the resist-free metal deposition tech-nique. Figures 4(a), 4(b) and 4(c) show the two probe re-sistances as a function of the backgate voltage ( VBG ) for different channel distances in the resist-free Ni, the re-sist-processed Ni, and the resist-free Au devices, respec-tively. The pair of numbers indicates the electrode number shown in Fig. 2(c). When the effect of the resist residue on the metal/graphene interaction is addressed by comparing both Ni electrode cases, there are two clear differences: (i) the DP shift due to the overlapping of the charge-transfer region ( e-doping) and (ii) the asymmetry of the resistance due to p-n junction formation. 29 For (i), Figure 4(d) shows
VDP as a function of the channel distance for both Ni elec-trode devices. These data are fitted by the exponential curve VDP = - A exp(-L CT /B), where L CT is the charge trans-fer length and A and B are constants. The value extrapolat-ed to zero VDP indicates the value of 2× L CT . The L CT for the resist-processed Ni device was ~0.6 m, roughly consistent with the previous report in the resist-processed device. 30 The L CT of ~1.3 m for the resist-free Ni device was much longer. For (ii), based on the CQ measurement, the EF of graphene underneath the Ni electrode can be modulated in the resist-processed case, while the modulation is very lim-ited in the resist-free case. The p-n junction formed near the contact always exists for VBG < 0 V for the resist-free Ni device, which results in the additional resistance. On the other hand, the p-n junction can be released for large nega-tive VBG for the resist-processed Ni device, as shown in Figure A3 of the supplementary material. 23 Therefore, the origins of the large DP shift and the large asymmetry of the resistance are the combination of the large charge transfer and the limited EF modulation. For the estimation of
c, TLM was used for the re-sist-free Au device. 23 However, TLM is not valid for the Ni device because the channel resistance is not equivalent for all channels with different lengths due to the large DP shift. Therefore, four probe measurements were performed for the Ni devices. Figure 4(e) summarizes the values of
c
obtained at VBG = ±30 V in the present study. Although a considerable improvement in
c was expected by employ-ing the resist-free process for Ni as a result of the increase in the DOS due to the
-d coupling, no significant im-provement was observed. In contrast,
c for the resist-free Au device was the lowest of all the previously reported values. 2Finally, the correlation between the DOS of graphene and
c is addressed. The DOS and
c at exactly the same E F
cannot be compared in this study due to the different SiO 2thickness for the C-V and I-V devices. Because the relation between EF and the carrier density ( n) is expressed as
p= F FE v n , the EF at VBG = 30 V for the device with the 90-nm thick SiO 2 layer is ~0.3 eV, suggesting that the gate voltage ranges for the C-V and I-V devices are compa-rable. Therefore, the highest DOS at a high EF was selected as shown by the arrows in Fig. 3(d), whereas the
c ob-tained at VBG = 30 V was used. Figure 4(f) shows the line-ar relationship between
c and the inverse of the DOS . The reduction of the distance between graphene and the metal due to the lack of resist residue generally leads to an in-crease in the transmission probability. Because the contri-bution to
c from the DOS of graphene and the distance is
-20 -15 -10 -5 050246810 Dirac point voltage (V) Channel distance ( m)
2×LCT
w/o resist wresist
CT DP
LV = ‐Aexp ‐ B
05000 10000 15000 20000 25000 -30 -20 -10 010 20 30 Resistance ( )Gate voltage (V) 05000 10000 15000 -30 -20 -10 010 20 30 Resistance ( )Gate voltage (V) 05000 10000 15000 20000 -30 -20 -10 010 20 30 Resistance ( )Gate voltage (V)
12 23 34 12 23 34
(a) Ni w/o resist (b) Ni w resist (c) Au w/o resist
12 23 34
Backgate voltage (V) Backgate voltage (V) Backgate voltage (V) 0500 1000 1500 2000 0.25 0.3 0.35 0.4 Contact resistivity ( m) DOS -1 (10 -13 eVcm 2)Ni w/o resist Ni w resist Au w/o resist Au Ni
C = R CW
~1500 ~1000 TLM 4P [ m] ~50 NA NA NA Resist ww/o w/o @ V G=30V
(e) (f) (d)
FIG 4 Two probe resistances as a function of VBG for different channel lengths for the (a) resist-free Ni, (b) resist-processed Ni and (c) resist-free Au devices. The pair of numbers indicates the electrode number shown in Fig. 2(c). (d) VDP as a function of the channel distance for Ni electrode de-vices. LCT is the charge transfer length. (e) Summary of the
c values determined in the present study. (f) Relationship between
c and the inverse of the DOS , where the largest DOS , shown by the arrows in Fig. 3(d), was selected. 5linked, the dominant factor that reduces
c is not clear yet. As observed from Fig. 4(f), however, it is true that the low c for the resist-free Au device is due to both the reduction of the graphene/Au distance and the largest DOS achieved by the large EF modulation. It should be emphasized that the low
c achieved in the resist-free Au device is not suitable for the topgate de-vice because the DOS of graphene underneath the metal electrode cannot be modulated by the topgate. Therefore, we initially hoped for a large increase in the DOS based on the strong
-d coupling for Ni because this can be used even for the topgate device structure. However, a large in-crease was not observed, suggesting that the selection of the contact metal might not help in reducing
c. The use of graphitic contact formation could be one solution for re-ducing
c because the DOS of graphite is much larger than that of graphene and because better contact properties have been reported for carbon nanotubes. 31
In this work, using the resist-free metal deposition technique, the DOS of graphene in the contact structure was estimated from the CQ measurement. The utilization of or-ganic resist in the device fabrication process was confirmed to weaken the graphene/metal interaction. For the re-sist-free metal/graphene contacts, the DOS of graphene is maintained for Au, while it is largely modulated for Ni. Although the dominant factor to reduce
c cannot be eluci-dated at present, it is definitely important to engineer the
DOS of graphene to reduce
c.
Acknowledgements
We thank Drs. Nabatame & Narushima, NIMS, for the fabrication of the SiN membrane masks. We are grateful to Covalent Materials for kindly providing us with the Kish graphite. This work was partly supported by the JSPS through its “Funding Program for World-Leading Innova-tive R&D on Science and Technology (FIRST Program)”, by a Grant-in-Aid for Scienti fi c Research from the Ministry of Education, Culture, Sports, Science and Technology, and by the Semiconductor Technology Academic Research Center (STARC).
References
K. Nagashio, T. Nishimura, K. Kita, A. Toriumi, Appl. Phys. Lett. 97 , 143514 (2010). 2. F. Xia, V. Perebeinos, Y. -M. Lin, Y. Wu, Ph. Avouris, Nature Nanotech. 6, 179 (2011). 3. E. L. Wolf, Principles of Electron Tunneling Spectroscopy, Oxford Univ. Press, NY, (2012). 4. S. M. Sze, K. K. Ng, Physics of Semiconductor Devices, John Wiley & Sons, Hoboken, NJ, (2007). 5. J. Tersoff, Appl. Phys. Lett. 74 , 2122 (1999). 6. K. Nagashio, T. Nishimura, K. Kita, A. Toriumi, IEDM Tech. Dig. 565 (2009). 7. K. Nagashio, A. Toriumi, Jpn. J. Appl. Phys. 50 , 070108 (2011). 8. U. Bangert, A. Bleloch, M. H. Gass, A. Seepujak, J. van den Berg, Phys. Rev. B 81 , 245423 (2010). 9. K. Yuge, Phys. Rev. B 79 , 144109 (2009). 10. G. Giovannetti, P. A. Khomyakov, G. Brocks, V. M. Karpan, J. van den Brink, P. J. Kelly, Phys. Rev. Lett. 101 , 026803 (2008). 11. J. H. Parq, J. Yu, Y. -K. Kwon, G. Kim, Phys. Rev. B 82 ,193406 (2010). 12. C. Oshima, A. Nagashima, J. Phys. Condens. Matter. 9, 1 (1997). 13. A. Varykhalov, J. Sanchez-Barriga, A. M. Shikin, C. Biswas, E. Vescovo, A. Rybkin, D. Marchenko, O. Rader, Phys. Rev. Lett.
101 , 157601 (2008). 14. B. Hammer, J. K. Norskov, Nature 376 , 238 (1995). 15. T. Moriyama, K. Nagashio, T. Nishimura, A. Toriumi, J. Appl. Phys (under review). 16. M. Ishigami, J. H. Chen, W. G. Cullen, M. S. Fuhrer, E. D. Willians, Nano Lett. 7, 1643 (2007). 17. Z. Cheng, Q. Zhou, C. Wang, Q. Li, C. Wang, Y. Fang, Nano lett. 11 , 767 (2011). 18. A. M. Goossens, V. E. Calado, A. Barreiro, K. Watanabe, T. Taniguchi, L. M. K. Vandersypen, Appl. Phys. Lett. 100 , 073110 (2012). 19. M. Acharya, M. S. Strano, J. P. Mathews, S. J. L. Billinge, V. Petkov, S. Subramoney, H. C. Foley, Philos. Mag. B 79 , 1499 (1999). 20. T. Fang, A. Konar, H. Xing, D. Jena, Appl. Phys. Lett. 91 ,092109 (2007). 21. K. Nagashio, T. Nishimura, A. Toriumi, Appl. Phys. Lett. 102 ,173507 (2013). 22. G. Li, A. Luican, E. Y. Andrei, Phys. Rev. Lett. 102 , 176804 (2009). 23. See supplementary material at [URL will be inserted by AIP] for the device fabrication. 24. Z. Chen, J. Appenzeller, IEDM Tech. Dig. 509 (2009). 25. L. A. Ponomarenko, R. Yang, R. V. Grobachev, P. Blake, A. S. Mayorov, K. S. Novoselov, M. I. Katsnelson, A. K. Geim, Phys. Rev. Lett. 105 , 136801 (2010). 26. H. Xu, Z. Zhang, Z. Wang, S. Wang, X. Liang, L. -M. Peng, ACS NANO 5, 2340 (2011). 27. K. Nagashio, T. Nishimura, K. Kita, A. Toriumi, Appl. Phys. Express 2, 025003 (2009). 28. J. Martin, N. Akerman, G. Ulbricht, T. Lohmann, J. H. Smet, K. von Klitzing, A. Yacoby, Nature phys. 4, 144 (2008) 29. B. Huard, N. Stander, J. A. Sulpizio, D. Goldhaber-Gordon, Phys. Rev. B 78 , 121402(R) (2008). 30. T. Mueller, F. Xia, M. Freitag, J. Tsang, Ph. Avouris, Phys. Rev. B 79 , 245430 (2009). 31. Y. Chai, A. Hazeghi, K. Takei,; H. -Y. Chen, P. C. H. Chan, A. Javay, H. -S. P. Wong, IEEE-ED, 59 , 12 (2012). 1supplementary material
The density of states of graphene underneath a metal electrode and its correlation with the contact resistivity
Ryota Ifuku, Kosuke Nagashio, Tomonori Nishimura, and Akira Toriumi
Department of Materials Engineering, The University of Tokyo, Tokyo, 113-8656, JAPAN Corresponding author: nagashio@material.t.u-tokyo.ac.jp
Figure A1(a) shows the typical optical image of graphene on ~ 3-nm SiO 2 /n +-Si with the RGB ad-justment. Graphene can be detected after the contrast is optimized by adjusting RGB values of the im-age (graphene on the TV monitor is more clear). Graphene is brighter than SiO 2 because of the negative optical contrast value, as shown in Fig. 1(b) . As shown in Figure A1(b) , its contours are clearly ob-served in the dark-field mode. By combining these two methods, the monolayer graphene was detected and confirmed by the Raman spectroscopy, as shown in Fig. A1(c) . Although the D peak of graphene is blinded by the peaks from the Si substrate (1200 - 1500 cm -1 ), no D band is generally observed in graphene transferred from Kish graphite.
Fig. A1 (a) Typical optical image of graphene on ~ 3-nm SiO 2/n +-Si, (b) Dark field mode of (a). (c) Raman data of graphene on ~ 3-nm SiO 2 /n + -Si.
For the resist-free metal deposition, the pre-patterned PMMA stencil mask supported by the Si substrate with the 200-m square window was fabricated, as shown in Fig. 2(b) . Figure A2 shows the fabrication sequence of this PMMA mask. (a) PVA and PMMA were spin-coated on the Si wafer with 1-cm square size. (b) The patterns required for C-V measurement were drawn by the electron beam li-
1200 1600 2000 2400 2800 Intensity / a.u. Raman shift / cm -1
G2D
(c)
graphene BLG tape residue
(a) (b) graphene 20 μmDark field RGB adjustment 20 μmThin graphite
Si
2thography. (c) The Si wafer with PMMA mask was soaked in water to dissolve PVA. (d) The floating PMMA was caught by the Si wafer with a 200-m square window. Finally, it was dried on the oven.
Fig. A2 Fabrication sequence of the PMMA mask.
The larger asymmetry of the resistance was observed especially for resist-free Ni device, as shown in Fig. 4 . Figure A3 shows the schematic of the band diagram showing (i) the charge transfer region, (ii) p-n junction & (iii) carrier density modulation just below Ni for (a) the resist-processed Ni device and (b) the resist-free Ni device. The p-n junction formed near the contact always exists for VBG < 0 V for resist-free Ni device, which results in the additional resistance. Therefore, the larger asymmetry of the resistance was observed for resist-free Ni device.
Figure A3 Schematics of the band diagram and I-V curve for (a) the resist-processed Ni device and (b) the resist-free Ni device.
Si PVA PMMA Si PMMA Si PMMA Si PMMA DIW Si PVA PMMA 1 m0.1 m500 m 1 cm 5 mm e-(a) (b) (c) (d) (e) Thickness
Floating PMMA
Backgate voltage Resistance
Ni: w/o reist Ni: w resist (a) (b) (ii) p-n junction SDEF(iii) Very weak modulation VBG <0 VBG >0 VBG =0 Trace of Dirac point (i) Charge transfer region (iii) Modulation SDVBG <0 VBG >0 VBG =0 EF(i) Charge transfer region Trace of Dirac point
3
Figure A4 shows TLM analysis at different VBG for resist-free Au device. The fitted contact re-sistance (2×R C ) was found to be ~20-60 for different back gate biases. Because the width of the graphene channel is 1.4 m, the unit length contact resistance is ~ 14-42 m for different back gate biases. It should be noted that the data for the Dirac point (~0 V) is neglected since the absolute errors around the Dirac point are significantly large due to the large resistance.
Fig. A4 TLM analysis at different V BG for resist-free Au device.
0500 1000 1500 2000 2500 3000 3500 01234Resistance ( )
Channel length ( m) Au w/o resist -10V -20V -30V @V BG |
5368 | https://openstax.org/books/college-algebra-2e/pages/4-key-concepts | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
College Algebra 2e
Key Concepts
College Algebra 2eKey Concepts
Search for key terms or text.
Key Concepts
## 4.1 Linear Functions
Linear functions can be represented in words, function notation, tabular form, and graphical form. See Example 1.
An increasing linear function results in a graph that slants upward from left to right and has a positive slope. A decreasing linear function results in a graph that slants downward from left to right and has a negative slope. A constant linear function results in a graph that is a horizontal line. See Example 2.
Slope is a rate of change. The slope of a linear function can be calculated by dividing the difference between y-values by the difference in corresponding x-values of any two points on the line. See Example 3 and Example 4.
An equation for a linear function can be written from a graph. See Example 5.
The equation for a linear function can be written if the slope and initial value are known. See Example 6 and Example 7.
A linear function can be used to solve real-world problems given information in different forms. See Example 8, Example 9, and Example 10.
Linear functions can be graphed by plotting points or by using the y-intercept and slope. See Example 11 and Example 12.
Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See Example 13.
The equation for a linear function can be written by interpreting the graph. See Example 14.
The x-intercept is the point at which the graph of a linear function crosses the x-axis. See Example 15.
Horizontal lines are written in the form, See Example 16.
Vertical lines are written in the form, See Example 17.
Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See Example 18.
A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x- and y-values of the given point into the equation, and using the that results. Similarly, the point-slope form of an equation can also be used. See Example 19.
A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope. See Example 20 and Example 21.
## 4.2 Modeling with Linear Functions
We can use the same problem strategies that we would use for any type of function.
When modeling and solving a problem, identify the variables and look for key values, including the slope and y-intercept. See Example 1.
Draw a diagram, where appropriate. See Example 2 and Example 3.
Check for reasonableness of the answer.
Linear models may be built by identifying or calculating the slope and using the y-intercept.
The x-intercept may be found by setting which is setting the expression equal to 0.
The point of intersection of a system of linear equations is the point where the x- and y-values are the same. See Example 4.
A graph of the system may be used to identify the points where one line falls below (or above) the other line.
## 4.3 Fitting Linear Models to Data
Scatter plots show the relationship between two sets of data. See Example 1.
Scatter plots may represent linear or non-linear models.
The line of best fit may be estimated or calculated, using a calculator or statistical software. See Example 2.
Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be used to predict values outside the domain and range of the data. See Example 3.
The correlation coefficient, indicates the degree of linear relationship between data. See Example 4.
A regression line best fits the data. See Example 5.
The least squares regression line is found by minimizing the squares of the distances of points from a line passing through the data and may be used to make predictions regarding either of the variables. See Example 6.
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5370 | https://brainly.com/question/59022781 | [FREE] Use the $t$-formulae to prove that: \frac{\sin x - \cos x + 1}{\sin x + \cos x - 1} = \frac{1 + \tan - brainly.com
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Use the t-formulae to prove that:
sin x+cos x−1 sin x−cos x+1=1−tan(2 x)1+tan(2 x)
where x=(4 n+1)2 π,n∈Z.
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To prove the identity s i n x+c o s x−1 s i n x−c o s x+1=1−t a n(2 x)1+t a n(2 x), we'll use the t-formulae to express sin x and cos x in terms of tan(2 x).
Express sin x and cos x using t=tan(2 x):
sin x=1+t 2 2 t
cos x=1+t 2 1−t 2
Substitute these expressions into the left-hand side (LHS) of the given identity:
LHS: s i n x+c o s x−1 s i n x−c o s x+1
Numerator: sin x−cos x+1=1+t 2 2 t−1+t 2 1−t 2+1
Simplifying:
1+t 2 2 t+(1−t 2)+(1+t 2)=1+t 2 2 t+1−t 2+1+t 2=1+t 2 2 t+2
Denominator: sin x+cos x−1=1+t 2 2 t+1+t 2 1−t 2−1
Simplifying:
1+t 2 2 t+1−t 2−(1+t 2)=1+t 2 2 t+1−t 2−1−t 2=1+t 2 2 t−2 t 2
Form the LHS expression:
LHS: 1+t 2 2 t−2 t 21+t 2 2 t+2
Simplify to:
2 t−2 t 2 2 t+2=2 t(1−t)(t+1)2=1−t t+1
Compare with the right-hand side (RHS):
RHS: 1−t 1+t
Since both sides simplify to 1−t t+1, the identity holds true for the given conditions.
Therefore, the identity s i n x+c o s x−1 s i n x−c o s x+1=1−t a n(2 x)1+t a n(2 x) is proven.
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The given identity is proven by expressing sin x and cos x in terms of tan(2 x) using the t-formulae, substituting these into the LHS, simplifying both sides, and showing they are equal. As a result, the identity holds true under the specified conditions. Thus, it demonstrates the equivalence of the two expressions.
Explanation
To prove the identity s i n x+c o s x−1 s i n x−c o s x+1=1−t a n(2 x)1+t a n(2 x), we will use the t-formulae to express sin x and cos x in terms of tan(2 x).
Express sin x and cos x using t=tan(2 x):
sin x=1+t 2 2 t
cos x=1+t 2 1−t 2
Substitute these expressions into the left-hand side (LHS) of the given identity:
LHS: s i n x+c o s x−1 s i n x−c o s x+1
Numerator:
sin x−cos x+1=1+t 2 2 t−1+t 2 1−t 2+1
Simplifying:
1+t 2 2 t+(1−t 2)+(1+t 2)=1+t 2 2 t+1−t 2+1+t 2=1+t 2 2 t+2
Denominator:
sin x+cos x−1=1+t 2 2 t+1+t 2 1−t 2−1
Simplifying:
1+t 2 2 t+1−t 2−(1+t 2)=1+t 2 2 t+1−t 2−1−t 2=1+t 2 2 t−2 t 2
Form the LHS expression:
LHS: 1+t 2 2 t−2 t 21+t 2 2 t+2
Simplify to:
2 t−2 t 2 2 t+2=2 t(1−t)(t+1)2=1−t t+1
Compare with the right-hand side (RHS):
RHS: 1−t 1+t
Since both sides simplify to 1−t t+1, the identity holds true for the given conditions.
Therefore, the identity s i n x+c o s x−1 s i n x−c o s x+1=1−t a n(2 x)1+t a n(2 x) is proven.
Examples & Evidence
For example, if you take a specific angle where x=4 π, you can substitute this into both sides to verify the identity mathematically, ensuring both sides yield the same numerical result. This provides a practical exercise to solidify your understanding of how the identities work in specific cases.
The steps detailed in the proof follow established trigonometric identities and transformations, which are commonly taught in high school mathematics, reinforcing their accuracy and reliability.
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Community Answer Verify that the equation is an identity. 2 1−cosx +cos 2 2 x =1 Choose the correct answer below. A. 2 1−cosx +cos 2 2 x = 2 1−cosx + sinx 1−cosx = sinx −sinx+1−cosx =1 B. 2 1−cosx +cos 2 2 x = 2 1−cosx + 1+cosx sinx = 1+cosx −1−cosx+sinx =1 C. 2 1−cosx +cos 2 2 x = 2 1−cosx + 2 1+cosx = 2 1−cosx+1+cosx =1 D. 2 1−cosx +cos 2 2 x = 2 1−cosx + 1−tan 2 x 2tanx = 1−tan 2 x −1+tan 2 x+2tanx =1
Community Answer Show that tan(/2)=1−cos/sin Hint: Use the fact that sin^2 x=1−cos2x/2 ; cos^2 x= 1+cos2x/2
Community Answer 3. Prove the following identities: a) [tex]\sin x \cdot \tan x + \cos x = \frac{1}{\cos x}[/tex] b) [tex]\left(\tan x + \frac{\cos x}{\sin x}\right) \sin^2 x = \tan x[/tex] c) [tex]\frac{\sin^2 x}{\cos x(\sin x + \cos x)} + \frac{\sin x}{\sin x + \cos x} = \tan x[/tex] d) [tex]\left(\frac{1}{1 + \sin x} + \frac{1}{1 - \sin x}\right) \cos^2 x = 2[/tex] e) [tex]\frac{1 - \cos x}{\sin x} \cdot \frac{\sin x}{1 + \cos x} = 0[/tex] f) [tex]\frac{1 + 2 \sin x \cos x}{\sin x + \cos x} = \sin x + \cos x[/tex] g) [tex]\frac{1 - \cos \beta}{\sin \beta} = \frac{\sin \beta}{1 + \cos \beta}[/tex]
Community Answer Using addition and subtraction formulas prove the following identities 1. sin(π /2 -x)=sin ( π/2+ x) 2. tan x - tan y= sin(x-y)/ cos x cos y
Community Answer 5.0 2 Given: tan(x/2) = 1-cos(x)/sin(x) prove: tan(x/2)= sin(x)/1/ cos(x) complete the steps for proof?
Community Answer Given: tangent (startfraction x over 2 endfraction) = startfraction 1 minus cosine (x) over sine (x) endfraction prove: tangent (startfraction x over 2 endfraction) = startfraction sine (x) over 1 cosine (x) endfraction complete the steps of the proof.
Community Answer [tex][ \begin{tabular}{l|l} \textbf{Expression} & \textbf{Justification} \ \hline $\tan \left(\frac{x}{2}\right)=\frac{1-\cos (x)}{\sin (x)}$ & \text{Given} \ \hline $\tan \left(\frac{x}{2}\right)=\frac{1-\cos (x)}{\sin (x)} \cdot \frac{1+\cos (x)}{1+\cos (x)}$ & \text{Multiply by a form of 1} \ \hline $\tan \left(\frac{x}{2}\right)=\frac{1-\cos ^2(x)}{\sin (x)[1+\cos (x)]}$ & \text{Multiply numerators and denominators} \ \hline $\tan \left(\frac{x}{2}\right)=\frac{\sin ^2(x)}{\sin (x)[1+\cos (x)]}$ & \text{Simplify using } 1-\cos^2(x) = \sin^2(x) \ \hline $\tan \left(\frac{x}{2}\right)=\frac{\sin (x)}{1+\cos (x)}$ & \text{Cancel } \sin(x) \text{ from numerator and denominator} \ \end{tabular} ][/tex] Given: [tex]$\tan \left(\frac{x}{2}\right)=\frac{1-\cos (x)}{\sin (x)}$[/tex] Prove: [tex]$\tan \left(\frac{x}{2}\right)=\frac{\sin (x)}{1+\cos (x)}$[/tex] Complete the steps of the proof.
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
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Find the equation of the line specified. The line passes through the points (−2,3) and (−4,7). A. y=−2 x−1 B. y=−2 x+3 C. y=−4 x−1 D. y=−2 x+7
Patty, Quinlan, and Rashad want to be club officers. The teacher who directs the club will place their names in a hat and choose two without looking. The student whose name is chosen first will be president and the student whose name is chosen second will be vice president. Which choice represents the sample space, S, for this event? bi g c i rc S=PQR bi g c i rc S=PQR,PRQ,QPR,QRP,RPQ,RQP bi g c i rc S=PQ,PR,QR bi g c i rc S=PQ,QP,PR,RP,QR,RQ
Determine the equation of the line that passes through the given points: (2,6) and (4,16). A. y=−4 x+5 B. y=5 x−16 C. y=5 x−4 D. y=−5 x+4
Two students from a group of eight boys and 12 girls are sent to represent the school in a parade. If the students are chosen at random, what is the probability that the students chosen are not both girls? A. 190 12 B. 95 33 C. 95 62 D. 190 178
Write an equation of the line that passes through a pair of points: A. y=x+3 B. y=x−3 C. y=−x+2 D. y=−x−2
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5371 | https://ibzmesstechnik.de/wp-content/uploads/2022/05/Chiral-compound-analysis.pdf | olecules that make up living things tend to be chiral: they have the property of “hand-edness” and a preference for one kind of mirror-image isomer, or enantiomer. Molecules that are metabolized by living things are often chiral as well, with specific chiral forms preferred by organ-isms. In medicine, for example, certain drugs are more effective in a preferred enantiomeric form and may produce fewer side effects. In the food indus-t ry, one form of a food supplement may be per-ceived as sweeter or less bitter, or may be otherwise more nutritious. In agriculture, the preferred chiral form of an insecticide may be more effective at lower application rates or be more specific toward a targeted pest. As the issue of chirality becomes more appreciated in bioscience research, tools to measure it increase in importance in analytical laboratories.
Chiral compounds, also known as optical iso-mers, are a specific case of stereoisomers. They are distinguished by having enantiomeric counterparts with identical chemical properties in nonchiral en-vironments but differing in the way they rotate a plane of polarized light.a As might be expected, po-l a r i m e t ry, the measurement of optical isomers by means of their specific rotations, can be a powerful analytical tool for analyses of chiral compounds.
Furthermore, information gained by the use of po-larimeters and polarimetric (LC) detectors is key to the full characterization of these compounds.
The property of optical rotation is specific to a compound, and it is related to its concentration in s o l u t i o n .b Through the use of polarimetry, one can determine a concentration through a compound’s known specific rotation, or vice versa. Using a po-larimetric detector in HPLC, one can differentiate between enantiomers, or one may simply find it more appropriate for the detection of certain sam-ples found in nature.
Before discussing how polarimetry can be used in practical applications, it might be well to refresh our knowledge of polarimeters and polarimetric detectors.
When plane polarized light is passed through a solution containing an optically active compound, A p p l i c a t i o n N o t e 20 / NOVEMBER 2001 Chiral compound analyses and Faraday polarimetry Peter Rozea it will be rotated in a clockwise or counterclockwise direction. In effect, two pieces of information are obtained: a direction (which way it rotates) and a magnitude (how many degrees). This information is unique and important, and translates into practical applications that will be discussed later.
In a simple polarimeter, a solution containing a chiral compound is placed into a measuring cell of a given length, typically, 10 cm. A light beam of the proper wavelength is polarized (via a polarizing lens), passed through the solution cell to a second polarizing lens (typically called an analyzer), and then to an eyepiece. In a standard procedure, the cell is first filled with a blank, i.e., containing every-thing but the compound to be measured, and the analyzer lens is rotated to obtain a maximum bright-ness, establishing a zero point or base measuring condition. The cell is then filled with the optically active solution, the analzyer rotated to obtain an equivalent light intensity, and the degrees from the base condition read. In a case in which the solution concentration is known and the specific rotation is not, two measurements of solutions having differing concentrations are necessary to establish the fact of rotation clockwise or counterc l o c k w i s e .c In next-generation automated polarimeters, a photocell or photomultiplier replaces the eyepiece and a feedback circuit controls a motor that rotates the analyzer lens. This arrangement is sufficient for a polarimeter, but not for a polarimetric (HPLC) de-t e c t o r, which must respond much more quickly than that permitted by a mechanical system. For this, a Faraday electronics scheme is necessary, which provides a nonmechanical way to turn the analyzer, or, more accurately, to compensate for the rotation from the sample and relate the compensa-tion to a measurement of degree and direction.
Michael Faraday determined that plane polarized light could also be rotated by an electric field as it passed through a transparent medium; further, the strength of the field is (reasonably) proportional to the rotation.d A nonmechanical polarimetric detec-tor (or polarimeter) could then be built in which light, rotated by a chiral compound in a cell, could be corrected to a base condition by allowing it to continue through some transparent medium (e.g., air or quartz) around which is a coil. Light then reaching a photocell is transformed to an electrical current. Using a feedback circuit, this current is ap-plied to the coil so that there is a correction, or compensation, to the base condition for a rotation from a sample in the measuring cell. The positive or negative voltage required to effect this compensa-tion is then related to a direction of rotation and a magnitude, in degrees, which is typically displayed on the face of the instrument and/or output to data acquisition elsewhere.
Simplified schematics of a polarimeter and a Fara-day polarimetric detector are shown in F i g u re 1. In c o m m e rcial polarimeters, a wavelength filter is placed in the path of a white light source to yield monochro-matic light at the required wavelength; polarimetric detectors use coherent light sources or may simply use white light of a standard spectral pattern.
Within the Faraday electronics schemes, there are two variations: one using light intensity as a base condition and the other using an applied elec-trical phase as a base condition. In the first, in a process known as Faraday compensation, a voltage is applied to the coil that is just sufficient to restore the light to the original (i.e., base) condition of in-t e n s i t y. In the second, in a process known as Fara-day modulation/compensation, an alternating cur-rent is first applied to the coil to establish a base condition of standard electrical frequency. Rotated light will be manifest as a second electrical phase; further application of a direct voltage to the coil is used to eliminate the phase anomaly, and this volt-age is then related to the rotation.e In either case of instruments using Faraday com-pensation electronics, there are no moving me-chanical parts; machines will not wear out or lose performance over time. In the case of the modula-tion/compensation scheme, measurements are achieved that are independent of light intensity, enabling more field-friendly (nonlaser) light s o u rces, an improved consistency of measurement, and, in the case of polarimetric detectors, fewer complications from coeluting, achiral peaks.
M As the issue of chirality becomes more appreciated in bioscience research, tools to measure it increase in importance in analytical laboratories.
Figure 1 Early polarimeter and polarimetric detector (simplified diagrams). Plane polarized light is rotated as it passes through a chiral solution. In the simple polarime -ter, a second polarizing filter (F-2) is realigned to the first so that the degree of rotation can be determined. Plane polarized light is also rotated as it passes through a transparent medium within an electromagnetic field—the basis for polarimetric detectors and some advanced po -larimeters. A photomultiplier provides feedback to the Faraday section, providing a compensating current that is proportional to the required angular correction.
Figure 2 C H I R A LYSER™ polarimetric detector made by I B Z M E S S T E C H N I K GMBH ( H a n n o v e r, Germany); sold in N o rth America by J M Science, Inc. (Grand Island, NY).
The detector uses the Faraday modulation/compensation electronic scheme.
aChiral: containing at least one molecular species that has one or more centers of symmetry, sometimes referenced as stereogenic centers or simply chiral centers. A molecule can generally be said to have a center of symmetry, or be chiral, if a component atom is singly bound only to different and distinct functional groups.
Chirality is usually associated with carbon atoms, but other atoms (most notably nitrogen, phosphorus, and sulfur) can also give rise to it. Additionally, chirality can arise through overall molecular structure, as is sometimes the case with multiple fused-ring molecules (which may “pucker” in more than one way).
bThis relationship is via Biot’s formula, [α]D = α/(l × C), where [α]D = specific rotation of the compound, α is the observed rotation in degrees, l is the light pathlength in dm, and C is the compound’s concentration in g/mL, under conditions of designated wave-length and temperature.
c For example, a rotation of –270° could otherwise be interpreted as +90°; the analyzer would be turned to the same apparent posi-tion. A second concentration, say, 50% of the first one, would in-dicate which of the two cases is accurate. In the case in which a specific rotation is known and the solution concentration is not, two measurements will also be necessary for the same reason.
dThis relationship can be represented by the formula, a ~ d × V × H, where a is the observed rotation, d is the light pathlength, V is the Verdet constant (a physical property of a specific transparent material), and H is the intensity of a magnetic field. The field is generated by a coil, which is defined by its number of windings and the current passing through them. In any particular instru-ment, the pathlength, Verdet constant, and coil windings are fixed; the observed rotation is therefore proportional to the ap-plied electric current.
eA rotation of the light produces a second, overlaid electrical phase having a frequency of twice that of the induced phase. This phase is generated as the ac current is modulated by the rotation.
Applications Applications for polarimeters are well known, with machines in common use in quality control laboratories and process areas as well as the researc h l a b o r a t o ry. Of significance may be that Faraday-type polarimeters are now commercially available that are equal to the best of the mechanical types in accuracy and precision, but that have no moving parts.
Less well known may be applications for polari-metric detectors; unlike Europe and Japan, these in-struments are not yet in wide use in the U.S. Before discussing applications for these detectors, however, we should first understand their limitations. Biot’s formula suggests a flow cell with the longest possible light pathlength; however, good chromatography de-mands the smallest possible cell volume. Compro-mise cells therefore yield chromatography with rela-tively wide bandspread.
Perhaps the greatest limiting factor in Faraday com-pensation systems is the earth’s magnetic field, which, at the highest sensitivities of these instruments, inter-feres with the generated fields of the instruments and influences their output signals. Field instruments, i.e., those used in the typical chemistry laboratory, must be detuned somewhat to prevent them from acting like compasses.f As a practical matter, although these detectors may be tuned to levels yielding rotations of 1 0– 6°, 2 × 1 0– 5° is a more realistic expectation.g A degree of rotation is reached through a combi-nation of the compound’s specific rotation and its solution concentration. As a rule of thumb, the fol-lowing applies to the better of today’s machines: 22 / NOVEMBER 2001 If: Specific rotation (degrees) × sample concentration (%) × injection volume (µL) ≥ 100, then: One will see one’s sample above baseline noise.
Some chiral compounds may have very small specific rotations and will therefore be below detect-able limits. Highly concentrated samples can some-times produce refractive index (RI) effects and dis-torted signals; column capacities may limit injection volumes. The presence of microbubbles in the flow cell can also yield RI effects, and degassing, a worth-while procedure in any HPLC work, is even more important when using polarimetric detection. In the final analysis, even the best of polarimetric detectors will have limits for practical applications.
Despite these limitations, polarimetric detectors are the only instruments that do what they do, and they are destined to come into greater use as their applications become better understood and as gen-eral interest in chirality grows.
How can these instruments be used? Applica-tions for polarimetric detectors arise as a result of the information they provide as discussed earlier: They tell the direction as well as the magnitude of rotation. Also, a built-in deficiency, viz., their non-response to nonoptical compounds, yields others.
Obviously, they are used to characterize optically active samples, but they are also used to differenti-ate optically active compounds from nonoptically active compounds in mixed samples. In a typical ar-rangement, the polarimetric detector is placed im-mediately downstream from a UV detector. Both traces are output simultaneously, and the peaks generated via the polarimetric detector point to the chiral compounds in the UV display.
Probably the biggest application area is the characterization of separated enantiomers, most useful in chiral HPLC methods development. A polarimetric detector will show elution order, i.e., which enantiomer elutes first,h which is of consid-Figure 3 Analysis of a proprietary, pure enantiomer sam -ple showing comparison of a UV and optical rotation chromatographic trace. The optical rotation detector (ORD) is placed downstream from the UV detector. Both outputs are superimposed on this two-channel display, with the signal from the ORD (top signal) offset so that it may display negative peaks. In this run, the combination of the high concentration of sample and its (likely) strong rotation has caused an overrange of the ORD peak—not an unusual occurrence in this preparative HPLC labora -tory. Since the purpose of the run was to merely deter -mine the sign of rotation of the sample, the overrange signal was not a problem. The negative ORD peak indi -cates a negative-rotating enantiomer. In this particular case, the ORD could be used to obtain quantitative infor -mation. However, in some cases, the combination of rota -tion strength and flow cell amount will be inadequate for quantitative analyses compared to the higher signal sen -sitivities available from UV detectors (assuming samples with chromophores). In such cases, the ORD is used to obtain additional, qualitative information and is consid -ered to be a secondary detector, with the UV detector considered the primary detector. Qualitative information from the ORD could be the indication of the presence of an optical sample (in a “dirty” process sample, for exam -ple) and in determining the sign of rotation. Especially if two-channel data acquisition is available, enabling a su -perimposable display, the ORD trace will point to the peaks of quantitative interest on the UV trace. (Graphic complements of Schering-Plough Research Institute, Kenil -worth, NJ.) Figure 4 A run to determine elution order of enantiomers in a separation of a racemate (50/50 mixture of two enan -tiomers), in this case, of chlorpheniramine maleate salt. The ORD (yielding the top signal) is downstream from the UV de -tector; both outputs are superimposed onto the same display by using two of the available channels. As may be seen by the signal-to-noise ratios of the two, the UV trace will yield the best quantitative information; thus it is used for this pur -pose. However, the ORD has established that the positive ro -tator has eluted first by this method. As the method is refined to suit the purposes of the chromatographer, the ORD will be used to track the elution order. For standard HPLC prepara -tion of a preferred enantiomer from a racemate, it is desir -able to have the preferred enantiomer elute first; thus a method is developed to accomplish this. Once the prepara -tive separation to the preferred enantiomer is accomplished, h o w e v e r, it is necessary to determine its chiral purity, i.e., the ratio of the preferred enantiomer to the other one. In this case, the opposite situation of elution order will be sought, with the lesser peak eluted first, so that it is not lost in the tail of the dominant peak. An ORD is used to obtain elution or -der in the development of both methods, i.e., the preparative method and the analytical method. a) The object of the anal -ysis was only to confirm that the separated product was in -deed the (+) enantiomer. (Graphics complements of Scher -ing-Plough Research Institute.) a b b a Figure 5 Quantitative analysis of a proprietary, pure enantiomer sample having no UV chromophore. The sample has a strong enough rotation to be detectable via the ORD; in this case, the amount of sample used was too much, since its signal (top trace) has overranged. If a chi -ral sample has too weak a chromophore for UV detec -tion, it may be detected via ORD if its combination of ro -tation strength and amount in the flow cell is adequate.
The ORD is in the role of primary detector. a) Related (proprietary) compound having a very weak chro -mophore (at ~5.5 min). The ORD trace might yield better quantitative information in this case also. (Graphics com -pliments of Schering-Plough Research Institute.) A P P L I C ATION NOTE c o n t i n u e d f An overly tuned polarimetric detector could be used for naviga-tion. Such a detector may yield a different result when its orienta-tion on the laboratory bench is changed; further, metal masses, such as belt buckles or cars passing on the roadway, could also af-fect measurements.
g2 × 10–5° corresponds to an arc of about 7 ft on a circle the size of the earth.
hSpecifically, the identification of the levo (–) or the dextro (+) forms. Note that these are directional designations and have no relation to absolute conformation—designated by “R” and “S.” The relation of absolute conformation to direction may be deter-mined through the use of circular dichromism instruments or other standard instrumental means. In most cases, however, these relationships will be found in the literature.
AMERICAN LABORATORY / 23 erable importance to chiral HPLC. Consider meth-ods development for a preparative separation. If the object of the separation is to prepare the maxi-mum amount of a preferred enantiomer, it is im-portant to discover a method in which the desired enantiomer elutes first, since efficiency can be sev-eral times greater. Once this enantiomer has been isolated, however, it is usually necessary to then determine its chiral purity,i i.e., that portion not contaminated by the undesired enantiomer. In this case, there will be a very large peak and a very small one, perhaps less than 1%. If the smaller peak elutes second, there is a likelihood that it could become lost in the tail of the dominant peak. For this reason, a separation method result-ing in the elution of the smaller peak first is desir-able. Identification of the enantiomers through the use of the polarimetric detector can therefore guide methods development.
Another feature of a polarimetric detector can be exploited. If there is an enantiomeric mixture in its flow cell, the detector will show a net signal that is proportional to the component enan-tiomers of the mixture. For example, a mixture of 25% of one enantiomer and 75% of its coenan-tiomer will result in a net signal of 50% of the co-e n a n t i o m e r.j Therefore, against a reference, pure enantiomer standard, it is possible to obtain a rough approximation of chiral purity using nonchiral HPLC columns.
In most cases in which adequate separations of enantiomers are found for chiral columns, an absorbance detector will yield better quantitative measurements, simply because of its greater sensi-t i v i t y. In some cases in which resolutions are poor, however, a polarimetric detector may yield more ac-curate data.1 Examples of real world chromatography using a chiral detector (F i g u re 2) placed just downstream from an absorbance detector are shown in Figures 3, 4, and 5. Chromatographic traces for both detectors are displayed simultaneously.
In certain preparative or even process chiral chro-matography situations, use of a polarimetric detec-tor could enable a signal-based, rather than a time-based, automation scheme, with the possibility of long-term, unattended operation. The positive and negative voltage outputs from the detector (result-ing from the detector’s seeing positive and negative peaks) afford the opportunity for definitive switch-ing. Provided a test run ensures that there are no ar-tifacts from RI effects, the detector will ignore any nonchiral species that may be present. Simulated moving bed adsorption (SMB) preparative LC—used to make large-scale separations of two-component systems (such as the chiral separation of enantio-mers)—could be especially suited to the definitive switching of valves.
In an application perhaps more theoretical than practical, a polarimetric detector could even serv e as a kind of universal detector, detecting nonchiral be seen. These instruments are beginning to claim their rightful place in North American laboratories.
Reference 1. Mannschrect A, Mintas M, Becher G, Stühler G. Liquid chromatography of enantiomers: determination of enantiomeric purity in spite of extensive peak overlap.
Angewandte Chemie 1980; 19(6):459–70.
as well as chiral compounds in a procedure some-times called indirect polarimetry. This technique utilizes a chiral mobile phase, chosen not to inter-fere with the species being measured. In effect, an artificial baseline is established, either above or be-low the real baseline. A chiral species will show in its usual fashion relative to the new baseline; non-chiral species will be displayed as displacement peaks, as a depression of the signal from the chiral mobile phase.
P o l a r i m e t ry is an obvious complement for work on chiral compounds. Instruments using Faraday electronics have made no-moving-part polarimeters and polarimetric LC detectors possible. New genera-tion machines have increased the range of applica-tions as well as the percentages of samples that can M r. Rozea is President, Synergetic Associates, Inc., P. O .
Box 7716, Lancaster, PA 17604, U.S.A.; tel.: 7 1 7 - 8 9 8 -4101; fax: 717-898-4106; e-mail: prozea@aol.com.
Perhaps the greatest limiting factor in Faraday compensation systems is the earth’s magnetic field, which, at the highest sensitivities of these instruments, interferes with the generated fields of the instruments and influences their output signals.
i Determinations of this type are often designated by the desired enantiomer’s enantiomeric excess, expressed as a percentage, a.k.a. percent e.e.
j A 50–50 enantiomeric mixture, i.e., a racemate, will have a net signal of zero—no deviation from the baseline. |
5372 | https://ssd.jpl.nasa.gov/planets/phys_par.html | Planetary Physical Parameters
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Planets
Planetary Physical Parameters
Planetary Physical Parameters
The following tables contain selected physical characteristics of the planets and dwarf planets, respectively. Table column headings are described below.
| Planet | Equatorial Radius | Mean Radius | Mass | Bulk Density | Sidereal Rotation Period | Sidereal Orbital Period | V(1,0) | Geometric Albedo | Equatorial Gravity | Escape Velocity |
--- --- --- --- ---
| | (km) | (km) | (×10 24 kg) | (g cm-3) | (d) | (y) | (mag) | | (m s-2) | (km s-1) |
| Mercury | 2440.53 [D] ±0.04 | 2439.4 [D] ±0.1 | 0.330103 [F] ±0.000021 | 5.4289 [] ±0.0007 | 58.6462 [C] | 0.2408467 [B] | -0.60 [E] ±0.10 | 0.106 [B] | 3.70 [] | 4.25 [] |
| Venus | 6051.8 [D] ±1.0 | 6051.8 [D] ±1.0 | 4.86731 [G] ±0.00023 | 5.243 [] ±0.003 | -243.018 [C] | 0.61519726 [B] | -4.47 [E] ±0.07 | 0.65 [B] | 8.87 [] | 10.36 [] |
| Earth | 6378.1366 [D] ±0.0001 | 6371.0084 [D] ±0.0001 | 5.97217 [H] ±0.00028 | 5.5134 [] ±0.0003 | 0.99726968 [B] | 1.0000174 [B] | -3.86 [B] | 0.367 [B] | 9.80 [] | 11.19 [] |
| Mars | 3396.19 [D] ±0.1 | 3389.50 [D] ±0.2 | 0.641691 [I] ±0.000030 | 3.9340 [] ±0.0007 | 1.02595676 [C] | 1.8808476 [B] | -1.52 [B] | 0.150 [B] | 3.71 [] | 5.03 [] |
| Jupiter | 71492 [D] ±4 | 69911 [D] ±6 | 1898.125 [J] ±0.088 | 1.3262 [] ±0.0003 | 0.41354 [C] | 11.862615 [B] | -9.40 [B] | 0.52 [B] | 24.79 [] | 60.20 [] |
| Saturn | 60268 [D] ±4 | 58232 [D] ±6 | 568.317 [K] ±0.026 | 0.6871 [] ±0.0002 | 0.44401 [C] | 29.447498 [B] | -8.88 [B] | 0.47 [B] | 10.44 [] | 36.09 [] |
| Uranus | 25559 [D] ±4 | 25362 [D] ±7 | 86.8099 [L] ±0.0040 | 1.270 [] ±0.001 | -0.71833 [C] | 84.016846 [B] | -7.19 [B] | 0.51 [B] | 8.87 [] | 21.38 [] |
| Neptune | 24764 [D] ±15 | 24622 [D] ±19 | 102.4092 [M] ±0.0048 | 1.638 [] ±0.004 | 0.67125 [C] | 164.79132 [B] | -6.87 [B] | 0.41 [B] | 11.15 [] | 23.56 [] |
| Dwarf Planet | Equatorial Radius | Mean Radius | Mass | Bulk Density | Sidereal Rotation Period | Sidereal Orbital Period | V(1,0) | Geometric Albedo | Equatorial Gravity | Escape Velocity |
--- --- --- --- ---
| | (km) | (km) | (×10 18 kg) | (g cm-3) | (d) | (y) | (mag) | | (m s-2) | (km s-1) |
| Ceres | 482.1 [O] | 469.7 [O] | 938.416 [O] ±0.013 | 2.162 [O] ±0.008 | 0.37809042 [Q] | 4.61 [Q] | 3.34 [R] | 0.090 [P] ±0.003 | 0.27 [] | 0.51 [] |
| Pluto | 1188.3 [D] ±1.6 | 1188.3 [D] ±1.6 | 13024.6 [N] ±6.0 | 1.853 [N] ±0.004 | -6.3872 [C] | 247.92065 [B] | -1.0 [Q] | 0.3 [B] | 0.62 [] | 1.21 [] |
| Eris | 1200 [S] ±50 | 1200 [S] ±50 | 16600 [S] ±200 | 2.3 [S] ±0.3 | 1.079 [Y] | 557.56 [Q] | -1.1 [Q] | 0.84 [] | 0.77 [] | 1.36 [] |
| Makemake | 717 [T] ±7 | 714 [T] ±7 | 3100 [U] | 2.1 [U] | 0.937 [Q] | 307.54 [Q] | -0.2 [Q] | 0.81 [T] ±0.02 | 0.40 [] | 0.76 [] |
| Haumea | 870 [W] | 715 [W] | 4006 [V] ±40 | 2.6 [W] | 0.1631 [X] | 284.81 [Q] | 0.2 [Q] | 0.72 [] | 0.35 [] | 0.78 [] |
Table Column Descriptions
| Equatorial Radius | Radius of the planet at the equator. |
| Mean Radius | Radius of a sphere with the equivalent volume of the planet. |
| Mass | Total mass of the planet. |
| Bulk Density | Density computed using the total volume and mass of the planet. |
| Sidereal Rotation Period | Time required for a full rotation of the planet relate to fixed stars. |
| Sidereal Orbital Period | Time required for the planet to make one complete orbit around the sun relative to fixed stars. |
| V(1,0) | The visual magitude of the planet as seen at a distance of 1 au from both the Sun and observer. |
| Geometric Albedo | See glossary definition of albedo. |
| Equatorial Gravity | The gravitational acceleration on the planet's surface at the equator. |
| Escape Velocity | The minimum velocity required for an object to escape the gravitational influence of the planet. |
References
[]Value and uncertainty derived from other referenced values and uncertainties in this table. Bulk Density computed based on the volume of a sphere with the published mean radius.
[B]Explanatory Supplement to the Astronomical Almanac. 1992. K. P. Seidelmann, Ed., p.706 (Table 15.8) and p.316 (Table 5.8.1), University Science Books, Mill Valley, California.
[C] Seidelmann, P.K. et al. 2007. "Report of the IAU/IAG Working Group on cartographic coordinates and rotational elements:2006" Celestial Mech. Dyn. Astr.98:155-180.
[D] Archinal, B.A. et al. 2018. "Report of the IAU/IAG Working Group on cartographic coordinates and rotational elements:2015" Celestial Mech. Dyn. Astr.130:22.
[E] Hilton, J.L. 2005. "Improving the Visual Magnitudes of the Planets in The Astronomical Almanac. I. Mercury and Venus" Astron. J129:2902-2906.
[F] Anderson, J.D., et al. 1987. "The mass, gravity field, and ephemeris of Mercury" Icarus71:337-349.
[G] Konopliv, A.S., et al. 1999. "Venus gravity: 180th degree and order model" Icarus139:3-18.
[H] Folkner, W.M. and Williams, J.G. 2008. "Mass parameters and uncertainties in planetary ephemeris DE421." Interoffice Memo. 343R-08-004 (internal document), Jet Propulsion Laboratory, Pasadena, CA.
[I] Jacobson, R.A. 2008. "Ephemerides of the Martian Satellites - MAR080" Interoffice Memo. 343R-08-006 (internal document), Jet Propulsion Laboratory, Pasadena, CA.
[J] Jacobson, R.A. 2013. "Jovian Satellite ephemeris - JUP310" private communication.
[K] Jacobson, R.A., et al. 2006. "The gravity field of the Saturnian system from satellite observations and spacecraft tracking data" AJ132(6):2520-2526.
[L] Jacobson, R.A. 2014. "The Orbits of the Uranian Satellites and Rings, the Gravity Field of the Uranian System, and the Orientation of the Pole of Uranus" AJ148:76-88.
[M] Jacobson, R.A. 2009. "The orbits of the Neptunian satellites and the orientation of the pole of Neptune" AJ137:4322.
[N] Brozovic, M. et al. 2024. "Post-new-horizons Orbits and Masses for the Satellites of Pluto" AJ167:256
[O] R.S. Park, et al. (2016) "A partially differentiated interior for (1) Ceres deduced from its gravity field and shape" Nature537:515-517
[P] Li et al. (2006) Icarus182:143-160
[Q] JPL Small-Body Database. 2019.
[R] IRAS-A-FPA-3-RDR-IMPS-V6.0
[S] Brown, Michael E.; Schaller, Emily L. (2007) Science316:1585.
[T] Brown, Michael E. (2013) ApJ767:L7.
[U] Parker, Alex et al. (2018) DPS meeting# 50 id. 509.02.
[V] Ragozzine, D.; Brown, M. E. (2009) AJ137:4766-4776.
[W] Lockwood, Alexandra C.; Brown, Michael E.; Stansberry, John. (2014) Earth, Moon, Planets111:127-137.
[X] Ortiz, J.L.; Santos-Sanz, P.; Sicardy, B.; Benedetti-Rossi, G.; et al. (2017) Nature550:219-223.
[Y] Roe, H. G.; Pike, R.E.; Brown, M.E. (2008) Icarus198:459-464.
Change Log
2019-Dec-12
A second table for dwarf planets was added. Mass values were updated from current estimates of GM (referenced above). The value of G (Newtonian gravitational constant) was taken from the current best estimate (CODATA 2018) available from the NIST website, G=6.67430 (±0.00015) × 10-11 kg-1 m 3 s-2.
2018-Mar-28
Mass values were updated from current estimates of GM (referenced above). The value of G (Newtonian gravitational constant) was taken from the current best estimate (CODATA 2014) available from the NIST website, G=6.67408 (±0.00031) × 10-11 kg-1 m 3 s-2. Bulk density values (and uncertainties) were computed based on updated mass values and computed volumes (based on a sphere of the published mean radius). Equatorial surface gravity values were computed from updated mass and equatorial radius values. Escape velocities were also updated based on computed values using updated masses and mean radii.
2008-Nov-05
Mass values were updated from current estimates of GM (referenced above). The value of G (Newtonian gravitational constant) was taken from the current best estimate (CODATA 2006) available from the NIST website, G=6.67428 (±0.00067) × 10-11 kg-1 m 3 s-2. Mean radius values were added. Bulk density values (and uncertainties) were computed based on updated mass values and computed volumes (based on a sphere of the published mean radius). Equatorial surface gravity values were computed from updated mass and equatorial radius values. Escape velocities were also updated based on computed values using updated masses and mean radii.
2008-Oct-24
Sidereal rotation periods were updated using improved values from the "Report of the IAU/IAG Working Group on cartographic coordinates and rotational elements: 2006" published in 2007. Rotation rates were significantly improved for Saturn, Uranus, and Neptune. Visual magnitude parameter V(1,0) values were improved for Mercury and Venus using reference [E] listed above.
2006-Mar-14
A previous reference ("Astrometric and Geodetic Properties of Earth and the Solar System" in Global Earth Physics, A Handbook of Physical Constants, AGU Reference Shelf 1, 1995, American Geophysical Union, Tables 6,7,10.) had a few erroneous values. Those values have been replaced using data from the current reference list.
2002-Nov-22
Pluto's mass and density were updated using a more current reference. It should be noted that Pluto's radius ranges from about 1150 to 1206 km in the literature.
2001-May-22
Sidereal orbit period data were corrected. Previous values were taken from p.704 in reference [B] where they were incorrectly labeled as Sidereal instead of Tropical. The corrected values are derived from the mean longitude rates shown in Table 5.8.1 in reference [B].
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5373 | https://prostatecanceruk.org/prostate-information-and-support/treatments/permanent-seed-brachytherapy/ | What is permanent seed brachytherapy?
Permanent seed brachytherapy, also known as low dose-rate (LDR) brachytherapy, is a type of radiotherapy where tiny radioactive seeds are put into your prostate. Each radioactive seed is the size and shape of a grain of rice. The seeds stay in the prostate forever and give a steady dose of radiation over a few months.
The radiation damages the prostate cells and stops them dividing and growing. The cancer cells can't recover from this damage and die. But healthy cells can repair themselves more easily.
The seeds release most of their radiation in the first three months after they’re put into the prostate. After around 8 to 10 months, almost all the radiation has been released. The amount of radiation left in the seeds is so small that it doesn’t have an effect on your body.
Permanent seed brachytherapy is as good at treating localised prostate cancer that has a low risk of spreading as surgery (radical prostatectomy) or external beam radiotherapy. Read more about other treatment options for localised prostate cancer.
Watch Chris' story below for one man's experience of brachytherapy.
Permanent seed brachytherapy fact sheet
This fact sheet is for anyone who is thinking about having permanent seed brachytherapy to treat their prostate cancer.
Download or order fact sheet
Who can have permanent seed brachytherapy?
Permanent seed brachytherapy on its own may be suitable for men whose hasn't spread outside the prostate (localised prostate cancer). This is because the radiation from the radioactive seeds doesn’t travel very far, so will only treat cancer that is still inside the prostate.
You may be able to have brachytherapy together with external beam radiotherapy if you have localised or locally advanced prostate cancer. This is sometimes called a brachytherapy boost. You may also have hormone therapy alongside your external beam radiotherapy and brachytherapy boost. Having these other treatments at the same time as permanent seed brachytherapy can help make the treatment more effective. But it can also increase the risk of side effects.
If you have localised or locally advanced prostate cancer, your Cambridge Prognostic Group (CPG) will help your doctor decide what treatment options are suitable for you. However, some doctors may use the old system of low, intermediate and high risk instead. You can read more in our booklet, Prostate cancer: A guide for men who’ve just been diagnosed.
When is permanent seed brachytherapy not suitable?
Permanent seed brachytherapy on its own won’t be suitable if your cancer has spread just outside your prostate (locally advanced prostate cancer ). But, if you have locally advanced prostate cancer, you may be offered brachytherapy together with external beam radiotherapy (see above).
You won't be able to have permanent seed brachytherapy at all if your cancer has spread to other parts of your body (advanced prostate cancer).
It may not be suitable if you have a very large prostate. If you do have a large prostate you may be able to have hormone therapy before treatment to shrink your prostate.
It may also not be suitable if you have severe problems urinating, such as those caused by an enlarged prostate or overactive bladder. These include needing to urinate more often, a weak urine flow or problems emptying your bladder. Permanent seed brachytherapy can make these problems worse. Before you have treatment, your doctor, nurse or radiographer will ask you about any urinary problems, and you may have some tests.
You may not be able to have permanent seed brachytherapy if you have some types of inflammatory bowel disease (IBD). This is because it could make your bowel problems worse. Brachytherapy won’t be suitable if you’ve had surgery to remove your rectum (back passage), because the treatment involves using an ultrasound probe in the back passage to make sure the seeds are put in the correct place. Your doctor or nurse will explain your treatment options to you.
If you’ve recently had surgery to treat an enlarged prostate, called a transurethral resection of the prostate (TURP), you may have to wait atleast three months before having permanent seed brachytherapy. Some hospitals don’t offer brachytherapy to men who’ve had a TURP because it can make the treatment more difficult to perform.
You will usually have a general anaesthetic while the brachytherapy seeds are put in place, so you’ll be asleep and won’t feel anything. This means permanent seed brachytherapy may only be an option if you are fit and healthy enough to have an anaesthetic. However, you may be able to have a spinal (epidural) anaesthetic instead. This may depend on what your hospital offers.
Not all hospitals offer permanent seed brachytherapy. If your hospital doesn’t do it, your doctor may refer you to one that does.
What are the advantages and disadvantages?
What may be important to one man might be less important to someone else. If you are offered permanent seed brachytherapy, speak to your doctor or nurse before deciding whether to have it. They will be able to help you think about which treatment option is right for you. There’s usually no rush to make a decision so give yourself time to think about whether permanent seed brachytherapy is right for you.
Advantages
Recovery is quick, so most men can return to their normal activities one or two days after treatment.
Permanent seed brachytherapy delivers radiation directly into the prostate, so there may be less damage to surrounding healthy tissue, and a lower risk of some side effects.
You will only be in hospital for one or two days.
If your cancer comes back, you may be able to have further treatment.
Disadvantages
Permanent seed brachytherapy can cause side effects such as urinary and erection problems. It can also cause bowel problems but this isn't common.
You will need a general or spinal anaesthetic, which can have side effects.
It may be some time before you know whether the treatment has been successful.
You will need to avoid sitting close to pregnant women or children during the first two months after treatment.
If you are having external beam radiotherapy or hormone therapy as well as permanent seed brachytherapy, think about the advantages and disadvantages of those treatments as well.
What does treatment involve?
You will be referred to a specialist who treats cancer with radiotherapy, called a clinical oncologist. The treatment itself may be planned and carried out by other specialists including a therapy radiographer, a radiologist, a urologist, a physicist and sometimes a specialist nurse.
Each hospital may do things slightly differently but you will usually have:
an appointment to check the treatment is suitable for you
a planning session, to plan the treatment
the treatment itself.
You will have the treatment during one or two hospital visits. Many hospitals offer treatment in just one visit, where your treatment will be planned and the seeds put in at the same time under the same anaesthetic. This is sometimes called a one-stage procedure. You may not need to stay in hospital overnight.
If your treatment is spread over two visits, you will have a planning session on your first visit. The radioactive seeds will be put in on the second visit, two to four weeks later. You may hear this called a two-stage procedure. Some men may be offered the two-stage procedure, for example, if they need treatment to reduce the size of the prostate. Some hospitals only offer the two-stage procedure.
Before the planning session, let your specialist know if you are taking any medicines, especially medicines that thin your blood such as aspirin, warfarin or clopidogrel. Don’t stop taking any medicines without speaking to your doctor or nurse.
Planning session
During the planning session, you will have an ultrasound scan to find out the size, shape and position of your prostate. This is called a volume study and is done by a clinical oncologist, physicist, radiographer or radiologist. They will use the scan to work out how many radioactive seeds you need.
It’s important that your bowel is empty so the scan shows clear images of your prostate. You may need to take a medicine called a laxative the day before the planning session to empty your bowels. Or you might be given an enema when you arrive at the hospital instead. An enema is a liquid medicine that is put inside your back passage (rectum). Your doctor, nurse or radiographer will give you more information about this.
You will usually have a general anaesthetic so that you’re asleep during the ultrasound scan. This will be given by a health professional called an anaesthetist. If you can’t have a general anaesthetic for health reasons, you may be able to have a spinal (epidural) anaesthetic. This is where anaesthetic is injected into your spine so that you can’t feel anything in your lower body. In some hospitals, the anaesthetist will talk through the different types of anaesthetic before deciding with you which is the best option.
A thin tube (catheter) may be passed up your penis into your bladder to drain urine, and an ultrasound probe is put inside your back passage. The probe is attached to an ultrasound machine that displays an image of the prostate. Your doctor, radiographer or physicist will use this image to work out how many radioactive seeds you need and where to put them.
The planning session is a final check that the treatment is suitable for you. If the scan shows that your prostate is too large, you may be offered hormone therapy for up to six months to shrink your prostate. You’ll then have another planning session before you have the seeds put in. Very occasionally, the scan may show that permanent seed brachytherapy isn’t possible because of the position of your prostate within the pelvic bones. If this happens, your specialist will discuss other treatment options with you.
The planning session usually takes about half an hour plus the time it takes for you to recover from the anaesthetic. You can go home the same day if you aren’t having the treatment straight away. Ask a friend or family member to take you home, as you shouldn’t drive for 24 to 48 hours after an anaesthetic.
Placing the seeds
The clinical oncologist will put the seeds into your prostate. If you have the treatment on the same day as your planning session, the seeds will be put in straight after the planning scan, under the same anaesthetic.
If you have the treatment on a different day to your planning session, you’ll need another anaesthetic on the day of your treatment. You may also need to take another laxative, or have another enema to empty your bowels for the treatment. You'll usually have a catheter to drain urine from your bladder.
An ultrasound probe is again put inside your back passage to take images of your prostate and make sure the seeds are put in the right place. In some hospitals, the clinical oncologist might put gel into your urethra (the tube you urinate through). This is instead of a catheter and helps the doctor see your urethra more clearly so they avoid putting any seeds into it.
The clinical oncologist then puts thin needles through your perineum (the area between the testicles and the back passage), and into your prostate. They pass the radioactive seeds through the needles into the prostate. The needles are then taken out, leaving the seeds behind.
Depending on the size of your prostate, between 60 and 120 seeds are put into the prostate using around 15 to 30 needles. The seeds can be loose individual seeds or linked together in a chain using material that slowly dissolves. Each hospital is different and the clinical oncologist will decide what type of seeds you will have. Treatment usually takes 45 to 90 minutes.
Where the seeds go in the prostate
After your treatment
You’ll wake up from the anaesthetic in the recovery room, before going back to the ward or discharge area. Most men feel fine after a general anaesthetic but a few men feel sick or dizzy. Your nurse may give you an ice pack to put between your legs to help prevent swelling.
If you have a catheter, it will usually be removed before you wake up but it may be left in for a few hours until you are fully awake. Very rarely, you might go home with your catheter and have to return within a few days to have it removed. Having the catheter removed may be uncomfortable, but it shouldn’t be painful.
You can go home when you’ve recovered from the anaesthetic and can pass urine. Most men go home on the same day as their treatment. But some men find it difficult to urinate at first, and need to stay in hospital overnight. You shouldn’t drive for 24 to 48 hours after the anaesthetic. Ask a family member or friend to take you home.
Your doctor or nurse will give you any medicines that you need at home. These may include drugs to help you urinate, such as tamsulosin, and antibiotics to prevent infection.
You may have some pain or bleeding from the area where the needles were put in. You can take pain-relieving drugs, such as paracetamol, for the first few days if you need to.
When to call your doctor, nurse or radiographer
Your doctor, nurse or radiographer will give you a telephone number to call if you have any questions or concerns. Contact them if any of the following things happen.
If your urine is very bloody or has large clots in it, you may have some bleeding in your prostate. Contact your doctor or nurse as soon as possible.
If you suddenly and painfully can’t urinate, you may have acute urine retention. Go to your local accident and emergency (A&E) department as this will need treatment as soon as possible. Take information about your cancer treatment with you, if you can.
If you have a high temperature (more than 38ºC or 101ºF), this may be a sign of infection. Contact your doctor or nurse or go to your local A&E department.
What happens after permanent seed brachytherapy?
The prostate absorbs most of the radiation, and it’s safe for you to be near other people or pets. But you should avoid sitting closer than 50 cm (20 inches) to pregnant women and children during the first two months after treatment. You can give children a cuddle (at chest level) for a few minutes each day, but avoid having them on your lap. If you have pets, try not to let them sit on your lap for the first two months after treatment. Your doctor or nurse will talk to you about this in more detail.
Although the seeds usually stay in the prostate it is possible, but rare, for seeds to come out in your semen when you ejaculate. To be on the safe side, don’t have sex for a few days after treatment, and use a condom the first five times you ejaculate. Double-wrap used condoms and put them in the bin.
It is also rare for a seed to come out in your urine. If this happens at the hospital, don’t try to pick it up. Leave it where it is and let the hospital staff know straight away. If this happens after you’ve left the hospital, don’t try to pick up the seed. Just flush it down the toilet.
Always tell your doctor, nurse or radiographer if you think you have passed a seed. Your treatment will still work, because there will still be enough radiation left in the prostate to treat your cancer.
It is possible for a seed to move into your bloodstream and travel to another part of your body, but this is rare. This shouldn’t do any harm and will often be picked up when you have a scan at your follow-up appointment. If you have any unusual symptoms, speak to your doctor or nurse.
Your radiographer will give you an advice card that says you’ve had treatment with internal radiation. You should carry this card with you for at least 20 months after your treatment.
If a man dies, for whatever reason, in the first 20 months after having treatment, it won’t be possible to have a cremation because of the radioactive seeds. Speak to your doctor or nurse if you are worried about this. Some men decide not to have permanent seed brachytherapy because of this – for personal or religious reasons.
Going back to normal activities
You should be able to return to your normal activities in a few days. You can go back to work as soon as you feel able. This will depend on how much physical effort your work involves. It’s best to avoid heavy lifting for a few days after having the seeds put in. Speak to your doctor, nurse or radiographer about your own situation.
Travel
Remember to take your advice card with you when you travel. The radiation in the seeds can occasionally set off metal or radiation sensors at the airport, train station or cruise port.
Speak to your doctor, nurse or radiographer if you plan to travel anywhere soon after having permanent seed brachytherapy, or if you have any concerns about holidays and travel plans. Read more about travelling with prostate cancer.
Your follow-up appointment
You’ll have an appointment with your doctor, nurse or radiographer a few weeks after your treatment. They will check how well you are recovering, your PSA level, and ask about any side effects you might have.
After your treatment you’ll have a computerised tomography (CT) or magnetic resonance imaging (MRI) scan to check the position of the seeds. This can happen on the same day as your treatment, but it may be up to six weeks after your treatment, depending on your hospital.
PSA test
This is a blood test that measures the amount of prostate specific antigen (PSA) in your blood. PSA is a protein produced by normal cells in your prostate, and also by prostate cancer cells.
You will have regular PSA tests after your treatment to check how well it has worked. You will also be asked about any side effects. In most hospitals, you’ll have a PSA test after six weeks of your treatment. Then for the next two years you will have a PSA test at least every six months, and then at least once a year after that. Each hospital will do things slightly differently, so ask your doctor or nurse how often you will have PSA tests.
Your PSA should drop to its lowest level (nadir) 18 months to two years after treatment. How quickly this happens, and how low your PSA level falls, varies from man to man, and will depend on how big your prostate is and whether you’re also having hormone therapy. Your PSA level won’t fall to zero as your healthy prostate cells will continue to produce some PSA.
Your PSA level may rise after your treatment, and then fall again. This is called ‘PSA bounce’. It could happen up to three years after treatment. This is more common in younger men and men with a large prostate. It is normal, and doesn’t mean your cancer has come back or that you need more treatment.
If your PSA level consistently rises, particularly in a short amount of time, this could be a sign that cancer has come back. If this happens, your doctor will talk to you about further tests and treatment options if you need them. Further treatment options may include hormone therapy, HIFU, cryotherapy, or high dose-rate brachytherapy. Surgery might also be an option, but there's a higher risk of side effects if you've already had brachytherapy.
Read more about follow-up appointments and further treatment options.
What are the side effects?
Like all treatments, permanent seed brachytherapy can cause side effects. These will affect each man differently, and you may not get all the possible side effects.
Side effects usually start to appear about a week after treatment, when radiation from the seeds starts to have an effect. They are generally at their worst a few weeks or months after treatment, when the swelling is at its worst and the radiation dose is strongest. They are often worse in men with a large prostate, as more seeds and needles are used during their treatment. Side effects should improve over the following months as the seeds lose their radiation and the swelling goes down.
You might have worse side effects if you have permanent seed brachytherapy together with external beam radiotherapy and hormone therapy.
You might also get more side effects if you had problems before the treatment. For example, if you already had urinary, erection or bowel problems, these may get worse after permanent seed brachytherapy.
After the treatment, you might get some of the following:
blood-stained urine or rusty or brown-coloured semen for a few days or weeks
bruising and pain in the area between your testicles and back passage which can spread to your inner thighs and penis – this will disappear in a week or two
discomfort when you urinate and a need to urinate more often, especially at night, and more urgently.
Some side effects may take several weeks to develop and may last for longer. These may include problems urinating, erection problems, bowel problems and tiredness. Sometimes long-term or late side effects after radiotherapy treatment are called pelvic radiation disease. The Pelvic Radiation Disease Association has more information.
Smoking
Researchers have been looking at whether smoking increases the chance of having long-term bowel and urinary problems after radiotherapy for prostate cancer. At the moment only a small number of studies have been done, so we need more research into this. If you’re thinking of stopping smoking there’s lots of information and support available on the NHS website.
Urinary problems
Permanent seed brachytherapy can irritate the bladder and urethra. You may hear this called radiation cystitis. Symptoms include:
needing to urinate more often or urgently
difficulty urinating
discomfort or a burning feeling when you urinate
blood in your urine.
In some men, permanent seed brachytherapy can cause the prostate to swell, narrowing the urethra and making it difficult to urinate.
A few men find they suddenly and painfully can’t urinate in the first few days or weeks after treatment. This is called acute urine retention. If this happens, contact your doctor or nurse straight away, or go to your nearest accident and emergency (A&E) department as soon as possible. They may need to put in a catheter to drain the urine. You may need to have the catheter in for several weeks until your symptoms have settled down.
Urinary problems may be worse in the first few weeks after brachytherapy, especially in men with a large prostate, but they usually start to improve after a few months.
Medicines called alpha-blockers may help with problems urinating. You can also help yourself by drinking liquid regularly (two litres or three to four pints a day) and by avoiding drinks that may irritate the bladder, such as alcohol, fizzy drinks, artificial sweeteners, and drinks with caffeine, such as tea and coffee.
Permanent seed brachytherapy can also cause scarring in your urethra, making it narrower over time. This is called a stricture, and can make it difficult to urinate. This is rare and may happen several months or years after treatment. If it happens, you might need an operation to widen your urethra or the opening of the bladder.
Some men leak urine (urinary incontinence) after permanent seed brachytherapy, but this isn’t common. It may be more likely if you’ve previously had surgery to treat an enlarged prostate, called a transurethral resection of the prostate (TURP). Problems with leaking urine may improve with time, and there are ways to manage them.
Read more about urinary problems and how to manage them, or visit our interactive online guide for lots more tips.
Bowel problems
Your bowel and back passage are close to the prostate. Permanent seed brachytherapy can irritate the lining of the bowel and back passage, which can cause bowel problems. The risk of bowel problems after permanent seed brachytherapy is low. But you're more likely to have problems if you’re also having external beam radiotherapy.
Bowel problems can include:
loose and watery bowel movements (diarrhoea)
passing more wind than usual
needing to empty your bowels more often
needing to empty your bowels urgently
bleeding from the back passage
feeling that you need to empty your bowels but not being able to go.
Bowel problems tend to be mild and are less common than after external beam radiotherapy. They often get better with time but a few men have problems a few years after treatment. Try not to be embarrassed to tell your hospital doctor or your GP about any bowel problems. There are treatments that can help.
A small number of men may have bleeding from the back passage after brachytherapy. This can also be a sign of other problems such as piles (haemorrhoids) or more serious problems such as bowel cancer, so tell your nurse or GP about any bleeding. They may do tests to find out what is causing it. They can also tell you about treatments that can help.
Using a rectal spacer to protect your back passage
Your doctor may suggest using a rectal spacer to help protect the inside of your back passage from radiation damage. The spacer is placed between your prostate and your back passage. This means that less radiation reaches your back passage, which may help to lower your risk of bowel problems during or after your treatment.
Rectal spacers aren’t commonly used in permanent seed brachytherapy alone. But you may have one if you’re also having external beam radiotherapy. Rectal spacers are still new and aren’t very common yet, so they might not be available at your hospital. Ask your doctor, nurse or radiographer for more information about rectal spacers, their side effects and other ways to manage bowel problems.
Screening for bowel cancer
If you’re invited to take part in the NHS bowel screening programme soon after having brachytherapy, the test may pick up some blood in your bowel movements, even if you can’t see any blood yourself. Your doctor, nurse or radiographer may suggest that you delay your bowel screening test for a few months if you’ve recently had brachytherapy. This will help to make sure you don’t get incorrect results.
A small number of men get blood in their bowel movements after permanent seed brachytherapy, and this shouldn’t be anything to worry about. But if you notice blood, you should always let your doctor, nurse or radiographer know.
Sexual side effects
Brachytherapy can affect the blood vessels and nerves that control erections. This may cause problems getting or keeping an erection (erectile dysfunction). Erection problems may not happen straight after treatment, but sometimes develop some time afterwards.
The risk of long-term erection problems after brachytherapy varies from man to man. You may be more likely to have problems if you had any erection problems before treatment, or if you are also having hormone therapy or external beam radiotherapy.
If you have anal sex and prefer being the penetrative partner you normally need a strong erection, so erection problems can be a particular issue.
There are ways to manage erection problems, including treatments that may help keep your erection hard enough for anal sex. Ask your doctor or nurse about these, or speak to our Specialist Nurses.
You may produce less semen than before the treatment, or none at all. This can be a permanent side effect of brachytherapy. Your orgasms may feel different or you may get some pain in your penis when you orgasm. You may also notice a small amount of blood in the semen. This usually isn’t a problem, but tell your doctor or nurse if it happens. Some men have weaker orgasms than before treatment, and a small number of men can no longer orgasm afterwards.
If you have anal sex and are the receptive partner, there’s a risk that your partner might be exposed to some radiation during sex in the first few months after treatment. Anal sex or stimulation using fingers or sex toys is unlikely to move the brachytherapy seeds out of the prostate. Your doctor may suggest you avoid receiving anal sex or anal play for the first six months after your treatment. Ask your doctor for more information about having anal sex after permanent seed brachytherapy. They may contact the local medical physics team who can give specific information tailored to you.
If you prefer to be the receptive partner during anal sex, then bowel problems or a sensitive anus after permanent seed brachytherapy may affect your sex life. Even when the risk of radiation to your partner has passed, wait until any bowel problems have improved before trying anal play or sex.
Read more about sexual side effects after prostate cancer treatment. We also have specific information if you're a gay or bisexual man. And there are lots of tips for managing sexual problems in our interactive online guide.
Having children
Brachytherapy may make you infertile, which means you may not be able to have children naturally. But it may still be possible to make someone pregnant after brachytherapy.
It’s possible that the radiation could change your sperm and this might affect any children you conceive. The risk of this is very low, but use contraception for at least a year after treatment if there’s a chance you could get someone pregnant. Ask your doctor or nurse for more information.
If you plan to have children in the future, you may be able to store your sperm before you start treatment so that you can use it later for fertility treatment. If this is relevant to you, ask your doctor, nurse or radiographer whether sperm storage is available locally.
Tiredness (fatigue)
You may feel tired for the first few days after treatment as you recover from the anaesthetic. The effect of radiation on the body may make you feel tired for longer, especially if you’re also having external beam radiotherapy or hormone therapy. If you get up a lot during the night to urinate, this can also make you feel tired during the day.
Fatigue is extreme tiredness that can affect your everyday life. It can affect your energy levels, motivation and emotions. Fatigue can continue after the treatment has finished and may last several months.
There are things you can do to help manage fatigue. For example, planning your day to make the most of when you have more energy. Read more about fatigue, or get tips for dealing with fatigue in our interactive online guide. Our Specialist Nurses also offer support that can help you improve your fatigue over time.
Questions to ask your doctor, radiographer or nurse
Will I have a planning session at a different time to the treatment, or immediately before?
Will I have external beam radiotherapy or hormone therapy as well?
What side effects might I get?
How will we know if the treatment has worked?
What should my PSA level be after treatment and how often will you test it?
If my PSA continues to rise, what other treatments are available?
References
Updated: October 2022 | Due for Review: October 2025
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• Andreyev HJN. GI Consequences of Cancer Treatment: A Clinical Perspective. Radiat Res. 2016 Mar 28;185(4):341–8.
• Avellino G, Theva D, Oates RD. Common urologic diseases in older men and their treatment: how they impact fertility. Fertil Steril. 2017 Feb;107(2):305–11.
• Awad MA, Gaither TW, Osterberg EC, Murphy GP, Baradaran N, Breyer BN. Prostate cancer radiation and urethral strictures: a systematic review and meta-analysis. Prostate Cancer Prostatic Dis. 2018 Jun;21(2):168–74.
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• Husson O, Mols F, Poll-Franse LV van de. The relation between information provision and health-related quality of life, anxiety and depression among cancer survivors: a systematic review. Ann Oncol. 2010 Sep 24;mdq413.
• Huyghe E, Delannes M, Wagner F, Delaunay B, Nohra J, Thoulouzan M, et al. Ejaculatory Function After Permanent 125I Prostate Brachytherapy for Localized Prostate Cancer. Int J Radiat Oncol. 2009 May;74(1):126–32.
• Ishiyama H, Hirayama T, Jhaveri P, Satoh T, Paulino AC, Xu B, et al. Is There an Increase in Genitourinary Toxicity in Patients Treated With Transurethral Resection of the Prostate and Radiotherapy?: A Systematic Review. Am J Clin Oncol. 2014 Jun;37(3):297–304.
• Jassem J. Tobacco smoking after diagnosis of cancer: clinical aspects. Transl Lung Cancer Res. 2019 May;8(S1):S50–8.
• Keilani M, Hasenoehrl T, Baumann L, Ristl R, Schwarz M, Marhold M, et al. Effects of resistance exercise in prostate cancer patients: a meta-analysis. Support Care Cancer. 2017 Jun 10;
• Keogh JWL, MacLeod RD. Body Composition, Physical Fitness, Functional Performance, Quality of Life, and Fatigue Benefits of Exercise for Prostate Cancer Patients: A Systematic Review. J Pain Symptom Manage. 2012 Jan;43(1):96–110.
• Keto CJ, Aronson WJ, Terris MK, Presti JC, Kane CJ, Amling CL, et al. Obesity is associated with castration-resistant disease and metastasis in men treated with androgen deprivation therapy after radical prostatectomy: results from the SEARCH database. BJU Int. 2011;110(4):492–8.
• King MT, Keyes M, Frank SJ, Crook JM, Butler WM, Rossi PJ, et al. Low dose rate brachytherapy for primary treatment of localized prostate cancer: A systemic review and executive summary of an evidence-based consensus statement. Brachytherapy. 2021 Nov;20(6):1114–29.
• Kishan A, Kupelian P. Late rectal toxicity after low-dose-rate brachytherapy: incidence, predictors, and management of side effects. Brachytherapy. 2014;14(2):148–59.
• Kubo K, Wadasaki K, Kimura T, Murakami Y, Kajiwara M, Teishima J, et al. Clinical features of prostate-specific antigen bounce after 125I brachytherapy for prostate cancer. J Radiat Res (Tokyo). 2018 Sep 1;59(5):649–55.
• Lardas M, Liew M, van den Bergh RC, De Santis M, Bellmunt J, Van den Broeck T, et al. Quality of Life Outcomes after Primary Treatment for Clinically Localised Prostate Cancer: A Systematic Review. Eur Urol. 2017 Dec;72(6):869–85.
• Larkin D, Lopez V, Aromataris E. Managing cancer-related fatigue in men with prostate cancer: A systematic review of non-pharmacological interventions. Int J Nurs Pract. 2014 Oct;20(5):549–60.
• Lehto US, Tenhola H, Taari K, Aromaa A. Patients’ perceptions of the negative effects following different prostate cancer treatments and the impact on psychological well-being: a nationwide survey. Br J Cancer. 2017 Mar;116(7):864–73.
• Lin PH, Aronson W, Freedland SJ. Nutrition, dietary interventions and prostate cancer: the latest evidence. BMC Med. 2015 Jan 8;13:3.
• Maletzki P, Schwab C, Markart P, Engeler D, Schiefer J, Plasswilm L, et al. Late seed migration after prostate brachytherapy with Iod-125 permanent implants. Prostate Int. 2018 Jun;6(2):66–70.
• McInnis MK, Pukall CF. Sex After Prostate Cancer in Gay and Bisexual Men: A Review of the Literature. Sex Med Rev. 2020 Jul;8(3):466–72.
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• Michalski J, Mutic S, Eichling J, Ahmed SN. Radiation exposure to family and household members after prostate brachytherapy. Int J Radiat Oncol. 2003 Jul;56(3):764–8.
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• Morris KA, Haboubi NY. Pelvic radiation therapy: Between delight and disaster. World J Gastrointest Surg. 2015 Nov 27;7(11):279–88.
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5374 | https://www.anhi.org/content/dam/an/anhi/pdf/muac-instructions-final-pdf.pdf | MUAC Z-SCORE TAPE INSTRUCTIONS FOR HEALTHCARE PROFESSIONALS MUAC Z-Score Tape INSTRUCTIONS FOR USE A HEALTHY, BALANCED DIET MAY HELP REDUCE THE RISK OF MALNUTRITION IN CHILDREN Page 2 of 2 4. Identify the color band corresponding to the specific age of the patient. In the example below, the MUAC is 16.6 cm. This measure correlates the age-based MUAC in centimeters with nutrition risk categories based on the z-score, visually represented by color coded lines. Example of MUAC z-score ranges when measured at 16.6 cm: Age (yrs.) MUAC z‐score range Risk classification 5 0 to ‐1 Normal 7 ‐1 to ‐2 Mild undernutrition 9 ‐2 to ‐3 Moderate undernutrition Note: Your MUAC z-score tapes are for multi-use. Discontinue use if degradation occurs. Reference table for z-score ranges on tape: Color/Pattern Key MUAC z‐score range Risk classification Solid orange 2 to 3 Moderate overnutrition Solid yellow 1 to 2 Mild overnutrition Solid green 0 to 1 Normal Hashed green ‐1 to 0 Normal Hashed yellow ‐2 to ‐1 Mild undernutrition Hashed orange ‐3 to ‐2 Moderate undernutrition Hashed red ‐4 to ‐3 Severe undernutrition Mid-Upper Arm Circumference (MUAC) measures muscle and fat mass and is an important indicator of nutritional status. The MUAC z-score tape is a valuable tool that helps healthcare providers assess nutritional status in both infants and children. You can measure MUAC in three simple steps.
2 to 3 1 to 2 0 to 1 -1 to 0 -2 to -1 -3 to -2 -4 to -3 MUAC Z-score Range Moderate Overnutrition Mild Overnutrition Normal Normal Mild Undernutrition Moderate Undernutrition Severe Undernutrition Risk Classification Solid Orange Solid Yellow Solid Green Hashed Green Hashed Yellow Hashed Orange Hashed Red Color/Pattern Key REFERENCE TABLE FOR Z-SCORE RANGES ON TAPE Age (yrs.) 5 7 9 MUAC Z-score Range -1 to 0 -2 to -1 -3 to -2 Risk Classificaton Normal Mild Undernutrition Moderate Undernutrition PRACTICE USING THE Z-SCORE TOOL In this example, the MUAC is 16.6 cm. The nutritional risk category is determined by using the age of the child to review the proper line. Here are three examples for this measurement.
anhi.org ©2023 Abbott Laboratories 202316161 / October 2023 LITHO IN USA STEP 1: PREPARE YOUR EQUIPMENT AND CREATE A LOOP STEP 2: IDENTIFY THE MIDPOINT OF THE UPPER ARM STEP 2: RECORD YOUR FINDINGS Your MUAC z-score tapes may arrive individually or in a set containing one 2-sided infant tool and one 2-sided child tool. If it comes as a set, you’ll want to separate the individual tapes. Locate the end of the tape and insert the end into slit “A” then through slit “B.” The selected arm should be straight and hanging down at the patient’s side with the elbow fully extended before measuring. The midpoint of the arm is located between the acromion and the olecranon process (or between the shoulder cap and elbow). The centimeter markings on the z-score tape can be used as a ruler to assist with finding the midpoint. Once identified, slide the loop you created up to the midpoint of the extended arm. Pull the tail end of the tape until it is snug but does not compress the skin.
Find and record the MUAC measurement in centimeters as found between the two green arrows while using the age of the child to select the proper measurement. Note both the color and the pattern of the line (solid or hashed). MUAC Z-Score Tape INSTRUCTIONS FOR USE A HEALTHY, BALANCED DIET MAY HELP REDUCE THE RISK OF MALNUTRITION IN CHILDREN Page 1 of 2 1. Your Mid-Upper Arm Circumference (MUAC) z-score tapes may arrive individually or in sets containing one 2-sided infant tool and one 2-sided child tool. If it comes as a set, you will want to separate the individual tapes. 2. Familiarize yourself with the key that orients you to the age group, the colors and the markings. Note that the 2 green arrow heads represent the zero-end of the tape measure, and this is where you read the patient’s MUAC measurement and z-score. To measure, slide the tail end of the tape into slit “A” and back out through slit “B” to create a loop. 3. Identify the midpoint of the upper arm between the acromion and the olecranon process (between the shoulder cap and elbow) and slide the loop of the device up to the midpoint (the ruler on the tape can be used to assist with this process). You should make sure that the elbow is fully extended (in other words, that the arm is straight and hanging down at side) before measuring. Pull the tail end of the tape until it is snug but does not compress the skin to read the MUAC measure and z-score. Continued next page. MUAC Z-Score Tape INSTRUCTIONS FOR USE THY, BALANCED DIET MAY HELP REDUCE THE RISK OF MALNUTRITION IN CHILDREN Page 1 of 2 d-Upper Arm Circumference (MUAC) z-score tapes may arrive individually or in sets ng one 2-sided infant tool and one 2-sided child tool. If it comes as a set, you will want ate the individual tapes. ize yourself with the key that orients you to the age group, the colors and the markings. t the 2 green arrow heads represent the zero-end of the tape measure, and this is ou read the patient’s MUAC measurement and z-score. To measure, slide the tail end pe into slit “A” and back out through slit “B” to create a loop. the midpoint of the upper arm between the acromion and the olecranon process n the shoulder cap and elbow) and slide the loop of the device up to the midpoint (the the tape can be used to assist with this process). You should make sure that the elbow xtended (in other words, that the arm is straight and hanging down at side) before ng. Pull the tail end of the tape until it is snug but does not compress the skin to read AC measure and z-score. ed next page. Note: Your MUAC z-score tapes are for multi-use. Discontinue use if degradation occurs.
Visit anhi.org to view additional MUAC resources Visit muac.abbott to order Z-Score tapes |
5375 | https://ui.adsabs.harvard.edu/abs/2024PhFl...36k5108K/abstract | Conjugate direct resistance heating of metallic plates. multiplicities and stability - Astrophysics Data System
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Conjugate direct resistance heating of metallic plates. multiplicities and stability
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Krikkis, Rizos N. (0)
Abstract
A numerical bifurcation analysis is presented for an industrial application where direct resistance heating through a DC is applied to a flat metallic plate, which is cooled by a turbulent boundary layer and radiation. The process is modeled with a conjugate heat transfer between the plate and the cooling air steam. The convective part of the heat transfer mechanism is formulated in the framework of an integral approach, considering a turbulent core based on power law velocity and temperature profiles and a thin laminar sublayer thermally coupled with axial conduction along the plate. The analysis reveals that the problem admits two solutions: one stable and one unstable, separated by a limit point. The existence of multiple solutions is a consequence of the nonlinear electric resistivity-temperature relationship, allowing thermal equilibrium between heat generation and heat dissipation in multiple points. The application of realistic boundary conditions at the wall-fluid interface shows that the thermal to the hydrodynamic boundary layer thicknesses ratio along the plate is no longer close to the value of 1.25, as it is the case with a constant wall temperature. Instead, significant deviations occur due to the thermal coupling between the wall and the cooling fluid. The multiplicity structure and, consequently, the limit points depend on the plate Reynolds number and on the conduction-convection parameter. The locus of the limit points defines an instability threshold beyond which any excess applied current will trigger a thermal runaway phenomenon. This is also an equivalent of the maximum current carrying capacity of the plate.
Publication:
Physics of Fluids, Volume 36, Issue 11, id.115108, 9 pp.
Pub Date:November 2024 DOI:
10.1063/5.0228617
Bibcode: 2024PhFl...36k5108K Copied!Keywords:
Turbulent Flows
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5376 | https://chemistryguru.com.sg/ideal-gas-law-and-applications | Ideal Gas Law and Applications
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Ideal Gas Law and Applications
In the topic of Gaseous State we have a few types of calculation questions that require manipulation of the Ideal Gas equation into various forms for easy application.
In a previous video we have discussed how to use Ideal Gas equation in graph sketching.
Before we do that let's do a recap on the Ideal Gas Equation, PV = nRT.
where:
P is pressure in Pa,
V is volume in m 3,
n is moles of gas,
R is gas constant 8.31 J K-1 mol-1, and
T is temperature in K.
When we are using the ideal gas equation itself, remember that all the terms have to be in SI units.
Let's move on to the 3 applications that we will be discussing:
1. Boyle's Law at Constant Temperature
Usually the question will give us a fixed mass of gas hence moles of gas will be constant.
There will be a before scenario where pressure and volume are given, and when the pressure (or volume) changes at constant temperature, we are required to determine the new volume (or pressure) in the after scenario.
We can start with the Ideal Gas equation and write out the variable terms on the left hand side of equation and put the constant terms on the right hand side of equation.
In this case at constant temp, PV is a constant hence P 1 V 1 = P 2 V 2, where P 1 and V 1 are the pressure and volume in Scenario 1 and P 2 and V 2 are the pressure and volume in Scenario 2.
Usually there is only 1 unknown and we can use the equation to solve for that unknown.
2. Charle's Law at Constant Pressure
We can do a similar rearrangement of the Ideal Gas equation given pressure is a constant.
Then we can use V 1 / T 1 = V 2 / T 2 to solve for an unknown volume or temperature for a before-and-after scenario type of question.
3. Determine Final Pressure on Mixing 2 Gases at Constant Temperature
Sometimes two different non-reacting gases are mixed and we are required to determine the final pressure.
The formula is not difficult to derive so let's go through that here.
When 2 gases are mixed, the total number of moles is just the summation of moles of the first gas and the moles of the second gas, ie:
n T = n 1 + n 2
We can write moles with respect to the other terms in Ideal Gas equation and do some rearrangement to get the final equation that is very simple to apply, ie:
P T V T = P 1 V 1 + P 2 V 2
We can use the equation to solve for this simple exercise to determine the final pressure to be 3 atm.
For the detailed step-by-step discussion on applications of Ideal Gas equation, check out this video!
Topic: Gaseous State, Physical Chemistry, A Level Chemistry, Singapore
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Checking for interval overlap
October 16, 2011
This’ll be a short post, based on a short exchange I recently had with a colleague at work (hey Per!). It’s about a few tricks for checking interval overlaps that aren’t as well-known as they should be. (Note: This post was originally written for AltDevBlogADay and posted there first).
The problem is very simple: We have two (closed) intervals and and want to find out if they overlap (i.e. have a non-empty intersection). So how would you do it? A simple solution can be obtained directly from the definition: we first compute the intersection of the two intervals, then check whether it’s non-empty.
The code for this is straightforward, and I’ve seen it many times in various codebases:
i0 = max(a0, b0); // lower bound of intersection interval i1 = min(a1, b1); // upper bound of intersection interval return i0 <= i1; // interval non-empty?
However, doing the check this way without a native “max” operation (which most architectures don’t provide for at least some types) requires three comparisons, and sometimes some branching (if the architecture also lacks conditional moves). It’s not a tragedy, but neither is it particularly pretty.
Center and Extent
If you’re dealing with floating-point numbers, one reasonably well-known technique is using center and extents (actually half-extents) for overlap checks; this is the standard technique to check for collisions using Separating Axis Tests, for example. For closed and open intervals, the intuition here is that they can be viewed as 1D balls; instead of describing them by two points on their diameter, you can equivalently express them using their center and radius. It boils down to this:
float center0 = 0.5f (a0 + a1); float center1 = 0.5f (b0 + b1); float radius0 = 0.5f (a1 - a0); float radius1 = 0.5f (b1 - b0); return fabsf(center1 - center0) <= (radius0 + radius1);
This looks like a giant step back. Lots of extra computation! However, if you’re storing (open or closed) intervals, you might as well use the Center-Extent form in the first place, in which case all but the last line go away: a bit of math and just one floating-point compare, which is attractive when FP compares are expensive (as they often are). However, this mechanism doesn’t look like it’s easy to extend to non-floats; what about the multiply by 0.5? The trick here is to realize that all of the expressions involved are linear, so we might as well get rid of the scale by 0.5 altogether:
center0_x2 = a0 + a1; center1_x2 = b0 + b1; radius0_x2 = a1 - a0; radius1_x2 = b1 - b0; return abs(center1_x2 - center0_x2) <= (radius0_x2 + radius1_x2);
No scaling anymore, so this method can be applied to integer types as well as floating-point ones, provided there’s no overflow. In case you’re wondering, there’s efficient branch-less ways to implement abs() for both floats and integers, so there’s no hidden branches here. Of course you can also use a branch-less min/max on the original code snippet, but the end result here will be a bit nicer than that. Because eliminating common terms yields:
t0 = b1 - a0; t1 = a1 - b0; return abs(t0 - t1) <= (t0 + t1);
I thought this was quite pretty when I stumbled upon it for the first time. It only does one comparison, so it’s optimal in that sense, but it still involves a bit of arithmetic.
Rephrasing the question
The key to a better approach is inverting the sense of the question: instead of asking whether two intervals overlap, try to find out when they don’t. Now, intervals don’t have holes. So if two intervals and don’t overlap, that means that must be either fully to the left or fully to the right of on the real number line. Now, if is fully to the left of , that means in particular that b’s rightmost point must be to the left of a – that is, smaller than . And again, vice versa for the right side. So the two intervals don’t overlap if either or . Applying that to our original problem (which involves negating the whole expression using de Morgan's laws), this gives the following version of the interval overlap check:
return a0 <= b1 && b0 <= a1;
Which is about as simple as it gets.
Again, it’s not earth-shattering, but it’s not at all obvious from the original snippet above that the test can be done this way, so the longer version tends to come up quite often – even when the max and min operations use branches. It’s generally worthwhile to know all three forms so you can use them when appropriate – hence this post.
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From → Coding, Maths
4 Comments
doynax permalink
Personally I always thought exploiting unsigned comparisons was cute, e.g.:(unsigned) b1 – a2 <= (b1 – a1) – (b2 – a2)
My fourteen-year-old self was amazingly pleased with himself for figuring this out and using it to speed up collision detection, while still doing the detection the N^2 way of course ;)
Reply
2. dithermaster permalink
That’s funny, I had exactly the same problem and solved it the same way. A friend was doing a scheduling application and was stuck trying to figure out if two appointments overlapped. After listing a bunch of ways they could overlap and considering what the comparison operators might look like, I had the flash of insight to instead consider the two simple case where they don’t overlap: one ends before or starts after the other. Done!
Reply
3. Bob Davis permalink
Check logic in last section re the right side, that is the comparison for second interval being right of the first; should be a1 > b0 or equivalently, b0 = a1 rather than b0 <= a1.
Reply
fgiesen permalink
That’s wrong, I’m explicitly using closed not half-open intervals.
Reply
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5378 | https://chem.libretexts.org/Courses/Chabot_College/Introduction_to_General_Organic_and_Biochemistry/09%3A_Solutions/9.10%3A_Osmosis_and_Osmotic_Pressure | Skip to main content
9.10: Osmosis and Osmotic Pressure
Last updated
: Aug 11, 2022
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9.9: Properties of Solutions
9.11: Dialysis
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: 402323
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Describe osmosis and how it relates to osmotic pressure.
Osmotic Pressure
The last colligative property of solutions we will consider is a very important one for biological systems. It involves osmosis, the process by which solvent molecules can pass through certain membranes but solute particles cannot. When two solutions of different concentration are present on either side of these membranes (called semipermeable membranes), there is a tendency for solvent molecules to move from the more dilute solution to the more concentrated solution until the concentrations of the two solutions are equal. This tendency is called osmotic pressure. External pressure can be exerted on a solution to counter the flow of solvent; the pressure required to halt the osmosis of a solvent is equal to the osmotic pressure of the solution.
Consider the apparatus illustrated in Figure 9.10.1, in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of solvent is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as osmosis.
When osmosis is carried out in an apparatus like that shown in Figure 9.10.1, the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure (π) of the solution. The osmotic pressure of a dilute solution can be determined in a similar way the pressure of an ideal gas is calculated using the ideal gas law:
π=(nV)RT
where n is the number of moles of particles in solution, V is the volume, and R is the universal gas constant.
Osmolarity (osmol) is a way of reporting the total number of particles in a solution to determine osmotic pressure. It is defined as the molarity of a solute times the number of particles a formula unit of the solute makes when it dissolves (represented by i):
osmol=M×i
If more than one solute is present in a solution, the individual osmolarities are additive to get the total osmolarity of the solution. Solutions that have the same osmolarity have the same osmotic pressure. If solutions of differing osmolarities are present on opposite sides of a semipermeable membrane, solvent will transfer from the lower-osmolarity solution to the higher-osmolarity solution. Counterpressure exerted on the high-osmolarity solution will reduce or halt the solvent transfer. An even higher pressure can be exerted to force solvent from the high-osmolarity solution to the low-osmolarity solution, a process called reverse osmosis. Reverse osmosis is used to make potable water from saltwater where sources of fresh water are scarce.
Example 9.10.1
A 0.50 M NaCl aqueous solution and a 0.30 M Ca(NO3)2 aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow.
Solution
The solvent will flow into the solution of higher osmolarity. The NaCl solute separates into two ions—Na+ and Cl−—when it dissolves, so its osmolarity is as follows:
osmol (NaCl) = 0.50 M × 2 = 1.0 osmol
The Ca(NO3)2 solute separates into three ions—one Ca2+ and two NO3−—when it dissolves, so its osmolarity is as follows:
osmol [Ca(NO3)2] = 0.30 M × 3 = 0.90 osmol
The osmolarity of the Ca(NO3)2 solution is lower than that of the NaCl solution, so water will transfer through the membrane from the Ca(NO3)2 solution to the NaCl solution.
Exercise 9.10.1
A 1.5 M C6H12O6 aqueous solution and a 0.40 M Al(NO3)3 aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow.
Answer
: osmol C6H12O6 = 1.5; osmol Al(NO3)3 = 1.6
The solvent flows from C6H12O6solution (lower osmolarity) to Al(NO3)3 solution (higher osmolarity).
Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture (Figure 9.10.2a). This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation (Figure 9.10.2c). Only if red blood cells are placed in isotonic solutions that have the same osmolarity as exists inside the cells are they unaffected by negative effects of osmotic pressure (Figure 9.10.2b). Glucose solutions of about 0.31 M, or sodium chloride (NaCl) solutions of about 0.16 M, are isotonic with blood plasma.
Note: Isotonic Solutions for Red Blood Cells
The concentration of an red blood cell isotonic solution made with sodium chloride (NaCl) is half that of an isotonic solution made with glucose (0.16 M and 0.31 M respectively). This is because NaCl produces two ions when a formula unit dissolves, while molecular glucose produces only one particle when a formula unit dissolves. The osmolarities are therefore the same even though the concentrations of the two solutions are different.
isotonic NaCl solution: osmol=0.16 M×2=0.32 osmol/L
isotonic glucose solution: osmol=0.31 M×1=0.31 osmol/L
Osmotic pressure explains why you should not drink seawater if you are abandoned in a life raft in the middle of the ocean. Its osmolarity is about three times higher than most bodily fluids. You would actually become thirstier as water from your cells was drawn out to dilute the salty ocean water you ingested. Our bodies do a better job coping with hypotonic solutions than with hypertonic ones. The excess water is collected by our kidneys and excreted.
Osmotic pressure effects are used in the food industry to make pickles from cucumbers and other vegetables and in brining meat to make corned beef. It is also a factor in the mechanism of getting water from the roots to the tops of trees!
9.9: Properties of Solutions
9.11: Dialysis |
5379 | https://www.math.ucdavis.edu/~deloera/RECENT_WORK/semesterberichte.pdf | Mathematische Semesterberichte manuscript No.
(will be inserted by the editor) Jes´ us A. De Loera The Many Aspects of Counting Lattice Points in Polytopes ⋆ Abstract. A wide variety of topics in pure and applied mathematics involve the problem of counting the number of lattice points inside a convex bounded polyhedron, for short called a polytope. Applications range from the very pure (number theory, toric Hilbert functions, Kostant’s partition function in representation theory) to the most applied (cryptography, integer programming, contingency tables). This paper is a survey of this problem and its applications. We review the basic structure theorems about this type of counting problem. Perhaps the most famous special case is the theory of Ehrhart polynomials, introduced in the 1960s by Eugene Ehrhart. These polynomials count the number of lattice points in the different integral dilations of an integral convex polytope. We discuss recent algorithmic solutions to this problem and conclude with a look at what happens when trying to count lattice points in more complicated regions of space.
1. Introduction The main actors of this story are convex polytopes (e.g. cubes, triangles, and their high dimensional analogues) and the points with integer coordinates that lie inside them, i.e. their lattice points. The convex polytopes we care about here will most likely be encoded in a computer, and for this reason they will be represented as the sets of non-negative solutions of a system of linear equations Ax = b, for suitable integral matrix A and vector b. This is not at all a restrictive formulation; any system of linear inequalities and equalities with rational coefficients can be put into this form (the books [7, 53,60] are good references for many concepts used here). We will start now setting up our mathematical tale: Given a polytope P = {x : Ax = b, x ≥0}, specified by d × n integral matrix A and an integral d-vector b, i.e. P = {x : Ax = b, x ≥0}, our goal ⋆The author gratefully acknowledges the financial support received from NSF grant DMS-0309694, an Alexander von Humboldt fellowship, and the UC Davis Chancellor’s fellow Award.
Jes´ us A. De Loera: Department of Mathematics, University of California, Davis CA 95616-8633, U.S.A. email: deloera@math.ucdavis.edu 2 J. A. De Loera is to count how many lattice points are inside P. We are in fact interested on the counting function φA(b) = #{x : Ax = b, x ≥0, x integral}.
This is the function that counts the number of lattice points inside con-vex polytopes given in terms of a fixed matrix A and a right-hand-side vector b that is possibly changing or consists of variables. In the first part of this survey I hope to convince you that the functions φA are very useful and natural. As we will see applied mathematicians often work with fixed b and would perhaps prefer fast evaluation of φA(b) for fixed values of b.
Pure mathematicians, on the other hand, would perhaps prefer to compute explicit exact formulas in terms of variables bi.
Take the following example. If A = [3, 5, 17] then we have a plane inter-secting the nonnegative orthant, as presented in Figure 1, the polytope P is a triangle (embedded in three-dimensional space). It gives us a function φA(n) = #{(x, y, z) : 3x + 5y + 17z = n, x ≥0, y ≥0, z ≥0}. Thus, for example, φA(58) = 9, φA(101) = 25, φA(1110) = 2471, etc. For us a fine general formula for φA(b) could be the generating function whose terms encode the different values of φA: ∞ X n=0 φA(n)tn = 1 (1 −t17) (1 −t5) (1 −t3).
x y z A=[3,5,17] Fig. 1. Lattice points on a spatial triangle.
In the second part of the survey, I will outline the structure theory of the counting functions φA(b) and some useful algorithms to compute φA(b) involving rational functions. By the way, whenever I say counting, I mean exact counting. There is a rich and exciting theory of estimation and The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 3 approximation, but that is a different story (see for example [27,38,59]).
Now we present 2. Applications 2.1. in Applied Mathematics: – Let us begin with two puzzles about coins and cars. A curious cashier may wonder: How many ways are there to give change of 10 dollars into pennies (p), nickels (n), dimes (d), quarters (q), using exactly 100 coins? The answer is equivalent to the number of solutions of the system of linear equations p + 5n + 10d + 25q = 10000, p + n + d + q = 100, where p, n, d, q are nonnegative integral numbers. This is a reasonable exercise, but do not be fooled into thinking that problems with few constraints are always easier than those with many equations. Even with just one single row for the matrix A there are hard instances that cannot be solved so by mere brute force enumeration (i.e. trying to list all possibilities). For instance, from the article , a very hard instance is {(x, y, z, w, v) ∈R5 ≥0 : 12223x+12224y +36674z +61119w+85569v = 89643481}.
We can imagine x, y, x, w, v are the five (integral) quantities of cars or trucks, from five different models, with weights (in kilograms) 12223, 12224, etc. and we are supposed to completely fill in a ferry boat with maximal capacity of 89643481 kilograms. How many ways are there to do this? This is like filling a sack with objects up to the maximum ca-pacity; this is why problems with a single-row matrix receive the name of knapsack problems. Knapsack problems are geometrically the simplest possible since their polytopes are simplices, the generalization of trian-gles for higher dimensions. Note that the counting problem can easily be turned into an optimization problem if we just ask: How can we fit the most cars?
– Continuing with optimization problems. Suppose you have a computer network between Berlin and Magdeburg. You are interested to send as many email messages as possible given its layout. In Figure 2 (A), Berlin and Magdeburg are represented as nodes 1 and 6 respectively. The other four nodes represent nearby cities within the network. Every time we send an e-mail, if it enters a node, it has to leave that node (conditions bi = 0 means no loss of information). There is also some capacity as-sociated to each connecting phone line or internet cable that limits the number of messages we can send (e.g. between node 2 and 4 there is a capacity c24 of 5 messages).
4 J. A. De Loera 1 b =v 6 b=−v 2 1 2 3 4 5 6 b = 0 8 3 b = 0 b = 0 b = 0 3 24 34 c = 23 13 c = 35 56 5 45 46 12 c = 7 c = 5 c =6 c = 9 c = 4 4 c =3 c =4 Fig. 2. A simple network.
1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 5 6 0 5 3 4 4 7 7 4 2 5 3 3 3 3 5 6 6 5 1 5 3 3 4 4 7 5 3 3 5 3 3 5 3 3 5 6 0 2 6 5 6 5 1 2 6 5 7 4 2 4 4 7 5 6 0 5 3 3 3 5 6 7 4 2 5 3 3 2 6 5 6 5 1 5 3 3 3 5 6 Fig. 3. All integral maximum flows in the network.
Classical and widely-known theorems indicate that the largest num-ber of email messages one can send for this example is 11. But how many distinct optimal solutions are there? It turns out that there are exactly 9 ways to route 11 emails in this small network (see Figure 3 (B)). In general, if G is a network with n nodes and m arcs, with integer-valued capacities for the arcs and excess function for the nodes (namely c : arcs(G) →Z≥0 and b : nodes(G) →Z). A flow is a function f : arcs(G) →Z≥0 so that, for any node x, the sum of flow values in outgoing arcs minus the sum of values in incoming arcs equals b(x), and 0 ≤f(i, j) ≤c(i, j). We may be interested on knowing all maxi-mal integral-valued flows or at least many of them. This is equivalent to counting the number of lattice points of a flow polytope (see and references therein). This may be relevant for the security of communica-tion networks, if we have only one or a few optimal solutions, this may indicate some kind of weakness in case of an attack .
– Suppose you are a compiler designer, then you would typically worry about how often is a certain instruction I of the computer code executed?
Take for instance the following toy example (this program has really no The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 5 particular meaning, it simply repeats instruction I a few times depending on values of N and M): void proc(int N, int M) { int i,j; for (i=2N-M; i<= 4N+M-min(N,M), i++) for(j=0; j<N-2i; j++) I; } Clearly, the number of times we reach instruction I would depend on N,M. In terms of these parameters the set of all possible solutions is given by the number of lattice points inside of parametric polygons. In our toy example these are described by the linear conditions {(i, j) ∈Z2 : i ≥2N−M, i ≤4N+M−min(N, M), j ≥0, j−2i ≤N−1}.
The answers to many questions about the computer performance or resource allocation available while compiling code with nested loops de-pends on counting lattice points of polytopes. Examples include how many memory locations are touched during a loop? How many cache misses does a loop generate? How much memory is dynamically allo-cated by a portion of code? See more details and references in [20,33, 45].
– Statistics is a topic where integral data tables are studied and generated a plenty. Take for instance the following 3-dimensional table of format 2×3×3 given below. Table 1 has been extracted from the 1990 decennial USA census. Visit the web page of the U.S. census
census.gov/ where you can extract your own contingency tables!
There are a variety of statistical questions where one may consider the “totals margins” (e.g. total number of white male respondents was 329, the number of chinese females is one, etc.) as fixed and we may wish to study the space of all integral tables that satisfy these sum restrictions on their entries. For example, counting those tables is closely related to the problem of sampling such a table uniformly at random or with respect to other probability distribution model. Another application is when we want to decide whether two of the traits of the statistic, say, income and race, are independent of each other. Several tests of inde-pendence depend on counting the integral tables in question. See [25, 30] and its references. The spaces of tables with fixed marginal totals are again polytopes, the so called transportation polytopes. The lattice points inside transportation polytopes receive the name of contingency tables. For example, if we consider our census table above, there are 18 entries possible depending on gender (f, m), race (w, b, c), and income 6 J. A. De Loera Gender = Male Income Level Race ≤$10, 000 > $10000 and ≤$25000 > $25000 Total White 96 72 161 329 Black 10 7 6 23 Chinese 1 1 2 4 Total 107 80 169 356 Gender = Female Income Level Race ≤$10, 000 > $10000 and ≤$25000 > $25000 Total White 186 127 51 364 Black 11 7 3 21 Chinese 0 1 0 1 Total 197 135 54 386 Table 1. Three-way cross-classification of Gender, Race, and Income for a selected U.S. census tract. Source: 1990 Census Public Use Microdata Files.
values (1, 2 or 3). Since we fixed the “line” sums, we have, for example, a linear equation xm,w,1 + xm,w,2 + xm,w,3 = 329 that explains that the number of white males is 329 when we add the three categories of income. There is a similar equation xm,b,2 + xf,b,2 = 14 that indicates the number of blacks (female and male) with middle income must be 14. Altogether we will have 21 linear equations and 18 non-negative variables (one for each combination of possible gender, race, income) describing the polytope. How many integral points does it contain?
2.2. in Pure Mathematics – Many combinatorial structures can be counted as lattice points of poly-topes. For example, matchings on graphs , t-designs , linear ex-tensions of posets , and of course magic squares! Magic squares are square arrays of nonnegative integer numbers where the sums along each row, column, or diagonal is a constant, the so called magic sum. In Figure 4 we see an example of a 4 × 4 magic square with magic sum 24.
How many 4×4 magic squares with sum s are there? Denote by f4×4(s) the number of 4 × 4 magic squares with magic sum s. Using the latest methods for counting lattice points we can present a nice formula for f4×4(s) (and other magic objects ): The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 7 12 0 5 7 0 12 7 5 7 5 0 12 5 7 12 0 Fig. 4. A 4 × 4 magic square.
1 480 s7 + 7 240 s6 + 89 480 s5 + 11 16 s4 + 49 30 s3 + 38 15 s2 + 71 30 s + 1 if 2|s, 1 480 s7 + 7 240 s6 + 89 480 s5 + 11 16 s4 + 779 480 s3 + 593 240 s2 + 1051 480 s + 13 16 if 2 ̸ |s.
In other words f4×4(1) = 8,f4×4(2) = 48, f4×4(3) = 200, f4×4(4) = 675, f4×4(5) = 1904, etc. All values can be read from the two-case formula, which is applied according with the parity of the magic sum.
This formula, given in finitely many polynomial pieces, is a good example of a quasipolynomial . We will come back to this later. The important thing to notice right now is that f4×4(s) is again one of our functions φA(b) for a polytope which is almost a transportation polytope: The matrix A is read offfrom the row, column, and diagonal sums for the entries of a 4 × 4 magic array. Our polytope lives in sixteen dimensional space, and b is just s(1, 1, . . . , 1).
– Researchers have observed that polytopes play a special role in represen-tation theory (e.g. [16,21,41,42]). One example of such polytopes comes from the Gelfand-Tsetlin patterns which arise in the representation the-ory of glnC. What is a Gelfand Tsetlin pattern? It is a triangular array of numbers x = (xij)1≤i≤j≤n satisfying that xij ≥0, for 1 ≤i ≤j ≤n; and xi,j+1 ≥xij ≥xi+1,j+1, for 1 ≤i ≤j ≤n −1. xin = λi if 1 ≤i ≤n, for λ ∈Rn. Given λ, µ ∈Zn, the Gelfand–Tsetlin polytope GT (λ, µ) ⊂Xn is the convex polytope of GT-patterns (xij)1≤i≤j≤n satisfying in addition that x11 = µ1; and Pj i=1 xij −Pj−1 i=1 xi,j−1 = µj, for 2 ≤j ≤n.
The importance of GT-polytopes stems from a classic result of I. M.
Gelfand and M. L. Tsetlin, which states that the number of integral lattice points in the Gelfand–Tsetlin polytope GT (λ, µ) equals the di-mension of the weight µ subspace of the irreducible representation of glnC with highest weight λ. If you are not familiar with representation theory let me also say that the lattice points of Gelfand-Tsetlin polytopes also have a nice combinatorial interpretation as semi-standard tableaux 8 J. A. De Loera of of shape λ and content µ. The number of such tableaux is a Kostka number.
1 1 2 2 3 1 1 2 2 3 Fig. 5. An example of two tableaux with content (2, 2, 1) and shape (3, 2).
It is well known that the structure of a complex semisimple Lie algebra g can be obtained from the discrete geometric information given by the root system . Polytopes derived from the root system of g and their lattice points appear in the study of a tensor product decomposition into irreducibles. If we consider a simple g-module Vλ with highest weight λ results of Berenstein and Zelevinsky indicate that the multiplicity of the simple module Vλ in the tensor product Vµ ⊗Vν for given dominant weights λ, µ, ν is equal to the number of lattice points inside certain polytopes. Also related are the Kostant’s partition function which ap-pear in many classical formulas such as Steinberg’s formula . They can also be seen naturally as counting lattice points of certain polytopes.
– Another area of application is computational commutative algebra .
Given a finite set of monomials m1,m2,..., mk in C[x1, x2, . . . , xn] one can consider the ring C[m1, m2, . . . , mk] generated by those monomials (i.e. like the usual polynomial ring but powers of the monomials mi replace powers of variables). Algebraists are interested on the Hilbert series of this ring with respect to various gradings (ways of measuring the degree of a monomial). The Hilbert series are generating functions for the vector space dimensions of the graded pieces of the ring. Barvinok and Woods have recently shown that Hilbert series of a monomial algebra for the finest Zn grading can be computed in polynomial time, via lattice point counting techniques, when the number of variables is fixed. Yet another connection to commutative algebra is the computation of Gr¨ obner bases of toric ideals . These, like usual vector space bases, allow many important calculations about the ideal.
3. The Main Structure Theory By now I hope you are convinced of the interesting nature of the count-ing function φA(b), and maybe you wish to learn more about it. I will present what I would call the structure theory of these functions. The functions φA(b) are so basic and natural that one can easily believe their The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 9 study goes back to Gauss, Euler or even earlier. Surprisingly the first di-rect investigations of φA(b) date back to the 1960’s work of Eug ene Ehrhart (1906-2000) [28,29] (with Pick’s theorem as an sporadic earlier particular case). Ehrhart was a man with artistic and mathematical talents. As a lyc´ ee teacher he did many of his investigations as an amateur mathemati-cian. To read more about him see the tribute written by Philippe Clauss
Fig. 6. Eugene Ehrhart (1906-2000).
Ehrhart began by considering the case when b changes as a single-parameter dilation, i.e. b = λv for a fixed vector v. Geometrically this can be interpreted as inflating or dilating the associated polytope while leaving the angles and proportions fixed. More precisely, let P be a convex poly-tope in Rd. For each integer n ≥1, define the dilation of P by n as the polyhedron nP = {nq : q ∈P} (see Figure 7). Thus for P a d-polytope, Ehrhart studied the particular case of φA(b) given by the counting func-tion i(P, n) = #(nP ∩Zd) = #{q ∈P : nq ∈Zd}. This is the number of lattice points in the dilation nP of P. Similarly, if P ◦denotes the relative interior of P, i.e. i(P ◦, n) = #{q ∈P −∂P : nq ∈Zd}. Let us take the simplest ever example, a unit square (see Figure 7). the reader can easily verify i(P, n) = (n+1)2 and i(P ◦, n) = (n−1)2. In fact, for a d-dimensional unit cube we have i(P, n) = (n + 1)d.
10 J. A. De Loera P 3P Fig. 7. Dilating a unit square.
But if we just slightly translate the square to have non-integer coordi-nates, say vertices with coordinates −1/2, 1/2, the task of computing i(P, n) becomes more complicated. The correct description is given by the following theorem (see [29,47,57] and the many references therein).
Theorem 1. Let P be a convex d-dimensional rational polytope. Then for the univariate function i(P, n) there exists an integer N > 0 and polynomi-als f0, f1, . . . , fN−1 such that i(P, n) = fi(n) if n ≡i modulo N. We say that i(P, n) is the Ehrhart quasipolynomial of P, in the dilation variable n.
Moreover the non-zero polynomial components fi have degree d = dim(P) and their leading term equals the volume of P. When the coordinates of the vertices of P are integers i(P, n) is given by a single polynomial.
Why is the leading coefficient equal to the volume? Intuitively, dilating the polytope P is equivalent to leaving P as it is but refining the grid size of the lattice points. When we count lattice points we are essentially adding the volumes of tiny unit cubes centered at those lattice points which, as we refine the grid, approximate better and better the volume of the polytope: the main principle of the Riemann integral! A lot of the more recent work about Ehrhart counting functions was motivated by trying to understand the lower degree coefficients of the polynomial components. See for instance [12,14,35,47,55]. It is also important to mention that it can certainly happen that a polytope with non-integral coordinates has an Ehrhart polynomial .
One possible generalization of Ehrhart’s result is to count the lattice points with weights coming from a weight function f instead of count-ing them as one, in other words one considers the function i(P, f, n) = P α∈nP∩Zd f(α). For example, if the polytope P is the unit square [0, 1]2, and the weight function f(x, y) is a polynomial of the form xkyk with 0 ≤k ≤3, the corresponding weighted Ehrhart functions are The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 11 i(P, n) =n2 + 2n + 1 = (n + 1)2, i(P, xy, n) = 1 4 n4 + 1 2 n3 + 1 4 n2, i(P, x2y2, n) = 1 9 n6 + 1 3 n5 + 13 36 n4 + 1 6 n3 + 1 36 n2, i(P, x3y3, n) = 1 16 n8 + 1 4 n7 + 3 8 n6 + 1 4 n5 + 1 16 n4.
The degree of the polynomial i(P, xkyk) is 2k + 2, in general, it will be the dimension of the polytope plus the degree of the weight function. What is the leading coefficient now? Can you guess? See for details.
Now we are interested on another generalization, when φA(b) is a mul-tivariable function, b can change differently in all entries. We present now a fairly general structure theorem (see [15,58]). To understand the statement the reader must think of cones. Given a finite set of vectors S, the cone cone(S) is the collection of all nonnegative linear combinations of elements of S. We call the set S the generators of the cone. Intuitively, cones look like ice cream cones, but with flat faces; see Figure 12 for a picture of a two-dimensional cone. When a cone does not contain a linear subspace we say the cone is pointed. In that case, we have a minimum set of genera-tors, which we call the rays of the cone. As practice to familiarize yourself with cones, observe that the parameter space for φA(b) is cone(A); only for b ∈cone(A) the function φA(b) can be non-zero.
Theorem 2. For a d × n integral matrix A and a parameter vector b ∈ cone(A), there exist a finite decomposition of Zd ∩cone(A) such that φA is a multivariate polynomial of degree n −d in each piece. The number n −d is the dimension of the polytope {x : Ax = bx ≥0}.
More precisely, cone(A) can be decomposed into convex polyhedral sub-cones of cone(A), called chambers, such that, for all integral vectors b inside a chamber the function φA(b) can be written as a fixed polynomial function of degree n −d in the variables b1, . . . , bd plus a “correction polynomial” of smaller degree. The correction terms depend periodically on the values of b1, b2, . . . , bd.
Now we present two examples to illustrate Theorem 2. Let us begin with the 3 × 6 matrix 12 J. A. De Loera 1 2 3 4 5 6 Fig. 8. A two-dimensional slice of cone(A), example one.
A = 2 1 1 0 0 0 0 1 0 2 1 0 0 0 1 0 1 2 .
It is not a surprise that some “periodic” or “modular” components must appear in the final formulas for φA. For example, observe that any vector b = (b1, b2, b3) which is a non-negative integer combination of the columns of A must satisfy b1 +b2 +b3 ≡0, mod 2 because all columns of A do. The cone cone(A) is three dimensional, corresponding to the three entries of b.
For the sake of being able to draw it, I showed only a 2-dimensional slice of the cone marking the 6 generating columns of A with the numbers 1 to 6.
See Figure 8.
The theorem says there should be finitely many polynomials describing φA. The cone is decomposed into 12 chambers, in this example the regions are triangular cones (in Figure 8 represented by triangles). Only 3 polynomi-als concern the formula of φA for those vectors inside the left-side chamber shaded in the picture; these are: φA(b1, b2, b3) = b2b3 2 + b2b2 3 8 −b2 3 24 + correction where correction = 1 + b2 2 + 2b3 3 if b1 ≡0 and b2 ≡0 mod 2, 1 2 + b2 2 + 5b3 12 if b1 ≡1 and b2 ≡1 mod 2, 1 2 + 3b2 8 + 13b3 24 otherwise.
The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 13 1 2 3 4 e−e e−e e−e e−e e−e e−e 1 3 2 3 1 4 1 2 3 2 4 4 Fig. 9. The digraph that gives matrix A for example two and the seven chambers for φA (a two-dimensional representation).
Figure 9 shows a second example where the matrix A is very special.
A is defined to be the node-arc incidence matrix of the digraph on the left of the figure. Thus A is given by the pairwise differences of standard unit vectors ei. For instance the last column of A is just e3 −e4 (in this way we marked the six columns of the matrix in Figure 9). Remarkably, in this example there is a single polynomial (no correction terms) in each chamber.
Therefore seven polynomials define the whole counting function φA, one per triangular cone in the figure (there only represented by a triangle). The reason for this is that the matrix A is totally unimodular . In particular, the vertices of the polytopes {x : Ax = b, x ≥0} are integral for all integral b. If you recall Theorem 1, you can see φA(tb) is actually a polynomial (not a quasipolynomial) for any fixed b in the variable t. Of the seven polynomials, take as an example the one for the biggest triangular cone region, with corners e1 −e2, e1 −e4, and e3 −e4, it is φA(b) = (b1 + b2 + 3)(b1 + b2 + 2)(b1 + b2 + 1)/6.
This is valid for all vectors integral b within this triangular cone.
What are these mysterious chambers? Their structure is rather pretty and worth studying further. Let ∆be the set of column vectors of the input matrix A (defining our polytope), this means we consider the column vectors of the matrix A without multiplicities. The chamber complex is a special polyhedral subdivision of the cone generated by the nonnegative linear combinations of elements in ∆, denoted by C(∆). It is defined as the common refinement of the simplicial cones C(σ) running over all possible subsets of ∆which form a basis for Rd. More precisely, denote by Σ∆the set of all simplicial cones whose extreme rays, are generated by elements of ∆, If σ is an element of Σ∆and ∂σ denotes its boundary and ∂Σ∆the union of the boundaries of all simplicial cones, then the complement of Σ∆ inside C(∆) consists of the disjoint union of open convex cells. See Figure 14 J. A. De Loera 10 for a two-dimensional slice view of a chamber complex of a pentagonal cone. The refinement we described gives eleven chambers or regions. One can think of them as the result of overlapping all the possible triangulations of the cone. One important fact is that finding the chamber that contains a particular vector b is equivalent to enumerating the vertices of the polytope {x : Ax = b, x ≥0}.
Fig. 10. The construction of the chamber complex of a pentagonal cone (shown as a 2-dimensional slice).
4. Actually Counting: Modern Algorithms and Software Counting lattice points is difficult: We know that when the dimension is an input variable the problem of detecting a lattice point in polyhedra is NP-hard . Already counting 2 × n contingency tables is #P-hard .
Branch-and-Bound techniques and exhaustive enumeration work to some extent, but the method gets stuck in surprisingly small problems, e.g. hard knapsack problems. Exhaustive enumeration techniques are not useful to derive formulas as they depend on the size of the right-hand-side vector b.
During the 1980’s and 1990’s a new breed of algorithms and counting formulas were created. They rely on deeper algebraic and analytic properties The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 15 of polyhedra. Some authors emphasize the more algebraic-geometric nature of the lattice points of polyhedra via the theory of toric varieties, while others make a much stronger use of complex analysis and convexity (see [9, 6,26,40,43,50,54,52] and the many references within).
Several researchers have even produced software for counting lattice points in polyhedra. For instance, Lisonek created a Maple package that works on knapsack type problems . The Beck-Pixton software can perform fast computations with two-dimensional transportation polytopes.
They have the best computation times for that family of polytopes. Vergne and collaborators (see ) have developed a package special for unimod-ular matrices which is very fast. Working with general polytopes we have two programs: one is ehrhart, developed by Clauss, Loechner and Wilde ( The first implementa-tion of Barvinok’s algorithm was the software LattE developed at UC Davis [23,24]. LattE counts lattice points, generates Ehrhart’s quasipolynomials, and even solves small integer programs. Now we wish to sketch some of the ideas involved in some the algorithms. As the reader will see we have a bias for the use of rational functions. We sketch Barvinok’s algorithm first, then we sketch and refer to other algorithms that use again rational functions. As of today, these rational-function-based algorithms are the fastest available.
4.1. Barvinok’s algorithm In 1993 A. Barvinok [8,9] found an algorithm to count integer points in-side polyhedra. When the dimension is fixed the algorithm can count the number of lattice points in a polytope in polynomial time on the size of the input (the size is given by the binary encoding of the data). The key ideas are using rational functions as efficient data structures and the uni-modular signed decomposition of polyhedra. Given a convex polyhedron P (not necessarily a polytope anymore!), we wish to compute the multivariate generating function f(P; x) = X α∈P∩Zd xα, where xα = xα1 1 xα2 2 . . . xαd d . This is an infinite formal power series if P is not bounded, but if P is a polytope it is a (Laurent) polynomial with one monomial per lattice point. For example, if we consider a rectangle with vertices V1 = (0, 0), V2 = (5000, 0), V3 = (0, 5000), and V4 = (5000, 5000) the generating function f(P) has over 25,000,000 monomials, f(P, z1, z2) = 1 + z1 + z2 + z1z2 2 + z2 1z2 + · · · + z5000 1 z5000 2 .
The representation of f(P; z1, z2) as monomials is clearly way too long to be of practical use. But Barvinok’s method instead rewrites it as a compact 16 J. A. De Loera sum of rational functions. For instance, only four rational functions suffice to represent the over 25 million monomials. Indeed f(P, z1, z2) equals 1 (1 −z1) (1 −z2)+ z1 5000 (1 −z1−1) (1 −z2)+ z2 5000 (1 −z2−1) (1 −z1)+ z1 5000z2 5000 (1 −z1−1) (1 −z2−1).
Note that if we wish to know the number of lattice points in P, and we knew f(P; z), we could compute it as the limit when the vector (x1, . . . , xn) goes to (1, 1, . . ., 1). Similarly the maximum of a linear functional over the lattice points equals the highest degree of the univariate polynomial one gets after doing a monomial substitution xi →tci (See ). These two calculations can be difficult because to the poles of the rational functions.
One uses complex analysis (residue calculations) to find the answer.
A beautiful theorem of M. Brion says that to compute the rational function representation of f(P; z) it is enough to do it for tangent cones at each vertex of P. Let P be a convex polytope and let V (P) be the vertex set of P. Let Kv be the tangent cone at v ∈V (P). This is the (possibly translated) cone defined by the facets touching vertex v (see Figure 11).
Then the following nice formula holds: Kv v Fig. 11. The tangent cone at vertex v f(P; z) = X v∈V (P) f(Kv; z).
Since it is enough to compute everything for cones, we first explain how to compute the rational function for the “smallest” cones, simple cones. A cone is said to be simple if its rays are linearly independent vectors. For The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 17 Fig. 12. A two dimensional cone (its vertex at the origin) and its fundamental parallelepiped instance all the tangent cones of the pentagon of Figure 11 are simple.
Obtaining the rational function representation of the lattice points of a simple cone K ⊂Rd, is easy (see all details in or in Chapter IV of ).
Here is the formula, that one can write directly from the coordinates of the rays of the cone and its fundamental parallelepiped Π: f(K; z) = P u∈Π∩Zd zu (1 −zc1)(1 −zc2) . . . (1 −zcd).
Here Π is the half open parallelepiped {x : x = α1c1 + · · · + αdcd, 0 ≤ αi < 1}. We can do a two-dimensional example shown in Figure 12: we have d = 2 and c1 = (1, 2), c2 = (4, −1). We have: f(K; z) = z4 1z2 + z3 1z2 + z2 1z2 + z1z2 + z4 1 + z3 1 + z2 1 + z1 + 1 (1 −z1z2 2)(1 −z4 1z−1 2 ) .
But what to do if the cone K is not simple? Break it into simple cones!
The wonderful idea of A. Barvinok was noting that, although triangulating the cone K may be an inefficient way to subdivide the cone (i.e. expo-nentially many pieces), if one is willing to add and substract cones for fix dimension d, there exists a polynomial time algorithm which decomposes a rational polyhedral cone K ⊂Rd into simple unimodular cones Ki. A simple cone is unimodular if its fundamental parallelepiped there is a single lattice point. See Figure 13. In fact, via the decomposition, with numbers ϵi ∈{−1, 1} we can write an expression 18 J. A. De Loera f(K; z) = X i∈I ϵif(Ki; z), |I| < ∞.
The index set I is of size polynomial in the input data as long as we work in fixed dimension.
+ + + + + cone, then First triangulate Each simple cone gets further decompose by a plus−minus sum of unimodular cones − Fig. 13. The signed decomposition of a cone into unimodular cones. Two general steps 4.2. More methods using rational functions Recall φA(b) = #{x : Ax = b, x ≥0, integer}. If we denote by Ai the columns of the matrix A, then it is easy to see that The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 19 X φA(b)zb = 1 Qn j=1(1 −zAj).
It is also fairly easy to see that: X b∈cone(A)∩Zn φA(b)e−⟨b,z⟩= 1 Q Ai∈A 1 −e−⟨Ai,z⟩.
How to invert this equations to recover φA? Some authors have taken a complex analytic view. For instance, in the integral is evaluated by repeated application of Cauchy’s integral theorem and clever manipulations.
φA(b) = 1 (2πi)m Z |z1|=ϵ1 · · · Z |zm|=ϵm z−b1−1 1 · · · z−bm−1 m (1 −zA1) · · · (1 −zAd) dz.
Here 0 < ϵ1, . . . , ϵm < 1 are different numbers such that we can expand all the 1 1 −zMk into the power series about 0. This method has been applied very successfully to determine the volumes of Birkhoffpolytopes in . The reader may recall from basic calculus that to integrate rational functions it was sometimes useful to use a partial fraction decomposition, in order to simplify the expressions. M. Vergne and collaborators have exploited the structure of the hyperplane arrangement associated to the columns of the matrix A, namely HA = [ Ai column of A {z ∈Cr|⟨Ai, z⟩= 0}.
This was used to do a nice “simpler fraction” decomposition of the for-mula 1 Qn j=1(1−zAj ). A subset σ of columns of A is called basic if the elements Ai ∈σ form a vector space basis for Rd. For such σ, set fσ(z) := 1 Q Ai∈σ⟨Ai, z⟩.
We consider the vector space SA the spanned by such “simple” elements fσ. Brion and Vergne proved that there is a direct sum decomposition of the space RA of rational functions whose poles are on the hyperplane 20 J. A. De Loera arrangement HA and SA is one of the summands. The projection map T R : RA − →SA is called the total residue map. This projection allows to survive only simple summands that come from special bases. See [6,19,54] and references within for more details and implementation.
5. Beyond Polytopes To conclude this survey we would like to comment on what happens when we try to extend counting lattice points inside other region of space besides polytopes. We will see very interesting effects. Consider first the region K(n) = {(x, y) : 1 ≤x ≤n, n x ≤y ≤n} bounded by a hyperbola and lines.
Let L be the open strip properly in between the hyperbolas y = n x and y = n+1 x , as denoted in Figure 14. Thus L does not include the boundary of K(n) nor that of K(n + 1). See Figure 14.
x y n+1 n 1 1 n n+1 L K(n) K(n+1) Fig. 14. The regions K(n), K(n + 1), and L.
Given all of this, we arrive at the following proposition: Proposition 1. The number of lattice points inside the planar non-convex region R = (K(n + 1)\K(n)) ∪(K(n)\K(n + 1)) equals 2n + 3 if and only if n is prime.
Here is why: observe that if n is not a prime, then there is at least one (a, b) such that (a, b) = n and (a, b) is on the lower boundary of K(n) but (a, b) ̸= (n, 1) or (1, n). L has no lattice points otherwise one such (c, d) will The Many Aspects of Counting Lattice Points in Polytopes ⋆⋆ 21 have n+1 c > d > n c , however, this implies that n < cd < n + 1 which is a contradiction since no integer can exist between consecutive integers. Thus the lattice points in the region R are the 2n+1 lattice points along the lines x = n + 1 and y = n + 1, plus at least four more (a, b), (b, a), (n, 1), (1, n) along the lower hyperbola. Next suppose n is prime. Again, as we argued before, we have that the region L does not contribute any lattice points. The lattice points in the region R are the 2n+1 points along the lines x = n+1 and y = n + 1, plus exactly two more (n, 1), (1, n).
Already a simple variation of the construction with the region K(a, b) = {(x, y) : 1 ≤x ≤a, 1 ≤y ≤b, xy ≥b} allows to count the number of integer divisors of b in the interval [1, a]. When the sets for which we count lattice points are arbitrary the problem gets harder and harder as the dimension of the set grows: Given (a, b, c) positive integers, deciding whether there is a lattice point in the set {x : ax2 + bx = c, x ≥0} is an NP-complete problem . Finally, deciding whether there is a non-negative integer root for arbitrary polynomials in Z[x1, . . . , x9] is undecidable, i.e.
roughly speaking there is no computer algorithm to solve this problem for all instances . Thus we clearly need to be less ambitious!
A convex set C is a set of Euclidean space such that for any pair of points in C the line segment joining x and y is completely inside C. Polyhedra are the simplest convex sets in a way. Thus it is natural to ask whether other convex sets besides polyhedra allow a nice theory of counting lattice points like the one we outlined in previous sections. Alas, this still contains some interesting surprises! In a way counting points inside general convex sets is an older problem. Already Gauss proposed a famous counting problems for circles : Fig. 15. Gauss and the circle area approximation problem Suppose you are given a circle of radius r centered at the origin. How good is the approximation of the number of lattice points N(r) to the area 22 J. A. De Loera of the circle as r grows? In other words, what is the exact order of magnitude for the error function N(r) −πr2? The answer is still not known exactly, although it is supposed to be in between r.5 and r.6301369863 (see ).
Counting lattice points in (four dimensional) convex bodies is something that credit card cyber-thieves would care about too! The reason is the RSA encryption algorithm relies on the difficulty of factorizing a number of the form pq with p, q large primes. Here is a way to factorize via counting lattice points: For an integer number n = pq consider the 4-dimensional ball B(n) = {x ∈R4 : x2 1 + x2 2 + x2 3 + x2 4 ≤n}. Jacobi (see [4,37]) proved that the number of ways in which a positive integer can be written as a sum of four squares is eight times the sum of its divisors that are not a multiple of four. So, for example there are 744 ways of writing 100 as sum of four squares because 1, 2, 5, 10, 25, 50 are the only divisors of 100 not divisible by 4. Thus if we know that n = pq, and |B(n)| denotes the number of lattice points in B(n) we have |B(n)| −|B(n −1)| = 8(1 + p + q + n). Therefore a factorization of n could be done fast if we knew how to compute |B(n)| fast.
Thus the RSA cryptosystems used in internet transactions could be broken if you knew how to quickly count lattice points inside four dimensional balls.
This is a very intriguing connection.
Acknowledgements: I thank the anonymous referees for the many cor-rections. I am grateful to G¨ unter M. Ziegler and J¨ urgen Wolfart for their help and suggestions for improvement. I thank Philippe Clauss who kindly provided references and the photograph we used of E. Ehrhart.
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5380 | https://math.stackexchange.com/questions/609157/sum-of-4-squares | number theory - sum of 4 squares - Mathematics Stack Exchange
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sum of 4 squares
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Is there any natural number A A which cannot be written as:
A=W 2+X 2+Y 2+Z 2 A=W 2+X 2+Y 2+Z 2
where W,X,Y,Z∈N∪0 W,X,Y,Z∈N∪0
I was considering the fact that a 2+b 2≠1 mod 4 a 2+b 2≠1 mod 4 and was attempting to determine simular results for 3 3 and 4 4 squares when I realised that I could not find a number which did not work for 4 4. I did a quick search and found that no numbers less than 1000 1000 contradict this. I have been trying to find a proof but have been, so far, unsuccessful. This is too general a concept to be new but I have not been exposed to it previously.
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asked Dec 16, 2013 at 15:26
kainekaine
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4 No, all numbers can be expressed as the sum of four squares. Check this out: en.wikipedia.org/wiki/Lagrange's_four-square_theorem 2013-12-16 15:27:19 +00:00 Commented Dec 16, 2013 at 15:27
If you post that as an answer, that is sufficient. It shows me where to look to learn more.kaine –kaine 2013-12-16 15:29:50 +00:00 Commented Dec 16, 2013 at 15:29
I should have assumed this was tied to Lagrange and Gauss; it seems like everything is.kaine –kaine 2013-12-16 15:33:42 +00:00 Commented Dec 16, 2013 at 15:33
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No, all numbers can be expressed as the sum of four squares. More information, and a proof using the Hurwitz integers, can be found here.
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answered Dec 16, 2013 at 15:32
Daniel RDaniel R
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This is Legendre's theorem on the sum of four squares. As the squares are congruent to 0, 1 or 4 (mod 8), anything integer congruent to 7 (mod 8) cannot be written as the sum of three squares. Perhaps the easiest way to prove this theorem is using Quaternions, an extension of the complex numbers in R 4 R 4.
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answered Dec 16, 2013 at 15:30
vukovvukov
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4 Lagrange, not Legendre.egreg –egreg 2013-12-16 15:47:34 +00:00 Commented Dec 16, 2013 at 15:47
Legendre did, apparently, extend this to develop the 3 squares situation (which is what vukov is describing) so it isn't completely off.kaine –kaine 2013-12-16 15:56:38 +00:00 Commented Dec 16, 2013 at 15:56
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Just for completion, here is an elementary proof using modular arithmetic, which I wrote up previously for a number theory course.
Claim For any n∈N n∈N there exists w,x,y,z∈Z w,x,y,z∈Z such that n=w 2+x 2+y 2+z 2 n=w 2+x 2+y 2+z 2.
Proof: First, we observe that sums of four squares are closed under multiplication. That is given a,b,c,d,e,f,g,h a,b,c,d,e,f,g,h integers, there are w,x,y,z w,x,y,z such that
(a 2+b 2+c 2+d 2)(e 2+f 2+g 2+h 2)=w 2+x 2+y 2+z 2.(a 2+b 2+c 2+d 2)(e 2+f 2+g 2+h 2)=w 2+x 2+y 2+z 2.
I will not prove this, it follows from expanding the terms and finding the right expressions for w,x,y,z w,x,y,z. Alternatively, there is a proof considering certain matrices over C C.
So, let n∈N n∈N. By prime decomposition, there are primes p 1,…,p k p 1,…,p k such that n=p e 1 1⋯p e k k n=p 1 e 1⋯p k e k. If one of the p i p i is 2 2 then notice that 2=1 2+1 2+0 2+0 2 2=1 2+1 2+0 2+0 2. So that any power of 2 2 is a sum of four squares by the previous result. Now, let p|n p|n such that p≡1 mod 4 p≡1 mod 4 or p≡3 mod 4 p≡3 mod 4.
Consider the sets S 1={u 2:u∈{0,1,…,p−1}}S 1={u 2:u∈{0,1,…,p−1}} and S 2={−1−v 2:v∈{0,1,…,p−1}}S 2={−1−v 2:v∈{0,1,…,p−1}}. Since p p is a prime, we have there are p−1 2 p−1 2 nozero squares modulo p p and p+1 2 p+1 2 including 0 0. S 1 S 1 and S 2 S 2 both contain precisely these numbers so that the cardinality of both is p+1 2 p+1 2. Thus, since there are only p p elements of the set {0,1,…,p−1}{0,1,…,p−1}, we see S 1∩S 2≠∅S 1∩S 2≠∅. Thus, there is u,v∈{0,1,…,p−1}u,v∈{0,1,…,p−1} such that u 2≡−1−v 2 mod p u 2≡−1−v 2 mod p. Thus, we have u 2+v 2+1≡0 mod p u 2+v 2+1≡0 mod p. So that there is an integer k k such that
u 2+v 2+1=k p.u 2+v 2+1=k p.
Now, if k=1 k=1 we are done, since p p is the sum of three squares, we can add 0 2 0 2 and the result holds. Suppose k≠1 k≠1. We can see immediately that k>1 k>1, since u 2+v 2+1≥1 u 2+v 2+1≥1. So, we have k∈N k∈N. We will then produce a strictly decreasing sequence of integers k 1,k 2,…k 1,k 2,… which are bounded below by 1 1. As a result, there is m m such that k m=1 k m=1. In which case we have that p p is the sum of four squares.
Now, we examine our k k. We have, by renaming u u and v v four integers a,b,c,d a,b,c,d such that a 2+b 2+c 2+d 2=k p a 2+b 2+c 2+d 2=k p. Now k k is either even or odd. If k k is even, then k p k p must be even. Since the square of an even number is even and an odd number is odd, we then can only have 0,2 or 4 of a,b,c,d a,b,c,d even. Without loss of generality, take a≡b mod 2 a≡b mod 2 and c≡d mod 2 c≡d mod 2. Then, let a 1=a+b 2,b 1=a−b 2,c 1=c+d 2,d 1=c−d 2 a 1=a+b 2,b 1=a−b 2,c 1=c+d 2,d 1=c−d 2 and observe that
a 1 2+b 1 2+c 1 2+d 1 2=a 2+b 2+c 2+d 2 2=k p 2.a 1 2+b 1 2+c 1 2+d 1 2=a 2+b 2+c 2+d 2 2=k p 2.
Thus, there is k 1=k/2 k 1=k/2 such that k 1 p k 1 p is the sum of four squares.
Now, by repeating the above argument, we can assume we have found k l k l such that k l k l is odd. Thus, k l p=a l 2+b l 2+c l 2+d l 2 k l p=a l 2+b l 2+c l 2+d l 2. So, we can find e,f,g,h∈{−k+1 2,…,k−1 2}e,f,g,h∈{−k+1 2,…,k−1 2} such that a l≡e,b l≡f,c l≡g,d l≡h mod k l a l≡e,b l≡f,c l≡g,d l≡h mod k l. Thus, e 2+f 2+g 2+h 2<4(k l 2)2=k 2 l e 2+f 2+g 2+h 2<4(k l 2)2=k l 2.
Thus
(a l 2+b l 2+c l 2+d l 2)(e 2+f 2+g 2+h 2)<k 2 l k l p.(a l 2+b l 2+c l 2+d l 2)(e 2+f 2+g 2+h 2)<k l 2 k l p.
So we can divide through by k 2 l k l 2 and find integers a l+1,b l+1,c l+1,d l+1 a l+1,b l+1,c l+1,d l+1 such that a l+1 2+b l+1 2+c l+1 2+d l+1 2=k l+1 p a l+1 2+b l+1 2+c l+1 2+d l+1 2=k l+1 p where k l+1<k l k l+1<k l.
Thus, by repeating the arguments above, we have constructed a decreasing sequence of natural numbers (k i)(k i). So, we must have some m m such that k m=1 k m=1. In which case, we see that there are integers w 0,x 0,y 0,z 0 w 0,x 0,y 0,z 0 such that
w 2 0+x 2 0+y 2 0+z 2 0=k m p=p.w 0 2+x 0 2+y 0 2+z 0 2=k m p=p.
As we've seen before, we have seen that sum of four squares are closed under multiplication, so that there are integers w,x,y,z w,x,y,z such that n=w 2+x 2+y 2+z 2.n=w 2+x 2+y 2+z 2. as desired. Thus any integer is the sum of four squares.
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answered Dec 16, 2013 at 16:15
doppzdoppz
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This can even be extended to higher powers:
Edward Waring, says that for every natural number n n there exists an associated positive integer s such that every natural number is the sum of at most s s k k th powers of natural numbers (for example, every number is the sum of at most 4 squares, or 9 cubes, or 19 fourth powers, etc.) from here.
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answered Dec 16, 2013 at 16:05
draks ...draks ...
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5381 | https://cards.algoreducation.com/it/content/VweB_ABf/temperatura-dilatazione-calore | La temperatura e il calore | Algor Cards
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La temperatura e il calore
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La temperatura e il calore sono concetti fondamentali in fisica, che descrivono l'agitazione termica delle particelle e il trasferimento di energia. La dilatazione termica dei solidi e dei liquidi è un fenomeno che mostra come cambiano le dimensioni dei corpi con la variazione di temperatura. Il calore si propaga attraverso conduzione, convezione e irraggiamento, essenziali per comprendere molti fenomeni naturali e processi tecnologici.
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Misurazione della Temperatura e Scale Termometriche
La temperatura è una grandezza fisica che esprime il livello di agitazione termica delle particelle di un sistema e si manifesta come calore o freddo percepibile. La misurazione della temperatura avviene tramite strumenti chiamati termometri, che possono essere calibrati secondo diverse scale termometriche. La scala Celsius è ampiamente utilizzata in ambito quotidiano e definisce il punto di congelamento dell'acqua a 0 °C e il punto di ebollizione a 100 °C a livello del mare e a pressione atmosferica standard. La scala Kelvin, invece, è la scala termometrica adottata nel Sistema Internazionale di unità di misura (SI) e parte dallo zero assoluto, che è la temperatura teorica più bassa possibile, corrispondente a -273,15 °C o 0 K. La conversione tra Celsius e Kelvin è data dalla relazione T(K) = t(°C) + 273,15. È importante notare che un incremento di un grado Celsius corrisponde esattamente a un incremento di un kelvin, rendendo le due scale direttamente proporzionali.
Dilatazione Termica dei Solidi e dei Liquidi
La dilatazione termica è il fenomeno per cui la dimensione di un corpo cambia in risposta a una variazione di temperatura. Nei solidi, si verifica principalmente una dilatazione lineare, descritta dalla legge: L = L0(1 + αLΔT), dove L è la lunghezza dopo la dilatazione, L0 la lunghezza originale, αL il coefficiente di dilatazione lineare e ΔT la variazione di temperatura. Per i liquidi e i solidi, si considera anche la dilatazione volumica, espressa dalla formula: ΔV = βV0ΔT, dove ΔV è la variazione di volume, V0 il volume iniziale, β il coefficiente di dilatazione volumica e ΔT la variazione di temperatura. Il coefficiente di dilatazione volumica è generalmente tre volte il coefficiente di dilatazione lineare per i solidi isotropi, in accordo con la relazione β ≈ 3αL, riflettendo il fatto che la dilatazione avviene in tutte e tre le dimensioni spaziali.
Calore e Trasferimento di Energia
Il calore è energia in transito che si trasferisce tra corpi o sistemi a differenti temperature, fino al raggiungimento dell'equilibrio termico. L'unità di misura del calore nel Sistema Internazionale è il joule (J). La quantità di calore scambiato in un processo può essere calcolata con la formula Q = mcΔT, dove Q è il calore assorbito o ceduto, m la massa del corpo, c il calore specifico (quantità di energia necessaria per aumentare di 1 °C la temperatura di 1 kg di una sostanza) e ΔT la variazione di temperatura. Il calore specifico è una proprietà intrinseca di ogni materiale e varia in base alla sua natura e alla temperatura a cui si trova.
Modalità di Propagazione del Calore
Il calore si propaga attraverso tre meccanismi principali: conduzione, convezione e irraggiamento. La conduzione è il trasferimento di energia termica attraverso un materiale solido per contatto diretto tra particelle adiacenti. La convezione è caratteristica dei fluidi (liquidi e gas) e avviene quando il calore provoca il movimento del fluido stesso, con le parti più calde che tendono a salire e quelle più fredde a scendere, generando correnti convettive. L'irraggiamento è il trasferimento di energia termica sotto forma di onde elettromagnetiche, che possono propagarsi anche nel vuoto, come avviene per il calore emesso dal Sole. Questi meccanismi possono avvenire simultaneamente e sono fondamentali per comprendere fenomeni naturali e processi tecnologici.
Mostra di più
La temperatura e il calore
La temperatura
Definizione di temperatura
La temperatura è una grandezza fisica che esprime il livello di agitazione termica delle particelle di un sistema
Strumenti per la misurazione della temperatura
Termometri
I termometri sono strumenti utilizzati per misurare la temperatura e possono essere calibrati secondo diverse scale termometriche
Scale termometriche
Le scale termometriche, come la scala Celsius e la scala Kelvin, sono utilizzate per misurare la temperatura in diversi contesti
Dilatazione termica
La dilatazione termica è il fenomeno per cui la dimensione di un corpo cambia in risposta a una variazione di temperatura
Il calore
Definizione di calore
Il calore è energia in transito che si trasferisce tra corpi o sistemi a differenti temperature, fino al raggiungimento dell'equilibrio termico
Unità di misura del calore
Il joule è l'unità di misura del calore nel Sistema Internazionale
Proprietà del calore
Calore specifico
Il calore specifico è la quantità di energia necessaria per aumentare di 1 °C la temperatura di 1 kg di una sostanza ed è una proprietà intrinseca di ogni materiale
Propagazione del calore
Il calore si propaga attraverso tre meccanismi principali: conduzione, convezione e irraggiamento
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00
La ______ è una misura del livello di agitazione termica delle particelle e si percepisce come calore o freddo.
temperatura
01
Il punto in cui l'acqua congela è definito a 0 °C sulla scala ______, mentre il punto di ebollizione è a 100 °C.
Celsius
02
La scala ______ inizia dallo zero assoluto, che equivale a -273,15 °C o 0 K, ed è usata nel Sistema Internazionale.
Kelvin
03
Dilatazione lineare nei solidi
L = L0(1 + αLΔT), dove L è la lunghezza finale, L0 la lunghezza iniziale, αL il coefficiente di dilatazione lineare, ΔT la variazione di temperatura.
04
Dilatazione volumica
ΔV = βV0ΔT, dove ΔV è la variazione di volume, V0 il volume iniziale, β il coefficiente di dilatazione volumica, ΔT la variazione di temperatura.
05
Relazione tra dilatazione lineare e volumica
Per solidi isotropi, β ≈ 3αL, indicando che la dilatazione volumica è circa tre volte quella lineare.
06
L'energia che si muove tra oggetti a temperature diverse fino a che non si stabilizza si chiama ______.
calore
07
Conduzione termica
Trasferimento energia tra particelle contigue in solidi.
08
Convezione termica
Movimento fluidi causa calore, parti calde salgono, fredde scendono.
09
Irraggiamento termico
Energia viaggia come onde elettromagnetiche, anche nel vuoto.
Q&A
Ecco un elenco delle domande più frequenti su questo argomento
Quali sono le principali scale termometriche e come si converte la temperatura da Celsius a Kelvin?
Come si esprime matematicamente la dilatazione termica di solidi e liquidi?
Come si calcola la quantità di calore scambiato in un processo termico?
Quali sono i meccanismi attraverso cui si propaga il calore?
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5382 | https://smg.quora.com/Is-there-any-reason-why-the-difference-between-square-numbers-is-odd-numbers-the-difference-between-odd-numbers-is-2s | Is there any reason why the difference between square numbers is odd numbers, the difference between odd numbers is 2s, and the difference between those 2s is 0? - Science and Math Geeks - Quora
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Is there any reason why the difference between square numbers is odd numbers, the difference between odd numbers is 2s, and the difference between those 2s is 0?
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Arbash Nazeer
Number theory aficionado
·Updated 5y
I'm guessing you meant difference between consecutive square numbers, odd numbers and so forth.
Let the difference function be D(n)D(n). Representing the question:
n 2 D(n)−−−→2 n+1 D(n)−−−→2 D(n)−−−→0 n 2→D(n)2 n+1→D(n)2→D(n)0
The immediate observation is that the degree reduces by one in every step, and in fact, the degree reduces by one regardless of what function you start from. This is because D(n)=P(n+1)−P(n)D(n)=P(n+1)−P(n), and applying this shows us why we go from squares to odd numbers to 2 2 s to 0 0.
Other interesting observation is that applying D(n)D(n) on P(n)P(n) gives us a unique polyn
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I'm guessing you meant difference between consecutive square numbers, odd numbers and so forth.
Let the difference function be D(n)D(n). Representing the question:
n 2 D(n)−−−→2 n+1 D(n)−−−→2 D(n)−−−→0 n 2→D(n)2 n+1→D(n)2→D(n)0
The immediate observation is that the degree reduces by one in every step, and in fact, the degree reduces by one regardless of what function you start from. This is because D(n)=P(n+1)−P(n)D(n)=P(n+1)−P(n), and applying this shows us why we go from squares to odd numbers to 2 2 s to 0 0.
Other interesting observation is that applying D(n)D(n) on P(n)P(n) gives us a unique polynomial, but the inverse, D−1(n)D−1(n) is not unique. (Applying D−1(n)D−1(n) on 2 2 s give us not only the set of odd numbers, but also the even numbers).
Try to prove that D−1(n)=(∑P(n))+k D−1(n)=(∑P(n))+k. In fact, D(n)D(n) is very similar to differentiation and D−1(n)D−1(n) is similar to integration.
9 3
Jubin Singh
Studied at Delhi Public School, Rohini (Graduated 2022)
·5y
So I think that the question you have asked is a bit ambiguous but I think I have got what you have asked..
so firstly difference between square numbers is an odd number is not necessary as for eg . 6 2−2 2=32 6 2−2 2=32(which is even)
Answer for second ques) difference between two odd numbers
(let us suppose 2 n+1 2 n+1 and 2 m+1 2 m+1) = 2 m+1−(2 n+1)=2(m−n)2 m+1−(2 n+1)=2(m−n) which is equivalent to 2 q 2 q(for some integer q)
Answer for 3rd ques) something subtracted from itself is always zero
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Mark Brown
Number 1 baller
·5y
The difference between square numbers is not always odd, 36–16=20. The differences of odd numbers is a multiple of two because odd numbers go up in increments of two, so you'll always be able to count by twos to the other number, and anything subtracted by itself is zero.
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5383 | https://www.aafp.org/pubs/afp/issues/2024/0400/letter-menstrual-cycle-length-infertility.html | Scheduled maintenance is planned for September 26–29. You may experience brief interruptions during this time.
Letters to the Editor
Longer Menstrual Cycle and Infertility Evaluation
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Am Fam Physician. 2024;109(4):296-298
assignment
Author disclosure: No relevant financial relationships.
To the Editor:
In the article written by Dr. Phillips and colleagues, they describe a suggested workup for women suspected of having anovulation.1 The article recommends that clinicians obtain a day-21 progesterone level, and for levels less than 3 ng per mL (9.54 nmol per L), conclude that ovulation has not occurred. This method leaves out a group of people with a longer menstrual cycle. If a woman has a 35-day menstrual cycle, she may not ovulate until day 21 or 22; therefore, a day-21 progesterone level would be falsely low. The algorithm should recommend a midluteal progesterone level that would allow clinicians to discuss cycle specifics with their patients and determine which day would be the most accurate time to obtain a progesterone level. The American Society for Reproductive Medicine states, “Given the range of normal variation in ovulatory cycles, a serum progesterone measurement generally should be obtained approximately 1 week before the expected onset of the next menses, rather than on any one specific cycle day (e.g., cycle-day 21).”2
The article also states that the American College of Obstetricians and Gynecologists no longer recommends the examination of cervical mucus in infertility evaluation. Upon review of the cited reference, I could not find any such recommendation.3 On the contrary, the American Society for Reproductive Medicine states that cervical fluid tracking is an inexpensive method with moderate-quality evidence for increasing the likelihood of achieving pregnancy: “A retrospective cohort study involving 1,681 cycles observed that pregnancy rates were highest (approximately 38%) when intercourse occurred on the day of peak mucus (day 0) and appreciably lower (approximately 15% to 20%) on the day before or after the peak.”4
During a time when many patients feel a loss of control and lack of understanding, we can tailor infertility care and empower patients to understand more about their bodies.
Anne Keenan, MD
Minneapolis, Minn.
hors0097@umn.edu
Author disclosure: No relevant financial relationships.
Phillips K, Olanrewaju RA, Omole F. Infertility: evaluation and management. Am Fam Physician. 2023;107(6):623-630.
Practice Committee of the American Society for Reproductive Medicine. Diagnostic evaluation of the infertile female: a committee opinion. Fertil Steril. 2015;103(6):e44-e50.
ACOG Committee Opinion 762: Prepregnancy counseling. Obstet Gynecol. 2019;133(1):e78-e89.
Practice Committee of the American Society for Reproductive Medicine and the Practice Committee of the Society for Reproductive Endocrinology and Infertility. Optimizing natural fertility: a committee opinion. Fertil Steril. 2022;117(1):53-63.
In Reply:
This article was intended to serve only as a framework for the infertility workup. We agree that a patient with a longer menstrual cycle may have spuriously low progesterone levels at day 21. There are always patient-specific characteristics for which few algorithms can adequately account. We advise that the recommendations be adjusted as needed to address specific situations.
The sentence, “However, evaluation of cervical mucus is no longer routinely used in infertility evaluation” is from information provided in reference 10 and not 11, as incorrectly cited in the article. The article has been corrected online. We did not intend to suggest that cervical mucus screening cannot be used for certain patients; however, routine postcoital cervical mucus testing is not recommended by the American Society for Reproductive Medicine in the initial workup of infertility.1
Kiwita Phillips, MD, FACOG
Atlanta, Ga.
kphillips@msm.edu
Raimot Olanrewaju, MD, FACOG
Atlanta, Ga.
Folashade Omole, MBChB, FAAFP
Atlanta, Ga.
Author disclosure: No relevant financial relationships.
Reference
Practice Committee of the American Society for Reproductive Medicine. Fertility evaluation of infertile women: a committee opinion. Fertil Steril. 2021;116(5):1255-1265.
Email letter submissions to afplet@aafp.org. Letters should be fewer than 400 words and limited to six references, one table or figure, and three authors. Letters submitted for publication in AFP must not be submitted to any other publication. Letters may be edited to meet style and space requirements.
This series is coordinated by Kenny Lin, MD, MPH, deputy editor.
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5384 | https://www.cambridge.org/core/services/aop-cambridge-core/content/view/E57E42E003AA848FE8B21BC7C650C14F/S001309150003162Xa.pdf/isogonals_of_a_triangle.pdf | 166 Isogonals of a Triangle, By J. S. MACKAY, M.A., LL.D.
DEFINITION.—If tivo angles /tare the same vertex and the same bisector, the sides of either angle are isogonal to each other with respect to the other angle.
Thus the isogonal of AP with respect to L. BAC is the image of AP in the bisector of L BAC. It is indifferent whether the bisector of the interior L BAC be taken, or the bisector of the angle adjacent to it; the isogonal of AP remains the same.
Tt follows from the definition that (1) 'Die internal and the external bisectors of L BAC are their own isogonals.
(2) The line joining the orthocentre of a triangle to any vertex is isogonal to the line joining the circumcentre to that vertex.
(3) Any internal median of a triangle is isogonal to the cor-„ responding symmedian.
(4) The tangents to the circumcircle of ABC at A, B, C are isogonal to the external medians.
[The external medians are the parallels to the sides of a triangle drawn through the opposite vertices.
The reason for the giving of this name will l>e found in the Proceedings of the Edinhurgh Mathematical Society, Vol. I., p. 16 (1S94).] This terminology was proposed by Mr G. de Longchamps in his Journal Ue Malhtmatiques Elanentaires, 2nd series, Vol. V., p. 245 (188G).
Published online by Cambridge University Press 167 Si-(a) If P, Q be any two points taken on a pair of lines isogonal with respect to angle BAC, the distances of P from AB, AC are in-versely proportional to those if Q from AB, AC.
FIGURE 29.
If the quadrilateral AQ2QQj be revolved through two right angles round the bisector of L B as an axis, it will become homo-thetic to the quadrilateral APjPP.;; therefore P P ^ P P . - Q Q ^ Q Q , (a') If P, Q be any two points and if the distances of P from AB, AC be inversely proportional to those of Q from. AB, AC, then AP, AQ are isogonal with respect to L. BAC.
This may be proved indirectly.
(1) The points /J a Ql Q.2 P2 are coitcyclicj Since PiP2 QJQJ are antiparallel with respect to c BAC; therefore P, P2 Qj Q2 are concyclic.
(2) The centre of the circle P-iQiQJ\ is the mid point of PQ.
For the perpendicular to PJQJ at its mid point goes through the centre of the circle; and this perpendicular bisects PQ.
So does the perpendicular to P2Q2 at its mid point.
(3) 1\P2 is perpendicular to AQ and Q,Q, „ „ ,,AP.
Sir James Ivory in Leybourn's Mathematical Repository, new series, Vol. I., Part II., p. 19 (1806). The mode of proof is due to Professor Neuberg. See his excellent memoir on the Recent Geometry of the Triangle in Rouche and Com-berousse's TraiUdc Geometric, First Part, p. 438 (1801).
t This and the two following theorems are due to Steiner. See Gergonne's Annales, XIX., 37-04 (182S), or Steiner's Gesammelte Wcrke, I., 191-210 (1881).
The proof given of (1) is Professor Neuberg's. See the reference in the preceding note, Published online by Cambridge University Press 168 For AP is a diameter of the circunicircle of AP,P2; therefore the isogonal of AP with respect to .^PjAP.jis the perpendicular from A to P1P,2 (4) The circumcentre of either of the triangles AP,P2 AQ,Q2 and the ortliocentre of the other are collinear with the point A.
(5) Triangle, PPlP2 is inverseli/ similar f to QQ-2Qi-This follows from the demonstration of S 1 ; or it may be thus proved: L PP,Po = L PAP, = L QAQi = L QQ2Q,.
Similarly L. PP,P, = L QQiQo.
(6) IfPP1 QQ2 meet at D and PP2 QQl „ „ E, tlien AD, AE are isogonals with respect to L JJAC.
FIGURE 30.
Join PjQ, P2Q,.
Since P,Q, P.Q2 are concyclic, therefore L AQ.,P, = L AQ,P., therefore tlieir complements are equal that is L P^-.D = L P,Q,E.
Similarly /.Q2P,D= L Q,P2E ; therefore triangles PiQ^D, P^QiE are similar; therefore PjQ,: PjD = P2Q,: P2E.
Now triangles APjQ.^ AP2Q, are similar; therefore AP, : P,Q2 = AP2: P2Q,.
Hence AP, : P, D = AP 2: P.,E and ^P,AD= ^.P.,AE.
The same result might be arrived at by revolving the quadri-lateral AQ.jQQj through two right angles round the bisector of L 13AC.
This mode of proof is given by Professor Fuhnnann in his Syntkclische Jicwcise planiiiictrisckcr Satic, p. !)3 (1890).
t See Ivory's paper already cited, p. 20.
Published online by Cambridge University Press 169 (a) If ABC he a triangle, and if AP, AQ he isogonal with respect to A, then BP-BQ : CP-CQ = AB"-: AC" FlGURE 31.
About APQ circumscribe a circle, cutting AB, AC in F, E ; join FE.
Because L BAP = L. CAQ therefore arc FP = arc EQ therefore FE is parallel to BC therefore AB : BF = AC :CE therefore AB : ABBF = AC2 : ACCE therefore AB2 : BPBQ = AC- : CPCQ.
A second demonstration will be found in C. Adams's Die merkwiirdigsten Eigenschafteri, des geradlinigen Dreiecks, p. 1 (1846), and a third in Professor Fuhrmann's Synthetische Betveise, p. 94 (1890).
(a) If ABC he a triangle and BC he divided at P and Q so that BPBQ : CP-CQ = AJP:AC-then f AP, AQ are isogonals with respect to A.
This may be proved indirectly.
(1) If AQ be the internal or the external median from A, then BQ = CQ, and the theorem becomes } BP:CP = AB2: AC2.
Pappus's Mathematical Collection, VI. 12. The same theorem differently stated is more than once proved in Book VII. among the lemmas which Pappus gives for Apollonins's treatise on Determinate Section. The proof in the text is taken from Pappus.
t In Pappus's Mathematical Collection, VI. 13, there is proved the theorem : If BPBQ : CPCQ > AB-: A C then L. BAP > L. CAQ.
X Adams (see the reference to him on this page) gives (l)-(4), (6), (8). His proof of (4) is different from that in the text.
Published online by Cambridge University Press 170 (2) If AQ be the internal or the external median from A and L. BAC be right, then AP is perpendicular to BC.
FIGURES 32, 33.
Since i_ ACB = L CAQ = u BAP therefore L ACB + j_ CAP = L BAP + L CAP = a right angle.
(3) If AP and AQ coincide, then AP is either the internal or the external bisector of L A, and the theorem becomes BP2:CP2 = AB2:AC2 or BP :CP =AB : AC a known result, namely, Euclid VI. 3, or the cognate theorem.
(4) BPC/': BQ-CQ = A1K : AQ2.
This follows from the theorem of § 2 by considering APQ as the triangle and AB, AC as the isogonals.
(5) If AP, AQ which are isogonal with respect to L BAC meet the circumcircle of ABC in H, S, then APAS = AQAR.
FIGURE 34.
Fqr triangles ACR, AQB are similar therefore AQ • AR = AB-AC.
Similarly AP-AS=ABAC.
(6) RS is parallel to BC.
(7) The distances from the mid point of any side of a triangle to the points where two isogonals from the opposite vertex meet the circumcircle are equal. For the perpendicular which bisects BC bisects RS.
Mr Emile Vigarie in the Journal Ac Mathematiques Mdmentaires, 2nd series, IV. 59 (1885).
Published online by Cambridge University Press 171 (8) If APR becomes the diameter of the circumcircle ABC then AQ becomes perpendicular to BC, and AQ-AR= AB-AC, a theorem of Brahmegupta's.
See Chasles's Aperru, 2nd ed., pp. 420-447.
(9) If AP, AQ coincide, then AP becomes either the internal or the external bisector of L. A.
Hence in the first cast; AB-AC = AP-AS = AP-PS + AP-= BP-PC + AP5 ; and in the second case AB AC = APAS = AP-PS - AP-= BP-PC - AP-.
(10) In triangle ABC, AP, AQ are isogonals with respect to A ; through B drato BE parallel to AP meeting CA in E; „ C „ CF „ „ AQ ., BA „ F; tlien EF is antiparallel to BC ivith respect to A.
FIGURE 35.
For /.ABE= ^BAP= £.CAQ= L ACF ; therefore the points E, B, C, F are concyclic.
The same thing would happen if BE, CF were drawn parallel to AQ, AP.
(11) In triangle ABC, AP, AQ are isogonal; from P and Q perpendiculars are drawn to BC; these perpendiculars are inter-sected at D, E by a perpendicular to AB at B, and at D1, E' by a perpendicular to AC at C. To prove f BDBE : CD' CE' = AB4: AC4.
Mr Emile Vigarie.
t Mr Emile Vigarie in the Journal de Mathimatiques Eleinentairei, 2nd series, IV. 224 (1885) say8 that this theorem was communicated to him by his friend Mi Tb. Valiech.
Published online by Cambridge University Press 172 FIGURE 36.
Draw AX perpendicular to BC.
The similar triangles BDP, BEQ, ABX give BD:BP = AB:AX BE:BQ = AB:AX BD • BE AB2 therefore Similarly therefore therefore BP CD' CP BD CD' BD BQ •CE' • CQ • BE • CE' • BE AX2 AC2 AX2 CP CQ BP-BQ AC2 AB2 AC-AB2 CD' • CE' AB2 AC (12) If in (11) AQ be the median from A, then BD : CD' = AB>: AC3.
FIGURE 36.
For BE : BQ = AB : AX and CE' : CQ = AC : AX ; therefore BE : CE' = AB : AC, whence the result follows.
8 3.
If three straight lines drawn through tlie vertices of a triangle are concurrent, their isogonals with respect to the angles of the triangle are also concur rent.\ Mr Emile Vigarie in the Journal de Mathematiques EUmcntaires, 2nd series, IV. 225 (1885).
t Steiner 'mGergonne's Annalcs, xix. 37-64 (1828), or Steiner's Gfcsammelte Werke, I. 193 (1831). Ivory in his paper previously cited proves the theorem : If the isogonals BO, BO' meet the bisector of I- A at O, 0', then B O : C 0 = B 0 ' : C 0 ' ; and he adds as a corollary that CO, CO' are isogonals with respect to 0.
Published online by Cambridge University Press 173 FIGURE 37.
Let BO, BO' be isogonals with respect to B and CO, CO' „ „ .„ „ C; then AO, AO' are ,, „ „ A.
Denote the distances of O from the sides by p1 pt p3 and those of 0' by g, ?2 q3 Then ;>, qx = p, q and px q, = p, q3 therefore Vi<li=Pi<li therefore AO, AO' are isogonals with respect to A.
Another demonstration will be found in C, Adams's Eigenschaften des...Dreiecks, pp. 7-8 (1846).
Points such as 0, 0' determined by the intersection of pairs'of isogonal lines will be called isogonal points, or simply isogonals, with respect to the triangle ABC.
They are sometimes called isoyonally conjugate points, or isogonal conjugates, but more frequently on the continent of Europe inverse points with respect to the triangle ABC.
The designation, inverse points, was suggested about the same time in Scotland and in France. See a paper read before the Royal Society of Edinburgh on 20th March 18G5, by the Rev. Hugh.
Martin, and printed in their Transactions, xxiv. 37-52 : and an article by Mr J. J. A. Mathieu in the Nouvelles Annales, 2nd series, IV. 393-407, 481-493, 529-537 (1865).
Perhaps the adoption of the nomenclature proposed by Mr G. de Longchamps in the Journal de Mathe'matiques Me'mentaires, 2nd series, V. 109 (1886) would be advantageous.
(1) L. BOG + L BO'C = ISO" + A.
FIGURE 37.
For ^BOC =A + AB0 + AC0, = A + CBO'+ BCO', and _B0'C = A-l-AB0' + AC0'; therefore L BOC + L BO'C = 2A + B + C, = 180° +A.
Professor J. Neuberg's Mimoire sur le TiStraidre, p. 10 (1884).
12 Vol. 13 Published online by Cambridge University Press 174 (2) In triangle ABC, APj BP2 CP3 are concurrent at O, and their isogonals AQ2 BQ, CQ3 are concurrent at 0'.
FIGURE 36.
Suppose BP2 BQ2 to form one straight line and CP3 CQ3 „ „ „ „ „ ; then the points O O' coincide. There are four cases.
(a) If BP2 CP3 bisect the interior angles B, C, then AP, bisects the interior angle A.
(b) If BP, CP3 bisect the exterior angles B, C, then AP, bisects the interior angle A.
(c) If BP2 bisects the interior angle B and CP3 „ „ exterior „ C, then AP! „ „ exterior „ A.
(d) If BP2 bisects the exterior angle B and CP3 „ „ interior „ C, then APi ,, „ exterior ,, A.
Hence the six bisectors of the angles of a triangle meet three by three in four points.
FIGURE 30.
(3) By considering A P ^ as the triangle, and AB, AC as the isogonals BP^CP, : BC^-CQ, = AP/2: AQ,2.
Similarly CP./AP.,: CQ.,-AQ2 = BP,2: BQ,2, and AP3-BP3: AQ,BQ3 -- CP,2: CQ3 2; BP1-CPs-AP,-CP1-APa.BP3 _ APrBP./ CP/ BQ1-CQ2-AQ3CQ1-AQ2 BQ3 AQ;-BQ2-CQ:J r Now BPfCPj-AP^CP^AP.-BP, and BQICQ2-AQ3 = CQ 1AQ 2BQ 3; therefore AP 1BP. jCP 3_ BP.CP.yAP, _ CP.-AP.-BP, AQI-BQ2-CQ3 BQ.CQ.AQ, CQ^AQ.-BQ," C. Adams's Eiyenschaften dcs...Dreiecks, p. 8 (184C). Adams gives also (3).
Published online by Cambridge University Press 175 § 4.
Positions of two isogonal points with reference to a triangle.
(1) Any point on a side has for isogonal point the opposite vertex.
(2) A vertex has for isogonal point any point on the opposite side.
(3) A point inside the triangle has its isogonal point also inside the triangle.
(4) If a point be outside the triangle and situated in the a'hgle vertically opposite to L. BAC, for example, its isogonal point will be outside the triangle and situated in that segment of the circumcircle (remote from A) cut off by BO.
(5) If a point be outside the circumcircle and situated within the angle BAC, for example, its isogonal point will be outside the circumcircle and situated within the same angle.
(6) If a point be on the circumference of the circumcircle, its isogonal point will be at infinity.
The trutli of these statements, which are not quite obvious, may be ascertained by the construction of a few figures. Of the last statement the following proof may be given : — FIGURE 39.
If AD, BE, OF, be three parallel lines drawn through the ver-tices of a triangle ABC, their three isogonals will be concurrent at a point on the circumference of the circumcircle. f Because AD, BE, CF are parallel, therefore arc AE = arc BD, arc BC==are EF.
Make arc CP equal to arc BD ; join AP, BP, CP.
They are all given by Mr J. J. A. Mathieu in Nouvdles Annalcs, 2nd series, IV. 403 (1865).
+ Professor Eugenio Beltrami in Mcmorie dd VAccademia delle Scienze del Istituto di Bologna, 2nd series, II., 383 (1863).
Published online by Cambridge University Press 176 Since arc CP = arc BD, therefore ^CAP= L. BAD, and AP is isogonal to AD.
Since arc CP = arc AE, therefore L CBP = L ABE, and BP is isogonal to BE.
Since arc BC = arc EF, arc CP = arc AE, therefore arc BP = arc AF ; therefore u BCP = L ACF, and CP is isogonal to CF.
Hence, if P be a point on the circumcircle of ABC, the point isogonal to it is the point of concurrency of AD, BE, CF.
(1) AD is perpendicular to the Wallace line P (ABC).
This follows from § 1, (3).
§ 5.
If three angular transversals cut the opposite sides in three collinear points, their isogonals will also cut the opposite sides in three oollinear points. FIGURE 40.
Let AD, AD'; BE, BE'; CF, CF' be pairs of isogonals ; then if D, E, F, be collinear, so will D', E', F'.
BD • BD' c2 F o r CD • CD' ~ ~¥' CE • CE' a AE • AE' c-AF • AF' V BF . BF' ~ a-x, . BD • CE • AF BD' • CE' • AF' t h e r 6 f 0 r e C D A E - B F • OB' • AE' • JBF = L Professor J. Neuberg in Rouche and Comberousse's Traiti de Geometrit, First Part, p. 439 (1891).
t Townsend's Modern Geometry, I. 181 (1863).
Published online by Cambridge University Press 177 Now BD-CE-AF _ .
CD AE BF ~ L ' therefore CD • AE • BF B D ' ' C E ' ' A F ' CD' • AE' • BF therefore D', E', F' are collinear.
§6-If 0 be any point in the plane of triangle ABC, and AO BO CO meet the circumcircle in A1 5, C1 and D E F be the projections of 0 on BC CA AB the triangles A1BlC1 DEF are directly similar, and the point 0 of triangle DEF corresponds to that point of A^BJJ^ which is isogonal to 0.
FIGURE 41.
For the points O F B D are concyclic ; therefore L FDO = L FBO = L B ^ O .
Similarly u EDO = L CJAJO.
The demonstration may be easily seen to apply to the more general case where Al Ht C1 are taken inverse to O with any other constant of inversion. (1) If O be the orthocentre of ABC, it must be the incentre or an excentre of DEF, and therefore the incentre or an excentre of A.B.C, (2) If O be the circumcentre of ABC, it must be the orthocentre of DEF, and therefore the circumcentre of AjB,^.
(3) If O be the incentre of ABC, it must be the circumcentre of DEF, and therefore the orthocentre of AJBJC!.
(4) If O be an excentre of ABC, it must be the circumcentre of DEF, and therefore the orthocentre of AJBJCJ.
Mr E. M. Langley and Professor Neuberg. (1)—(4) are Mr Langley's. See the Seventeenth, General Report of the Association for the Improrcment of Geometrical Teaching, p. 45 (1891.) Published online by Cambridge University Press 178 If two points be isoyonal with respect to a triangle their six projections on the sides of the triangle are concyclic FIGURE 42.
Let O, O' be isogonal with respect to ABC, and let D, E, F, D', E', F' be their projections on the sides BC, CA, AB.
Then EF is antiparallel to E'F' with respect to A ; therefore E,E' F,F', are concyclic.
Similarly F,F' D,D' „ and D,D' E,E' „ „ ; therefore the six points are concyclic.
If 0 0' be isogonal points with respect to ABC, and D E F D'E'F' be their respective projections on ISC CA A IS, then AO BO CO are perpendicular to the sides of D'E'F' AOBO'CO' „ „ „ DEF.
FIGURE 42.
This has been established in § 1, (3).
The application of the preceding properties of isogonals to the particular case of medians and symmedians will be taken up in a succeeding paper.
Steiner in Gergonne's Animles, xix. 37-64 (1828).
In the same article will be found also the property of § 8.
Published online by Cambridge University Press |
5385 | https://jila.colorado.edu/~ajsh/bh/hawk.html | Hawking Radiation
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The Extremal Reissner-Nordström Geometry
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--- ---
| Hawking radiation | | | | | | Classically, black holes are black. Quantum mechanically, black holes radiate, with a radiation known as Hawking radiation, after the British physicist Stephen Hawking who first proposed it. The animation at top left cartoons the Hawking radiation from a black hole of the size shown at bottom left. The blobs are supposed to be individual photons. Notice, first, that the photons have ‘sizes’ (wavelengths) comparable to the size of the black hole, and, second, that the Hawking radiation is not very bright — the black hole emits roughly one photon every light crossing time of the black hole. So a black hole observed by its Hawking radiation looks fuzzy, a quantum mechanical object. This is one animation that I did not compute mathematically. How do you draw a quantum mechanical object, whose appearance depends not only on the object but also on the way the observer chooses to observe it? I figured my impressionism was good enough here. Hawking radiation has a blackbody (Planck) spectrum with a temperature T given by kT=ℏg2πc=ℏc4πrs , where k is Boltzmann’s constant, ℏ=h/(2π) is Planck’s constant divided by 2π, and g=GM/r2s is the surface gravity at the horizon, the Schwarzschild radius rs, of the black hole of mass M. Numerically, the Hawking temperature is T=4×10−20gKelvin if the gravitational acceleration g is measured in Earth gravities (gees). The Hawking luminosity L of the black hole is given by the usual Stefan-Boltzmann blackbody formula L=AσT4 where A=4πr2s is the surface area of the black hole, and σ=π2k4/(60c2ℏ3) is the Stefan-Boltzmann constant. If the Hawking temperature exceeds the rest mass energy of a particle type, then the black hole radiates particles and antiparticles of that type, in addition to photons, and the Hawking luminosity of the black hole rises to L=A(neff/2)σT4 , where neff is the effective number of relativistic particle types, including the two helicity types (polarizations) of the photon. Black holes for which astronomical evidence exists have masses ranging from stellar-sized black holes of a few solar masses, up to supermassive black holes in the nuclei of galaxies, such as the 6×109 solar mass black hole at the centre of the galaxy Messier 87. The Hawking radiation from such black holes is minuscule. The Hawking temperature of a 30 solar mass black hole is a tiny 2×10−9Kelvin, and its Hawking luminosity a miserable 10−31Watts. Bigger black holes are colder and dimmer: the Hawking temperature is inversely proportional to the mass, while the Hawking luminosity is inversely proportional to the square of the mass. Answer to the quiz question 7: No, the x-ray emission from x-ray binary star systems is not Hawking radiation. The x-rays come not from the black hole (or neutron star), but from a circling accretion disk of hot gas. The accretion disk is heated to x-ray emitting temperatures by the release of gravitational energy as the gas spirals toward the black hole (or neutron star). The Hawking radiation from the black hole is exquisitely tiny by comparison. Claus Kiefer (1998) ‘‘Towards a Full Quantum Theory of Black Holes’’ (gr-qc/9803049) gives a pedagogical review of Hawking radiation and other quantum aspects of black holes. 15 Apr 2003 update. Adam Helfer (2003) ‘‘Do black holes radiate?’’ (gr-qc/0304042) opens with the statement: ‘‘The prediction that black holes radiate due to quantum effects is often considered one of the most secure in quantum field theory in curved space-time. Yet this prediction rests on two dubious assumptions ...’’. This delightfully readable review paper does an excellent job of convincing the reader that Hawking radiation is still far from being an established prediction of the quantum physics of black holes. The paper gives the clearest exposition of Hawking radiation that I know of, emphasizing the physical concepts while simplifying the mathematics to its barest essentials (not that the mathematics is simple even in stripped form). 14 Mar 2025 update. At last a definitive calculation. arXiv:2401.03098 “Hawking radiation inside a rotating black hole” calculates the spectrum of Hawking radiation seen by various observers at various positions inside as well as outside a rotating black hole. The Hawking spectrum is in general nonthermal, and it diverges at the inner horizon. See Tyler McMaken's PhD thesis defense July 2024: Quantum Effects inside Rotating, Accreting Black Holes. --- | | | Evaporation of a mini black hole Black holes get the energy to radiate Hawking radiation from their rest mass energy. So if a black hole is not accreting mass from outside, it will lose mass by Hawking radiation, and will eventually evaporate. For astronomical black holes, the evaporation time is prodigiously long — about 1061 times the age of the Universe for a 30 solar mass black hole. However, the evaporation time is shorter for smaller black holes (evaporation time t is proportional to M3), and black holes with masses less than about 1011kg (the mass of a small mountain) can evaporate in less than the age of the Universe. The Hawking temperature of such mini black holes is high: a 1011kg black hole has a temperature of about 1012Kelvin, equivalent to the rest mass energy of a proton. The gravitational pull of such a mini black hole would be about 1 gee at a distance of 1 metre. It is not well established what an evaporating mini black hole would actually look like in realistic detail. The Hawking radiation itself would consist of fiercely energetic particles, antiparticles, and gamma rays. Such radiation is invisible to the human eye, so optically the evaporating black hole might look like a dud. However, it is also possible that the Hawking radiation, rather than emerging directly, might power a hadronic fireball that would degrade the radiation into particles and gamma rays of less extreme energy, possibly making the evaporating black hole visible to the eye. Whatever the case, you would not want to go near an evaporating mini black hole, which would be a source of lethal gamma rays and energetic particles, even if it didn’t look like much visually. The animation at left is a fanciful depiction of the final moments of the evaporation of a hypothetical mini black hole. In the final second of its existence, the mini black hole radiates about 1000 tonnes of rest mass energy. Such an explosion is large by human standards, but modest by astronomical standards. An evaporating black hole would be detectable from Earth only if it went off within the solar system, or at best no further away than the nearest star. What’s the cross shape? Telescopic diffraction spikes, added for artistic effect. Compare these beautiful Hubble Space Telescope pictures of the Orion nebula or of stars in the Messier 4 globular star cluster. | --- Back to The Extremal Reissner-Nordström Geometry Forward to Black Hole Quiz Andrew Hamilton’s Homepage Other Relativity and Black Hole links | | | index | movies | approach | orbit | singularity | dive | Schwarzschild | wormhole | collapse | Reissner-Nordström | extremal RN | Hawking | quiz | home | links | Updated 19 Apr 1998; converted to mathjax 3 Feb 2018 | |
5386 | https://www.intmath.com/matrices-determinants/eigenvalues-eigenvectors-calculator.php | Eigenvalues and eigenvectors calculator
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Eigenvalues and eigenvectors calculator
This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. It will find the eigenvalues of that matrix, and also outputs the corresponding eigenvectors.
For background on these concepts, see 7. Eigenvalues and Eigenvectors
Instructions
First, choose the matrix size you want to enter. You will see a randomly generated matrix to give you an idea of what your output will look like.
Then, enter your own numbers in the boxes that appear. You can enter integers or decimals. (More advanced entry and output is in the works, but not available yet.)
On a keyboard, you can use the tab key to easily move to the next matrix entry box.
Click calculate when ready.
The output will involve either real and/or complex eigenvalues and eigenvector entries.
You can change the precision (number of significant digits) of the answers, using the pull-down menu.
Eigenvalues and eigenvectors calculator
Matrix size: Precision:
calculate
3 real eigenvalues:−4.7775,9.2613,6.6162
For real eigenvalue λ 1 = −4.7775, the eigenvector is:
v1 = [−1.172, 0.55778, 1]T
For real eigenvalue λ 2 = 9.2613, the eigenvector is:
v2 = [ 0.79942, 0.47821, 1]T
For real eigenvalue λ 3 = 6.6162, the eigenvector is:
v3 = [ 6.3373, 14.282, 1]T
NOTE 1: The eigenvector output you see here may not be the same as what you obtain on paper. Remember, you can have any scalar multiple of the eigenvector, and it will still be an eigenvector. The convention used here is eigenvectors have been scaled so the final entry is 1.
NOTE 2: The larger matrices involve a lot of calculation, so expect the answer to take a bit longer.
NOTE 3: Eigenvectors are usually column vectors, but the larger ones would take up a lot of vertical space, so they are written horizontally, with a "T" superscript (known as the transpose of the matrix).
NOTE 4: When there are complex eigenvalues, there's always an even number of them, and they always appear as a complex conjugate pair, e.g. 3 + 5 i and 3 − 5 i.
NOTE 5: When there are eigenvectors with complex elements, there's always an even number of such eigenvectors, and the corresponding elements always appear as complex conjugate pairs. (It may take some manipulating by multiplying each element by a complex number to see this is so in some cases.)
Credit: This calculator was built using the Numeric.js library.
8. Applications of Eigenvalues and Eigenvectors
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5387 | https://www.tax.ny.gov/pit/file/it201i-tax-tables.htm | Tax tables for Form IT-201
Disponible en español
2022 New York State Tax Table
If your New York adjusted gross income, Form IT-201, line 33 is more than $107,650, you cannot use these tables. To compute your tax, see Tax computation - New York AGI of more than $107,650. If you fail to follow these instructions, you may have to pay interest and penalty if the income tax you report on your return is less than the correct amount.
Example: A married couple are filing a joint return on Form IT-201. Their taxable income on line 38 is $38,275. First, they find the 38,250 - 38,300 income line. Next, they find the column for Married filing jointly and read down the column. The amount shown where the income line and filing status column meet is $1,809. This is the tax amount they must write on Form IT-201, line 39.
$0—$5,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
| Your New York State tax is:
$0 | $13 | $0 | $0 | $0
13 | 25 | 1 | 1 | 1
25 | 50 | 2 | 2 | 2
50 | 100 | 3 | 3 | 3
100 | 150 | 5 | 5 | 5
150 | 200 | 7 | 7 | 7
200 | 250 | 9 | 9 | 9
250 | 300 | 11 | 11 | 11
300 | 350 | 13 | 13 | 13
350 | 400 | 15 | 15 | 15
400 | 450 | 17 | 17 | 17
450 | 500 | 19 | 19 | 19
500 | 550 | 21 | 21 | 21
550 | 600 | 23 | 23 | 23
600 | 650 | 25 | 25 | 25
650 | 700 | 27 | 27 | 27
700 | 750 | 29 | 29 | 29
750 | 800 | 31 | 31 | 31
800 | 850 | 33 | 33 | 33
850 | 900 | 35 | 35 | 35
900 | 950 | 37 | 37 | 37
950 | 1,000 | 39 | 39 | 39
1,000 | Your New York State tax is:
1,000 | 1,050 | 41 | 41 | 41
1,050 | 1,100 | 43 | 43 | 43
1,100 | 1,150 | 45 | 45 | 45
1,150 | 1,200 | 47 | 47 | 47
1,200 | 1,250 | 49 | 49 | 49
1,250 | 1,300 | 51 | 51 | 51
1,300 | 1,350 | 53 | 53 | 53
1,350 | 1,400 | 55 | 55 | 55
1,400 | 1,450 | 57 | 57 | 57
1,450 | 1,500 | 59 | 59 | 59
1,500 | 1,550 | 61 | 61 | 61
1,550 | 1,600 | 63 | 63 | 63
1,600 | 1,650 | 65 | 65 | 65
1,650 | 1,700 | 67 | 67 | 67
1,700 | 1,750 | 69 | 69 | 69
1,750 | 1,800 | 71 | 71 | 71
1,800 | 1,850 | 73 | 73 | 73
1,850 | 1,900 | 75 | 75 | 75
1,900 | 1,950 | 77 | 77 | 77
1,950 | 2,000 | 79 | 79 | 79
This column must also be used by a qualifying surviving spouse
$6,000—$14,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
2,000 | Your New York State tax is:
2,000 | 2,050 | 81 | 81 | 81
2,050 | 2,100 | 83 | 83 | 83
2,100 | 2,150 | 85 | 85 | 85
2,150 | 2,200 | 87 | 87 | 87
2,200 | 2,250 | 89 | 89 | 89
2,250 | 2,300 | 91 | 91 | 91
2,300 | 2,350 | 93 | 93 | 93
2,350 | 2,400 | 95 | 95 | 95
2,400 | 2,450 | 97 | 97 | 97
2,450 | 2,500 | 99 | 99 | 99
2,500 | 2,550 | 101 | 101 | 101
2,550 | 2,600 | 103 | 103 | 103
2,600 | 2,650 | 105 | 105 | 105
2,650 | 2,700 | 107 | 107 | 107
2,700 | 2,750 | 109 | 109 | 109
2,750 | 2,800 | 111 | 111 | 111
2,800 | 2,850 | 113 | 113 | 113
2,850 | 2,900 | 115 | 115 | 115
2,900 | 2,950 | 117 | 117 | 117
2,950 | 3,000 | 119 | 119 | 119
3,000 | Your New York State tax is:
3,000 | 3,050 | 121 | 121 | 121
3,050 | 3,100 | 123 | 123 | 123
3,100 | 3,150 | 125 | 125 | 125
3,150 | 3,200 | 127 | 127 | 127
3,200 | 3,250 | 129 | 129 | 129
3,250 | 3,300 | 131 | 131 | 131
3,300 | 3,350 | 133 | 133 | 133
3,350 | 3,400 | 135 | 135 | 135
3,400 | 3,450 | 137 | 137 | 137
3,450 | 3,500 | 139 | 139 | 139
3,500 | 3,550 | 141 | 141 | 141
3,550 | 3,600 | 143 | 143 | 143
3,600 | 3,650 | 145 | 145 | 145
3,650 | 3,700 | 147 | 147 | 147
3,700 | 3,750 | 149 | 149 | 149
3,750 | 3,800 | 151 | 151 | 151
3,800 | 3,850 | 153 | 153 | 153
3,850 | 3,900 | 155 | 155 | 155
3,900 | 3,950 | 157 | 157 | 157
3,950 | 4,000 | 159 | 159 | 159
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
4,000 | Your New York State tax is:
4,000 | 4,050 | 161 | 161 | 161
4,050 | 4,100 | 163 | 163 | 163
4,100 | 4,150 | 165 | 165 | 165
4,150 | 4,200 | 167 | 167 | 167
4,200 | 4,250 | 169 | 169 | 169
4,250 | 4,300 | 171 | 171 | 171
4,300 | 4,350 | 173 | 173 | 173
4,350 | 4,400 | 175 | 175 | 175
4,400 | 4,450 | 177 | 177 | 177
4,450 | 4,500 | 179 | 179 | 179
4,500 | 4,550 | 181 | 181 | 181
4,550 | 4,600 | 183 | 183 | 183
4,600 | 4,650 | 185 | 185 | 185
4,650 | 4,700 | 187 | 187 | 187
4,700 | 4,750 | 189 | 189 | 189
4,750 | 4,800 | 191 | 191 | 191
4,800 | 4,850 | 193 | 193 | 193
4,850 | 4,900 | 195 | 195 | 195
4,900 | 4,950 | 197 | 197 | 197
4,950 | 5,000 | 199 | 199 | 199
5,000 | Your New York State tax is:
5,000 | 5,050 | 201 | 201 | 201
5,050 | 5,100 | 203 | 203 | 203
5,100 | 5,150 | 205 | 205 | 205
5,150 | 5,200 | 207 | 207 | 207
5,200 | 5,250 | 209 | 209 | 209
5,250 | 5,300 | 211 | 211 | 211
5,300 | 5,350 | 213 | 213 | 213
5,350 | 5,400 | 215 | 215 | 215
5,400 | 5,450 | 217 | 217 | 217
5,450 | 5,500 | 219 | 219 | 219
5,500 | 5,550 | 221 | 221 | 221
5,550 | 5,600 | 223 | 223 | 223
5,600 | 5,650 | 225 | 225 | 225
5,650 | 5,700 | 227 | 227 | 227
5,700 | 5,750 | 229 | 229 | 229
5,750 | 5,800 | 231 | 231 | 231
5,800 | 5,850 | 233 | 233 | 233
5,850 | 5,900 | 235 | 235 | 235
5,900 | 5,950 | 237 | 237 | 237
5,950 | 6,000 | 239 | 239 | 239
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
6,000 | Your New York State tax is:
6,000 | 6,050 | 241 | 241 | 241
6,050 | 6,100 | 243 | 243 | 243
6,100 | 6,150 | 245 | 245 | 245
6,150 | 6,200 | 247 | 247 | 247
6,200 | 6,250 | 249 | 249 | 249
6,250 | 6,300 | 251 | 251 | 251
6,300 | 6,350 | 253 | 253 | 253
6,350 | 6,400 | 255 | 255 | 255
6,400 | 6,450 | 257 | 257 | 257
6,450 | 6,500 | 259 | 259 | 259
6,500 | 6,550 | 261 | 261 | 261
6,550 | 6,600 | 263 | 263 | 263
6,600 | 6,650 | 265 | 265 | 265
6,650 | 6,700 | 267 | 267 | 267
6,700 | 6,750 | 269 | 269 | 269
6,750 | 6,800 | 271 | 271 | 271
6,800 | 6,850 | 273 | 273 | 273
6,850 | 6,900 | 275 | 275 | 275
6,900 | 6,950 | 277 | 277 | 277
6,950 | 7,000 | 279 | 279 | 279
7,000 | Your New York State tax is:
7,000 | 7,050 | 281 | 281 | 281
7,050 | 7,100 | 283 | 283 | 283
7,100 | 7,150 | 285 | 285 | 285
7,150 | 7,200 | 287 | 287 | 287
7,200 | 7,250 | 289 | 289 | 289
7,250 | 7,300 | 291 | 291 | 291
7,300 | 7,350 | 293 | 293 | 293
7,350 | 7,400 | 295 | 295 | 295
7,400 | 7,450 | 297 | 297 | 297
7,450 | 7,500 | 299 | 299 | 299
7,500 | 7,550 | 301 | 301 | 301
7,550 | 7,600 | 303 | 303 | 303
7,600 | 7,650 | 305 | 305 | 305
7,650 | 7,700 | 307 | 307 | 307
7,700 | 7,750 | 309 | 309 | 309
7,750 | 7,800 | 311 | 311 | 311
7,800 | 7,850 | 313 | 313 | 313
7,850 | 7,900 | 315 | 315 | 315
7,900 | 7,950 | 317 | 317 | 317
7,950 | 8,000 | 319 | 319 | 319
8,000 | Your New York State tax is:
8,000 | 8,050 | 321 | 321 | 321
8,050 | 8,100 | 323 | 323 | 323
8,100 | 8,150 | 325 | 325 | 325
8,150 | 8,200 | 327 | 327 | 327
8,200 | 8,250 | 329 | 329 | 329
8,250 | 8,300 | 331 | 331 | 331
8,300 | 8,350 | 333 | 333 | 333
8,350 | 8,400 | 335 | 335 | 335
8,400 | 8,450 | 337 | 337 | 337
8,450 | 8,500 | 339 | 339 | 339
8,500 | 8,550 | 341 | 341 | 341
8,550 | 8,600 | 343 | 343 | 343
8,600 | 8,650 | 346 | 345 | 345
8,650 | 8,700 | 348 | 347 | 347
8,700 | 8,750 | 350 | 349 | 349
8,750 | 8,800 | 352 | 351 | 351
8,800 | 8,850 | 355 | 353 | 353
8,850 | 8,900 | 357 | 355 | 355
8,900 | 8,950 | 359 | 357 | 357
8,950 | 9,000 | 361 | 359 | 359
This column must also be used by a qualifying surviving spouse
$15,000—$23,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
9,000 | Your New York State tax is:
9,000 | 9,050 | 364 | 361 | 361
9,050 | 9,100 | 366 | 363 | 363
9,100 | 9,150 | 368 | 365 | 365
9,150 | 9,200 | 370 | 367 | 367
9,200 | 9,250 | 373 | 369 | 369
9,250 | 9,300 | 375 | 371 | 371
9,300 | 9,350 | 377 | 373 | 373
9,350 | 9,400 | 379 | 375 | 375
9,400 | 9,450 | 382 | 377 | 377
9,450 | 9,500 | 384 | 379 | 379
9,500 | 9,550 | 386 | 381 | 381
9,550 | 9,600 | 388 | 383 | 383
9,600 | 9,650 | 391 | 385 | 385
9,650 | 9,700 | 393 | 387 | 387
9,700 | 9,750 | 395 | 389 | 389
9,750 | 9,800 | 397 | 391 | 391
9,800 | 9,850 | 400 | 393 | 393
9,850 | 9,900 | 402 | 395 | 395
9,900 | 9,950 | 404 | 397 | 397
9,950 | 10,000 | 406 | 399 | 399
10,000 | Your New York State tax is:
10,000 | 10,050 | 409 | 401 | 401
10,050 | 10,100 | 411 | 403 | 403
10,100 | 10,150 | 413 | 405 | 405
10,150 | 10,200 | 415 | 407 | 407
10,200 | 10,250 | 418 | 409 | 409
10,250 | 10,300 | 420 | 411 | 411
10,300 | 10,350 | 422 | 413 | 413
10,350 | 10,400 | 424 | 415 | 415
10,400 | 10,450 | 427 | 417 | 417
10,450 | 10,500 | 429 | 419 | 419
10,500 | 10,550 | 431 | 421 | 421
10,550 | 10,600 | 433 | 423 | 423
10,600 | 10,650 | 436 | 425 | 425
10,650 | 10,700 | 438 | 427 | 427
10,700 | 10,750 | 440 | 429 | 429
10,750 | 10,800 | 442 | 431 | 431
10,800 | 10,850 | 445 | 433 | 433
10,850 | 10,900 | 447 | 435 | 435
10,900 | 10,950 | 449 | 437 | 437
10,950 | 11,000 | 451 | 439 | 439
11,000 | Your New York State tax is:
11,000 | 11,050 | 454 | 441 | 441
11,050 | 11,100 | 456 | 443 | 443
11,100 | 11,150 | 458 | 445 | 445
11,150 | 11,200 | 460 | 447 | 447
11,200 | 11,250 | 463 | 449 | 449
11,250 | 11,300 | 465 | 451 | 451
11,300 | 11,350 | 467 | 453 | 453
11,350 | 11,400 | 469 | 455 | 455
11,400 | 11,450 | 472 | 457 | 457
11,450 | 11,500 | 474 | 459 | 459
11,500 | 11,550 | 476 | 461 | 461
11,550 | 11,600 | 478 | 463 | 463
11,600 | 11,650 | 481 | 465 | 465
11,650 | 11,700 | 483 | 467 | 467
11,700 | 11,750 | 485 | 469 | 469
11,750 | 11,800 | 488 | 471 | 471
11,800 | 11,850 | 491 | 473 | 473
11,850 | 11,900 | 493 | 475 | 475
11,900 | 11,950 | 496 | 477 | 477
11,950 | 12,000 | 498 | 479 | 479
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
12,000 | Your New York State tax is:
12,000 | 12,050 | 501 | 481 | 481
12,050 | 12,100 | 504 | 483 | 483
12,100 | 12,150 | 506 | 485 | 485
12,150 | 12,200 | 509 | 487 | 487
12,200 | 12,250 | 512 | 489 | 489
12,250 | 12,300 | 514 | 491 | 491
12,300 | 12,350 | 517 | 493 | 493
12,350 | 12,400 | 519 | 495 | 495
12,400 | 12,450 | 522 | 497 | 497
12,450 | 12,500 | 525 | 499 | 499
12,500 | 12,550 | 527 | 501 | 501
12,550 | 12,600 | 530 | 503 | 503
12,600 | 12,650 | 533 | 505 | 505
12,650 | 12,700 | 535 | 507 | 507
12,700 | 12,750 | 538 | 509 | 509
12,750 | 12,800 | 540 | 511 | 511
12,800 | 12,850 | 543 | 513 | 513
12,850 | 12,900 | 546 | 515 | 515
12,900 | 12,950 | 548 | 517 | 518
12,950 | 13,000 | 551 | 519 | 520
13,000 | Your New York State tax is:
13,000 | 13,050 | 554 | 521 | 522
13,050 | 13,100 | 556 | 523 | 524
13,100 | 13,150 | 559 | 525 | 527
13,150 | 13,200 | 561 | 527 | 529
13,200 | 13,250 | 564 | 529 | 531
13,250 | 13,300 | 567 | 531 | 533
13,300 | 13,350 | 569 | 533 | 536
13,350 | 13,400 | 572 | 535 | 538
13,400 | 13,450 | 575 | 537 | 540
13,450 | 13,500 | 577 | 539 | 542
13,500 | 13,550 | 580 | 541 | 545
13,550 | 13,600 | 582 | 543 | 547
13,600 | 13,650 | 585 | 545 | 549
13,650 | 13,700 | 588 | 547 | 551
13,700 | 13,750 | 590 | 549 | 554
13,750 | 13,800 | 593 | 551 | 556
13,800 | 13,850 | 596 | 553 | 558
13,850 | 13,900 | 598 | 555 | 560
13,900 | 13,950 | 601 | 557 | 563
13,950 | 14,000 | 604 | 559 | 565
14,000 | Your New York State tax is:
14,000 | 14,050 | 607 | 561 | 567
14,050 | 14,100 | 610 | 563 | 569
14,100 | 14,150 | 613 | 565 | 572
14,150 | 14,200 | 616 | 567 | 574
14,200 | 14,250 | 619 | 569 | 576
14,250 | 14,300 | 622 | 571 | 578
14,300 | 14,350 | 625 | 573 | 581
14,350 | 14,400 | 628 | 575 | 583
14,400 | 14,450 | 631 | 577 | 585
14,450 | 14,500 | 634 | 579 | 587
14,500 | 14,550 | 637 | 581 | 590
14,550 | 14,600 | 639 | 583 | 592
14,600 | 14,650 | 642 | 585 | 594
14,650 | 14,700 | 645 | 587 | 596
14,700 | 14,750 | 648 | 589 | 599
14,750 | 14,800 | 651 | 591 | 601
14,800 | 14,850 | 654 | 593 | 603
14,850 | 14,900 | 657 | 595 | 605
14,900 | 14,950 | 660 | 597 | 608
14,950 | 15,000 | 663 | 599 | 610
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
15,000 | Your New York State tax is:
15,000 | 15,050 | 666 | 601 | 612
15,050 | 15,100 | 669 | 603 | 614
15,100 | 15,150 | 672 | 605 | 617
15,150 | 15,200 | 675 | 607 | 619
15,200 | 15,250 | 678 | 609 | 621
15,250 | 15,300 | 680 | 611 | 623
15,300 | 15,350 | 683 | 613 | 626
15,350 | 15,400 | 686 | 615 | 628
15,400 | 15,450 | 689 | 617 | 630
15,450 | 15,500 | 692 | 619 | 632
15,500 | 15,550 | 695 | 621 | 635
15,550 | 15,600 | 698 | 623 | 637
15,600 | 15,650 | 701 | 625 | 639
15,650 | 15,700 | 704 | 627 | 641
15,700 | 15,750 | 707 | 629 | 644
15,750 | 15,800 | 710 | 631 | 646
15,800 | 15,850 | 713 | 633 | 648
15,850 | 15,900 | 716 | 635 | 650
15,900 | 15,950 | 718 | 637 | 653
15,950 | 16,000 | 721 | 639 | 655
16,000 | Your New York State tax is:
16,000 | 16,050 | 724 | 641 | 657
16,050 | 16,100 | 727 | 643 | 659
16,100 | 16,150 | 730 | 645 | 662
16,150 | 16,200 | 733 | 647 | 664
16,200 | 16,250 | 736 | 649 | 666
16,250 | 16,300 | 739 | 651 | 668
16,300 | 16,350 | 742 | 653 | 671
16,350 | 16,400 | 745 | 655 | 673
16,400 | 16,450 | 748 | 657 | 675
16,450 | 16,500 | 751 | 659 | 677
16,500 | 16,550 | 754 | 661 | 680
16,550 | 16,600 | 756 | 663 | 682
16,600 | 16,650 | 759 | 665 | 684
16,650 | 16,700 | 762 | 667 | 686
16,700 | 16,750 | 765 | 669 | 689
16,750 | 16,800 | 768 | 671 | 691
16,800 | 16,850 | 771 | 673 | 693
16,850 | 16,900 | 774 | 675 | 695
16,900 | 16,950 | 777 | 677 | 698
16,950 | 17,000 | 780 | 679 | 700
17,000 | Your New York State tax is:
17,000 | 17,050 | 783 | 681 | 702
17,050 | 17,100 | 786 | 683 | 704
17,100 | 17,150 | 789 | 685 | 707
17,150 | 17,200 | 792 | 687 | 709
17,200 | 17,250 | 795 | 689 | 711
17,250 | 17,300 | 797 | 692 | 713
17,300 | 17,350 | 800 | 694 | 716
17,350 | 17,400 | 803 | 696 | 718
17,400 | 17,450 | 806 | 698 | 720
17,450 | 17,500 | 809 | 701 | 722
17,500 | 17,550 | 812 | 703 | 725
17,550 | 17,600 | 815 | 705 | 727
17,600 | 17,650 | 818 | 707 | 729
17,650 | 17,700 | 821 | 710 | 731
17,700 | 17,750 | 824 | 712 | 734
17,750 | 17,800 | 827 | 714 | 737
17,800 | 17,850 | 830 | 716 | 739
17,850 | 17,900 | 833 | 719 | 742
17,900 | 17,950 | 835 | 721 | 744
17,950 | 18,000 | 838 | 723 | 747
This column must also be used by a qualifying surviving spouse
$24,000—$32,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
18,000 | Your New York State tax is:
18,000 | 18,050 | 841 | 725 | 750
18,050 | 18,100 | 844 | 728 | 752
18,100 | 18,150 | 847 | 730 | 755
18,150 | 18,200 | 850 | 732 | 758
18,200 | 18,250 | 853 | 734 | 760
18,250 | 18,300 | 856 | 737 | 763
18,300 | 18,350 | 859 | 739 | 765
18,350 | 18,400 | 862 | 741 | 768
18,400 | 18,450 | 865 | 743 | 771
18,450 | 18,500 | 868 | 746 | 773
18,500 | 18,550 | 871 | 748 | 776
18,550 | 18,600 | 873 | 750 | 779
18,600 | 18,650 | 876 | 752 | 781
18,650 | 18,700 | 879 | 755 | 784
18,700 | 18,750 | 882 | 757 | 786
18,750 | 18,800 | 885 | 759 | 789
18,800 | 18,850 | 888 | 761 | 792
18,850 | 18,900 | 891 | 764 | 794
18,900 | 18,950 | 894 | 766 | 797
18,950 | 19,000 | 897 | 768 | 800
19,000 | Your New York State tax is:
19,000 | 19,050 | 900 | 770 | 802
19,050 | 19,100 | 903 | 773 | 805
19,100 | 19,150 | 906 | 775 | 807
19,150 | 19,200 | 909 | 777 | 810
19,200 | 19,250 | 912 | 779 | 813
19,250 | 19,300 | 914 | 782 | 815
19,300 | 19,350 | 917 | 784 | 818
19,350 | 19,400 | 920 | 786 | 821
19,400 | 19,450 | 923 | 788 | 823
19,450 | 19,500 | 926 | 791 | 826
19,500 | 19,550 | 929 | 793 | 828
19,550 | 19,600 | 932 | 795 | 831
19,600 | 19,650 | 935 | 797 | 834
19,650 | 19,700 | 938 | 800 | 836
19,700 | 19,750 | 941 | 802 | 839
19,750 | 19,800 | 944 | 804 | 842
19,800 | 19,850 | 947 | 806 | 844
19,850 | 19,900 | 950 | 809 | 847
19,900 | 19,950 | 952 | 811 | 849
19,950 | 20,000 | 955 | 813 | 852
20,000 | Your New York State tax is:
20,000 | 20,050 | 958 | 815 | 855
20,050 | 20,100 | 961 | 818 | 857
20,100 | 20,150 | 964 | 820 | 860
20,150 | 20,200 | 967 | 822 | 863
20,200 | 20,250 | 970 | 824 | 865
20,250 | 20,300 | 973 | 827 | 868
20,300 | 20,350 | 976 | 829 | 870
20,350 | 20,400 | 979 | 831 | 873
20,400 | 20,450 | 982 | 833 | 876
20,450 | 20,500 | 985 | 836 | 878
20,500 | 20,550 | 988 | 838 | 881
20,550 | 20,600 | 990 | 840 | 884
20,600 | 20,650 | 993 | 842 | 886
20,650 | 20,700 | 996 | 845 | 889
20,700 | 20,750 | 999 | 847 | 891
20,750 | 20,800 | 1,002 | 849 | 894
20,800 | 20,850 | 1,005 | 851 | 897
20,850 | 20,900 | 1,008 | 854 | 899
20,900 | 20,950 | 1,011 | 856 | 902
20,950 | 21,000 | 1,014 | 858 | 905
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
21,000 | Your New York State tax is:
21,000 | 21,050 | 1,017 | 860 | 908
21,050 | 21,100 | 1,020 | 863 | 911
21,100 | 21,150 | 1,023 | 865 | 914
21,150 | 21,200 | 1,026 | 867 | 917
21,200 | 21,250 | 1,029 | 869 | 920
21,250 | 21,300 | 1,031 | 872 | 923
21,300 | 21,350 | 1,034 | 874 | 926
21,350 | 21,400 | 1,037 | 876 | 929
21,400 | 21,450 | 1,040 | 878 | 932
21,450 | 21,500 | 1,043 | 881 | 935
21,500 | 21,550 | 1,046 | 883 | 938
21,550 | 21,600 | 1,049 | 885 | 940
21,600 | 21,650 | 1,052 | 887 | 943
21,650 | 21,700 | 1,055 | 890 | 946
21,700 | 21,750 | 1,058 | 892 | 949
21,750 | 21,800 | 1,061 | 894 | 952
21,800 | 21,850 | 1,064 | 896 | 955
21,850 | 21,900 | 1,067 | 899 | 958
21,900 | 21,950 | 1,069 | 901 | 961
21,950 | 22,000 | 1,072 | 903 | 964
22,000 | Your New York State tax is:
22,000 | 22,050 | 1,075 | 905 | 967
22,050 | 22,100 | 1,078 | 908 | 970
22,100 | 22,150 | 1,081 | 910 | 973
22,150 | 22,200 | 1,084 | 912 | 976
22,200 | 22,250 | 1,087 | 914 | 979
22,250 | 22,300 | 1,090 | 917 | 981
22,300 | 22,350 | 1,093 | 919 | 984
22,350 | 22,400 | 1,096 | 921 | 987
22,400 | 22,450 | 1,099 | 923 | 990
22,450 | 22,500 | 1,102 | 926 | 993
22,500 | 22,550 | 1,105 | 928 | 996
22,550 | 22,600 | 1,107 | 930 | 999
22,600 | 22,650 | 1,110 | 932 | 1,002
22,650 | 22,700 | 1,113 | 935 | 1,005
22,700 | 22,750 | 1,116 | 937 | 1,008
22,750 | 22,800 | 1,119 | 939 | 1,011
22,800 | 22,850 | 1,122 | 941 | 1,014
22,850 | 22,900 | 1,125 | 944 | 1,017
22,900 | 22,950 | 1,128 | 946 | 1,019
22,950 | 23,000 | 1,131 | 948 | 1,022
23,000 | Your New York State tax is:
23,000 | 23,050 | 1,134 | 950 | 1,025
23,050 | 23,100 | 1,137 | 953 | 1,028
23,100 | 23,150 | 1,140 | 955 | 1,031
23,150 | 23,200 | 1,143 | 957 | 1,034
23,200 | 23,250 | 1,146 | 959 | 1,037
23,250 | 23,300 | 1,148 | 962 | 1,040
23,300 | 23,350 | 1,151 | 964 | 1,043
23,350 | 23,400 | 1,154 | 966 | 1,046
23,400 | 23,450 | 1,157 | 968 | 1,049
23,450 | 23,500 | 1,160 | 971 | 1,052
23,500 | 23,550 | 1,163 | 973 | 1,055
23,550 | 23,600 | 1,166 | 975 | 1,057
23,600 | 23,650 | 1,169 | 978 | 1,060
23,650 | 23,700 | 1,172 | 980 | 1,063
23,700 | 23,750 | 1,175 | 983 | 1,066
23,750 | 23,800 | 1,178 | 985 | 1,069
23,800 | 23,850 | 1,181 | 988 | 1,072
23,850 | 23,900 | 1,184 | 991 | 1,075
23,900 | 23,950 | 1,186 | 993 | 1,078
23,950 | 24,000 | 1,189 | 996 | 1,081
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
24,000 | Your New York State tax is:
24,000 | 24,050 | 1,192 | 999 | 1,084
24,050 | 24,100 | 1,195 | 1,001 | 1,087
24,100 | 24,150 | 1,198 | 1,004 | 1,090
24,150 | 24,200 | 1,201 | 1,006 | 1,093
24,200 | 24,250 | 1,204 | 1,009 | 1,096
24,250 | 24,300 | 1,207 | 1,012 | 1,098
24,300 | 24,350 | 1,210 | 1,014 | 1,101
24,350 | 24,400 | 1,213 | 1,017 | 1,104
24,400 | 24,450 | 1,216 | 1,020 | 1,107
24,450 | 24,500 | 1,219 | 1,022 | 1,110
24,500 | 24,550 | 1,222 | 1,025 | 1,113
24,550 | 24,600 | 1,224 | 1,027 | 1,116
24,600 | 24,650 | 1,227 | 1,030 | 1,119
24,650 | 24,700 | 1,230 | 1,033 | 1,122
24,700 | 24,750 | 1,233 | 1,035 | 1,125
24,750 | 24,800 | 1,236 | 1,038 | 1,128
24,800 | 24,850 | 1,239 | 1,041 | 1,131
24,850 | 24,900 | 1,242 | 1,043 | 1,134
24,900 | 24,950 | 1,245 | 1,046 | 1,136
24,950 | 25,000 | 1,248 | 1,048 | 1,139
25,000 | Your New York State tax is:
25,000 | 25,050 | 1,251 | 1,051 | 1,142
25,050 | 25,100 | 1,254 | 1,054 | 1,145
25,100 | 25,150 | 1,257 | 1,056 | 1,148
25,150 | 25,200 | 1,260 | 1,059 | 1,151
25,200 | 25,250 | 1,263 | 1,062 | 1,154
25,250 | 25,300 | 1,265 | 1,064 | 1,157
25,300 | 25,350 | 1,268 | 1,067 | 1,160
25,350 | 25,400 | 1,271 | 1,069 | 1,163
25,400 | 25,450 | 1,274 | 1,072 | 1,166
25,450 | 25,500 | 1,277 | 1,075 | 1,169
25,500 | 25,550 | 1,280 | 1,077 | 1,172
25,550 | 25,600 | 1,283 | 1,080 | 1,174
25,600 | 25,650 | 1,286 | 1,083 | 1,177
25,650 | 25,700 | 1,289 | 1,085 | 1,180
25,700 | 25,750 | 1,292 | 1,088 | 1,183
25,750 | 25,800 | 1,295 | 1,090 | 1,186
25,800 | 25,850 | 1,298 | 1,093 | 1,189
25,850 | 25,900 | 1,301 | 1,096 | 1,192
25,900 | 25,950 | 1,303 | 1,098 | 1,195
25,950 | 26,000 | 1,306 | 1,101 | 1,198
26,000 | Your New York State tax is:
26,000 | 26,050 | 1,309 | 1,104 | 1,201
26,050 | 26,100 | 1,312 | 1,106 | 1,204
26,100 | 26,150 | 1,315 | 1,109 | 1,207
26,150 | 26,200 | 1,318 | 1,111 | 1,210
26,200 | 26,250 | 1,321 | 1,114 | 1,213
26,250 | 26,300 | 1,324 | 1,117 | 1,215
26,300 | 26,350 | 1,327 | 1,119 | 1,218
26,350 | 26,400 | 1,330 | 1,122 | 1,221
26,400 | 26,450 | 1,333 | 1,125 | 1,224
26,450 | 26,500 | 1,336 | 1,127 | 1,227
26,500 | 26,550 | 1,339 | 1,130 | 1,230
26,550 | 26,600 | 1,341 | 1,132 | 1,233
26,600 | 26,650 | 1,344 | 1,135 | 1,236
26,650 | 26,700 | 1,347 | 1,138 | 1,239
26,700 | 26,750 | 1,350 | 1,140 | 1,242
26,750 | 26,800 | 1,353 | 1,143 | 1,245
26,800 | 26,850 | 1,356 | 1,146 | 1,248
26,850 | 26,900 | 1,359 | 1,148 | 1,251
26,900 | 26,950 | 1,362 | 1,151 | 1,253
26,950 | 27,000 | 1,365 | 1,153 | 1,256
This column must also be used by a qualifying surviving spouse
$33,000—$41,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
27,000 | Your New York State tax is:
27,000 | 27,050 | 1,368 | 1,156 | 1,259
27,050 | 27,100 | 1,371 | 1,159 | 1,262
27,100 | 27,150 | 1,374 | 1,161 | 1,265
27,150 | 27,200 | 1,377 | 1,164 | 1,268
27,200 | 27,250 | 1,380 | 1,167 | 1,271
27,250 | 27,300 | 1,382 | 1,169 | 1,274
27,300 | 27,350 | 1,385 | 1,172 | 1,277
27,350 | 27,400 | 1,388 | 1,174 | 1,280
27,400 | 27,450 | 1,391 | 1,177 | 1,283
27,450 | 27,500 | 1,394 | 1,180 | 1,286
27,500 | 27,550 | 1,397 | 1,182 | 1,289
27,550 | 27,600 | 1,400 | 1,185 | 1,291
27,600 | 27,650 | 1,403 | 1,188 | 1,294
27,650 | 27,700 | 1,406 | 1,190 | 1,297
27,700 | 27,750 | 1,409 | 1,193 | 1,300
27,750 | 27,800 | 1,412 | 1,195 | 1,303
27,800 | 27,850 | 1,415 | 1,198 | 1,306
27,850 | 27,900 | 1,418 | 1,201 | 1,309
27,900 | 27,950 | 1,420 | 1,203 | 1,312
27,950 | 28,000 | 1,423 | 1,206 | 1,315
28,000 | Your New York State tax is:
28,000 | 28,050 | 1,426 | 1,209 | 1,318
28,050 | 28,100 | 1,429 | 1,212 | 1,321
28,100 | 28,150 | 1,432 | 1,215 | 1,324
28,150 | 28,200 | 1,435 | 1,218 | 1,327
28,200 | 28,250 | 1,438 | 1,221 | 1,330
28,250 | 28,300 | 1,441 | 1,224 | 1,332
28,300 | 28,350 | 1,444 | 1,227 | 1,335
28,350 | 28,400 | 1,447 | 1,230 | 1,338
28,400 | 28,450 | 1,450 | 1,233 | 1,341
28,450 | 28,500 | 1,453 | 1,236 | 1,344
28,500 | 28,550 | 1,456 | 1,239 | 1,347
28,550 | 28,600 | 1,458 | 1,241 | 1,350
28,600 | 28,650 | 1,461 | 1,244 | 1,353
28,650 | 28,700 | 1,464 | 1,247 | 1,356
28,700 | 28,750 | 1,467 | 1,250 | 1,359
28,750 | 28,800 | 1,470 | 1,253 | 1,362
28,800 | 28,850 | 1,473 | 1,256 | 1,365
28,850 | 28,900 | 1,476 | 1,259 | 1,368
28,900 | 28,950 | 1,479 | 1,262 | 1,370
28,950 | 29,000 | 1,482 | 1,265 | 1,373
29,000 | Your New York State tax is:
29,000 | 29,050 | 1,485 | 1,268 | 1,376
29,050 | 29,100 | 1,488 | 1,271 | 1,379
29,100 | 29,150 | 1,491 | 1,274 | 1,382
29,150 | 29,200 | 1,494 | 1,277 | 1,385
29,200 | 29,250 | 1,497 | 1,280 | 1,388
29,250 | 29,300 | 1,499 | 1,282 | 1,391
29,300 | 29,350 | 1,502 | 1,285 | 1,394
29,350 | 29,400 | 1,505 | 1,288 | 1,397
29,400 | 29,450 | 1,508 | 1,291 | 1,400
29,450 | 29,500 | 1,511 | 1,294 | 1,403
29,500 | 29,550 | 1,514 | 1,297 | 1,406
29,550 | 29,600 | 1,517 | 1,300 | 1,408
29,600 | 29,650 | 1,520 | 1,303 | 1,411
29,650 | 29,700 | 1,523 | 1,306 | 1,414
29,700 | 29,750 | 1,526 | 1,309 | 1,417
29,750 | 29,800 | 1,529 | 1,312 | 1,420
29,800 | 29,850 | 1,532 | 1,315 | 1,423
29,850 | 29,900 | 1,535 | 1,318 | 1,426
29,900 | 29,950 | 1,537 | 1,320 | 1,429
29,950 | 30,000 | 1,540 | 1,323 | 1,432
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
30,000 | Your New York State tax is:
30,000 | 30,050 | 1,543 | 1,326 | 1,435
30,050 | 30,100 | 1,546 | 1,329 | 1,438
30,100 | 30,150 | 1,549 | 1,332 | 1,441
30,150 | 30,200 | 1,552 | 1,335 | 1,444
30,200 | 30,250 | 1,555 | 1,338 | 1,447
30,250 | 30,300 | 1,558 | 1,341 | 1,449
30,300 | 30,350 | 1,561 | 1,344 | 1,452
30,350 | 30,400 | 1,564 | 1,347 | 1,455
30,400 | 30,450 | 1,567 | 1,350 | 1,458
30,450 | 30,500 | 1,570 | 1,353 | 1,461
30,500 | 30,550 | 1,573 | 1,356 | 1,464
30,550 | 30,600 | 1,575 | 1,358 | 1,467
30,600 | 30,650 | 1,578 | 1,361 | 1,470
30,650 | 30,700 | 1,581 | 1,364 | 1,473
30,700 | 30,750 | 1,584 | 1,367 | 1,476
30,750 | 30,800 | 1,587 | 1,370 | 1,479
30,800 | 30,850 | 1,590 | 1,373 | 1,482
30,850 | 30,900 | 1,593 | 1,376 | 1,485
30,900 | 30,950 | 1,596 | 1,379 | 1,487
30,950 | 31,000 | 1,599 | 1,382 | 1,490
31,000 | Your New York State tax is:
31,000 | 31,050 | 1,602 | 1,385 | 1,493
31,050 | 31,100 | 1,605 | 1,388 | 1,496
31,100 | 31,150 | 1,608 | 1,391 | 1,499
31,150 | 31,200 | 1,611 | 1,394 | 1,502
31,200 | 31,250 | 1,614 | 1,397 | 1,505
31,250 | 31,300 | 1,616 | 1,399 | 1,508
31,300 | 31,350 | 1,619 | 1,402 | 1,511
31,350 | 31,400 | 1,622 | 1,405 | 1,514
31,400 | 31,450 | 1,625 | 1,408 | 1,517
31,450 | 31,500 | 1,628 | 1,411 | 1,520
31,500 | 31,550 | 1,631 | 1,414 | 1,523
31,550 | 31,600 | 1,634 | 1,417 | 1,525
31,600 | 31,650 | 1,637 | 1,420 | 1,528
31,650 | 31,700 | 1,640 | 1,423 | 1,531
31,700 | 31,750 | 1,643 | 1,426 | 1,534
31,750 | 31,800 | 1,646 | 1,429 | 1,537
31,800 | 31,850 | 1,649 | 1,432 | 1,540
31,850 | 31,900 | 1,652 | 1,435 | 1,543
31,900 | 31,950 | 1,654 | 1,437 | 1,546
31,950 | 32,000 | 1,657 | 1,440 | 1,549
32,000 | Your New York State tax is:
32,000 | 32,050 | 1,660 | 1,443 | 1,552
32,050 | 32,100 | 1,663 | 1,446 | 1,555
32,100 | 32,150 | 1,666 | 1,449 | 1,558
32,150 | 32,200 | 1,669 | 1,452 | 1,561
32,200 | 32,250 | 1,672 | 1,455 | 1,564
32,250 | 32,300 | 1,675 | 1,458 | 1,566
32,300 | 32,350 | 1,678 | 1,461 | 1,569
32,350 | 32,400 | 1,681 | 1,464 | 1,572
32,400 | 32,450 | 1,684 | 1,467 | 1,575
32,450 | 32,500 | 1,687 | 1,470 | 1,578
32,500 | 32,550 | 1,690 | 1,473 | 1,581
32,550 | 32,600 | 1,692 | 1,475 | 1,584
32,600 | 32,650 | 1,695 | 1,478 | 1,587
32,650 | 32,700 | 1,698 | 1,481 | 1,590
32,700 | 32,750 | 1,701 | 1,484 | 1,593
32,750 | 32,800 | 1,704 | 1,487 | 1,596
32,800 | 32,850 | 1,707 | 1,490 | 1,599
32,850 | 32,900 | 1,710 | 1,493 | 1,602
32,900 | 32,950 | 1,713 | 1,496 | 1,604
32,950 | 33,000 | 1,716 | 1,499 | 1,607
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
33,000 | Your New York State tax is:
33,000 | 33,050 | 1,719 | 1,502 | 1,610
33,050 | 33,100 | 1,722 | 1,505 | 1,613
33,100 | 33,150 | 1,725 | 1,508 | 1,616
33,150 | 33,200 | 1,728 | 1,511 | 1,619
33,200 | 33,250 | 1,731 | 1,514 | 1,622
33,250 | 33,300 | 1,733 | 1,516 | 1,625
33,300 | 33,350 | 1,736 | 1,519 | 1,628
33,350 | 33,400 | 1,739 | 1,522 | 1,631
33,400 | 33,450 | 1,742 | 1,525 | 1,634
33,450 | 33,500 | 1,745 | 1,528 | 1,637
33,500 | 33,550 | 1,748 | 1,531 | 1,640
33,550 | 33,600 | 1,751 | 1,534 | 1,642
33,600 | 33,650 | 1,754 | 1,537 | 1,645
33,650 | 33,700 | 1,757 | 1,540 | 1,648
33,700 | 33,750 | 1,760 | 1,543 | 1,651
33,750 | 33,800 | 1,763 | 1,546 | 1,654
33,800 | 33,850 | 1,766 | 1,549 | 1,657
33,850 | 33,900 | 1,769 | 1,552 | 1,660
33,900 | 33,950 | 1,771 | 1,554 | 1,663
33,950 | 34,000 | 1,774 | 1,557 | 1,666
34,000 | Your New York State tax is:
34,000 | 34,050 | 1,777 | 1,560 | 1,669
34,050 | 34,100 | 1,780 | 1,563 | 1,672
34,100 | 34,150 | 1,783 | 1,566 | 1,675
34,150 | 34,200 | 1,786 | 1,569 | 1,678
34,200 | 34,250 | 1,789 | 1,572 | 1,681
34,250 | 34,300 | 1,792 | 1,575 | 1,683
34,300 | 34,350 | 1,795 | 1,578 | 1,686
34,350 | 34,400 | 1,798 | 1,581 | 1,689
34,400 | 34,450 | 1,801 | 1,584 | 1,692
34,450 | 34,500 | 1,804 | 1,587 | 1,695
34,500 | 34,550 | 1,807 | 1,590 | 1,698
34,550 | 34,600 | 1,809 | 1,592 | 1,701
34,600 | 34,650 | 1,812 | 1,595 | 1,704
34,650 | 34,700 | 1,815 | 1,598 | 1,707
34,700 | 34,750 | 1,818 | 1,601 | 1,710
34,750 | 34,800 | 1,821 | 1,604 | 1,713
34,800 | 34,850 | 1,824 | 1,607 | 1,716
34,850 | 34,900 | 1,827 | 1,610 | 1,719
34,900 | 34,950 | 1,830 | 1,613 | 1,721
34,950 | 35,000 | 1,833 | 1,616 | 1,724
35,000 | Your New York State tax is:
35,000 | 35,050 | 1,836 | 1,619 | 1,727
35,050 | 35,100 | 1,839 | 1,622 | 1,730
35,100 | 35,150 | 1,842 | 1,625 | 1,733
35,150 | 35,200 | 1,845 | 1,628 | 1,736
35,200 | 35,250 | 1,848 | 1,631 | 1,739
35,250 | 35,300 | 1,850 | 1,633 | 1,742
35,300 | 35,350 | 1,853 | 1,636 | 1,745
35,350 | 35,400 | 1,856 | 1,639 | 1,748
35,400 | 35,450 | 1,859 | 1,642 | 1,751
35,450 | 35,500 | 1,862 | 1,645 | 1,754
35,500 | 35,550 | 1,865 | 1,648 | 1,757
35,550 | 35,600 | 1,868 | 1,651 | 1,759
35,600 | 35,650 | 1,871 | 1,654 | 1,762
35,650 | 35,700 | 1,874 | 1,657 | 1,765
35,700 | 35,750 | 1,877 | 1,660 | 1,768
35,750 | 35,800 | 1,880 | 1,663 | 1,771
35,800 | 35,850 | 1,883 | 1,666 | 1,774
35,850 | 35,900 | 1,886 | 1,669 | 1,777
35,900 | 35,950 | 1,888 | 1,671 | 1,780
35,950 | 36,000 | 1,891 | 1,674 | 1,783
This column must also be used by a qualifying surviving spouse
$42,000—$50,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
36,000 | Your New York State tax is:
36,000 | 36,050 | 1,894 | 1,677 | 1,786
36,050 | 36,100 | 1,897 | 1,680 | 1,789
36,100 | 36,150 | 1,900 | 1,683 | 1,792
36,150 | 36,200 | 1,903 | 1,686 | 1,795
36,200 | 36,250 | 1,906 | 1,689 | 1,798
36,250 | 36,300 | 1,909 | 1,692 | 1,800
36,300 | 36,350 | 1,912 | 1,695 | 1,803
36,350 | 36,400 | 1,915 | 1,698 | 1,806
36,400 | 36,450 | 1,918 | 1,701 | 1,809
36,450 | 36,500 | 1,921 | 1,704 | 1,812
36,500 | 36,550 | 1,924 | 1,707 | 1,815
36,550 | 36,600 | 1,926 | 1,709 | 1,818
36,600 | 36,650 | 1,929 | 1,712 | 1,821
36,650 | 36,700 | 1,932 | 1,715 | 1,824
36,700 | 36,750 | 1,935 | 1,718 | 1,827
36,750 | 36,800 | 1,938 | 1,721 | 1,830
36,800 | 36,850 | 1,941 | 1,724 | 1,833
36,850 | 36,900 | 1,944 | 1,727 | 1,836
36,900 | 36,950 | 1,947 | 1,730 | 1,838
36,950 | 37,000 | 1,950 | 1,733 | 1,841
37,000 | Your New York State tax is:
37,000 | 37,050 | 1,953 | 1,736 | 1,844
37,050 | 37,100 | 1,956 | 1,739 | 1,847
37,100 | 37,150 | 1,959 | 1,742 | 1,850
37,150 | 37,200 | 1,962 | 1,745 | 1,853
37,200 | 37,250 | 1,965 | 1,748 | 1,856
37,250 | 37,300 | 1,967 | 1,750 | 1,859
37,300 | 37,350 | 1,970 | 1,753 | 1,862
37,350 | 37,400 | 1,973 | 1,756 | 1,865
37,400 | 37,450 | 1,976 | 1,759 | 1,868
37,450 | 37,500 | 1,979 | 1,762 | 1,871
37,500 | 37,550 | 1,982 | 1,765 | 1,874
37,550 | 37,600 | 1,985 | 1,768 | 1,876
37,600 | 37,650 | 1,988 | 1,771 | 1,879
37,650 | 37,700 | 1,991 | 1,774 | 1,882
37,700 | 37,750 | 1,994 | 1,777 | 1,885
37,750 | 37,800 | 1,997 | 1,780 | 1,888
37,800 | 37,850 | 2,000 | 1,783 | 1,891
37,850 | 37,900 | 2,003 | 1,786 | 1,894
37,900 | 37,950 | 2,005 | 1,788 | 1,897
37,950 | 38,000 | 2,008 | 1,791 | 1,900
38,000 | Your New York State tax is:
38,000 | 38,050 | 2,011 | 1,794 | 1,903
38,050 | 38,100 | 2,014 | 1,797 | 1,906
38,100 | 38,150 | 2,017 | 1,800 | 1,909
38,150 | 38,200 | 2,020 | 1,803 | 1,912
38,200 | 38,250 | 2,023 | 1,806 | 1,915
38,250 | 38,300 | 2,026 | 1,809 | 1,917
38,300 | 38,350 | 2,029 | 1,812 | 1,920
38,350 | 38,400 | 2,032 | 1,815 | 1,923
38,400 | 38,450 | 2,035 | 1,818 | 1,926
38,450 | 38,500 | 2,038 | 1,821 | 1,929
38,500 | 38,550 | 2,041 | 1,824 | 1,932
38,550 | 38,600 | 2,043 | 1,826 | 1,935
38,600 | 38,650 | 2,046 | 1,829 | 1,938
38,650 | 38,700 | 2,049 | 1,832 | 1,941
38,700 | 38,750 | 2,052 | 1,835 | 1,944
38,750 | 38,800 | 2,055 | 1,838 | 1,947
38,800 | 38,850 | 2,058 | 1,841 | 1,950
38,850 | 38,900 | 2,061 | 1,844 | 1,953
38,900 | 38,950 | 2,064 | 1,847 | 1,955
38,950 | 39,000 | 2,067 | 1,850 | 1,958
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
39,000 | Your New York State tax is:
39,000 | 39,050 | 2,070 | 1,853 | 1,961
39,050 | 39,100 | 2,073 | 1,856 | 1,964
39,100 | 39,150 | 2,076 | 1,859 | 1,967
39,150 | 39,200 | 2,079 | 1,862 | 1,970
39,200 | 39,250 | 2,082 | 1,865 | 1,973
39,250 | 39,300 | 2,084 | 1,867 | 1,976
39,300 | 39,350 | 2,087 | 1,870 | 1,979
39,350 | 39,400 | 2,090 | 1,873 | 1,982
39,400 | 39,450 | 2,093 | 1,876 | 1,985
39,450 | 39,500 | 2,096 | 1,879 | 1,988
39,500 | 39,550 | 2,099 | 1,882 | 1,991
39,550 | 39,600 | 2,102 | 1,885 | 1,993
39,600 | 39,650 | 2,105 | 1,888 | 1,996
39,650 | 39,700 | 2,108 | 1,891 | 1,999
39,700 | 39,750 | 2,111 | 1,894 | 2,002
39,750 | 39,800 | 2,114 | 1,897 | 2,005
39,800 | 39,850 | 2,117 | 1,900 | 2,008
39,850 | 39,900 | 2,120 | 1,903 | 2,011
39,900 | 39,950 | 2,122 | 1,905 | 2,014
39,950 | 40,000 | 2,125 | 1,908 | 2,017
40,000 | Your New York State tax is:
40,000 | 40,050 | 2,128 | 1,911 | 2,020
40,050 | 40,100 | 2,131 | 1,914 | 2,023
40,100 | 40,150 | 2,134 | 1,917 | 2,026
40,150 | 40,200 | 2,137 | 1,920 | 2,029
40,200 | 40,250 | 2,140 | 1,923 | 2,032
40,250 | 40,300 | 2,143 | 1,926 | 2,034
40,300 | 40,350 | 2,146 | 1,929 | 2,037
40,350 | 40,400 | 2,149 | 1,932 | 2,040
40,400 | 40,450 | 2,152 | 1,935 | 2,043
40,450 | 40,500 | 2,155 | 1,938 | 2,046
40,500 | 40,550 | 2,158 | 1,941 | 2,049
40,550 | 40,600 | 2,160 | 1,943 | 2,052
40,600 | 40,650 | 2,163 | 1,946 | 2,055
40,650 | 40,700 | 2,166 | 1,949 | 2,058
40,700 | 40,750 | 2,169 | 1,952 | 2,061
40,750 | 40,800 | 2,172 | 1,955 | 2,064
40,800 | 40,850 | 2,175 | 1,958 | 2,067
40,850 | 40,900 | 2,178 | 1,961 | 2,070
40,900 | 40,950 | 2,181 | 1,964 | 2,072
40,950 | 41,000 | 2,184 | 1,967 | 2,075
41,000 | Your New York State tax is:
41,000 | 41,050 | 2,187 | 1,970 | 2,078
41,050 | 41,100 | 2,190 | 1,973 | 2,081
41,100 | 41,150 | 2,193 | 1,976 | 2,084
41,150 | 41,200 | 2,196 | 1,979 | 2,087
41,200 | 41,250 | 2,199 | 1,982 | 2,090
41,250 | 41,300 | 2,201 | 1,984 | 2,093
41,300 | 41,350 | 2,204 | 1,987 | 2,096
41,350 | 41,400 | 2,207 | 1,990 | 2,099
41,400 | 41,450 | 2,210 | 1,993 | 2,102
41,450 | 41,500 | 2,213 | 1,996 | 2,105
41,500 | 41,550 | 2,216 | 1,999 | 2,108
41,550 | 41,600 | 2,219 | 2,002 | 2,110
41,600 | 41,650 | 2,222 | 2,005 | 2,113
41,650 | 41,700 | 2,225 | 2,008 | 2,116
41,700 | 41,750 | 2,228 | 2,011 | 2,119
41,750 | 41,800 | 2,231 | 2,014 | 2,122
41,800 | 41,850 | 2,234 | 2,017 | 2,125
41,850 | 41,900 | 2,237 | 2,020 | 2,128
41,900 | 41,950 | 2,239 | 2,022 | 2,131
41,950 | 42,000 | 2,242 | 2,025 | 2,134
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
42,000 | Your New York State tax is:
42,000 | 42,050 | 2,245 | 2,028 | 2,137
42,050 | 42,100 | 2,248 | 2,031 | 2,140
42,100 | 42,150 | 2,251 | 2,034 | 2,143
42,150 | 42,200 | 2,254 | 2,037 | 2,146
42,200 | 42,250 | 2,257 | 2,040 | 2,149
42,250 | 42,300 | 2,260 | 2,043 | 2,151
42,300 | 42,350 | 2,263 | 2,046 | 2,154
42,350 | 42,400 | 2,266 | 2,049 | 2,157
42,400 | 42,450 | 2,269 | 2,052 | 2,160
42,450 | 42,500 | 2,272 | 2,055 | 2,163
42,500 | 42,550 | 2,275 | 2,058 | 2,166
42,550 | 42,600 | 2,277 | 2,060 | 2,169
42,600 | 42,650 | 2,280 | 2,063 | 2,172
42,650 | 42,700 | 2,283 | 2,066 | 2,175
42,700 | 42,750 | 2,286 | 2,069 | 2,178
42,750 | 42,800 | 2,289 | 2,072 | 2,181
42,800 | 42,850 | 2,292 | 2,075 | 2,184
42,850 | 42,900 | 2,295 | 2,078 | 2,187
42,900 | 42,950 | 2,298 | 2,081 | 2,189
42,950 | 43,000 | 2,301 | 2,084 | 2,192
43,000 | Your New York State tax is:
43,000 | 43,050 | 2,304 | 2,087 | 2,195
43,050 | 43,100 | 2,307 | 2,090 | 2,198
43,100 | 43,150 | 2,310 | 2,093 | 2,201
43,150 | 43,200 | 2,313 | 2,096 | 2,204
43,200 | 43,250 | 2,316 | 2,099 | 2,207
43,250 | 43,300 | 2,318 | 2,101 | 2,210
43,300 | 43,350 | 2,321 | 2,104 | 2,213
43,350 | 43,400 | 2,324 | 2,107 | 2,216
43,400 | 43,450 | 2,327 | 2,110 | 2,219
43,450 | 43,500 | 2,330 | 2,113 | 2,222
43,500 | 43,550 | 2,333 | 2,116 | 2,225
43,550 | 43,600 | 2,336 | 2,119 | 2,227
43,600 | 43,650 | 2,339 | 2,122 | 2,230
43,650 | 43,700 | 2,342 | 2,125 | 2,233
43,700 | 43,750 | 2,345 | 2,128 | 2,236
43,750 | 43,800 | 2,348 | 2,131 | 2,239
43,800 | 43,850 | 2,351 | 2,134 | 2,242
43,850 | 43,900 | 2,354 | 2,137 | 2,245
43,900 | 43,950 | 2,356 | 2,139 | 2,248
43,950 | 44,000 | 2,359 | 2,142 | 2,251
44,000 | Your New York State tax is:
44,000 | 44,050 | 2,362 | 2,145 | 2,254
44,050 | 44,100 | 2,365 | 2,148 | 2,257
44,100 | 44,150 | 2,368 | 2,151 | 2,260
44,150 | 44,200 | 2,371 | 2,154 | 2,263
44,200 | 44,250 | 2,374 | 2,157 | 2,266
44,250 | 44,300 | 2,377 | 2,160 | 2,268
44,300 | 44,350 | 2,380 | 2,163 | 2,271
44,350 | 44,400 | 2,383 | 2,166 | 2,274
44,400 | 44,450 | 2,386 | 2,169 | 2,277
44,450 | 44,500 | 2,389 | 2,172 | 2,280
44,500 | 44,550 | 2,392 | 2,175 | 2,283
44,550 | 44,600 | 2,394 | 2,177 | 2,286
44,600 | 44,650 | 2,397 | 2,180 | 2,289
44,650 | 44,700 | 2,400 | 2,183 | 2,292
44,700 | 44,750 | 2,403 | 2,186 | 2,295
44,750 | 44,800 | 2,406 | 2,189 | 2,298
44,800 | 44,850 | 2,409 | 2,192 | 2,301
44,850 | 44,900 | 2,412 | 2,195 | 2,304
44,900 | 44,950 | 2,415 | 2,198 | 2,306
44,950 | 45,000 | 2,418 | 2,201 | 2,309
This column must also be used by a qualifying surviving spouse
$51,000—$59,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
45,000 | Your New York State tax is:
45,000 | 45,050 | 2,421 | 2,204 | 2,312
45,050 | 45,100 | 2,424 | 2,207 | 2,315
45,100 | 45,150 | 2,427 | 2,210 | 2,318
45,150 | 45,200 | 2,430 | 2,213 | 2,321
45,200 | 45,250 | 2,433 | 2,216 | 2,324
45,250 | 45,300 | 2,435 | 2,218 | 2,327
45,300 | 45,350 | 2,438 | 2,221 | 2,330
45,350 | 45,400 | 2,441 | 2,224 | 2,333
45,400 | 45,450 | 2,444 | 2,227 | 2,336
45,450 | 45,500 | 2,447 | 2,230 | 2,339
45,500 | 45,550 | 2,450 | 2,233 | 2,342
45,550 | 45,600 | 2,453 | 2,236 | 2,344
45,600 | 45,650 | 2,456 | 2,239 | 2,347
45,650 | 45,700 | 2,459 | 2,242 | 2,350
45,700 | 45,750 | 2,462 | 2,245 | 2,353
45,750 | 45,800 | 2,465 | 2,248 | 2,356
45,800 | 45,850 | 2,468 | 2,251 | 2,359
45,850 | 45,900 | 2,471 | 2,254 | 2,362
45,900 | 45,950 | 2,473 | 2,256 | 2,365
45,950 | 46,000 | 2,476 | 2,259 | 2,368
46,000 | Your New York State tax is:
46,000 | 46,050 | 2,479 | 2,262 | 2,371
46,050 | 46,100 | 2,482 | 2,265 | 2,374
46,100 | 46,150 | 2,485 | 2,268 | 2,377
46,150 | 46,200 | 2,488 | 2,271 | 2,380
46,200 | 46,250 | 2,491 | 2,274 | 2,383
46,250 | 46,300 | 2,494 | 2,277 | 2,385
46,300 | 46,350 | 2,497 | 2,280 | 2,388
46,350 | 46,400 | 2,500 | 2,283 | 2,391
46,400 | 46,450 | 2,503 | 2,286 | 2,394
46,450 | 46,500 | 2,506 | 2,289 | 2,397
46,500 | 46,550 | 2,509 | 2,292 | 2,400
46,550 | 46,600 | 2,511 | 2,294 | 2,403
46,600 | 46,650 | 2,514 | 2,297 | 2,406
46,650 | 46,700 | 2,517 | 2,300 | 2,409
46,700 | 46,750 | 2,520 | 2,303 | 2,412
46,750 | 46,800 | 2,523 | 2,306 | 2,415
46,800 | 46,850 | 2,526 | 2,309 | 2,418
46,850 | 46,900 | 2,529 | 2,312 | 2,421
46,900 | 46,950 | 2,532 | 2,315 | 2,423
46,950 | 47,000 | 2,535 | 2,318 | 2,426
47,000 | Your New York State tax is:
47,000 | 47,050 | 2,538 | 2,321 | 2,429
47,050 | 47,100 | 2,541 | 2,324 | 2,432
47,100 | 47,150 | 2,544 | 2,327 | 2,435
47,150 | 47,200 | 2,547 | 2,330 | 2,438
47,200 | 47,250 | 2,550 | 2,333 | 2,441
47,250 | 47,300 | 2,552 | 2,335 | 2,444
47,300 | 47,350 | 2,555 | 2,338 | 2,447
47,350 | 47,400 | 2,558 | 2,341 | 2,450
47,400 | 47,450 | 2,561 | 2,344 | 2,453
47,450 | 47,500 | 2,564 | 2,347 | 2,456
47,500 | 47,550 | 2,567 | 2,350 | 2,459
47,550 | 47,600 | 2,570 | 2,353 | 2,461
47,600 | 47,650 | 2,573 | 2,356 | 2,464
47,650 | 47,700 | 2,576 | 2,359 | 2,467
47,700 | 47,750 | 2,579 | 2,362 | 2,470
47,750 | 47,800 | 2,582 | 2,365 | 2,473
47,800 | 47,850 | 2,585 | 2,368 | 2,476
47,850 | 47,900 | 2,588 | 2,371 | 2,479
47,900 | 47,950 | 2,590 | 2,373 | 2,482
47,950 | 48,000 | 2,593 | 2,376 | 2,485
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
48,000 | Your New York State tax is:
48,000 | 48,050 | 2,596 | 2,379 | 2,488
48,050 | 48,100 | 2,599 | 2,382 | 2,491
48,100 | 48,150 | 2,602 | 2,385 | 2,494
48,150 | 48,200 | 2,605 | 2,388 | 2,497
48,200 | 48,250 | 2,608 | 2,391 | 2,500
48,250 | 48,300 | 2,611 | 2,394 | 2,502
48,300 | 48,350 | 2,614 | 2,397 | 2,505
48,350 | 48,400 | 2,617 | 2,400 | 2,508
48,400 | 48,450 | 2,620 | 2,403 | 2,511
48,450 | 48,500 | 2,623 | 2,406 | 2,514
48,500 | 48,550 | 2,626 | 2,409 | 2,517
48,550 | 48,600 | 2,628 | 2,411 | 2,520
48,600 | 48,650 | 2,631 | 2,414 | 2,523
48,650 | 48,700 | 2,634 | 2,417 | 2,526
48,700 | 48,750 | 2,637 | 2,420 | 2,529
48,750 | 48,800 | 2,640 | 2,423 | 2,532
48,800 | 48,850 | 2,643 | 2,426 | 2,535
48,850 | 48,900 | 2,646 | 2,429 | 2,538
48,900 | 48,950 | 2,649 | 2,432 | 2,540
48,950 | 49,000 | 2,652 | 2,435 | 2,543
49,000 | Your New York State tax is:
49,000 | 49,050 | 2,655 | 2,438 | 2,546
49,050 | 49,100 | 2,658 | 2,441 | 2,549
49,100 | 49,150 | 2,661 | 2,444 | 2,552
49,150 | 49,200 | 2,664 | 2,447 | 2,555
49,200 | 49,250 | 2,667 | 2,450 | 2,558
49,250 | 49,300 | 2,669 | 2,452 | 2,561
49,300 | 49,350 | 2,672 | 2,455 | 2,564
49,350 | 49,400 | 2,675 | 2,458 | 2,567
49,400 | 49,450 | 2,678 | 2,461 | 2,570
49,450 | 49,500 | 2,681 | 2,464 | 2,573
49,500 | 49,550 | 2,684 | 2,467 | 2,576
49,550 | 49,600 | 2,687 | 2,470 | 2,578
49,600 | 49,650 | 2,690 | 2,473 | 2,581
49,650 | 49,700 | 2,693 | 2,476 | 2,584
49,700 | 49,750 | 2,696 | 2,479 | 2,587
49,750 | 49,800 | 2,699 | 2,482 | 2,590
49,800 | 49,850 | 2,702 | 2,485 | 2,593
49,850 | 49,900 | 2,705 | 2,488 | 2,596
49,900 | 49,950 | 2,707 | 2,490 | 2,599
49,950 | 50,000 | 2,710 | 2,493 | 2,602
50,000 | Your New York State tax is:
50,000 | 50,050 | 2,713 | 2,496 | 2,605
50,050 | 50,100 | 2,716 | 2,499 | 2,608
50,100 | 50,150 | 2,719 | 2,502 | 2,611
50,150 | 50,200 | 2,722 | 2,505 | 2,614
50,200 | 50,250 | 2,725 | 2,508 | 2,617
50,250 | 50,300 | 2,728 | 2,511 | 2,619
50,300 | 50,350 | 2,731 | 2,514 | 2,622
50,350 | 50,400 | 2,734 | 2,517 | 2,625
50,400 | 50,450 | 2,737 | 2,520 | 2,628
50,450 | 50,500 | 2,740 | 2,523 | 2,631
50,500 | 50,550 | 2,743 | 2,526 | 2,634
50,550 | 50,600 | 2,745 | 2,528 | 2,637
50,600 | 50,650 | 2,748 | 2,531 | 2,640
50,650 | 50,700 | 2,751 | 2,534 | 2,643
50,700 | 50,750 | 2,754 | 2,537 | 2,646
50,750 | 50,800 | 2,757 | 2,540 | 2,649
50,800 | 50,850 | 2,760 | 2,543 | 2,652
50,850 | 50,900 | 2,763 | 2,546 | 2,655
50,900 | 50,950 | 2,766 | 2,549 | 2,657
50,950 | 51,000 | 2,769 | 2,552 | 2,660
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
51,000 | Your New York State tax is:
51,000 | 51,050 | 2,772 | 2,555 | 2,663
51,050 | 51,100 | 2,775 | 2,558 | 2,666
51,100 | 51,150 | 2,778 | 2,561 | 2,669
51,150 | 51,200 | 2,781 | 2,564 | 2,672
51,200 | 51,250 | 2,784 | 2,567 | 2,675
51,250 | 51,300 | 2,786 | 2,569 | 2,678
51,300 | 51,350 | 2,789 | 2,572 | 2,681
51,350 | 51,400 | 2,792 | 2,575 | 2,684
51,400 | 51,450 | 2,795 | 2,578 | 2,687
51,450 | 51,500 | 2,798 | 2,581 | 2,690
51,500 | 51,550 | 2,801 | 2,584 | 2,693
51,550 | 51,600 | 2,804 | 2,587 | 2,695
51,600 | 51,650 | 2,807 | 2,590 | 2,698
51,650 | 51,700 | 2,810 | 2,593 | 2,701
51,700 | 51,750 | 2,813 | 2,596 | 2,704
51,750 | 51,800 | 2,816 | 2,599 | 2,707
51,800 | 51,850 | 2,819 | 2,602 | 2,710
51,850 | 51,900 | 2,822 | 2,605 | 2,713
51,900 | 51,950 | 2,824 | 2,607 | 2,716
51,950 | 52,000 | 2,827 | 2,610 | 2,719
52,000 | Your New York State tax is:
52,000 | 52,050 | 2,830 | 2,613 | 2,722
52,050 | 52,100 | 2,833 | 2,616 | 2,725
52,100 | 52,150 | 2,836 | 2,619 | 2,728
52,150 | 52,200 | 2,839 | 2,622 | 2,731
52,200 | 52,250 | 2,842 | 2,625 | 2,734
52,250 | 52,300 | 2,845 | 2,628 | 2,736
52,300 | 52,350 | 2,848 | 2,631 | 2,739
52,350 | 52,400 | 2,851 | 2,634 | 2,742
52,400 | 52,450 | 2,854 | 2,637 | 2,745
52,450 | 52,500 | 2,857 | 2,640 | 2,748
52,500 | 52,550 | 2,860 | 2,643 | 2,751
52,550 | 52,600 | 2,862 | 2,645 | 2,754
52,600 | 52,650 | 2,865 | 2,648 | 2,757
52,650 | 52,700 | 2,868 | 2,651 | 2,760
52,700 | 52,750 | 2,871 | 2,654 | 2,763
52,750 | 52,800 | 2,874 | 2,657 | 2,766
52,800 | 52,850 | 2,877 | 2,660 | 2,769
52,850 | 52,900 | 2,880 | 2,663 | 2,772
52,900 | 52,950 | 2,883 | 2,666 | 2,774
52,950 | 53,000 | 2,886 | 2,669 | 2,777
53,000 | Your New York State tax is:
53,000 | 53,050 | 2,889 | 2,672 | 2,780
53,050 | 53,100 | 2,892 | 2,675 | 2,783
53,100 | 53,150 | 2,895 | 2,678 | 2,786
53,150 | 53,200 | 2,898 | 2,681 | 2,789
53,200 | 53,250 | 2,901 | 2,684 | 2,792
53,250 | 53,300 | 2,903 | 2,686 | 2,795
53,300 | 53,350 | 2,906 | 2,689 | 2,798
53,350 | 53,400 | 2,909 | 2,692 | 2,801
53,400 | 53,450 | 2,912 | 2,695 | 2,804
53,450 | 53,500 | 2,915 | 2,698 | 2,807
53,500 | 53,550 | 2,918 | 2,701 | 2,810
53,550 | 53,600 | 2,921 | 2,704 | 2,812
53,600 | 53,650 | 2,924 | 2,707 | 2,815
53,650 | 53,700 | 2,927 | 2,710 | 2,818
53,700 | 53,750 | 2,930 | 2,713 | 2,821
53,750 | 53,800 | 2,933 | 2,716 | 2,824
53,800 | 53,850 | 2,936 | 2,719 | 2,827
53,850 | 53,900 | 2,939 | 2,722 | 2,830
53,900 | 53,950 | 2,941 | 2,724 | 2,833
53,950 | 54,000 | 2,944 | 2,727 | 2,836
This column must also be used by a qualifying surviving spouse
$60,000+
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
54,000 | Your New York State tax is:
54,000 | 54,050 | 2,947 | 2,730 | 2,839
54,050 | 54,100 | 2,950 | 2,733 | 2,842
54,100 | 54,150 | 2,953 | 2,736 | 2,845
54,150 | 54,200 | 2,956 | 2,739 | 2,848
54,200 | 54,250 | 2,959 | 2,742 | 2,851
54,250 | 54,300 | 2,962 | 2,745 | 2,853
54,300 | 54,350 | 2,965 | 2,748 | 2,856
54,350 | 54,400 | 2,968 | 2,751 | 2,859
54,400 | 54,450 | 2,971 | 2,754 | 2,862
54,450 | 54,500 | 2,974 | 2,757 | 2,865
54,500 | 54,550 | 2,977 | 2,760 | 2,868
54,550 | 54,600 | 2,979 | 2,762 | 2,871
54,600 | 54,650 | 2,982 | 2,765 | 2,874
54,650 | 54,700 | 2,985 | 2,768 | 2,877
54,700 | 54,750 | 2,988 | 2,771 | 2,880
54,750 | 54,800 | 2,991 | 2,774 | 2,883
54,800 | 54,850 | 2,994 | 2,777 | 2,886
54,850 | 54,900 | 2,997 | 2,780 | 2,889
54,900 | 54,950 | 3,000 | 2,783 | 2,891
54,950 | 55,000 | 3,003 | 2,786 | 2,894
55,000 | Your New York State tax is:
55,000 | 55,050 | 3,006 | 2,789 | 2,897
55,050 | 55,100 | 3,009 | 2,792 | 2,900
55,100 | 55,150 | 3,012 | 2,795 | 2,903
55,150 | 55,200 | 3,015 | 2,798 | 2,906
55,200 | 55,250 | 3,018 | 2,801 | 2,909
55,250 | 55,300 | 3,020 | 2,803 | 2,912
55,300 | 55,350 | 3,023 | 2,806 | 2,915
55,350 | 55,400 | 3,026 | 2,809 | 2,918
55,400 | 55,450 | 3,029 | 2,812 | 2,921
55,450 | 55,500 | 3,032 | 2,815 | 2,924
55,500 | 55,550 | 3,035 | 2,818 | 2,927
55,550 | 55,600 | 3,038 | 2,821 | 2,929
55,600 | 55,650 | 3,041 | 2,824 | 2,932
55,650 | 55,700 | 3,044 | 2,827 | 2,935
55,700 | 55,750 | 3,047 | 2,830 | 2,938
55,750 | 55,800 | 3,050 | 2,833 | 2,941
55,800 | 55,850 | 3,053 | 2,836 | 2,944
55,850 | 55,900 | 3,056 | 2,839 | 2,947
55,900 | 55,950 | 3,058 | 2,841 | 2,950
55,950 | 56,000 | 3,061 | 2,844 | 2,953
56,000 | Your New York State tax is:
56,000 | 56,050 | 3,064 | 2,847 | 2,956
56,050 | 56,100 | 3,067 | 2,850 | 2,959
56,100 | 56,150 | 3,070 | 2,853 | 2,962
56,150 | 56,200 | 3,073 | 2,856 | 2,965
56,200 | 56,250 | 3,076 | 2,859 | 2,968
56,250 | 56,300 | 3,079 | 2,862 | 2,970
56,300 | 56,350 | 3,082 | 2,865 | 2,973
56,350 | 56,400 | 3,085 | 2,868 | 2,976
56,400 | 56,450 | 3,088 | 2,871 | 2,979
56,450 | 56,500 | 3,091 | 2,874 | 2,982
56,500 | 56,550 | 3,094 | 2,877 | 2,985
56,550 | 56,600 | 3,096 | 2,879 | 2,988
56,600 | 56,650 | 3,099 | 2,882 | 2,991
56,650 | 56,700 | 3,102 | 2,885 | 2,994
56,700 | 56,750 | 3,105 | 2,888 | 2,997
56,750 | 56,800 | 3,108 | 2,891 | 3,000
56,800 | 56,850 | 3,111 | 2,894 | 3,003
56,850 | 56,900 | 3,114 | 2,897 | 3,006
56,900 | 56,950 | 3,117 | 2,900 | 3,008
56,950 | 57,000 | 3,120 | 2,903 | 3,011
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
57,000 | Your New York State tax is:
57,000 | 57,050 | 3,123 | 2,906 | 3,014
57,050 | 57,100 | 3,126 | 2,909 | 3,017
57,100 | 57,150 | 3,129 | 2,912 | 3,020
57,150 | 57,200 | 3,132 | 2,915 | 3,023
57,200 | 57,250 | 3,135 | 2,918 | 3,026
57,250 | 57,300 | 3,137 | 2,920 | 3,029
57,300 | 57,350 | 3,140 | 2,923 | 3,032
57,350 | 57,400 | 3,143 | 2,926 | 3,035
57,400 | 57,450 | 3,146 | 2,929 | 3,038
57,450 | 57,500 | 3,149 | 2,932 | 3,041
57,500 | 57,550 | 3,152 | 2,935 | 3,044
57,550 | 57,600 | 3,155 | 2,938 | 3,046
57,600 | 57,650 | 3,158 | 2,941 | 3,049
57,650 | 57,700 | 3,161 | 2,944 | 3,052
57,700 | 57,750 | 3,164 | 2,947 | 3,055
57,750 | 57,800 | 3,167 | 2,950 | 3,058
57,800 | 57,850 | 3,170 | 2,953 | 3,061
57,850 | 57,900 | 3,173 | 2,956 | 3,064
57,900 | 57,950 | 3,175 | 2,958 | 3,067
57,950 | 58,000 | 3,178 | 2,961 | 3,070
58,000 | Your New York State tax is:
58,000 | 58,050 | 3,181 | 2,964 | 3,073
58,050 | 58,100 | 3,184 | 2,967 | 3,076
58,100 | 58,150 | 3,187 | 2,970 | 3,079
58,150 | 58,200 | 3,190 | 2,973 | 3,082
58,200 | 58,250 | 3,193 | 2,976 | 3,085
58,250 | 58,300 | 3,196 | 2,979 | 3,087
58,300 | 58,350 | 3,199 | 2,982 | 3,090
58,350 | 58,400 | 3,202 | 2,985 | 3,093
58,400 | 58,450 | 3,205 | 2,988 | 3,096
58,450 | 58,500 | 3,208 | 2,991 | 3,099
58,500 | 58,550 | 3,211 | 2,994 | 3,102
58,550 | 58,600 | 3,213 | 2,996 | 3,105
58,600 | 58,650 | 3,216 | 2,999 | 3,108
58,650 | 58,700 | 3,219 | 3,002 | 3,111
58,700 | 58,750 | 3,222 | 3,005 | 3,114
58,750 | 58,800 | 3,225 | 3,008 | 3,117
58,800 | 58,850 | 3,228 | 3,011 | 3,120
58,850 | 58,900 | 3,231 | 3,014 | 3,123
58,900 | 58,950 | 3,234 | 3,017 | 3,125
58,950 | 59,000 | 3,237 | 3,020 | 3,128
59,000 | Your New York State tax is:
59,000 | 59,050 | 3,240 | 3,023 | 3,131
59,050 | 59,100 | 3,243 | 3,026 | 3,134
59,100 | 59,150 | 3,246 | 3,029 | 3,137
59,150 | 59,200 | 3,249 | 3,032 | 3,140
59,200 | 59,250 | 3,252 | 3,035 | 3,143
59,250 | 59,300 | 3,254 | 3,037 | 3,146
59,300 | 59,350 | 3,257 | 3,040 | 3,149
59,350 | 59,400 | 3,260 | 3,043 | 3,152
59,400 | 59,450 | 3,263 | 3,046 | 3,155
59,450 | 59,500 | 3,266 | 3,049 | 3,158
59,500 | 59,550 | 3,269 | 3,052 | 3,161
59,550 | 59,600 | 3,272 | 3,055 | 3,163
59,600 | 59,650 | 3,275 | 3,058 | 3,166
59,650 | 59,700 | 3,278 | 3,061 | 3,169
59,700 | 59,750 | 3,281 | 3,064 | 3,172
59,750 | 59,800 | 3,284 | 3,067 | 3,175
59,800 | 59,850 | 3,287 | 3,070 | 3,178
59,850 | 59,900 | 3,290 | 3,073 | 3,181
59,900 | 59,950 | 3,292 | 3,075 | 3,184
59,950 | 60,000 | 3,295 | 3,078 | 3,187
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
60,000 | Your New York State tax is:
60,000 | 60,050 | 3,298 | 3,081 | 3,190
60,050 | 60,100 | 3,301 | 3,084 | 3,193
60,100 | 60,150 | 3,304 | 3,087 | 3,196
60,150 | 60,200 | 3,307 | 3,090 | 3,199
60,200 | 60,250 | 3,310 | 3,093 | 3,202
60,250 | 60,300 | 3,313 | 3,096 | 3,204
60,300 | 60,350 | 3,316 | 3,099 | 3,207
60,350 | 60,400 | 3,319 | 3,102 | 3,210
60,400 | 60,450 | 3,322 | 3,105 | 3,213
60,450 | 60,500 | 3,325 | 3,108 | 3,216
60,500 | 60,550 | 3,328 | 3,111 | 3,219
60,550 | 60,600 | 3,330 | 3,113 | 3,222
60,600 | 60,650 | 3,333 | 3,116 | 3,225
60,650 | 60,700 | 3,336 | 3,119 | 3,228
60,700 | 60,750 | 3,339 | 3,122 | 3,231
60,750 | 60,800 | 3,342 | 3,125 | 3,234
60,800 | 60,850 | 3,345 | 3,128 | 3,237
60,850 | 60,900 | 3,348 | 3,131 | 3,240
60,900 | 60,950 | 3,351 | 3,134 | 3,242
60,950 | 61,000 | 3,354 | 3,137 | 3,245
61,000 | Your New York State tax is:
61,000 | 61,050 | 3,357 | 3,140 | 3,248
61,050 | 61,100 | 3,360 | 3,143 | 3,251
61,100 | 61,150 | 3,363 | 3,146 | 3,254
61,150 | 61,200 | 3,366 | 3,149 | 3,257
61,200 | 61,250 | 3,369 | 3,152 | 3,260
61,250 | 61,300 | 3,371 | 3,154 | 3,263
61,300 | 61,350 | 3,374 | 3,157 | 3,266
61,350 | 61,400 | 3,377 | 3,160 | 3,269
61,400 | 61,450 | 3,380 | 3,163 | 3,272
61,450 | 61,500 | 3,383 | 3,166 | 3,275
61,500 | 61,550 | 3,386 | 3,169 | 3,278
61,550 | 61,600 | 3,389 | 3,172 | 3,280
61,600 | 61,650 | 3,392 | 3,175 | 3,283
61,650 | 61,700 | 3,395 | 3,178 | 3,286
61,700 | 61,750 | 3,398 | 3,181 | 3,289
61,750 | 61,800 | 3,401 | 3,184 | 3,292
61,800 | 61,850 | 3,404 | 3,187 | 3,295
61,850 | 61,900 | 3,407 | 3,190 | 3,298
61,900 | 61,950 | 3,409 | 3,192 | 3,301
61,950 | 62,000 | 3,412 | 3,195 | 3,304
This column must also be used by a qualifying surviving spouse
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
62,000 | Your New York State tax is:
62,000 | 62,050 | 3,415 | 3,198 | 3,307
62,050 | 62,100 | 3,418 | 3,201 | 3,310
62,100 | 62,150 | 3,421 | 3,204 | 3,313
62,150 | 62,200 | 3,424 | 3,207 | 3,316
62,200 | 62,250 | 3,427 | 3,210 | 3,319
62,250 | 62,300 | 3,430 | 3,213 | 3,321
62,300 | 62,350 | 3,433 | 3,216 | 3,324
62,350 | 62,400 | 3,436 | 3,219 | 3,327
62,400 | 62,450 | 3,439 | 3,222 | 3,330
62,450 | 62,500 | 3,442 | 3,225 | 3,333
62,500 | 62,550 | 3,445 | 3,228 | 3,336
62,550 | 62,600 | 3,447 | 3,230 | 3,339
62,600 | 62,650 | 3,450 | 3,233 | 3,342
62,650 | 62,700 | 3,453 | 3,236 | 3,345
62,700 | 62,750 | 3,456 | 3,239 | 3,348
62,750 | 62,800 | 3,459 | 3,242 | 3,351
62,800 | 62,850 | 3,462 | 3,245 | 3,354
62,850 | 62,900 | 3,465 | 3,248 | 3,357
62,900 | 62,950 | 3,468 | 3,251 | 3,359
62,950 | 63,000 | 3,471 | 3,254 | 3,362
63,000 | Your New York State tax is:
63,000 | 63,050 | 3,474 | 3,257 | 3,365
63,050 | 63,100 | 3,477 | 3,260 | 3,368
63,100 | 63,150 | 3,480 | 3,263 | 3,371
63,150 | 63,200 | 3,483 | 3,266 | 3,374
63,200 | 63,250 | 3,486 | 3,269 | 3,377
63,250 | 63,300 | 3,488 | 3,271 | 3,380
63,300 | 63,350 | 3,491 | 3,274 | 3,383
63,350 | 63,400 | 3,494 | 3,277 | 3,386
63,400 | 63,450 | 3,497 | 3,280 | 3,389
63,450 | 63,500 | 3,500 | 3,283 | 3,392
63,500 | 63,550 | 3,503 | 3,286 | 3,395
63,550 | 63,600 | 3,506 | 3,289 | 3,397
63,600 | 63,650 | 3,509 | 3,292 | 3,400
63,650 | 63,700 | 3,512 | 3,295 | 3,403
63,700 | 63,750 | 3,515 | 3,298 | 3,406
63,750 | 63,800 | 3,518 | 3,301 | 3,409
63,800 | 63,850 | 3,521 | 3,304 | 3,412
63,850 | 63,900 | 3,524 | 3,307 | 3,415
63,900 | 63,950 | 3,526 | 3,309 | 3,418
63,950 | 64,000 | 3,529 | 3,312 | 3,421
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
64,000 | Your New York State tax is:
64,000 | 64,050 | 3,532 | 3,315 | 3,424
64,050 | 64,100 | 3,535 | 3,318 | 3,427
64,100 | 64,150 | 3,538 | 3,321 | 3,430
64,150 | 64,200 | 3,541 | 3,324 | 3,433
64,200 | 64,250 | 3,544 | 3,327 | 3,436
64,250 | 64,300 | 3,547 | 3,330 | 3,438
64,300 | 64,350 | 3,550 | 3,333 | 3,441
64,350 | 64,400 | 3,553 | 3,336 | 3,444
64,400 | 64,450 | 3,556 | 3,339 | 3,447
64,450 | 64,500 | 3,559 | 3,342 | 3,450
64,500 | 64,550 | 3,562 | 3,345 | 3,453
64,550 | 64,600 | 3,564 | 3,347 | 3,456
64,600 | 64,650 | 3,567 | 3,350 | 3,459
64,650 | 64,700 | 3,570 | 3,353 | 3,462
64,700 | 64,750 | 3,573 | 3,356 | 3,465
64,750 | 64,800 | 3,576 | 3,359 | 3,468
64,800 | 64,850 | 3,579 | 3,362 | 3,471
64,850 | 64,900 | 3,582 | 3,365 | 3,474
64,900 | 64,950 | 3,585 | 3,368 | 3,476
64,950 | 65,000 | 3,588 | 3,371 | 3,479
65,000 or more:
$65,000 or more—compute your New York State tax usingNew York State tax rate schedule.If the amount on Form IT-201, line 33, is more than $107,650, seeTax computation – New York adjusted gross income of more than $107,650.
2022 New York City Tax Table
Example: A married couple are filing a joint return on Form IT-201. Their taxable income on line 47 is $38,275. First, they find the 38,250 - 38,300 income line. Next, they find the column for Married filing jointly and read down the column. The amount shown where the income line and filing status column meet is $1,292. This is the tax amount they must write on Form IT-201, line 47a.
$0—$5,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
| Your New York City tax is:
0 | 18 | 0 | 0 | 0
18 | 25 | 1 | 1 | 1
25 | 50 | 1 | 1 | 1
50 | 100 | 2 | 2 | 2
100 | 150 | 4 | 4 | 4
150 | 200 | 5 | 5 | 5
200 | 250 | 7 | 7 | 7
250 | 300 | 8 | 8 | 8
300 | 350 | 10 | 10 | 10
350 | 400 | 12 | 12 | 12
400 | 450 | 13 | 13 | 13
450 | 500 | 15 | 15 | 15
500 | 550 | 16 | 16 | 16
550 | 600 | 18 | 18 | 18
600 | 650 | 19 | 19 | 19
650 | 700 | 21 | 21 | 21
700 | 750 | 22 | 22 | 22
750 | 800 | 24 | 24 | 24
800 | 850 | 25 | 25 | 25
850 | 900 | 27 | 27 | 27
900 | 950 | 28 | 28 | 28
950 | 1000 | 30 | 30 | 30
1,000 | Your New York City tax is:
1,000 | 1,050 | 32 | 32 | 32
1,050 | 1,100 | 33 | 33 | 33
1,100 | 1,150 | 35 | 35 | 35
1,150 | 1,200 | 36 | 36 | 36
1,200 | 1,250 | 38 | 38 | 38
1,250 | 1,300 | 39 | 39 | 39
1,300 | 1,350 | 41 | 41 | 41
1,350 | 1,400 | 42 | 42 | 42
1,400 | 1,450 | 44 | 44 | 44
1,450 | 1,500 | 45 | 45 | 45
1,500 | 1,550 | 47 | 47 | 47
1,550 | 1,600 | 48 | 48 | 48
1,600 | 1,650 | 50 | 50 | 50
1,650 | 1,700 | 52 | 52 | 52
1,700 | 1,750 | 53 | 53 | 53
1,750 | 1,800 | 55 | 55 | 55
1,800 | 1,850 | 56 | 56 | 56
1,850 | 1,900 | 58 | 58 | 58
1,900 | 1,950 | 59 | 59 | 59
1,950 | 2,000 | 61 | 61 | 61
This column must also be used by a qualifying surviving spouse
$6,000—$14,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
2,000 | Your New York City tax is:
2,000 | 2,050 | 62 | 62 | 62
2,050 | 2,100 | 64 | 64 | 64
2,100 | 2,150 | 65 | 65 | 65
2,150 | 2,200 | 67 | 67 | 67
2,200 | 2,250 | 68 | 68 | 68
2,250 | 2,300 | 70 | 70 | 70
2,300 | 2,350 | 72 | 72 | 72
2,350 | 2,400 | 73 | 73 | 73
2,400 | 2,450 | 75 | 75 | 75
2,450 | 2,500 | 76 | 76 | 76
2,500 | 2,550 | 78 | 78 | 78
2,550 | 2,600 | 79 | 79 | 79
2,600 | 2,650 | 81 | 81 | 81
2,650 | 2,700 | 82 | 82 | 82
2,700 | 2,750 | 84 | 84 | 84
2,750 | 2,800 | 85 | 85 | 85
2,800 | 2,850 | 87 | 87 | 87
2,850 | 2,900 | 88 | 88 | 88
2,900 | 2,950 | 90 | 90 | 90
2,950 | 3,000 | 92 | 92 | 92
3,000 | Your New York City tax is:
3,000 | 3,050 | 93 | 93 | 93
3,050 | 3,100 | 95 | 95 | 95
3,100 | 3,150 | 96 | 96 | 96
3,150 | 3,200 | 98 | 98 | 98
3,200 | 3,250 | 99 | 99 | 99
3,250 | 3,300 | 101 | 101 | 101
3,300 | 3,350 | 102 | 102 | 102
3,350 | 3,400 | 104 | 104 | 104
3,400 | 3,450 | 105 | 105 | 105
3,450 | 3,500 | 107 | 107 | 107
3,500 | 3,550 | 108 | 108 | 108
3,550 | 3,600 | 110 | 110 | 110
3,600 | 3,650 | 112 | 112 | 112
3,650 | 3,700 | 113 | 113 | 113
3,700 | 3,750 | 115 | 115 | 115
3,750 | 3,800 | 116 | 116 | 116
3,800 | 3,850 | 118 | 118 | 118
3,850 | 3,900 | 119 | 119 | 119
3,900 | 3,950 | 121 | 121 | 121
3,950 | 4,000 | 122 | 122 | 122
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
4,000 | Your New York City tax is:
4,000 | 4,050 | 124 | 124 | 124
4,050 | 4,100 | 125 | 125 | 125
4,100 | 4,150 | 127 | 127 | 127
4,150 | 4,200 | 129 | 129 | 129
4,200 | 4,250 | 130 | 130 | 130
4,250 | 4,300 | 132 | 132 | 132
4,300 | 4,350 | 133 | 133 | 133
4,350 | 4,400 | 135 | 135 | 135
4,400 | 4,450 | 136 | 136 | 136
4,450 | 4,500 | 138 | 138 | 138
4,500 | 4,550 | 139 | 139 | 139
4,550 | 4,600 | 141 | 141 | 141
4,600 | 4,650 | 142 | 142 | 142
4,650 | 4,700 | 144 | 144 | 144
4,700 | 4,750 | 145 | 145 | 145
4,750 | 4,800 | 147 | 147 | 147
4,800 | 4,850 | 149 | 149 | 149
4,850 | 4,900 | 150 | 150 | 150
4,900 | 4,950 | 152 | 152 | 152
4,950 | 5,000 | 153 | 153 | 153
5,000 | Your New York City tax is:
5,000 | 5,050 | 155 | 155 | 155
5,050 | 5,100 | 156 | 156 | 156
5,100 | 5,150 | 158 | 158 | 158
5,150 | 5,200 | 159 | 159 | 159
5,200 | 5,250 | 161 | 161 | 161
5,250 | 5,300 | 162 | 162 | 162
5,300 | 5,350 | 164 | 164 | 164
5,350 | 5,400 | 165 | 165 | 165
5,400 | 5,450 | 167 | 167 | 167
5,450 | 5,500 | 169 | 169 | 169
5,500 | 5,550 | 170 | 170 | 170
5,550 | 5,600 | 172 | 172 | 172
5,600 | 5,650 | 173 | 173 | 173
5,650 | 5,700 | 175 | 175 | 175
5,700 | 5,750 | 176 | 176 | 176
5,750 | 5,800 | 178 | 178 | 178
5,800 | 5,850 | 179 | 179 | 179
5,850 | 5,900 | 181 | 181 | 181
5,900 | 5,950 | 182 | 182 | 182
5,950 | 6,000 | 184 | 184 | 184
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
6,000 | Your New York City tax is:
6,000 | 6,050 | 185 | 185 | 185
6,050 | 6,100 | 187 | 187 | 187
6,100 | 6,150 | 189 | 189 | 189
6,150 | 6,200 | 190 | 190 | 190
6,200 | 6,250 | 192 | 192 | 192
6,250 | 6,300 | 193 | 193 | 193
6,300 | 6,350 | 195 | 195 | 195
6,350 | 6,400 | 196 | 196 | 196
6,400 | 6,450 | 198 | 198 | 198
6,450 | 6,500 | 199 | 199 | 199
6,500 | 6,550 | 201 | 201 | 201
6,550 | 6,600 | 202 | 202 | 202
6,600 | 6,650 | 204 | 204 | 204
6,650 | 6,700 | 205 | 205 | 205
6,700 | 6,750 | 207 | 207 | 207
6,750 | 6,800 | 209 | 209 | 209
6,800 | 6,850 | 210 | 210 | 210
6,850 | 6,900 | 212 | 212 | 212
6,900 | 6,950 | 213 | 213 | 213
6,950 | 7,000 | 215 | 215 | 215
7,000 | Your New York City tax is:
7,000 | 7,050 | 216 | 216 | 216
7,050 | 7,100 | 218 | 218 | 218
7,100 | 7,150 | 219 | 219 | 219
7,150 | 7,200 | 221 | 221 | 221
7,200 | 7,250 | 222 | 222 | 222
7,250 | 7,300 | 224 | 224 | 224
7,300 | 7,350 | 225 | 225 | 225
7,350 | 7,400 | 227 | 227 | 227
7,400 | 7,450 | 229 | 229 | 229
7,450 | 7,500 | 230 | 230 | 230
7,500 | 7,550 | 232 | 232 | 232
7,550 | 7,600 | 233 | 233 | 233
7,600 | 7,650 | 235 | 235 | 235
7,650 | 7,700 | 236 | 236 | 236
7,700 | 7,750 | 238 | 238 | 238
7,750 | 7,800 | 239 | 239 | 239
7,800 | 7,850 | 241 | 241 | 241
7,850 | 7,900 | 242 | 242 | 242
7,900 | 7,950 | 244 | 244 | 244
7,950 | 8,000 | 245 | 245 | 245
8,000 | Your New York City tax is:
8,000 | 8,050 | 247 | 247 | 247
8,050 | 8,100 | 249 | 249 | 249
8,100 | 8,150 | 250 | 250 | 250
8,150 | 8,200 | 252 | 252 | 252
8,200 | 8,250 | 253 | 253 | 253
8,250 | 8,300 | 255 | 255 | 255
8,300 | 8,350 | 256 | 256 | 256
8,350 | 8,400 | 258 | 258 | 258
8,400 | 8,450 | 259 | 259 | 259
8,450 | 8,500 | 261 | 261 | 261
8,500 | 8,550 | 262 | 262 | 262
8,550 | 8,600 | 264 | 264 | 264
8,600 | 8,650 | 265 | 265 | 265
8,650 | 8,700 | 267 | 267 | 267
8,700 | 8,750 | 269 | 269 | 269
8,750 | 8,800 | 270 | 270 | 270
8,800 | 8,850 | 272 | 272 | 272
8,850 | 8,900 | 273 | 273 | 273
8,900 | 8,950 | 275 | 275 | 275
8,950 | 9,000 | 276 | 276 | 276
This column must also be used by a qualifying surviving spouse
$15,000—$23,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
9,000 | Your New York City tax is:
9,000 | 9,050 | 278 | 278 | 278
9,050 | 9,100 | 279 | 279 | 279
9,100 | 9,150 | 281 | 281 | 281
9,150 | 9,200 | 282 | 282 | 282
9,200 | 9,250 | 284 | 284 | 284
9,250 | 9,300 | 285 | 285 | 285
9,300 | 9,350 | 287 | 287 | 287
9,350 | 9,400 | 289 | 289 | 289
9,400 | 9,450 | 290 | 290 | 290
9,450 | 9,500 | 292 | 292 | 292
9,500 | 9,550 | 293 | 293 | 293
9,550 | 9,600 | 295 | 295 | 295
9,600 | 9,650 | 296 | 296 | 296
9,650 | 9,700 | 298 | 298 | 298
9,700 | 9,750 | 299 | 299 | 299
9,750 | 9,800 | 301 | 301 | 301
9,800 | 9,850 | 302 | 302 | 302
9,850 | 9,900 | 304 | 304 | 304
9,900 | 9,950 | 305 | 305 | 305
9,950 | 10,000 | 307 | 307 | 307
10,000 | Your New York City tax is:
10,000 | 10,050 | 309 | 309 | 309
10,050 | 10,100 | 310 | 310 | 310
10,100 | 10,150 | 312 | 312 | 312
10,150 | 10,200 | 313 | 313 | 313
10,200 | 10,250 | 315 | 315 | 315
10,250 | 10,300 | 316 | 316 | 316
10,300 | 10,350 | 318 | 318 | 318
10,350 | 10,400 | 319 | 319 | 319
10,400 | 10,450 | 321 | 321 | 321
10,450 | 10,500 | 322 | 322 | 322
10,500 | 10,550 | 324 | 324 | 324
10,550 | 10,600 | 325 | 325 | 325
10,600 | 10,650 | 327 | 327 | 327
10,650 | 10,700 | 329 | 329 | 329
10,700 | 10,750 | 330 | 330 | 330
10,750 | 10,800 | 332 | 332 | 332
10,800 | 10,850 | 333 | 333 | 333
10,850 | 10,900 | 335 | 335 | 335
10,900 | 10,950 | 336 | 336 | 336
10,950 | 11,000 | 338 | 338 | 338
11,000 | Your New York City tax is:
11,000 | 11,050 | 339 | 339 | 339
11,050 | 11,100 | 341 | 341 | 341
11,100 | 11,150 | 342 | 342 | 342
11,150 | 11,200 | 344 | 344 | 344
11,200 | 11,250 | 346 | 346 | 346
11,250 | 11,300 | 347 | 347 | 347
11,300 | 11,350 | 349 | 349 | 349
11,350 | 11,400 | 350 | 350 | 350
11,400 | 11,450 | 352 | 352 | 352
11,450 | 11,500 | 353 | 353 | 353
11,500 | 11,550 | 355 | 355 | 355
11,550 | 11,600 | 356 | 356 | 356
11,600 | 11,650 | 358 | 358 | 358
11,650 | 11,700 | 359 | 359 | 359
11,700 | 11,750 | 361 | 361 | 361
11,750 | 11,800 | 362 | 362 | 362
11,800 | 11,850 | 364 | 364 | 364
11,850 | 11,900 | 366 | 366 | 366
11,900 | 11,950 | 367 | 367 | 367
11,950 | 12,000 | 369 | 369 | 369
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
12,000 | Your New York City tax is:
12,000 | 12,050 | 370 | 370 | 370
12,050 | 12,100 | 372 | 372 | 372
12,100 | 12,150 | 374 | 373 | 373
12,150 | 12,200 | 376 | 375 | 375
12,200 | 12,250 | 377 | 376 | 376
12,250 | 12,300 | 379 | 378 | 378
12,300 | 12,350 | 381 | 379 | 379
12,350 | 12,400 | 383 | 381 | 381
12,400 | 12,450 | 385 | 382 | 382
12,450 | 12,500 | 387 | 384 | 384
12,500 | 12,550 | 389 | 386 | 386
12,550 | 12,600 | 391 | 387 | 387
12,600 | 12,650 | 393 | 389 | 389
12,650 | 12,700 | 394 | 390 | 390
12,700 | 12,750 | 396 | 392 | 392
12,750 | 12,800 | 398 | 393 | 393
12,800 | 12,850 | 400 | 395 | 395
12,850 | 12,900 | 402 | 396 | 396
12,900 | 12,950 | 404 | 398 | 398
12,950 | 13,000 | 406 | 399 | 399
13,000 | Your New York City tax is:
13,000 | 13,050 | 408 | 401 | 401
13,050 | 13,100 | 409 | 402 | 402
13,100 | 13,150 | 411 | 404 | 404
13,150 | 13,200 | 413 | 406 | 406
13,200 | 13,250 | 415 | 407 | 407
13,250 | 13,300 | 417 | 409 | 409
13,300 | 13,350 | 419 | 410 | 410
13,350 | 13,400 | 421 | 412 | 412
13,400 | 13,450 | 423 | 413 | 413
13,450 | 13,500 | 424 | 415 | 415
13,500 | 13,550 | 426 | 416 | 416
13,550 | 13,600 | 428 | 418 | 418
13,600 | 13,650 | 430 | 419 | 419
13,650 | 13,700 | 432 | 421 | 421
13,700 | 13,750 | 434 | 422 | 422
13,750 | 13,800 | 436 | 424 | 424
13,800 | 13,850 | 438 | 426 | 426
13,850 | 13,900 | 440 | 427 | 427
13,900 | 13,950 | 441 | 429 | 429
13,950 | 14,000 | 443 | 430 | 430
14,000 | Your New York City tax is:
14,000 | 14,050 | 445 | 432 | 432
14,050 | 14,100 | 447 | 433 | 433
14,100 | 14,150 | 449 | 435 | 435
14,150 | 14,200 | 451 | 436 | 436
14,200 | 14,250 | 453 | 438 | 438
14,250 | 14,300 | 455 | 439 | 439
14,300 | 14,350 | 456 | 441 | 441
14,350 | 14,400 | 458 | 442 | 442
14,400 | 14,450 | 460 | 444 | 444
14,450 | 14,500 | 462 | 446 | 446
14,500 | 14,550 | 464 | 447 | 448
14,550 | 14,600 | 466 | 449 | 450
14,600 | 14,650 | 468 | 450 | 451
14,650 | 14,700 | 470 | 452 | 453
14,700 | 14,750 | 472 | 453 | 455
14,750 | 14,800 | 473 | 455 | 457
14,800 | 14,850 | 475 | 456 | 459
14,850 | 14,900 | 477 | 458 | 461
14,900 | 14,950 | 479 | 459 | 463
14,950 | 15,000 | 481 | 461 | 465
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
15,000 | Your New York City tax is:
15,000 | 15,050 | 483 | 462 | 467
15,050 | 15,100 | 485 | 464 | 468
15,100 | 15,150 | 487 | 466 | 470
15,150 | 15,200 | 488 | 467 | 472
15,200 | 15,250 | 490 | 469 | 474
15,250 | 15,300 | 492 | 470 | 476
15,300 | 15,350 | 494 | 472 | 478
15,350 | 15,400 | 496 | 473 | 480
15,400 | 15,450 | 498 | 475 | 482
15,450 | 15,500 | 500 | 476 | 483
15,500 | 15,550 | 502 | 478 | 485
15,550 | 15,600 | 503 | 479 | 487
15,600 | 15,650 | 505 | 481 | 489
15,650 | 15,700 | 507 | 482 | 491
15,700 | 15,750 | 509 | 484 | 493
15,750 | 15,800 | 511 | 486 | 495
15,800 | 15,850 | 513 | 487 | 497
15,850 | 15,900 | 515 | 489 | 498
15,900 | 15,950 | 517 | 490 | 500
15,950 | 16,000 | 519 | 492 | 502
16,000 | Your New York City tax is:
16,000 | 16,050 | 520 | 493 | 504
16,050 | 16,100 | 522 | 495 | 506
16,100 | 16,150 | 524 | 496 | 508
16,150 | 16,200 | 526 | 498 | 510
16,200 | 16,250 | 528 | 499 | 512
16,250 | 16,300 | 530 | 501 | 514
16,300 | 16,350 | 532 | 502 | 515
16,350 | 16,400 | 534 | 504 | 517
16,400 | 16,450 | 535 | 506 | 519
16,450 | 16,500 | 537 | 507 | 521
16,500 | 16,550 | 539 | 509 | 523
16,550 | 16,600 | 541 | 510 | 525
16,600 | 16,650 | 543 | 512 | 527
16,650 | 16,700 | 545 | 513 | 529
16,700 | 16,750 | 547 | 515 | 530
16,750 | 16,800 | 549 | 516 | 532
16,800 | 16,850 | 551 | 518 | 534
16,850 | 16,900 | 552 | 519 | 536
16,900 | 16,950 | 554 | 521 | 538
16,950 | 17,000 | 556 | 522 | 540
17,000 | Your New York City tax is:
17,000 | 17,050 | 558 | 524 | 542
17,050 | 17,100 | 560 | 526 | 544
17,100 | 17,150 | 562 | 527 | 546
17,150 | 17,200 | 564 | 529 | 547
17,200 | 17,250 | 566 | 530 | 549
17,250 | 17,300 | 567 | 532 | 551
17,300 | 17,350 | 569 | 533 | 553
17,350 | 17,400 | 571 | 535 | 555
17,400 | 17,450 | 573 | 536 | 557
17,450 | 17,500 | 575 | 538 | 559
17,500 | 17,550 | 577 | 539 | 561
17,550 | 17,600 | 579 | 541 | 562
17,600 | 17,650 | 581 | 542 | 564
17,650 | 17,700 | 582 | 544 | 566
17,700 | 17,750 | 584 | 546 | 568
17,750 | 17,800 | 586 | 547 | 570
17,800 | 17,850 | 588 | 549 | 572
17,850 | 17,900 | 590 | 550 | 574
17,900 | 17,950 | 592 | 552 | 576
17,950 | 18,000 | 594 | 553 | 577
This column must also be used by a qualifying surviving spouse
$24,000—$32,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
18,000 | Your New York City tax is:
18,000 | 18,050 | 596 | 555 | 579
18,050 | 18,100 | 598 | 556 | 581
18,100 | 18,150 | 599 | 558 | 583
18,150 | 18,200 | 601 | 559 | 585
18,200 | 18,250 | 603 | 561 | 587
18,250 | 18,300 | 605 | 563 | 589
18,300 | 18,350 | 607 | 564 | 591
18,350 | 18,400 | 609 | 566 | 593
18,400 | 18,450 | 611 | 567 | 594
18,450 | 18,500 | 613 | 569 | 596
18,500 | 18,550 | 614 | 570 | 598
18,550 | 18,600 | 616 | 572 | 600
18,600 | 18,650 | 618 | 573 | 602
18,650 | 18,700 | 620 | 575 | 604
18,700 | 18,750 | 622 | 576 | 606
18,750 | 18,800 | 624 | 578 | 608
18,800 | 18,850 | 626 | 579 | 609
18,850 | 18,900 | 628 | 581 | 611
18,900 | 18,950 | 630 | 583 | 613
18,950 | 19,000 | 631 | 584 | 615
19,000 | Your New York City tax is:
19,000 | 19,050 | 633 | 586 | 617
19,050 | 19,100 | 635 | 587 | 619
19,100 | 19,150 | 637 | 589 | 621
19,150 | 19,200 | 639 | 590 | 623
19,200 | 19,250 | 641 | 592 | 625
19,250 | 19,300 | 643 | 593 | 626
19,300 | 19,350 | 645 | 595 | 628
19,350 | 19,400 | 646 | 596 | 630
19,400 | 19,450 | 648 | 598 | 632
19,450 | 19,500 | 650 | 599 | 634
19,500 | 19,550 | 652 | 601 | 636
19,550 | 19,600 | 654 | 603 | 638
19,600 | 19,650 | 656 | 604 | 640
19,650 | 19,700 | 658 | 606 | 641
19,700 | 19,750 | 660 | 607 | 643
19,750 | 19,800 | 661 | 609 | 645
19,800 | 19,850 | 663 | 610 | 647
19,850 | 19,900 | 665 | 612 | 649
19,900 | 19,950 | 667 | 613 | 651
19,950 | 20,000 | 669 | 615 | 653
20,000 | Your New York City tax is:
20,000 | 20,050 | 671 | 616 | 655
20,050 | 20,100 | 673 | 618 | 656
20,100 | 20,150 | 675 | 619 | 658
20,150 | 20,200 | 677 | 621 | 660
20,200 | 20,250 | 678 | 623 | 662
20,250 | 20,300 | 680 | 624 | 664
20,300 | 20,350 | 682 | 626 | 666
20,350 | 20,400 | 684 | 627 | 668
20,400 | 20,450 | 686 | 629 | 670
20,450 | 20,500 | 688 | 630 | 672
20,500 | 20,550 | 690 | 632 | 673
20,550 | 20,600 | 692 | 633 | 675
20,600 | 20,650 | 693 | 635 | 677
20,650 | 20,700 | 695 | 636 | 679
20,700 | 20,750 | 697 | 638 | 681
20,750 | 20,800 | 699 | 639 | 683
20,800 | 20,850 | 701 | 641 | 685
20,850 | 20,900 | 703 | 643 | 687
20,900 | 20,950 | 705 | 644 | 688
20,950 | 21,000 | 707 | 646 | 690
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
21,000 | Your New York City tax is:
21,000 | 21,050 | 709 | 647 | 692
21,050 | 21,100 | 710 | 649 | 694
21,100 | 21,150 | 712 | 650 | 696
21,150 | 21,200 | 714 | 652 | 698
21,200 | 21,250 | 716 | 653 | 700
21,250 | 21,300 | 718 | 655 | 702
21,300 | 21,350 | 720 | 656 | 704
21,350 | 21,400 | 722 | 658 | 705
21,400 | 21,450 | 724 | 659 | 707
21,450 | 21,500 | 725 | 661 | 709
21,500 | 21,550 | 727 | 663 | 711
21,550 | 21,600 | 729 | 664 | 713
21,600 | 21,650 | 731 | 666 | 715
21,650 | 21,700 | 733 | 668 | 717
21,700 | 21,750 | 735 | 670 | 719
21,750 | 21,800 | 737 | 672 | 720
21,800 | 21,850 | 739 | 673 | 722
21,850 | 21,900 | 740 | 675 | 724
21,900 | 21,950 | 742 | 677 | 726
21,950 | 22,000 | 744 | 679 | 728
22,000 | Your New York City tax is:
22,000 | 22,050 | 746 | 681 | 730
22,050 | 22,100 | 748 | 683 | 732
22,100 | 22,150 | 750 | 685 | 734
22,150 | 22,200 | 752 | 687 | 735
22,200 | 22,250 | 754 | 689 | 737
22,250 | 22,300 | 756 | 690 | 739
22,300 | 22,350 | 757 | 692 | 741
22,350 | 22,400 | 759 | 694 | 743
22,400 | 22,450 | 761 | 696 | 745
22,450 | 22,500 | 763 | 698 | 747
22,500 | 22,550 | 765 | 700 | 749
22,550 | 22,600 | 767 | 702 | 751
22,600 | 22,650 | 769 | 704 | 752
22,650 | 22,700 | 771 | 705 | 754
22,700 | 22,750 | 772 | 707 | 756
22,750 | 22,800 | 774 | 709 | 758
22,800 | 22,850 | 776 | 711 | 760
22,850 | 22,900 | 778 | 713 | 762
22,900 | 22,950 | 780 | 715 | 764
22,950 | 23,000 | 782 | 717 | 766
23,000 | Your New York City tax is:
23,000 | 23,050 | 784 | 719 | 767
23,050 | 23,100 | 786 | 720 | 769
23,100 | 23,150 | 788 | 722 | 771
23,150 | 23,200 | 789 | 724 | 773
23,200 | 23,250 | 791 | 726 | 775
23,250 | 23,300 | 793 | 728 | 777
23,300 | 23,350 | 795 | 730 | 779
23,350 | 23,400 | 797 | 732 | 781
23,400 | 23,450 | 799 | 734 | 783
23,450 | 23,500 | 801 | 736 | 784
23,500 | 23,550 | 803 | 737 | 786
23,550 | 23,600 | 804 | 739 | 788
23,600 | 23,650 | 806 | 741 | 790
23,650 | 23,700 | 808 | 743 | 792
23,700 | 23,750 | 810 | 745 | 794
23,750 | 23,800 | 812 | 747 | 796
23,800 | 23,850 | 814 | 749 | 798
23,850 | 23,900 | 816 | 751 | 799
23,900 | 23,950 | 818 | 752 | 801
23,950 | 24,000 | 819 | 754 | 803
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
24,000 | Your New York City tax is:
24,000 | 24,050 | 821 | 756 | 805
24,050 | 24,100 | 823 | 758 | 807
24,100 | 24,150 | 825 | 760 | 809
24,150 | 24,200 | 827 | 762 | 811
24,200 | 24,250 | 829 | 764 | 813
24,250 | 24,300 | 831 | 766 | 814
24,300 | 24,350 | 833 | 768 | 816
24,350 | 24,400 | 835 | 769 | 818
24,400 | 24,450 | 836 | 771 | 820
24,450 | 24,500 | 838 | 773 | 822
24,500 | 24,550 | 840 | 775 | 824
24,550 | 24,600 | 842 | 777 | 826
24,600 | 24,650 | 844 | 779 | 828
24,650 | 24,700 | 846 | 781 | 830
24,700 | 24,750 | 848 | 783 | 831
24,750 | 24,800 | 850 | 784 | 833
24,800 | 24,850 | 851 | 786 | 835
24,850 | 24,900 | 853 | 788 | 837
24,900 | 24,950 | 855 | 790 | 839
24,950 | 25,000 | 857 | 792 | 841
25,000 | Your New York City tax is:
25,000 | 25,050 | 859 | 794 | 843
25,050 | 25,100 | 861 | 796 | 845
25,100 | 25,150 | 863 | 798 | 846
25,150 | 25,200 | 865 | 799 | 848
25,200 | 25,250 | 867 | 801 | 850
25,250 | 25,300 | 869 | 803 | 852
25,300 | 25,350 | 870 | 805 | 854
25,350 | 25,400 | 872 | 807 | 856
25,400 | 25,450 | 874 | 809 | 858
25,450 | 25,500 | 876 | 811 | 860
25,500 | 25,550 | 878 | 813 | 862
25,550 | 25,600 | 880 | 815 | 863
25,600 | 25,650 | 882 | 816 | 865
25,650 | 25,700 | 884 | 818 | 867
25,700 | 25,750 | 886 | 820 | 869
25,750 | 25,800 | 888 | 822 | 871
25,800 | 25,850 | 890 | 824 | 873
25,850 | 25,900 | 891 | 826 | 875
25,900 | 25,950 | 893 | 828 | 877
25,950 | 26,000 | 895 | 830 | 878
26,000 | Your New York City tax is:
26,000 | 26,050 | 897 | 831 | 880
26,050 | 26,100 | 899 | 833 | 882
26,100 | 26,150 | 901 | 835 | 884
26,150 | 26,200 | 903 | 837 | 886
26,200 | 26,250 | 905 | 839 | 888
26,250 | 26,300 | 907 | 841 | 890
26,300 | 26,350 | 909 | 843 | 892
26,350 | 26,400 | 911 | 845 | 893
26,400 | 26,450 | 912 | 847 | 895
26,450 | 26,500 | 914 | 848 | 897
26,500 | 26,550 | 916 | 850 | 899
26,550 | 26,600 | 918 | 852 | 901
26,600 | 26,650 | 920 | 854 | 903
26,650 | 26,700 | 922 | 856 | 905
26,700 | 26,750 | 924 | 858 | 907
26,750 | 26,800 | 926 | 860 | 909
26,800 | 26,850 | 928 | 862 | 910
26,850 | 26,900 | 930 | 863 | 912
26,900 | 26,950 | 932 | 865 | 914
26,950 | 27,000 | 933 | 867 | 916
This column must also be used by a qualifying surviving spouse
$33,000—$41,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
27,000 | Your New York City tax is:
27,000 | 27,050 | 935 | 869 | 918
27,050 | 27,100 | 937 | 871 | 920
27,100 | 27,150 | 939 | 873 | 922
27,150 | 27,200 | 941 | 875 | 924
27,200 | 27,250 | 943 | 877 | 925
27,250 | 27,300 | 945 | 878 | 927
27,300 | 27,350 | 947 | 880 | 929
27,350 | 27,400 | 949 | 882 | 931
27,400 | 27,450 | 951 | 884 | 933
27,450 | 27,500 | 953 | 886 | 935
27,500 | 27,550 | 954 | 888 | 937
27,550 | 27,600 | 956 | 890 | 939
27,600 | 27,650 | 958 | 892 | 941
27,650 | 27,700 | 960 | 894 | 942
27,700 | 27,750 | 962 | 895 | 944
27,750 | 27,800 | 964 | 897 | 946
27,800 | 27,850 | 966 | 899 | 948
27,850 | 27,900 | 968 | 901 | 950
27,900 | 27,950 | 970 | 903 | 952
27,950 | 28,000 | 972 | 905 | 954
28,000 | Your New York City tax is:
28,000 | 28,050 | 974 | 907 | 956
28,050 | 28,100 | 975 | 909 | 957
28,100 | 28,150 | 977 | 910 | 959
28,150 | 28,200 | 979 | 912 | 961
28,200 | 28,250 | 981 | 914 | 963
28,250 | 28,300 | 983 | 916 | 965
28,300 | 28,350 | 985 | 918 | 967
28,350 | 28,400 | 987 | 920 | 969
28,400 | 28,450 | 989 | 922 | 971
28,450 | 28,500 | 991 | 924 | 973
28,500 | 28,550 | 993 | 926 | 974
28,550 | 28,600 | 995 | 927 | 976
28,600 | 28,650 | 996 | 929 | 978
28,650 | 28,700 | 998 | 931 | 980
28,700 | 28,750 | 1,000 | 933 | 982
28,750 | 28,800 | 1,002 | 935 | 984
28,800 | 28,850 | 1,004 | 937 | 986
28,850 | 28,900 | 1,006 | 939 | 988
28,900 | 28,950 | 1,008 | 941 | 989
28,950 | 29,000 | 1,010 | 942 | 991
29,000 | Your New York City tax is:
29,000 | 29,050 | 1,012 | 944 | 993
29,050 | 29,100 | 1,014 | 946 | 995
29,100 | 29,150 | 1,016 | 948 | 997
29,150 | 29,200 | 1,017 | 950 | 999
29,200 | 29,250 | 1,019 | 952 | 1,001
29,250 | 29,300 | 1,021 | 954 | 1,003
29,300 | 29,350 | 1,023 | 956 | 1,004
29,350 | 29,400 | 1,025 | 957 | 1,006
29,400 | 29,450 | 1,027 | 959 | 1,008
29,450 | 29,500 | 1,029 | 961 | 1,010
29,500 | 29,550 | 1,031 | 963 | 1,012
29,550 | 29,600 | 1,033 | 965 | 1,014
29,600 | 29,650 | 1,035 | 967 | 1,016
29,650 | 29,700 | 1,037 | 969 | 1,018
29,700 | 29,750 | 1,038 | 971 | 1,020
29,750 | 29,800 | 1,040 | 973 | 1,021
29,800 | 29,850 | 1,042 | 974 | 1,023
29,850 | 29,900 | 1,044 | 976 | 1,025
29,900 | 29,950 | 1,046 | 978 | 1,027
29,950 | 30,000 | 1,048 | 980 | 1,029
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
30,000 | Your New York City tax is:
30,000 | 30,050 | 1,050 | 982 | 1,031
30,050 | 30,100 | 1,052 | 984 | 1,033
30,100 | 30,150 | 1,054 | 986 | 1,035
30,150 | 30,200 | 1,056 | 988 | 1,037
30,200 | 30,250 | 1,058 | 989 | 1,039
30,250 | 30,300 | 1,059 | 991 | 1,041
30,300 | 30,350 | 1,061 | 993 | 1,042
30,350 | 30,400 | 1,063 | 995 | 1,044
30,400 | 30,450 | 1,065 | 997 | 1,046
30,450 | 30,500 | 1,067 | 999 | 1,048
30,500 | 30,550 | 1,069 | 1,001 | 1,050
30,550 | 30,600 | 1,071 | 1,003 | 1,052
30,600 | 30,650 | 1,073 | 1,005 | 1,054
30,650 | 30,700 | 1,075 | 1,006 | 1,056
30,700 | 30,750 | 1,077 | 1,008 | 1,058
30,750 | 30,800 | 1,079 | 1,010 | 1,060
30,800 | 30,850 | 1,080 | 1,012 | 1,062
30,850 | 30,900 | 1,082 | 1,014 | 1,063
30,900 | 30,950 | 1,084 | 1,016 | 1,065
30,950 | 31,000 | 1,086 | 1,018 | 1,067
31,000 | Your New York City tax is:
31,000 | 31,050 | 1,088 | 1,020 | 1,069
31,050 | 31,100 | 1,090 | 1,021 | 1,071
31,100 | 31,150 | 1,092 | 1,023 | 1,073
31,150 | 31,200 | 1,094 | 1,025 | 1,075
31,200 | 31,250 | 1,096 | 1,027 | 1,077
31,250 | 31,300 | 1,098 | 1,029 | 1,079
31,300 | 31,350 | 1,100 | 1,031 | 1,081
31,350 | 31,400 | 1,101 | 1,033 | 1,083
31,400 | 31,450 | 1,103 | 1,035 | 1,084
31,450 | 31,500 | 1,105 | 1,036 | 1,086
31,500 | 31,550 | 1,107 | 1,038 | 1,088
31,550 | 31,600 | 1,109 | 1,040 | 1,090
31,600 | 31,650 | 1,111 | 1,042 | 1,092
31,650 | 31,700 | 1,113 | 1,044 | 1,094
31,700 | 31,750 | 1,115 | 1,046 | 1,096
31,750 | 31,800 | 1,117 | 1,048 | 1,098
31,800 | 31,850 | 1,119 | 1,050 | 1,100
31,850 | 31,900 | 1,121 | 1,052 | 1,102
31,900 | 31,950 | 1,122 | 1,053 | 1,104
31,950 | 32,000 | 1,124 | 1,055 | 1,105
32,000 | Your New York City tax is:
32,000 | 32,050 | 1,126 | 1,057 | 1,107
32,050 | 32,100 | 1,128 | 1,059 | 1,109
32,100 | 32,150 | 1,130 | 1,061 | 1,111
32,150 | 32,200 | 1,132 | 1,063 | 1,113
32,200 | 32,250 | 1,134 | 1,065 | 1,115
32,250 | 32,300 | 1,136 | 1,067 | 1,117
32,300 | 32,350 | 1,138 | 1,068 | 1,119
32,350 | 32,400 | 1,140 | 1,070 | 1,121
32,400 | 32,450 | 1,142 | 1,072 | 1,123
32,450 | 32,500 | 1,143 | 1,074 | 1,125
32,500 | 32,550 | 1,145 | 1,076 | 1,126
32,550 | 32,600 | 1,147 | 1,078 | 1,128
32,600 | 32,650 | 1,149 | 1,080 | 1,130
32,650 | 32,700 | 1,151 | 1,082 | 1,132
32,700 | 32,750 | 1,153 | 1,084 | 1,134
32,750 | 32,800 | 1,155 | 1,085 | 1,136
32,800 | 32,850 | 1,157 | 1,087 | 1,138
32,850 | 32,900 | 1,159 | 1,089 | 1,140
32,900 | 32,950 | 1,161 | 1,091 | 1,142
32,950 | 33,000 | 1,163 | 1,093 | 1,144
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
33,000 | Your New York City tax is:
33,000 | 33,050 | 1,164 | 1,095 | 1,146
33,050 | 33,100 | 1,166 | 1,097 | 1,147
33,100 | 33,150 | 1,168 | 1,099 | 1,149
33,150 | 33,200 | 1,170 | 1,100 | 1,151
33,200 | 33,250 | 1,172 | 1,102 | 1,153
33,250 | 33,300 | 1,174 | 1,104 | 1,155
33,300 | 33,350 | 1,176 | 1,106 | 1,157
33,350 | 33,400 | 1,178 | 1,108 | 1,159
33,400 | 33,450 | 1,180 | 1,110 | 1,161
33,450 | 33,500 | 1,182 | 1,112 | 1,163
33,500 | 33,550 | 1,184 | 1,114 | 1,165
33,550 | 33,600 | 1,185 | 1,115 | 1,167
33,600 | 33,650 | 1,187 | 1,117 | 1,168
33,650 | 33,700 | 1,189 | 1,119 | 1,170
33,700 | 33,750 | 1,191 | 1,121 | 1,172
33,750 | 33,800 | 1,193 | 1,123 | 1,174
33,800 | 33,850 | 1,195 | 1,125 | 1,176
33,850 | 33,900 | 1,197 | 1,127 | 1,178
33,900 | 33,950 | 1,199 | 1,129 | 1,180
33,950 | 34,000 | 1,201 | 1,131 | 1,182
34,000 | Your New York City tax is:
34,000 | 34,050 | 1,203 | 1,132 | 1,184
34,050 | 34,100 | 1,205 | 1,134 | 1,186
34,100 | 34,150 | 1,206 | 1,136 | 1,188
34,150 | 34,200 | 1,208 | 1,138 | 1,189
34,200 | 34,250 | 1,210 | 1,140 | 1,191
34,250 | 34,300 | 1,212 | 1,142 | 1,193
34,300 | 34,350 | 1,214 | 1,144 | 1,195
34,350 | 34,400 | 1,216 | 1,146 | 1,197
34,400 | 34,450 | 1,218 | 1,147 | 1,199
34,450 | 34,500 | 1,220 | 1,149 | 1,201
34,500 | 34,550 | 1,222 | 1,151 | 1,203
34,550 | 34,600 | 1,224 | 1,153 | 1,205
34,600 | 34,650 | 1,226 | 1,155 | 1,207
34,650 | 34,700 | 1,227 | 1,157 | 1,209
34,700 | 34,750 | 1,229 | 1,159 | 1,210
34,750 | 34,800 | 1,231 | 1,161 | 1,212
34,800 | 34,850 | 1,233 | 1,163 | 1,214
34,850 | 34,900 | 1,235 | 1,164 | 1,216
34,900 | 34,950 | 1,237 | 1,166 | 1,218
34,950 | 35,000 | 1,239 | 1,168 | 1,220
35,000 | Your New York City tax is:
35,000 | 35,050 | 1,241 | 1,170 | 1,222
35,050 | 35,100 | 1,243 | 1,172 | 1,224
35,100 | 35,150 | 1,245 | 1,174 | 1,226
35,150 | 35,200 | 1,247 | 1,176 | 1,228
35,200 | 35,250 | 1,248 | 1,178 | 1,230
35,250 | 35,300 | 1,250 | 1,179 | 1,231
35,300 | 35,350 | 1,252 | 1,181 | 1,233
35,350 | 35,400 | 1,254 | 1,183 | 1,235
35,400 | 35,450 | 1,256 | 1,185 | 1,237
35,450 | 35,500 | 1,258 | 1,187 | 1,239
35,500 | 35,550 | 1,260 | 1,189 | 1,241
35,550 | 35,600 | 1,262 | 1,191 | 1,243
35,600 | 35,650 | 1,264 | 1,193 | 1,245
35,650 | 35,700 | 1,266 | 1,195 | 1,247
35,700 | 35,750 | 1,268 | 1,196 | 1,249
35,750 | 35,800 | 1,269 | 1,198 | 1,251
35,800 | 35,850 | 1,271 | 1,200 | 1,252
35,850 | 35,900 | 1,273 | 1,202 | 1,254
35,900 | 35,950 | 1,275 | 1,204 | 1,256
35,950 | 36,000 | 1,277 | 1,206 | 1,258
This column must also be used by a qualifying surviving spouse
$42,000—$50,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
36,000 | Your New York City tax is:
36,000 | 36,050 | 1,279 | 1,208 | 1,260
36,050 | 36,100 | 1,281 | 1,210 | 1,262
36,100 | 36,150 | 1,283 | 1,211 | 1,264
36,150 | 36,200 | 1,285 | 1,213 | 1,266
36,200 | 36,250 | 1,287 | 1,215 | 1,268
36,250 | 36,300 | 1,289 | 1,217 | 1,270
36,300 | 36,350 | 1,291 | 1,219 | 1,272
36,350 | 36,400 | 1,292 | 1,221 | 1,273
36,400 | 36,450 | 1,294 | 1,223 | 1,275
36,450 | 36,500 | 1,296 | 1,225 | 1,277
36,500 | 36,550 | 1,298 | 1,226 | 1,279
36,550 | 36,600 | 1,300 | 1,228 | 1,281
36,600 | 36,650 | 1,302 | 1,230 | 1,283
36,650 | 36,700 | 1,304 | 1,232 | 1,285
36,700 | 36,750 | 1,306 | 1,234 | 1,287
36,750 | 36,800 | 1,308 | 1,236 | 1,289
36,800 | 36,850 | 1,310 | 1,238 | 1,291
36,850 | 36,900 | 1,312 | 1,240 | 1,293
36,900 | 36,950 | 1,313 | 1,242 | 1,294
36,950 | 37,000 | 1,315 | 1,243 | 1,296
37,000 | Your New York City tax is:
37,000 | 37,050 | 1,317 | 1,245 | 1,298
37,050 | 37,100 | 1,319 | 1,247 | 1,300
37,100 | 37,150 | 1,321 | 1,249 | 1,302
37,150 | 37,200 | 1,323 | 1,251 | 1,304
37,200 | 37,250 | 1,325 | 1,253 | 1,306
37,250 | 37,300 | 1,327 | 1,255 | 1,308
37,300 | 37,350 | 1,329 | 1,257 | 1,310
37,350 | 37,400 | 1,331 | 1,258 | 1,312
37,400 | 37,450 | 1,333 | 1,260 | 1,314
37,450 | 37,500 | 1,334 | 1,262 | 1,315
37,500 | 37,550 | 1,336 | 1,264 | 1,317
37,550 | 37,600 | 1,338 | 1,266 | 1,319
37,600 | 37,650 | 1,340 | 1,268 | 1,321
37,650 | 37,700 | 1,342 | 1,270 | 1,323
37,700 | 37,750 | 1,344 | 1,272 | 1,325
37,750 | 37,800 | 1,346 | 1,274 | 1,327
37,800 | 37,850 | 1,348 | 1,275 | 1,329
37,850 | 37,900 | 1,350 | 1,277 | 1,331
37,900 | 37,950 | 1,352 | 1,279 | 1,333
37,950 | 38,000 | 1,354 | 1,281 | 1,335
38,000 | Your New York City tax is:
38,000 | 38,050 | 1,355 | 1,283 | 1,336
38,050 | 38,100 | 1,357 | 1,285 | 1,338
38,100 | 38,150 | 1,359 | 1,287 | 1,340
38,150 | 38,200 | 1,361 | 1,289 | 1,342
38,200 | 38,250 | 1,363 | 1,290 | 1,344
38,250 | 38,300 | 1,365 | 1,292 | 1,346
38,300 | 38,350 | 1,367 | 1,294 | 1,348
38,350 | 38,400 | 1,369 | 1,296 | 1,350
38,400 | 38,450 | 1,371 | 1,298 | 1,352
38,450 | 38,500 | 1,373 | 1,300 | 1,354
38,500 | 38,550 | 1,375 | 1,302 | 1,356
38,550 | 38,600 | 1,376 | 1,304 | 1,357
38,600 | 38,650 | 1,378 | 1,305 | 1,359
38,650 | 38,700 | 1,380 | 1,307 | 1,361
38,700 | 38,750 | 1,382 | 1,309 | 1,363
38,750 | 38,800 | 1,384 | 1,311 | 1,365
38,800 | 38,850 | 1,386 | 1,313 | 1,367
38,850 | 38,900 | 1,388 | 1,315 | 1,369
38,900 | 38,950 | 1,390 | 1,317 | 1,371
38,950 | 39,000 | 1,392 | 1,319 | 1,373
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
39,000 | Your New York City tax is:
39,000 | 39,050 | 1,394 | 1,321 | 1,375
39,050 | 39,100 | 1,396 | 1,322 | 1,377
39,100 | 39,150 | 1,397 | 1,324 | 1,378
39,150 | 39,200 | 1,399 | 1,326 | 1,380
39,200 | 39,250 | 1,401 | 1,328 | 1,382
39,250 | 39,300 | 1,403 | 1,330 | 1,384
39,300 | 39,350 | 1,405 | 1,332 | 1,386
39,350 | 39,400 | 1,407 | 1,334 | 1,388
39,400 | 39,450 | 1,409 | 1,336 | 1,390
39,450 | 39,500 | 1,411 | 1,337 | 1,392
39,500 | 39,550 | 1,413 | 1,339 | 1,394
39,550 | 39,600 | 1,415 | 1,341 | 1,396
39,600 | 39,650 | 1,417 | 1,343 | 1,398
39,650 | 39,700 | 1,418 | 1,345 | 1,399
39,700 | 39,750 | 1,420 | 1,347 | 1,401
39,750 | 39,800 | 1,422 | 1,349 | 1,403
39,800 | 39,850 | 1,424 | 1,351 | 1,405
39,850 | 39,900 | 1,426 | 1,353 | 1,407
39,900 | 39,950 | 1,428 | 1,354 | 1,409
39,950 | 40,000 | 1,430 | 1,356 | 1,411
40,000 | Your New York City tax is:
40,000 | 40,050 | 1,432 | 1,358 | 1,413
40,050 | 40,100 | 1,434 | 1,360 | 1,415
40,100 | 40,150 | 1,436 | 1,362 | 1,417
40,150 | 40,200 | 1,438 | 1,364 | 1,419
40,200 | 40,250 | 1,439 | 1,366 | 1,420
40,250 | 40,300 | 1,441 | 1,368 | 1,422
40,300 | 40,350 | 1,443 | 1,369 | 1,424
40,350 | 40,400 | 1,445 | 1,371 | 1,426
40,400 | 40,450 | 1,447 | 1,373 | 1,428
40,450 | 40,500 | 1,449 | 1,375 | 1,430
40,500 | 40,550 | 1,451 | 1,377 | 1,432
40,550 | 40,600 | 1,453 | 1,379 | 1,434
40,600 | 40,650 | 1,455 | 1,381 | 1,436
40,650 | 40,700 | 1,457 | 1,383 | 1,438
40,700 | 40,750 | 1,459 | 1,384 | 1,440
40,750 | 40,800 | 1,460 | 1,386 | 1,441
40,800 | 40,850 | 1,462 | 1,388 | 1,443
40,850 | 40,900 | 1,464 | 1,390 | 1,445
40,900 | 40,950 | 1,466 | 1,392 | 1,447
40,950 | 41,000 | 1,468 | 1,394 | 1,449
41,000 | Your New York City tax is:
41,000 | 41,050 | 1,470 | 1,396 | 1,451
41,050 | 41,100 | 1,472 | 1,398 | 1,453
41,100 | 41,150 | 1,474 | 1,400 | 1,455
41,150 | 41,200 | 1,476 | 1,401 | 1,457
41,200 | 41,250 | 1,478 | 1,403 | 1,459
41,250 | 41,300 | 1,480 | 1,405 | 1,461
41,300 | 41,350 | 1,481 | 1,407 | 1,463
41,350 | 41,400 | 1,483 | 1,409 | 1,464
41,400 | 41,450 | 1,485 | 1,411 | 1,466
41,450 | 41,500 | 1,487 | 1,413 | 1,468
41,500 | 41,550 | 1,489 | 1,415 | 1,470
41,550 | 41,600 | 1,491 | 1,416 | 1,472
41,600 | 41,650 | 1,493 | 1,418 | 1,474
41,650 | 41,700 | 1,495 | 1,420 | 1,476
41,700 | 41,750 | 1,497 | 1,422 | 1,478
41,750 | 41,800 | 1,499 | 1,424 | 1,480
41,800 | 41,850 | 1,501 | 1,426 | 1,482
41,850 | 41,900 | 1,502 | 1,428 | 1,484
41,900 | 41,950 | 1,504 | 1,430 | 1,485
41,950 | 42,000 | 1,506 | 1,432 | 1,487
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
42,000 | Your New York City tax is:
42,000 | 42,050 | 1,508 | 1,433 | 1,489
42,050 | 42,100 | 1,510 | 1,435 | 1,491
42,100 | 42,150 | 1,512 | 1,437 | 1,493
42,150 | 42,200 | 1,514 | 1,439 | 1,495
42,200 | 42,250 | 1,516 | 1,441 | 1,497
42,250 | 42,300 | 1,518 | 1,443 | 1,499
42,300 | 42,350 | 1,520 | 1,445 | 1,501
42,350 | 42,400 | 1,522 | 1,447 | 1,503
42,400 | 42,450 | 1,523 | 1,448 | 1,505
42,450 | 42,500 | 1,525 | 1,450 | 1,506
42,500 | 42,550 | 1,527 | 1,452 | 1,508
42,550 | 42,600 | 1,529 | 1,454 | 1,510
42,600 | 42,650 | 1,531 | 1,456 | 1,512
42,650 | 42,700 | 1,533 | 1,458 | 1,514
42,700 | 42,750 | 1,535 | 1,460 | 1,516
42,750 | 42,800 | 1,537 | 1,462 | 1,518
42,800 | 42,850 | 1,539 | 1,463 | 1,520
42,850 | 42,900 | 1,541 | 1,465 | 1,522
42,900 | 42,950 | 1,543 | 1,467 | 1,524
42,950 | 43,000 | 1,544 | 1,469 | 1,526
43,000 | Your New York City tax is:
43,000 | 43,050 | 1,546 | 1,471 | 1,527
43,050 | 43,100 | 1,548 | 1,473 | 1,529
43,100 | 43,150 | 1,550 | 1,475 | 1,531
43,150 | 43,200 | 1,552 | 1,477 | 1,533
43,200 | 43,250 | 1,554 | 1,479 | 1,535
43,250 | 43,300 | 1,556 | 1,480 | 1,537
43,300 | 43,350 | 1,558 | 1,482 | 1,539
43,350 | 43,400 | 1,560 | 1,484 | 1,541
43,400 | 43,450 | 1,562 | 1,486 | 1,543
43,450 | 43,500 | 1,564 | 1,488 | 1,545
43,500 | 43,550 | 1,565 | 1,490 | 1,547
43,550 | 43,600 | 1,567 | 1,492 | 1,548
43,600 | 43,650 | 1,569 | 1,494 | 1,550
43,650 | 43,700 | 1,571 | 1,495 | 1,552
43,700 | 43,750 | 1,573 | 1,497 | 1,554
43,750 | 43,800 | 1,575 | 1,499 | 1,556
43,800 | 43,850 | 1,577 | 1,501 | 1,558
43,850 | 43,900 | 1,579 | 1,503 | 1,560
43,900 | 43,950 | 1,581 | 1,505 | 1,562
43,950 | 44,000 | 1,583 | 1,507 | 1,564
44,000 | Your New York City tax is:
44,000 | 44,050 | 1,585 | 1,509 | 1,566
44,050 | 44,100 | 1,586 | 1,511 | 1,568
44,100 | 44,150 | 1,588 | 1,512 | 1,569
44,150 | 44,200 | 1,590 | 1,514 | 1,571
44,200 | 44,250 | 1,592 | 1,516 | 1,573
44,250 | 44,300 | 1,594 | 1,518 | 1,575
44,300 | 44,350 | 1,596 | 1,520 | 1,577
44,350 | 44,400 | 1,598 | 1,522 | 1,579
44,400 | 44,450 | 1,600 | 1,524 | 1,581
44,450 | 44,500 | 1,602 | 1,526 | 1,583
44,500 | 44,550 | 1,604 | 1,527 | 1,585
44,550 | 44,600 | 1,606 | 1,529 | 1,587
44,600 | 44,650 | 1,607 | 1,531 | 1,589
44,650 | 44,700 | 1,609 | 1,533 | 1,590
44,700 | 44,750 | 1,611 | 1,535 | 1,592
44,750 | 44,800 | 1,613 | 1,537 | 1,594
44,800 | 44,850 | 1,615 | 1,539 | 1,596
44,850 | 44,900 | 1,617 | 1,541 | 1,598
44,900 | 44,950 | 1,619 | 1,542 | 1,600
44,950 | 45,000 | 1,621 | 1,544 | 1,602
This column must also be used by a qualifying surviving spouse
$51,000—$59,999
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
45,000 | Your New York City tax is:
45,000 | 45,050 | 1,623 | 1,546 | 1,604
45,050 | 45,100 | 1,625 | 1,548 | 1,606
45,100 | 45,150 | 1,627 | 1,550 | 1,608
45,150 | 45,200 | 1,628 | 1,552 | 1,610
45,200 | 45,250 | 1,630 | 1,554 | 1,611
45,250 | 45,300 | 1,632 | 1,556 | 1,613
45,300 | 45,350 | 1,634 | 1,557 | 1,615
45,350 | 45,400 | 1,636 | 1,559 | 1,617
45,400 | 45,450 | 1,638 | 1,561 | 1,619
45,450 | 45,500 | 1,640 | 1,563 | 1,621
45,500 | 45,550 | 1,642 | 1,565 | 1,623
45,550 | 45,600 | 1,644 | 1,567 | 1,625
45,600 | 45,650 | 1,646 | 1,569 | 1,627
45,650 | 45,700 | 1,648 | 1,571 | 1,629
45,700 | 45,750 | 1,649 | 1,573 | 1,631
45,750 | 45,800 | 1,651 | 1,575 | 1,632
45,800 | 45,850 | 1,653 | 1,577 | 1,634
45,850 | 45,900 | 1,655 | 1,578 | 1,636
45,900 | 45,950 | 1,657 | 1,580 | 1,638
45,950 | 46,000 | 1,659 | 1,582 | 1,640
46,000 | Your New York City tax is:
46,000 | 46,050 | 1,661 | 1,584 | 1,642
46,050 | 46,100 | 1,663 | 1,586 | 1,644
46,100 | 46,150 | 1,665 | 1,588 | 1,646
46,150 | 46,200 | 1,667 | 1,590 | 1,648
46,200 | 46,250 | 1,669 | 1,592 | 1,650
46,250 | 46,300 | 1,670 | 1,594 | 1,652
46,300 | 46,350 | 1,672 | 1,596 | 1,653
46,350 | 46,400 | 1,674 | 1,598 | 1,655
46,400 | 46,450 | 1,676 | 1,599 | 1,657
46,450 | 46,500 | 1,678 | 1,601 | 1,659
46,500 | 46,550 | 1,680 | 1,603 | 1,661
46,550 | 46,600 | 1,682 | 1,605 | 1,663
46,600 | 46,650 | 1,684 | 1,607 | 1,665
46,650 | 46,700 | 1,686 | 1,609 | 1,667
46,700 | 46,750 | 1,688 | 1,611 | 1,669
46,750 | 46,800 | 1,690 | 1,613 | 1,671
46,800 | 46,850 | 1,691 | 1,615 | 1,673
46,850 | 46,900 | 1,693 | 1,617 | 1,674
46,900 | 46,950 | 1,695 | 1,619 | 1,676
46,950 | 47,000 | 1,697 | 1,620 | 1,678
47,000 | Your New York City tax is:
47,000 | 47,050 | 1,699 | 1,622 | 1,680
47,050 | 47,100 | 1,701 | 1,624 | 1,682
47,100 | 47,150 | 1,703 | 1,626 | 1,684
47,150 | 47,200 | 1,705 | 1,628 | 1,686
47,200 | 47,250 | 1,707 | 1,630 | 1,688
47,250 | 47,300 | 1,709 | 1,632 | 1,690
47,300 | 47,350 | 1,711 | 1,634 | 1,692
47,350 | 47,400 | 1,713 | 1,636 | 1,694
47,400 | 47,450 | 1,714 | 1,638 | 1,695
47,450 | 47,500 | 1,716 | 1,640 | 1,697
47,500 | 47,550 | 1,718 | 1,641 | 1,699
47,550 | 47,600 | 1,720 | 1,643 | 1,701
47,600 | 47,650 | 1,722 | 1,645 | 1,703
47,650 | 47,700 | 1,724 | 1,647 | 1,705
47,700 | 47,750 | 1,726 | 1,649 | 1,707
47,750 | 47,800 | 1,728 | 1,651 | 1,709
47,800 | 47,850 | 1,730 | 1,653 | 1,711
47,850 | 47,900 | 1,732 | 1,655 | 1,713
47,900 | 47,950 | 1,734 | 1,657 | 1,715
47,950 | 48,000 | 1,735 | 1,659 | 1,716
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
48,000 | Your New York City tax is:
48,000 | 48,050 | 1,737 | 1,661 | 1,718
48,050 | 48,100 | 1,739 | 1,662 | 1,720
48,100 | 48,150 | 1,741 | 1,664 | 1,722
48,150 | 48,200 | 1,743 | 1,666 | 1,724
48,200 | 48,250 | 1,745 | 1,668 | 1,726
48,250 | 48,300 | 1,747 | 1,670 | 1,728
48,300 | 48,350 | 1,749 | 1,672 | 1,730
48,350 | 48,400 | 1,751 | 1,674 | 1,732
48,400 | 48,450 | 1,753 | 1,676 | 1,734
48,450 | 48,500 | 1,755 | 1,678 | 1,736
48,500 | 48,550 | 1,756 | 1,680 | 1,737
48,550 | 48,600 | 1,758 | 1,682 | 1,739
48,600 | 48,650 | 1,760 | 1,683 | 1,741
48,650 | 48,700 | 1,762 | 1,685 | 1,743
48,700 | 48,750 | 1,764 | 1,687 | 1,745
48,750 | 48,800 | 1,766 | 1,689 | 1,747
48,800 | 48,850 | 1,768 | 1,691 | 1,749
48,850 | 48,900 | 1,770 | 1,693 | 1,751
48,900 | 48,950 | 1,772 | 1,695 | 1,753
48,950 | 49,000 | 1,774 | 1,697 | 1,755
49,000 | Your New York City tax is:
49,000 | 49,050 | 1,776 | 1,699 | 1,757
49,050 | 49,100 | 1,777 | 1,701 | 1,758
49,100 | 49,150 | 1,779 | 1,703 | 1,760
49,150 | 49,200 | 1,781 | 1,704 | 1,762
49,200 | 49,250 | 1,783 | 1,706 | 1,764
49,250 | 49,300 | 1,785 | 1,708 | 1,766
49,300 | 49,350 | 1,787 | 1,710 | 1,768
49,350 | 49,400 | 1,789 | 1,712 | 1,770
49,400 | 49,450 | 1,791 | 1,714 | 1,772
49,450 | 49,500 | 1,793 | 1,716 | 1,774
49,500 | 49,550 | 1,795 | 1,718 | 1,776
49,550 | 49,600 | 1,797 | 1,720 | 1,778
49,600 | 49,650 | 1,798 | 1,722 | 1,779
49,650 | 49,700 | 1,800 | 1,724 | 1,781
49,700 | 49,750 | 1,802 | 1,725 | 1,783
49,750 | 49,800 | 1,804 | 1,727 | 1,785
49,800 | 49,850 | 1,806 | 1,729 | 1,787
49,850 | 49,900 | 1,808 | 1,731 | 1,789
49,900 | 49,950 | 1,810 | 1,733 | 1,791
49,950 | 50,000 | 1,812 | 1,735 | 1,793
50,000 | Your New York City tax is:
50,000 | 50,050 | 1,814 | 1,737 | 1,795
50,050 | 50,100 | 1,816 | 1,739 | 1,797
50,100 | 50,150 | 1,818 | 1,741 | 1,799
50,150 | 50,200 | 1,820 | 1,743 | 1,800
50,200 | 50,250 | 1,822 | 1,745 | 1,802
50,250 | 50,300 | 1,824 | 1,746 | 1,804
50,300 | 50,350 | 1,826 | 1,748 | 1,806
50,350 | 50,400 | 1,828 | 1,750 | 1,808
50,400 | 50,450 | 1,829 | 1,752 | 1,810
50,450 | 50,500 | 1,831 | 1,754 | 1,812
50,500 | 50,550 | 1,833 | 1,756 | 1,814
50,550 | 50,600 | 1,835 | 1,758 | 1,816
50,600 | 50,650 | 1,837 | 1,760 | 1,818
50,650 | 50,700 | 1,839 | 1,762 | 1,820
50,700 | 50,750 | 1,841 | 1,764 | 1,821
50,750 | 50,800 | 1,843 | 1,766 | 1,823
50,800 | 50,850 | 1,845 | 1,767 | 1,825
50,850 | 50,900 | 1,847 | 1,769 | 1,827
50,900 | 50,950 | 1,849 | 1,771 | 1,829
50,950 | 51,000 | 1,851 | 1,773 | 1,831
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
51,000 | Your New York City tax is:
51,000 | 51,050 | 1,853 | 1,775 | 1,833
51,050 | 51,100 | 1,855 | 1,777 | 1,835
51,100 | 51,150 | 1,857 | 1,779 | 1,837
51,150 | 51,200 | 1,859 | 1,781 | 1,839
51,200 | 51,250 | 1,860 | 1,783 | 1,841
51,250 | 51,300 | 1,862 | 1,785 | 1,842
51,300 | 51,350 | 1,864 | 1,787 | 1,844
51,350 | 51,400 | 1,866 | 1,788 | 1,846
51,400 | 51,450 | 1,868 | 1,790 | 1,848
51,450 | 51,500 | 1,870 | 1,792 | 1,850
51,500 | 51,550 | 1,872 | 1,794 | 1,852
51,550 | 51,600 | 1,874 | 1,796 | 1,854
51,600 | 51,650 | 1,876 | 1,798 | 1,856
51,650 | 51,700 | 1,878 | 1,800 | 1,858
51,700 | 51,750 | 1,880 | 1,802 | 1,860
51,750 | 51,800 | 1,882 | 1,804 | 1,862
51,800 | 51,850 | 1,884 | 1,806 | 1,863
51,850 | 51,900 | 1,886 | 1,808 | 1,865
51,900 | 51,950 | 1,888 | 1,809 | 1,867
51,950 | 52,000 | 1,890 | 1,811 | 1,869
52,000 | Your New York City tax is:
52,000 | 52,050 | 1,891 | 1,813 | 1,871
52,050 | 52,100 | 1,893 | 1,815 | 1,873
52,100 | 52,150 | 1,895 | 1,817 | 1,875
52,150 | 52,200 | 1,897 | 1,819 | 1,877
52,200 | 52,250 | 1,899 | 1,821 | 1,879
52,250 | 52,300 | 1,901 | 1,823 | 1,881
52,300 | 52,350 | 1,903 | 1,825 | 1,883
52,350 | 52,400 | 1,905 | 1,827 | 1,885
52,400 | 52,450 | 1,907 | 1,829 | 1,886
52,450 | 52,500 | 1,909 | 1,830 | 1,888
52,500 | 52,550 | 1,911 | 1,832 | 1,890
52,550 | 52,600 | 1,913 | 1,834 | 1,892
52,600 | 52,650 | 1,915 | 1,836 | 1,894
52,650 | 52,700 | 1,917 | 1,838 | 1,896
52,700 | 52,750 | 1,919 | 1,840 | 1,898
52,750 | 52,800 | 1,921 | 1,842 | 1,900
52,800 | 52,850 | 1,922 | 1,844 | 1,902
52,850 | 52,900 | 1,924 | 1,846 | 1,904
52,900 | 52,950 | 1,926 | 1,848 | 1,906
52,950 | 53,000 | 1,928 | 1,850 | 1,907
53,000 | Your New York City tax is:
53,000 | 53,050 | 1,930 | 1,851 | 1,909
53,050 | 53,100 | 1,932 | 1,853 | 1,911
53,100 | 53,150 | 1,934 | 1,855 | 1,913
53,150 | 53,200 | 1,936 | 1,857 | 1,915
53,200 | 53,250 | 1,938 | 1,859 | 1,917
53,250 | 53,300 | 1,940 | 1,861 | 1,919
53,300 | 53,350 | 1,942 | 1,863 | 1,921
53,350 | 53,400 | 1,944 | 1,865 | 1,923
53,400 | 53,450 | 1,946 | 1,867 | 1,925
53,450 | 53,500 | 1,948 | 1,869 | 1,927
53,500 | 53,550 | 1,950 | 1,871 | 1,928
53,550 | 53,600 | 1,952 | 1,872 | 1,930
53,600 | 53,650 | 1,954 | 1,874 | 1,932
53,650 | 53,700 | 1,955 | 1,876 | 1,934
53,700 | 53,750 | 1,957 | 1,878 | 1,936
53,750 | 53,800 | 1,959 | 1,880 | 1,938
53,800 | 53,850 | 1,961 | 1,882 | 1,940
53,850 | 53,900 | 1,963 | 1,884 | 1,942
53,900 | 53,950 | 1,965 | 1,886 | 1,944
53,950 | 54,000 | 1,967 | 1,888 | 1,946
This column must also be used by a qualifying surviving spouse
$60,000+
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
54,000 | Your New York City tax is:
54,000 | 54,050 | 1,969 | 1,890 | 1,948
54,050 | 54,100 | 1,971 | 1,892 | 1,949
54,100 | 54,150 | 1,973 | 1,893 | 1,951
54,150 | 54,200 | 1,975 | 1,895 | 1,953
54,200 | 54,250 | 1,977 | 1,897 | 1,955
54,250 | 54,300 | 1,979 | 1,899 | 1,957
54,300 | 54,350 | 1,981 | 1,901 | 1,959
54,350 | 54,400 | 1,983 | 1,903 | 1,961
54,400 | 54,450 | 1,985 | 1,905 | 1,963
54,450 | 54,500 | 1,986 | 1,907 | 1,965
54,500 | 54,550 | 1,988 | 1,909 | 1,967
54,550 | 54,600 | 1,990 | 1,911 | 1,969
54,600 | 54,650 | 1,992 | 1,913 | 1,970
54,650 | 54,700 | 1,994 | 1,914 | 1,972
54,700 | 54,750 | 1,996 | 1,916 | 1,974
54,750 | 54,800 | 1,998 | 1,918 | 1,976
54,800 | 54,850 | 2,000 | 1,920 | 1,978
54,850 | 54,900 | 2,002 | 1,922 | 1,980
54,900 | 54,950 | 2,004 | 1,924 | 1,982
54,950 | 55,000 | 2,006 | 1,926 | 1,984
55,000 | Your New York City tax is:
55,000 | 55,050 | 2,008 | 1,928 | 1,986
55,050 | 55,100 | 2,010 | 1,930 | 1,988
55,100 | 55,150 | 2,012 | 1,932 | 1,990
55,150 | 55,200 | 2,014 | 1,934 | 1,991
55,200 | 55,250 | 2,016 | 1,935 | 1,993
55,250 | 55,300 | 2,017 | 1,937 | 1,995
55,300 | 55,350 | 2,019 | 1,939 | 1,997
55,350 | 55,400 | 2,021 | 1,941 | 1,999
55,400 | 55,450 | 2,023 | 1,943 | 2,001
55,450 | 55,500 | 2,025 | 1,945 | 2,003
55,500 | 55,550 | 2,027 | 1,947 | 2,005
55,550 | 55,600 | 2,029 | 1,949 | 2,007
55,600 | 55,650 | 2,031 | 1,951 | 2,009
55,650 | 55,700 | 2,033 | 1,953 | 2,011
55,700 | 55,750 | 2,035 | 1,955 | 2,012
55,750 | 55,800 | 2,037 | 1,956 | 2,014
55,800 | 55,850 | 2,039 | 1,958 | 2,016
55,850 | 55,900 | 2,041 | 1,960 | 2,018
55,900 | 55,950 | 2,043 | 1,962 | 2,020
55,950 | 56,000 | 2,045 | 1,964 | 2,022
56,000 | Your New York City tax is:
56,000 | 56,050 | 2,047 | 1,966 | 2,024
56,050 | 56,100 | 2,048 | 1,968 | 2,026
56,100 | 56,150 | 2,050 | 1,970 | 2,028
56,150 | 56,200 | 2,052 | 1,972 | 2,030
56,200 | 56,250 | 2,054 | 1,974 | 2,032
56,250 | 56,300 | 2,056 | 1,976 | 2,033
56,300 | 56,350 | 2,058 | 1,978 | 2,035
56,350 | 56,400 | 2,060 | 1,979 | 2,037
56,400 | 56,450 | 2,062 | 1,981 | 2,039
56,450 | 56,500 | 2,064 | 1,983 | 2,041
56,500 | 56,550 | 2,066 | 1,985 | 2,043
56,550 | 56,600 | 2,068 | 1,987 | 2,045
56,600 | 56,650 | 2,070 | 1,989 | 2,047
56,650 | 56,700 | 2,072 | 1,991 | 2,049
56,700 | 56,750 | 2,074 | 1,993 | 2,051
56,750 | 56,800 | 2,076 | 1,995 | 2,053
56,800 | 56,850 | 2,078 | 1,997 | 2,054
56,850 | 56,900 | 2,079 | 1,999 | 2,056
56,900 | 56,950 | 2,081 | 2,000 | 2,058
56,950 | 57,000 | 2,083 | 2,002 | 2,060
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
57,000 | Your New York City tax is:
57,000 | 57,050 | 2,085 | 2,004 | 2,062
57,050 | 57,100 | 2,087 | 2,006 | 2,064
57,100 | 57,150 | 2,089 | 2,008 | 2,066
57,150 | 57,200 | 2,091 | 2,010 | 2,068
57,200 | 57,250 | 2,093 | 2,012 | 2,070
57,250 | 57,300 | 2,095 | 2,014 | 2,072
57,300 | 57,350 | 2,097 | 2,016 | 2,074
57,350 | 57,400 | 2,099 | 2,018 | 2,075
57,400 | 57,450 | 2,101 | 2,020 | 2,077
57,450 | 57,500 | 2,103 | 2,021 | 2,079
57,500 | 57,550 | 2,105 | 2,023 | 2,081
57,550 | 57,600 | 2,107 | 2,025 | 2,083
57,600 | 57,650 | 2,109 | 2,027 | 2,085
57,650 | 57,700 | 2,110 | 2,029 | 2,087
57,700 | 57,750 | 2,112 | 2,031 | 2,089
57,750 | 57,800 | 2,114 | 2,033 | 2,091
57,800 | 57,850 | 2,116 | 2,035 | 2,093
57,850 | 57,900 | 2,118 | 2,037 | 2,095
57,900 | 57,950 | 2,120 | 2,039 | 2,096
57,950 | 58,000 | 2,122 | 2,041 | 2,098
58,000 | Your New York City tax is:
58,000 | 58,050 | 2,124 | 2,042 | 2,100
58,050 | 58,100 | 2,126 | 2,044 | 2,102
58,100 | 58,150 | 2,128 | 2,046 | 2,104
58,150 | 58,200 | 2,130 | 2,048 | 2,106
58,200 | 58,250 | 2,132 | 2,050 | 2,108
58,250 | 58,300 | 2,134 | 2,052 | 2,110
58,300 | 58,350 | 2,136 | 2,054 | 2,112
58,350 | 58,400 | 2,138 | 2,056 | 2,114
58,400 | 58,450 | 2,140 | 2,058 | 2,116
58,450 | 58,500 | 2,141 | 2,060 | 2,117
58,500 | 58,550 | 2,143 | 2,062 | 2,119
58,550 | 58,600 | 2,145 | 2,063 | 2,121
58,600 | 58,650 | 2,147 | 2,065 | 2,123
58,650 | 58,700 | 2,149 | 2,067 | 2,125
58,700 | 58,750 | 2,151 | 2,069 | 2,127
58,750 | 58,800 | 2,153 | 2,071 | 2,129
58,800 | 58,850 | 2,155 | 2,073 | 2,131
58,850 | 58,900 | 2,157 | 2,075 | 2,133
58,900 | 58,950 | 2,159 | 2,077 | 2,135
58,950 | 59,000 | 2,161 | 2,079 | 2,137
59,000 | Your New York City tax is:
59,000 | 59,050 | 2,163 | 2,081 | 2,138
59,050 | 59,100 | 2,165 | 2,083 | 2,140
59,100 | 59,150 | 2,167 | 2,084 | 2,142
59,150 | 59,200 | 2,169 | 2,086 | 2,144
59,200 | 59,250 | 2,171 | 2,088 | 2,146
59,250 | 59,300 | 2,172 | 2,090 | 2,148
59,300 | 59,350 | 2,174 | 2,092 | 2,150
59,350 | 59,400 | 2,176 | 2,094 | 2,152
59,400 | 59,450 | 2,178 | 2,096 | 2,154
59,450 | 59,500 | 2,180 | 2,098 | 2,156
59,500 | 59,550 | 2,182 | 2,100 | 2,158
59,550 | 59,600 | 2,184 | 2,102 | 2,159
59,600 | 59,650 | 2,186 | 2,104 | 2,161
59,650 | 59,700 | 2,188 | 2,105 | 2,163
59,700 | 59,750 | 2,190 | 2,107 | 2,165
59,750 | 59,800 | 2,192 | 2,109 | 2,167
59,800 | 59,850 | 2,194 | 2,111 | 2,169
59,850 | 59,900 | 2,196 | 2,113 | 2,171
59,900 | 59,950 | 2,198 | 2,115 | 2,173
59,950 | 60,000 | 2,200 | 2,117 | 2,175
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
60,000 | Your New York City tax is:
60,000 | 60,050 | 2,202 | 2,119 | 2,177
60,050 | 60,100 | 2,204 | 2,121 | 2,179
60,100 | 60,150 | 2,205 | 2,123 | 2,181
60,150 | 60,200 | 2,207 | 2,125 | 2,183
60,200 | 60,250 | 2,209 | 2,126 | 2,185
60,250 | 60,300 | 2,211 | 2,128 | 2,187
60,300 | 60,350 | 2,213 | 2,130 | 2,189
60,350 | 60,400 | 2,215 | 2,132 | 2,191
60,400 | 60,450 | 2,217 | 2,134 | 2,192
60,450 | 60,500 | 2,219 | 2,136 | 2,194
60,500 | 60,550 | 2,221 | 2,138 | 2,196
60,550 | 60,600 | 2,223 | 2,140 | 2,198
60,600 | 60,650 | 2,225 | 2,142 | 2,200
60,650 | 60,700 | 2,227 | 2,144 | 2,202
60,700 | 60,750 | 2,229 | 2,146 | 2,204
60,750 | 60,800 | 2,231 | 2,147 | 2,206
60,800 | 60,850 | 2,233 | 2,149 | 2,208
60,850 | 60,900 | 2,235 | 2,151 | 2,210
60,900 | 60,950 | 2,236 | 2,153 | 2,212
60,950 | 61,000 | 2,238 | 2,155 | 2,214
61,000 | Your New York City tax is:
61,000 | 61,050 | 2,240 | 2,157 | 2,216
61,050 | 61,100 | 2,242 | 2,159 | 2,218
61,100 | 61,150 | 2,244 | 2,161 | 2,220
61,150 | 61,200 | 2,246 | 2,163 | 2,222
61,200 | 61,250 | 2,248 | 2,165 | 2,223
61,250 | 61,300 | 2,250 | 2,167 | 2,225
61,300 | 61,350 | 2,252 | 2,168 | 2,227
61,350 | 61,400 | 2,254 | 2,170 | 2,229
61,400 | 61,450 | 2,256 | 2,172 | 2,231
61,450 | 61,500 | 2,258 | 2,174 | 2,233
61,500 | 61,550 | 2,260 | 2,176 | 2,235
61,550 | 61,600 | 2,262 | 2,178 | 2,237
61,600 | 61,650 | 2,264 | 2,180 | 2,239
61,650 | 61,700 | 2,266 | 2,182 | 2,241
61,700 | 61,750 | 2,267 | 2,184 | 2,243
61,750 | 61,800 | 2,269 | 2,186 | 2,245
61,800 | 61,850 | 2,271 | 2,188 | 2,247
61,850 | 61,900 | 2,273 | 2,189 | 2,249
61,900 | 61,950 | 2,275 | 2,191 | 2,251
61,950 | 62,000 | 2,277 | 2,193 | 2,253
This column must also be used by a qualifying surviving spouse
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
62,000 | Your New York City tax is:
62,000 | 62,050 | 2,279 | 2,195 | 2,254
62,050 | 62,100 | 2,281 | 2,197 | 2,256
62,100 | 62,150 | 2,283 | 2,199 | 2,258
62,150 | 62,200 | 2,285 | 2,201 | 2,260
62,200 | 62,250 | 2,287 | 2,203 | 2,262
62,250 | 62,300 | 2,289 | 2,205 | 2,264
62,300 | 62,350 | 2,291 | 2,207 | 2,266
62,350 | 62,400 | 2,293 | 2,209 | 2,268
62,400 | 62,450 | 2,295 | 2,210 | 2,270
62,450 | 62,500 | 2,297 | 2,212 | 2,272
62,500 | 62,550 | 2,298 | 2,214 | 2,274
62,550 | 62,600 | 2,300 | 2,216 | 2,276
62,600 | 62,650 | 2,302 | 2,218 | 2,278
62,650 | 62,700 | 2,304 | 2,220 | 2,280
62,700 | 62,750 | 2,306 | 2,222 | 2,282
62,750 | 62,800 | 2,308 | 2,224 | 2,284
62,800 | 62,850 | 2,310 | 2,226 | 2,285
62,850 | 62,900 | 2,312 | 2,228 | 2,287
62,900 | 62,950 | 2,314 | 2,230 | 2,289
62,950 | 63,000 | 2,316 | 2,231 | 2,291
63,000 | Your New York City tax is:
63,000 | 63,050 | 2,318 | 2,233 | 2,293
63,050 | 63,100 | 2,320 | 2,235 | 2,295
63,100 | 63,150 | 2,322 | 2,237 | 2,297
63,150 | 63,200 | 2,324 | 2,239 | 2,299
63,200 | 63,250 | 2,326 | 2,241 | 2,301
63,250 | 63,300 | 2,328 | 2,243 | 2,303
63,300 | 63,350 | 2,329 | 2,245 | 2,305
63,350 | 63,400 | 2,331 | 2,247 | 2,307
63,400 | 63,450 | 2,333 | 2,249 | 2,309
63,450 | 63,500 | 2,335 | 2,251 | 2,311
63,500 | 63,550 | 2,337 | 2,252 | 2,313
63,550 | 63,600 | 2,339 | 2,254 | 2,315
63,600 | 63,650 | 2,341 | 2,256 | 2,317
63,650 | 63,700 | 2,343 | 2,258 | 2,318
63,700 | 63,750 | 2,345 | 2,260 | 2,320
63,750 | 63,800 | 2,347 | 2,262 | 2,322
63,800 | 63,850 | 2,349 | 2,264 | 2,324
63,850 | 63,900 | 2,351 | 2,266 | 2,326
63,900 | 63,950 | 2,353 | 2,268 | 2,328
63,950 | 64,000 | 2,355 | 2,270 | 2,330
If yourtaxableincome is: | And you are:
Atleast | But lessthan | SingleorMarriedfilingseparately | Married filing jointly | Headof a household
64,000 | Your New York City tax is:
64,000 | 64,050 | 2,357 | 2,272 | 2,332
64,050 | 64,100 | 2,359 | 2,273 | 2,334
64,100 | 64,150 | 2,360 | 2,275 | 2,336
64,150 | 64,200 | 2,362 | 2,277 | 2,338
64,200 | 64,250 | 2,364 | 2,279 | 2,340
64,250 | 64,300 | 2,366 | 2,281 | 2,342
64,300 | 64,350 | 2,368 | 2,283 | 2,344
64,350 | 64,400 | 2,370 | 2,285 | 2,346
64,400 | 64,450 | 2,372 | 2,287 | 2,348
64,450 | 64,500 | 2,374 | 2,289 | 2,349
64,500 | 64,550 | 2,376 | 2,291 | 2,351
64,550 | 64,600 | 2,378 | 2,293 | 2,353
64,600 | 64,650 | 2,380 | 2,294 | 2,355
64,650 | 64,700 | 2,382 | 2,296 | 2,357
64,700 | 64,750 | 2,384 | 2,298 | 2,359
64,750 | 64,800 | 2,386 | 2,300 | 2,361
64,800 | 64,850 | 2,388 | 2,302 | 2,363
64,850 | 64,900 | 2,390 | 2,304 | 2,365
64,900 | 64,950 | 2,391 | 2,306 | 2,367
64,950 | 65,000 | 2,393 | 2,308 | 2,369
65,000 or more:
$65,000 or more—compute your New York City tax usingNew York City tax rate schedule.
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5388 | https://www.webqc.org/molecular-weight-of-KClO.html | KClO (Potassium hypochlorite) molar mass
Printed from
Molar Mass, Molecular Weight and Elemental Composition Calculator
Enter a chemical formula to calculate its molar mass and elemental composition:
Did you mean KClO3?
Molar mass of KClO (Potassium hypochlorite) is 90.5507 g/mol
Convert between KClO weight and moles
| Compound | Moles | Weight, g |
---
| KClO | | | | |
Elemental composition of KClO
| Element | Symbol | Atomic weight | Atoms | Mass percent |
--- ---
| Potassium | K | 39.0983 | 1 | 43.1784 |
| Chlorine | Cl | 35.453 | 1 | 39.1527 |
| Oxygen | O | 15.9994 | 1 | 17.6690 |
| Computing molar mass step by step |
| First, compute the number of each atom in KClO: K: 1, Cl: 1, O: 1 Then, lookup atomic weights for each element in periodic table: K: 39.0983, Cl: 35.453, O: 15.9994 Now, compute the sum of products of number of atoms to the atomic weight: Molar mass (KClO) = ∑ Count i Weight i = Count(K) Weight(K) + Count(Cl) Weight(Cl) + Count(O) Weight(O) = 1 39.0983 + 1 35.453 + 1 15.9994 = 90.5507 g/mol |
| Mass Percent Composition | Atomic Percent Composition |
--- |
| K Potassium (43.18%) Cl Chlorine (39.15%) O Oxygen (17.67%) | K Potassium (33.33%) Cl Chlorine (33.33%) O Oxygen (33.33%) |
| Mass Percent Composition |
| K Potassium (43.18%) Cl Chlorine (39.15%) O Oxygen (17.67%) |
| Atomic Percent Composition |
| K Potassium (33.33%) Cl Chlorine (33.33%) O Oxygen (33.33%) |
| Chemical structure |
| |
| Lewis structure |
| 3D molecular structure |
| Appearance |
| Potassium hypochlorite has a light grey color and a strong chlorine smell. |
Related compounds
| Formula | Compound name |
--- |
| KClO 3 | Potassium chlorate |
| KClO 4 | Potassium perchlorate |
| KClO 2 | Potassium chlorite |
| Related |
| Oxidation state calculator |
| Compound properties |
Computing molar mass (molar weight)
To calculate molar mass of a chemical compound enter its formula and click 'Compute'. In chemical formula you may use:
Any chemical element. Capitalize the first letter in chemical symbol and use lower case for the remaining letters: Ca, Fe, Mg, Mn, S, O, H, C, N, Na, K, Cl, Al.
Functional groups: D, T, Ph, Me, Et, Bu, AcAc, For, Tos, Bz, TMS, tBu, Bzl, Bn, Dmg
parenthesis () or brackets [].
Common compound names.
Examples of molar mass computations: NaCl, Ca(OH)2, K4[Fe(CN)6], CuSO45H2O, nitric acid, potassium permanganate, ethanol, fructose, caffeine, water.
Molar mass calculator also displays common compound name, Hill formula, elemental composition, mass percent composition, atomic percent compositions and allows to convert from weight to number of moles and vice versa.
Computing molecular weight (molecular mass)
To calculate molecular weight of a chemical compound enter it's formula, specify its isotope mass number after each element in square brackets.
Examples of molecular weight computations: CO2, SO2.
Definitions
Molecular mass (molecular weight) is the mass of one molecule of a substance and is expressed in the unified atomic mass units (u). (1 u is equal to 1/12 the mass of one atom of carbon-12)
Molar mass (molar weight) is the mass of one mole of a substance and is expressed in g/mol.
Mole is a standard scientific unit for measuring large quantities of very small entities such as atoms and molecules. One mole contains exactly 6.022 ×10 23 particles (Avogadro's number)
Steps to calculate molar mass
Identify the compound: write down the chemical formula of the compound. For example, water is H 2 O, meaning it contains two hydrogen atoms and one oxygen atom.
Find atomic masses: look up the atomic masses of each element present in the compound. The atomic mass is usually found on the periodic table and is given in atomic mass units (amu).
Calculate molar mass of each element: multiply the atomic mass of each element by the number of atoms of that element in the compound.
Add them together: add the results from step 3 to get the total molar mass of the compound.
Example: calculating molar mass
Let's calculate the molar mass of carbon dioxide (CO 2):
Carbon (C) has an atomic mass of about 12.01 amu.
Oxygen (O) has an atomic mass of about 16.00 amu.
CO 2 has one carbon atom and two oxygen atoms.
The molar mass of carbon dioxide is 12.01 + (2 × 16.00) = 44.01 g/mol.
Lesson on computing molar mass
Practice what you learned:
Practice calculating molar mass
Weights of atoms and isotopes are from NIST article.
Related: Molecular weights of amino acids
molecular weights calculated today
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5389 | https://learn.microsoft.com/en-us/dotnet/csharp/programming-guide/classes-and-structs/restricting-accessor-accessibility | Restricting Accessor Accessibility - C# | Microsoft Learn
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Restricting Accessor Accessibility (C# Programming Guide)
10/30/2024
Feedback
In this article
Restrictions on Access Modifiers on Accessors
Access Modifiers on Overriding Accessors
Implementing Interfaces
Accessor Accessibility Domain
Example
Comments
See also
Show 3 more
The get and set portions of a property or indexer are called accessors. By default these accessors have the same visibility or access level of the property or indexer to which they belong. For more information, see accessibility levels. However, it's sometimes useful to restrict access to one of these accessors. Typically, you restrict the accessibility of the set accessor, while keeping the get accessor publicly accessible. For example:
C#Copy
```csharp
private string _name = "Hello";
public string Name
{
get
{
return _name;
}
protected set
{
_name = value;
}
}
```
In this example, a property called Name defines a get and set accessor. The get accessor receives the accessibility level of the property itself, public in this case, while the set accessor is explicitly restricted by applying the protected access modifier to the accessor itself.
Note
The examples in this article don't use automatically implemented properties. Automatically implemented properties provide a concise syntax for declaring properties when a custom backing field isn't required.
Restrictions on Access Modifiers on Accessors
Using the accessor modifiers on properties or indexers is subject to these conditions:
You can't use accessor modifiers on an interface or an explicit interface member implementation.
You can use accessor modifiers only if the property or indexer has both set and get accessors. In this case, the modifier is permitted on only one of the two accessors.
If the property or indexer has an override modifier, the accessor modifier must match the accessor of the overridden accessor, if any.
The accessibility level on the accessor must be more restrictive than the accessibility level on the property or indexer itself.
Access Modifiers on Overriding Accessors
When you override a property or indexer, the overridden accessors must be accessible to the overriding code. Also, the accessibility of both the property/indexer and its accessors must match the corresponding overridden property/indexer and its accessors. For example:
C#Copy
```csharp
public class Parent
{
public virtual int TestProperty
{
// Notice the accessor accessibility level.
protected set { }
// No access modifier is used here.
get { return 0; }
}
}
public class Kid : Parent
{
public override int TestProperty
{
// Use the same accessibility level as in the overridden accessor.
protected set { }
// Cannot use access modifier here.
get { return 0; }
}
}
```
Implementing Interfaces
When you use an accessor to implement an interface, the accessor may not have an access modifier. However, if you implement the interface using one accessor, such as get, the other accessor can have an access modifier, as in the following example:
C#Copy
```csharp
public interface ISomeInterface
{
int TestProperty
{
// No access modifier allowed here
// because this is an interface.
get;
}
}
public class TestClass : ISomeInterface
{
public int TestProperty
{
// Cannot use access modifier here because
// this is an interface implementation.
get { return 10; }
// Interface property does not have set accessor,
// so access modifier is allowed.
protected set { }
}
}
```
Accessor Accessibility Domain
If you use an access modifier on the accessor, the accessibility domain of the accessor is determined by this modifier.
If you didn't use an access modifier on the accessor, the accessibility domain of the accessor is determined by the accessibility level of the property or indexer.
Example
The following example contains three classes, BaseClass, DerivedClass, and MainClass. There are two properties on the BaseClass, Name and Id on both classes. The example demonstrates how the property Id on DerivedClass can be hidden by the property Id on BaseClass when you use a restrictive access modifier such as protected or private. Therefore, when you assign values to this property, the property on the BaseClass class is called instead. Replacing the access modifier by public will make the property accessible.
The example also demonstrates that a restrictive access modifier, such as private or protected, on the set accessor of the Name property in DerivedClass prevents access to the accessor in the derived class. It generates an error when you assign to it, or accesses the base class property of the same name, if it's accessible.
C#Copy
```csharp
public class BaseClass
{
private string _name = "Name-BaseClass";
private string _id = "ID-BaseClass";
public string Name
{
get { return _name; }
set { }
}
public string Id
{
get { return _id; }
set { }
}
}
public class DerivedClass : BaseClass
{
private string _name = "Name-DerivedClass";
private string _id = "ID-DerivedClass";
new public string Name
{
get
{
return _name;
}
// Using "protected" would make the set accessor not accessible.
set
{
_name = value;
}
}
// Using private on the following property hides it in the Main Class.
// Any assignment to the property will use Id in BaseClass.
new private string Id
{
get
{
return _id;
}
set
{
_id = value;
}
}
}
class MainClass
{
static void Main()
{
BaseClass b1 = new BaseClass();
DerivedClass d1 = new DerivedClass();
b1.Name = "Mary";
d1.Name = "John";
b1.Id = "Mary123";
d1.Id = "John123"; // The BaseClass.Id property is called.
System.Console.WriteLine("Base: {0}, {1}", b1.Name, b1.Id);
System.Console.WriteLine("Derived: {0}, {1}", d1.Name, d1.Id);
// Keep the console window open in debug mode.
System.Console.WriteLine("Press any key to exit.");
System.Console.ReadKey();
}
}
/ Output:
Base: Name-BaseClass, ID-BaseClass
Derived: John, ID-BaseClass
/
```
Comments
Notice that if you replace the declaration new private string Id by new public string Id, you get the output:
Name and ID in the base class: Name-BaseClass, ID-BaseClass``Name and ID in the derived class: John, John123
See also
Properties
Indexers
Access Modifiers
Init only properties
Required properties
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In this article
Restrictions on Access Modifiers on Accessors
Access Modifiers on Overriding Accessors
Implementing Interfaces
Accessor Accessibility Domain
Example
Comments
See also
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5390 | https://www.quora.com/Why-does-k-k+1-2k+1-+6-k+1-2-k+1-big-k-2k+1-+6-k+1-big | Something went wrong. Wait a moment and try again.
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To understand why the equation
k(k+1)(2k+1)+6(k+1)2=(k+1)(k(2k+1)+6(k+1))
is true, we can start by factoring the left-hand side and simplifying it to see if it matches the right-hand side.
Factor out (k+1) from the left-hand side:
The left-hand side can be rewritten as:
k(k+1)(2k+1)+6(k+1)2=(k+1)(k(2k+1)+6(k+1))
Here, we factor out (k+1) from both terms:
The first term, k(k+1)(2k+1), clearly has (k+1) as a factor.
The second term, 6(k+1)2, can also be expressed as 6(k+1)(k+1), which also has (k+1) as a factor.
Combine the terms inside the parentheses:
To understand why the equation
k(k+1)(2k+1)+6(k+1)2=(k+1)(k(2k+1)+6(k+1))
is true, we can start by factoring the left-hand side and simplifying it to see if it matches the right-hand side.
Factor out (k+1) from the left-hand side:
The left-hand side can be rewritten as:
k(k+1)(2k+1)+6(k+1)2=(k+1)(k(2k+1)+6(k+1))
Here, we factor out (k+1) from both terms:
The first term, k(k+1)(2k+1), clearly has (k+1) as a factor.
The second term, 6(k+1)2, can also be expressed as 6(k+1)(k+1), which also has (k+1) as a factor.
Combine the terms inside the parentheses:
Now, we can express the left-hand side as:
(k+1)(k(2k+1)+6(k+1))
This matches the right-hand side exactly.
Simplify the expression inside the parentheses:
To verify the equality, you can expand the right-hand side:
k(2k+1)+6(k+1)=2k2+k+6k+6=2k2+7k+6
Thus, both sides simplify to:
(k+1)(2k2+7k+6)
This shows that both sides of the equation are equal, confirming that the original equation holds true.
So, the equation is valid due to the common factor (k+1) being factored out from both terms on the left-hand side, leading to the same expression as on the right-hand side.
Related questions
How would one get
(
k
)
(
k
+
1
)
(
2
k
+
1
)
6
+
(
k
+
1
)
2
to equal
(
k
+
1
)
(
k
+
2
)
(
2
k
+
3
)
6
without manipulating the right side?
What is k(k+1) (2k+1) /6+(k+1) ^2?
Why does
i
2
=
−
1
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Why does
x
y
+
x
y
2
+
x
y
3
+
⋯
=
x
y
−
1
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Why does
−
1
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≠
√
i
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?
Marco Wong
Student of life first, student of math in UBC next
·
8y
Try not to look at each part of the equation by their respective positions. Instead try see what is happening.
OK, so originally the equation is k(k+1)(2k+1) + 6(k+1)^2. What factors are common across both the first and second part? It’s (k+1). It’s the most basic part of factoring.
Lets take us (k+1) from both sides. We are left with k(2k+1) + 6(k+1). But this is not the same as the original equation, since we divided (took out) k+1. So we have to multiply it back in. k(2k+1) + 6(k+1) times (k+1). Which is, (k+1)(k(2k+1) + 6(k+1)).
It’s easy to try and just match each position of the left hand s
Try not to look at each part of the equation by their respective positions. Instead try see what is happening.
OK, so originally the equation is k(k+1)(2k+1) + 6(k+1)^2. What factors are common across both the first and second part? It’s (k+1). It’s the most basic part of factoring.
Lets take us (k+1) from both sides. We are left with k(2k+1) + 6(k+1). But this is not the same as the original equation, since we divided (took out) k+1. So we have to multiply it back in. k(2k+1) + 6(k+1) times (k+1). Which is, (k+1)(k(2k+1) + 6(k+1)).
It’s easy to try and just match each position of the left hand side and right hand side, and sometimes it works. But, in this case, it requires more than that.
Hope this helps!
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Colin Finger
Took higher maths in high school and got quite good marks :)
·
8y
What I did was resolve both sides and the result is 2k³ + 9k² + 13k + 6, which proves that this statement is true.
I’m afraid I find it hard to explain the reason for this in words, but I’ll at least try to show how both sides are equal. Also if I may ask, where does this equasion come from? Just it might help me to explain the equasion or the reason for it a little better :)
—
Left side: (k)(k + 1)(2k + 1) + 6(k + 1)² = (k² + k)(2k + 1) + 6(k² + 2k + 1) = 2k³ + k² + 2k² + k + 6k² + 12k + 6 = 2k³ + 9k² + 13k + 6
—
Right side: (k + 1)((k)(2k + 1) + 6(k + 1)) = (k + 1)(2k² + k + 6(k + 1)) = (k +
What I did was resolve both sides and the result is 2k³ + 9k² + 13k + 6, which proves that this statement is true.
I’m afraid I find it hard to explain the reason for this in words, but I’ll at least try to show how both sides are equal. Also if I may ask, where does this equasion come from? Just it might help me to explain the equasion or the reason for it a little better :)
—
Left side: (k)(k + 1)(2k + 1) + 6(k + 1)² = (k² + k)(2k + 1) + 6(k² + 2k + 1) = 2k³ + k² + 2k² + k + 6k² + 12k + 6 = 2k³ + 9k² + 13k + 6
—
Right side: (k + 1)((k)(2k + 1) + 6(k + 1)) = (k + 1)(2k² + k + 6(k + 1)) = (k + 1)(2k² + 7k + 6) = 2k³ + 7k² + 6k + 2k² + 7k + 6 = 2k³ + 9k² + 13k + 6
I hope this helps you :)
Padmavathy Venkatesan
Works at Amazon India (amazon.in)
·
8y
Originally Answered: Why does (k) (k+1) (2k+1) +6(k+1) ^2 = (k+1) ((k) (2k+1) +6(k+1))?
·
We have ,
k(k+1)(2k+1)+6(k+1)^2
We can group it as follows
{k(k+1)(2k+1)}+6(k+1)^2
Now since (k+1) is common, we can take it out
(k+1){k(2k+1)+6(k+1)}
Therefore,
k(k+1)(2k+1)+6(k+1)^2=(k+1){k(2k+1)+6(k+1)}
This is because of the distributive property:
a(b+c)=ab+ac
Related questions
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(
√
2
√
2
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√
2
=
2
?
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(
a
+
b
)
2
=
a
2
+
b
2
+
2
a
b
?
Why does
tan
π
8
=
√
2
−
1
?
Why does 2^k+2^k=2^(k+1)?
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cos
2
x
+
sin
2
x
=
1
?
Weon A Domicilio
Knows Spanish
·
Wed
You could fully develop both sides, but it’s just a simple factorization.
k(k+1)(2k+1)+6(k+1)2
=k(k+1)(2k+1)+6(k+1)(k+1)
=(1+k)[k(2k+1)+6(k+1)]
I guess you were studying something related to the sum of natural squares.
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Krishna Raj
Lives in Bengaluru, Karnataka, India
·
Author has 136 answers and 356.8K answer views
·
Updated 7y
Related
How can
k
2
+
2
k
+
1
=
(
k
+
1
)
2
?
Here’s another way of looking at this. Take
a
=
k
and
b
=
1
.
The expressions written inside the rectangles are their corresponding area and the expressions written outside the rectangles are the corresponding lengths of the curly braces pointing towards.
Here’s another way of looking at this. Take
a
=
k
and
b
=
1
.
The expressions written inside the rectangles are their corresponding area and the expressions written outside the rectangles are the corresponding lengths of the curly braces pointing towards.
Kyle Weatherly
Still in High school, so I have no credentials yet, but math is one hobby I have
·
Author has 110 answers and 166.9K answer views
·
8y
This relies on some quick algebra to solve. (k)(k+1)(2k+1)+6(k+1)^2 has a (k+1) in both of its expressions (“expressions” being multiplication here), so you can factor out the (k+1) and quickly get (k+1)((k)(2k+1)+6(k+1)).
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Thiru Mangai
Studied at Pondicherry Engineering College
·
8y
(k)(k+1)(2k+1) + 6(k+1)^2 = k(k+1)(2k+1) + 6(k+1)(k+1)
Taking (k+1) as common since it is a common factor for both,
= (k+1)((k)(2k+1) + 6(k+1))
Akashdeep Deb
Summer Analyst at Goldman Sachs (company)
·
Author has 63 answers and 422.8K answer views
·
8y
Originally Answered: Why does (k) (k+1) (2k+1) +6(k+1) ^2 = (k+1) ((k) (2k+1) +6(k+1))?
·
k(k+1)(2k+1)+6(k+1)2=(k+1)[k(2k+1)+6(k+1)]
This is the distributive property. (k+1) here is a common factor.
ab+bc=b(a+c)
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Douglas Magowan
Private Pilot
·
Author has 1.2K answers and 1.3M answer views
·
8y
Originally Answered: Why does (k) (k+1) (2k+1) +6(k+1) ^2 = (k+1) ((k) (2k+1) +6(k+1))?
·
What have I done so far. I have swapped the (k) and the (k+1) in the first term to show how to get it to the left hand side. And, I have broken out [math] (k+1)^2[/math] into (k+1)(k+1), because that is what squaring means.
Ever term has a (k+1) in it. I can factor that out.
[math] (k+1)\big((k)(2k+1) + 6(k+1)\big)[/math]
Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006)
·
Author has 8.5K answers and 21.1M answer views
·
Updated 5y
Related
How do you prove
[math]\sum_{k=1}^{\infty}\Big( \frac{2k+1}{k+1} - 2k (\ln(k+1) - \ln{k})\Big) = \ln(2 \pi) - 1 - \gamma[/math]
(calculus, integration, sequences, series, definite integrals, math)?
We want to prove that
[math]\displaystyle \sum_{k=1}^{\infty} \Big(\frac{2k+1}{k+1} - 2k (\ln(k+1) - \ln{k})\Big) = \ln(2 \pi) - 1 - \gamma. \tag{}[/math]
To this end, observe that
[math]k (\ln(k+1) - \ln{k}) = (k+1) \ln(k+1) - k \ln{k} - \ln(k+1). \tag{}[/math]
This allows us for a partial telescoping series:
[math]\begin{align} \displaystyle \sum_{k=1}^n k (\ln(k+1) - \ln{k}) &= (n+1)\ln(n+1) - \sum_{k=1}^n \ln(k+1)\ &= \displaystyle (n+1)\ln(n+1) - \ln((n+1)!). \end{align} \tag{}[/math]
Applying Stirling’s formula
[math]\displaystyle \ln(n!) = n\ln{n} - n + \frac{1}{2} \ln(2\pi n) + O\Big(\frac{1}{n}\Big) \tag{}[/math]
to the series above, w
We want to prove that
[math]\displaystyle \sum_{k=1}^{\infty} \Big(\frac{2k+1}{k+1} - 2k (\ln(k+1) - \ln{k})\Big) = \ln(2 \pi) - 1 - \gamma. \tag{}[/math]
To this end, observe that
[math]k (\ln(k+1) - \ln{k}) = (k+1) \ln(k+1) - k \ln{k} - \ln(k+1). \tag{}[/math]
This allows us for a partial telescoping series:
[math]\begin{align} \displaystyle \sum_{k=1}^n k (\ln(k+1) - \ln{k}) &= (n+1)\ln(n+1) - \sum_{k=1}^n \ln(k+1)\ &= \displaystyle (n+1)\ln(n+1) - \ln((n+1)!). \end{align} \tag{}[/math]
Applying Stirling’s formula
[math]\displaystyle \ln(n!) = n\ln{n} - n + \frac{1}{2} \ln(2\pi n) + O\Big(\frac{1}{n}\Big) \tag{}[/math]
to the series above, we then obtain
[math]\displaystyle \sum_{k=1}^n k (\ln(k+1) - \ln{k}) = n+1 - \frac{1}{2} \ln(2\pi(n+1)) + O\Big(\frac{1}{n}\Big). \tag{}[/math]
Next, we see that
[math]\begin{align} \displaystyle \sum_{k=1}^n \frac{2k+1}{k+1} &= \sum_{k=1}^n \Big(2 - \frac{1}{k+1}\Big)\ &= \displaystyle 2n - \sum_{k=1}^n \frac{1}{k+1}\ &= \displaystyle 2n + 1 - \sum_{k=1}^{n+1} \frac{1}{k}, \text{ via re-indexing}. \end{align} \tag{}[/math]
Then, applying the classic estimate
[math]\displaystyle H_n = \sum_{k=1}^n \frac{1}{k} = \ln{n} + \gamma + O\Big(\frac{1}{n}\Big), \tag{}[/math]
we obtain
[math]\displaystyle \sum_{k=1}^n \frac{2k+1}{k+1} = 2n + 1 - \ln(n+1) - \gamma + O\Big(\frac{1}{n}\Big). \tag{} [/math]
Putting this all together, we obtain
[math]\displaystyle \sum_{k=1}^n \Big(\frac{2k+1}{k+1} - 2k (\ln(k+1) - \ln{k})\Big) = \ln\Big(\frac{2\pi(n+1)}{n+1}\Big) - 1 - \gamma + O\Big(\frac{1}{n}\Big). \tag{} [/math]
Letting [math]n \to \infty[/math], we conclude that
[math]\displaystyle \sum_{k=1}^{\infty} \Big(\frac{2k+1}{k+1} - 2k (\ln(k+1) - \ln{k})\Big) = \ln(2\pi) - 1 - \gamma. \tag{}[/math]
Umang Mittal
Application Development Analyst at Accenture (company) (2015–present)
·
8y
math(k+1)(2k+1) + 6(k+1)^2=(k+1((k)(2k+1)+6(k+1))[/math]
[math] (k)(k+1)(2k+1) + 6(k+1)(k+1)[/math]
essentially i have just taken (k+1) common on the LHS to make it equal to the RHS
Andrew Hendrickson
PhD candidate in mathematics
·
Author has 193 answers and 620.3K answer views
·
9y
Related
Why does
math\times(\frac{c}{d}) = \frac{(a\times b)}{(c\times d)}[/math]
?
I like David Joyce's answer. It's a nice concrete way to prove it. Here is a visual way to think about it. Suppose you have a rectangle of length a and width c. Now chop that rectangle along the length into b equal pieces, so that you get b rectangles of length (a/b) and width c. Next chop each of those rectangles along the width into d equal pieces. That gives you bd rectangles of length (a/b) and width (c/d). What is the area of each of the remaining rectangles?
On the one hand, we can calculate the area directly by multiplying the length times the width: (a/b)(c/d).
On the other ha
I like David Joyce's answer. It's a nice concrete way to prove it. Here is a visual way to think about it. Suppose you have a rectangle of length a and width c. Now chop that rectangle along the length into b equal pieces, so that you get b rectangles of length (a/b) and width c. Next chop each of those rectangles along the width into d equal pieces. That gives you bd rectangles of length (a/b) and width (c/d). What is the area of each of the remaining rectangles?
On the one hand, we can calculate the area directly by multiplying the length times the width: (a/b)(c/d).
On the other hand, we could calculate the area of the original rectangle, ac, and divide it by the number of smaller rectangles, bd, to get an area of (ac)/(bd).
These areas must be equal, so we conclude (a/b)(c/d)=(ac)/(bd). Try drawing a picture with some actual values of a,b,c, and d to see this.
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(
k
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(
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+
(
k
+
1
)
2
to equal
(
k
+
1
)
(
k
+
2
)
(
2
k
+
3
)
6
without manipulating the right side?
What is k(k+1) (2k+1) /6+(k+1) ^2?
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3
+
⋯
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5391 | https://static.bigideasmath.com/protected/content/pe/hs/hscc_alg2_pe_04.pdf | 4 Polynomial Functions 4.1 Graphing Polynomial Functions 4.2 Adding, Subtracting, and Multiplying Polynomials 4.3 Dividing Polynomials 4.4 Factoring Polynomials 4.5 Solving Polynomial Equations 4.6 The Fundamental Theorem of Algebra 4.7 Transformations of Polynomial Functions 4.8 Analyzing Graphs of Polynomial Functions 4.9 Modeling with Polynomial Functions Basketball (p. 178) Ruins of Caesarea (p. 195) Quonset Hut (p. 218) Electric Vehicles (p. 161) Zebra Mussels (p. 203) Zebra Mussels (p. 203) Ruins of Caesarea (p. 195) g y Quonset Hut (p. 21 218) 8) SEE the Big Idea 155 Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Simplifying Algebraic Expressions Example 1 Simplify the expression 9x + 4x. 9x + 4x = (9 + 4)x Distributive Property = 13x Add coefficients.
Example 2 Simplify the expression 2(x + 4) + 3(6 − x).
2(x + 4) + 3(6 − x) = 2(x) + 2(4) + 3(6) + 3(−x) Distributive Property = 2x + 8 + 18 − 3x Multiply.
= 2x − 3x + 8 + 18 Group like terms.
= −x + 26 Combine like terms.
Simplify the expression. 1. 6x − 4x 2. 12m − m − 7m + 3 3. 3( y + 2) − 4y 4. 9x − 4(2x − 1) 5. −(z + 2) − 2(1 − z) 6. −x2 + 5x + x2 Finding Volume Example 3 Find the volume of a rectangular prism with length 10 centimeters, width 4 centimeters, and height 5 centimeters.
Volume = ℓ wh Write the volume formula.
= (10)(4)(5) Substitute 10 forℓ , 4 for w, and 5 for h.
= 200 Multiply. The volume is 200 cubic centimeters.
Find the volume of the solid. 7. cube with side length 4 inches 8. sphere with radius 2 feet 9. rectangular prism with length 4 feet, width 2 feet, and height 6 feet 10. right cylinder with radius 3 centimeters and height 5 centimeters 11. ABSTRACT REASONING Does doubling the volume of a cube have the same effect on the side length? Explain your reasoning.
10 cm 4 cm 5 cm Dynamic Solutions available at BigIdeasMath.com 156 Chapter 4 Polynomial Functions Mathematical Mathematical Practices Practices Using Technology to Explore Concepts Mathematically profi cient students use technological tools to explore concepts.
Monitoring Progress Monitoring Progress Use a graphing calculator to determine whether the function is continuous. Explain your reasoning. 1. f(x) = x2 − x — x 2. f(x) = x3 − 3 3. f(x) = √— x2 + 1 4. f(x) = ∣ x + 2 ∣ 5. f(x) = 1 — x 6. f(x) = 1 — √— x2 − 1 7. f(x) = x 8. f(x) = 2x − 3 9. f(x) = x — x Determining Whether Functions Are Continuous Use a graphing calculator to compare the two functions. What can you conclude? Which function is not continuous? f(x) = x2 g(x) = x3 − x2 — x − 1 SOLUTION The graphs appear to be identical, but g is not defi ned when x = 1. There is a hole in the graph of g at the point (1, 1). Using the table feature of a graphing calculator, you obtain an error for g(x) when x = 1. So, g is not continuous.
Continuous Functions A function is continuous when its graph has no breaks, holes, or gaps.
Core Core Concept Concept x y Graph of a continuous function x y Graph of a function that is not continuous 3 −2 −3 2 f(x) = x2 X Y1 Y1=1 1 0 1 4 9 16 25 0 -1 1 2 3 4 5 3 −2 −3 2 3 hole X Y1 Y1=ERROR 1 0 ERROR 4 9 16 25 0 -1 1 2 3 4 5 g(x) = x3 − x2 x − 1 Section 4.1 Graphing Polynomial Functions 157 Graphing Polynomial Functions 4.1 Identifying Graphs of Polynomial Functions Work with a partner. Match each polynomial function with its graph. Explain your reasoning. Use a graphing calculator to verify your answers.
a. f(x) = x3 − x b. f(x) = −x3 + x c. f(x) = −x4 + 1 d. f(x) = x4 e. f(x) = x3 f. f(x) = x4 − x2 A. 6 −4 −6 4 B. 6 −4 −6 4 C. 6 −4 −6 4 D. 6 −4 −6 4 E. 6 −4 −6 4 F. 6 −4 −6 4 Identifying x-Intercepts of Polynomial Graphs Work with a partner. Each of the polynomial graphs in Exploration 1 has x-intercept(s) of −1, 0, or 1. Identify the x-intercept(s) of each graph. Explain how you can verify your answers.
Communicate Your Answer Communicate Your Answer 3. What are some common characteristics of the graphs of cubic and quartic polynomial functions? 4. Determine whether each statement is true or false. Justify your answer. a. When the graph of a cubic polynomial function rises to the left, it falls to the right. b. When the graph of a quartic polynomial function falls to the left, it rises to the right.
CONSTRUCTING VIABLE ARGUMENTS To be profi cient in math, you need to justify your conclusions and communicate them to others.
Essential Question Essential Question What are some common characteristics of the graphs of cubic and quartic polynomial functions?
A polynomial function of the form f(x) = an x n + an – 1x n – 1 + . . . + a1x + a0 where an ≠ 0, is cubic when n = 3 and quartic when n = 4.
158 Chapter 4 Polynomial Functions 4.1 Lesson Identifying Polynomial Functions Decide whether each function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coeffi cient.
a. f(x) = −2x3 + 5x + 8 b. g(x) = −0.8x3 + √ — 2 x4 − 12 c. h(x) = −x2 + 7x−1 + 4x d. k(x) = x2 + 3x SOLUTION a. The function is a polynomial function that is already written in standard form. It has degree 3 (cubic) and a leading coeffi cient of −2.
b. The function is a polynomial function written as g(x) = √ — 2 x4 − 0.8x3 − 12 in standard form. It has degree 4 (quartic) and a leading coeffi cient of √ — 2 .
c. The function is not a polynomial function because the term 7x−1 has an exponent that is not a whole number.
d. The function is not a polynomial function because the term 3x does not have a variable base and an exponent that is a whole number.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coeffi cient. 1. f(x) = 7 − 1.6x2 − 5x 2. p(x) = x + 2x−2 + 9.5 3. q(x) = x3 − 6x + 3x4 What You Will Learn What You Will Learn Identify polynomial functions. Graph polynomial functions using tables and end behavior.
Polynomial Functions Recall that a monomial is a number, a variable, or the product of a number and one or more variables with whole number exponents. A polynomial is a monomial or a sum of monomials. A polynomial function is a function of the form f(x) = an xn + an−1xn−1 + ⋅ ⋅ ⋅ + a1x + a0 where an ≠ 0, the exponents are all whole numbers, and the coeffi cients are all real numbers. For this function, an is the leading coeffi cient, n is the degree, and a0 is the constant term. A polynomial function is in standard form when its terms are written in descending order of exponents from left to right.
You are already familiar with some types of polynomial functions, such as linear and quadratic. Here is a summary of common types of polynomial functions.
Common Polynomial Functions Degree Type Standard Form Example 0 Constant f(x) = a0 f(x) = −14 1 Linear f(x) = a1x + a0 f(x) = 5x − 7 2 Quadratic f(x) = a2x2 + a1x + a0 f(x) = 2x2 + x − 9 3 Cubic f(x) = a3x3 + a2x2 + a1x + a0 f(x) = x3 − x2 + 3x 4 Quartic f(x) = a4x4 + a3x3 + a2x2 + a1x + a0 f(x) = x4 + 2x − 1 polynomial, p. 158 polynomial function, p. 158 end behavior, p. 159 Previous monomial linear function quadratic function Core Vocabulary Core Vocabulary Section 4.1 Graphing Polynomial Functions 159 Describing End Behavior Describe the end behavior of the graph of f(x) = −0.5x4 + 2.5x2 + x − 1.
SOLUTION The function has degree 4 and leading coeffi cient −0.5. Because the degree is even and the leading coeffi cient is negative, f(x) → −∞ as x → −∞ and f(x) → −∞ as x → +∞. Check this by graphing the function on a graphing calculator, as shown.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Evaluate the function for the given value of x. 4. f(x) = −x3 + 3x2 + 9; x = 4 5. f(x) = 3x5 − x4 − 6x + 10; x = −2 6. Describe the end behavior of the graph of f(x) = 0.25x3 − x2 − 1. Check Evaluating a Polynomial Function Evaluate f(x) = 2x4 − 8x2 + 5x − 7 when x = 3.
SOLUTION f(x) = 2x4 − 8x2 + 5x − 7 Write original equation. f(3) = 2(3)4 − 8(3)2 + 5(3) − 7 Substitute 3 for x. = 162 − 72 + 15 − 7 Evaluate powers and multiply.
= 98 Simplify.
The end behavior of a function’s graph is the behavior of the graph as x approaches positive infi nity (+∞) or negative infi nity (−∞). For the graph of a polynomial function, the end behavior is determined by the function’s degree and the sign of its leading coeffi cient.
Core Core Concept Concept End Behavior of Polynomial Functions Degree: odd Leading coeffi cient: positive x y f(x) +∞ as x +∞ f(x) −∞ as x −∞ Degree: odd Leading coeffi cient: negative x y f(x) −∞ as x +∞ f(x) +∞ as x −∞ Degree: even Leading coeffi cient: positive x y f(x) +∞ as x +∞ f(x) +∞ as x −∞ Degree: even Leading coeffi cient: negative x y f(x) −∞ as x +∞ f(x) −∞ as x −∞ READING The expression “x → +∞” is read as “x approaches positive infi nity.” 10 −10 −10 10 160 Chapter 4 Polynomial Functions Graphing Polynomial Functions To graph a polynomial function, fi rst plot points to determine the shape of the graph’s middle portion. Then connect the points with a smooth continuous curve and use what you know about end behavior to sketch the graph. Graphing Polynomial Functions Graph (a) f(x) = −x3 + x2 + 3x − 3 and (b) f(x) = x4 − x3 − 4x2 + 4.
SOLUTION a. To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior.
x −2 −1 0 1 2 f(x) 3 −4 −3 0 −1 The degree is odd and the leading coeffi cient is negative. So, f(x) → +∞ as x → −∞ and f(x) → −∞ as x → +∞.
b. To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior.
x −2 −1 0 1 2 f(x) 12 2 4 0 −4 The degree is even and the leading coeffi cient is positive. So, f(x) → +∞ as x → −∞ and f(x) → +∞ as x → +∞. Sketching a Graph Sketch a graph of the polynomial function f having these characteristics.
• f is increasing when x < 0 and x > 4.
• f is decreasing when 0 < x < 4.
• f(x) > 0 when −2 < x < 3 and x > 5.
• f(x) < 0 when x < −2 and 3 < x < 5.
Use the graph to describe the degree and leading coeffi cient of f.
SOLUTION x y increasing increasing decreasing 3 −2 4 5 The graph is above the x-axis when f(x) > 0.
The graph is below the x-axis when f(x) < 0. From the graph, f(x) → −∞ as x → −∞ and f(x) → +∞ as x → +∞. So, the degree is odd and the leading coeffi cient is positive.
x y 3 1 −1 3 5 −3 (1, 0) (2, −1) (−1, −4) (−2, 3) (0, −3) x y −3 1 5 −1 −3 (0, 4) (1, 0) (2, −4) (−1, 2) Section 4.1 Graphing Polynomial Functions 161 Solving a Real-Life Problem The estimated number V (in thousands) of electric vehicles in use in the United States can be modeled by the polynomial function V(t) = 0.151280t3 − 3.28234t2 + 23.7565t − 2.041 where t represents the year, with t = 1 corresponding to 2001.
a. Use a graphing calculator to graph the function for the interval 1 ≤ t ≤ 10. Describe the behavior of the graph on this interval.
b. What was the average rate of change in the number of electric vehicles in use from 2001 to 2010?
c. Do you think this model can be used for years before 2001 or after 2010? Explain your reasoning.
SOLUTION a. Using a graphing calculator and a viewing window of 1 ≤ x ≤ 10 and 0 ≤ y ≤ 65, you obtain the graph shown. From 2001 to 2004, the numbers of electric vehicles in use increased. Around 2005, the growth in the numbers in use slowed and started to level off. Then the numbers in use started to increase again in 2009 and 2010.
b. The years 2001 and 2010 correspond to t = 1 and t = 10. Average rate of change over 1 ≤ t ≤ 10: V(10) − V(1) —— 10 − 1 = 58.57 − 18.58444 —— 9 ≈ 4.443 The average rate of change from 2001 to 2010 is about 4.4 thousand electric vehicles per year.
c. Because the degree is odd and the leading coeffi cient is positive, V(t) → −∞ as t → −∞ and V(t) → +∞ as t → +∞. The end behavior indicates that the model has unlimited growth as t increases. While the model may be valid for a few years after 2010, in the long run, unlimited growth is not reasonable. Notice in 2000 that V(0) = −2.041. Because negative values of V(t) do not make sense given the context (electric vehicles in use), the model should not be used for years before 2001.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Graph the polynomial function. 7. f(x) = x4 + x2 − 3 8. f(x) = 4 − x3 9. f(x) = x3 − x2 + x − 1 10. Sketch a graph of the polynomial function f having these characteristics. • f is decreasing when x < −1.5 and x > 2.5; f is increasing when −1.5 < x < 2.5. • f(x) > 0 when x < −3 and 1 < x < 4; f(x) < 0 when −3 < x < 1 and x > 4. Use the graph to describe the degree and leading coeffi cient of f. 11. WHAT IF? Repeat Example 6 using the alternative model for electric vehicles of V(t) = −0.0290900t4 + 0.791260t3 − 7.96583t2 + 36.5561t − 12.025. 10 0 1 65 T c w a b c S a 162 Chapter 4 Polynomial Functions Exercises 4.1 Dynamic Solutions available at BigIdeasMath.com 1. WRITING Explain what is meant by the end behavior of a polynomial function. 2. WHICH ONE DOESN’T BELONG? Which function does not belong with the other three? Explain your reasoning.
f(x) = 7x5 + 3x2 − 2x g(x) = 3x3 − 2x8 + 3 — 4 h(x) = −3x4 + 5x−1 − 3x2 k(x) = √ — 3 x + 8x4 + 2x + 1 Vocabulary and Core Concept Check Vocabulary and Core Concept Check In Exercises 3–8, decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coeffi cient. (See Example 1.) 3. f(x) = −3x + 5x3 − 6x2 + 2 4.
p(x) = 1 — 2 x2 + 3x − 4x3 + 6x4 − 1 5. f(x) = 9x4 + 8x3 − 6x−2 + 2x 6. g(x) = √ — 3 − 12x + 13x2 7. h(x) = 5 — 3 x2 − √ — 7 x4 + 8x3 − 1 — 2 + x 8. h(x) = 3x4 + 2x − 5 — x + 9x3 − 7 ERROR ANALYSIS In Exercises 9 and 10, describe and correct the error in analyzing the function. 9. f(x) = 8x3 − 7x4 − 9x − 3x2 + 11 f is a polynomial function.
The degree is 3 and f is a cubic function. The leading coeffi cient is 8.
✗ 10. f(x) = 2x4 + 4x − 9 √ — x + 3x2 − 8 f is a polynomial function.
The degree is 4 and f is a quartic function. The leading coeffi cient is 2.
✗ In Exercises 11–16, evaluate the function for the given value of x. (See Example 2.) 11. h(x) = −3x4 + 2x3 − 12x − 6; x = −2 12. f(x) = 7x4 − 10x2 + 14x − 26; x = −7 13. g(x) = x6 − 64x4 + x2 − 7x − 51; x = 8 14. g(x) = −x3 + 3x2 + 5x + 1; x = −12 15. p(x) = 2x3 + 4x2 + 6x + 7; x = 1 — 2 16. h(x) = 5x3 − 3x2 + 2x + 4; x = − 1 — 3 In Exercises 17–20, describe the end behavior of the graph of the function. (See Example 3.) 17. h(x) = −5x4 + 7x3 − 6x2 + 9x + 2 18.
g(x) = 7x7 + 12x5 − 6x3 − 2x − 18 19. f(x) = −2x4 + 12x8 + 17 + 15x2 20. f(x) = 11 − 18x2 − 5x5 − 12x4 − 2x In Exercises 21 and 22, describe the degree and leading coeffi cient of the polynomial function using the graph. 21. x y 22. x y Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics Section 4.1 Graphing Polynomial Functions 163 23. USING STRUCTURE Determine whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coeffi cient.
f(x) = 5x3x + 5 — 2x3 − 9x4 + √ — 2 x2 + 4x −1 −x−5x5 − 4 24. WRITING Let f(x) = 13. State the degree, type, and leading coeffi cient. Describe the end behavior of the function. Explain your reasoning.
In Exercises 25–32, graph the polynomial function.
(See Example 4.) 25. p(x) = 3 −x4 26. g(x) = x3 + x + 3 27. f(x) = 4x − 9 −x3 28. p(x) = x5 − 3x3 + 2 29. h(x) = x4 − 2x3 + 3x 30. h(x) = 5 + 3x2 −x4 31. g(x) = x5 − 3x4 + 2x − 4 32. p(x) = x6 − 2x5 − 2x3 + x + 5 ANALYZING RELATIONSHIPS In Exercises 33–36, describe the x-values for which (a) f is increasing or decreasing, (b) f(x) > 0, and (c) f(x) < 0. 33. x y 4 −8 −4 4 6 2 f 34. f x y 4 4 −4 −8 35. f x y 1 4 2 −2 36. f x y 2 −4 −2 −2 −4 In Exercises 37–40, sketch a graph of the polynomial function f having the given characteristics. Use the graph to describe the degree and leading coeffi cient of the function f. (See Example 5.) 37. • f is increasing when x > 0.5; f is decreasing when x < 0.5.
• f(x) > 0 when x < −2 and x > 3; f(x) < 0 when −2 < x < 3.
38. • f is increasing when −2 < x < 3; f is decreasing when x < −2 and x > 3.
• f(x) > 0 when x < −4 and 1 < x < 5; f(x) < 0 when −4 < x < 1 and x > 5.
39. • f is increasing when −2 < x < 0 and x > 2; f is decreasing when x < −2 and 0 < x < 2.
• f(x) > 0 when x < −3, −1 < x < 1, and x > 3; f(x) < 0 when −3 < x < −1 and 1 < x < 3.
40. • f is increasing when x < −1 and x > 1; f is decreasing when −1 < x < 1.
• f(x) > 0 when −1.5 < x < 0 and x > 1.5; f(x) < 0 when x < −1.5 and 0 < x < 1.5.
41. MODELING WITH MATHEMATICS From 1980 to 2007 the number of drive-in theaters in the United States can be modeled by the function d(t) = −0.141t3 + 9.64t2 − 232.5t + 2421 where d(t) is the number of open theaters and t is the number of years after 1980. (See Example 6.) a. Use a graphing calculator to graph the function for the interval 0 ≤ t ≤ 27. Describe the behavior of the graph on this interval.
b. What is the average rate of change in the number of drive-in movie theaters from 1980 to 1995 and from 1995 to 2007? Interpret the average rates of change.
c. Do you think this model can be used for years before 1980 or after 2007? Explain. 42. PROBLEM SOLVING The weight of an ideal round-cut diamond can be modeled by w = 0.00583d 3 − 0.0125d 2 + 0.022d − 0.01 where w is the weight of the diamond (in carats) and d is the diameter (in millimeters). According to the model, what is the weight of a diamond with a diameter of 12 millimeters?
diameter 164 Chapter 4 Polynomial Functions 43. ABSTRACT REASONING Suppose f(x) → ∞ as x → −∞ and f(x) → −∞ as x → ∞. Describe the end behavior of g(x) = −f(x). Justify your answer. 44. THOUGHT PROVOKING Write an even degree polynomial function such that the end behavior of f is given by f(x) → −∞ as x → −∞ and f(x) → −∞ as x → ∞. Justify your answer by drawing the graph of your function. 45. USING TOOLS When using a graphing calculator to graph a polynomial function, explain how you know when the viewing window is appropriate. 46. MAKING AN ARGUMENT Your friend uses the table to speculate that the function f is an even degree polynomial and the function g is an odd degree polynomial. Is your friend correct? Explain your reasoning.
x f(x) g(x) −8 4113 497 −2 21 5 0 1 1 2 13 −3 8 4081 −495 47. DRAWING CONCLUSIONS The graph of a function is symmetric with respect to the y-axis if for each point (a, b) on the graph, (−a, b) is also a point on the graph. The graph of a function is symmetric with respect to the origin if for each point (a, b) on the graph, (−a, −b) is also a point on the graph. a. Use a graphing calculator to graph the function y = xn when n = 1, 2, 3, 4, 5, and 6. In each case, identify the symmetry of the graph. b. Predict what symmetry the graphs of y = x10 and y = x11 each have. Explain your reasoning and then confi rm your predictions by graphing.
48. HOW DO YOU SEE IT? The graph of a polynomial function is shown.
x f y 4 6 −2 2 −2 −6 a. Describe the degree and leading coeffi cient of f. b. Describe the intervals where the function is increasing and decreasing. c. What is the constant term of the polynomial function? 49. REASONING A cubic polynomial function f has a leading coeffi cient of 2 and a constant term of −5. When f(1) = 0 and f(2) = 3, what is f(−5)? Explain your reasoning. 50. CRITICAL THINKING The weight y (in pounds) of a rainbow trout can be modeled by y = 0.000304x3, where x is the length (in inches) of the trout.
a. Write a function that relates the weight y and length x of a rainbow trout when y is measured in kilograms and x is measured in centimeters. Use the fact that 1 kilogram ≈ 2.20 pounds and 1 centimeter ≈ 0.394 inch.
b. Graph the original function and the function from part (a) in the same coordinate plane. What type of transformation can you apply to the graph of y = 0.000304x3 to produce the graph from part (a)?
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Simplify the expression. (Skills Review Handbook) 51. xy + x2 + 2xy + y2 − 3x2 52. 2h3g + 3hg3 + 7h2g2 + 5h3g + 2hg3 53. −wk + 3kz − 2kw + 9zk − kw 54. a2(m − 7a3) − m(a2 − 10) 55. 3x(xy − 4) + 3(4xy + 3) − xy(x2y − 1) 56. cv(9 − 3c) + 2c(v − 4c) + 6c Reviewing what you learned in previous grades and lessons Section 4.2 Adding, Subtracting, and Multiplying Polynomials 165 Adding, Subtracting, and Multiplying Polynomials 4.2 Cubing Binomials Work with a partner. Find each product. Show your steps.
a. (x + 1)3 = (x + 1)(x + 1)2 Rewrite as a product of fi rst and second powers.
= (x + 1) Multiply second power.
= Multiply binomial and trinomial.
= Write in standard form, ax3 + bx2 + cx + d.
b. (a + b)3 = (a + b)(a + b)2 Rewrite as a product of fi rst and second powers.
= (a + b) Multiply second power.
= Multiply binomial and trinomial.
= Write in standard form.
c. (x − 1)3 = (x − 1)(x − 1)2 Rewrite as a product of fi rst and second powers.
= (x − 1) Multiply second power.
= Multiply binomial and trinomial.
= Write in standard form.
d. (a −b)3 = (a −b)(a −b)2 Rewrite as a product of fi rst and second powers.
= (a −b) Multiply second power.
= Multiply binomial and trinomial.
= Write in standard form. Generalizing Patterns for Cubing a Binomial Work with a partner.
a. Use the results of Exploration 1 to describe a pattern for the coeffi cients of the terms when you expand the cube of a binomial. How is your pattern related to Pascal’s Triangle, shown at the right?
b. Use the results of Exploration 1 to describe a pattern for the exponents of the terms in the expansion of a cube of a binomial.
c. Explain how you can use the patterns you described in parts (a) and (b) to fi nd the product (2x − 3)3. Then find this product.
Communicate Your Answer Communicate Your Answer 3. How can you cube a binomial? 4. Find each product. a. (x + 2)3 b. (x − 2)3 c. (2x − 3)3 d. (x − 3)3 e. (−2x + 3)3 f. (3x − 5)3 LOOKING FOR STRUCTURE To be profi cient in math, you need to look closely to discern a pattern or structure.
Essential Question Essential Question How can you cube a binomial?
1 4 6 4 1 1 3 3 1 1 2 1 1 1 1 166 Chapter 4 Polynomial Functions 4.2 Lesson What You Will Learn What You Will Learn Add and subtract polynomials. Multiply polynomials. Use Pascal’s Triangle to expand binomials.
Adding and Subtracting Polynomials Recall that the set of integers is closed under addition and subtraction because every sum or difference results in an integer. To add or subtract polynomials, you add or subtract the coeffi cients of like terms. Because adding or subtracting polynomials results in a polynomial, the set of polynomials is closed under addition and subtraction. Adding Polynomials Vertically and Horizontally a. Add 3x3 + 2x2 − x − 7 and x3 − 10x2 + 8 in a vertical format.
b. Add 9y3 + 3y2 − 2y + 1 and −5y2 + y − 4 in a horizontal format.
SOLUTION a. 3x3 + 2x2 − x − 7 + x3 − 10x2 + 8 4x3 − 8x2 − x + 1 b. (9y3 + 3y2 − 2y + 1) + (−5y2 + y − 4) = 9y3 + 3y2 − 5y2 − 2y + y + 1 − 4 = 9y3 − 2y2 − y − 3 To subtract one polynomial from another, add the opposite. To do this, change the sign of each term of the subtracted polynomial and then add the resulting like terms. Subtracting Polynomials Vertically and Horizontally a. Subtract 2x3 + 6x2 − x + 1 from 8x3 − 3x2 − 2x + 9 in a vertical format.
b. Subtract 3z2 + z − 4 from 2z2 + 3z in a horizontal format.
SOLUTION a. Align like terms, then add the opposite of the subtracted polynomial. 8x3 − 3x2 − 2x + 9 − (2x3 + 6x2 − x + 1) 8x3 − 3x2 − 2x + 9 + −2x3 − 6x2 + x − 1 6x3 − 9x2 − x + 8 b. Write the opposite of the subtracted polynomial, then add like terms. (2z2 + 3z) − (3z2 + z − 4) = 2z2 + 3z − 3z2 − z + 4 = −z2 + 2z + 4 Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Find the sum or difference. 1. (2x2 − 6x + 5) + (7x2 − x − 9) 2. (3t3 + 8t2 − t − 4) − (5t3 − t2 + 17) Pascal’s Triangle, p. 169 Previous like terms identity Core Vocabulary Core Vocabulary COMMON ERROR A common mistake is to forget to change signs correctly when subtracting one polynomial from another. Be sure to add the opposite of every term of the subtracted polynomial. Section 4.2 Adding, Subtracting, and Multiplying Polynomials 167 Multiplying Polynomials To multiply two polynomials, you multiply each term of the fi rst polynomial by each term of the second polynomial. As with addition and subtraction, the set of polynomials is closed under multiplication. Multiplying Polynomials Vertically and Horizontally a. Multiply −x2 + 2x + 4 and x − 3 in a vertical format.
b. Multiply y + 5 and 3y2 − 2y + 2 in a horizontal format.
SOLUTION a. −x2 + 2x + 4 × x − 3 3x2 − 6x − 12 Multiply −x2 + 2x + 4 by −3. −x3 + 2x2 + 4x Multiply −x2 + 2x + 4 by x. −x3 + 5x2 − 2x − 12 Combine like terms.
b. ( y + 5)(3y2 − 2y + 2) = ( y + 5)3y2 − ( y + 5)2y + ( y + 5)2 = 3y3 + 15y2 − 2y2 − 10y + 2y + 10 = 3y3 + 13y2 − 8y + 10 Multiplying Three Binomials Multiply x − 1, x + 4, and x + 5 in a horizontal format.
SOLUTION (x − 1)(x + 4)(x + 5) = (x2 + 3x − 4)(x + 5) = (x2 + 3x − 4)x + (x2 + 3x − 4)5 = x3 + 3x2 − 4x + 5x2 + 15x − 20 = x3 + 8x2 + 11x − 20 Some binomial products occur so frequently that it is worth memorizing their patterns. You can verify these polynomial identities by multiplying.
REMEMBER Product of Powers Property am ⋅ an = am+n a is a real number and m and n are integers.
Core Core Concept Concept Special Product Patterns Sum and Difference Example (a + b)(a − b) = a2 − b2 (x + 3)(x − 3) = x2 − 9 Square of a Binomial Example (a + b)2 = a2 + 2ab + b2 (y + 4)2 = y2 + 8y + 16 (a − b)2 = a2 − 2ab + b2 (2t − 5)2 = 4t2 − 20t + 25 Cube of a Binomial Example (a + b)3 = a3 + 3a2b + 3ab2 + b3 (z + 3)3 = z3 + 9z2 + 27z + 27 (a − b)3 = a3 − 3a2b + 3ab2 − b3 (m − 2)3 = m3 − 6m2 + 12m − 8 COMMON ERROR In general, (a ± b)2 ≠ a2 ± b2 and (a ± b)3 ≠ a3 ± b3.
168 Chapter 4 Polynomial Functions Proving a Polynomial Identity a. Prove the polynomial identity for the cube of a binomial representing a sum: (a + b)3 = a3 + 3a2b + 3ab2 + b3.
b. Use the cube of a binomial in part (a) to calculate 113.
SOLUTION a. Expand and simplify the expression on the left side of the equation. (a + b)3 = (a + b)(a + b)(a + b) = (a2 + 2ab + b2)(a + b) = (a2 + 2ab + b2)a + (a2 + 2ab + b2)b = a3 + a2b + 2a2b + 2ab2 + ab2 + b3 = a3 + 3a2b + 3ab2 + b3 ✓ The simplifi ed left side equals the right side of the original identity. So, the identity (a + b)3 = a3 + 3a2b + 3ab2 + b3 is true.
b. To calculate 113 using the cube of a binomial, note that 11 = 10 + 1. 113 = (10 + 1)3 Write 11 as 10 + 1. = 103 + 3(10)2(1) + 3(10)(1)2 + 13 Cube of a binomial = 1000 + 300 + 30 + 1 Expand. = 1331 Simplify. Using Special Product Patterns Find each product.
a. (4n + 5)(4n − 5) b. (9y − 2)2 c. (ab + 4)3 SOLUTION a. (4n + 5)(4n − 5) = (4n)2 − 52 Sum and difference = 16n2 − 25 Simplify.
b. (9y − 2)2 = (9y)2 − 2(9y)(2) + 22 Square of a binomial = 81y2 − 36y + 4 Simplify.
c. (ab + 4)3 = (ab)3 + 3(ab)2(4) + 3(ab)(4)2 + 43 Cube of a binomial = a3b3 + 12a2b2 + 48ab + 64 Simplify.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Find the product. 3. (4x2 + x − 5)(2x + 1) 4. ( y − 2)(5y2 + 3y − 1) 5. (m − 2)(m − 1)(m + 3) 6. (3t − 2)(3t + 2) 7. (5a + 2)2 8. (xy − 3)3 9. (a) Prove the polynomial identity for the cube of a binomial representing a difference: (a − b)3 = a3 − 3a2b + 3ab2 − b3. (b) Use the cube of a binomial in part (a) to calculate 93. REMEMBER Power of a Product Property (ab)m = ambm a and b are real numbers and m is an integer. Section 4.2 Adding, Subtracting, and Multiplying Polynomials 169 In general, the nth row in Pascal’s Triangle gives the coeffi cients of (a + b)n. Here are some other observations about the expansion of (a + b)n. 1. An expansion has n + 1 terms. 2. The power of a begins with n, decreases by 1 in each successive term, and ends with 0. 3. The power of b begins with 0, increases by 1 in each successive term, and ends with n. 4. The sum of the powers of each term is n. Using Pascal’s Triangle to Expand Binomials Use Pascal’s Triangle to expand (a) (x − 2)5 and (b) (3y + 1)3.
SOLUTION a. The coeffi cients from the fi fth row of Pascal’s Triangle are 1, 5, 10, 10, 5, and 1. (x − 2)5 = 1x5 + 5x4(−2) + 10x3(−2)2 + 10x2(−2)3 + 5x(−2)4 + 1(−2)5 = x5 − 10x4 + 40x3 − 80x2 + 80x − 32 b. The coeffi cients from the third row of Pascal’s Triangle are 1, 3, 3, and 1. (3y + 1)3 = 1(3y)3 + 3(3y)2(1) + 3(3y)(1)2 + 1(1)3 = 27y3 + 27y2 + 9y + 1 Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 10. Use Pascal’s Triangle to expand (a) (z + 3)4 and (b) (2t − 1)5. Pascal’s Triangle Consider the expansion of the binomial (a + b)n for whole number values of n. When you arrange the coeffi cients of the variables in the expansion of (a + b)n, you will see a special pattern called Pascal’s Triangle. Pascal’s Triangle is named after French mathematician Blaise Pascal (1623−1662).
Core Core Concept Concept Pascal’s Triangle In Pascal’s Triangle, the fi rst and last numbers in each row are 1. Every number other than 1 is the sum of the closest two numbers in the row directly above it. The numbers in Pascal’s Triangle are the same numbers that are the coeffi cients of binomial expansions, as shown in the fi rst six rows. n (a + b)n Binomial Expansion Pascal’s Triangle 0th row 0 (a + b)0 = 1 1 1st row 1 (a + b)1 = 1a + 1b 1 1 2nd row 2 (a + b)2 = 1a2 + 2ab + 1b2 1 2 1 3rd row 3 (a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3 1 3 3 1 4th row 4 (a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4 1 4 6 4 1 5th row 5 (a + b)5 = 1a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + 1b5 1 5 10 10 5 1 170 Chapter 4 Polynomial Functions Exercises 4.2 Dynamic Solutions available at BigIdeasMath.com 1. WRITING Describe three different methods to expand (x + 3)3. 2. WRITING Is (a + b)(a − b) = a2 − b2 an identity? Explain your reasoning.
Vocabulary and Core Concept Check Vocabulary and Core Concept Check In Exercises 3–8, fi nd the sum. (See Example 1.) 3. (3x2 + 4x − 1) + (−2x2 − 3x + 2) 4. (−5x2 + 4x − 2) + (−8x2 + 2x + 1) 5. (12x5 − 3x4 + 2x − 5) + (8x4 − 3x3 + 4x + 1) 6. (8x4 + 2x2 − 1) + (3x3 − 5x2 + 7x + 1) 7. (7x6 + 2x5 − 3x2 + 9x) + (5x5 + 8x3 − 6x2 + 2x − 5) 8. (9x4 − 3x3 + 4x2 + 5x + 7) + (11x4 − 4x2 − 11x − 9) In Exercises 9–14, fi nd the difference. (See Example 2.) 9. (3x3 − 2x2 + 4x − 8) − (5x3 + 12x2 − 3x − 4) 10. (7x4 − 9x3 − 4x2 + 5x + 6) − (2x4 + 3x3 − x2 + x − 4) 11. (5x6 − 2x4 + 9x3 + 2x − 4) − (7x5 − 8x4 + 2x − 11) 12. (4x5 − 7x3 − 9x2 + 18) − (14x5 − 8x4 + 11x2 + x) 13. (8x5 + 6x3 − 2x2 + 10x) − (9x5 − x3 − 13x2 + 4) 14. (11x4 − 9x2 + 3x + 11) − (2x4 + 6x3 + 2x − 9) 15. MODELING WITH MATHEMATICS During a recent period of time, the numbers (in thousands) of males M and females F that attend degree-granting institutions in the United States can be modeled by M = 19.7t2 + 310.5t + 7539.6 F = 28t2 + 368t + 10127.8 where t is time in years. Write a polynomial to model the total number of people attending degree-granting institutions. Interpret its constant term.
16. MODELING WITH MATHEMATICS A farmer plants a garden that contains corn and pumpkins. The total area (in square feet) of the garden is modeled by the expression 2x2 + 5x + 4. The area of the corn is modeled by the expression x2 − 3x + 2. Write an expression that models the area of the pumpkins.
In Exercises 17–24, fi nd the product. (See Example 3.) 17. 7x3(5x2 + 3x + 1) 18. −4x5(11x3 + 2x2 + 9x + 1) 19. (5x2 − 4x + 6)(−2x + 3) 20. (−x − 3)(2x2 + 5x + 8) 21. (x2 − 2x − 4)(x2 − 3x − 5) 22. (3x2 + x − 2)(−4x2 − 2x − 1) 23. (3x3 − 9x + 7)(x2 − 2x + 1) 24. (4x2 − 8x − 2)(x4 + 3x2 + 4x) ERROR ANALYSIS In Exercises 25 and 26, describe and correct the error in performing the operation.
25. (x2 − 3x + 4) − (x3 + 7x − 2) = x2 − 3x + 4 − x3 + 7x − 2 = −x3 + x2 + 4x + 2 ✗ 26. (2x − 7)3 = (2x)3 − 73 = 8x3 − 343 ✗ Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics Section 4.2 Adding, Subtracting, and Multiplying Polynomials 171 In Exercises 27–32, fi nd the product of the binomials. (See Example 4.) 27. (x − 3)(x + 2)(x + 4) 28. (x − 5)(x + 2)(x − 6) 29. (x − 2)(3x + 1)(4x − 3) 30. (2x + 5)(x − 2)(3x + 4) 31. (3x − 4)(5 − 2x)(4x + 1) 32. (4 − 5x)(1 − 2x)(3x + 2) 33. REASONING Prove the polynomial identity (a + b)(a − b) = a2 − b2. Then give an example of two whole numbers greater than 10 that can be multiplied using mental math and the given identity. Justify your answers. (See Example 5.) 34. NUMBER SENSE You have been asked to order textbooks for your class. You need to order 29 textbooks that cost $31 each. Explain how you can use the polynomial identity (a + b)(a − b) = a2 − b2 and mental math to fi nd the total cost of the textbooks.
In Exercises 35–42, fi nd the product. (See Example 6.) 35. (x − 9)(x + 9) 36. (m + 6)2 37. (3c − 5)2 38. (2y − 5)(2y + 5) 39. (7h + 4)2 40. (9g − 4)2 41. (2k + 6)3 42. (4n − 3)3 In Exercises 43–48, use Pascal’s Triangle to expand the binomial. (See Example 7.) 43. (2t + 4)3 44. (6m + 2)2 45. (2q − 3)4 46. (g + 2)5 47. (yz + 1)5 48. (np − 1)4 49. COMPARING METHODS Find the product of the expression (a2 + 4b2)2(3a2 − b2)2 using two different methods. Which method do you prefer? Explain.
50. THOUGHT PROVOKING Adjoin one or more polygons to the rectangle to form a single new polygon whose perimeter is double that of the rectangle. Find the perimeter of the new polygon.
2x + 3 x + 1 MATHEMATICAL CONNECTIONS In Exercises 51 and 52, write an expression for the volume of the fi gure as a polynomial in standard form. 51. V =ℓ wh 52. V = πr2h 2x + 2 x + 1 x + 3 3x − 4 x − 2 53. MODELING WITH MATHEMATICS Two people make three deposits into their bank accounts earning the same simple interest rate r.
Person A 2-5384100608 01/01/2012 01/01/2013 Deposit Deposit $2000.00 $3000.00 01/01/2014 Deposit $1000.00 Transaction Amount Person B 1-5233032905 01/01/2012 01/01/2013 Deposit Deposit $5000.00 $1000.00 01/01/2014 Deposit $4000.00 Transaction Amount Person A’s account is worth 2000(1 + r)3 + 3000(1 + r)2 + 1000(1 + r) on January 1, 2015. a. Write a polynomial for the value of Person B’s account on January 1, 2015. b. Write the total value of the two accounts as a polynomial in standard form. Then interpret the coeffi cients of the polynomial. c. Suppose their interest rate is 0.05. What is the total value of the two accounts on January 1, 2015?
172 Chapter 4 Polynomial Functions 54. PROBLEM SOLVING The sphere is centered in the cube. Find an expression for the volume of the cube outside the sphere. 55. MAKING AN ARGUMENT Your friend claims the sum of two binomials is always a binomial and the product of two binomials is always a trinomial. Is your friend correct? Explain your reasoning. 56. HOW DO YOU SEE IT? You make a tin box by cutting x-inch-by-x-inch pieces of tin off the corners of a rectangle and folding up each side. The plan for your box is shown.
12 − 2x 6 − 2x x x x x x x x x a. What are the dimensions of the original piece of tin?
b. Write a function that represents the volume of the box. Without multiplying, determine its degree.
USING TOOLS In Exercises 57–60, use a graphing calculator to make a conjecture about whether the two functions are equivalent. Explain your reasoning.
57. f(x) = (2x − 3)3; g(x) = 8x3 − 36x2 + 54x − 27 58. h(x) = (x + 2)5; k(x) = x5 + 10x4 + 40x3 + 80x2 + 64x 59. f(x) = (−x − 3)4; g(x) = x4 + 12x3 + 54x2 + 108x + 80 60. f(x) = (−x + 5)3; g(x) = −x3 + 15x2 − 75x + 125 61. REASONING Copy Pascal’s Triangle and add rows for n = 6, 7, 8, 9, and 10. Use the new rows to expand (x + 3)7 and (x − 5)9.
62. ABSTRACT REASONING You are given the function f(x) = (x + a)(x + b)(x + c)(x + d). When f(x) is written in standard form, show that the coeffi cient of x3 is the sum of a, b, c, and d, and the constant term is the product of a, b, c, and d.
63. DRAWING CONCLUSIONS Let g(x) = 12x4 + 8x + 9 and h(x) = 3x5 + 2x3 − 7x + 4.
a. What is the degree of the polynomial g(x) + h(x)? b. What is the degree of the polynomial g(x) − h(x)? c. What is the degree of the polynomial g(x) ⋅ h(x)? d. In general, if g(x) and h(x) are polynomials such that g(x) has degree m and h(x) has degree n, and m > n, what are the degrees of g(x) + h(x), g(x) − h(x), and g(x) ⋅ h(x)?
64. FINDING A PATTERN In this exercise, you will explore the sequence of square numbers. The fi rst four square numbers are represented below.
1 4 16 9 a. Find the differences between consecutive square numbers. Explain what you notice.
b. Show how the polynomial identity (n + 1)2 −n2 = 2n + 1 models the differences between square numbers.
c. Prove the polynomial identity in part (b).
65. CRITICAL THINKING Recall that a Pythagorean triple is a set of positive integers a, b, and c such that a2 + b2 = c2. The numbers 3, 4, and 5 form a Pythagorean triple because 32 + 42 = 52. You can use the polynomial identity (x2 −y2)2 + (2xy)2 = (x2 + y2)2 to generate other Pythagorean triples.
a. Prove the polynomial identity is true by showing that the simplifi ed expressions for the left and right sides are the same. b. Use the identity to generate the Pythagorean triple when x = 6 and y = 5.
c. Verify that your answer in part (b) satisfi es a2 + b2 = c2.
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Perform the operation. Write the answer in standard form. (Section 3.2) 66. (3 − 2i) + (5 + 9i) 67. (12 + 3i) − (7 − 8i) 68. (7i)(−3i) 69. (4 + i)(2 − i) Reviewing what you learned in previous grades and lessons x + 2 3 Section 4.3 Dividing Polynomials 173 Dividing Polynomials 4.3 Dividing Polynomials Work with a partner. Match each division statement with the graph of the related cubic polynomial f(x). Explain your reasoning. Use a graphing calculator to verify your answers.
a. f(x) — x = (x − 1)(x + 2) b. f(x) — x − 1 = (x − 1)(x + 2) c. f(x) — x + 1 = (x − 1)(x + 2) d. f(x) — x − 2 = (x − 1)(x + 2) e. f(x) — x + 2 = (x − 1)(x + 2) f. f(x) — x − 3 = (x − 1)(x + 2) A. 6 −4 −6 4 B. 8 −4 −8 8 C. 6 −4 −6 4 D. 8 −2 −8 6 E. 6 −4 −6 4 F. 6 −4 −6 4 Dividing Polynomials Work with a partner. Use the results of Exploration 1 to fi nd each quotient. Write your answers in standard form. Check your answers by multiplying.
a. (x3 + x2 − 2x) ÷ x b. (x3 − 3x + 2) ÷ (x − 1) c. (x3 + 2x2 − x − 2) ÷ (x + 1) d. (x3 − x2 − 4x + 4) ÷ (x − 2) e. (x3 + 3x2 − 4) ÷ (x + 2) f. (x3 − 2x2 − 5x + 6) ÷ (x − 3) Communicate Your Answer Communicate Your Answer 3. How can you use the factors of a cubic polynomial to solve a division problem involving the polynomial?
REASONING ABSTRACTLY To be profi cient in math, you need to understand a situation abstractly and represent it symbolically.
Essential Question Essential Question How can you use the factors of a cubic polynomial to solve a division problem involving the polynomial?
174 Chapter 4 Polynomial Functions 4.3 Lesson What You Will Learn What You Will Learn Use long division to divide polynomials by other polynomials. Use synthetic division to divide polynomials by binomials of the form x − k. Use the Remainder Theorem.
Long Division of Polynomials When you divide a polynomial f(x) by a nonzero polynomial divisor d(x), you get a quotient polynomial q(x) and a remainder polynomial r(x). f(x) — d(x) = q(x) + r(x) — d(x) The degree of the remainder must be less than the degree of the divisor. When the remainder is 0, the divisor divides evenly into the dividend. Also, the degree of the divisor is less than or equal to the degree of the dividend f(x). One way to divide polynomials is called polynomial long division. Using Polynomial Long Division Divide 2x4 + 3x3 + 5x − 1 by x2 + 3x + 2.
SOLUTION Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coeffi cient of x2 in the dividend. At each stage, divide the term with the highest power in what is left of the dividend by the fi rst term of the divisor. This gives the next term of the quotient. 2x2 − 3x + 5 quotient x2 + 3x + 2 ) ‾‾‾ 2x4 + 3x3 + 0x2 + 5x − 1 2x4 + 6x3 + 4x2 Multiply divisor by 2x4 — x2 = 2x2. −3x3 − 4x2 + 5x Subtract. Bring down next term. −3x3 − 9x2 − 6x Multiply divisor by −3x3 — x2 = −3x. 5x2 + 11x − 1 Subtract. Bring down next term.
5x2 + 15x + 10 Multiply divisor by 5x2 — x2 = 5.
−4x − 11 remainder 2x4 + 3x3 + 5x − 1 —— x2 + 3x + 2 = 2x2 − 3x + 5 + −4x − 11 — x2 + 3x + 2 Check You can check the result of a division problem by multiplying the quotient by the divisor and adding the remainder. The result should be the dividend.
(2x2 − 3x + 5)(x2 + 3x + 2) + (−4x − 11) = (2x2)(x2 + 3x + 2) − (3x)(x2 + 3x + 2) + (5)(x2 + 3x + 2) − 4x − 11 = 2x4 + 6x3 + 4x2 − 3x3 − 9x2 − 6x + 5x2 + 15x + 10 − 4x − 11 = 2x4 + 3x3 + 5x − 1 ✓ Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Divide using polynomial long division. 1. (x3 − x2 − 2x + 8) ÷ (x − 1) 2. (x4 + 2x2 − x + 5) ÷ (x2 − x + 1) COMMON ERROR The expression added to the quotient in the result of a long division problem is r(x) — d(x) , not r(x).
polynomial long division, p. 174 synthetic division, p. 175 Previous long division divisor quotient remainder dividend Core Vocabulary Core Vocabulary Section 4.3 Dividing Polynomials 175 Synthetic Division There is a shortcut for dividing polynomials by binomials of the form x − k. This shortcut is called synthetic division. This method is shown in the next example. Using Synthetic Division Divide −x3 + 4x2 + 9 by x − 3.
SOLUTION Step 1 Write the coeffi cients of the dividend in order of descending exponents. Include a “0” for the missing x-term. Because the divisor is x − 3, use k = 3. Write the k-value to the left of the vertical bar. 3 −1 4 0 9 Step 2 Bring down the leading coeffi cient. Multiply the leading coeffi cient by the k-value. Write the product under the second coeffi cient. Add. 3 −1 4 0 9 −3 −1 1 Step 3 Multiply the previous sum by the k-value. Write the product under the third coeffi cient. Add. Repeat this process for the remaining coeffi cient. The fi rst three numbers in the bottom row are the coeffi cients of the quotient, and the last number is the remainder. 3 −1 4 0 9 −3 3 9 −1 1 3 18 −x3 + 4x2 + 9 —— x − 3 = −x2 + x + 3 + 18 — x − 3 Using Synthetic Division Divide 3x3 − 2x2 + 2x − 5 by x + 1.
SOLUTION Use synthetic division. Because the divisor is x + 1 = x − (−1), k = −1. −1 3 −2 2 −5 −3 5 −7 3 −5 7 −12 3x3 − 2x2 + 2x − 5 —— x + 1 = 3x2 − 5x + 7 − 12 — x + 1 Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Divide using synthetic division. 3. (x3 − 3x2 − 7x + 6) ÷ (x − 2) 4. (2x3 − x − 7) ÷ (x + 3) k-value coeffi cients of −x3 + 4x2 + 9 coeffi cients of quotient remainder STUDY TIP Note that dividing polynomials does not always result in a polynomial. This means that the set of polynomials is not closed under division.
176 Chapter 4 Polynomial Functions The Remainder Theorem tells you that synthetic division can be used to evaluate a polynomial function. So, to evaluate f(x) when x = k, divide f(x) by x − k. The remainder will be f(k). Evaluating a Polynomial Use synthetic division to evaluate f(x) = 5x3 − x2 + 13x + 29 when x = −4.
SOLUTION −4 5 −1 13 29 −20 84 −388 5 −21 97 −359 The remainder is −359. So, you can conclude from the Remainder Theorem that f(−4) = −359.
Check Check this by substituting x = −4 in the original function. f(−4) = 5(−4)3 − (−4)2 + 13(−4) + 29 = −320 − 16 − 52 + 29 = −359 ✓ Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Use synthetic division to evaluate the function for the indicated value of x. 5. f(x) = 4x2 − 10x − 21; x = 5 6. f(x) = 5x4 + 2x3 − 20x − 6; x = 2 The Remainder Theorem The remainder in the synthetic division process has an important interpretation. When you divide a polynomial f(x) by d(x) = x − k, the result is f(x) — d(x) = q(x) + r(x) — d(x) Polynomial division f(x) — x − k = q(x) + r(x) — x − k Substitute x − k for d(x). f(x) = (x − k)q(x) + r(x). Multiply both sides by x − k.
Because either r(x) = 0 or the degree of r(x) is less than the degree of x − k, you know that r(x) is a constant function. So, let r(x) = r, where r is a real number, and evaluate f(x) when x = k. f(k) = (k − k)q(k) + r Substitute k for x and r for r(x).
f(k) = r Simplify.
This result is stated in the Remainder Theorem.
Core Core Concept Concept The Remainder Theorem If a polynomial f(x) is divided by x − k, then the remainder is r = f(k). Section 4.3 Dividing Polynomials 177 Exercises 4.3 In Exercises 5–10, divide using polynomial long division. (See Example 1.) 5. ( x2 + x − 17 ) ÷ ( x − 4 ) 6. ( 3x2 − 14x − 5 ) ÷ ( x − 5 ) 7. ( x3 + x2 + x + 2 ) ÷ ( x2 − 1 ) 8. ( 7x3 + x2 + x ) ÷ ( x2 + 1 ) 9. ( 5x4 − 2x3 − 7x2 − 39 ) ÷ ( x2 + 2x − 4 ) 10. ( 4x4 + 5x − 4 ) ÷ ( x2 − 3x − 2 ) In Exercises 11–18, divide using synthetic division. (See Examples 2 and 3.) 11. ( x2 + 8x + 1 ) ÷ ( x − 4 ) 12. ( 4x2 − 13x − 5 ) ÷ ( x − 2 ) 13. ( 2x2 − x + 7 ) ÷ ( x + 5 ) 14. ( x3 − 4x + 6 ) ÷ ( x + 3 ) 15. ( x2 + 9 ) ÷ ( x − 3 ) 16. ( 3x3 − 5x2 − 2 ) ÷ ( x − 1 ) 17. ( x4 − 5x3 − 8x2 + 13x − 12 ) ÷ ( x − 6 ) 18. ( x4 + 4x3 + 16x − 35 ) ÷ ( x + 5 ) ANALYZING RELATIONSHIPS In Exercises 19–22, match the equivalent expressions. Justify your answers. 19. ( x2 + x − 3 ) ÷ ( x − 2 ) 20. ( x2 − x − 3 ) ÷ ( x − 2 ) 21. ( x2 − x + 3 ) ÷ ( x − 2 ) 22. ( x2 + x + 3 ) ÷ ( x − 2 ) A. x + 1 − 1 — x − 2 B. x + 3 + 9 — x − 2 C. x + 1 + 5 — x − 2 D. x + 3 + 3 — x − 2 ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in using synthetic division to divide x3 − 5x + 3 by x − 2.
23. 2 1 0 −5 3 2 4 −2 1 2 −1 1 x3 − 5x + 3 —— x − 2 = x3 + 2x2 − x + 1 ✗ 24. 2 1 −5 3 2 −6 1 −3 −3 x3 − 5x + 3 —— x − 2 = x2 − 3x − 3 — x − 2 ✗ Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics 1. WRITING Explain the Remainder Theorem in your own words. Use an example in your explanation. 2. VOCABULARY What form must the divisor have to make synthetic division an appropriate method for dividing a polynomial? Provide examples to support your claim. 3. VOCABULARY Write the polynomial divisor, dividend, and quotient functions −3 1 −2 −9 18 represented by the synthetic division shown at the right. −3 15 −18 4. WRITING Explain what the colored numbers represent in the 1 −5 6 0 synthetic division in Exercise 3.
Vocabulary and Core Concept Check Vocabulary and Core Concept Check Dynamic Solutions available at BigIdeasMath.com 178 Chapter 4 Polynomial Functions In Exercises 25–32, use synthetic division to evaluate the function for the indicated value of x. (See Example 4.) 25. f(x) = −x2 − 8x + 30; x = −1 26. f(x) = 3x2 + 2x − 20; x = 3 27. f(x) = x3 − 2x2 + 4x + 3; x = 2 28. f(x) = x3 + x2 − 3x + 9; x = −4 29. f(x) = x3 − 6x + 1; x = 6 30. f(x) = x3 − 9x − 7; x = 10 31. f(x) = x4 + 6x2 − 7x + 1; x = 3 32. f(x) = −x4 − x3 − 2; x = 5 33. MAKING AN ARGUMENT You use synthetic division to divide f(x) by (x −a) and fi nd that the remainder equals 15. Your friend concludes that f(15) = a. Is your friend correct? Explain your reasoning.
34. THOUGHT PROVOKING A polygon has an area represented by A = 4x2 + 8x + 4. The fi gure has at least one dimension equal to 2x + 2. Draw the fi gure and label its dimensions.
35. USING TOOLS The total attendance A (in thousands) at NCAA women’s basketball games and the number T of NCAA women’s basketball teams over a period of time can be modeled by A = −1.95x3 + 70.1x2 − 188x + 2150 T = 14.8x + 725 where x is in years and 0 < x < 18. Write a function for the average attendance per team over this period of time.
36. COMPARING METHODS The profi t P (in millions of dollars) for a DVD manufacturer can be modeled by P = −6x3 + 72x, where x is the number (in millions) of DVDs produced. Use synthetic division to show that the company yields a profi t of $96 million when 2 million DVDs are produced. Is there an easier method? Explain.
37. CRITICAL THINKING What is the value of k such that ( x3 − x2 + kx − 30 ) ÷ ( x − 5 ) has a remainder of zero?
○ A −14 ○ B −2 ○ C 26 ○ D 32 38. HOW DO YOU SEE IT? The graph represents the polynomial function f(x) = x3 + 3x2 − x − 3.
x y 10 −20 −10 4 2 −2 −4 a. The expression f(x) ÷ (x − k) has a remainder of −15. What is the value of k?
b. Use the graph to compare the remainders of ( x3 + 3x2 − x − 3 ) ÷ ( x + 3 ) and ( x3 + 3x2 − x − 3 ) ÷ ( x + 1 ) .
39. MATHEMATICAL CONNECTIONS The volume V of the rectangular prism is given by V = 2x3 + 17x2 + 46x + 40. Find an expression for the missing dimension.
x + 4 ?
x + 2 40. USING STRUCTURE You divide two polynomials and obtain the result 5x2 − 13x + 47 − 102 — x + 2 . What is the dividend? How did you fi nd it?
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Find the zero(s) of the function. (Sections 3.1 and 3.2) 41. f(x) = x2 − 6x + 9 42. g(x) = 3(x + 6)(x − 2) 43. g(x) = x2 + 14x + 49 44. h(x) = 4x2 + 36 Reviewing what you learned in previous grades and lessons Section 4.4 Factoring Polynomials 179 Factoring Polynomials 4.4 Factoring Polynomials Work with a partner. Match each polynomial equation with the graph of its related polynomial function. Use the x-intercepts of the graph to write each polynomial in factored form. Explain your reasoning. a. x2 + 5x + 4 = 0 b. x3 − 2x2 − x + 2 = 0 c. x3 + x2 − 2x = 0 d. x3 − x = 0 e. x4 − 5x2 + 4 = 0 f. x4 − 2x3 − x2 + 2x = 0 A. 6 −4 −6 4 B. 6 −4 −6 4 C. 6 −4 −6 4 D. 6 −4 −6 4 E. 6 −4 −6 4 F. 6 −4 −6 4 Factoring Polynomials Work with a partner. Use the x-intercepts of the graph of the polynomial function to write each polynomial in factored form. Explain your reasoning. Check your answers by multiplying.
a. f(x) = x2 − x − 2 b. f(x) = x3 − x2 − 2x c. f(x) = x3 − 2x2 − 3x d. f(x) = x3 − 3x2 − x + 3 e. f(x) = x 4 + 2x3 − x2 − 2x f. f(x) = x 4 − 10x2 + 9 Communicate Your Answer Communicate Your Answer 3. How can you factor a polynomial? 4. What information can you obtain about the graph of a polynomial function written in factored form?
MAKING SENSE OF PROBLEMS To be profi cient in math, you need to check your answers to problems and continually ask yourself, “Does this make sense?” Essential Question Essential Question How can you factor a polynomial?
180 Chapter 4 Polynomial Functions 4.4 Lesson What You Will Learn What You Will Learn Factor polynomials. Use the Factor Theorem.
Factoring Polynomials Previously, you factored quadratic polynomials. You can also factor polynomials with degree greater than 2. Some of these polynomials can be factored completely using techniques you have previously learned. A factorable polynomial with integer coeffi cients is factored completely when it is written as a product of unfactorable polynomials with integer coeffi cients. Finding a Common Monomial Factor Factor each polynomial completely.
a. x3 − 4x2 − 5x b. 3y5 − 48y3 c. 5z4 + 30z3 + 45z2 SOLUTION a. x3 − 4x2 − 5x = x(x2 − 4x − 5) Factor common monomial. = x(x − 5)(x + 1) Factor trinomial.
b. 3y5 − 48y3 = 3y3(y2 − 16) Factor common monomial. = 3y3(y − 4)(y + 4) Difference of Two Squares Pattern c. 5z4 + 30z3 + 45z2 = 5z2(z2 + 6z + 9) Factor common monomial. = 5z2(z + 3)2 Perfect Square Trinomial Pattern Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Factor the polynomial completely. 1. x3 − 7x2 + 10x 2. 3n7 − 75n5 3. 8m5 − 16m4 + 8m3 In part (b) of Example 1, the special factoring pattern for the difference of two squares was used to factor the expression completely. There are also factoring patterns that you can use to factor the sum or difference of two cubes.
factored completely, p. 180 factor by grouping, p. 181 quadratic form, p. 181 Previous zero of a function synthetic division Core Vocabulary Core Vocabulary Core Core Concept Concept Special Factoring Patterns Sum of Two Cubes Example a3 + b3 = (a + b)(a2 − ab + b2) 64x3 + 1 = (4x)3 + 13 = (4x + 1)(16x2 − 4x + 1) Difference of Two Cubes Example a3 − b3 = (a − b)(a2 + ab + b2) 27x3 − 8 = (3x)3 − 23 = (3x − 2)(9x2 + 6x + 4) Section 4.4 Factoring Polynomials 181 Factoring the Sum or Difference of Two Cubes Factor (a) x3 − 125 and (b) 16s5 + 54s2 completely.
SOLUTION a. x3 − 125 = x3 − 53 Write as a3 − b3.
= (x − 5)(x2 + 5x + 25) Difference of Two Cubes Pattern b. 16s5 + 54s2 = 2s2(8s3 + 27) Factor common monomial.
= 2s2 [(2s)3 + 33] Write 8s3 + 27 as a3 + b3.
= 2s2(2s + 3)(4s2 − 6s + 9) Sum of Two Cubes Pattern For some polynomials, you can factor by grouping pairs of terms that have a common monomial factor. The pattern for factoring by grouping is shown below. ra + rb + sa + sb = r(a + b) + s(a + b) = (r + s)(a + b) Factoring by Grouping Factor z3 + 5z2 − 4z − 20 completely.
SOLUTION z3 + 5z2 − 4z − 20 = z2(z + 5) − 4(z + 5) Factor by grouping.
= (z2 − 4)(z + 5) Distributive Property = (z − 2)(z + 2)(z + 5) Difference of Two Squares Pattern An expression of the form au2 + bu + c, where u is an algebraic expression, is said to be in quadratic form. The factoring techniques you have studied can sometimes be used to factor such expressions. Factoring Polynomials in Quadratic Form Factor (a) 16x4 − 81 and (b) 3p8 + 15p5 + 18p2 completely.
SOLUTION a. 16x4 − 81 = (4x2)2 − 92 Write as a2 − b2. = (4x2 + 9)(4x2 − 9) Difference of Two Squares Pattern = (4x2 + 9)(2x − 3)(2x + 3) Difference of Two Squares Pattern b. 3p8 + 15p5 + 18p2 = 3p2( p6 + 5p3 + 6) Factor common monomial. = 3p2( p3 + 3)( p3 + 2) Factor trinomial in quadratic form.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Factor the polynomial completely. 4. a3 + 27 5. 6z5 − 750z2 6. x3 + 4x2 − x − 4 7. 3y3 + y2 + 9y + 3 8. −16n4 + 625 9. 5w6 − 25w4 + 30w2 LOOKING FOR STRUCTURE The expression 16x 4 − 81 is in quadratic form because it can be written as u2 − 81 where u = 4x2.
182 Chapter 4 Polynomial Functions Determining Whether a Linear Binomial Is a Factor Determine whether (a) x − 2 is a factor of f(x) = x2 + 2x − 4 and (b) x + 5 is a factor of f(x) = 3x4 + 15x3 − x2 + 25.
SOLUTION a. Find f(2) by direct substitution. b. Find f(−5) by synthetic division. f(2) = 22 + 2(2) − 4 −5 3 15 −1 0 25 = 4 + 4 − 4 −15 0 5 −25 = 4 3 0 −1 5 0 Because f(2) ≠ 0, the binomial Because f(−5) = 0, the binomial x − 2 is not a factor of x + 5 is a factor of f(x) = x2 + 2x − 4. f(x) = 3x4 + 15x3 − x2 + 25. Factoring a Polynomial Show that x + 3 is a factor of f(x) = x4 + 3x3 − x − 3. Then factor f(x) completely.
SOLUTION Show that f(−3) = 0 by synthetic division. −3 1 3 0 −1 −3 −3 0 0 3 1 0 0 −1 0 Because f(−3) = 0, you can conclude that x + 3 is a factor of f(x) by the Factor Theorem. Use the result to write f(x) as a product of two factors and then factor completely. f(x) = x4 + 3x3 − x − 3 Write original polynomial. = (x + 3)(x3 − 1) Write as a product of two factors. = (x + 3)(x − 1)(x2 + x + 1) Difference of Two Cubes Pattern STUDY TIP In part (b), notice that direct substitution would have resulted in more diffi cult computations than synthetic division.
The Factor Theorem When dividing polynomials in the previous section, the examples had nonzero remainders. Suppose the remainder is 0 when a polynomial f(x) is divided by x − k. Then, f(x) — x − k = q(x) + 0 — x − k = q(x) where q(x) is the quotient polynomial. Therefore, f(x) = (x − k) ⋅ q(x), so that x − k is a factor of f(x). This result is summarized by the Factor Theorem, which is a special case of the Remainder Theorem.
Core Core Concept Concept The Factor Theorem A polynomial f(x) has a factor x − k if and only if f(k) = 0.
READING In other words, x − k is a factor of f (x) if and only if k is a zero of f.
ANOTHER WAY Notice that you can factor f (x) by grouping.
f (x) = x3(x + 3) − 1(x + 3) = (x3 − 1)(x + 3) = (x + 3)(x − 1) ⋅ (x2 + x + 1) Section 4.4 Factoring Polynomials 183 Because the x-intercepts of the graph of a function are the zeros of the function, you can use the graph to approximate the zeros. You can check the approximations using the Factor Theorem. Real-Life Application During the fi rst 5 seconds of a roller coaster ride, the function h(t) = 4t 3 − 21t 2 + 9t + 34 represents the height h (in feet) of the roller coaster after t seconds. How long is the roller coaster at or below ground level in the fi rst 5 seconds?
SOLUTION 1. Understand the Problem You are given a function rule that represents the height of a roller coaster. You are asked to determine how long the roller coaster is at or below ground during the fi rst 5 seconds of the ride.
2. Make a Plan Use a graph to estimate the zeros of the function and check using the Factor Theorem. Then use the zeros to describe where the graph lies below the t-axis.
3. Solve the Problem From the graph, two of the zeros appear to be −1 and 2. The third zero is between 4 and 5.
Step 1 Determine whether −1 is a zero using synthetic division. −1 4 −21 9 34 −4 25 −34 4 −25 34 0 Step 2 Determine whether 2 is a zero. If 2 is also a zero, then t − 2 is a factor of the resulting quotient polynomial. Check using synthetic division.
2 4 −25 34 8 −34 4 −17 0 So, h(t) = (t + 1)(t − 2)(4t − 17). The factor 4t − 17 indicates that the zero between 4 and 5 is 17 — 4 , or 4.25. The zeros are −1, 2, and 4.25. Only t = 2 and t = 4.25 occur in the fi rst 5 seconds. The graph shows that the roller coaster is at or below ground level for 4.25 − 2 = 2.25 seconds.
4. Look Back Use a table of values to verify the positive zeros and heights between the zeros.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 10. Determine whether x − 4 is a factor of f(x) = 2x2 + 5x − 12. 11. Show that x − 6 is a factor of f(x) = x3 − 5x2 − 6x. Then factor f(x) completely. 12. In Example 7, does your answer change when you fi rst determine whether 2 is a zero and then whether −1 is a zero? Justify your answer. STUDY TIP You could also check that 2 is a zero using the original function, but using the quotient polynomial helps you fi nd the remaining factor.
The remainder is 0, so t − 2 is a factor of h(t) and 2 is a zero of h.
h(−1) = 0, so −1 is a zero of h and t + 1 is a factor of h(t).
D f h H l S 1 2 3 t h 80 40 1 h(t) = 4t3 − 21t2 + 9t + 34 5 X Y1 X=2 33.75 20.25 0 -16.88 -20.25 0 54 1.25 .5 2 2.75 3.5 4.25 5 negative zero zero 184 Chapter 4 Polynomial Functions Exercises 4.4 Dynamic Solutions available at BigIdeasMath.com In Exercises 5–12, factor the polynomial completely. (See Example 1.) 5. x3 − 2x2 − 24x 6. 4k5 − 100k3 7. 3p5 − 192p3 8. 2m6 − 24m5 + 64m4 9. 2q4 + 9q3 − 18q2 10. 3r6 − 11r5 − 20r4 11. 10w10 − 19w9 + 6w8 12. 18v9 + 33v8 + 14v7 In Exercises 13–20, factor the polynomial completely. (See Example 2.) 13. x3 + 64 14. y3 + 512 15. g3 − 343 16. c3 − 27 17. 3h9 − 192h6 18. 9n6 − 6561n3 19. 16t 7 + 250t4 20. 135z11 − 1080z8 ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in factoring the polynomial.
21. 3x3 + 27x = 3x(x2 + 9) = 3x(x + 3)(x − 3) ✗ 22. x9 + 8x3 = (x3)3 + (2x)3 = (x3 + 2x)[(x3)2 − (x3)(2x) + (2x)2] = (x3 + 2x)(x6 − 2x4 + 4x2) ✗ In Exercises 23–30, factor the polynomial completely. (See Example 3.) 23. y3 − 5y2 + 6y − 30 24. m3 − m2 + 7m − 7 25. 3a3 + 18a2 + 8a + 48 26. 2k3 − 20k2 + 5k − 50 27. x3 − 8x2 − 4x + 32 28. z3 − 5z2 − 9z + 45 29. 4q3 − 16q2 − 9q + 36 30. 16n3 + 32n2 − n − 2 In Exercises 31–38, factor the polynomial completely. (See Example 4.) 31. 49k4 − 9 32. 4m4 − 25 33. c4 + 9c2 + 20 34. y4 − 3y2 − 28 35. 16z4 − 81 36. 81a4 − 256 37. 3r8 + 3r5 − 60r2 38. 4n12 − 32n7 + 48n2 In Exercises 39–44, determine whether the binomial is a factor of the polynomial. (See Example 5.) 39. f(x) = 2x3 + 5x2 − 37x − 60; x − 4 40. g(x) = 3x3 − 28x2 + 29x + 140; x + 7 41. h(x) = 6x5 − 15x4 − 9x3; x + 3 42. g(x) = 8x5 − 58x4 + 60x3 + 140; x − 6 43. h(x) = 6x4 − 6x3 − 84x2 + 144x; x + 4 44. t(x) = 48x4 + 36x3 − 138x2 − 36x; x + 2 Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics Vocabulary and Core Concept Check 1. COMPLETE THE SENTENCE The expression 9x4 − 49 is in _ form because it can be written as u2 − 49 where u = _____. 2. VOCABULARY Explain when you should try factoring a polynomial by grouping. 3. WRITING How do you know when a polynomial is factored completely? 4. WRITING Explain the Factor Theorem and why it is useful.
p Section 4.4 Factoring Polynomials 185 In Exercises 45–50, show that the binomial is a factor of the polynomial. Then factor the polynomial completely. (See Example 6.) 45. g(x) = x3 − x2 − 20x; x + 4 46. t(x) = x3 − 5x2 − 9x + 45; x − 5 47. f(x) = x4 − 6x3 − 8x + 48; x − 6 48. s(x) = x4 + 4x3 − 64x − 256; x + 4 49. r(x) = x3 − 37x + 84; x + 7 50. h(x) = x3 − x2 − 24x − 36; x + 2 ANALYZING RELATIONSHIPS In Exercises 51–54, match the function with the correct graph. Explain your reasoning. 51. f(x) = (x − 2)(x − 3)(x + 1) 52. g(x) = x(x + 2)(x + 1)(x − 2) 53. h(x) = (x + 2)(x + 3)(x − 1) 54. k(x) = x(x − 2)(x − 1)(x + 2) A. x y 4 4 −4 B. x y 4 4 −4 C. x y 4 −4 4 −4 D. x y 6 4 −4 55. MODELING WITH MATHEMATICS The volume (in cubic inches) of a shipping box is modeled by V = 2x3 − 19x2 + 39x, where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning. (See Example 7.) x V 40 20 4 2 8 6 56. MODELING WITH MATHEMATICS The volume (in cubic inches) of a rectangular birdcage can be modeled by V = 3x3 − 17x2 + 29x − 15, where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning.
x V 2 −4 −2 4 −2 USING STRUCTURE In Exercises 57–64, use the method of your choice to factor the polynomial completely. Explain your reasoning. 57. a6 + a5 − 30a4 58. 8m3 − 343 59. z3 − 7z2 − 9z + 63 60. 2p8 − 12p5 + 16p2 61. 64r3 + 729 62. 5x5 − 10x4 − 40x3 63. 16n4 − 1 64. 9k3 − 24k2 + 3k − 8 65. REASONING Determine whether each polynomial is factored completely. If not, factor completely. a. 7z4(2z2 − z − 6) b. (2 − n)(n2 + 6n)(3n − 11) c. 3(4y − 5)(9y2 − 6y − 4) 66. PROBLEM SOLVING The profi t P (in millions of dollars) for a T-shirt manufacturer can be modeled by P = −x3 + 4x2 + x, where x is the number (in millions) of T-shirts produced. Currently the company produces 4 million T-shirts and makes a profi t of $4 million. What lesser number of T-shirts could the company produce and still make the same profi t? 67. PROBLEM SOLVING The profi t P (in millions of dollars) for a shoe manufacturer can be modeled by P = −21x3 + 46x, where x is the number (in millions) of shoes produced. The company now produces 1 million shoes and makes a profi t of $25 million, but it would like to cut back production. What lesser number of shoes could the company produce and still make the same profi t?
186 Chapter 4 Polynomial Functions 68. THOUGHT PROVOKING Find a value of k such that f(x) — x − k has a remainder of 0. Justify your answer. f(x) = x3 − 3x2 − 4x 69. COMPARING METHODS You are taking a test where calculators are not permitted. One question asks you to evaluate g(7) for the function g(x) = x3 − 7x2 − 4x + 28. You use the Factor Theorem and synthetic division and your friend uses direct substitution. Whose method do you prefer? Explain your reasoning. 70. MAKING AN ARGUMENT You divide f(x) by (x − a) and fi nd that the remainder does not equal 0. Your friend concludes that f(x) cannot be factored. Is your friend correct? Explain your reasoning. 71. CRITICAL THINKING What is the value of k such that x − 7 is a factor of h(x) = 2x3 − 13x2 − kx + 105? Justify your answer. 72. HOW DO YOU SEE IT? Use the graph to write an equation of the cubic function in factored form. Explain your reasoning.
x y 4 −4 −2 4 −4 73. ABSTRACT REASONING Factor each polynomial completely. a. 7ac2 + bc2 −7ad 2 − bd 2 b. x2n − 2x n + 1 c. a5b2 − a2b4 + 2a4b − 2ab3 + a3 − b2 74. REASONING The graph of the function f(x) = x4 + 3x3 + 2x2 + x + 3 is shown. Can you use the Factor Theorem to factor f(x)? Explain.
75. MATHEMATICAL CONNECTIONS The standard equation of a circle with radius r and center (h, k) is (x − h)2 + (y − k)2 = r2. Rewrite each equation of a circle in standard form. Identify the center and radius of the circle. Then graph the circle.
x y (h, k) (x, y) r a. x2 + 6x + 9 + y2 = 25 b. x2 − 4x + 4 + y2 = 9 c. x2 − 8x + 16 + y2 + 2y + 1 = 36 76. CRITICAL THINKING Use the diagram to complete parts (a)–(c).
a. Explain why a3 − b3 is equal to the sum of the volumes of the solids I, II, and III.
b. Write an algebraic expression for the volume of each of the three solids. Leave your expressions in factored form.
c. Use the results from part (a) and part (b) to derive the factoring pattern a3 − b3.
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Solve the quadratic equation by factoring. (Section 3.1) 77. x2 − x − 30 = 0 78. 2x2 − 10x − 72 = 0 79. 3x2 − 11x + 10 = 0 80. 9x2 − 28x + 3 = 0 Solve the quadratic equation by completing the square. (Section 3.3) 81. x2 − 12x + 36 = 144 82. x2 − 8x − 11 = 0 83. 3x2 + 30x + 63 = 0 84. 4x2 + 36x − 4 = 0 Reviewing what you learned in previous grades and lessons x y 4 2 −4 −2 4 2 −2 −4 III II I b b b a a a 187 187 4.1–4.4 What Did You Learn?
Core Vocabulary Core Vocabulary polynomial, p. 158 polynomial function, p. 158 end behavior, p. 159 Pascal’s Triangle, p. 169 polynomial long division, p. 174 synthetic division, p. 175 factored completely, p. 180 factor by grouping, p. 181 quadratic form, p. 181 Core Concepts Core Concepts Section 4.1 Common Polynomial Functions, p. 158 End Behavior of Polynomial Functions, p. 159 Graphing Polynomial Functions, p. 160 Section 4.2 Operations with Polynomials, p. 166 Special Product Patterns, p. 167 Pascal’s Triangle, p. 169 Section 4.3 Polynomial Long Division, p. 174 Synthetic Division, p. 175 The Remainder Theorem, p. 176 Section 4.4 Factoring Polynomials, p. 180 Special Factoring Patterns, p. 180 The Factor Theorem, p. 182 Mathematical Practices Mathematical Practices 1. Describe the entry points you used to analyze the function in Exercise 43 on page 164.
2. Describe how you maintained oversight in the process of factoring the polynomial in Exercise 49 on page 185.
• When you sit down at your desk, review your notes from the last class.
• Repeat in your mind what you are writing in your notes.
• When a mathematical concept is particularly diffi cult, ask your teacher for another example.
Study Skills Keeping Your Mind Focused 188 Chapter 4 Polynomial Functions 4.1–4.4 Quiz Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coeffi cient. (Section 4.1) 1. f(x) = 5 + 2x2 − 3x4 − 2x − x3 2. g(x) = 1 — 4 x3 + 2x − 3x2 + 1 3. h(x) = 3 − 6x3 + 4x−2 + 6x 4.
Describe the x-values for which (a) f is increasing or decreasing, (b) f(x) > 0, and (c) f(x) < 0. (Section 4.1) 5.
Write an expression for the area and perimeter for the fi gure shown. (Section 4.2) Perform the indicated operation. (Section 4.2) 6. (7x2 − 4) − (3x2 − 5x + 1) 7. (x2 − 3x + 2)(3x − 1) 8. (x − 1)(x + 3)(x − 4) 9. Use Pascal’s Triangle to expand (x + 2)5. (Section 4.2) 10. Divide 4x4 − 2x3 + x2 − 5x + 8 by x2 − 2x − 1. (Section 4.3) Factor the polynomial completely. (Section 4.4) 11. a3 − 2a2 − 8a 12. 8m3 + 27 13. z3 + z2 − 4z − 4 14. 49b4 − 64 15. Show that x + 5 is a factor of f(x) = x3 − 2x2 − 23x + 60. Then factor f(x) completely. (Section 4.4) 16. The estimated price P (in cents) of stamps in the United States can be modeled by the polynomial function P(t) = 0.007t3 − 0.16t2 + 1t + 17, where t represents the number of years since 1990. (Section 4.1) a. Use a graphing calculator to graph the function for the interval 0 ≤ t ≤ 20. Describe the behavior of the graph on this interval.
b. What was the average rate of change in the price of stamps from 1990 to 2010? 17. The volume V (in cubic feet) of a rectangular wooden crate is modeled by the function V(x) = 2x3 − 11x2 + 12x, where x is the width (in feet) of the crate. Determine the values of x for which the model makes sense. Explain your reasoning. (Section 4.4) x y 4 2 −4 −2 4 6 2 −2 (2, 3) (3, 0) (1, 0) f x V 4 −2 V(x) = 2x3 − 11x2 + 12x x + 1 x + 3 x x Section 4.5 Solving Polynomial Equations 189 Solving Polynomial Equations 4.5 Cubic Equations and Repeated Solutions Work with a partner. Some cubic equations have three distinct solutions. Others have repeated solutions. Match each cubic polynomial equation with the graph of its related polynomial function. Then solve each equation. For those equations that have repeated solutions, describe the behavior of the related function near the repeated zero using the graph or a table of values.
a. x3 − 6x2 + 12x − 8 = 0 b. x3 + 3x2 + 3x + 1 = 0 c. x3 − 3x + 2 = 0 d. x3 + x2 − 2x = 0 e. x3 − 3x − 2 = 0 f. x3 − 3x2 + 2x = 0 A.
6 −4 −6 4 B.
6 −4 −6 4 C.
6 −4 −6 4 D.
6 −4 −6 4 E.
6 −4 −6 4 F.
6 −4 −6 4 Quartic Equations and Repeated Solutions Work with a partner. Determine whether each quartic equation has repeated solutions using the graph of the related quartic function or a table of values. Explain your reasoning. Then solve each equation.
a. x4 − 4x3 + 5x2 − 2x = 0 b. x4 − 2x3 −x2 + 2x = 0 c. x4 − 4x3 + 4x2 = 0 d. x4 + 3x3 = 0 Communicate Your Answer Communicate Your Answer 3. How can you determine whether a polynomial equation has a repeated solution? 4. Write a cubic or a quartic polynomial equation that is different from the equations in Explorations 1 and 2 and has a repeated solution.
USING TOOLS STRATEGICALLY To be profi cient in math, you need to use technological tools to explore and deepen your understanding of concepts.
Essential Question Essential Question How can you determine whether a polynomial equation has a repeated solution?
190 Chapter 4 Polynomial Functions 4.5 Lesson What You Will Learn What You Will Learn Find solutions of polynomial equations and zeros of polynomial functions. Use the Rational Root Theorem. Use the Irrational Conjugates Theorem.
Finding Solutions and Zeros You have used the Zero-Product Property to solve factorable quadratic equations. You can extend this technique to solve some higher-degree polynomial equations. Solving a Polynomial Equation by Factoring Solve 2x3 − 12x2 + 18x = 0.
SOLUTION 2x3 − 12x2 + 18x = 0 Write the equation. 2x(x2 − 6x + 9) = 0 Factor common monomial. 2x(x − 3)2 = 0 Perfect Square Trinomial Pattern 2x = 0 or (x − 3)2 = 0 Zero-Product Property x = 0 or x = 3 Solve for x. The solutions, or roots, are x = 0 and x = 3.
In Example 1, the factor x − 3 appears more than once. This creates a repeated solution of x = 3. Note that the graph of the related function touches the x-axis (but does not cross the x-axis) at the repeated zero x = 3, and crosses the x-axis at the zero x = 0. This concept can be generalized as follows.
• When a factor x − k of f(x) is raised to an odd power, the graph of f crosses the x-axis at x = k.
• When a factor x − k of f(x) is raised to an even power, the graph of f touches the x-axis (but does not cross the x-axis) at x = k. Finding Zeros of a Polynomial Function Find the zeros of f(x) = −2x4 + 16x2 − 32. Then sketch a graph of the function.
SOLUTION 0 = −2x4 + 16x2 − 32 Set f(x) equal to 0. 0 = −2(x4 − 8x2 + 16) Factor out −2. 0 = −2(x2 − 4)(x2 − 4) Factor trinomial in quadratic form. 0 = −2(x + 2)(x − 2)(x + 2)(x − 2) Difference of Two Squares Pattern 0 = −2(x + 2)2(x − 2)2 Rewrite using exponents.
Because both factors x + 2 and x − 2 are raised to an even power, the graph of f touches the x-axis at the zeros x = −2 and x = 2. By analyzing the original function, you can determine that the y-intercept is −32. Because the degree is even and the leading coeffi cient is negative, f(x) → −∞ as x → −∞ and f(x) → −∞ as x → +∞. Use these characteristics to sketch a graph of the function.
Check STUDY TIP Because the factor x − 3 appears twice, the root x = 3 has a multiplicity of 2.
repeated solution, p. 190 Previous roots of an equation real numbers conjugates Core Vocabulary Core Vocabulary 6 −6 −2 12 Zero X=3 Y=0 x y −40 4 −4 (2, 0) (−2, 0) (0, −32) Section 4.5 Solving Polynomial Equations 191 Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Solve the equation. 1. 4x4 − 40x2 + 36 = 0 2. 2x5 + 24x = 14x3 Find the zeros of the function. Then sketch a graph of the function.
3. f(x) = 3x4 − 6x2 + 3 4. f(x) = x3 + x2 − 6x The Rational Root Theorem The solutions of the equation 64x3 + 152x2 − 62x − 105 = 0 are − 5 — 2 , − 3 — 4 , and 7 — 8 . Notice that the numerators (5, 3, and 7) of the zeros are factors of the constant term, −105. Also notice that the denominators (2, 4, and 8) are factors of the leading coeffi cient, 64. These observations are generalized by the Rational Root Theorem.
The Rational Root Theorem can be a starting point for fi nding solutions of polynomial equations. However, the theorem lists only possible solutions. In order to fi nd the actual solutions, you must test values from the list of possible solutions. Using the Rational Root Theorem Find all real solutions of x3 − 8x2 + 11x + 20 = 0.
SOLUTION The polynomial f(x) = x3 − 8x2 + 11x + 20 is not easily factorable. Begin by using the Rational Root Theorem.
Step 1 List the possible rational solutions. The leading coeffi cient of f(x) is 1 and the constant term is 20. So, the possible rational solutions of f(x) = 0 are x = ± 1 — 1 , ± 2 — 1 , ± 4 — 1 , ± 5 — 1 , ± 10 — 1 , ± 20 — 1 .
Step 2 Test possible solutions using synthetic division until a solution is found. Test x = 1: Test x = −1: 1 1 −8 11 20 −1 1 −8 11 20 1 −7 4 −1 9 −20 1 −7 4 24 1 −9 20 0 Step 3 Factor completely using the result of the synthetic division. (x + 1)(x2 − 9x + 20) = 0 Write as a product of factors. (x + 1)(x − 4)(x − 5) = 0 Factor the trinomial. So, the solutions are x = −1, x = 4, and x = 5.
STUDY TIP Notice that you can use the Rational Root Theorem to list possible zeros of polynomial functions.
ANOTHER WAY You can use direct substitution to test possible solutions, but synthetic division helps you identify other factors of the polynomial.
f(1) ≠ 0, so x − 1 is not a factor of f(x).
f(−1) = 0, so x + 1 is a factor of f(x).
Core Core Concept Concept The Rational Root Theorem If f(x) = an x n + ∙ ∙ ∙ + a1x + a0 has integer coeffi cients, then every rational solution of f(x) = 0 has the following form: p — q = factor of constant term a0 ——— factor of leading coeffi cient an 192 Chapter 4 Polynomial Functions In Example 3, the leading coeffi cient of the polynomial is 1. When the leading coeffi cient is not 1, the list of possible rational solutions or zeros can increase dramatically. In such cases, the search can be shortened by using a graph. Finding Zeros of a Polynomial Function Find all real zeros of f(x) = 10x4 − 11x3 − 42x2 + 7x + 12.
SOLUTION Step 1 List the possible rational zeros of f : ± 1 — 1 , ± 2 — 1 , ± 3 — 1 , ± 4 — 1 , ± 6 — 1 , ± 12 — 1 , ± 1 — 2 , ± 3 — 2 , ± 1 — 5 , ± 2 — 5 , ± 3 — 5 , ± 4 — 5 , ± 6 — 5 , ± 12 — 5 , ± 1 — 10 , ± 3 — 10 Step 2 Choose reasonable values from the list above to test using the graph of the function. For f, the values x = − 3 — 2 , x = − 1 — 2 , x = 3 — 5 , and x = 12 — 5 are reasonable based on the graph shown at the right.
Step 3 Test the values using synthetic division until a zero is found. − 3 — 2 10 −11 −42 7 12 − 1 — 2 10 −11 −42 7 12 −15 39 9 — 2 − 69 — 4 −5 8 17 −12 10 −26 −3 23 — 2 − 21 — 4 10 −16 −34 24 0 − 1 — 2 is a zero.
Step 4 Factor out a binomial using the result of the synthetic division. f(x) = ( x + 1 — 2 ) (10x3 − 16x2 − 34x + 24) Write as a product of factors. = ( x + 1 — 2 ) (2)(5x3 − 8x2 − 17x + 12) Factor 2 out of the second factor. = (2x + 1)(5x3 − 8x2 − 17x + 12) Multiply the fi rst factor by 2.
Step 5 Repeat the steps above for g(x) = 5x3 − 8x2 − 17x +12. Any zero of g will also be a zero of f. The possible rational zeros of g are: x = ±1, ±2, ±3, ±4, ±6, ±12, ± 1 — 5 , ± 2 — 5 , ± 3 — 5 , ± 4 — 5 , ± 6 — 5 , ± 12 — 5 The graph of g shows that 3 — 5 may be a zero. Synthetic division shows that 3 — 5 is a zero and g(x) = ( x − 3 — 5 ) (5x2 − 5x − 20) = (5x − 3)(x2 − x − 4). It follows that: f(x) = (2x + 1) ⋅ g(x) = (2x + 1)(5x − 3)(x2 − x − 4) Step 6 Find the remaining zeros of f by solving x2 − x − 4 = 0. x = −(−1) ± √—— (−1)2 − 4(1)(−4) ——— 2(1) Substitute 1 for a, −1 for b, and −4 for c in the Quadratic Formula. x = 1 ± √ — 17 — 2 Simplify. The real zeros of f are − 1 — 2 , 3 — 5 , 1 + √ — 17 — 2 ≈ 2.56, and 1 − √ — 17 — 2 ≈ −1.56.
5 −100 −5 100 f 5 −25 −5 25 g Section 4.5 Solving Polynomial Equations 193 Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 5. Find all real solutions of x3 − 5x2 − 2x + 24 = 0. 6. Find all real zeros of f(x) = 3x4 − 2x3 − 37x2 + 24x + 12. The Irrational Conjugates Theorem In Example 4, notice that the irrational zeros are conjugates of the form a + √ — b and a − √ — b . This illustrates the theorem below. Using Zeros to Write a Polynomial Function Write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the zeros 3 and 2 + √ — 5 .
SOLUTION Because the coeffi cients are rational and 2 + √ — 5 is a zero, 2 − √ — 5 must also be a zero by the Irrational Conjugates Theorem. Use the three zeros and the Factor Theorem to write f(x) as a product of three factors. f(x) = (x − 3) [ x − ( 2 + √ — 5 ) ] [ x − ( 2 − √ — 5 ) ] Write f(x) in factored form. = (x − 3) [ (x − 2) − √ — 5 ] [ (x − 2) + √ — 5 ] Regroup terms.
= (x − 3) [ (x − 2)2 − 5 ] Multiply.
= (x − 3) [ (x2 − 4x + 4) − 5 ] Expand binomial. = (x − 3)(x2 − 4x − 1) Simplify. = x3 − 4x2 − x − 3x2 + 12x + 3 Multiply. = x3 − 7x2 + 11x + 3 Combine like terms.
Check You can check this result by evaluating f at each of its three zeros.
f(3) = 33 − 7(3)2 + 11(3) + 3 = 27 − 63 + 33 + 3 = 0 ✓ f ( 2 + √ — 5 ) = ( 2 + √ — 5 ) 3 − 7 ( 2 + √ — 5 ) 2 + 11 ( 2 + √ — 5 ) + 3 = 38 + 17 √ — 5 − 63 − 28 √ — 5 + 22 + 11 √ — 5 + 3 = 0 ✓ Because f ( 2 + √ — 5 ) = 0, by the Irrational Conjugates Theorem f ( 2 − √ — 5 ) = 0. ✓ Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 7. Write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the zeros 4 and 1 − √ — 5 . Core Core Concept Concept The Irrational Conjugates Theorem Let f be a polynomial function with rational coeffi cients, and let a and b be rational numbers such that √ — b is irrational. If a + √ — b is a zero of f, then a − √ — b is also a zero of f.
194 Chapter 4 Polynomial Functions Exercises 4.5 Dynamic Solutions available at BigIdeasMath.com In Exercises 3–12, solve the equation. (See Example 1.) 3. z3 − z2 − 12z = 0 4. a3 − 4a2 + 4a = 0 5. 2x4 − 4x3 = −2x2 6. v3 − 2v2 − 16v = − 32 7. 5w3 = 50w 8. 9m5 = 27m3 9. 2c4 − 6c3 = 12c2 − 36c 10. p4 + 40 = 14p2 11. 12n2 + 48n = −n3 − 64 12. y3 − 27 = 9y2 − 27y In Exercises 13–20, fi nd the zeros of the function. Then sketch a graph of the function. (See Example 2.) 13. h(x) = x4 + x3 − 6x2 14. f(x) = x4 − 18x2 + 81 15. p(x) = x6 − 11x5 + 30x4 16. g(x) = −2x5 + 2x4 + 40x3 17. g(x) = −4x4 + 8x3 + 60x2 18. h(x) = −x3 − 2x2 + 15x 19. h(x) = −x3 − x2 + 9x + 9 20. p(x) = x3 − 5x2 − 4x + 20 21. USING EQUATIONS According to the Rational Root Theorem, which is not a possible solution of the equation 2x4 − 5x3 + 10x2 − 9 = 0?
○ A −9 ○ B − 1 — 2 ○ C 5 — 2 ○ D 3 22. USING EQUATIONS According to the Rational Root Theorem, which is not a possible zero of the function f(x) = 40x5 − 42x4 − 107x3 + 107x2 + 33x − 36?
○ A − 2 — 3 ○ B − 3 — 8 ○ C 3 — 4 ○ D 4 — 5 ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in listing the possible rational zeros of the function. 23. f(x) = x3 + 5x2 − 9x − 45 Possible rational zeros of f : 1, 3, 5, 9, 15, 45 ✗ 24. f(x) = 3x3 + 13x2 − 41x + 8 Possible rational zeros of f: ±1, ±3, ± 1 — 2 , ± 1 — 4 , ± 1 — 8 , ± 3 — 2 , ± 3 — 4 , ± 3 — 8 ✗ In Exercises 25–32, fi nd all the real solutions of the equation. (See Example 3.) 25. x3 + x2 − 17x + 15 = 0 26. x3 − 2x2 − 5x + 6 = 0 Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics Vocabulary and Core Concept Check 1. COMPLETE THE SENTENCE If a polynomial function f has integer coeffi cients, then every rational solution of f(x) = 0 has the form p — q , where p is a factor of the _____________ and q is a factor of the _____________. 2. DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
Find all the real solutions of x3 − 2x2 − x + 2 = 0.
Find the x-intercepts of the graph of y = x3 − 2x2 − x + 2.
Find the y-intercept of the graph of y = x3 − 2x2 − x + 2.
Find the real zeros of f(x) = x3 − 2x2 − x + 2.
p Section 4.5 Solving Polynomial Equations 195 27. x3 − 10x2 + 19x + 30 = 0 28. x3 + 4x2 − 11x − 30 = 0 29. x3 − 6x2 − 7x + 60 = 0 30. x3 − 16x2 + 55x + 72 = 0 31. 2x3 − 3x2 − 50x − 24 = 0 32. 3x3 + x2 − 38x + 24 = 0 In Exercises 33–38, fi nd all the real zeros of the function. (See Example 4.) 33. f(x) = x3 − 2x2 − 23x + 60 34. g(x) = x3 − 28x − 48 35. h(x) = x3 + 10x2 + 31x + 30 36. f(x) = x3 − 14x2 + 55x − 42 37. p(x) = 2x3 − x2 − 27x + 36 38. g(x) = 3x3 − 25x2 + 58x − 40 USING TOOLS In Exercises 39 and 40, use the graph to shorten the list of possible rational zeros of the function. Then fi nd all real zeros of the function. 39. f(x) = 4x3 − 20x + 16 40. f(x) = 4x3 − 49x − 60 x y −20 4 2 −4 40 x y −80 −120 2 −4 In Exercises 41–46, write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the given zeros. (See Example 5.) 41. −2, 3, 6 42. −4, −2, 5 43. −2, 1 + √ — 7 44. 4, 6 − √ — 7 45. −6, 0, 3 − √ — 5 46. 0, 5, −5 + √ — 8 47. COMPARING METHODS Solve the equation x3 − 4x2 − 9x + 36 = 0 using two different methods. Which method do you prefer? Explain your reasoning.
48. REASONING Is it possible for a cubic function to have more than three real zeros? Explain.
49. PROBLEM SOLVING At a factory, molten glass is poured into molds to make paperweights. Each mold is a rectangular prism with a height 3 centimeters greater than the length of each side of its square base. Each mold holds 112 cubic centimeters of glass. What are the dimensions of the mold?
50. MATHEMATICAL CONNECTIONS The volume of the cube shown is 8 cubic centimeters.
a. Write a polynomial equation that you can use to fi nd the value of x.
b. Identify the possible rational solutions of the equation in part (a).
c. Use synthetic division to fi nd a rational solution of the equation. Show that no other real solutions exist.
d. What are the dimensions of the cube?
51. PROBLEM SOLVING Archaeologists discovered a huge hydraulic concrete block at the ruins of Caesarea with a volume of 945 cubic meters. The block is x meters high by 12x − 15 meters long by 12x − 21 meters wide. What are the dimensions of the block? 52. MAKING AN ARGUMENT Your friend claims that when a polynomial function has a leading coeffi cient of 1 and the coeffi cients are all integers, every possible rational zero is an integer. Is your friend correct? Explain your reasoning.
53. MODELING WITH MATHEMATICS During a 10-year period, the amount (in millions of dollars) of athletic equipment E sold domestically can be modeled by E(t) = −20t 3 + 252t 2 − 280t + 21,614, where t is in years. a. Write a polynomial equation to fi nd the year when about $24,014,000,000 of athletic equipment is sold. b. List the possible whole-number solutions of the equation in part (a). Consider the domain when making your list of possible solutions. c. Use synthetic division to fi nd when $24,014,000,000 of athletic equipment is sold.
x − 3 x − 3 x − 3 196 Chapter 4 Polynomial Functions 54. THOUGHT PROVOKING Write a third or fourth degree polynomial function that has zeros at ± 3 — 4 . Justify your answer. 55. MODELING WITH MATHEMATICS You are designing a marble basin that will hold a fountain for a city park. The sides and bottom of the basin should be 1 foot thick. Its outer length should be twice its outer width and outer height. What should the outer dimensions of the basin be if it is to hold 36 cubic feet of water?
x x 2x 1 ft 56. HOW DO YOU SEE IT? Use the information in the graph to answer the questions.
x y 4 6 2 4 2 −2 −4 f a. What are the real zeros of the function f ? b. Write an equation of the quartic function in factored form. 57. REASONING Determine the value of k for each equation so that the given x-value is a solution. a. x3 − 6x2 − 7x + k = 0; x = 4 b. 2x3 + 7x2 − kx − 18 = 0; x = −6 c. kx3 − 35x2 + 19x + 30 = 0; x = 5 58. WRITING EQUATIONS Write a polynomial function g of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the zeros −2 + √ — 7 and 3 + √ — 2 .
In Exercises 59–62, solve f(x) = g(x) by graphing and algebraic methods.
59. f(x) = x3 + x2 − x − 1; g(x) = −x + 1 60. f(x) = x4 − 5x3 + 2x2 + 8x; g(x) = −x2 + 6x − 8 61. f(x) = x3 − 4x2 + 4x; g(x) = −2x + 4 62. f(x) = x4 + 2x3 − 11x2 − 12x + 36; g(x) = −x2 − 6x − 9 63. MODELING WITH MATHEMATICS You are building a pair of ramps for a loading platform. The left ramp is twice as long as the right ramp. If 150 cubic feet of concrete are used to build the ramps, what are the dimensions of each ramp?
21x + 6 3x x 3x 64. MODELING WITH MATHEMATICS Some ice sculptures are made by fi lling a mold and then freezing it. You are making an ice mold for a school dance. It is to be shaped like a pyramid with a height 1 foot greater than the length of each side of its square base. The volume of the ice sculpture is 4 cubic feet. What are the dimensions of the mold? 65. ABSTRACT REASONING Let an be the leading coeffi cient of a polynomial function f and a0 be the constant term. If an has r factors and a0 has s factors, what is the greatest number of possible rational zeros of f that can be generated by the Rational Zero Theorem? Explain your reasoning.
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coeffi cient. (Section 4.1) 66. h(x) = −3x2 + 2x − 9 + √ — 4 x3 67. g(x) =2x3 − 7x2 − 3x−1 + x 68. f(x) = 1 — 3 x2 + 2x3 − 4x4 − √ — 3 69. p(x) = 2x − 5x3 + 9x2 + 4 √ — x + 1 Find the zeros of the function. (Section 3.2) 70. f(x) = 7x2 + 42 71. g(x) = 9x2 + 81 72. h(x) = 5x2 + 40 73. f(x) = 8x2 − 1 Reviewing what you learned in previous grades and lessons x x + 1 x Section 4.6 The Fundamental Theorem of Algebra 197 The Fundamental Theorem of Algebra 4.6 Cubic Equations and Imaginary Solutions Work with a partner. Match each cubic polynomial equation with the graph of its related polynomial function. Then fi nd all solutions. Make a conjecture about how you can use a graph or table of values to determine the number and types of solutions of a cubic polynomial equation.
a. x3 − 3x2 + x + 5 = 0 b. x3 − 2x2 −x + 2 = 0 c. x3 −x2 − 4x + 4 = 0 d. x3 + 5x2 + 8x + 6 = 0 e. x3 − 3x2 + x − 3 = 0 f. x3 − 3x2 + 2x = 0 A.
6 −6 −6 2 B. 6 −2 −6 6 C.
6 −4 −6 4 D. 6 −2 −6 6 E.
6 −4 −6 4 F. 6 −2 −6 6 Quartic Equations and Imaginary Solutions Work with a partner. Use the graph of the related quartic function, or a table of values, to determine whether each quartic equation has imaginary solutions. Explain your reasoning. Then fi nd all solutions.
a. x4 − 2x3 −x2 + 2x = 0 b. x4 − 1 = 0 c. x4 + x3 −x − 1 = 0 d. x4 − 3x3 + x2 + 3x − 2 = 0 Communicate Your Answer Communicate Your Answer 3. How can you determine whether a polynomial equation has imaginary solutions? 4. Is it possible for a cubic equation to have three imaginary solutions? Explain your reasoning.
USING TOOLS STRATEGICALLY To be profi cient in math, you need to use technology to enable you to visualize results and explore consequences.
Essential Question Essential Question How can you determine whether a polynomial equation has imaginary solutions?
198 Chapter 4 Polynomial Functions 4.6 Lesson What You Will Learn What You Will Learn Use the Fundamental Theorem of Algebra. Find conjugate pairs of complex zeros of polynomial functions. Use Descartes’s Rule of Signs.
The Fundamental Theorem of Algebra The table shows several polynomial equations and their solutions, including repeated solutions. Notice that for the last equation, the repeated solution x = −1 is counted twice.
Equation Degree Solution(s) Number of solutions 2x − 1 = 0 1 1 — 2 1 x2 − 2 = 0 2 ± √ — 2 2 x3 − 8 = 0 3 2, −1 ± i √ — 3 3 x3 + x2 − x − 1 = 0 3 −1, −1, 1 3 In the table, note the relationship between the degree of the polynomial f(x) and the number of solutions of f(x) = 0. This relationship is generalized by the Fundamental Theorem of Algebra, fi rst proven by German mathematician Carl Friedrich Gauss (1777−1855).
complex conjugates, p. 199 Previous repeated solution degree of a polynomial solution of an equation zero of a function conjugates Core Vocabulary Core Vocabulary The corollary to the Fundamental Theorem of Algebra also means that an nth-degree polynomial function f has exactly n zeros. Finding the Number of Solutions or Zeros a. How many solutions does the equation x3 + 3x2 + 16x + 48 = 0 have?
b. How many zeros does the function f(x) = x4 + 6x3 + 12x2 + 8x have?
SOLUTION a. Because x3 + 3x2 + 16x + 48 = 0 is a polynomial equation of degree 3, it has three solutions. (The solutions are −3, 4i, and −4i.) b. Because f(x) = x4 + 6x3 + 12x2 + 8x is a polynomial function of degree 4, it has four zeros. (The zeros are −2, −2, −2, and 0.) STUDY TIP The statements “the polynomial equation f (x) = 0 has exactly n solutions” and “the polynomial function f has exactly n zeros” are equivalent.
Core Core Concept Concept The Fundamental Theorem of Algebra Theorem If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has at least one solution in the set of complex numbers.
Corollary If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has exactly n solutions provided each solution repeated twice is counted as two solutions, each solution repeated three times is counted as three solutions, and so on. Section 4.6 The Fundamental Theorem of Algebra 199 Finding the Zeros of a Polynomial Function Find all zeros of f(x) = x5 + x3 − 2x2 − 12x − 8.
SOLUTION Step 1 Find the rational zeros of f. Because f is a polynomial function of degree 5, it has fi ve zeros. The possible rational zeros are ±1, ±2, ±4, and ±8. Using synthetic division, you can determine that −1 is a zero repeated twice and 2 is also a zero. Step 2 Write f(x) in factored form. Dividing f(x) by its known factors x + 1, x + 1, and x − 2 gives a quotient of x2 + 4. So, f(x) = (x + 1)2(x − 2)(x2 + 4).
Step 3 Find the complex zeros of f. Solving x2 + 4 = 0, you get x = ±2i. This means x2 + 4 = (x + 2i )(x − 2i ). f(x) = (x + 1)2(x − 2)(x + 2i )(x − 2i ) From the factorization, there are fi ve zeros. The zeros of f are −1, −1, 2, −2i, and 2i. The graph of f and the real zeros are shown. Notice that only the real zeros appear as x-intercepts. Also, the graph of f touches the x-axis at the repeated zero x = −1 and crosses the x-axis at x = 2.
5 −25 −5 5 Zero X=-1 Y=0 5 −25 −5 5 Zero X=2 Y=0 Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 1. How many solutions does the equation x4 + 7x2 − 144 = 0 have? 2. How many zeros does the function f(x) = x3 − 5x2 − 8x + 48 have?
Find all zeros of the polynomial function. 3. f(x) = x3 + 7x2 + 16x + 12 4. f(x) = x5 − 3x4 + 5x3 − x2 − 6x + 4 Complex Conjugates Pairs of complex numbers of the forms a + bi and a − bi, where b ≠ 0, are called complex conjugates. In Example 2, notice that the zeros 2i and −2i are complex conjugates. This illustrates the next theorem.
STUDY TIP Notice that you can use imaginary numbers to write (x2 + 4) as (x + 2i )(x − 2i ). In general, (a2 + b2) = (a + bi )(a − bi ).
Core Core Concept Concept The Complex Conjugates Theorem If f is a polynomial function with real coeffi cients, and a + bi is an imaginary zero of f, then a − bi is also a zero of f.
200 Chapter 4 Polynomial Functions Using Zeros to Write a Polynomial Function Write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the zeros 2 and 3 + i.
SOLUTION Because the coeffi cients are rational and 3 + i is a zero, 3 − i must also be a zero by the Complex Conjugates Theorem. Use the three zeros and the Factor Theorem to write f(x) as a product of three factors.
f(x) = (x − 2)[x − (3 + i)][x − (3 − i)] Write f(x) in factored form. = (x − 2)[(x − 3) − i][(x − 3) + i] Regroup terms. = (x − 2)[(x − 3)2 − i2] Multiply. = (x − 2)[(x2 − 6x + 9) − (−1)] Expand binomial and use i2 = −1. = (x − 2)(x2 − 6x + 10) Simplify. = x3 − 6x2 + 10x − 2x2 + 12x − 20 Multiply. = x3 − 8x2 + 22x − 20 Combine like terms.
Check You can check this result by evaluating f at each of its three zeros.
f(2) = (2)3 − 8(2)2 + 22(2) − 20 = 8 − 32 + 44 − 20 = 0 ✓ f(3 + i) = (3 + i)3 − 8(3 + i)2 + 22(3 + i) − 20 = 18 + 26i − 64 − 48i + 66 + 22i − 20 = 0 ✓ Because f(3 + i) = 0, by the Complex Conjugates Theorem f(3 − i) = 0. ✓ Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the given zeros. 5. −1, 4i 6. 3, 1 + i √ — 5 7. √ — 2 , 1 − 3i 8. 2, 2i, 4 − √ — 6 Descartes’s Rule of Signs French mathematician René Descartes (1596−1650) found the following relationship between the coeffi cients of a polynomial function and the number of positive and negative zeros of the function.
Core Core Concept Concept Descartes’s Rule of Signs Let f(x) = anxn + an−1xn−1 + ⋅ ⋅ ⋅ + a2x2 + a1x + a0 be a polynomial function with real coeffi cients. • The number of positive real zeros of f is equal to the number of changes in sign of the coeffi cients of f(x) or is less than this by an even number.
• The number of negative real zeros of f is equal to the number of changes in sign of the coeffi cients of f(−x) or is less than this by an even number. Section 4.6 The Fundamental Theorem of Algebra 201 Using Descartes’s Rule of Signs Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for f(x) = x6 − 2x5 + 3x4 − 10x3 − 6x2 − 8x − 8.
SOLUTION f(x) = x6 − 2x5 + 3x4 − 10x3 − 6x2 − 8x − 8.
The coeffi cients in f(x) have 3 sign changes, so f has 3 or 1 positive real zero(s).
f(−x) = (−x)6 − 2(−x)5 + 3(−x)4 − 10(−x)3 − 6(−x)2 − 8(−x) − 8 = x6 + 2x5 + 3x4 + 10x3 − 6x2 + 8x − 8 The coeffi cients in f(−x) have 3 sign changes, so f has 3 or 1 negative zero(s). The possible numbers of zeros for f are summarized in the table below.
Positive real zeros Negative real zeros Imaginary zeros Total zeros 3 3 0 6 3 1 2 6 1 3 2 6 1 1 4 6 Real-Life Application A tachometer measures the speed (in revolutions per minute, or RPMs) at which an engine shaft rotates. For a certain boat, the speed x (in hundreds of RPMs) of the engine shaft and the speed s (in miles per hour) of the boat are modeled by s(x) = 0.00547x3 − 0.225x2 + 3.62x − 11.0.
What is the tachometer reading when the boat travels 15 miles per hour?
SOLUTION Substitute 15 for s(x) in the function. You can rewrite the resulting equation as 0 = 0.00547x3 − 0.225x2 + 3.62x − 26.0.
The related function to this equation is f(x) = 0.00547x3 − 0.225x2 + 3.62x − 26.0. By Descartes’s Rule of Signs, you know f has 3 or 1 positive real zero(s). In the context of speed, negative real zeros and imaginary zeros do not make sense, so you do not need to check for them. To approximate the positive real zeros of f, use a graphing calculator. From the graph, there is 1 real zero, x ≈ 19.9. The tachometer reading is about 1990 RPMs.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function. 9. f(x) = x3 + 9x − 25 10. f(x) = 3x4 − 7x3 + x2 − 13x + 8 11. WHAT IF? In Example 5, what is the tachometer reading when the boat travels 20 miles per hour? 10 0 20 20 30 40 50 0 60 70 70 80 RPM x100 40 −60 −10 40 Y=0 X=19.863247 Zero 202 Chapter 4 Polynomial Functions Exercises 4.6 Dynamic Solutions available at BigIdeasMath.com In Exercises 3–8, identify the number of solutions or zeros. (See Example 1.) 3. x4 + 2x3 − 4x2 + x = 0 4. 5y3 − 3y2 + 8y = 0 5. 9t6 − 14t3 + 4t − 1 = 0 6. f(z) = −7z4 + z2 − 25 7. g(s) = 4s5 − s3 + 2s7 − 2 8. h(x) = 5x4 + 7x8 − x12 In Exercises 9–16, fi nd all zeros of the polynomial function. (See Example 2.) 9. f(x) = x4 − 6x3 + 7x2 + 6x − 8 10. f(x) = x4 + 5x3 − 7x2 − 29x + 30 11. g(x) = x4 − 9x2 − 4x + 12 12. h(x) = x3 + 5x2 − 4x − 20 13. g(x) = x4 + 4x3 + 7x2 + 16x + 12 14. h(x) = x4 − x3 + 7x2 − 9x − 18 15. g(x) = x5 + 3x4 − 4x3 − 2x2 − 12x − 16 16. f(x) = x5 − 20x3 + 20x2 − 21x + 20 ANALYZING RELATIONSHIPS In Exercises 17–20, determine the number of imaginary zeros for the function with the given degree and graph. Explain your reasoning. 17. Degree: 4 18. Degree: 5 x y 40 20 −20 4 −4 x y 40 20 −20 4 2 −4 19. Degree: 2 20. Degree: 3 x y 6 −6 4 2 −2 −4 x y 40 20 −20 4 −2 −4 In Exercises 21–28, write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the given zeros. (See Example 3.) 21. −5, −1, 2 22.
−2, 1, 3 23. 3, 4 + i 24. 2, 5 − i 25. 4, − √ — 5 26. 3i, 2 − i 27. 2, 1 + i, 2 − √ — 3 28. 3, 4 + 2i, 1 + √ — 7 ERROR ANALYSIS In Exercises 29 and 30, describe and correct the error in writing a polynomial function with rational coeffi cients and the given zero(s).
29. Zeros: 2, 1 + i f(x) = (x − 2) [ x − (1 + i ) ] = x(x − 1 − i ) − 2(x − 1 − i ) = x2 − x − ix − 2x + 2 + 2i = x2 − (3 + i ) x + (2 + 2i) ✗ 30. Zero: 2 + i f(x) = [ x − (2 + i ) ] [ x + (2 + i ) ] = (x − 2 − i )(x + 2 + i ) = x2 + 2x + ix − 2x − 4 − 2i − ix − 2i − i2 = x2 − 4i − 3 ✗ Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics 1. COMPLETE THE SENTENCE The expressions 5 + i and 5 − i are _____________. 2. WRITING How many solutions does the polynomial equation (x + 8)3(x − 1) = 0 have? Explain.
Vocabulary and Core Concept Check Vocabulary and Core Concept Check Section 4.6 The Fundamental Theorem of Algebra 203 31. OPEN-ENDED Write a polynomial function of degree 6 with zeros 1, 2, and −i. Justify your answer.
32. REASONING Two zeros of f(x) = x3 − 6x2 − 16x + 96 are 4 and −4. Explain why the third zero must also be a real number.
In Exercises 33–40, determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function. (See Example 4.) 33. g(x) = x4 − x2 − 6 34. g(x) = −x3 + 5x2 + 12 35. g(x) = x3 − 4x2 + 8x + 7 36. g(x) = x5 − 2x3 − x2 + 6 37. g(x) = x5 − 3x3 + 8x − 10 38. g(x) = x5 + 7x4 − 4x3 − 3x2 + 9x − 15 39. g(x) = x6 + x5 − 3x4 + x3 + 5x2 + 9x − 18 40. g(x) = x7 + 4x4 − 10x + 25 41. REASONING Which is not a possible classifi cation of zeros for f(x) = x5 − 4x3 + 6x2 + 2x − 6? Explain.
○ A three positive real zeros, two negative real zeros, and zero imaginary zeros ○ B three positive real zeros, zero negative real zeros, and two imaginary zeros ○ C one positive real zero, four negative real zeros, and zero imaginary zeros ○ D one positive real zero, two negative real zeros, and two imaginary zeros 42. USING STRUCTURE Use Descartes’s Rule of Signs to determine which function has at least 1 positive real zero.
○ A f(x) = x4 + 2x3 − 9x2 − 2x − 8 ○ B f(x) = x4 + 4x3 + 8x2 + 16x + 16 ○ C f(x) = −x4 − 5x2 − 4 ○ D f(x) = x4 + 4x3 + 7x2 + 12x + 12 43. MODELING WITH MATHEMATICS From 1890 to 2000, the American Indian, Eskimo, and Aleut population P (in thousands) can be modeled by the function P = 0.004t3 − 0.24t2 + 4.9t + 243, where t is the number of years since 1890. In which year did the population fi rst reach 722,000? (See Example 5.) 44. MODELING WITH MATHEMATICS Over a period of 14 years, the number N of inland lakes infested with zebra mussels in a certain state can be modeled by N = −0.0284t4 + 0.5937t3 − 2.464t2 + 8.33t − 2.5 where t is time (in years). In which year did the number of infested inland lakes fi rst reach 120?
45. MODELING WITH MATHEMATICS For the 12 years that a grocery store has been open, its annual revenue R (in millions of dollars) can be modeled by the function R = 0.0001(−t 4 + 12t 3 − 77t 2 + 600t + 13,650) where t is the number of years since the store opened. In which year(s) was the revenue $1.5 million? 46. MAKING AN ARGUMENT Your friend claims that 2 − i is a complex zero of the polynomial function f(x) = x3 − 2x2 + 2x + 5i, but that its conjugate is not a zero. You claim that both 2 − i and its conjugate must be zeros by the Complex Conjugates Theorem. Who is correct? Justify your answer. 47. MATHEMATICAL CONNECTIONS A solid monument with the dimensions shown is to be built using 1000 cubic feet of marble. What is the value of x?
3 ft 3 ft 3 ft 3 ft 2x x x 2x 204 Chapter 4 Polynomial Functions 48. THOUGHT PROVOKING Write and graph a polynomial function of degree 5 that has all positive or negative real zeros. Label each x-intercept. Then write the function in standard form. 49. WRITING The graph of the constant polynomial function f(x) = 2 is a line that does not have any x-intercepts. Does the function contradict the Fundamental Theorem of Algebra? Explain. 50. HOW DO YOU SEE IT? The graph represents a polynomial function of degree 6. x y y = f(x) a. How many positive real zeros does the function have? negative real zeros? imaginary zeros? b. Use Descartes’s Rule of Signs and your answers in part (a) to describe the possible sign changes in the coeffi cients of f(x). 51. FINDING A PATTERN Use a graphing calculator to graph the function f(x) = (x + 3)n for n = 2, 3, 4, 5, 6, and 7. a. Compare the graphs when n is even and n is odd. b. Describe the behavior of the graph near the zero x = −3 as n increases. c. Use your results from parts (a) and (b) to describe the behavior of the graph of g(x) = (x − 4)20 near x = 4.
52. DRAWING CONCLUSIONS Find the zeros of each function. f(x) = x2 − 5x + 6 g(x) = x3 − 7x + 6 h(x) = x4 + 2x3 + x2 + 8x − 12 k(x) = x5 − 3x4 − 9x3 + 25x2 − 6x a. Describe the relationship between the sum of the zeros of a polynomial function and the coeffi cients of the polynomial function.
b. Describe the relationship between the product of the zeros of a polynomial function and the coeffi cients of the polynomial function.
53. PROBLEM SOLVING You want to save money so you can buy a used car in four years. At the end of each summer, you deposit $1000 earned from summer jobs into your bank account. The table shows the value of your deposits over the four-year period. In the table, g is the growth factor 1 + r, where r is the annual interest rate expressed as a decimal.
Deposit Year 1 Year 2 Year 3 Year 4 1st Deposit 1000 1000g 1000g2 1000g3 2nd Deposit − 1000 3rd Deposit − − 1000 4th Deposit − − − 1000 a. Copy and complete the table.
b. Write a polynomial function that gives the value v of your account at the end of the fourth summer in terms of g. c. You want to buy a car that costs about $4300. What growth factor do you need to obtain this amount? What annual interest rate do you need?
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Describe the transformation of f(x) = x2 represented by g. Then graph each function. (Section 2.1) 54. g(x) = −3x2 55. g(x) = (x − 4)2 + 6 56. g(x) = −(x − 1)2 57. g(x) = 5(x + 4)2 Write a function g whose graph represents the indicated transformation of the graph of f. (Sections 1.2 and 2.1) 58. f(x) = x; vertical shrink by a factor of 1 — 3 and a refl ection in the y-axis 59. f(x) = ∣ x + 1 ∣ − 3; horizontal stretch by a factor of 9 60. f(x) = x2; refl ection in the x-axis, followed by a translation 2 units right and 7 units up Reviewing what you learned in previous grades and lessons Section 4.7 Transformations of Polynomial Functions 205 Transformations of Polynomial Functions 4.7 Transforming the Graph of a Cubic Function Work with a partner. The graph of the cubic function f(x) = x3 is shown. The graph of each cubic function g represents a transformation of the graph of f. Write a rule for g. Use a graphing calculator to verify your answers.
a. 6 −4 −6 4 g b. 6 −4 −6 4 g c. 6 −4 −6 4 g d. 6 −4 −6 4 g Transforming the Graph of a Quartic Function Work with a partner. The graph of the quartic function f(x) = x4 is shown. The graph of each quartic function g represents a transformation of the graph of f. Write a rule for g. Use a graphing calculator to verify your answers.
a. 6 −4 −6 4 g b. 6 −4 −6 4 g Communicate Your Answer Communicate Your Answer 3. How can you transform the graph of a polynomial function? 4. Describe the transformation of f(x) = x4 represented by g(x) = (x + 1)4 + 3. Then graph g.
LOOKING FOR STRUCTURE To be profi cient in math, you need to see complicated things, such as some algebraic expressions, as being single objects or as being composed of several objects.
Essential Question Essential Question How can you transform the graph of a polynomial function?
6 −4 −6 4 f 6 −4 −6 4 f 206 Chapter 4 Polynomial Functions 4.7 Lesson What You Will Learn What You Will Learn Describe transformations of polynomial functions. Write transformations of polynomial functions.
Describing Transformations of Polynomial Functions You can transform graphs of polynomial functions in the same way you transformed graphs of linear functions, absolute value functions, and quadratic functions. Examples of transformations of the graph of f(x) = x4 are shown below.
Previous polynomial function transformations Core Vocabulary Core Vocabulary Translating a Polynomial Function Describe the transformation of f(x) = x3 represented by g(x) = (x + 5)3 + 2. Then graph each function.
SOLUTION Notice that the function is of the form g(x) = (x − h)3 + k. Rewrite the function to identify h and k.
g(x) = ( x − (−5) ) 3 + 2 h k Because h = −5 and k = 2, the graph of g is a translation 5 units left and 2 units up of the graph of f.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 1. Describe the transformation of f(x) = x4 represented by g(x) = (x − 3)4 − 1. Then graph each function. x y 4 2 −2 2 −2 −4 f g Core Core Concept Concept Transformation f(x) Notation Examples Horizontal Translation Graph shifts left or right.
f(x − h) g(x) = (x − 5)4 5 units right g(x) = (x + 2)4 2 units left Vertical Translation Graph shifts up or down.
f(x) + k g(x) = x4 + 1 1 unit up g(x) = x4 − 4 4 units down Refl ection Graph fl ips over x- or y-axis.
f(−x) −f(x) g(x) = (−x)4 = x4 over y-axis g(x) = −x4 over x-axis Horizontal Stretch or Shrink Graph stretches away from or shrinks toward y-axis.
f(ax) g(x) = (2x)4 shrink by a factor of 1 — 2 g(x) = ( 1 — 2 x ) 4 stretch by a factor of 2 Vertical Stretch or Shrink Graph stretches away from or shrinks toward x-axis.
a ⋅ f(x) g(x) = 8x4 stretch by a factor of 8 g(x) = 1 — 4 x4 shrink by a factor of 1 — 4 Section 4.7 Transformations of Polynomial Functions 207 Transforming Polynomial Functions Describe the transformation of f represented by g. Then graph each function.
a. f(x) = x4, g(x) = − 1 — 4 x4 b. f(x) = x5, g(x) = (2x)5 − 3 SOLUTION a. Notice that the function is of b. Notice that the function is of the form g(x) = −ax4, where the form g(x) = (ax)5 + k, where a = 1 — 4 . a = 2 and k = −3. So, the graph of g is a So, the graph of g is a refl ection in the x-axis and a horizontal shrink by a factor of vertical shrink by a factor of 1 — 2 and a translation 3 units down 1 — 4 of the graph of f. of the graph of f. x y 4 −4 2 −2 f g x y 2 2 −2 f g Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 2. Describe the transformation of f(x) = x3 represented by g(x) = 4(x + 2)3. Then graph each function. Writing Transformations of Polynomial Functions Writing Transformed Polynomial Functions Let f(x) = x3 + x2 + 1. Write a rule for g and then graph each function. Describe the graph of g as a transformation of the graph of f.
a. g(x) = f(−x) b. g(x) = 3f(x) SOLUTION a. g(x) = f(−x) b. g(x) = 3f(x) = (−x)3 + (−x)2 + 1 = 3(x3 + x2 + 1) = −x3 + x2 + 1 = 3x3 + 3x2 + 3 f g x y 4 −2 2 −2 f g x y 8 4 −4 2 The graph of g is a refl ection The graph of g is a vertical stretch in the y-axis of the graph of f. by a factor of 3 of the graph of f.
REMEMBER Vertical stretches and shrinks do not change the x-intercept(s) of a graph. You can observe this using the graph in Example 3(b). 208 Chapter 4 Polynomial Functions Writing a Transformed Polynomial Function Let the graph of g be a vertical stretch by a factor of 2, followed by a translation 3 units up of the graph of f(x) = x4 − 2x2. Write a rule for g.
SOLUTION Step 1 First write a function h that represents the vertical stretch of f.
h(x) = 2 ⋅ f(x) Multiply the output by 2.
= 2(x4 − 2x2) Substitute x4 − 2x2 for f(x).
= 2x4 − 4x2 Distributive Property Step 2 Then write a function g that represents the translation of h.
g(x) = h(x) + 3 Add 3 to the output.
= 2x4 − 4x2 + 3 Substitute 2x4 − 4x2 for h(x). The transformed function is g(x) = 2x4 − 4x2 + 3. Modeling with Mathematics The function V(x) = 1 — 3 x3 − x2 represents the volume (in cubic feet) of the square pyramid shown. The function W(x) = V(3x) represents the volume (in cubic feet) when x is measured in yards. Write a rule for W. Find and interpret W(10).
SOLUTION 1. Understand the Problem You are given a function V whose inputs are in feet and whose outputs are in cubic feet. You are given another function W whose inputs are in yards and whose outputs are in cubic feet. The horizontal shrink shown by W(x) = V(3x) makes sense because there are 3 feet in 1 yard. You are asked to write a rule for W and interpret the output for a given input.
2. Make a Plan Write the transformed function W(x) and then fi nd W(10).
3. Solve the Problem W(x) = V(3x) = 1 — 3 (3x)3 − (3x)2 Replace x with 3x in V(x). = 9x3 − 9x2 Simplify.
Next, fi nd W(10). W(10) = 9(10)3 − 9(10)2 = 9000 − 900 = 8100 When x is 10 yards, the volume of the pyramid is 8100 cubic feet.
4. Look Back Because W(10) = V(30), you can check that your solution is correct by verifying that V(30) = 8100. V(30) = 1 — 3 (30)3 − (30)2 = 9000 − 900 = 8100 ✓ Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 3. Let f(x) = x5 − 4x + 6 and g(x) = −f(x). Write a rule for g and then graph each function. Describe the graph of g as a transformation of the graph of f. 4. Let the graph of g be a horizontal stretch by a factor of 2, followed by a translation 3 units to the right of the graph of f(x) = 8x3 + 3. Write a rule for g. 5. WHAT IF? In Example 5, the height of the pyramid is 6x, and the volume (in cubic feet) is represented by V(x) = 2x3. Write a rule for W. Find and interpret W(7). Check 2 −3 −2 5 g h f x ft x ft (x − 3) ft Section 4.7 Transformations of Polynomial Functions 209 1. COMPLETE THE SENTENCE The graph of f(x) = (x + 2)3 is a ____________ translation of the graph of f(x) = x3. 2. VOCABULARY Describe how the vertex form of quadratic functions is similar to the form f(x) = a(x − h)3 + k for cubic functions.
Exercises 4.7 Vocabulary and Core Concept Check Vocabulary and Core Concept Check In Exercises 3–6, describe the transformation of f represented by g. Then graph each function. (See Example 1.) 3. f(x) = x4, g(x) = x4 + 3 4. f(x) = x4, g(x) = (x − 5)4 5. f(x) = x5, g(x) = (x − 2)5 − 1 6. f(x) = x6, g(x) = (x + 1)6 − 4 ANALYZING RELATIONSHIPS In Exercises 7–10, match the function with the correct transformation of the graph of f. Explain your reasoning.
x y f 7. y = f(x − 2) 8. y = f(x + 2) + 2 9. y = f(x − 2) + 2 10. y = f(x) − 2 A.
x y B.
x y C.
x y D. x y In Exercises 11–16, describe the transformation of f represented by g. Then graph each function. (See Example 2.) 11. f(x) = x4, g(x) = −2x4 12. f(x) = x6, g(x) = −3x6 13. f(x) = x3, g(x) = 5x3 + 1 14. f(x) = x4, g(x) = 1 — 2 x4 + 1 15. f(x) = x5, g(x) = 3 — 4 (x + 4)5 16. f(x) = x4, g(x) = (2x)4 − 3 In Exercises 17–20, write a rule for g and then graph each function. Describe the graph of g as a transformation of the graph of f. (See Example 3.) 17. f(x) = x4 + 1, g(x) = f(x + 2) 18. f(x) = x5 − 2x + 3, g(x) = 3f(x) 19. f(x) = 2x3 − 2x2 + 6, g(x) = − 1 — 2 f(x) 20. f(x) = x4 + x3 − 1, g(x) = f(−x) − 5 21. ERROR ANALYSIS Describe and correct the error in graphing the function g(x) = (x + 2)4 − 6.
x y 4 −4 2 4 ✗ Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics Dynamic Solutions available at BigIdeasMath.com 210 Chapter 4 Polynomial Functions 22. ERROR ANALYSIS Describe and correct the error in describing the transformation of the graph of f(x) = x5 represented by the graph of g(x) = (3x)5 − 4.
The graph of g is a horizontal shrink by a factor of 3, followed by a translation 4 units down of the graph of f.
✗ In Exercises 23–26, write a rule for g that represents the indicated transformations of the graph of f. (See Example 4.) 23. f(x) = x3 − 6; translation 3 units left, followed by a refl ection in the y-axis 24. f(x) = x4 + 2x + 6; vertical stretch by a factor of 2, followed by a translation 4 units right 25. f(x) = x3 + 2x2 − 9; horizontal shrink by a factor of 1 — 3 and a translation 2 units up, followed by a refl ection in the x-axis 26. f(x) = 2x5 − x3 + x2 + 4; refl ection in the y-axis and a vertical stretch by a factor of 3, followed by a translation 1 unit down 27. MODELING WITH MATHEMATICS The volume V (in cubic feet) of the pyramid is given by V(x) = x3 − 4x. The function W(x) = V(3x) gives the volume (in cubic feet) of the pyramid when x is measured in yards. Write a rule for W. Find and interpret W(5). (See Example 5.) 28. MAKING AN ARGUMENT The volume of a cube with side length x is given by V(x) = x3. Your friend claims that when you divide the volume in half, the volume decreases by a greater amount than when you divide each side length in half. Is your friend correct? Justify your answer. 29. OPEN-ENDED Describe two transformations of the graph of f(x) = x5 where the order in which the transformations are performed is important. Then describe two transformations where the order is not important. Explain your reasoning.
30. THOUGHT PROVOKING Write and graph a transformation of the graph of f(x) = x5 − 3x4 + 2x − 4 that results in a graph with a y-intercept of −2.
31. PROBLEM SOLVING A portion of the path that a hummingbird fl ies while feeding can be modeled by the function f(x) = − 1 — 5 x(x − 4)2(x − 7), 0 ≤ x ≤ 7 where x is the horizontal distance (in meters) and f(x) is the height (in meters). The hummingbird feeds each time it is at ground level. a. At what distances does the hummingbird feed? b. A second hummingbird feeds 2 meters farther away than the fi rst hummingbird and fl ies twice as high. Write a function to model the path of the second hummingbird. 32. HOW DO YOU SEE IT? Determine the real zeros of each function. Then describe the transformation of the graph of f that results in the graph of g.
33. MATHEMATICAL CONNECTIONS Write a function V for the volume (in cubic yards) of the right circular cone shown. Then write a function W that gives the volume (in cubic yards) of the cone when x is measured in feet. Find and interpret W(3).
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Find the minimum value or maximum value of the function. Describe the domain and range of the function, and where the function is increasing and decreasing. (Section 2.2) 34. h(x) = (x + 5)2 − 7 35. f(x) = 4 − x2 36. f(x) = 3(x − 10)(x + 4) 37. g(x) = −(x + 2)(x + 8) 38. h(x) = 1 — 2 (x − 1)2 − 3 39. f(x) = −2x2 + 4x − 1 Reviewing what you learned in previous grades and lessons (2x − 4) ft (3x + 6) ft x ft 3x yd (x + 3) yd x y 8 4 8 −4 −8 f g Section 4.8 Analyzing Graphs of Polynomial Functions 211 Analyzing Graphs of Polynomial Functions 4.8 Approximating Turning Points Work with a partner. Match each polynomial function with its graph. Explain your reasoning. Then use a graphing calculator to approximate the coordinates of the turning points of the graph of the function. Round your answers to the nearest hundredth.
a. f(x) = 2x2 + 3x − 4 b. f(x) = x2 + 3x + 2 c. f(x) = x3 − 2x2 − x + 1 d. f(x) = −x3 + 5x − 2 e. f(x) = x4 − 3x2 + 2x − 1 f. f(x) = −2x5 − x2 + 5x + 3 A. 6 −4 −6 4 B. 6 −6 −6 2 C. 6 −6 −6 2 D. 6 −7 −6 3 E. 6 −2 −6 6 F. 6 −4 −6 4 Communicate Your Answer Communicate Your Answer 2. How many turning points can the graph of a polynomial function have? 3. Is it possible to sketch the graph of a cubic polynomial function that has no turning points? Justify your answer.
ATTENDING TO PRECISION To be profi cient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.
Essential Question Essential Question How many turning points can the graph of a polynomial function have?
A turning point of the graph of a polynomial function is a point on the graph at which the function changes from • increasing to decreasing, or • decreasing to increasing.
x y 1 −1 1.5 1 turning point turning point 212 Chapter 4 Polynomial Functions 4.8 Lesson What You Will Learn What You Will Learn Use x-intercepts to graph polynomial functions. Use the Location Principle to identify zeros of polynomial functions. Find turning points and identify local maximums and local minimums of graphs of polynomial functions. Identify even and odd functions.
Graphing Polynomial Functions In this chapter, you have learned that zeros, factors, solutions, and x-intercepts are closely related concepts. Here is a summary of these relationships.
local maximum, p. 214 local minimum, p. 214 even function, p. 215 odd function, p. 215 Previous end behavior increasing decreasing symmetric about the y-axis Core Vocabulary Core Vocabulary Using x-Intercepts to Graph a Polynomial Function Graph the function f(x) = 1 — 6 (x + 3)(x − 2)2.
SOLUTION Step 1 Plot the x-intercepts. Because −3 and 2 are zeros of f, plot (−3, 0) and (2, 0).
Step 2 Plot points between and beyond the x-intercepts.
x −2 −1 0 1 3 y 8 — 3 3 2 2 — 3 1 Step 3 Determine end behavior. Because f(x) has three factors of the form x − k and a constant factor of 1 — 6 , f is a cubic function with a positive leading coeffi cient. So, f(x) → −∞ as x → −∞ and f(x) → +∞ as x → +∞.
Step 4 Draw the graph so that it passes through the plotted points and has the appropriate end behavior.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Graph the function. 1. f(x) = 1 — 2 (x + 1)(x − 4)2 2. f(x) = 1 — 4 (x + 2)(x − 1)(x − 3) Zeros, Factors, Solutions, and Intercepts Let f(x) = anxn + an−1xn−1 + ⋅ ⋅ ⋅ + a1x + a0 be a polynomial function.
The following statements are equivalent. Zero: k is a zero of the polynomial function f. Factor: x − k is a factor of the polynomial f(x). Solution: k is a solution (or root) of the polynomial equation f(x) = 0. x-Intercept: If k is a real number, then k is an x-intercept of the graph of the polynomial function f. The graph of f passes through (k, 0).
Concept Summary Concept Summary x y 4 2 −4 −2 4 −2 −4 (2, 0) (−3, 0) Section 4.8 Analyzing Graphs of Polynomial Functions 213 To use this principle to locate real zeros of a polynomial function, fi nd a value a at which the polynomial function is negative and another value b at which the function is positive. You can conclude that the function has at least one real zero between a and b. Locating Real Zeros of a Polynomial Function Find all real zeros of f(x) = 6x3 + 5x2 − 17x − 6.
SOLUTION Step 1 Use a graphing calculator to make a table.
Step 2 Use the Location Principle. From the table shown, you can see that f(1) < 0 and f(2) > 0. So, by the Location Principle, f has a zero between 1 and 2. Because f is a polynomial function of degree 3, it has three zeros. The only possible rational zero between 1 and 2 is 3 — 2 . Using synthetic division, you can confi rm that 3 — 2 is a zero.
Step 3 Write f(x) in factored form. Dividing f(x) by its known factor x − 3 — 2 gives a quotient of 6x2 + 14x + 4. So, you can factor f(x) as f(x) = ( x − 3 — 2 ) (6x2 + 14x + 4) = 2 ( x − 3 — 2 ) (3x2 + 7x + 2) = 2 ( x − 3 — 2 ) (3x + 1)(x + 2). From the factorization, there are three zeros. The zeros of f are 3 — 2 , − 1 — 3 , and −2. Check this by graphing f.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 3. Find all real zeros of f(x) = 18x3 + 21x2 − 13x − 6. Check The Location Principle You can use the Location Principle to help you fi nd real zeros of polynomial functions.
Core Core Concept Concept The Location Principle If f is a polynomial function, and a and b are two real numbers such that f(a) < 0 and f(b) > 0, then f has at least one real zero between a and b.
x y f(b) f(a) a b zero X Y1 X=1 -6 -12 28 150 390 784 1368 1 0 2 3 4 5 6 5 −20 −5 20 Zero X=1.5 Y=0 214 Chapter 4 Polynomial Functions Turning Points Another important characteristic of graphs of polynomial functions is that they have turning points corresponding to local maximum and minimum values.
• The y-coordinate of a turning point is a local maximum of the function when the point is higher than all nearby points.
• The y-coordinate of a turning point is a local minimum of the function when the point is lower than all nearby points.
The turning points of a graph help determine the intervals for which a function is increasing or decreasing.
READING Local maximum and local minimum are sometimes referred to as relative maximum and relative minimum. Finding Turning Points Graph each function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which each function is increasing or decreasing.
a. f(x) = x3 − 3x2 + 6 b. g(x) = x4 − 6x3 + 3x2 + 10x − 3 SOLUTION a. Use a graphing calculator to graph the function. The graph of f has one x-intercept and two turning points. Use the graphing calculator’s zero, maximum, and minimum features to approximate the coordinates of the points. The x-intercept of the graph is x ≈ −1.20. The function has a local maximum at (0, 6) and a local minimum at (2, 2). The function is increasing when x < 0 and x > 2 and decreasing when 0 < x < 2.
b. Use a graphing calculator to graph the function. The graph of g has four x-intercepts and three turning points. Use the graphing calculator’s zero, maximum, and minimum features to approximate the coordinates of the points. The x-intercepts of the graph are x ≈ −1.14, x ≈ 0.29, x ≈ 1.82, and x ≈ 5.03. The function has a local maximum at (1.11, 5.11) and local minimums at (−0.57, −6.51) and (3.96, −43.04). The function is increasing when −0.57 < x < 1.11 and x > 3.96 and decreasing when x < −0.57 and 1.11 < x < 3.96.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 4. Graph f(x) = 0.5x3 + x2 − x + 2. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing. Core Core Concept Concept Turning Points of Polynomial Functions 1. The graph of every polynomial function of degree n has at most n − 1 turning points.
2. If a polynomial function of degree n has n distinct real zeros, then its graph has exactly n − 1 turning points.
x y local maximum local minimum function is increasing function is decreasing function is increasing 5 −10 −3 25 Y=6 Maximum X=0 6 −70 −4 40 Y=-6.50858 Minimum X=-0.569071 Section 4.8 Analyzing Graphs of Polynomial Functions 215 Identifying Even and Odd Functions Determine whether each function is even, odd, or neither.
a. f(x) = x3 − 7x b. g(x) = x4 + x2 − 1 c. h(x) = x3 + 2 SOLUTION a. Replace x with −x in the equation for f, and then simplify. f(−x) = (−x)3 − 7(−x) = −x3 + 7x = −(x3 − 7x) = −f(x) Because f(−x) = −f(x), the function is odd.
b. Replace x with −x in the equation for g, and then simplify. g(−x) = (−x)4 + (−x)2 − 1 = x4 + x2 − 1 = g(x) Because g(−x) = g(x), the function is even.
c. Replacing x with −x in the equation for h produces h(−x) = (−x)3 + 2 = −x3 + 2. Because h(x) = x3 + 2 and −h(x) = −x3 − 2, you can conclude that h(−x) ≠ h(x) and h(−x) ≠ −h(x). So, the function is neither even nor odd.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Determine whether the function is even, odd, or neither. 5. f(x) = −x2 + 5 6. f(x) = x4 − 5x3 7. f(x) = 2x5 Even and Odd Functions Core Core Concept Concept Even and Odd Functions A function f is an even function when f(−x) = f(x) for all x in its domain. The graph of an even function is symmetric about the y-axis. A function f is an odd function when f(−x) = −f(x) for all x in its domain. The graph of an odd function is symmetric about the origin. One way to recognize a graph that is symmetric about the origin is that it looks the same after a 180° rotation about the origin. Even Function Odd Function x y (x, y) (−x, y) x y (x, y) (−x, −y) For an even function, if (x, y) is on the For an odd function, if (x, y) is on the graph, then (−x, y) is also on the graph. graph, then (−x, −y) is also on the graph.
216 Chapter 4 Polynomial Functions Exercises 4.8 Dynamic Solutions available at BigIdeasMath.com ANALYZING RELATIONSHIPS In Exercises 3–6, match the function with its graph.
3. f(x) = (x − 1)(x −2)(x + 2) 4. h(x) = (x + 2)2(x + 1) 5. g(x) = (x + 1)(x − 1)(x + 2) 6. f(x) = (x − 1)2(x + 2) A. x y 3 1 3 −1 −3 B. x y 2 −2 2 C. x y −2 2 D. x y 2 2 In Exercises 7–14, graph the function. (See Example 1.) 7. f(x) = (x − 2)2(x + 1) 8. f(x) = (x + 2)2(x + 4)2 9. h(x) = (x + 1)2(x − 1)(x − 3) 10. g(x) = 4(x + 1)(x + 2)(x − 1) 11. h(x) = 1 — 3 (x − 5)(x + 2)(x − 3) 12. g(x) = 1 — 12 (x + 4)(x + 8)(x − 1) 13. h(x) = (x − 3)(x2 + x + 1) 14. f(x) = (x − 4)(2x2 − 2x + 1) ERROR ANALYSIS In Exercises 15 and 16, describe and correct the error in using factors to graph f.
15. f(x) = (x + 2)(x − 1)2 ✗ 16. f(x) = x2(x − 3)3 ✗ In Exercises 17–22, fi nd all real zeros of the function. (See Example 2.) 17. f(x) = x3 − 4x2 − x + 4 18. f(x) = x3 − 3x2 − 4x + 12 19. h(x) = 2x3 + 7x2 − 5x − 4 20. h(x) = 4x3 − 2x2 − 24x − 18 21. g(x) = 4x3 + x2 − 51x + 36 22. f(x) = 2x3 − 3x2 − 32x − 15 Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics 1. COMPLETE THE SENTENCE A local maximum or local minimum of a polynomial function occurs at a ______________ point of the graph of the function. 2. WRITING Explain what a local maximum of a function is and how it may be different from the maximum value of the function.
x y −4 −2 4 −2 −4 x y 4 2 4 6 2 −2 Vocabulary and Core Concept Check Vocabulary and Core Concept Check Section 4.8 Analyzing Graphs of Polynomial Functions 217 In Exercises 23–30, graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.
(See Example 3.) 23. g(x) = 2x3 + 8x2 − 3 24. g(x) = −x4 + 3x 25. h(x) = x4 − 3x2 + x 26. f(x) = x5 − 4x3 + x2 + 2 27. f(x) = 0.5x3 − 2x + 2.5 28. f(x) = 0.7x4 − 3x3 + 5x 29. h(x) = x5 + 2x2 − 17x − 4 30. g(x) = x4 − 5x3 + 2x2 + x − 3 In Exercises 31–36, estimate the coordinates of each turning point. State whether each corresponds to a local maximum or a local minimum. Then estimate the real zeros and fi nd the least possible degree of the function. 31. x y 2 2 −2 32. x y 4 4 −4 33. x y −2 −4 −6 2 4 34. x y 10 2 −2 35. x y −4 2 36. x y 6 2 3 1 −1 −3 OPEN-ENDED In Exercises 37 and 38, sketch a graph of a polynomial function f having the given characteristics.
37. • The graph of f has x-intercepts at x = −4, x = 0, and x = 2.
• f has a local maximum value when x = 1. • f has a local minimum value when x = −2. 38. • The graph of f has x-intercepts at x = −3, x = 1, and x = 5. • f has a local maximum value when x = 1.
• f has a local minimum value when x = −2 and when x = 4.
In Exercises 39–46, determine whether the function is even, odd, or neither. (See Example 4.) 39. h(x) = 4x7 40. g(x) = −2x6 + x2 41. f(x) = x4 + 3x2 − 2 42. f(x) = x5 + 3x3 −x 43. g(x) = x2 + 5x + 1 44. f(x) = −x3 + 2x − 9 45. f(x) = x4 − 12x2 46. h(x) = x5 + 3x4 47. USING TOOLS When a swimmer does the breaststroke, the function S = −241t7 + 1060t 6 − 1870t 5 + 1650t 4 − 737t 3 + 144t 2 − 2.43t models the speed S (in meters per second) of the swimmer during one complete stroke, where t is the number of seconds since the start of the stroke and 0 ≤ t ≤ 1.22. Use a graphing calculator to graph the function. At what time during the stroke is the swimmer traveling the fastest? 48. USING TOOLS During a recent period of time, the number S (in thousands) of students enrolled in public schools in a certain country can be modeled by S = 1.64x3 − 102x2 + 1710x + 36,300, where x is time (in years). Use a graphing calculator to graph the function for the interval 0 ≤ x ≤ 41. Then describe how the public school enrollment changes over this period of time. 49. WRITING Why is the adjective local, used to describe the maximums and minimums of cubic functions, sometimes not required for quadratic functions?
218 Chapter 4 Polynomial Functions 50. HOW DO YOU SEE IT? The graph of a polynomial function is shown. x y 10 −10 2 −2 −4 y = f(x) a. Find the zeros, local maximum, and local minimum values of the function. b. Compare the x-intercepts of the graphs of y = f(x) and y = −f(x). c. Compare the maximum and minimum values of the functions y = f(x) and y = −f(x). 51. MAKING AN ARGUMENT Your friend claims that the product of two odd functions is an odd function. Is your friend correct? Explain your reasoning. 52. MODELING WITH MATHEMATICS You are making a rectangular box out of a 16-inch-by-20-inch piece of cardboard. The box will be formed by making the cuts shown in the diagram and folding up the sides. You want the box to have the greatest volume possible. 20 in.
16 in.
x x x x x x x x a. How long should you make the cuts? b. What is the maximum volume? c. What are the dimensions of the fi nished box?
53. PROBLEM SOLVING Quonset huts are temporary, all-purpose structures shaped like half-cylinders. You have 1100 square feet of material to build a quonset hut. a. The surface area S of a quonset hut is given by S = πr2 + πr ℓ . Substitute 1100 for S and then write an expression forℓ in terms of r.
b. The volume V of a quonset hut is given by V = 1 — 2 πr2 ℓ . Write an equation that gives V as a function in terms of r only.
c. Find the value of r that maximizes the volume of the hut.
54. THOUGHT PROVOKING Write and graph a polynomial function that has one real zero in each of the intervals −2 < x < −1, 0 < x < 1, and 4 < x < 5. Is there a maximum degree that such a polynomial function can have? Justify your answer.
55. MATHEMATICAL CONNECTIONS A cylinder is inscribed in a sphere of radius 8 inches. Write an equation for the volume of the cylinder as a function of h. Find the value of h that maximizes the volume of the inscribed cylinder. What is the maximum volume of the cylinder?
h 8 in.
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency State whether the table displays linear data, quadratic data, or neither. Explain. (Section 2.4) 56. Months, x 0 1 2 3 Savings (dollars), y 100 150 200 250 57. Time (seconds), x 0 1 2 3 Height (feet), y 300 284 236 156 Reviewing what you learned in previous grades and lessons Section 4.9 Modeling with Polynomial Functions 219 Modeling with Polynomial Functions 4.9 Modeling Real-Life Data Work with a partner. The distance a baseball travels after it is hit depends on the angle at which it was hit and the initial speed. The table shows the distances a baseball hit at an angle of 35° travels at various initial speeds.
Initial speed, x (miles per hour) 80 85 90 95 100 105 110 115 Distance, y (feet) 194 220 247 275 304 334 365 397 a. Recall that when data have equally-spaced x-values, you can analyze patterns in the differences of the y-values to determine what type of function can be used to model the data. If the fi rst differences are constant, then the set of data fi ts a linear model. If the second differences are constant, then the set of data fi ts a quadratic model. Find the fi rst and second differences of the data. Are the data linear or quadratic? Explain your reasoning. 194 220 247 275 304 334 365 397 b. Use a graphing calculator to draw a scatter plot of the data. Do the data appear linear or quadratic? Use the regression feature of the graphing calculator to fi nd a linear or quadratic model that best fi ts the data.
120 190 75 400 c. Use the model you found in part (b) to fi nd the distance a baseball travels when it is hit at an angle of 35° and travels at an initial speed of 120 miles per hour.
d. According to the Baseball Almanac, “Any drive over 400 feet is noteworthy. A blow of 450 feet shows exceptional power, as the majority of major league players are unable to hit a ball that far. Anything in the 500-foot range is genuinely historic.” Estimate the initial speed of a baseball that travels a distance of 500 feet.
Communicate Your Answer Communicate Your Answer 2. How can you fi nd a polynomial model for real-life data? 3. How well does the model you found in Exploration 1(b) fi t the data? Do you think the model is valid for any initial speed? Explain your reasoning.
USING TOOLS STRATEGICALLY To be profi cient in math, you need to use technological tools to explore and deepen your understanding of concepts.
Essential Question Essential Question How can you fi nd a polynomial model for real-life data?
220 Chapter 4 Polynomial Functions 4.9 Lesson What You Will Learn What You Will Learn Write polynomial functions for sets of points. Write polynomial functions using fi nite differences. Use technology to fi nd models for data sets.
Writing Polynomial Functions for a Set of Points You know that two points determine a line and three points not on a line determine a parabola. In Example 1, you will see that four points not on a line or a parabola determine the graph of a cubic function. Writing a Cubic Function Write the cubic function whose graph is shown.
SOLUTION Step 1 Use the three x-intercepts to write the function in factored form.
f(x) = a(x + 4)(x − 1)(x − 3) Step 2 Find the value of a by substituting the coordinates of the point (0, −6).
−6 = a(0 + 4)(0 − 1)(0 − 3) −6 = 12a − 1 — 2 = a The function is f(x) = − 1 — 2 (x + 4)(x − 1)(x − 3).
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Write a cubic function whose graph passes through the given points. 1. (−4, 0), (0, 10), (2, 0), (5, 0) 2. (−1, 0), (0, −12), (2, 0), (3, 0) Finite Differences When the x-values in a data set are equally spaced, the differences of consecutive y-values are called fi nite differences. Recall from Section 2.4 that the fi rst and second differences of y = x2 are: equally-spaced x-values fi rst differences: −5 −3 −1 1 3 5 second differences: 2 2 2 2 2 x −3 −2 −1 0 1 2 3 y 9 4 1 0 1 4 9 Notice that y = x2 has degree two and that the second differences are constant and nonzero. This illustrates the fi rst of the two properties of fi nite differences shown on the next page.
Check Check the end behavior of f. The degree of f is odd and a < 0. So, f(x) → +∞ as x → −∞ and f(x) → −∞ as x → +∞, which matches the graph. ✓ fi nite differences, p. 220 Previous scatter plot Core Vocabulary Core Vocabulary x y −4 −16 4 2 −2 (−4, 0) (3, 0) (1, 0) (0, −6) Section 4.9 Modeling with Polynomial Functions 221 The second property of fi nite differences allows you to write a polynomial function that models a set of equally-spaced data. Writing a Function Using Finite Differences Use fi nite differences to determine the degree of the polynomial function that fi ts the data. Then use technology to fi nd the polynomial function.
SOLUTION Step 1 Write the function values. Find the fi rst differences by subtracting consecutive values. Then fi nd the second differences by subtracting consecutive fi rst differences. Continue until you obtain differences that are nonzero and constant.
f(1) f(2) f(3) f(4) f(5) f(6) f(7) 1 4 10 20 35 56 84 3 6 10 15 21 28 3 4 5 6 7 1 1 1 1 Because the third differences are nonzero and constant, you can model the data exactly with a cubic function.
Step 2 Enter the data into a graphing calculator and use cubic regression to obtain a polynomial function. Because 1 — 6 ≈ 0.1666666667, 1 — 2 = 0.5, and 1 — 3 ≈ 0.333333333, a polynomial function that fi ts the data exactly is f(x) = 1 — 6 x3 + 1 — 2 x2 + 1 — 3 x.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 3. Use fi nite differences to determine the degree of the polynomial function that fi ts the data. Then use technology to fi nd the polynomial function. x 1 2 3 4 5 6 7 f (x) 1 4 10 20 35 56 84 x −3 −2 −1 0 1 2 f(x) 6 15 22 21 6 −29 Core Core Concept Concept Properties of Finite Differences 1. If a polynomial function y = f(x) has degree n, then the nth differences of function values for equally-spaced x-values are nonzero and constant.
2. Conversely, if the nth differences of equally-spaced data are nonzero and constant, then the data can be represented by a polynomial function of degree n.
y=ax 3+bx 2+cx+d b=.5 a=.1666666667 c=.3333333333 d=0 R 2=1 CubicReg Write function values for equally-spaced x-values.
First differences Second differences Third differences 222 Chapter 4 Polynomial Functions Finding Models Using Technology In Examples 1 and 2, you found a cubic model that exactly fi ts a set of data. In many real-life situations, you cannot fi nd models to fi t data exactly. Despite this limitation, you can still use technology to approximate the data with a polynomial model, as shown in the next example. Real-Life Application The table shows the total U.S. biomass energy consumptions y (in trillions of British thermal units, or Btus) in the year t, where t = 1 corresponds to 2001. Find a polynomial model for the data. Use the model to estimate the total U.S. biomass energy consumption in 2013.
t 1 2 3 4 5 6 y 2622 2701 2807 3010 3117 3267 t 7 8 9 10 11 12 y 3493 3866 3951 4286 4421 4316 SOLUTION Step 1 Enter the data into a graphing calculator and make a scatter plot. The data suggest a cubic model.
Step 2 Use the cubic regression feature. The polynomial model is y = −2.545t3 + 51.95t2 − 118.1t + 2732.
13 2500 0 4500 y=ax 3+bx 2+cx+d b=51.95376845 a=-2.545325045 c=-118.1139601 d=2732.141414 R 2=.9889472257 CubicReg Step 3 Check the model by graphing it and the data in the same viewing window.
Step 4 Use the trace feature to estimate the value of the model when t = 13.
13 2500 0 4500 14 2000 0 5000 X=13 Y=4384.7677 Y 1=-2.5453250453256x^3+ The approximate total U.S. biomass energy consumption in 2013 was about 4385 trillion Btus.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Use a graphing calculator to fi nd a polynomial function that fi ts the data. 4. x 1 2 3 4 5 6 y 5 13 17 11 11 56 5. x 0 2 4 6 8 10 y 8 0 15 69 98 87 According to the U.S. Department of Energy, biomass includes “agricultural and forestry residues, municipal solid wastes, industrial wastes, and terrestrial and aquatic crops grown solely for energy purposes.” Among the uses for biomass is production of electricity and liquid fuels such as ethanol.
T B a e S S Section 4.9 Modeling with Polynomial Functions 223 1. COMPLETE THE SENTENCE When the x-values in a set of data are equally spaced, the differences of consecutive y-values are called _______. 2. WRITING Explain how you know when a set of data could be modeled by a cubic function.
Exercises 4.9 Vocabulary and Core Concept Check Vocabulary and Core Concept Check In Exercises 3–6, write a cubic function whose graph is shown. (See Example 1.) 3. 4. x y 4 4 −2 −4 (1, 0) (2, 0) (0, 2) (−1, 0) x y 4 −8 4 −2 −4 (−2, 4) (2, 0) (−3, 0) (−1, 0) 5. 6. x y −8 −4 8 −8 (−5, 0) (2, −2) (4, 0) (1, 0) x y 4 8 4 −4 −8 (−3, 0) (−6, 0) (0, −9) (3, 0) In Exercises 7–12, use fi nite differences to determine the degree of the polynomial function that fi ts the data. Then use technology to fi nd the polynomial function. (See Example 2.) 7. x −6 −3 0 3 6 9 f(x) −2 15 −4 49 282 803 8. x −1 0 1 2 3 4 f(x) −14 −5 −2 7 34 91 9. (−4, −317), (−3, −37), (−2, 21), (−1, 7), (0, −1), (1, 3), (2, −47), (3, −289), (4, −933) 10. (−6, 744), (−4, 154), (−2, 4), (0, −6), (2, 16), (4, 154), (6, 684), (8, 2074), (10, 4984) 11. (−2, 968), (−1, 422), (0, 142), (1, 26), (2, −4), (3, −2), (4, 2), (5, 2), (6, 16) 12. (1, 0), (2, 6), (3, 2), (4, 6), (5, 12), (6, −10), (7, −114), (8, −378), (9, −904) 13. ERROR ANALYSIS Describe and correct the error in writing a cubic function whose graph passes through the given points. (−6, 0), (1, 0), (3, 0), (0, 54) 54 = a(0 − 6)(0 + 1)(0 + 3) 54 = −18a a = −3 f(x) = −3(x − 6)(x + 1)(x + 3) ✗ 14. MODELING WITH MATHEMATICS The dot patterns show pentagonal numbers. The number of dots in the nth pentagonal number is given by f(n) = 1 — 2 n(3n − 1). Show that this function has constant second-order differences.
15. OPEN-ENDED Write three different cubic functions that pass through the points (3, 0), (4, 0), and (2, 6). Justify your answers.
16. MODELING WITH MATHEMATICS The table shows the ages of cats and their corresponding ages in human years. Find a polynomial model for the data for the fi rst 8 years of a cat’s life. Use the model to estimate the age (in human years) of a cat that is 3 years old. (See Example 3.) Age of cat, x 1 2 4 6 7 8 Human years, y 15 24 32 40 44 48 Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics Dynamic Solutions available at BigIdeasMath.com 224 Chapter 4 Polynomial Functions 17. MODELING WITH MATHEMATICS The data in the table show the average speeds y (in miles per hour) of a pontoon boat for several different engine speeds x (in hundreds of revolutions per minute, or RPMs). Find a polynomial model for the data. Estimate the average speed of the pontoon boat when the engine speed is 2800 RPMs.
x 10 20 25 30 45 55 y 4.5 8.9 13.8 18.9 29.9 37.7 18. HOW DO YOU SEE IT? The graph shows typical speeds y (in feet per second) of a space shuttle x seconds after it is launched.
Space Launch Shuttle speed (feet per second) 0 1000 2000 Time (seconds) 60 80 100 40 20 x y a. What type of polynomial function models the data? Explain. b. Which nth-order fi nite difference should be constant for the function in part (a)? Explain. 19. MATHEMATICAL CONNECTIONS The table shows the number of diagonals for polygons with n sides. Find a polynomial function that fi ts the data. Determine the total number of diagonals in the decagon shown.
Number of sides, n 3 4 5 6 7 8 Number of diagonals, d 0 2 5 9 14 20 20. MAKING AN ARGUMENT Your friend states that it is not possible to determine the degree of a function given the fi rst-order differences. Is your friend correct? Explain your reasoning.
21. WRITING Explain why you cannot always use fi nite differences to fi nd a model for real-life data sets. 22. THOUGHT PROVOKING A, B, and C are zeros of a cubic polynomial function. Choose values for A, B, and C such that the distance from A to B is less than or equal to the distance from A to C. Then write the function using the A, B, and C values you chose.
23. MULTIPLE REPRESENTATIONS Order the polynomial functions according to their degree, from least to greatest.
A. f(x) = −3x + 2x2 + 1 B.
x y 2 −2 4 2 −2 g C.
x −2 −1 0 1 2 3 h(x) 8 6 4 2 0 −2 D. x −2 −1 0 1 2 3 k(x) 25 6 7 4 −3 10 24. ABSTRACT REASONING Substitute the expressions z, z + 1, z + 2, ⋅ ⋅ ⋅ , z + 5 for x in the function f(x) = ax3 + bx2 + cx + d to generate six equally-spaced ordered pairs. Then show that the third-order differences are constant.
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Solve the equation using square roots. (Section 3.1) 25. x2 − 6 = 30 26. 5x2 − 38 = 187 27. 2(x − 3)2 = 24 28. 4 — 3 (x + 5)2 = 4 Solve the equation using the Quadratic Formula. (Section 3.4) 29. 2x2 + 3x = 5 30. 2x2 + 1 — 2 = 2x 31. 2x2 + 3x = −3x2 + 1 32. 4x − 20 = x2 Reviewing what you learned in previous grades and lessons diagonal 225 4.5–4.9 What Did You Learn?
Core Vocabulary Core Vocabulary repeated solution, p. 190 complex conjugates, p. 199 local maximum, p. 214 local minimum, p. 214 even function, p. 215 odd function, p. 215 fi nite differences, p. 220 Core Concepts Core Concepts Section 4.5 The Rational Root Theorem, p. 191 The Irrational Conjugates Theorem, p. 193 Section 4.6 The Fundamental Theorem of Algebra, p. 198 The Complex Conjugates Theorem, p. 199 Descartes’s Rule of Signs, p. 200 Section 4.7 Transformations of Polynomial Functions, p. 206 Writing Transformed Polynomial Functions, p. 207 Section 4.8 Zeros, Factors, Solutions, and Intercepts, p. 212 The Location Principle, p. 213 Turning Points of Polynomial Functions, p. 214 Even and Odd Functions, p. 215 Section 4.9 Writing Polynomial Functions for Data Sets, p. 220 Properties of Finite Differences, p. 221 Mathematical Practices Mathematical Practices 1. Explain how understanding the Complex Conjugates Theorem allows you to construct your argument in Exercise 46 on page 203.
2. Describe how you use structure to accurately match each graph with its transformation in Exercises 7–10 on page 209.
Performance Task For the Birds --Wildlife Management How does the presence of humans affect the population of sparrows in a park? Do more humans mean fewer sparrows? Or does the presence of humans increase the number of sparrows up to a point? Are there a minimum number of sparrows that can be found in a park, regardless of how many humans there are? What can a mathematical model tell you?
To explore the answers to these questions and more, go to BigIdeasMath.com.
225 226 Chapter 4 Polynomial Functions 4 Chapter Review Graphing Polynomial Functions (pp. 157–164) 4.1 Graph f(x) = x3 + 3x2 − 3x − 10.
To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior.
x −3 −2 −1 0 1 2 3 f(x) −1 0 −5 −10 −9 4 35 The degree is odd and the leading coeffi cient is positive. So, f(x) → −∞ as x → −∞ and f(x) → +∞ as x → +∞.
Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coeffi cient. 1. h(x) = −x3 + 2x2 − 15x7 2. p(x) = x3 − 5x0.5 + 13x2 + 8 Graph the polynomial function. 3. h(x) = x2 + 6x5 − 5 4. f(x) = 3x4 − 5x2 + 1 5. g(x) = −x4 + x + 2 Adding, Subtracting, and Multiplying Polynomials (pp. 165–172) 4.2 a. Multiply (x − 2), (x − 1), and (x + 3) in a horizontal format.
(x − 2)(x − 1)(x + 3) = (x2 − 3x + 2)(x + 3) = (x2 − 3x + 2)x + (x2 − 3x + 2)3 = x3 − 3x2 + 2x + 3x2 − 9x + 6 = x3 − 7x + 6 b. Use Pascal’s Triangle to expand (4x + 2)4.
The coeffi cients from the fourth row of Pascal’s Triangle are 1, 4, 6, 4, and 1. (4x + 2)4 = 1(4x)4 + 4(4x)3(2) + 6(4x)2(2)2 + 4(4x)(2)3 + 1(2)4 = 256x4 + 512x3 + 384x2 + 128x + 16 Find the sum or difference. 6. (4x3 − 12x2 − 5) − (−8x2 + 4x + 3) 7. (x4 + 3x3 − x2 + 6) + (2x4 − 3x + 9) 8. (3x2 + 9x + 13) − (x2 − 2x + 12) Find the product. 9. (2y2 + 4y − 7)(y + 3) 10. (2m + n)3 11. (s + 2)(s + 4)(s − 3) Use Pascal’s Triangle to expand the binomial. 12. (m + 4)4 13. (3s + 2)5 14. (z + 1)6 x y 4 −12 4 −4 (2, 4) (−3, −1) (−2, 0) (0, −10) (1, −9) Dynamic Solutions available at BigIdeasMath.com Chapter 4 Chapter Review 227 Dividing Polynomials (pp. 173–178) 4.3 Use synthetic division to evaluate f(x) = −2x3 + 4x2 + 8x + 10 when x = −3. −3 −2 4 8 10 6 −30 66 −2 10 −22 76 The remainder is 76. So, you can conclude from the Remainder Theorem that f(−3) = 76. You can check this by substituting x = −3 in the original function.
Check f(−3) = −2(−3)3 + 4(−3)2 + 8(−3) + 10 = 54 + 36 − 24 + 10 = 76 ✓ Divide using polynomial long division or synthetic division. 15. (x3 + x2 + 3x − 4) ÷ (x2 + 2x + 1) 16. (x4 + 3x3 − 4x2 + 5x + 3) ÷ (x2 + x + 4) 17. (x4 − x2 − 7) ÷ (x + 4) 18. Use synthetic division to evaluate g(x) = 4x3 + 2x2 − 4 when x = 5.
Factoring Polynomials (pp. 179–186) 4.4 a. Factor x4 + 8x completely. x4 + 8x = x(x3 + 8) Factor common monomial.
= x(x3 + 23) Write x3 + 8 as a3 + b3.
= x(x + 2)(x2 − 2x + 4) Sum of Two Cubes Pattern b. Determine whether x + 4 is a factor of f(x) = x5 + 4x4 + 2x + 8.
Find f(−4) by synthetic division. −4 1 4 0 0 2 8 −4 0 0 0 −8 1 0 0 0 2 0 Because f(−4) = 0, the binomial x + 4 is a factor of f(x) = x5 + 4x4 + 2x + 8. Factor the polynomial completely. 19. 64x3 − 8 20. 2z5 − 12z3 + 10z 21. 2a3 − 7a2 − 8a + 28 22. Show that x + 2 is a factor of f(x) = x4 + 2x3 − 27x − 54. Then factor f(x) completely.
228 Chapter 4 Polynomial Functions Solving Polynomial Equations (pp. 189–196) 4.5 a. Find all real solutions of x3 + x2 − 8x − 12 = 0.
Step 1 List the possible rational solutions. The leading coeffi cient of the polynomial f(x) = x3 + x2 − 8x − 12 is 1, and the constant term is −12. So, the possible rational solutions of f(x) = 0 are x = ± 1 — 1 , ± 2 — 1 , ± 3 — 1 , ± 4 — 1 , ± 6 — 1 , ± 12 — 1 .
Step 2 Test possible solutions using synthetic division until a solution is found. 2 1 1 −8 −12 −2 1 1 −8 −12 2 6 −4 −2 2 12 1 3 −2 −16 1 −1 −6 0 f(2) ≠ 0, so x − 2 is not a factor of f(x). f(−2) = 0, so x + 2 is a factor of f(x).
Step 3 Factor completely using the result of synthetic division. (x + 2)(x2 − x − 6) = 0 Write as a product of factors. (x + 2)(x + 2)(x − 3) = 0 Factor the trinomial. So, the solutions are x = −2 and x = 3.
b. Write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the zeros −4 and 1 + √ — 2 . By the Irrational Conjugates Theorem, 1 − √ — 2 must also be a zero of f. f(x) = (x + 4) [ x − ( 1 + √ — 2 ) ] [ x − ( 1 − √ — 2 ) ] Write f(x) in factored form. = (x + 4) [ (x − 1) − √ — 2 ] [ (x − 1) + √ — 2 ] Regroup terms. = (x + 4) [ (x − 1)2 − 2 ] Multiply. = (x + 4) [ (x2 − 2x + 1) − 2 ] Expand binomial. = (x + 4)(x2 − 2x − 1) Simplify. = x3 − 2x2 − x + 4x2 − 8x − 4 Multiply. = x3 + 2x2 − 9x − 4 Combine like terms.
Find all real solutions of the equation. 23. x3 + 3x2 − 10x − 24 = 0 24. x3 + 5x2 − 2x − 24 = 0 Write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the given zeros. 25. 1, 2 − √ — 3 26. 2, 3, √ — 5 27. −2, 5, 3 + √ — 6 28. You use 240 cubic inches of clay to make a sculpture shaped as a rectangular prism. The width is 4 inches less than the length and the height is 2 inches more than three times the length. What are the dimensions of the sculpture? Justify your answer.
Chapter 4 Chapter Review 229 The Fundamental Theorem of Algebra (pp. 197–204) 4.6 Find all zeros of f(x) = x4 + 2x3 + 6x2 + 18x − 27.
Step 1 Find the rational zeros of f. Because f is a polynomial function of degree 4, it has four zeros. The possible rational zeros are ±1, ±3, ±9, and ± 27. Using synthetic division, you can determine that 1 is a zero and −3 is also a zero. Step 2 Write f(x) in factored form. Dividing f(x) by its known factors x − 1 and x + 3 gives a quotient of x2 + 9. So, f(x) = (x − 1)(x + 3)(x2 + 9).
Step 3 Find the complex zeros of f. Solving x2 + 9 = 0, you get x = ±3i. This means x2 + 9 = (x + 3i)(x − 3i). f(x) = (x − 1)(x + 3)(x + 3i)(x − 3i) From the factorization, there are four zeros. The zeros of f are 1, −3, −3i, and 3i. Write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the given zeros. 29. 3, 1 + 2i 30. −1, 2, 4i 31. −5, −4, 1 − i √ — 3 Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function. 32. f(x) = x4 − 10x + 8 33. f(x) = −6x4 − x3 + 3x2 + 2x + 18 Transformations of Polynomial Functions (pp. 205–210) 4.7 Describe the transformation of f(x) = x3 represented by g(x) = (x − 6)3 − 2. Then graph each function.
Notice that the function is of the form g(x) = (x − h)3 + k. Rewrite the function to identify h and k.
g(x) = (x − 6)3 + (−2) h k Because h = 6 and k = −2, the graph of g is a translation 6 units right and 2 units down of the graph of f.
Describe the transformation of f represented by g. Then graph each function. 34. f(x) = x3, g(x) = (−x)3 + 2 35. f(x) = x4, g(x) = −(x + 9)4 Write a rule for g. 36. Let the graph of g be a horizontal stretch by a factor of 4, followed by a translation 3 units right and 5 units down of the graph of f(x) = x5 + 3x. 37. Let the graph of g be a translation 5 units up, followed by a refl ection in the y-axis of the graph of f(x) = x4 − 2x3 − 12.
x y 4 8 4 −4 f g 230 Chapter 4 Polynomial Functions Analyzing Graphs of Polynomial Functions (pp. 211–218) 4.8 Graph the function f(x) = x(x + 2)(x − 2). Then estimate the points where the local maximums and local minimums occur. Step 1 Plot the x-intercepts. Because −2, 0, and 2 are zeros of f, plot (−2, 0), (0, 0), and (2, 0).
Step 2 Plot points between and beyond the x-intercepts.
x −3 −2 −1 0 1 2 3 y −15 0 3 0 −3 0 15 Step 3 Determine end behavior. Because f(x) has three factors of the form x − k and a constant factor of 1, f is a cubic function with a positive leading coeffi cient. So f(x) → −∞ as x → −∞ and f(x) → +∞ as x → +∞.
Step 4 Draw the graph so it passes through the plotted points and has the appropriate end behavior. The function has a local maximum at (−1.15, 3.08) and a local minimum at (1.15, −3.08).
Graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing. 38. f(x) = −2x3 − 3x2 − 1 39. f(x) = x4 + 3x3 − x2 − 8x + 2 Determine whether the function is even, odd, or neither. 40. f(x) = 2x3 + 3x 41. g(x) = 3x2 − 7 42. h(x) = x6 + 3x5 Modeling with Polynomial Functions (pp. 219–224) 4.9 Write the cubic function whose graph is shown.
Step 1 Use the three x-intercepts to write the function in factored form.
f(x) = a(x + 3)(x + 1)(x − 2) Step 2 Find the value of a by substituting the coordinates of the point (0, −12).
−12 = a(0 + 3)(0 + 1)(0 − 2) −12 = −6a 2 = a The function is f(x) = 2(x + 3)(x + 1)(x − 2). 43. Write a cubic function whose graph passes through the points (−4, 0), (4, 0), (0, 6), and (2, 0). 44. Use fi nite differences to determine the degree of the polynomial function that fi ts the data. Then use technology to fi nd the polynomial function.
6 −5 −6 4 Y=-3.08 X=1.15 Minimum x y 8 4 −12 −16 4 −2 −4 (2, 0) (−1, 0) (−3, 0) (0, −12) x 1 2 3 4 5 6 7 f(x) −11 −24 −27 −8 45 144 301 x y 4 −4 −2 4 −4 (2, 0) (0, 0) (−2, 0) (1, −3) Chapter 4 Chapter Test 231 Chapter Test 4 Write a polynomial function f of least degree that has rational coeffi cients, a leading coeffi cient of 1, and the given zeros. 1. 3, 1 − √ — 2 2. −2, 4, 3i Find the product or quotient. 3. (x6 − 4)(x2 − 7x + 5) 4. (3x4 − 2x3 − x − 1) ÷ (x2 − 2x + 1) 5. (2x3 − 3x2 + 5x − 1) ÷ (x + 2) 6. (2x + 3)3 7. The graphs of f(x) = x4 and g(x) = (x − 3)4 are shown.
a. How many zeros does each function have? Explain.
b. Describe the transformation of f represented by g.
c. Determine the intervals for which the function g is increasing or decreasing. 8. The volume V (in cubic feet) of an aquarium is modeled by the polynomial function V(x) = x3 + 2x2 − 13x + 10, where x is the length of the tank.
a. Explain how you know x = 4 is not a possible rational zero.
b. Show that x − 1 is a factor of V(x). Then factor V(x) completely.
c. Find the dimensions of the aquarium shown. 9. One special product pattern is (a − b)2 = a2 − 2ab + b2. Using Pascal’s Triangle to expand (a − b)2 gives 1a2 + 2a(−b) + 1(−b)2. Are the two expressions equivalent? Explain. 10. Can you use the synthetic division procedure that you learned in this chapter to divide any two polynomials? Explain. 11. Let T be the number (in thousands) of new truck sales. Let C be the number (in thousands) of new car sales. During a 10-year period, T and C can be modeled by the following equations where t is time (in years). T = 23t4 − 330t3 + 3500t2 − 7500t + 9000 C = 14t4 − 330t3 + 2400t2 − 5900t + 8900 a. Find a new model S for the total number of new vehicle sales.
b. Is the function S even, odd, or neither? Explain your reasoning. 12. Your friend has started a golf caddy business. The table shows the profi ts p (in dollars) of the business in the fi rst 5 months. Use fi nite differences to fi nd a polynomial model for the data. Then use the model to predict the profi t after 7 months.
Month, t 1 2 3 4 5 Profi t, p 4 2 6 22 56 x y 2 4 2 4 −2 f g Volume = 3 ft3 232 Chapter 4 Polynomial Functions 4 Cumulative Assessment 1. The synthetic division below represents f(x) ÷ (x − 3). Choose a value for m so that x − 3 is a factor of f(x). Justify your answer.
3 1 −3 m 3 3 0 1 0 2. Analyze the graph of the polynomial function to determine the sign of the leading coeffi cient, the degree of the function, and the number of real zeros. Explain.
x y 4 2 −4 2 −4 3. Which statement about the graph of the equation 12(x − 6) = −( y + 4)2 is not true?
○ A The vertex is (6, −4).
○ B The axis of symmetry is y = −4.
○ C The focus is (3, −4).
○ D The graph represents a function. 4. A parabola passes through the point shown in the graph. The equation of the axis of symmetry is x = −a. Which of the given points could lie on the parabola? If the axis of symmetry was x = a, then which points could lie on the parabola? Explain your reasoning.
−1 −2 1 3 2 −3 (0, 1) (1, 1) (−2, 1) (−4, 1) (−3, 1) (−5, 1) x y x = −a (3, 1) Chapter 4 Cumulative Assessment 233 5. Select values for the function to model each transformation of the graph of f(x) = x.
g(x) = ( x − ) + a. The graph is a translation 2 units up and 3 units left. b. The graph is a translation 2 units right and 3 units down. c. The graph is a vertical stretch by a factor of 2, followed by a translation 2 units up. d. The graph is a translation 3 units right and a vertical shrink by a factor of 1 — 2 , followed by a translation 4 units down. 6. The diagram shows a circle inscribed in a square. The area of the shaded region is 21.5 square meters. To the nearest tenth of a meter, how long is each side of the square?
r ○ A 4.6 meters ○ B 8.7 meters ○ C 9.7 meters ○ D 10.0 meters 7. Classify each function as even, odd, or neither. Justify your answer. a. f(x) = 3x5 b. f(x) = 4x3 + 8x c. f(x) = 3x3 + 12x2 + 1 d. f(x) = 2x4 e. f(x) = x11 − x7 f. f(x) = 2x8 + 4x4 + x2 − 5 8. The volume of the rectangular prism shown is given by V = 2x3 + 7x2 − 18x − 63. Which polynomial represents the area of the base of the prism?
○ A 2x2 + x − 21 ○ B 2x2 + 21 − x ○ C 13x + 21 + 2x2 ○ D 2x2 − 21 − 13x 9. The number R (in tens of thousands) of retirees receiving Social Security benefi ts is represented by the function R = 0.286t 3 − 4.68t2 + 8.8t + 403, 0 ≤ t ≤ 10 where t represents the number of years since 2000. Identify any turning points on the given interval. What does a turning point represent in this situation?
x − 3 |
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Functional equation
A functional equation, roughly speaking, is an equation in which some of the unknowns to be solved for are functions. For example, the following are functional equations:
Contents
1 Introductory Topics
1.1 The Inverse of a Function
2 Intermediate Topics
2.1 Cyclic Functions
2.2 Problem Examples
3 Advanced Topics
3.1 Functions and Relations
3.2 Injectivity and Surjectivity
4 See Also
Introductory Topics
The Inverse of a Function
The inverse of a function is a function that "undoes" a function. For an example, consider the function: . The function has the property that . In this case, is called the (right) inverse function. (Similarly, a function so that is called the left inverse function. Typically the right and left inverses coincide on a suitable domain, and in this case we simply call the right and left inverse function the inverse function.) Often the inverse of a function is denoted by .
Intermediate Topics
Cyclic Functions
A cyclic function is a function that has the property that:
A classic example of such a function is because . Cyclic functions can significantly help in solving functional identities. Consider this problem:
Find such that . Let and in this functional equation. This yields two new equations:
Now, if we multiply the first equation by 3 and the second equation by 4, and add the two equations, we have:
So, clearly,
Problem Examples
2006 AMC 12A Problem 18
2007 AIME II Problem 14
Advanced Topics
Functions and Relations
Given a set and , the Cartesian Product of these sets (denoted ) gives all ordered pairs with and . Symbolically,
A relation is a subset of . A function is a special time of relation where for every in the ordered pair , there exists a unique .
Injectivity and Surjectivity
Consider a function be a function from the set to the set , i.e., is the domain of and is the codomain of .
The function is injective (or one-to-one) if for all in the domain , if and only if . Symbolically, f(x)is injective⟺(∀a,b∈X,f(a)=f(b)⟹a=b).
The function is surjective (or onto) if for all in the codomain there exists a in the domain such that . Symbolically, f(x)is surjective⟺∀a∈Y,∃b∈X:f(b)=a.
The function is bijective (or one-to-one and onto) if it is both injective and subjective. Symbolically, f(x)is bijective⟺∀a∈Y,∃!b∈X:f(b)=a.
The function has an inverse function , where , if and only if it is a bijective function.
See Also
Functions
Cauchy Functional Equation
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Tema 1 Teoria Del Precio en Competencia Perfecta
Este documento describe los conceptos clave de un mercado de competencia perfecta y la teoría del precio en este tipo de mercado. Explica que en un mercado de competencia perfecta, el precio…
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International Stratigraphic Chart: International Commission On Stratigraphy
Geologie
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Famennian Givetian Eifelian Lochkovian Pragian Frasnian Emsian Ludfordian Gorstian Homerian
Sheinwoodian
Telychian Aeronian Rhuddanian
P h a n e r o z o i c P a l e o z o i c
Pridoli
Ludlow Wenlock
Llandovery
Upper Middle Lower
D e v o n i a n
411.2 ±2.8 416.0 ±2.8 422.9 ±2.5 428.2 ±2.3
Lopingian
Guadalupian
Cisuralian Upper Upper Middle Middle Lower Lower Upper
Upper
Middle Middle Lower Lower
P h a n e r o z o i c P a l e o z o i c M e s o z o i c C a r b o n i f e r o u s P e r m i a n J u r a s s i c T r i a s s i c
150.8 ±4.0 155.6 161.2 ±4.0 164.7 ±4.0 167.7 ±3.5 171.6 ±3.0 175.6 ±2.0 183.0 ±1.5 189.6 ±1.5 196.5 ±1.0 199.6 ±0.6 203.6 ±1.5 216.5 ±2.0 228.7 237.0 ±2.0 245.9 249.7 ±0.7 251.0 ±0.4 318.1 ±1.3 260.4 ±0.7 253.8 ±0.7 265.8 ±0.7 268.0 ±0.7 270.6 ±0.7 275.6 ±0.7 284.4 ±0.7 294.6 ±0.8 299.0 ±0.8 303.9 ±0.9 306.5 ±1.0 311.7 ±1.1
Tithonian
Kimmeridgian
Oxfordian Callovian Bajocian Bathonian Aalenian Toarcian
Pliensbachian
Sinemurian Hettangian Rhaetian Norian Carnian Ladinian Anisian
Olenekian
Induan
Changhsingian Wuchiapingian
Capitanian Wordian Roadian Kungurian Artinskian Sakmarian Asselian Gzhelian Kasimovian Moscovian Bashkirian
Serpukhovian
Visean Tournaisian
P e n n-s y l v a n i a n M i s s i s-s i p p i a n
359.2 ±2.5 345.3 ±2.1 326.4 ±1.6 374.5 ±2.6 385.3 ±2.6 391.8 ±2.7 397.5 ±2.7 407.0 ±2.8 443.7 ±1.5 436.0 ±1.9 439.0 ±1.8 418.7 ±2.7
Ediacaran Cryogenian Tonian Stenian Ectasian Calymmian Statherian Orosirian Rhyacian Siderian Neo-proterozoic Neoarchean
Mesoarchean Paleoarchean
Eoarchean Meso-proterozoic Paleo-proterozoic
A r c h e a n P r o t e r o z o i c P r e c a m b r i a n
~630 850 1000 1200 1400 1600 1800 2050 2300 2500 2800 3200 3600
Upper Middle Lower Selandian Tortonian Serravallian Langhian Burdigalian Aquitanian Ypresian Chattian Rupelian Priabonian Danian Thanetian Bartonian Lutetian Campanian Santonian Turonian Coniacian Albian Aptian Berriasian Barremian Valanginian Hauterivian
Maastrichtian
Cenomanian
Piacenzian Messinian
Zanclean
Gelasian
P h a n e r o z o i c C e n o z o i c M e s o z o i c
1.806 0.126 0.781 2.588 5.332 7.246 11.608 13.82 15.97 20.43 70.6 ±0.6 65.5 ±0.3 83.5 ±0.7 85.8 ±0.7 88.6 93.5 ±0.8 40.4 ±0.2 37.2 ±0.1 33.9 ±0.1 28.4 ±0.1 23.03 48.6 ±0.2 55.8 ±0.2 58.7 ±0.2 61.1 99.6 ±0.9 112.0 ±1.0 125.0 ±1.0 130.0 ±1.5 133.9 140.2 ±3.0 145.5 ±4.0
C r e t a c e o u s P a l e o g e n e N e o g e n e
S y s t e m P e r i o d E o n o t h e m E o n E r a t h e m E r a S t a g e A g e A g e M a G S S P E p o c h S e r i e s E o n o t h e m E o n E r a t h e m E r a S y s t e m S t a g e A g e A g e M a G S S P E p o c h S e r i e s P e r i o d E o n o t h e m E o n E r a t h e m E r a A g e M a G S S P G S S A S y s t e m P e r i o d E o n o t h e m E o n E r a S t a g e A g e A g e M a G S S P E p o c h S e r i e s P e r i o d
3.600
Oligocene Eocene Paleocene Pliocene Miocene Pleistocene Upper Lower Holocene
0.0118
INTERNATIONAL STRA TIGRAPH IC CHA RT
International Commission on S tratigraphy
421.3 ±2.6 426.2 ±2.4
Tremadocian
Darriwilian Hirnantian Paibian Upper Middle Lower Series 3
Stage 3 Stage 2
Fortunian
Stage 5
Drumian Guzhangian
Stage 9 Stage 10
Floian Dapingian Sandbian Katian
Stage 4
Series 2 Terreneuvian Furongian
C a m b r i a n O r d o v i c i a n
~ 503.0 ~ 506.5 ~ 510.0 ~ 517.0 ~ 521.0 ~ 534.6 460.9 ±1.6 471.8 ±1.6 488.3 ±1.7~ 492.0 ~ 496.0 501.0 ±2.0 542.0 ±1.0 455.8 ±1.6 445.6 ±1.5 468.1 ±1.6 478.6 ±1.7
Lower limit is not defined
S i l u r i a n
E r a t h e m S y s t e m
542 359.2 ±2.5 145.5 ±4.0
Q u a t e r n a r y
ICS
Copyright © 2007 International Commission on Stratigraphy
Subdivisions of the global geologi c record are formally defined by their lower boundary. Each unit of the Phanerozoic (~542 Ma to Present) and the base of Ediacaran are defined by a basal Global Standard Section and Point (GSSP ), whereas Precambrian units are formally subdivided by absolute age (Global Standard Stratigraphic Age,GSSA). Details of each GSSP are posted on the ICS website (
www.stratigraphy.org
).International chronostratigraphic units, rank,names and formal status are approved by the International Commission on Stratigraphy (ICS)and ratified by the International Union of Geological Sciences (IUGS).Numerical ages of the uni t boundaries in the Phanerozoic are subject to revision. Some stages within the Ordovician and Cambrian will be formally named upon international agreement on their GSSP limits. Most sub-Series boundaries (e.g., Middle and Upper Aptian) are not formally defined.Colors are according to the Commi ssion for the Geological Map of the World (
www.cgmw.org
). The listed numerical ages are from 'A Geologic Time Scale 2004', by F.M. Gradstein, J.G. Ogg,A.G. Smith, et al. (2004; Cambridge U niversity Press).This chart was drafted by Gabi Ogg. Intra Cambrian unit ages with are informal, and awaiting ratified definitions.
The status of the Quaternary is not yet decided. Its base may be assigned as the base of the Gelasian and extend the base of the Pleistocene to 2.6 Ma.
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5395 | https://www.youtube.com/watch?v=ZySl5U-i9nI | Logarithm Example – Geometric Sequence
Mr. Hudak
1020 subscribers
2 likes
Description
211 views
Posted: 2 Oct 2020
Find the number of generation before a population which grows exponentially reachers a specified level. This examples requires the use of logarithms.
Transcript:
here we have an example that asks how many generations will it take so the following geometric sequence will reach or exceed six thousand to solve a problem like this first we need to write the explicit function we already know what p0 is but we need to solve for r our common ratio and r can be solved by taking the next term and dividing it by the previous term it doesn't matter which two terms you grab from the sequence i can take the first two i can take the last two if i take the last two we would say that the next term is 1 9 and it's being divided by the previous term which is 1 over 27 and we know that when we divide by a fraction it is the same as multiplying by the reciprocal so we can say that r is 27 over 9 or that r is 3. and you can check this if you want you can take a different pair of numbers and see that you'll get the same thing if you were to take 1 over 243 and divide by 1 over 729 you will get 3 again and i'll leave that for you to explore so now we can write our explicit function and the question is asking how many generations will it take so that we reach or exceed 6000 so we're going to replace pn with 6000. now our goal is to isolate the n so we want to isolate 3 to the n and that means get rid of this 1 over 729 if i multiply both sides of the equation by 729 over 1 we see that fractions when multiplied by the reciprocals give us 1. and we're left just with three to the n on the right hand side but we need to multiply this together seven twenty nine times six thousand and we get four three seven four zero zero zero equals three to the n just want to check that i copied that right so now our exponent is where our variable lies and to get that out we need to use logarithms and the logarithm has to have a base that matches the exponential base so we're going to take log base 3 of 3 to the n to extract the end variable but anything you do to one side of an equation you need to do to both sides so we're going to take log base 3 of this 4 million 374 000. now the right hand side we know is going to cancel and leave us with just n but we still have to evaluate the other side of the equation [Music] and to do this we use something called the change of base formula so n is equal to ln of 4 374 000 divided by ln of 3. the change of base formula says you can take a log of any base and take the argument that's what's inside the parentheses so i took natural log of 4 million 374 000 and that goes in the numerator and then you take the log of the same base ln has to match top and bottom and we put the base of the log we get ln of 3. so we want to evaluate this now we go to our calculator ln 474 000 close the parenthesis divided by ln of 3. and when we hit enter we get that n is equal to 13.91864 once we pass a number so once we pass 13.0 then n is considered the next generation that means it would take 14 generations to meet or exceed 6 000. if i plug in 13.91864 and all of the decimals exactly i should get a number very close to exactly six thousand but when we speak in terms of generation there's the first generation second generation once you pass the whole number you go to the next generation it's kind of like how we do with years once we get to the year 2001 it's the 21st century so we can plug this back in to our model to see that we meet or exceed 6000 so again here's our model if we evaluate p of 14 that would be 3 to the 14th power over 729 we see we get six five six one so we meet or we exceeded six thousand and in this case we exceeded if you were to plug in 13 so i changed the 14 exponent to a 13 you see we have one hundred and eighty-three so once we multiply by three again we get that fourteenth generation we get that six thousand five hundred and sixty-one and someplace in between the thirteenth and fourteenth generation it'll be exactly 6000 so there's your example using logarithms with geometric sequences i hope this helps you |
5396 | https://www.webassign.net/labsgraceperiod/ncsugenchem202labv1/lab_9/manual.pdf | Titration Curves PURPOSE To determine the equivalence points of two titrations from plots of pH versus mL of titrant added.
GOALS 1 To gain experience performing acid-base titrations with a pH meter.
2 To plot titration curves of pH vs mL of titrant added.
3 To determine the equivalence point of a titration from a titration curve.
4 To determine the pKa of an analyte from a titration curve.
INTRODUCTION A titration1 is an analytical procedure in which a reaction is run under carefully controlled conditions. The stoichiometric volume of one reactant of known concentration, the titrant, that is required to react with another reactant of unknown concentration, the analyte2, is measured. The concentration of the analyte is determined from the concentration and volume of titrant and the stoichiometry of the reaction between them.
The experimental setup is shown in Figure 1. A buret, which contains the titrant, is calibrated so the volume of solution that it delivers can be determined with high accuracy and precision.
Titrant is added to the analyte until the stoichiometric volume of titrant has been added. This is called the equivalence point3, at which the volume of titrant delivered by the buret is read. Usually, the volume readings are estimated to the nearest 0.01 mL. The delivery of the titrant is adjusted with the stopcock on the buret. With practice, one can dispense fractions of a drop of titrant and control the procedure well enough that replicated titrations agree within 0.10 mL.
1 2 3 point ©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 1 Figure 1: Titration Setup Often, the equivalence point is determined visually as a color change of an indicator.
The indicator, a substance that changes color near the equivalence point, is added to the analyte solution.
Since the color change is near but not exactly at the equivalence point, the point at which the color change occurs is called the end point.4 Indicators are chosen so the endpoint is very close to the equivalence point.
It is important to keep a titration well mixed, so the titrant and analyte can contact each other and react rapidly. Either manual swirling of the flask or mechanical stirring can be used. You will use mechanical stirring in this experiment to ensure complete mixing of the solutions.
In this experiment, you will compare the predicted equivalence point to the experimentally determined equivalence point of two titrations. The first will be a monoprotic acid, potassium hydrogen phthalate (KHP), titrated with sodium hydroxide (NaOH) and the second will be a disodium salt, sodium carbonate (Na2CO3), titrated with hydrochloric acid (HCl). In the first, you will observe only one equivalence point according to the following reaction where P refers to the phthalate ion (C8H4O42-).
HP−(aq) + OH−(aq) →P2−(aq) + H2O(l) (1) In the second, two distinct equivalence points will be observed as two protons are added stepwise to the carbonate anion (CO32-) according to the following reactions.
4 %28chemistry%29 ©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 2 CO2− 3 (aq) + H3O+(aq) →HCO− 3 (aq) + H2O (2) HCO− 3 (aq) + H3O+(aq) →H2CO3(aq) + H2O(l) (3) The pH will be recorded for the analyte solution as increasing amounts of titrant are added.
The plot of pH (y-axis) vs volume of titrant added (x-axis) is called a titration curve.5 Figure 2: Predicted titration curve of 0.1 M NaC2H3O2 titrated with 0.20 M HCl Although indicators will be used in each of the titrations, they are simply for visual assistance.
The pH will begin to change rapidly as the equivalence points are approached. Thus, the equivalence points can be detected simply by monitoring the pH of the analyte solution. Also, the equivalence point of the titration can be determined from the titration curve. It is the volume of titrant where the slope of the titration curve is the greatest. You will be asked to compare the endpoints predicted by the indicators to the equivalence points determined from the titration curves.
You will also compare the predicted pH for various amounts of titrant added in each titration to the experimentally determined pH’s. One of these points is the midpoint. This is where one half of the equivalence volume has been added. It is special because at this volume of titrant added, exactly half of the analyte has been converted into its conjugate pair. At this solution composition, the Henderson-Hasselbalch equation predicts that the pH will be equal to the pKa of the acid species of the conjugate pair. Thus, you will also be able to determine the pKa’s of the analytes from the titration curves. These will be equal to the pH of the analyte solution at one half the equivalence volume of titrant added. You will compare the pKa values for both phthalic acid and carbonic acid found in reference tables to those determined from your titration curves.
Another point where the calculated pH will be compared to the measured pH is the first 5 curve ©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 3 equivalence point of the diprotic titration curve. At this point, the amphoteric6 intermediate species is the only species present. In the case of the carbonate (CO32-) titration, the intermediate species is the hydrogen carbonate or bicarbonate (HCO3-) ion. Because this species can act as both an acid and a base, both equilibria must be considered.
HCO− 3 (aq) + H2O(l) →CO2− 3 (aq) + H3O(aq) (4) HCO− 3 (aq) + H2O(l) →H2CO3(aq) + OH−(aq) (5) Comprehensively solving this system mathematically would involve solving five equations and five unknowns simultaneously! However, with a few assumptions, the result is quite simple. At the first equivalence point of a diprotic titration curve, the pH is the average of the pKa’s for that diprotic acid.
pH = 1/2(pKa1 + pKa2) (6) Use this equation for any solution containing only the intermediate form of a diprotic acid. For example, a solution of KHP, which you begin with in Part B. Phthalic acid is H2P and HP- is the intermediate form. Remember this when you are doing your prelab calculations!
As a reminder on concentration units: molarity is defined as the number of moles of solute in a liter of solution. This is numerically equal to the number of millimoles of solute in a milliliter of solution. It is often convenient to use this second definition of molarity in titrations and other work where small quantities are involved.
Note that the volume measurements in titrations are usually reported to four significant figures, so the concentrations are usually reported to four significant figures as well. Watch this in your work.
EQUIPMENT 1 MicroLab Interface 1 MicroLab pH Measurement Instruction Sheet 6 ©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 4 1 pH electrode in pH 7.00 buffer 2 30 mL beakers 2 50 mL beakers 1 250 mL beaker 1 100 mL graduated cylinder 1 20.00 mL volumetric pipet 1 pipet bulb 1 magnetic stir plate 1 magnetic stir bar 1 25 mL buret 1 ring stand 1 clamp 1 buret clamp 1 250 mL beaker for electrode rinsings 1 deionized water squirt bottle 1 box of Kimwipes REAGENTS ∼25 mL 0.20 M potassium hydrogen phthalate (KHP) ∼30 mL 0.20 M sodium hydroxide (NaOH) 3 drops phenolphthalein solution ∼25 mL 0.10 M sodium carbonate (Na2CO3) ∼30 mL 0.20 M hydrochloric acid (HCl) 3 drops methyl orange solution ∼15 mL pH 4.00 buffer ∼15 mL pH 7.00 buffer ∼15 mL pH 10.00 buffer ∼200 mL deionized water ©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 5 SAFETY Potassium hydrogen phthalate is listed as an irritant. Sodium carbonate is listed as both toxic and an irritant. However, the concentrations of these materials are quite low. If you spill any of these chemicals on skin or clothing, rinse the area immediately with water.
HCl and NaOH are corrosive. They can attack the skin and cause permanent damage to the eyes. If either solution splashes into your eyes, use the eyewash immediately. Hold your eyes open and flush with water for at least 15 minutes. If contact with skin or clothing occurs, immediately rinse the affected area with water for at least 15 minutes.
Have your lab partner notify your teaching assistant about the spill and your exposure.
The chemicals used in this experiment are reasonably dilute so gloves will not be available. The NaOH solution has an alkalinity similar to that of oven cleaner or bleach solution. The HCl solution has an acidity close to that of stomach acid. Remember to wash your hands with soap and water when the experiment is completed.
WASTE DISPOSAL All solutions can be disposed of down the sink drain followed by flushing with plenty of water.
PRIOR TO CLASS Please read the following sections of the Introductory Material: Volumetric Glassware7 Please review the following videos: Cleaning, conditioning and filling a buret8 Performing a titration9 Cleaning and storing a buret10 Pipeting Techniques11 Please complete your WebAssign prelab assignment. Check your WebAssign account for due dates. Students who do not complete the WebAssign prelab assignment are required to print and hand in the prelab worksheet.
LAB PROCEDURE Please print the worksheet for this lab. You will need this sheet to record your data.
In this experiment, you will be using pH electrodes connected to the MicroLab Interface. pH electrodes have a thin glass bulb at the tip. They break easily and are costly to replace.
7../equipment/manual.html#volumetric glassware 8 9 10 Z4F3MHdCws 11 ©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 6 Be careful not to shove the electrode into the bottom of a beaker or drop the electrode. There is a protective guard around the tip, which should remain in place at all times. The guard will not protect against careless treatment. Please use extreme care when using this equipment.
Best results in using the electrodes are obtained if: • Electrodes are kept in standard pH 7 buffer solution when not in use.
• Immediately prior to use, the electrodes are rinsed with deionized water and gently blotted with a Kimwipe, then placed in the test solution.
• The electrodes are rinsed and blotted again after the measurement and returned to the pH 7 buffer solution.
Part A: Calibrating the MicroLab pH Electrode 1 Open the MicroLab program.
2 Make sure the pH electrode is plugged into the interface.
3 Calibrate the pH electrode using the MicroLab instructions provided in the lab.
4 After the calibration is complete, measure the pH of each of the three buffer solutions of pH = 4.00 (red), pH = 7.00 (yellow), and pH = 10.00 (blue). Record the value in the digital display in WebAssign as a record of how accurately the probe is calibrated. Make sure the electrode is immersed in the solution and allow for a few seconds equilibration.
5 After the calibration is complete, continue with the MicroLab instructions provided.
The program will be configured to accept a volume of titrant, read from the buret, for each pH measurement.
6 For each pH reading, you will be prompted for a keyboard entry; enter the total volume of titrant as read from the buret. Make sure the electrode is immersed in the solution and allow for a few seconds equilibration.
Part B: Titration of KHP with NaOH 1 Using a clean, dry 50 mL beaker, obtain about 25 mL of 0.20 M potassium hydrogen phthalate (KHP) solution. Record the exact concentration of the KHP in Data Table A.
2 Condition a 20.00 mL volumetric pipet with KHP solution as shown in the Pipeting Techniques video linked to in the PRIOR TO CLASS section above and as described in the ”Volumetric Glassware” section of the Introductory Material in this lab manual, which is also linked to in PRIOR TO CLASS above.
3 Pipet 20.00 mL of the potassium hydrogen phthalate (KHP) solution into a 250 mL beaker, and add 100 mL of deionized water (graduated cylinder) and 3 drops of phenolphthalein indicator.
4 Obtain about 30 mL of 0.20 M NaOH solution in a clean, dry 50 mL beaker. Record the exact concentration of NaOH in Data Table A.
©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 7 5 Condition the 25.0 mL buret with NaOH solution as shown in the ”Conditioning a Buret” video linked to in the PRIOR TO CLASS section above and as described in the ”Volumetric Glassware” section of the Introductory Material in this lab manual.
6 Fill the buret with NaOH and carefully clamp it with the buret clamp to the ring stand.
7 Carefully slide the stir bar into the 250 mL beaker while tilted to avoid splashing or damage to the beaker. Position the stir plate under the 250 mL beaker and begin stirring slowly.
8 Carefully position the pH electrode in the 250 mL breaker until about 1/2 inch of the tip is in the solution. Clamp to the ring stand with the clamp provided. Be sure that the stir bar will not strike the pH electrode. If necessary, add more water from a graduated cylinder. See Figure 3 for the complete setup.
Figure 3: Experimental Setup 9 Position the buret so that the tip of the buret is just inside the beaker. Refer to Figure 3 Table A: Titration of KHP Question 1: From the values in Data Table A, calculate the theoretical equivalence volume (Veq) for your KHP titration. Record this value in Data Table A.
10 Take an initial pH reading by entering the initial buret reading in the MicroLab software window and hitting return. You should also record all of your data by hand in Data Table B just in ©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 8 case something goes wrong with the computer. Remember to read the buret to the nearest 0.01 mL. Reading a buret to this accuracy is tricky, so do the best you can, but recall that the last significant figure is expected to be an estimate.
11 Add the titrant in 2 mL increments from the buret, stirring for about 15 seconds after each incremental addition.
Then read the exact volume on the buret, enter this value into the MicroLab software and take a pH reading. Remember to record your measurements by hand in Data Table B. Please be sure to record the pH at 10.00, 15.00, 20.00 and 22.00 mL titrant, as you will need to enter those pH values into Data Table C on your InLab assignment.
12 When the pH begins to change more rapidly (or when you are within 2 mL of the predicted equivalence point), the increments of titrant should be changed to 0.5 mL. Note the pH range over which the indicator changes color in the ”Observations’ column of Data Table B.
13 Return to 2 mL increments of titrant as the changes in pH decrease beyond the equivalence point. Do not stop the titration until you have added approximately 5 mL of titrant beyond the equivalence point.
Table B: Volume of Titrant Added to KHP vs pH 14 When you are finished with your titration, stop the MicroLab data collection program. Carefully remove the pH electrode from the solution, rinse it off and place it in the pH 7 buffer until you are ready to use it in Part C.
15 Drain the remaining titrant into the analyte beaker. Discard all remaining solutions down the drain, followed by flushing with plenty of water.
16 Rinse and dry all of the glassware for use in Part C. Simply rinse the buret and pipet a few times with deionized water. It is not necessary to dry them since you will be conditioning them before use in Part C.
17 Use the MicroLab software to generate a graph. Ensure this plot has a title, axis labels and units where appropriate. Show this plot to your TA for manual grading of your work.
Question 2a: From your titration curve, what is the experimental Veq for your KHP titration?
Do not forget to subtract the initial buret reading when determining your Veq.
Question 2b: How do your theoretical and experimental equivalence volumes compare? What is their percent error?
%error = calculated - measured calculated x 100 18 Complete Data Table C below with pH values from your prelab assignment and pH values from your KHP titration curve.
Table C: Calculated vs Measured pH’s for KHP Titration ©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 9 Question 3a: What is the experimental pKa value for hydrogen phthalate (HP- or HC8H4O4-) that you found at the midpoint of your KHP titration curve?
Question 3b: The accepted value for the pKa of HP- is 5.408. How does this compare to your experimental value? What is their percent difference?
Question 4: How did the endpoint indicated by the phenolphthalein compare to the equivalence point determined by the titration curve? What conclusion can you make about the need for an indicator in a pH titration?
Part C: Titration of Na2CO3 with HCl 1 Rinse the 20.00 mL volumetric pipet from Part B with deionized water a few times to remove any remaining KHP solution.
2 Using a clean, dry 50 mL beaker, obtain about 25 mL of 0.10 M sodium carbonate (Na2CO3) solution. Record the exact concentration of the Na2CO3 in Data Table D.
3 Condition the 20.00 mL volumetric pipet with Na2CO3 solution as shown in the Pipeting Techniques video linked to in the PRIOR TO CLASS section above and as described in the ”Volumetric Glassware” section of the Introductory Material in this lab manual.
4 Pipet 20.00 mL of the sodium carbonate (Na2CO3) solution into a 250 mL beaker, and add 100 mL of deionized water (graduated cylinder) and 3 drops of methyl orange indicator.
5 Obtain about 30 mL of 0.20 M HCl solution in a clean, dry 50 mL beaker. Record the exact concentration of HCl in Data Table D.
6 Condition the 25.0 mL buret with HCl solution as shown in the ”Conditioning a Buret” video linked to in the PRIOR TO CLASS section above and as described in the ”Volumetric Glassware” section of the Introductory Material in this lab manual.
7 Fill the buret with HCl and carefully clamp it with the buret clamp to the ring stand.
8 Carefully slide the stir bar into the 250 mL beaker while tilted to avoid splashing or damage to the beaker. Position the stir plate under the 250 mL beaker and begin stirring slowly.
9 Carefully position the pH electrode in the 250 mL beaker until about 1/2 inch of the tip is in the solution. Clamp to the ring stand with the clamp provided. Be sure that the stir bar will not strike the pH electrode. If necessary, add more water from a graduated cylinder. See Figure 3 for the complete setup.
10 Position the buret so that the tip of the buret is just inside the beaker. Refer to Figure 3.
Table D: Titration of Na2CO3 Question 5: From the values in Data Table D, calculate the theoretical first and second equivalence volumes (Veq’s) for your Na2CO3 titration.
©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 10 11 Take an initial pH reading by entering the initial buret reading in the MicroLab software window and hitting return. You should also record all of your data by hand in Data Table E just in case something goes wrong with the computer.
12 Add the titrant in 2 mL increments from the buret, stir about 15 seconds. Then read the exact volume on the buret, enter this value into the MicroLab software and take a pH reading.
Remember to record your measurements by hand in Data Table E. Please be sure to record the pH at 5.00, 10.00, 15.00, 20.00, and 22.00 mL of titrant, as you will need to enter those pH values into Data Table C on your InLab assignment.
13 When the pH begins to change more rapidly (or when you are within 2 mL of the predicted first equivalence point), the increments of titrant should be changed to 0.5 mL.
14 Return to 2 mL increments of titrant as the changes in pH decrease beyond the first equivalence point.
15 When the pH begins to change more rapidly (or when you are within 2 mL of the predicted second equivalence point), the increments of titrant should be changed to 0.5 mL. Note the pH range over which the indicator changes color in the observations column of Data Table E.
16 Return to 2 mL increments of titrant as the changes in pH decrease beyond the second equiv-alence point. Do not stop the titration until you have added approximately 5 mL of titrant beyond the second equivalence point.
17 When you are finished with your titration, stop the MicroLab data collection program. Carefully remove the pH electrode from the solution, rinse it off and place it in the pH 7 buffer as you found it at the beginning of the lab period.
18 Drain the remaining titrant into the analyte beaker. Discard all remaining solutions down the drain followed by flushing with plenty of water.
19 Rinse and dry all of the glassware and return it to the setup area. Simply rinse the buret and pipet a few times with deionized water. It is not necessary to manually dry them. Simply clamp them upside down in the buret clamp with the stopcock of the buret open so they will dry on their own. Please refer to the ”Cleaning and Storing a Buret” video linked to in the PRIOR TO CLASS section above for more details.
Table E: Volume of Titrant Added to Na2CO3 vs pH 20 Use the MicroLab software to generate a graph. Ensure this plot has a title, axis labels, and units where appropriate. Show this plot to your TA for manual grading of your work.
21 When you are finished, close MicroLab.
Question 6a: From your titration curve, what are the experimental first and second Veq’s for your Na2CO3 titration? Do not forget to subtract the initial buret reading when determining your Veq’s.
Question 6b: How do your theoretical and experimental equivalence volumes compare? What are their percent error?
©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 11 22 Complete Data Table F in the Data Collection section of the lab manual with pH values from your prelab assignment and pH values from your Na2CO3 titration curve.
Table F: Calculated vs Measured pH’s for Na2CO3 Titration Question 7a: What are the experimental pKa values for carbonic acid (H2CO3) and hydrogen carbonate (HCO3-) that you found at the midpoints of your Na2CO3 titration curve?
Question 7b: The accepted values for the pKa’s of H2CO3 and HCO3- are 6.352 and 10.329, respectively. How do these compare to your experimental values? What are their percent error?
Question 8: How did the endpoint indicated by the methyl orange compare to the equivalence points determined by the titration curve? What conclusion can you make about the need for an indicator in a pH titration?
23 Before leaving, enter your results in the InLab assignment. If all results are scored as correct, log out. If not all results are correct, try to find the error or consult with your teaching assistant.
When all results are correct, note them and log out of WebAssign. The InLab assignment must be completed by the end of the lab period. If additional time is required, please consult with your teaching assistant.
©2010-2013 Advanced Instructional Systems, Inc. and North Carolina State University 12 |
5397 | https://content.next.westlaw.com/practical-law/document/I8d74c59cef2a11e28578f7ccc38dcbee/Metes-and-Bounds?viewType=FullText&transitionType=Default&contextData=(sc.Default) | Metes and Bounds | Practical Law
Metes and Bounds
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Metes and Bounds
Practical Law Glossary Item 5-507-1527
(Approx. 2 pages)
Glossary
Metes and Bounds
The legal description of a parcel of land that is measured in distances, angles, and directions. A surveyor uses directions and distances from physical monuments to define and describe the boundaries of the parcel of land. The monuments used by the surveyor can be either exclusive or a combination of:
Natural monuments existing on the land, such as trees, streams, or a neighbor's adjoining parcel of land.
Artificial monuments placed on the property by the surveyor, such as an identifiable stake in the ground.
A metes and bounds legal description starts from a point of beginning (POB). After the POB, the legal description traces the outline of the property's boundary lines until there is closure in the legal description. To achieve closure in the legal description, the last course and distance call of the legal description should end at the POB.
Metes and bounds legal descriptions are most often used in state land states. |
5398 | https://ecampusontario.pressbooks.pub/physicsfundamentals/chapter/12-1-concepts-transmitting-rotational-motion-through-gears/ | Skip to content
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12.1. Concepts: Transmitting rotational motion through gears
Introduction
Gears are wheels with teeth on their periphery and are used to transfer motion and power between machine components. The teeth prevent the slippage that occurs when two simple disks transmit motion from one another through friction. Two gears that mesh with each other can be classified as drivers and driven.
The driver gear will transmit motion to the driven one. If their diameters are different, while motion is transmitted, there will be a change in the rotational speed from one gear to another. Gears are used to either decrease or increase the rotational speed, depending on the application in which they are used.
Concepts
If we consider two disks, one of diameter D rotating at a speed of N rpm and one of diameter d, rotating at n rpm, then we can say that the point of contact between them will rotate at the same tangential speed, [latex]\boldsymbol v[/latex].
Expressing the tangential speed from both the disks, we get:
[latex]\omega_1 r_1 = \omega_2 r_2[/latex] (1)
If we substitute both angular velocities using the conversion of units below:
[latex]\omega = N \times {\large \frac{2\pi}{60}}[/latex] (2)
and the radii by the respective diameters, formula (1) becomes:
[latex]N \times D = n \times d[/latex] (3)
In the case of gears meshing with each other, the contact point is positioned on the reference diameter (reference circle) – see picture below.
The module is a characteristic of gears, and it is equal for gears that mesh with each other. The module is defined as:
[latex]m = {\large \frac{\text{pitch}}{\pi}}[/latex] (4)
The reference diameter is equal to the circumference/ [latex]\pi[/latex], which means:
[latex]D = {\large \frac{\text{pitch} \times T}{\pi}} = m \times T[/latex] (5)
If we substitute formula (5) into formula (3), we get the equation of gears:
[latex]N \times T = n \times t[/latex] (6)
where:
[latex]N[/latex] – represents the number of revolutions per minute for the driver gear;
[latex]T[/latex] – represents the number of teeth for the driver gear;
[latex]n[/latex] – represents the number of revolutions per minute for the driven gear;
[latex]t[/latex] – represents the number of teeth for the driven gear.
Gear trains. When there are more than two gears involved in the transmission of motion, they are called gear trains.
Gear trains can be simple gear trains (Figure 12.3 a) or compound gear trains (Figure 12.3 b).
When the number of gears is even, the first and the last gear will rotate in the same direction; when the number of gears is odd, the first and the last gear will rotate in opposite directions.
The general formula for the gear trains can be adapted from the general gear equation (formula 6), as follows
[latex]N \times T_1 \times T_2 \times T_3 \dots = n \times t_1 \times t_2 \times t_3 \dots \text{ (7)}[/latex]
where:
[latex]N[/latex] – represents the number of revolutions per minute for the first driver gear;
[latex]T_1, T_2, T_3, \dots[/latex] – represents the number of teeth for the driver gears;
[latex]n[/latex] – represents the number of revolutions per minute for the last driven gear;
[latex]t_1, t_2, t_3, \dots[/latex] – represents the number of teeth for the driven gears.
Image Attributions
Figure 12.1 adapted from:
Illustration of gear doodle icon by rawpixel.com courtesy of Freepik
Figure 12.2 adapted from:
Illustration of a cogwheel by rawpixel.com courtesy of Freepik
Figure 12.3 adapted from:
Gears V2.0 by Helindu courtesy of Sketchfab, CC-BY 4.0
License
College Physics – Fundamentals and Applications Copyright © by Daniela Stanescu, Centennial College is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.
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Advance Ratio
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Definition
Advance Ratio is the maximum percentage of the value of an asset or assets by reference to which a finance provider is prepared to make a loan or make available credit.
The value of the asset or assets is reduced in the calculation by a Security Margin, which is the inverse of the advance ratio. Where the advance ratio is 100%, the underlying finance may be termed 100% financing and may arise from a Commitment To Pay
The term is typically used in Supply Chain Finance
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