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Law of Sines
The Law of Sines is a useful identity in a triangle, which, along with the law of cosines and the law of tangents can be used to determine sides and angles. The law of sines can also be used to determine the circumradius, another useful function.
Contents
1 Statement
2 Proof
2.1 Method 1
2.2 Method 2
3 Method 3
4 Problems
4.1 Introductory
4.2 Intermediate
4.3 Olympiad
5 See Also
Statement
In triangle , where is the side opposite to , opposite to , opposite to , and where is the circumradius:
Proof
Method 1
In the diagram above, point is the circumcenter of . Point is on such that is perpendicular to . Since , and . But making . We can use simple trigonometry in right triangle to find that
The same holds for and , thus establishing the identity.
Method 2
This method only works to prove the regular (and not extended) Law of Sines.
The formula for the area of a triangle is .
Since it doesn't matter which sides are chosen as , , and , the following equality holds:
Assuming the triangle in question is nondegenerate, . Multiplying the equation by yields:
Method 3
We can circumvent some of the work in Method 1 by setting up the circle in a different way. Let be a diameter and be the center of the circle, and let be on . Furthermore, let , and let , , and .
We have that is a right angle, as is a diameter. Therefore, , so, rearranging, we have , or . Likewise, . Finally, we observe that , so evidently . Combining all three equalities,
Problems
Introductory
If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
(Source)
Intermediate
Triangle has sides , , and of length 43, 13, and 48, respectively. Let be the circlecircumscribed around and let be the intersection of and the perpendicular bisector of that is not on the same side of as . The length of can be expressed as , where and are positive integers and is not divisible by the square of any prime. Find the greatest integer less than or equal to .
(Source)
Olympiad
Let be a convex quadrilateral with , , and let be the intersection point of its diagonals. Prove that if and only if .
(Source)
See Also
Trigonometry
Trigonometric identities
Geometry
Law of Cosines
Retrieved from "
Categories:
Theorems
Trigonometry
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Class 12 math (India)
Course: Class 12 math (India)>Unit 7
Lesson 9: Review: Concavity & points of inflection
Concavity & inflection points challenge
Math>
Class 12 math (India)>
Playing with graphs (using differentiation)>
Review: Concavity & points of inflection
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Concavity & inflection points challenge
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Problem
The function f (not shown) is continuous and differentiable for all real numbers. The graphs of f′ and f′′ are shown.
Which of the following statements best describes what is happening on the graph of f at x=−2?
Choose 1 answer:
Choose 1 answer:
(Choice A) f(−2) is a relative minimum value of f. A f(−2) is a relative minimum value of f.
(Choice B) f(−2) is a relative maximum value of f. B f(−2) is a relative maximum value of f.
(Choice C) f(−2) is the absolute minimum value of f. C f(−2) is the absolute minimum value of f.
(Choice D) f(−2) is the absolute maximum value of f. D f(−2) is the absolute maximum value of f.
(Choice E) The point (−2,f(−2)) is a point of inflection of f. E The point (−2,f(−2)) is a point of inflection of f.
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3502 | https://oeis.org/A000142 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Factorial numbers: n! = 1234...n (order of symmetric group S_n, number of permutations of n letters).
(Formerly M1675 N0659)
1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000, 6402373705728000, 121645100408832000, 2432902008176640000, 51090942171709440000, 1124000727777607680000
(list;
graph;
refs;
listen;
history;
text;
internal format)
The earliest publication that discusses this sequence appears to be the Sepher Yezirah [Book of Creation], circa AD 300. (See Knuth, also the Zeilberger link.) - N. J. A. Sloane, Apr 07 2014
For n >= 1, a(n) is the number of n X n (0,1) matrices with each row and column containing exactly one entry equal to 1.
This sequence is the BinomialMean transform of A000354. (See A075271 for definition.) - John W. Layman, Sep 12 2002 [This is easily verified from the Paul Barry formula for A000354, by interchanging summations and using the formula: Sum_k (-1)^k C(n-i, k) = KroneckerDelta(i,n). - David Callan, Aug 31 2003]
Number of distinct subsets of T(n-1) elements with 1 element A, 2 elements B, ..., n - 1 elements X (e.g., at n = 5, we consider the distinct subsets of ABBCCCDDDD and there are 5! = 120). - Jon Perry, Jun 12 2003
n! is the smallest number with that prime signature. E.g., 720 = 2^4 3^2 5. - Amarnath Murthy, Jul 01 2003
a(n) is the permanent of the n X n matrix M with M(i, j) = 1. - Philippe Deléham, Dec 15 2003
Given n objects of distinct sizes (e.g., areas, volumes) such that each object is sufficiently large to simultaneously contain all previous objects, then n! is the total number of essentially different arrangements using all n objects. Arbitrary levels of nesting of objects are permitted within arrangements. (This application of the sequence was inspired by considering leftover moving boxes.) If the restriction exists that each object is able or permitted to contain at most one smaller (but possibly nested) object at a time, the resulting sequence begins 1,2,5,15,52 (Bell Numbers?). Sets of nested wooden boxes or traditional nested Russian dolls come to mind here. - Rick L. Shepherd, Jan 14 2004
From Michael Somos, Mar 04 2004; edited by M. F. Hasler, Jan 02 2015: (Start)
Stirling transform of [2, 2, 6, 24, 120, ...] is A052856 = [2, 2, 4, 14, 76, ...].
Stirling transform of [1, 2, 6, 24, 120, ...] is A000670 = [1, 3, 13, 75, ...].
Stirling transform of [0, 2, 6, 24, 120, ...] is A052875 = [0, 2, 12, 74, ...].
Stirling transform of [1, 1, 2, 6, 24, 120, ...] is A000629 = [1, 2, 6, 26, ...].
Stirling transform of [0, 1, 2, 6, 24, 120, ...] is A002050 = [0, 1, 5, 25, 140, ...].
Stirling transform of (A165326A089064)(1...) = [1, 0, 1, -1, 8, -26, 194, ...] is [1, 1, 2, 6, 24, 120, ...] (this sequence). (End)
First Eulerian transform of 1, 1, 1, 1, 1, 1... The first Eulerian transform transforms a sequence s to a sequence t by the formula t(n) = Sum_{k=0..n} e(n, k)s(k), where e(n, k) is a first-order Eulerian number [A008292]. - Ross La Haye, Feb 13 2005
Conjecturally, 1, 6, and 120 are the only numbers which are both triangular and factorial. - Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Mar 30 2005
n! is the n-th finite difference of consecutive n-th powers. E.g., for n = 3, [0, 1, 8, 27, 64, ...] -> [1, 7, 19, 37, ...] -> [6, 12, 18, ...] -> [6, 6, ...]. - Bryan Jacobs (bryanjj(AT)gmail.com), Mar 31 2005
a(n+1) = (n+1)! = 1, 2, 6, ... has e.g.f. 1/(1-x)^2. - Paul Barry, Apr 22 2005
Write numbers 1 to n on a circle. Then a(n) = sum of the products of all n - 2 adjacent numbers. E.g., a(5) = 123 + 234 + 345 + 451 +512 = 120. - Amarnath Murthy, Jul 10 2005
The number of chains of maximal length in the power set of {1, 2, ..., n} ordered by the subset relation. - Rick L. Shepherd, Feb 05 2006
The number of circular permutations of n letters for n >= 0 is 1, 1, 1, 2, 6, 24, 120, 720, 5040, 40320, ... - Xavier Noria (fxn(AT)hashref.com), Jun 04 2006
a(n) is the number of deco polyominoes of height n (n >= 1; see definitions in the Barcucci et al. references). - Emeric Deutsch, Aug 07 2006
a(n) is the number of partition tableaux of size n. See Steingrimsson/Williams link for the definition. - David Callan, Oct 06 2006
Consider the n! permutations of the integer sequence [n] = 1, 2, ..., n. The i-th permutation consists of ncycle(i) permutation cycles. Then, if the Sum_{i=1..n!} 2^ncycle(i) runs from 1 to n!, we have Sum_{i=1..n!} 2^ncycle(i) = (n+1)!. E.g., for n = 3 we have ncycle(1) = 3, ncycle(2) = 2, ncycle(3) = 1, ncycle(4) = 2, ncycle(5) = 1, ncycle(6) = 2 and 2^3 + 2^2 + 2^1 + 2^2 + 2^1 + 2^2 = 8 + 4 + 2 + 4 + 2 + 4 = 24 = (n+1)!. - Thomas Wieder, Oct 11 2006
a(n) is the number of set partitions of {1, 2, ..., 2n - 1, 2n} into blocks of size 2 (perfect matchings) in which each block consists of one even and one odd integer. For example, a(3) = 6 counts 12-34-56, 12-36-45, 14-23-56, 14-25-36, 16-23-45, 16-25-34. - David Callan, Mar 30 2007
Consider the multiset M = [1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...] = [1, 2, 2, ..., n x 'n'] and form the set U (where U is a set in the strict sense) of all subsets N (where N may be a multiset again) of M. Then the number of elements |U| of U is equal to (n+1)!. E.g. for M = [1, 2, 2] we get U = and |U| = 3! = 6. This observation is a more formal version of the comment given already by Rick L. Shepherd, Jan 14 2004. - Thomas Wieder, Nov 27 2007
For n >= 1, a(n) = 1, 2, 6, 24, ... are the positions corresponding to the 1's in decimal expansion of Liouville's constant (A012245). - Paul Muljadi, Apr 15 2008
Triangle A144107 has n! for row sums (given n > 0) with right border n! and left border A003319, the INVERTi transform of (1, 2, 6, 24, ...). - Gary W. Adamson, Sep 11 2008
Equals INVERT transform of A052186 and row sums of triangle A144108. - Gary W. Adamson, Sep 11 2008
From Abdullahi Umar, Oct 12 2008: (Start)
a(n) is also the number of order-decreasing full transformations (of an n-chain).
a(n-1) is also the number of nilpotent order-decreasing full transformations (of an n-chain). (End)
n! is also the number of optimal broadcast schemes in the complete graph K_{n}, equivalent to the number of binomial trees embedded in K_{n} (see Calin D. Morosan, Information Processing Letters, 100 (2006), 188-193). - Calin D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
Let S_{n} denote the n-star graph. The S_{n} structure consists of n S_{n-1} structures. This sequence gives the number of edges between the vertices of any two specified S_{n+1} structures in S_{n+2} (n >= 1). - K.V.Iyer, Mar 18 2009
Chromatic invariant of the sun graph S_{n-2}.
It appears that a(n+1) is the inverse binomial transform of A000255. - Timothy Hopper, Aug 20 2009
a(n) is also the determinant of a square matrix, An, whose coefficients are the reciprocals of beta function: a{i, j} = 1/beta(i, j), det(An) = n!. - Enrique Pérez Herrero, Sep 21 2009
The asymptotic expansions of the exponential integrals E(x, m = 1, n = 1) ~ exp(-x)/x(1 - 1/x + 2/x^2 - 6/x^3 + 24/x^4 + ...) and E(x, m = 1, n = 2) ~ exp(-x)/x(1 - 2/x + 6/x^2 - 24/x^3 + ...) lead to the factorial numbers. See A163931 and A130534 for more information. - Johannes W. Meijer, Oct 20 2009
Satisfies A(x)/A(x^2), A(x) = A173280. - Gary W. Adamson, Feb 14 2010
a(n) = G^n where G is the geometric mean of the first n positive integers. - Jaroslav Krizek, May 28 2010
Increasing colored 1-2 trees with choice of two colors for the rightmost branch of nonleaves. - Wenjin Woan, May 23 2011
Number of necklaces with n labeled beads of 1 color. - Robert G. Wilson v, Sep 22 2011
The sequence 1!, (2!)!, ((3!)!)!, (((4!)!)!)!, ..., ((...(n!)!)...)! (n times) grows too rapidly to have its own entry. See Hofstadter.
The e.g.f. of 1/a(n) = 1/n! is BesselI(0, 2sqrt(x)). See Abramowitz-Stegun, p. 375, 9.3.10. - Wolfdieter Lang, Jan 09 2012
a(n) is the length of the n-th row which is the sum of n-th row in triangle A170942. - Reinhard Zumkeller, Mar 29 2012
Number of permutations of elements 1, 2, ..., n + 1 with a fixed element belonging to a cycle of length r does not depend on r and equals a(n). - Vladimir Shevelev, May 12 2012
a(n) is the number of fixed points in all permutations of 1, ..., n: in all n! permutations, 1 is first exactly (n-1)! times, 2 is second exactly (n-1)! times, etc., giving (n-1)!n = n!. - Jon Perry, Dec 20 2012
For n >= 1, a(n-1) is the binomial transform of A000757. See Moreno-Rivera. - Luis Manuel Rivera Martínez, Dec 09 2013
Each term is divisible by its digital root (A010888). - Ivan N. Ianakiev, Apr 14 2014
For m >= 3, a(m-2) is the number hp(m) of acyclic Hamiltonian paths in a simple graph with m vertices, which is complete except for one missing edge. For m < 3, hp(m)=0. - Stanislav Sykora, Jun 17 2014
a(n) is the number of increasing forests with n nodes. - Brad R. Jones, Dec 01 2014
The factorial numbers can be calculated by means of the recurrence n! = (floor(n/2)!)^2 sf(n) where sf(n) are the swinging factorials A056040. This leads to an efficient algorithm if sf(n) is computed via prime factorization. For an exposition of this algorithm see the link below. - Peter Luschny, Nov 05 2016
Treeshelves are ordered (plane) binary (0-1-2) increasing trees where the nodes of outdegree 1 come in 2 colors. There are n! treeshelves of size n, and classical Françon's bijection maps bijectively treeshelves into permutations. - Sergey Kirgizov, Dec 26 2016
Satisfies Benford's law [Diaconis, 1977; Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017
a(n) = Sum((d_p)^2), where d_p is the number of standard tableaux in the Ferrers board of the integer partition p and summation is over all integer partitions p of n. Example: a(3) = 6. Indeed, the partitions of 3 are , [2,1], and [1,1,1], having 1, 2, and 1 standard tableaux, respectively; we have 1^2 + 2^2 + 1^2 = 6. - Emeric Deutsch, Aug 07 2017
a(n) is the n-th derivative of x^n. - Iain Fox, Nov 19 2017
a(n) is the number of maximum chains in the n-dimensional Boolean cube {0,1}^n in respect to the relation "precedes". It is defined as follows: for arbitrary vectors u, v of {0,1}^n, such that u = (u_1, u_2, ..., u_n) and v = (v_1, v_2, ..., v_n), "u precedes v" if u_i <= v_i, for i=1, 2, ..., n. - Valentin Bakoev, Nov 20 2017
a(n) is the number of shortest paths (for example, obtained by Breadth First Search) between the nodes (0,0,...,0) (i.e., the all-zeros vector) and (1,1,...,1) (i.e., the all-ones vector) in the graph H_n, corresponding to the n-dimensional Boolean cube {0,1}^n. The graph is defined as H_n = (V_n, E_n), where V_n is the set of all vectors of {0,1}^n, and E_n contains edges formed by each pair adjacent vectors. - Valentin Bakoev, Nov 20 2017
a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = sigma(gcd(i,j)) for 1 <= i,j <= n. - Bernard Schott, Dec 05 2018
a(n) is also the number of inversion sequences of length n. A length n inversion sequence e_1, e_2, ..., e_n is a sequence of n integers such that 0 <= e_i < i. - Juan S. Auli, Oct 14 2019
The term "factorial" ("factorielle" in French) was coined by the French mathematician Louis François Antoine Arbogast (1759-1803) in 1800. The notation "!" was first used by the French mathematician Christian Kramp (1760-1826) in 1808. - Amiram Eldar, Apr 16 2021
Also the number of signotopes of rank 2, i.e., mappings X:{{1..n} choose 2}->{+,-} such that for any three indices a < b < c, the sequence X(a,b), X(a,c), X(b,c) changes its sign at most once (see Felsner-Weil reference). - Manfred Scheucher, Feb 09 2022
a(n) is also the number of labeled commutative semisimple rings with n elements. As an example the only commutative semisimple rings with 4 elements are F_4 and F_2 X F_2. They both have exactly 2 automorphisms, hence a(4)=24/2+24/2=24. - Paul Laubie, Mar 05 2024
a(n) is the number of extremely unlucky Stirling permutations of order n+1; i.e., the number of Stirling permutations of order n+1 that have exactly one lucky car. - Bridget Tenner, Apr 09 2024
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 125; also p. 90, ex. 3.
Florian Cajori, A History of Mathematical Notations, Dover edition (2012), pars. 448-449.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 64-66.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.1 Symbols Galore, p. 106.
Douglas R. Hofstadter, Fluid concepts & creative analogies: computer models of the fundamental mechanisms of thought, Basic Books, 1995, pages 44-46.
A. N. Khovanskii. The Application of Continued Fractions and Their Generalizations to Problem in Approximation Theory. Groningen: Noordhoff, Netherlands, 1963. See p. 141 (10.19).
D. E. Knuth, The Art of Computer Programming, Vol. 3, Section 5.1.2, p. 23. [From N. J. A. Sloane, Apr 07 2014]
J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 693 pp. 90, 297, Ellipses Paris 2004.
A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
Sepher Yezirah [Book of Creation], circa AD 300. See verse 52.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 2, pages 19-24.
D. Stanton and D. White, Constructive Combinatorics, Springer, 1986; see p. 91.
Carlo Suares, Sepher Yetsira, Shambhala Publications, 1976. See verse 52.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 102.
LINKS
N. J. A. Sloane, The first 100 factorials: Table of n, n! for n = 0..100
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
L. F. A. Arbogast, Du calcul des dérivations, Strasbourg: Levrault, 1800.
S. B. Akers and B. Krishnamurthy, A group-theoretic model for symmetric interconnection networks, IEEE Trans. Comput., 38(4), April 1989, 555-566.
Masanori Ando, Odd number and Trapezoidal number, arXiv:1504.04121 [math.CO], 2015.
David Applegate and N. J. A. Sloane, Table giving cycle index of S_0 through S_40 in Maple format. [gzipped]
C. Banderier, M. Bousquet-Mélou, A. Denise, P. Flajolet, D. Gardy, and D. Gouyou-Beauchamps, Generating Functions for Generating Trees, Discrete Mathematics 246(1-3), March 2002, pp. 29-55.
Stefano Barbero, Umberto Cerruti, and Nadir Murru, On the operations of sequences in rings and binomial type sequences, Ricerche di Matematica (2018), pp 1-17., also arXiv:1805.11922 [math.NT], 2018.
E. Barcucci, A. Del Lungo, and R. Pinzani, "Deco" polyominoes, permutations and random generation, Theoretical Computer Science, 159, 1996, 29-42.
E. Barcucci, A. Del Lungo, R. Pinzani, and R. Sprugnoli, La hauteur des polyominos dirigés verticalement convexes, Actes du 31e Séminaire Lotharingien de Combinatoire, Publ. IRMA, Université Strasbourg I (1993).
Jean-Luc Baril, Sergey Kirgizov, and Vincent Vajnovszki, Patterns in treeshelves, Discrete Mathematics, Vol. 340, No. 12 (2017), 2946-2954, arXiv:1611.07793 [cs.DM], 2016.
A. Berger and T. P. Hill, What is Benford's Law?, Notices, Amer. Math. Soc., 64:2 (2017), 132-134.
M. Bhargava, The factorial function and generalizations, Amer. Math. Monthly, 107 (Nov. 2000), 783-799.
Natasha Blitvić and Einar Steingrímsson, Permutations, moments, measures, arXiv:2001.00280 [math.CO], 2020.
Henry Bottomley, Illustration of initial terms.
Douglas Butler, Factorials!.
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Index entries for "core" sequences.
Index to divisibility sequences.
Index entries for sequences related to factorial numbers.
Index entries for sequences related to Benford's law.
FORMULA
Exp(x) = Sum_{m >= 0} x^m/m!. - Mohammad K. Azarian, Dec 28 2010
Sum_{i=0..n} (-1)^i i^n binomial(n, i) = (-1)^n n!. - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
Sum_{i=0..n} (-1)^i (n-i)^n binomial(n, i) = n!. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 10 2007
The sequence trivially satisfies the recurrence a(n+1) = Sum_{k=0..n} binomial(n,k) a(k)a(n-k). - Robert FERREOL, Dec 05 2009
D-finite with recurrence: a(n) = na(n-1), n >= 1. n! ~ sqrt(2Pi) n^(n+1/2) / e^n (Stirling's approximation).
a(0) = 1, a(n) = subs(x = 1, (d^n/dx^n)(1/(2-x))), n = 1, 2, ... - Karol A. Penson, Nov 12 2001
E.g.f.: 1/(1-x). - Michael Somos, Mar 04 2004
a(n) = Sum_{k=0..n} (-1)^(n-k)A000522(k)binomial(n, k) = Sum_{k=0..n} (-1)^(n-k)(x+k)^nbinomial(n, k). - Philippe Deléham, Jul 08 2004
Binomial transform of A000166. - Ross La Haye, Sep 21 2004
a(n) = Sum_{i=1..n} ((-1)^(i-1) sum of 1..n taken n - i at a time) - e.g., 4! = (123 + 124 + 134 + 234) - (12 + 13 + 14 + 23 + 24 + 34) + (1 + 2 + 3 + 4) - 1 = (6 + 8 + 12 + 24) - (2 + 3 + 4 + 6 + 8 + 12) + 10 - 1 = 50 - 35 + 10 - 1 = 24. - Jon Perry, Nov 14 2005
a(n) = (n-1)(a(n-1) + a(n-2)), n >= 2. - Matthew J. White, Feb 21 2006
1 / a(n) = determinant of matrix whose (i,j) entry is (i+j)!/(i!(j+1)!) for n > 0. This is a matrix with Catalan numbers on the diagonal. - Alexander Adamchuk, Jul 04 2006
Hankel transform of A074664. - Philippe Deléham, Jun 21 2007
For n >= 2, a(n-2) = (-1)^nSum_{j=0..n-1} (j+1)Stirling1(n,j+1). - Milan Janjic, Dec 14 2008
From Paul Barry, Jan 15 2009: (Start)
G.f.: 1/(1-x-x^2/(1-3x-4x^2/(1-5x-9x^2/(1-7x-16x^2/(1-9x-25x^2... (continued fraction), hence Hankel transform is A055209.
G.f. of (n+1)! is 1/(1-2x-2x^2/(1-4x-6x^2/(1-6x-12x^2/(1-8x-20x^2... (continued fraction), hence Hankel transform is A059332. (End)
a(n) = Product_{p prime} p^(Sum_{k > 0} floor(n/p^k)) by Legendre's formula for the highest power of a prime dividing n!. - Jonathan Sondow, Jul 24 2009
a(n) = A053657(n)/A163176(n) for n > 0. - Jonathan Sondow, Jul 26 2009
It appears that a(n) = (1/0!) + (1/1!)n + (3/2!)n(n-1) + (11/3!)n(n-1)(n-2) + ... + (b(n)/n!)n(n-1)...21, where a(n) = (n+1)! and b(n) = A000255. - Timothy Hopper, Aug 12 2009
Sum_{n >= 0} 1/a(n) = e. - Jaume Oliver Lafont, Mar 03 2009
a(n) = a(n-1)^2/a(n-2) + a(n-1), n >= 2. - Jaume Oliver Lafont, Sep 21 2009
a(n) = Gamma(n+1). - Enrique Pérez Herrero, Sep 21 2009
a(n) = A173333(n,1). - Reinhard Zumkeller, Feb 19 2010
a(n) = A_{n}(1) where A_{n}(x) are the Eulerian polynomials. - Peter Luschny, Aug 03 2010
a(n) = n(2a(n-1) - (n-1)a(n-2)), n > 1. - Gary Detlefs, Sep 16 2010
1/a(n) = -Sum_{k=1..n+1} (-2)^k(n+k+2)a(k)/(a(2k+1)a(n+1-k)). - Groux Roland, Dec 08 2010
From Vladimir Shevelev, Feb 21 2011: (Start)
a(n) = Product_{p prime, p <= n} p^(Sum_{i >= 1} floor(n/p^i)).
The infinitary analog of this formula is: a(n) = Product_{q terms of A050376 <= n} q^((n)_q), where (n)_q denotes the number of those numbers <= n for which q is an infinitary divisor (for the definition see comment in A037445). (End)
The terms are the denominators of the expansion of sinh(x) + cosh(x). - Arkadiusz Wesolowski, Feb 03 2012
G.f.: 1 / (1 - x / (1 - x / (1 - 2x / (1 - 2x / (1 - 3x / (1 - 3x / ... )))))). - Michael Somos, May 12 2012
G.f. 1 + x/(G(0)-x) where G(k) = 1 - (k+1)x/(1 - x(k+2)/G(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Aug 14 2012
G.f.: W(1,1;-x)/(W(1,1;-x) - xW(1,2;-x)), where W(a,b,x) = 1 - abx/1! + a(a+1)b(b+1)x^2/2! - ... + a(a+1)...(a+n-1)b(b+1)...(b+n-1)x^n/n! + ...; see [A. N. Khovanskii, p. 141 (10.19)]. - Sergei N. Gladkovskii, Aug 15 2012
From Sergei N. Gladkovskii, Dec 26 2012: (Start)
G.f.: A(x) = 1 + x/(G(0) - x) where G(k) = 1 + (k+1)x - x(k+2)/G(k+1); (continued fraction).
Let B(x) be the g.f. for A051296, then A(x) = 2 - 1/B(x). (End)
G.f.: 1 + x(G(0) - 1)/(x-1) where G(k) = 1 - (2k+1)/(1-x/(x - 1/(1 - (2k+2)/(1-x/(x - 1/G(k+1) ))))); (continued fraction). - Sergei N. Gladkovskii, Jan 15 2013
G.f.: 1 + x(1 - G(0))/(sqrt(x)-x) where G(k) = 1 - (k+1)sqrt(x)/(1-sqrt(x)/(sqrt(x)-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 25 2013
G.f.: 1 + x/G(0) where G(k) = 1 - x(k+2)/( 1 - x(k+1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 23 2013
a(n) = det(S(i+1, j), 1 <= i, j <=n ), where S(n,k) are Stirling numbers of the second kind. - Mircea Merca, Apr 04 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x(k+1)/(x(k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 1/(1 - 1/(2x(k+1)) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 29 2013
G.f.: G(0), where G(k) = 1 + x(2k+1)/(1 - x(2k+2)/(x(2k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 07 2013
a(n) = P(n-1, floor(n/2)) floor(n/2)! (n - (n-2)((n+1) mod 2)), where P(n, k) are the k-permutations of n objects, n > 0. - Wesley Ivan Hurt, Jun 07 2013
a(n) = a(n-2)(n-1)^2 + a(n-1), n > 1. - Ivan N. Ianakiev, Jun 18 2013
a(n) = a(n-2)(n^2-1) - a(n-1), n > 1. - Ivan N. Ianakiev, Jun 30 2013
G.f.: 1 + x/Q(0), m=+2, where Q(k) = 1 - 2x(2k+1) - mx^2(k+1)(2k+1)/( 1 - 2x(2k+2) - mx^2(k+1)(2k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Sep 24 2013
a(n) = A245334(n,n). - Reinhard Zumkeller, Aug 31 2014
a(n) = Product_{i = 1..n} A014963^floor(n/i) = Product_{i = 1..n} A003418(floor(n/i)). - Matthew Vandermast, Dec 22 2014
a(n) = round(Sum_{k>=1} log(k)^n/k^2), for n>=1, which is related to the n-th derivative of the Riemann zeta function at x=2 as follows: round((-1)^n zeta^(n)(2)). Also see A073002. - Richard R. Forberg, Dec 30 2014
a(n) ~ Sum_{j>=0} j^n/e^j, where e = A001113. When substituting a generic variable for "e" this infinite sum is related to Eulerian polynomials. See A008292. This approximation of n! is within 0.4% at n = 2. See A255169. Accuracy, as a percentage, improves rapidly for larger n. - Richard R. Forberg, Mar 07 2015
a(n) = Product_{k=1..n} (C(n+1, 2)-C(k, 2))/(2k-1); see Masanori Ando link. - Michel Marcus, Apr 17 2015
Sum_{n>=0} a(n)/(a(n + 1)a(n + 2)) = Sum_{n>=0} 1/((n + 2)(n + 1)^2a(n)) = 2 - exp(1) - gamma + Ei(1) = 0.5996203229953..., where gamma = A001620, Ei(1) = A091725. - Ilya Gutkovskiy, Nov 01 2016
a(2^n) = 2^(2^n - 1) 1!! 3!! 7!! ... (2^n - 1)!!. For example, 16! = 2^15(13)(1357)(13579111315) = 20922789888000. - Peter Bala, Nov 01 2016
a(n) = sum(prod(B)), where the sum is over all subsets B of {1,2,...,n-1} and where prod(B) denotes the product of all the elements of set B. If B is a singleton set with element b, then we define prod(B)=b, and, if B is the empty set, we define prod(B) to be 1. For example, a(4)=(123)+(12)+(13)+(23)+(1)+(2)+(3)+1=24. - Dennis P. Walsh, Oct 23 2017
Sum_{n >= 0} 1/(a(n)(n+2)) = 1. - Multiplying the denominator by (n+2) in Jaume Oliver Lafont's entry above creates a telescoping sum. - Fred Daniel Kline, Nov 08 2020
O.g.f.: Sum_{k >= 0} k!x^k = Sum_{k >= 0} (k+y)^kx^k/(1 + (k+y)x)^(k+1) for arbitrary y. - Peter Bala, Mar 21 2022
E.g.f.: 1/(1 + LambertW(-xexp(-x))) = 1/(1-x), see A258773. -(1/x)substitute(z = xexp(-x), z(d/dz)LambertW(-z)) = 1/(1 - x). See A075513. Proof: Use the compositional inverse (xexp(-x))^[-1] = -LambertW(-z). See A000169 or A152917, and Richard P. Stanley: Enumerative Combinatorics, vol. 2, p. 37, eq. (5.52). - Wolfdieter Lang, Oct 17 2022
Sum_{k >= 1} 1/10^a(k) = A012245 (Liouville constant). - Bernard Schott, Dec 18 2022
From David Ulgenes, Sep 19 2023: (Start)
1/a(n) = (e/(2Pin)Integral_{x=-oo..oo} cos(x-narctan(x))/(1+x^2)^(n/2) dx). Proof: take the real component of Laplace's integral for 1/Gamma(x).
a(n) = Integral_{x=0..1} e^(-t)LerchPhi(1/e, -n, t) dt. Proof: use the relationship Gamma(x+1) = Sum_{n >= 0} Integral_{t=n..n+1} e^(-t)t^x dt = Sum_{n >= 0} Integral_{t=0..1} e^(-(t+n))(t+n)^x dt and interchange the order of summation and integration.
Conjecture: a(n) = 1/(2Pi)Integral_{x=-oo..oo}(n+ix+1)!/(ix+1)-(n+ix-1)!/(ix-1)dx. (End)
a(n) = floor(b(n)^n / (floor(((2^b(n) + 1) / 2^n)^b(n)) mod 2^b(n))), where b(n) = (n + 1)^(n + 2) = A007778(n+1). Joint work with Mihai Prunescu. - Lorenzo Sauras Altuzarra, Oct 18 2023
a(n) = e^(Integral_{x=1..n+1} Psi(x) dx) where Psi(x) is the digamma function. - Andrea Pinos, Jan 10 2024
a(n) = Integral_{x=0..oo} e^(-x^(1/n)) dx, for n > 0. - Ridouane Oudra, Apr 20 2024
O.g.f.: N(x) = hypergeometric([1,1], [], x) = LaplaceTransform(x/(1-x))/x, satisfying x^2N'(x) + (x-1)N(x) + 1 = 0, with N(0) = 1. - Wolfdieter Lang, May 31 2025
EXAMPLE
There are 3! = 123 = 6 ways to arrange 3 letters {a, b, c}, namely abc, acb, bac, bca, cab, cba.
Let n = 2. Consider permutations of {1, 2, 3}. Fix element 3. There are a(2) = 2 permutations in each of the following cases: (a) 3 belongs to a cycle of length 1 (permutations (1, 2, 3) and (2, 1, 3)); (b) 3 belongs to a cycle of length 2 (permutations (3, 2, 1) and (1, 3, 2)); (c) 3 belongs to a cycle of length 3 (permutations (2, 3, 1) and (3, 1, 2)). - Vladimir Shevelev, May 13 2012
G.f. = 1 + x + 2x^2 + 6x^3 + 24x^4 + 120x^5 + 720x^6 + 5040x^7 + ...
MAPLE
A000142 := n -> n!; seq(n!, n=0..20);
spec := [ S, {S=Sequence(Z) }, labeled ]; seq(combstructcount, n=0..20);
Maple program for computing cycle indices of symmetric groups
M:=6: f:=array(0..M): f:=1: print(n=, 0); print(f); f:=x: print(n=, 1); print(f); for n from 2 to M do f[n]:=expand((1/n)add( x[l]f[n-l], l=1..n)); print(n=, n); print(f[n]); od:
with(combstruct):ZL0:=[S, {S=Set(Cycle(Z, card>0))}, labeled]: seq(count(ZL0, size=n), n=0..20); # Zerinvary Lajos, Sep 26 2007
MATHEMATICA
Table[Factorial[n], {n, 0, 20}] ( Stefan Steinerberger, Mar 30 2006 )
FoldList[#1 #2 &, 1, Range@ 20] ( Robert G. Wilson v, May 07 2011 )
Range! ( Harvey P. Dale, Nov 19 2011 )
RecurrenceTable[{a[n] == na[n - 1], a == 1}, a, {n, 0, 22}] ( Ray Chandler, Jul 30 2015 )
PROG
(Axiom) [factorial(n) for n in 0..10]
(Magma) a:= func< n | Factorial(n) >; [ a(n) : n in [0..10]];
(Haskell)
a000142 :: (Enum a, Num a, Integral t) => t -> a
a000142 n = product [1 .. fromIntegral n]
a000142_list = 1 : zipWith () [1..] a000142_list
-- Reinhard Zumkeller, Mar 02 2014, Nov 02 2011, Apr 21 2011
for i in range(1, 1000):
y = i
for j in range(1, i):
y = i - j
print(y, "\n")
(Python)
import math
for i in range(1, 1000):
Ruskin Harding, Feb 22 2013
(PARI) a(n)=prod(i=1, n, i) \ Felix Fröhlich, Aug 17 2014
(PARI) {a(n) = if(n<0, 0, n!)}; / Michael Somos, Mar 04 2004 /
(Sage) [factorial(n) for n in (1..22)] # Giuseppe Coppoletta, Dec 05 2014
(GAP) List([0..22], Factorial); # Muniru A Asiru, Dec 05 2018
(Scala) (1: BigInt).to(24: BigInt).scanLeft(1: BigInt)(_ _) // Alonso del Arte, Mar 02 2019
(Julia) print([factorial(big(n)) for n in 0:28]) # Paul Muljadi, May 01 2024
Cf. A000165, A001044, A001563, A003422, A009445, A010050, A012245, A033312, A034886, A038507, A047920, A048631.
Factorial base representation: A007623.
Cf. A003319, A052186, A144107, A144108. - Gary W. Adamson, Sep 11 2008
Complement of A063992. - Reinhard Zumkeller, Oct 11 2008
Cf. A053657, A163176. - Jonathan Sondow, Jul 26 2009
Cf. A173280. - Gary W. Adamson, Feb 14 2010
Boustrophedon transforms: A230960, A230961.
Cf. A233589.
Cf. A245334.
A row of the array in A249026.
Cf. A001013 (multiplicative closure).
For factorials with initial digit d (1 <= d <= 9) see A045509, A045510, A045511, A045516, A045517, A045518, A282021, A045519; A045520, A045521, A045522, A045523, A045524, A045525, A045526, A045527, A045528, A045529.
Cf. A000169, A075513, A152917, A258773.
Sequence in context: A133942 A159333 A165233 A104150 A358185 A370360
Adjacent sequences: A000139 A000140 A000141 A000143 A000144 A000145
core,easy,nonn,nice,changed
N. J. A. Sloane |
3503 | https://cdn.kutasoftware.com/Worksheets/Alg1/Systems%20of%20Equations%20Substitution.pdf | ©P 2280S1i2G GKquhtlaY oSWo1fwtZwGalrUen SLCLWCr.J a CAVlolr GrUiqg9hetDsg OryewsdegrGvkeDdz.J H OMlaAdkeT LwqiUtphO eIGnffpiYn0i5tZeX 4AvlQgRe2bIrSaR f1W.y Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Period_ Date___ Solving Systems of Equations by Substitution Solve each system by substitution.
1) y = 6 x − 11 −2 x − 3 y = −7 2) 2 x − 3 y = −1 y = x − 1 3) y = −3 x + 5 5 x − 4 y = −3 4) −3 x − 3 y = 3 y = −5 x − 17 5) y = −2 4 x − 3 y = 18 6) y = 5 x − 7 −3 x − 2 y = −12 7) −4 x + y = 6 −5 x − y = 21 8) −7 x − 2 y = −13 x − 2 y = 11 9) −5 x + y = −2 −3 x + 6 y = −12 10) −5 x + y = −3 3 x − 8 y = 24 -1-©5 T2t0G1h2s AKGuqtbak FSDoafRtuwaalrKeR vL0LUCq.E n hAol8lw NrkiJgVhPt2sb VrDexs8e9rYvxeFdS.e d jM4aNdJew rwqi9tThU jI9n9fPilnCi4tAeZ GAulCgpeRbFrdae g1N.D Worksheet by Kuta Software LLC 11) x + 3 y = 1 −3 x − 3 y = −15 12) −3 x − 8 y = 20 −5 x + y = 19 13) −3 x + 3 y = 4 − x + y = 3 14) −3 x + 3 y = 3 −5 x + y = 13 15) 6 x + 6 y = −6 5 x + y = −13 16) 2 x + y = 20 6 x − 5 y = 12 17) −3 x − 4 y = 2 3 x + 3 y = −3 18) −2 x + 6 y = 6 −7 x + 8 y = −5 19) −5 x − 8 y = 17 2 x − 7 y = −17 20) −2 x − y = −9 5 x − 2 y = 18 -2-©8 y2J0J132e NKaujt0aM ySeoCfCt7wVayrleK 2LzLxCf.t w oAMlZlq wrZiugOh8tYsl CrGeMsYeNrUvgendm.R 4 3MEaFdMeW hwIiftBhQ 5IOnxfFisnfirtbeX nAmligBeAbxrhaD s1v.F Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Period Date__ Solving Systems of Equations by Substitution Solve each system by substitution.
1) y = 6 x − 11 −2 x − 3 y = −7 (2, 1) 2) 2 x − 3 y = −1 y = x − 1 (4, 3) 3) y = −3 x + 5 5 x − 4 y = −3 (1, 2) 4) −3 x − 3 y = 3 y = −5 x − 17 (−4, 3) 5) y = −2 4 x − 3 y = 18 (3, −2) 6) y = 5 x − 7 −3 x − 2 y = −12 (2, 3) 7) −4 x + y = 6 −5 x − y = 21 (−3, −6) 8) −7 x − 2 y = −13 x − 2 y = 11 (3, −4) 9) −5 x + y = −2 −3 x + 6 y = −12 (0, −2) 10) −5 x + y = −3 3 x − 8 y = 24 (0, −3) -1-©L 9200M1s2A dK1uYtmah 1SgoffFtlwhaxrkeL 6LDLQCM.z H lA7ldlB oroihg5hbtMsW 3rge9sTe3rcvBeldz.5 R eMcabdyeF owTiMtbhZ VIOnGf9ien4ittSeK iAGlAgYebbzr3ad A10.I Worksheet by Kuta Software LLC 11) x + 3 y = 1 −3 x − 3 y = −15 (7, −2) 12) −3 x − 8 y = 20 −5 x + y = 19 (−4, −1) 13) −3 x + 3 y = 4 − x + y = 3 No solution 14) −3 x + 3 y = 3 −5 x + y = 13 (−3, −2) 15) 6 x + 6 y = −6 5 x + y = −13 (−3, 2) 16) 2 x + y = 20 6 x − 5 y = 12 (7, 6) 17) −3 x − 4 y = 2 3 x + 3 y = −3 (−2, 1) 18) −2 x + 6 y = 6 −7 x + 8 y = −5 (3, 2) 19) −5 x − 8 y = 17 2 x − 7 y = −17 (−5, 1) 20) −2 x − y = −9 5 x − 2 y = 18 (4, 1) -2-Create your own worksheets like this one with Infinite Algebra 1. Free trial available at KutaSoftware.com |
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Cyclotron is a type of particle accelerator used to accelerate charged particles to high speeds. It was invented in 1929 by Ernest O. Lawrence. Cyclotrons are widely used in scientific research, medicine, and industry.
The basic principle of a cyclotron involves using a combination of electric and magnetic fields to accelerate charged particles along a circular path. This article covers the basics of cyclotron, including its definition, working, types, and other details related to it.
Table of Content
What is a Cyclotron?
Components and Operation of a Cyclotron
Working Principle of Cyclotron
Types of Cyclotrons
Advantages of Cyclotron
Limitations Of Cyclotron
What is a Cyclotron?
A cyclotron is a type of particle accelerator, a device used to accelerate charged particles to high speeds. Ernest O. Lawrence invented the cyclotron in 1929. Since then, it has become a fundamental tool in various scientific disciplines, including nuclear physics, particle physics, and medicine.
The basic principle of a cyclotron involves the use of electric and magnetic fields to accelerate charged particles in a spiral path. These particles oscillate and gain energy after being subjected to a magnetic field of a specific frequency. This phenomenon is called the ion cyclotron resonance. This depends on the mass and charge of the particle and the strength of the magnetic field. As the particles spiral outward, they gain energy with each revolution.
Properties of Cyclotron
Cyclotrons have several properties that make them valuable tools in scientific research, medicine, and industry. Some of these properties include:
Acceleration of Charged Particles: Cyclotrons are capable of accelerating charged particles, such as protons or alpha particles, to very high energies. This property is essential for conducting experiments in particle physics, and nuclear physics, and for various practical applications.
High Precision and Control: Cyclotrons can accelerate particles with high precision and control, allowing researchers to manipulate the energy and trajectory of the particles with accuracy.
High Efficiency: Cyclotrons are typically highly efficient in accelerating particles, with a significant fraction of the input energy being transferred to the particles as kinetic energy.
Relatively Compact Size: Compared to other types of particle accelerators, such as linear accelerators or synchrotrons, cyclotrons can be relatively compact.
Continuous Operation: Cyclotrons can operate continuously and this is advantageous for applications such as medical isotope production, where a constant supply of radioisotopes is required.
Versatility: Cyclotrons can accelerate a wide range of charged particles, including protons, deuterons(deuterium nucleus), alpha particles, and heavy ions. This versatility allows researchers to conduct a diverse array of experiments.
Reliability: With proper maintenance and care, cyclotrons can operate for many years, providing consistent access to accelerated particles.
Ion Cyclotron Resonance
Ion Cyclotron Resonance (ICR) is a phenomenon in which charged particles, such as ions, oscillate and gain energy when subjected to a magnetic field of a specific frequency. This resonance occurs when the frequency of the applied magnetic field matches the natural frequency of gyration of the ions around the magnetic field lines.
Components and Operation of a Cyclotron
Cyclotron comprises several key components, each playing a vital role in its operation. Here is an overview of these components and how they function together:
Magnet: Magnet is a crucial component of the cyclotron, providing a uniform and perpendicular magnetic field necessary to bend the path of charged particles.
Dees: Dees are hollow, D-shaped electrodes positioned within the magnetic field. They create an electric field that alternates in polarity as the particles move between them. This alternating electric field serves to accelerate the particles each time they pass through the gap between the dees.
RF (Radio Frequency) Oscillator: RF oscillator generates a high-frequency alternating electric field between the dees.
Vacuum Chamber: Entire cyclotron operates within a vacuum chamber to prevent particles from colliding with air molecules and losing energy.
Charged particles are injected into the central region. As the particles spiral outward due to the magnetic field, they pass through the gap between the dees repeatedly. Each time they cross the gap, they experience an electric field that accelerates them. After reaching the desired energy level, the particles are extracted from the cyclotron for further use.
Working Principle of Cyclotron
Cyclotron operates on the basis of the magnetic Lorentz force experienced by a charged particle travelling normal to a magnetic field. This force is perpendicular to both the particle's motion and the magnetic field. The particle travels in a circular motion as a result.
Working of a cyclotron is stated below,
A charged particle beam is accelerated in a cyclotron's vacuum chamber by applying a high frequency alternating voltage, between two hollow 'D'-shaped sheet metal electrodes called Dees. Particles move within the dees because they are positioned face to face with a small gap between them. The central region of this space is filled with particles.
The Dees, located between the electromagnet's poles, applies a static magnetic field B perpendicular to the electrode plane.
Because of the Lorentz force which acts perpendicular to the particle's direction of travel, the magnetic field causes the particle's path to bend in a circle.
An alternating voltage of several thousand volts is supplied between the dees which produces a pulsing sound.
The varying electric field created by the voltage in the area between the dees causes the particles to accelerate.
The frequency of the voltage is changed such that particles create a single circuit in a single voltage cycle. To satisfy this criterion, the frequency must be tuned to the particle's cyclotron frequency.
Cyclotron Frequency
Cyclotron frequency, also known as gyrofrequency, denotes the frequency of a charged particle's motion perpendicular to a uniform magnetic field B, which maintains a constant magnitude and direction. Due to the circular nature of this motion, the cyclotron frequency is determined by the equilibrium between the centripetal force and the Lorentz force
mv2/r = qvB
Where q is the charge,
m is the mass,
v is the velocity of the particle,
r is the radius of the circular path also called gyroradius.
From this,
v/r = qB/m
The cyclotron frequency is related to the angular frequency as fc = ω/2π = v/ 2πr
Now the cyclotron frequency becomes,
fc = qB/2πm
Energy of a Particle
During Ion Cyclotron Resonance, energy can be transferred from the oscillating magnetic field to the ions. As we have calculated in the previous section
v = qBr/m
Kinetic energy therefore becomes,
E = q2B2r2/2m
Types of Cyclotrons
Cyclotrons can be classified into different types based on various criteria such as size, energy range, and application. Here are some common types of cyclotrons:
Isochronous cyclotrons are designed to maintain constant particle velocities regardless of the particle's energy. This is achieved by varying the magnetic field strength as the particles gain energy, compensating for the relativistic increase in mass.
Superconducting cyclotrons use superconducting magnets to generate the magnetic field. These magnets can produce much higher magnetic fields than conventional magnets, allowing for higher particle energies.
Cyclotrons for Positron Emission Tomography (PET) are specifically designed to produce radioisotopes used in PET imaging. They typically accelerate protons to bombard a target material, producing radioisotopes such as fluorine-18, which are then used to label radiotracers for PET scans.
Heavy ion cyclotrons are designed to accelerate heavy ions, such as carbon, nitrogen, or oxygen ions, to high energies. They are used in nuclear physics research to study the properties of atomic nuclei.
Difference between Cyclotron and Betatron
The differences between Cyclotron and Betatron are stated below:
| Point of Difference | Cyclotron | Betatron |
---
| Definition | A cyclotron is a kind of particle accelerator that uses a spiral path to accelerate particles like electrons. | A betatron is a kind of particle accelerator designed primarily for the purpose of accelerating electrons or beta particles. |
| Path | It follows semicircular or spiral path. | It follows a circular path. |
| Type of Particle | It accelerates atomic and subatomic particles. | It accelerates electrons. |
| Structure | It contains two electrodes called dees attached back to back. | It contains an circular evacuated tube embedded in an electromagnet. |
Uses of Cyclotron
The uses of Cyclotron are as follows:
Experiments in Nuclear Physics: Atomic nuclei are bombarded by charged particles that are accelerated in cyclotrons.
Radiation Treatment: Cancer patients are treated with cyclotrons. Cyclotron ion beams have the ability to enter the body and destroy tumours by radiation damage, with the least amount of damage to healthy tissue.
Nuclear Transmutation: The nuclear structure can be altered with the use of cyclotrons.
Nuclear Medicine: Medical radioisotopes are created in nuclear medicine cyclotrons. Protons are the charged particles that cyclotron beams in a circular direction.
Advantages of Cyclotron
Cyclotrons offer several advantages over other types of particle accelerators and methods of particle production. Some of the key advantages include:
Cyclotrons can achieve high energies in relatively compact designs compared to linear accelerators (linacs) or synchrotrons.
Cyclotrons can operate continuously, providing a constant supply of radioisotopes for medical isotope production.
Cyclotrons are highly efficient in accelerating particles, with a significant fraction of the input energy being transferred to the particles as kinetic energy.
Cyclotrons are often known for their reliability and long operational lifetimes.
Cyclotrons are widely used in medicine for various applications, including producing radioisotopes for diagnostic imaging (such as PET scans) and cancer therapy.
Limitations Of Cyclotron
While cyclotrons offer numerous advantages, they also have several limitations and challenges that need to be considered. Some of the key limitations include:
Cyclotrons cannot accelerate neutral particles, such as neutrons or atoms, as they do not interact with electric or magnetic fields.
Cyclotrons have practical limitations on the maximum energy they can achieve due to relativistic effects. As particles approach the speed of light, their mass increases, requiring stronger magnetic fields to maintain circular orbits.
Electrons cannot be efficiently accelerated as they have a much higher charge to mass(e/m) ratio, meaning they are less affected by the magnetic field and experience weaker acceleration in a cyclotron.
Achieving high beam intensity can be challenging due to space charge effects, beam losses, and other factors, which can limit the usefulness of cyclotrons for certain applications requiring intense beams of particles.
Conclusion: Cyclotron
Cyclotron is a particle accelerator invented by Ernest O. Lawrence in 1929, used to accelerate charged particles to high speeds by employing electric and magnetic fields. Components of Cyclotron include a magnet, dees, radio oscillator, and vacuum chamber.
Cyclotrons can be classified into various types, such as isochronous cyclotrons, superconducting cyclotrons, cyclotrons for PET, and heavy ion cyclotrons. They offer advantages like achieving high energies in compact designs, efficiency, reliability, making them valuable tools in scientific research and medical applications.
Also Read,
Kinetic Energy
Motion of a Charged Particle in a Magnetic Field
Solved Examples on Cyclotron
Example 1: In a cyclotron the frequency of alternating current is 12 MHz. What should be the operating magnetic field to accelerate protons? Given mass of proton = 1.67 × 10-27 kg.
Solution:
Using formula for Cyclotron Frequency:
f = qB/2πm
Upon rearranging,
B = 2πmf/q
B = 2π × 1.67 × 10-27×12 × 106/1.6 × 10-19
B = 0.79 T
Example 2: For the above problem, what is the kinetic energy of the proton beam produced by the cyclotron if the radius of the dee is 0.53 m?
Solution:
Using formula for Kinetic Energy:
E = q2B2r2/2m
E = (1.6 × 10-19)2(0.79)2(0.53)2/2 × 1.67 × 10-27
E = 1.34 × 10-12 J
In eV units,
E = 1.34 × 10-12 /1.6 × 10-19
E = 8.38 MeV
Example 3: The magnetic field inside a cyclotron is 0.8 T. At what maximum radius should a proton beam be extracted so that its energy is 10 MeV?
Solution:
Using formula for Kinetic Energy:
E = q2B2r2/2m
Upon rearranging,
r = √2mE/qB
r = √(2 × 1.67 × 10-27 × 107 × 1.6 × 10-19 )/1.6 × 10-19 × 0.8
r = 7.31 × 10-19/1.28 × 10-19
r = 5.71 m
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3505 | https://www.youtube.com/watch?v=0L6DBNYNSFg | Teaching The Number Line | Middle School Math | Integers | Opposites | Sub Plans
Kacie Travis - The Efficient Classroom
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Posted: 28 Feb 2025
Join me for this lesson about properties of the number line. You will also explore the vocabulary integers and opposites. You will learn how to compare and order integers on a number line.
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Transcript:
hi there today we're going to be learning about the number line and just some of the basics about what the number line is and how it works so first things first the properties of a number line a number line is a straight line extending forever in opposite directions the number line can go up and down like this one here to the bottom right or left and right like this one here to the bottom left they are marked at intervals that indicate different um numbers on the number line we call these tick marks and they um should be spaced evenly as long as the space is representing the same distance like one um integer zero is in the middle of the actual number line now sometimes when it's drawn it's not always in the middle because you may want to be looking at some numbers um above or below zero it may not even be on the number line that you are seeing at all um it may be zoomed into a certain section of the number line but just know that in the big scheme of the real number system zero is in the middle um and then the numbers that are positive are going to be above or to the right of the zero like these and the negative numbers will be below or to the left of the zero like these and you can see this example of this vertical number line on the right um the zero is not in the middle of this drawn number line but it is in the middle of the set of positives and negative numbers opposites what are opposites they are two numbers that are at the same distance from zero on a number line but are on opposite sides of zero so for example if we have 15 the opposite of 15 would be the same distance that 15 is from zero but on the opposite side of zero so it would be -5 integers are sets of positive whole numbers they're opposites and zero so you won't have any decim decimals you won't have any fractions um you will have whole numbers they opposites and zero so you'll get negatives positives and zeros so we've got for example 915 5 345 0 -28 and then a couple more examples -14 893 are both examples of integers if you're ordering or comparing integers on a number line think about as you move from left to right on the number line the number value gets larger so if you're comparing the numbers -4 and -1 -4 is farther left which means -1 is going to be further right which means it's bigger so 4 is less than -1 As you move from right to left on the number line the value gets smaller so if we're comparing 9 to a -5 nine is further right on the number line and as you travel left it gets smaller which means nine is the bigger number so 9 is greater than -5 let's do some practice try a b and c on your own using the greater than or less than to Compare the numbers you can pause the video to practice and then we'll come together okay so we've got -4 compared to -10 -4 is further to the right on the number line so -4 is greater than -10 8 18 compared to -21 18 is further to the right on the number line then -21 so-8 is greater than -21 -3 compared to 03 is further to the left on the number line so it is neg it is less than zero now for d and e we're going to give the opposite of each number you can pause the video to write these down for D we want the opposite of 13 so the same distance from zero it's 13 units from 0er but on the other side of 0o so we have -3 109 is sorry NE 109 is 109 units away from zero which means we want it on the positive side 109 units away from zero so be what positive 109 for f and g try ordering these from least to greatest all right so for f we're going to start with the integer that is furthest to the left that'll be our least integer which is -14 and then moving along our number line and if you need to draw it out this is a good opportunity to visually see where these numbers go then we'd have -3 then we're going to we're done with our negative numbers to now we're moving to our positive numbers our least positive number is one then we have two then 16 for G we want our least number so the furthest left on the number line we have -10 then we have -9 then we have -5 then -4 then -1 try these on your own all right a we're comparing 15 to -16 15 is greater than -16 B -12 compared to -14 -12 is greater than -14 C comparing 8 to 58 is less than 5 give the opposite of each number d-36 the opposite of - 36 is 36 e the opposite of 8 18 is -8 order the integers from least to greatest for f we have 78 -10 3 and -3 our least integer is -108 -3 3 and 7 G we have integers -78 -63 82- 55 and -65 our least integer is -82 then going up to our more greater integers -78 -65 -63 - 55 practicing again we're going to use the number line shown to answer the questions below number one which point is three units to the right of zero so looking at zero we count up three units it's posit 3 number two which point is s units to the left of three so we're starting at three and we go seven units to the left so counting down that gets us to -4 number three which point is one unit left of -1 we start at -1 and we count one unit to the left of -1 it takes us to -2 let's put the this into context of some real world situations the elevator started at the third floor so imagine a vertical number line starting at a positive3 it went up four floors to let people on so it's gone up four floors from three so we land at the seventh floor then it comes down two floors so it goes down two Which floor did it stop on so this would be the fifth floor this morning the temperature was -10° as the sun came up it Rose 14° what is the temperature now so thinking about a number line it can be a vertical or a horizontal number line start at -10 and go up or to the right 14 so we're counting past zero and up four we land at positive4 degrees if you would like guided notes to go along with this PowerPoint they are linked Below in the form of Internet active notebook notes or um they are also in um full page guided notes um linked below um as always please like And subscribe have a wonderful day bye-bye |
3506 | https://math.stackexchange.com/questions/2360642/lagrange-interpolation-with-zeros-at-pm-infty | pointwise convergence - Lagrange interpolation with zeros at $\pm \infty$ - Mathematics Stack Exchange
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Lagrange interpolation with zeros at ±∞±∞
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Suppose I have a set of n n points on the plane at (x i,y i)(x i,y i), and I want to construct the polynomial interpolating them. This can be done with Lagrange interpolation in the straightforward manner.
Now suppose I add two points at (a,0)(a,0) and (b,0)(b,0), where amax(x i)b>max(x i). In other words, I add two zeros, one on either side of the set of points. We can again perform Lagrange interpolation on this augmented set of points to obtain a polynomial that approaches 0 as you approach a a and b b.
This defines a family of polynomials for each choice of a a and b b. I suspect this ought to converge pointwise to some well-defined function (though I haven't proven this).
Does this converge pointwise to something, and if so, what?
In effect we end up with Lagrange interpolation, but with extra zeros "at infinity." Does this lead to a well-defined interpolation method?
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edited Jul 16, 2017 at 16:22
Mike BattagliaMike Battaglia
asked Jul 16, 2017 at 16:15
Mike BattagliaMike Battaglia
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At first glance I highly doubt that this gives a convergent sequence of functions.Redundant Aunt –Redundant Aunt 2017-07-16 16:17:45 +00:00 Commented Jul 16, 2017 at 16:17
Sorry, editing error - meant pointwise. Changed it Mike Battaglia –Mike Battaglia 2017-07-16 16:22:55 +00:00 Commented Jul 16, 2017 at 16:22
If you have just one point (x 0,y 0)=(0,0)(x 0,y 0)=(0,0), it seems clear that it doesn't converge pointwise.Friedrich –Friedrich 2017-07-16 16:28:26 +00:00 Commented Jul 16, 2017 at 16:28
I don't see a sequence, only a family parameterized by a,b a,b, so you should make clear (in the question, not in a comment) what kind of limit you mean. From the title, one could guess a→−∞,b→∞a→−∞,b→∞.user436658 –user436658 2017-07-16 16:41:09 +00:00 Commented Jul 16, 2017 at 16:41
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If we write the Lagrange polynomials for the set of points including a a and b b as L a,b j(x)L j a,b(x) and L j(x)L j(x) for those without, then they are given by
L a,b j(x)=(x−a)(x−b)(x j−a)(x j−b)∏i≠j x−x i x j−x i=(x−a)(x−b)(x j−a)(x j−b)L j(x).()()L j a,b(x)=(x−a)(x−b)(x j−a)(x j−b)∏i≠j x−x i x j−x i=(x−a)(x−b)(x j−a)(x j−b)L j(x).
Now, since the function values at a a and b b are zero, the interpolating polynomial is
∑j y j L a,b j(x),∑j y j L j a,b(x),
not including L a L a or L b L b. Therefore, every term is of the form in (), and simple results about rational functions tell us that for fixed x x,
lim a→−∞b→∞(x−a)(x−b)(x j−a)(x j−b)=1,lim a→−∞b→∞(x−a)(x−b)(x j−a)(x j−b)=1,
so L a,b j(x)→L j(x)L j a,b(x)→L j(x) pointwise.
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answered Jul 16, 2017 at 17:25
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3507 | https://math.stackexchange.com/questions/4759173/how-many-3-element-subsets-of-1-2-3-19-20-have-product-divisible-by | combinatorics - How many $3$-element subsets of ${1,2,3,...,19,20}$ have product divisible by $4$? - Mathematics Stack Exchange
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How many 3 3-element subsets of {1,2,3,...,19,20}{1,2,3,...,19,20} have product divisible by 4 4?
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Same question :- Where am I overcounting?
How many 3 3 element subsets of the set {1,2,3,...,19,20}{1,2,3,...,19,20} are there such that the product of the three numbers in the subset is divisible by 4 4?
My attempt:-
I divided this into broadly 2 cases :-
Case 1:-
Subsets containing atleast 1 number of type 4k :-
4 k,4 k,4 k 4 k,4 k,4 k =(5 3)(5 3)
4 k,4 k,4 k+1 4 k,4 k,4 k+1 =(5 2)∗(5 1)(5 2)∗(5 1)
4 k,4 k,4 k+2 4 k,4 k,4 k+2 =(5 2)∗(5 1)(5 2)∗(5 1)
4 k,4 k,4 k+3 4 k,4 k,4 k+3 =(5 2)∗(5 1)(5 2)∗(5 1)
4 k,4 k+1,4 k+1 4 k,4 k+1,4 k+1 =(5 1)∗(5 2)(5 1)∗(5 2)
4 k,4 k+1,4 k+2 4 k,4 k+1,4 k+2 =(5 1)∗(5 1)∗(5 1)(5 1)∗(5 1)∗(5 1)
4 k,4 k+1,4 k+3 4 k,4 k+1,4 k+3 =(5 1)∗(5 1)∗(5 1)(5 1)∗(5 1)∗(5 1)
4 k,4 k+2,4 k+2 4 k,4 k+2,4 k+2 =(5 1)∗(5 2)(5 1)∗(5 2)
4 k,4 k+2,4 k+3 4 k,4 k+2,4 k+3 =(5 1)∗(5 1)∗(5 1)(5 1)∗(5 1)∗(5 1)
Case 2:- without any 4k type of number
4 k+2,4 k+2,4 k+2 4 k+2,4 k+2,4 k+2 =(5 3)(5 3)
I cant figure out what all cases am I missing ? I am getting 685 cases however total cases are 795
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edited Aug 27, 2023 at 16:11
user21820
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asked Aug 27, 2023 at 0:09
Vasu GuptaVasu Gupta
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4 Try complementary counting instead. Now there are just two cases: All numbers are odd or two numbers are odd and one is even.TheBestMagician –TheBestMagician 2023-08-27 00:15:03 +00:00 Commented Aug 27, 2023 at 0:15
2 You missed 4 k,4 k+3,4 k+3 4 k,4 k+3,4 k+3,and 4 k+2,4 k+2,2 k+1 4 k+2,4 k+2,2 k+1 Empy2 –Empy2 2023-08-27 00:31:58 +00:00 Commented Aug 27, 2023 at 0:31
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We use complementary counting. For a set to have the product of its elements not divisible by 4 4, there are two cases:
All elements are odd. There are 10 10 odd numbers, so there are (10 3)(10 3) ways here.
Two elements are odd and one is even, but not divisible by 4 4. This is (5 1)⋅(10 2)(5 1)⋅(10 2).
Thus the answer is (20 3)−(10 3)−(5 1)(10 2)=795(20 3)−(10 3)−(5 1)(10 2)=795.
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answered Aug 27, 2023 at 0:20
TheBestMagicianTheBestMagician
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Although complementary counting provides the most elegant approach, the approach taken by the OP (i.e. original poster) is still viable.
Case 1: At least one number that is a multiple of 4.4.
There are (20 3)(20 3) ways of choosing three numbers, and there are (15 3)(15 3) ways of choosing three numbers, none of which is an element in {4,8,12,16,20}.{4,8,12,16,20}.
Therefore, the Case 1 enumeration is
(20 3)−(15 3)=685.(20 3)−(15 3)=685.
Case 2: None of the numbers are a multiple of 4.4.
So, you either have exactly two even numbers, none of which is a multiple of 4,4, or three such even numbers.
Case 2a: Exactly two even numbers, neither of which is a multiple of 4.4.
So, you have exactly one element from {1,3,5,⋯,19},{1,3,5,⋯,19}, and exactly two elements from {2,6,10,14,18}.{2,6,10,14,18}.
So, the Case 2a enumeration is
(10 1)×(5 2)=100.(10 1)×(5 2)=100.
Case 2b: Exactly three even numbers, neither of which is a multiple of 4.4.
You then have exactly three elements from {2,6,10,14,18}.{2,6,10,14,18}.
So, the Case 2b enumeration is
(5 3)=10.(5 3)=10.
Final Total:
685+100+10=795.685+100+10=795.
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answered Aug 27, 2023 at 3:01
user2661923user2661923
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1Where am I overcounting?
Related
2Let S={1,2,3,...,1992}S={1,2,3,...,1992} find the number of subsets {a,b,c}{a,b,c} such that 3∣(a+b+c)3∣(a+b+c).
1Where am I overcounting?
1Let F F be a set of subsets of the set {1,2,3,...,n}{1,2,3,...,n} such that:
3Romanian IMO TST 2006, day 4, problem 3
0Forming n n-element subsets, with restrictive condition
7How many ways are there to choose subsets S S and T T of A={1,2,3,,.....,n}A={1,2,3,,.....,n} so that S S contains T T?
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3508 | https://www.teachingwithkayleeb.com/how-the-commutative-property-of-addition-can-help-students-memorize-math-facts/ | Select Page
How the Commutative Property of Addition Can Help Students Memorize Math Facts
by | | Math Fact Fluency, Uncategorized
With memorizing math facts, there are so many facts that students need to commit to memory. But with the commutative property of addition, students can automatically know twice as many facts. So it’s important to teach 1st grade and 2nd grade students the commutative property of addition.
Learning the commutative property is just one step students need on their way to math fact fluency. I have a free guide for students where I share 7 steps to help their students get to math fact mastery. Download the free guide for 1st and 2nd grade teachers here: The 7 Steps to Ensure Math Fact Fluency.
The Commutative Property of Addition
What is the commutative property of addition? It is when you can rearrange the addends in an addition equation and the sum stays the same.
So let me give you some commutative property of addition examples. If 2+3=5, then 3+2=5. If 4+6=10, then 6+4=10.
Knowing and applying this knowledge for math facts, can be so helpful to students. Students now have cut the amount of math fact equations to learn in half if they understand this concept! This is huge, especially for struggling students. But what’s the best way to teach this to students?
How to Teach the Commutative Property of Addition
For teaching this concept to students, I like to first make it very visual. I like to use counters and lay out a certain number of yellow and red counters. Then I ask students if there is an equation that matches the counters.
For example, if I lay out 2 red counters and 5 yellow. Student’s should come up with the equation 2+5=7. Then I ask, what if we switch around the red and yellow counters? What if I now put yellow on this side and red on this side? Is the answer different? No, it’s the same. So 5+2=7. Continue to practice this until students understand the pattern.
Then I like to make things less visual. I’ll write up a bunch of equations on the board. Then students have to come up with the answer and then the commutative pair.
I also like to take this a step further. I teach students the inverse operation as well. This is where students work with fact families. I help students see how addition and subtraction facts are related. I get them lots of practice with my Fact Family Triangles Task Cards. Find these here.
Then I like to set out my Fact Family Puzzles for students to work with. Find these here. I have a whole blog post going more in depth about how to teach fact families. Read it here.
Help students learn math facts faster by teaching them the commutative property of addition. This allows students to see patterns and relationships between math facts, and in return, help them to math fact mastery easier. For more help for helping students memorize addition and subtraction math facts download my free guide: The 7 Steps to Ensure Math Fact Fluency.
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3509 | https://simple.wikipedia.org/wiki/Imaginary_unit | Jump to content
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Contents
Beginning
1 History
2 Definition
3 Square root of i
4 Powers of i
5 Related pages
6 References
Imaginary unit
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From Simple English Wikipedia, the free encyclopedia
In math, the imaginary unit (written as "i" or "j") is a mathematical constant that only exists outside of the real numbers and is used in algebra. When we multiply the imaginary unit by a real number, we call the result an imaginary number. Though imaginary numbers can be used to solve a lot of mathematical problems, they cannot be represented by an amount of real life objects.
History
[change | change source]
Imaginary units were invented to answer the polynomial equation , which normally has no solution (see below). The term "imaginary" comes from by René Descartes and was meant to be insulting as, like zero and negative numbers at other times in history, imaginary numbers were thought to be useless as they are not natural. It wasn't until later centuries that the work of mathematicians like Leonhard Euler, Augustin-Louis Cauchy and Carl Friedrich Gauss would prove that imaginary numbers were very important for some areas of algebra.
Definition
[change | change source]
A common rule for multiplying and dividing numbers is that if the signs are different then the result is negative (e.g. ), but if both numbers have the same sign then the result will be positive (e.g. and ). However, this leads to problems with square root numbers of negatives, as two negative numbers will always make a positive number:
: so
: but
: as
To fill in this value gap the imaginary unit was made, which is defined as and . Using imaginary numbers we can solve our last example:
: and
Square root of i
[change | change source]
Although the imaginary unit comes from solving a quadratic equation (an equation where the unknown appears squared), we could ask whether we need to create new number values like the imaginary unit to solve equations where higher powers of like and appear. For example, the equation has a fourth power of the unknown variable . Do we need new units like to solve this equation?
We could also ask a similar question: we needed to create a new number to find the square root of -1, and we called this new number . Do we need to create a new number to find the square root(s) of ?
It turns out the answer to both these questions is no. For the second question, the square roots of can be written in terms of a real part and an imaginary part. Specifically, the square roots of can be written as: . We can check that these are really the square roots of by squaring them and seeing if we get :
: | | |
--- |
| | |
| | |
| | |
| | |
| | |
We can also notice that , so solves the equation , partially answering our first question-- for the equation , the solutions are still complex numbers (the result of adding a real number and an imaginary number). There are two more solutions for this particular equation, , and they are also complex numbers. No new numbers like the imaginary unit are needed to solve the equation.
In general, every equation where the unknown appears with whole number powers can be solved by complex numbers, so once we know about the imaginary unit, we can solve any equation of this form. This result is so important that it is called the fundamental theorem of algebra.
Powers of i
[change | change source]
The powers of or follow a regular and predictable pattern:
As shown, each time we multiply by another the values are and then repeat.
Related pages
[change | change source]
Complex number
Euler's Identity
Imaginary number
Mathematical constant
Real number
References
[change | change source]
↑ "Compendium of Mathematical Symbols". Math Vault. 2020-03-01. Retrieved 2020-08-10.
↑ Weisstein, Eric W. "Imaginary Unit". mathworld.wolfram.com. Retrieved 2020-08-10.
↑ Dunham, William. "Euler and the Fundamental Theorem of Algebra" (PDF). Retrieved 24 April 2022.[permanent dead link]
Retrieved from "
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3510 | https://www.youtube.com/watch?v=XchHLcCCUk8 | Laws of Logarithm | Logₐ1=0 | Logₐa=1
Tambuwal Maths Class
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Transcript:
hello good day viewers in this tutorial we are going to prove these two laws of logarithm the first one we have log of one with the base of a equal to zero this law stated that log of 1 to any base will give us zero while the second law stated that log of a number having the same base as the number is forever equal to one so now let us start with the first one here I would like to let the whole of this to be equal to a certain variable so whatever the variable is is said to be the solution and we would like to obtain that variable as 0. all right so we would like to late log of 1 with the base of a to be equal to X so whatever the value of x is is said to be the solution to this problem and we wish to get 0 for that let us apply definition of logarithm here e raised to the power of X will give us 1. so a raised to the power of X this is equal to 1. and you know according to the laws of indices any number raised to the power of 0 is equal to one provided that that number is not zero all right therefore we can transform this one into e to the power of 0. so we have a to the power of x equal to a to the power of 0. and since we have common bases it implies that the exponents are exactly the same therefore x equal to zero and remember what is X x is nothing but log of 1 with the base of a therefore log of 1 with the base of a is not him but zero and hence proved let us take the second law I would like to let the love of a base a to be equal to Y so let log of a base a b equal to Y therefore e to the power of y equal to e according to the definition of logarithm right and you know that this a to the right hand side can be raised to the power of 1 right it doesn't matter it will not change anything and you can see we have the same basis which implies that even the exponents are the same therefore Y is equal to 1. and you can clearly see that Y is nothing but log of a to base a and hence we see that log of a number having the same base as the number is equal to one and hence proof thank you for watching do share to your learning colleagues and don't forget to subscribe to my YouTube channel for more exciting videos remember we are going to prove all the laws of logarithm bye |
3511 | https://www.worldscientific.com/doi/full/10.1142/S1793042117500580?srsltid=AfmBOopoo8rMbXKCwfmpA15-g6I-p6zJl0TMrTZZAIhOx9pPu5UOXht_ | A congruence involving harmonic sums modulo pαqβ | International Journal of Number Theory
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International Journal of Number TheoryVol. 13, No. 05, pp. 1083-1094 (2017)Research Papers No Access
A congruence involving harmonic sums modulo p α q β p α q β
Tianxin Cai,
Zhongyan Shen,and
Lirui Jia
Tianxin Cai
Department of Mathematics, Zhejiang University, Hangzhou 310027, P.R. China
E-mail Address: txcai@zju.edu.cn
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,
Zhongyan Shen
Department of Mathematics, Zhejiang International Studies University, Hangzhou 310012, P.R. China
E-mail Address: huanchenszyan@163.com
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,and
Lirui Jia
Department of Mathematics, Zhejiang University, Hangzhou 310027, P.R. China
E-mail Address: jialirui@126.com
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Abstract
In 2014, Wang and Cai established the following harmonic congruence for any odd prime p p and positive integer r r,
Z(p r)≡−2 p r−1 B p−3(mod p r),Z(p r)≡−2 p r−1 B p−3(mod p r),
where Z(n)=∑i+j+k=n i,j,k∈P n 1 i j k Z(n)=∑i+j+k=n i,j,k∈𝒫 n 1 i j k and P n 𝒫 n denote the set of positive integers which are prime to n n. In this paper, we obtain an unexpected congruence for distinct odd primes p p, q q and positive integers α,β α,β,
Z(p α q β)≡2(2−q)(1−1 q 3)p α−1 q β−1 B p−3(mod p α)Z(p α q β)≡2(2−q)1−1 q 3 p α−1 q β−1 B p−3(mod p α)
and the necessary and sufficient condition for
Z(p α q β)≡0(mod p α q β).Z(p α q β)≡0(mod p α q β).
Finally, we raise a conjecture that for n>1 n>1 and odd prime power p α∥n p α∥n, α≥1 α≥1,
Z(n)≡∏q|n q≠p(1−2 q)(1−1 q 3)(−2 n p)B p−3(mod p α).Z(n)≡∏q|n q≠p 1−2 q 1−1 q 3−2 n p B p−3(mod p α).
However, we fail to prove it even for n n with three distinct prime factors.
Keywords:
Bernoulli numbers
harmonic sums
congruences
Chinese remainder theorem
AMSC:11A07, 11A41
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W. W. Chao, Problem 2981, Crux Mathematicorum30 (2004) p. 430. Google Scholar
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C. G. Ji, A simple proof of a curious congruence by Zhao, Proc. Amer. Math. Soc.133 (2005) 3469–3472. Crossref, Web of Science,Google Scholar
R. Meštrović, Wolstenholme’s theorem: Its generalizations and extensions in the last hundred and fifty years (1862–2012); preprint (2011); arXiv:1111.3057v2. Google Scholar
Š. Porubsky, Further congruences involving Bernoulli numbers, J. Number Theory16 (1983) 87–94. Crossref, Web of Science,Google Scholar
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Cited By 3Cited by lists all citing articles based on Crossref citation.
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25 April 2022 | International Journal of Number Theory, Vol. 18, No. 08
Congruences involving alternating harmonic sums modulo pαqβ Zhongyan Shen and Tianxin Cai 20 October 2018 | Mathematica Slovaca, Vol. 68, No. 5
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History
Received 14 December 2015
Accepted 12 May 2016
Published: 7 December 2016
Keywords
Bernoulli numbers
harmonic sums
congruences
Chinese remainder theorem
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3512 | https://www.centerforfaith.com/blog/sexual-orientation-and-gender-identity-change-efforts-part-4-the-diagnostic-problem | Sexual Orientation and Gender Identity Change Efforts, Part 4: The Diagnostic Problem | The Center for Faith, Sexuality & Gender
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Sexual Orientation and Gender Identity Change Efforts, Part 4: The Diagnostic Problem
Sexual Orientation and Gender Identity Change Efforts, Part 4: The Diagnostic Problem
August 30, 2021
This blog series is drawing from the research paper: “The Problems with Correlating Sexual Orientation Change Efforts and Gender Identity Change Efforts,” available for freehere. Written by Drs. Paul Eddy and Preston Sprinkle.
Sexual Orientation Change Efforts (SOCE) and so-called Gender Identity Change Efforts (GICE) might overlap in some ways, as I pointed out in the first post, but they present many other differences, which we looked at in the previous two posts. In our second post, we looked at the language problem inherent in correlating SOCE and GICE, and then in our third post, we looked at the ontological problem. In this fourth post, I want us to consider areas of difference between SOCE and GICE that constitute what I call the diagnostic problem.
Put simply: gender dysphoria is listed in the DSM-5 as a diagnosable mental condition while homosexuality is not. Richard Green, a lawyer and one of the most experienced researchers in the area of childhood gender dysphoria, draws attention to these diagnostic differences:
Whereas homosexuality per se was dropped by the APA as a disorder in 1973, in 1980, gender identity disorder was added…gender dysphoria remains in the current DSM (Fifth Edition). Therefore, the argument against attempting to modify sexual orientation because it is not a disorder is not symmetrical with attempts to modify or treat gender dysphoria.
Green goes on to question the conflation of SOCE and GICE in recent legislative bans on conversion therapy: “recent legislation with its conflation of sexual orientation and gender identity remains psychologically incoherent.”
Buy Now!
Embodied: The Latest from Preston Sprinkle
There are some, of course, who would love to see gender dysphoria removed from the DSM altogether. They say that trans-related diagnosis should never have been placed in the DSM to begin with and should be removed as soon as possible. Anything less is seen a direct assault on the depathologization of trans people.
However, not everyone within the trans community supports the removal of GD from the DSM. In fact, it was a sector of the trans community and their allies who were putting pressure on the psychiatric powers-that-be to add a trans-related diagnosis into the DSMin the first place. And many within the trans community today continue to argue that retaining a trans related diagnosis in the DSM is important. In a cross-cultural survey of 201 organizations dedicate to the well-being of transgender people—representing perspectives from North America, Latin America, Europe, Africa, Asia, and Oceana—55.8% stated that trans-related diagnoses should be removed from the DSM. This means that 44.2% believed that a diagnosis should be retained. For those defending retention, the primary reason was “healthcare reimbursement.”
And that’s the sticking point in this debate: In the world of medical insurance, diagnostic codes are required for such things as indicating diagnoses and treatments, and—importantly—determining financial costs, insurance coverage, and reimbursements. This means that removing trans-related diagnostic codes from the DSM could threaten the ability of trans people to submit valid insurance claims for coverage of medical transition-related costs. In light of this, one could argue that trans people don’t actually see gender dysphoria as a mental disorder; they simply need it to be viewed as such in order to get insurance coverage for surgeries.
But there’s another sticking point has to do with the depathologization of trans people mentioned above. Some believe that listing gender dysphoria in the Diagnostic and Statistical Manual of Mental Disorders paints trans people with the stigma of being mentally ill. This concern, though, has its own catch-22: reducing trans-related stigma (by removing GD from the DSM) comes at the expense of “the perpetuation of existing stigma and prejudices against the mentally ill.” In other words, the more that trans activists and allies strive to legitimate trans experience by distancing it from mental illness, the more they inadvertently stigmatizes and marginalizes people with mental illnesses.
It's beyond the scope of our blog series to sort all this out. Obviously, the removal vs. retention debate is complex and comes with potentially significant ramifications either way. Our point is simply what we noted at the beginning: gender dysphoria is listed in the DSM as a mental disorder while homosexuality is not. This indicates that SOCE are attempting to treat something that, from a psychological standpoint, needs no treatment. But gender dysphoria does need treatment. Labeling such treatment as GICE—the assumed cousin of SOCE—confuses the issue and helps no one.
Some will say, though, that the only reason trans people are distressed and have dysphoria is due lack of societal acceptance. It’s our transphobic society that causes someone to even have gender dysphoria.
This is a reasonable line of thinking, but fails to appreciate the inherently distressing nature of gender dysphoria itself.
Gender Dysphoria Is Inherently Distressing
Both gay and trans people might experience a kind of stress from living in a homophobic and transphobic society. Scholars have developed what’s called the Minority Stress Model for making sense of this socially induced stress. The data shows that the MSM is able to explain a good deal of the mental and emotional distress and illness experienced by sexual minority populations. When applied to sexual orientation, the MSM argument is pretty straightforward: If we rid the culture of all social stigma and pathologization concerning homosexuality and bisexuality, and if all same-sex relationships were accepted and celebrated in the way that heterosexual relationships are today, then LGB people would not have any higher rates of mental health problems than any other sector of the populace. Take away the socially induced minority stress, then you take away all distress that comes with being same-sex attracted.
However, can this same logic be applied to trans experiences? Certainly the answer is, in part, yes. Whatever minority stress that gay people experience will also be experienced by trans people. But the MSM doesn’t explain another feature of trans experiences: the mere experience of gender incongruence/dysphoria itself is distressing.
A research team led by M. Paz Galupo has recognized that lack of social acceptance (the MSM) is not the only cause of distress. Gender dysphoria is itself inherently distressing, apart from social factors. How did they discover this? They listened to actual trans people. Another research team did the same. They listened to trans people describe the inherent distress they experience, apart from any clear social factors. For instance:
Have you ever tried putting together a puzzle and attempted to shove a piece in that doesn’t fit? . . . So now imagine that your body is the puzzle and almost none of the pieces fit together no matter how hard you press or how many different combinations you try. That’s what gender dysphoria feels like to me. (24-year-old Afro-Indigenous transgender man)
Seeing genitalia that does not match my gender identity beats me down. A painful war between my mind and body goes on [Hispanic trans man, age 18]
Something is always missing, or something ever feels right . . . . I’m a Lego set that came with the wrong pieces. I do not feel real. I do not want to be what I am. (18-year-old White transgender woman).
being trapped in the rotting carcass of some stage creature (37-year-old transgender woman).
It feels like a knife in me (18-year-old nonbinary participant).
. . . excruciatingly painful. That is all I can say . . . . [l]ike I’m being tortured in my own body, every day. As if there is no end to my suffering [20-year-old White transgender man]
Gay people don’t describe their sexual orientation in this way. Whatever shame or distress they experience is due solely to social factors. Trans people (who have gender dysphoria) experience both—social stress and the inherent suffering that comes with gender dysphoria.
This brings us back to our main point: no “change” is needed for gay people, even if some aspects of society should change. But there is a change that gender dysphoric people want—the reduction or elimination of their dysphoria. And this is why correlating SOCE with GICE is misinformed and can lead to the kind of harm some people say they’re trying to combat. By scaring off therapeutic care with warnings of SOCE, people with gender dysphoria are unable to explore less invasive ways of treating their dysphoria.
For a helpful survey of the history and evolution of gender identity-related diagnoses in both the DSM and the ICD, see Jack Drescher, “Gender Identity Diagnoses: History and Controversies,” in Kreukels, Steensma, and de Vries, eds., Gender Dysphoria and Disorders of Sex Development, 137-50 (see esp. 140-44).
Richard Green, “Banning Therapy to Change Sexual Orientation or Gender Identity in Patients Under 18,” Journal of the American Academy of Psychiatry Law 45/1 (2017), 7–11 (here p. 8, 10).
Robert D. Davies and Madeline D. Davies, “The (Slow) Depathologizing of Gender Incongruence,” Journal of Nervous and Mental Disease 208/2 (2020), 152-54.
See Friedemann Pfäfflin, “Identity: A Historical and Political Reflection,” in Kreukels, Steensma, and de Vries, eds., Gender Dysphoria and Disorders of Sex Development, 131-46
S. R. Vance, P. T. Cohen-Kettenis, J. Drescher, H. F. L. Meyer-Bahlburg, F. Pfäfflin, and K. J. Zucker, “Opinions about the DSM Gender Identity Diagnosis: Results from an International Survey Administered to Organizations Concerned with the Welfare of Transgender People,” International Journal of Transgenderism 12/1 (2010), 1-24. E.g., C. Richards, J. Arcelus, J. Barrett, W. Pierre Bouman, P. Lenihan, S. Lorimer, S. Murjan, and L. Seal, “Trans is Not a Disorder – But Should Still Receive Funding,” Sexual and Relationship Therapy 30/3 (2015), 309-13.
Drescher, “Gender Identity Diagnoses: History and Controversies,” 146.
E.g., Catelan, Costa, and de Macedo Lisboa, “Psychological Interventions for Transgender Persons”; Jody L. Herman, Taylor N. T. Brown, and Ann P. Haas, Suicide Thoughts and Attempts Among Transgender Adults: Findings from the 2015 U.S. Transgender Survey (Los Angeles: Williams Institute, 2019), 30.
Again, this is in keeping with Ilan Meyer’s own reminder that his MSM was never intended to function as an all-encompassing theory; see Meyer, Pachankis, and Klein, “Do Genes Explain Sexual Minority Mental Health Disparities?” Meyer himself has critically reflected on some of the ways in which others have used his MSM in a haphazard and methodologically troubling fashion. See Sharon Schwartz and Ilan H. Meyer, “Mental Health Disparities Research: The Impact of Within and Between Group Analyses on Tests of Social Stress Hypotheses,” Social Science & Medicine 70/8 (2010), 1111–18. See also Frost, “Benefits and Challenges of Health Disparities and Social Stress Frameworks for Research on Sexual and Gender Minority Health,” 462.
Ibid., 4.
Ibid.
Ibid., 5.
Ibid.
Ibid.
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3513 | https://www.wyzant.com/resources/answers/921805/toggle-follow | WYZANT TUTORING
Woojin L.
Given a sequence defined recursively and its closed form, use strong mathematical induction to prove that the definitions are equivalent.
Suppose that a0, a1, a2, . . . is a sequence defined as follows:
a0 = 2, a1 = 1, an = an−1 + 12an−2 for every integer n ≥ 2.
Prove that an = (−3)n + 4n for every integer n ≥ 0.\
p(0): (-3)^0+4^0= 2
p(1): (-3)^1+4^1=1
so true
Inductive step:
Suppose that ai=(-3)^i+4^i for every integer i from 0 through k, for some integer k>1.
In particular, ak-1=(-3)^k-1 + 4^k-1 and ak=(-3)^k + 4^k. Since k>1, we have k+1>2, and therefore
ak+1=an+ 12an-1 by given recurrence relation
((-3)^k+4^k) + 12((-3)^k-1 +4^k-1)
-3^k+4^k + 12-3^k-1 + 124^k-1
-4-3-3^k-1+ 344^k-1
4-3^k + 34^k
(-3^k+4^k) + (-4-3^k + 34^k)
I got this far, but this doesn't seem right.
1 Expert Answer
Dalton P.
answered • 04/28/24
Instructor Access To Webassign Assignments With 10+ Years Of Tutoring!
Base case: If n=0, then a_0=(-3)^0+4^0=2 which matches the given a_0=2.
Inductive hypothesis: Assume a_n=(-3)^n+4^n holds for every n.
Prove the n+1 case: We have a_{n+1}=a_n+12a_{n-1}=(-3)^n+4^n+12((-3)^{n-1}+4^{n-1})=(-3)^n+4^n+12(-3)^{n-1}+12(4)^{n-1}=(-3)^n+4^n-(-12)(-3)^{n-1}+12(4)^{n-1}=(-3)^n+4^n-(4)(-3)^{n}+(3)(4)^{n}=(-3)^n-4(-3)^n + 4^n+(3)4^n= -3(-3)^n+4(4)^n=(-3)^{n+1}+4^{n+1}, thus proving the n+1 case, finishing induction.
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3514 | https://www.youtube.com/watch?v=q_nu5bB2z5w | Differential Equation-Clairaut's Form of Differential Equation & Bernoulli Equation |S Chand Academy
S Chand Academy
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This video describes the Clairaut's and Bernoulli's form of First Order Differential equation
For E-Books:
Transcript:
[Music] as John presents educational video lectures as per the aicte curriculum difficult Concepts made easy study anywhere anytime [Music] hi I'm Dr Neelam from Delhi Technological University we are covering solution technique for differential equation in today's lecture we will cover Bernoulli differential equation and Clorox differential equation so for more details you can refer to the book from ishan publishing details of which is given in this site and Link is given in the description box so in the first part of the video we will start with the Bernoulli differential equation Bernoulli differential equation so uh this is after the name of the mathematician Bernoulli and uh the differential equation takes the form Y dash plus P1 P X Plus p x y is equals to Y raised to the power n so if if any differential equation is given in this form this is called as Bernoulli differential equation so how to solve a Bernoulli differential equation for the solution we will divide this differential equation with y raised to the power n so this will be step number one so after dividing it y raise to the power minus N Y dash plus p x y raised to the power 1 minus n and this side it becomes 1. and then we will make a substitution for this differential equation so step two that we will make the substitution foreign that V is equals to Y raised to the power 1 minus n right we will differentiate this V and so V Dash will become 1 minus n y raised to the power minus n and since it is a function of X therefore this needs to be differentiated with respect to X so one more term will be there and on substituting this expression for V Dash and v in this form let me call this as one this has two so on substituting the expression for V and V Dash into this gets converted into a differential equation of first order which can be solved easily by the techniques by any of the techniques that is integrating factors separable variable form by any techniques which we know for solution of first order differential equation so after this 2 becomes so we can we can easily see that y raised to the power minus n and Y dash so from here we get Y dash over 1 minus n because this needs to come to this side plus p x V is equals to 1. so if we multiply it with v 1 minus n so it becomes 1 minus n p x V and this equals to 1 minus n since n is going to be a constant therefore this will be a constant and this will also be a constant so we can see that it has been converted into a differential equation of first order which can be solved by using as I said by using any techniques integrating factor or a separable variable form whichever is applicable so we will take one question for this so we'll take one example so let us say Y dash plus 4 by X Y is equals to X Cube y Square we need to find out whether it is a Bernal equation or not so we can see that it involves y Square therefore it is a Bernoulli equation right so this is the first order non-linear differential equation which is in Bernoulli form and that is why we will apply the solution technique which is given for Bernoulli form so if we compare it with the uh standard form of Bernoulli differential equation so here y n is y Square right so let me call this as equation number one so divide the equation number one divide equation 1 by y Square so it becomes y raised to the power minus 2 Y dash plus 4 by X Y raised to the power minus 1 and this equals to X Cube so this is the step number one that we need to divide the whole the both side of the differential equation by y raised to the power n so now next step is to make the substitution so let us make the substitution let V is equals to Y raised to the power minus 1. because y raised to the power 1 minus n to be substituted as y v so if we get this V dash for this so it becomes minus V raised to the Y raise to the power minus 2 and Y dash right so if we if we substitute this V and V Dash in equation number two let me call this as three so on substituting V and V Dash into this 2 becomes minus of V Dash plus 4 by X of V is equals to X cubed or if we write this as minus 4 by x v is equals to minus X cubed so we can see that it has become a first order linear differential equation which can be solved by integrating Factor so we will solve it by integrating Factor technique so for this integrating Factor can be found out as e raised to the power minus 4 by X DX so it becomes e raised to the power minus 4 natural log X and we need to write down the solution as so let me call this as 4 so solution of foreign can be written as can be written as V into integrating Factor so V into e raised to the power minus 4 Ln X is equals to e integration e raised to the power minus 4 Ln X and minus X Cube DX and plus there will be a constant of integration so we can simplify this thing and it becomes V is equals to e raised to the power 4 Ln X if this will go to this side and integration minus 4 Ln x minus X cubed DX plus c so this this can be easily simplified so we can uh we are not bothered about this expression because this is easily solvable now my um uh the the solution of the given differential equation should have been written in terms of the dependent variable and dependent variable was y there but we have written the solution here in terms of V therefore we must convert this solution in terms of Y so the relation between v and y was V was considered as V was has been substituted as y raised to the power minus 1 therefore the solution can be written as y raise to the power minus 1 is equals to e raised to the power 4 Ln X e raised to the power minus 4 Ln X minus X Cube DX plus c so that's how the solution of the given differential equation which is in Bernoulli form will be written so this becomes the solution of the given differential equation which was in Bernoulli form now I'm leaving this expression in this form because this is easily solvable so I'm leaving this expression for you and you can solve it and we you can let us know the answer in the comment so that's how the solution of a Bernal equation can be written so in this part of the video that when a differential equation is called in the Bernoulli form of differential equation we have learned how to solve a Bernoulli differential equation so in the second part we will learn about the clearance form of differential equation which will be of first order but not of first degree for details you can go through the book from s Chand publishing details of which is given here link is given in the description box please like share and subscribe the video and press the Bell icon for uh whenever for notification as and when new video will be uploaded thank you very much All rights resolved this video has been prepared for educational purposes only no part of it may be reproduce or copied without the permission of the copyright holder |
3515 | https://www.mathsgenie.co.uk/resources/3-circles-area-and-circumference.pdf | GCSE (1 – 9) Area and Circumference of Circles Name: _________ Instructions • Use black ink or ball-point pen.
• Answer all Questions.
• Answer the Questions in the spaces provided – there may be more space than you need.
• Diagrams are NOT accurately drawn, unless otherwise indicated.
• You must show all your working out.
Information • The marks for each Question are shown in brackets – use this as a guide as to how much time to spend on each Question.
Advice • Read each Question carefully before you start to answer it.
• Keep an eye on the time.
• Try to answer every Question.
• Check your answers if you have time at the end mathsgenie.co.uk 1 (Total for question 1 is 2 marks) (a) On the diagram below, draw a radius of the circle. (b) On the diagram below, draw a sector of the circle.
Shade the sector. 2 (Total for question 2 is 2 marks) (a) Write down the mathematical name for the straight line touching the circle.
(b) Write down the mathematical name for the straight line shown in the diagram. …....................................................
…....................................................
….............………………..
(Total for question 3 is 3 marks) 3 A circle has a radius of 6.5 cm. Work out the circumference of the circle.
Give your answer correct to 2 decimal places. ….............………………..
(Total for question 4 is 3 marks) 4 A circle has a diameter of 9 m. Work out the area of the circle.
Give your answer correct to 1 decimal place. 5 ….............………………..
(Total for question 5 is 3 marks) A circle has a diameter of 12 mm. Work out the circumference of the circle.
Give your answer in terms of π ….............………………..
(Total for question 6 is 3 marks) 6 A circle has a radius of 8 cm. Work out the area of the circle.
Give your answer in terms of π 6.5 cm 9 m 12 mm 8 cm 7 ….................................................… m (Total for question 7 is 4 marks) A semi-circle has an area of 50 m2. Find the perimeter of the semi-circle.
Give your answer correct to one decimal place. 8 £….............………………..
(Total for question 8 is 3 marks) A circular field has a diameter of 32 metres. A farmer wants to build a fence around the edge of the field. Each metre of fence will cost £15.95 Work out the total cost of the fence. 9 …..........................
(Total for question 9 is 5 marks) An area is formed by a square, ABCD, and a semi circle. BD is the diameter of the semi circle. The radius of the semi circle is 4m. The area is going to be covered completely with lawn seed. A box of lawn seed covers 25 m². How many boxes of lawn seed will be needed? You must show your working. A B C D 10 The diagram shows a shaded ring formed by cutting a smaller circle out of a larger circle. The radius of the smaller circle is 6 cm.
The diameter of the larger circle is 15 cm. Find the area of the shaded ring. (Total for question 10 is 3 marks) ….............…..........................cm2 11 The diagram shows three quarters of a circle with a radius of 12 metres. Find the perimeter of the shape. (Total for question 11 is 3 marks) ….............….......................… m 12 m 12 The diagram shows a semi circle inside a sector of a circle, ABC. AB is the diameter of the semi circle.
Angle BAC = 90° AB = 12 cm Find the area of the shaded region. (Total for question 12 is 3 marks) ….............…..........................cm2 A B C 13 A circle is enclosed by a square as shown in the diagram. Each side of the square measures 8cm. Find the area of the shaded region. Give your answer correct to 1 decimal place. (Total for question 13 is 3 marks) ….............…..........................cm2 14 (Total for question 14 is 3 marks) Shape A is a semi-circle which has a radius of 12 cm. Shape B is a circle. The area of shape A is 8 times the area of shape B. Show that the radius of shape B is 3 cm.
A B |
3516 | https://mathresearch.utsa.edu/wiki/index.php?title=Base_10,_Base_2_%26_Base_5 | Base 10, Base 2 & Base 5
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In a positional numeral system, the radix or base is the number of unique digits, including the digit zero, used to represent numbers. For example, for the decimal/denary system (the most common system in use today) the radix (base number) is ten, because it uses the ten digits from 0 through 9.
In any standard positional numeral system, a number is conventionally written as (x)y with x as the string of digits and y as its base, although for base ten the subscript is usually assumed (and omitted, together with the pair of parentheses), as it is the most common way to express value. For example, (100)10 is equivalent to 100 (the decimal system is implied in the latter) and represents the number one hundred, while (100)2 (in the binary system with base 2) represents the number four.
1 In numeral systems
2 Base 2
2.1 Representation
2.2 Counting in binary
2.2.1 Decimal counting
2.2.2 Binary counting
2.3 Fractions
2.4 Binary arithmetic
2.4.1 Addition
2.4.1.1 Long carry method
2.4.1.2 Addition table
2.4.2 Subtraction
2.4.3 Multiplication
2.4.3.1 Multiplication table
2.4.4 Division
2.4.5 Square root
2.5 Conversion to and from other numeral systems
2.5.1 Decimal to Binary
2.5.2 Binary to Decimal
2.5.3 Hexadecimal
2.5.4 Octal
2.6 Representing real numbers
3 Base 5
3.1 Comparison to other radices
3.2 Usage
3.3 Biquinary
4 Base 10
4.1 Origin
4.2 Decimal notation
4.3 Decimal fractions
4.4 Real number approximation
4.5 Infinite decimal expansion
4.5.1 Rational numbers
4.6 Decimal computation
5 Licensing
In numeral systems
In the system with radix 13, for example, a string of digits such as 398 denotes the (decimal) number 3 × 132 + 9 × 131 + 8 × 130 = 632.
More generally, in a system with radix b (b > 1), a string of digits d1 … dn denotes the number d1bn−1 + d2bn−2 + … + dnb0, where 0 ≤ di < b. In contrast to decimal, or radix 10, which has a ones' place, tens' place, hundreds' place, and so on, radix b would have a ones' place, then a b1s' place, a b2s' place, etc.
Commonly used numeral systems include:
| Base/radix | Name | Description |
---
| 2 | Binary numeral system | Used internally by nearly all computers, is base 2. The two digits are "0" and "1", expressed from switches displaying OFF and ON, respectively. Used in most electric counters. |
| 8 | Octal system | Used occasionally in computing. The eight digits are "0"–"7" and represent 3 bits (23). |
| 10 | Decimal system | Used by humans in the vast majority of cultures. Its ten digits are "0"–"9". Used in most mechanical counters. |
| 12 | Duodecimal (dozenal) system | Sometimes advocated due to divisibility by 2, 3, 4, and 6. It was traditionally used as part of quantities expressed in dozens and grosses. |
| 16 | Hexadecimal system | Often used in computing as a more compact representation of binary (1 hex digit per 4 bits). The sixteen digits are "0"–"9" followed by "A"–"F" or "a"–"f". |
| 20 | Vigesimal system | Traditional numeral system in several cultures, still used by some for counting. Historically also known as the score system in English, now most famous in the phrase "four score and seven years ago" in the Gettysburg Address. |
| 60 | Sexagesimal system | Originated in ancient Sumer and passed to the Babylonians. Used today as the basis of modern circular coordinate system (degrees, minutes, and seconds) and time measuring (minutes, and seconds) by analogy to the rotation of the Earth. |
Base 2
A binary number is a number expressed in the base-2 numeral system or binary numeral system, a method of mathematical expression which uses only two symbols: typically "0" (zero) and "1" (one).
The base-2 numeral system is a positional notation with a radix of 2. Each digit is referred to as a bit, or binary digit. Because of its straightforward implementation in digital electronic circuitry using logic gates, the binary system is used by almost all modern computers and computer-based devices, as a preferred system of use, over various other human techniques of communication, because of the simplicity of the language.
Representation
Any number can be represented by a sequence of bits (binary digits), which in turn may be represented by any mechanism capable of being in two mutually exclusive states. Any of the following rows of symbols can be interpreted as the binary numeric value of 667:
| | | | | | | | | | |
--- --- --- --- --- |
| 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| | | ― | | | ― | ― | | | | | ― | | | | |
| ☒ | ☐ | ☒ | ☐ | ☐ | ☒ | ☒ | ☐ | ☒ | ☒ |
| y | n | y | n | n | y | y | n | y | y |
A binary clock might use LEDs to express binary values. In this clock, each column of LEDs shows a binary-coded decimal numeral of the traditional sexagesimal time.
The numeric value represented in each case is dependent upon the value assigned to each symbol. In the earlier days of computing, switches, punched holes and punched paper tapes were used to represent binary values. In a modern computer, the numeric values may be represented by two different voltages; on a magnetic disk, magnetic polarities may be used. A "positive", "yes", or "on" state is not necessarily equivalent to the numerical value of one; it depends on the architecture in use.
In keeping with customary representation of numerals using Arabic numerals, binary numbers are commonly written using the symbols 0 and 1. When written, binary numerals are often subscripted, prefixed or suffixed in order to indicate their base, or radix. The following notations are equivalent:
100101 binary (explicit statement of format)
100101b (a suffix indicating binary format; also known as Intel convention)
100101B (a suffix indicating binary format)
bin 100101 (a prefix indicating binary format)
1001012 (a subscript indicating base-2 (binary) notation)
%100101 (a prefix indicating binary format; also known as Motorola convention)
0b100101 (a prefix indicating binary format, common in programming languages)
6b100101 (a prefix indicating number of bits in binary format, common in programming languages)
b100101 (a prefix indicating binary format, common in Lisp programming languages)
When spoken, binary numerals are usually read digit-by-digit, in order to distinguish them from decimal numerals. For example, the binary numeral 100 is pronounced one zero zero, rather than one hundred, to make its binary nature explicit, and for purposes of correctness. Since the binary numeral 100 represents the value four, it would be confusing to refer to the numeral as one hundred (a word that represents a completely different value, or amount). Alternatively, the binary numeral 100 can be read out as "four" (the correct value), but this does not make its binary nature explicit.
Counting in binary
Counting in binary is similar to counting in any other number system. Beginning with a single digit, counting proceeds through each symbol, in increasing order. Before examining binary counting, it is useful to briefly discuss the more familiar decimal counting system as a frame of reference.
Decimal counting
Decimal counting uses the ten symbols 0 through 9. Counting begins with the incremental substitution of the least significant digit (rightmost digit) which is often called the first digit. When the available symbols for this position are exhausted, the least significant digit is reset to 0, and the next digit of higher significance (one position to the left) is incremented (overflow), and incremental substitution of the low-order digit resumes. This method of reset and overflow is repeated for each digit of significance. Counting progresses as follows:
: 000, 001, 002, ... 007, 008, 009, (rightmost digit is reset to zero, and the digit to its left is incremented)
: 010, 011, 012, ...
: ...
: 090, 091, 092, ... 097, 098, 099, (rightmost two digits are reset to zeroes, and next digit is incremented)
: 100, 101, 102, ...
Binary counting
This counter shows how to count in binary from numbers zero through thirty-one.
A party trick to guess a number from which cards it is printed on uses the bits of the binary representation of the number. In the SVG file, click a card to toggle it
Binary counting follows the same procedure, except that only the two symbols 0 and 1 are available. Thus, after a digit reaches 1 in binary, an increment resets it to 0 but also causes an increment of the next digit to the left:
: 0000,
: 0001, (rightmost digit starts over, and next digit is incremented)
: 0010, 0011, (rightmost two digits start over, and next digit is incremented)
: 0100, 0101, 0110, 0111, (rightmost three digits start over, and the next digit is incremented)
: 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111 ...
In the binary system, each digit represents an increasing power of 2, with the rightmost digit representing 20, the next representing 21, then 22, and so on. The value of a binary number is the sum of the powers of 2 represented by each "1" digit. For example, the binary number 100101 is converted to decimal form as follows:
: 1001012 = [ ( 1 ) × 25 ] + [ ( 0 ) × 24 ] + [ ( 0 ) × 23 ] + [ ( 1 ) × 22 ] + [ ( 0 ) × 21 ] + [ ( 1 ) × 20 ]
: 1001012 = [ 1 × 32 ] + [ 0 × 16 ] + [ 0 × 8 ] + [ 1 × 4 ] + [ 0 × 2 ] + [ 1 × 1 ]
: 1001012 = 3710
Fractions
Fractions in binary arithmetic terminate only if 2 is the only prime factor in the denominator. As a result, 1/10 does not have a finite binary representation (10 has prime factors 2 and 5). This causes 10 × 0.1 not to precisely equal 1 in floating-point arithmetic. As an example, to interpret the binary expression for 1/3 = .010101..., this means: 1/3 = 0 × 2−1 + 1 × 2−2 + 0 × 2−3 + 1 × 2−4 + ... = 0.3125 + ... An exact value cannot be found with a sum of a finite number of inverse powers of two, the zeros and ones in the binary representation of 1/3 alternate forever.
| Fraction | Decimal | Binary | Fractional approximation |
--- --- |
| 1/1 | 1 or 0.999... | 1 or 0.111... | 1/2 + 1/4 + 1/8... |
| 1/2 | 0.5 or 0.4999... | 0.1 or 0.0111... | 1/4 + 1/8 + 1/16 . . . |
| 1/3 | 0.333... | 0.010101... | 1/4 + 1/16 + 1/64 . . . |
| 1/4 | 0.25 or 0.24999... | 0.01 or 0.00111... | 1/8 + 1/16 + 1/32 . . . |
| 1/5 | 0.2 or 0.1999... | 0.00110011... | 1/8 + 1/16 + 1/128 . . . |
| 1/6 | 0.1666... | 0.0010101... | 1/8 + 1/32 + 1/128 . . . |
| 1/7 | 0.142857142857... | 0.001001... | 1/8 + 1/64 + 1/512 . . . |
| 1/8 | 0.125 or 0.124999... | 0.001 or 0.000111... | 1/16 + 1/32 + 1/64 . . . |
| 1/9 | 0.111... | 0.000111000111... | 1/16 + 1/32 + 1/64 . . . |
| 1/10 | 0.1 or 0.0999... | 0.000110011... | 1/16 + 1/32 + 1/256 . . . |
| 1/11 | 0.090909... | 0.00010111010001011101... | 1/16 + 1/64 + 1/128 . . . |
| 1/12 | 0.08333... | 0.00010101... | 1/16 + 1/64 + 1/256 . . . |
| 1/13 | 0.076923076923... | 0.000100111011000100111011... | 1/16 + 1/128 + 1/256 . . . |
| 1/14 | 0.0714285714285... | 0.0001001001... | 1/16 + 1/128 + 1/1024 . . . |
| 1/15 | 0.0666... | 0.00010001... | 1/16 + 1/256 . . . |
| 1/16 | 0.0625 or 0.0624999... | 0.0001 or 0.0000111... | 1/32 + 1/64 + 1/128 . . . |
Binary arithmetic
Arithmetic in binary is much like arithmetic in other numeral systems. Addition, subtraction, multiplication, and division can be performed on binary numerals.
Addition
The circuit diagram for a binary half adder, which adds two bits together, producing sum and carry bits
The simplest arithmetic operation in binary is addition. Adding two single-digit binary numbers is relatively simple, using a form of carrying:
: 0 + 0 → 0
: 0 + 1 → 1
: 1 + 0 → 1
: 1 + 1 → 0, carry 1 (since 1 + 1 = 2 = 0 + (1 × 21) )
Adding two "1" digits produces a digit "0", while 1 will have to be added to the next column. This is similar to what happens in decimal when certain single-digit numbers are added together; if the result equals or exceeds the value of the radix (10), the digit to the left is incremented:
: 5 + 5 → 0, carry 1 (since 5 + 5 = 10 = 0 + (1 × 101) )
: 7 + 9 → 6, carry 1 (since 7 + 9 = 16 = 6 + (1 × 101) )
This is known as carrying. When the result of an addition exceeds the value of a digit, the procedure is to "carry" the excess amount divided by the radix (that is, 10/10) to the left, adding it to the next positional value. This is correct since the next position has a weight that is higher by a factor equal to the radix. Carrying works the same way in binary:
```
1 1 1 1 1 (carried digits)
0 1 1 0 1
+ 1 0 1 1 1
= 1 0 0 1 0 0 = 36
```
In this example, two numerals are being added together: 011012 (1310) and 101112 (2310). The top row shows the carry bits used. Starting in the rightmost column, 1 + 1 = 102. The 1 is carried to the left, and the 0 is written at the bottom of the rightmost column. The second column from the right is added: 1 + 0 + 1 = 102 again; the 1 is carried, and 0 is written at the bottom. The third column: 1 + 1 + 1 = 112. This time, a 1 is carried, and a 1 is written in the bottom row. Proceeding like this gives the final answer 1001002 (3610).
When computers must add two numbers, the rule that:
x xor y = (x + y) mod 2
for any two bits x and y allows for very fast calculation, as well.
Long carry method
A simplification for many binary addition problems is the Long Carry Method or Brookhouse Method of Binary Addition. This method is generally useful in any binary addition in which one of the numbers contains a long "string" of ones. It is based on the simple premise that under the binary system, when given a "string" of digits composed entirely of n ones (where n is any integer length), adding 1 will result in the number 1 followed by a string of n zeros. That concept follows, logically, just as in the decimal system, where adding 1 to a string of n 9s will result in the number 1 followed by a string of n 0s:
```
Binary Decimal
1 1 1 1 1 likewise 9 9 9 9 9
+ 1 + 1
——————————— ———————————
1 0 0 0 0 0 1 0 0 0 0 0
```
Such long strings are quite common in the binary system. From that one finds that large binary numbers can be added using two simple steps, without excessive carry operations. In the following example, two numerals are being added together: 1 1 1 0 1 1 1 1 1 02 (95810) and 1 0 1 0 1 1 0 0 1 12 (69110), using the traditional carry method on the left, and the long carry method on the right:
```
Traditional Carry Method Long Carry Method
vs.
1 1 1 1 1 1 1 1 (carried digits) 1 ← 1 ← carry the 1 until it is one digit past the "string" below
1 1 1 0 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 0 cross out the "string",
+ 1 0 1 0 1 1 0 0 1 1 + 1 0 1 0 1 1 0 0 1 1 and cross out the digit that was added to it
——————————————————————— ——————————————————————
= 1 1 0 0 1 1 1 0 0 0 1 1 1 0 0 1 1 1 0 0 0 1
```
The top row shows the carry bits used. Instead of the standard carry from one column to the next, the lowest-ordered "1" with a "1" in the corresponding place value beneath it may be added and a "1" may be carried to one digit past the end of the series. The "used" numbers must be crossed off, since they are already added. Other long strings may likewise be cancelled using the same technique. Then, simply add together any remaining digits normally. Proceeding in this manner gives the final answer of 1 1 0 0 1 1 1 0 0 0 12 (164910). In our simple example using small numbers, the traditional carry method required eight carry operations, yet the long carry method required only two, representing a substantial reduction of effort.
Addition table
| | 0 | 1 |
---
| 0 | 0 | 1 |
| 1 | 1 | 10 |
The binary addition table is similar, but not the same, as the truth table of the logical disjunction operation . The difference is that , while .
Subtraction
Subtraction works in much the same way:
: 0 − 0 → 0
: 0 − 1 → 1, borrow 1
: 1 − 0 → 1
: 1 − 1 → 0
Subtracting a "1" digit from a "0" digit produces the digit "1", while 1 will have to be subtracted from the next column. This is known as borrowing. The principle is the same as for carrying. When the result of a subtraction is less than 0, the least possible value of a digit, the procedure is to "borrow" the deficit divided by the radix (that is, 10/10) from the left, subtracting it from the next positional value.
```
(starred columns are borrowed from)
1 1 0 1 1 1 0
− 1 0 1 1 1
= 1 0 1 0 1 1 1
```
```
(starred columns are borrowed from)
1 0 1 1 1 1 1
- 1 0 1 0 1 1
= 0 1 1 0 1 0 0
```
Subtracting a positive number is equivalent to adding a negative number of equal absolute value. Computers use signed number representations to handle negative numbers—most commonly the two's complement notation. Such representations eliminate the need for a separate "subtract" operation. Using two's complement notation subtraction can be summarized by the following formula:
: A − B = A + not B + 1
Multiplication
Multiplication in binary is similar to its decimal counterpart. Two numbers A and B can be multiplied by partial products: for each digit in B, the product of that digit in A is calculated and written on a new line, shifted leftward so that its rightmost digit lines up with the digit in B that was used. The sum of all these partial products gives the final result.
Since there are only two digits in binary, there are only two possible outcomes of each partial multiplication:
If the digit in B is 0, the partial product is also 0
If the digit in B is 1, the partial product is equal to A
For example, the binary numbers 1011 and 1010 are multiplied as follows:
```
1 0 1 1 (A)
× 1 0 1 0 (B)
0 0 0 0 ← Corresponds to the rightmost 'zero' in B
+ 1 0 1 1 ← Corresponds to the next 'one' in B
+ 0 0 0 0
+ 1 0 1 1
= 1 1 0 1 1 1 0
```
Binary numbers can also be multiplied with bits after a binary point:
```
1 0 1 . 1 0 1 A (5.625 in decimal)
× 1 1 0 . 0 1 B (6.25 in decimal)
1 . 0 1 1 0 1 ← Corresponds to a 'one' in B
+ 0 0 . 0 0 0 0 ← Corresponds to a 'zero' in B
+ 0 0 0 . 0 0 0
+ 1 0 1 1 . 0 1
+ 1 0 1 1 0 . 1
= 1 0 0 0 1 1 . 0 0 1 0 1 (35.15625 in decimal)
```
See also Booth's multiplication algorithm.
Multiplication table
| | 0 | 1 |
---
| 0 | 0 | 0 |
| 1 | 0 | 1 |
The binary multiplication table is the same as the truth table of the logical conjunction operation .
Division
Long division in binary is again similar to its decimal counterpart.
In the example below, the divisor is 1012, or 5 in decimal, while the dividend is 110112, or 27 in decimal. The procedure is the same as that of decimal long division; here, the divisor 1012 goes into the first three digits 1102 of the dividend one time, so a "1" is written on the top line. This result is multiplied by the divisor, and subtracted from the first three digits of the dividend; the next digit (a "1") is included to obtain a new three-digit sequence:
```
1
_____
1 0 1 ) 1 1 0 1 1
− 1 0 1
0 0 1
```
The procedure is then repeated with the new sequence, continuing until the digits in the dividend have been exhausted:
```
1 0 1
_____
1 0 1 ) 1 1 0 1 1
− 1 0 1
1 1 1
− 1 0 1
0 1 0
```
Thus, the quotient of 110112 divided by 1012 is 1012, as shown on the top line, while the remainder, shown on the bottom line, is 102. In decimal, this corresponds to the fact that 27 divided by 5 is 5, with a remainder of 2.
Aside from long division, one can also devise the procedure so as to allow for over-subtracting from the partial remainder at each iteration, thereby leading to alternative methods which are less systematic, but more flexible as a result.
Square root
The process of taking a binary square root digit by digit is the same as for a decimal square root and is explained here. An example is:
```
1 0 0 1
√ 1010001
1
101 01
0
1001 100
0
10001 10001
10001
0
```
Conversion to and from other numeral systems
Decimal to Binary
Conversion of (357)10 to binary notation results in (101100101)
To convert from a base-10 integer to its base-2 (binary) equivalent, the number is divided by two. The remainder is the least-significant bit. The quotient is again divided by two; its remainder becomes the next least significant bit. This process repeats until a quotient of one is reached. The sequence of remainders (including the final quotient of one) forms the binary value, as each remainder must be either zero or one when dividing by two. For example, (357)10 is expressed as (101100101)2.
Binary to Decimal
Conversion from base-2 to base-10 simply inverts the preceding algorithm. The bits of the binary number are used one by one, starting with the most significant (leftmost) bit. Beginning with the value 0, the prior value is doubled, and the next bit is then added to produce the next value. This can be organized in a multi-column table. For example, to convert 100101011012 to decimal:
| Prior value | × 2 + | Next bit | Next value |
--- --- |
| 0 | × 2 + | 1 | = 1 |
| 1 | × 2 + | 0 | = 2 |
| 2 | × 2 + | 0 | = 4 |
| 4 | × 2 + | 1 | = 9 |
| 9 | × 2 + | 0 | = 18 |
| 18 | × 2 + | 1 | = 37 |
| 37 | × 2 + | 0 | = 74 |
| 74 | × 2 + | 1 | = 149 |
| 149 | × 2 + | 1 | = 299 |
| 299 | × 2 + | 0 | = 598 |
| 598 | × 2 + | 1 | = 1197 |
The result is 119710. The first Prior Value of 0 is simply an initial decimal value. This method is an application of the Horner scheme.
| | | | | | | | | | | | | |
--- --- --- --- --- ---
| Binary | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | |
| Decimal | 1×210 + | 0×29 + | 0×28 + | 1×27 + | 0×26 + | 1×25 + | 0×24 + | 1×23 + | 1×22 + | 0×21 + | 1×20 = | 1197 |
The fractional parts of a number are converted with similar methods. They are again based on the equivalence of shifting with doubling or halving.
In a fractional binary number such as 0.110101101012, the first digit is , the second , etc. So if there is a 1 in the first place after the decimal, then the number is at least , and vice versa. Double that number is at least 1. This suggests the algorithm: Repeatedly double the number to be converted, record if the result is at least 1, and then throw away the integer part.
For example, 10, in binary, is:
| Converting | Result |
--- |
| | 0. |
| | 0.0 |
| | 0.01 |
| | 0.010 |
| | 0.0101 |
Thus the repeating decimal fraction ... is equivalent to the repeating binary fraction ... .
Or for example, 0.110, in binary, is:
| Converting | Result |
--- |
| 0.1 | 0. |
| 0.1 × 2 = 0.2 < 1 | 0.0 |
| 0.2 × 2 = 0.4 < 1 | 0.00 |
| 0.4 × 2 = 0.8 < 1 | 0.000 |
| 0.8 × 2 = 1.6 ≥ 1 | 0.0001 |
| 0.6 × 2 = 1.2 ≥ 1 | 0.00011 |
| 0.2 × 2 = 0.4 < 1 | 0.000110 |
| 0.4 × 2 = 0.8 < 1 | 0.0001100 |
| 0.8 × 2 = 1.6 ≥ 1 | 0.00011001 |
| 0.6 × 2 = 1.2 ≥ 1 | 0.000110011 |
| 0.2 × 2 = 0.4 < 1 | 0.0001100110 |
This is also a repeating binary fraction ... . It may come as a surprise that terminating decimal fractions can have repeating expansions in binary. It is for this reason that many are surprised to discover that 0.1 + ... + 0.1, (10 additions) differs from 1 in floating point arithmetic. In fact, the only binary fractions with terminating expansions are of the form of an integer divided by a power of 2, which 1/10 is not.
The final conversion is from binary to decimal fractions. The only difficulty arises with repeating fractions, but otherwise the method is to shift the fraction to an integer, convert it as above, and then divide by the appropriate power of two in the decimal base. For example:
Another way of converting from binary to decimal, often quicker for a person familiar with hexadecimal, is to do so indirectly—first converting ( in binary) into ( in hexadecimal) and then converting ( in hexadecimal) into ( in decimal).
For very large numbers, these simple methods are inefficient because they perform a large number of multiplications or divisions where one operand is very large. A simple divide-and-conquer algorithm is more effective asymptotically: given a binary number, it is divided by 10k, where k is chosen so that the quotient roughly equals the remainder; then each of these pieces is converted to decimal and the two are concatenated. Given a decimal number, it can be split into two pieces of about the same size, each of which is converted to binary, whereupon the first converted piece is multiplied by 10k and added to the second converted piece, where k is the number of decimal digits in the second, least-significant piece before conversion.
Hexadecimal
Binary may be converted to and from hexadecimal more easily. This is because the radix of the hexadecimal system (16) is a power of the radix of the binary system (2). More specifically, 16 = 24, so it takes four digits of binary to represent one digit of hexadecimal, as shown in the adjacent table.
To convert a hexadecimal number into its binary equivalent, simply substitute the corresponding binary digits:
: 3A16 = 0011 10102
: E716 = 1110 01112
To convert a binary number into its hexadecimal equivalent, divide it into groups of four bits. If the number of bits isn't a multiple of four, simply insert extra 0 bits at the left (called padding). For example:
: 10100102 = 0101 0010 grouped with padding = 5216
: 110111012 = 1101 1101 grouped = DD16
To convert a hexadecimal number into its decimal equivalent, multiply the decimal equivalent of each hexadecimal digit by the corresponding power of 16 and add the resulting values:
: C0E716 = (12 × 163) + (0 × 162) + (14 × 161) + (7 × 160) = (12 × 4096) + (0 × 256) + (14 × 16) + (7 × 1) = 49,38310
Octal
Binary is also easily converted to the octal numeral system, since octal uses a radix of 8, which is a power of two (namely, 23, so it takes exactly three binary digits to represent an octal digit). The correspondence between octal and binary numerals is the same as for the first eight digits of hexadecimal in the table above. Binary 000 is equivalent to the octal digit 0, binary 111 is equivalent to octal 7, and so forth.
| Octal | Binary |
--- |
| 0 | 000 |
| 1 | 001 |
| 2 | 010 |
| 3 | 011 |
| 4 | 100 |
| 5 | 101 |
| 6 | 110 |
| 7 | 111 |
Converting from octal to binary proceeds in the same fashion as it does for hexadecimal:
: 658 = 110 1012
: 178 = 001 1112
And from binary to octal:
: 1011002 = 101 1002 grouped = 548
: 100112 = 010 0112 grouped with padding = 238
And from octal to decimal:
: 658 = (6 × 81) + (5 × 80) = (6 × 8) + (5 × 1) = 5310
: 1278 = (1 × 82) + (2 × 81) + (7 × 80) = (1 × 64) + (2 × 8) + (7 × 1) = 8710
Representing real numbers
Non-integers can be represented by using negative powers, which are set off from the other digits by means of a radix point (called a decimal point in the decimal system). For example, the binary number 11.012 means:
| | | |
---
| 1 × 21 | (1 × 2 = 2) | plus |
| 1 × 20 | (1 × 1 = 1) | plus |
| 0 × 2−1 | (0 × = 0) | plus |
| 1 × 2−2 | (1 × = 0.25) |
For a total of 3.25 decimal.
All dyadic rational numbers have a terminating binary numeral—the binary representation has a finite number of terms after the radix point. Other rational numbers have binary representation, but instead of terminating, they recur, with a finite sequence of digits repeating indefinitely. For instance
The phenomenon that the binary representation of any rational is either terminating or recurring also occurs in other radix-based numeral systems. See, for instance, the explanation in decimal. Another similarity is the existence of alternative representations for any terminating representation, relying on the fact that 0.111111... is the sum of the geometric series 2−1 + 2−2 + 2−3 + ... which is 1.
Binary numerals which neither terminate nor recur represent irrational numbers. For instance,
0.10100100010000100000100... does have a pattern, but it is not a fixed-length recurring pattern, so the number is irrational
1.0110101000001001111001100110011111110... is the binary representation of , the square root of 2, another irrational. It has no discernible pattern.
Base 5
Quinary /ˈkwaɪnəri/ (base-5 or pental) is a numeral system with five as the base. A possible origination of a quinary system is that there are five digits on either hand.
In the quinary place system, five numerals, from 0 to 4, are used to represent any real number. According to this method, five is written as 10, twenty-five is written as 100 and sixty is written as 220.
As five is a prime number, only the reciprocals of the powers of five terminate, although its location between two highly composite numbers (4 and 6) guarantees that many recurring fractions have relatively short periods.
Today, the main usage of base 5 is as a biquinary system, which is decimal using five as a sub-base. Another example of a sub-base system, is sexagesimal, base 60, which used 10 as a sub-base.
Each quinary digit can hold log25 (approx. 2.32) bits of information.
Comparison to other radices
A quinary multiplication table
| × | 1 | 2 | 3 | 4 | 10 | 11 | 12 | 13 | 14 | 20 |
| 1 | 1 | 2 | 3 | 4 | 10 | 11 | 12 | 13 | 14 | 20 |
| 2 | 2 | 4 | 11 | 13 | 20 | 22 | 24 | 31 | 33 | 40 |
| 3 | 3 | 11 | 14 | 22 | 30 | 33 | 41 | 44 | 102 | 110 |
| 4 | 4 | 13 | 22 | 31 | 40 | 44 | 103 | 112 | 121 | 130 |
| 10 | 10 | 20 | 30 | 40 | 100 | 110 | 120 | 130 | 140 | 200 |
| 11 | 11 | 22 | 33 | 44 | 110 | 121 | 132 | 143 | 204 | 220 |
| 12 | 12 | 24 | 41 | 103 | 120 | 132 | 144 | 211 | 223 | 240 |
| 13 | 13 | 31 | 44 | 112 | 130 | 143 | 211 | 224 | 242 | 310 |
| 14 | 14 | 33 | 102 | 121 | 140 | 204 | 223 | 242 | 311 | 330 |
| 20 | 20 | 40 | 110 | 130 | 200 | 220 | 240 | 310 | 330 | 400 |
Numbers zero to twenty-five in standard quinary
| Quinary | 0 | 1 | 2 | 3 | 4 | 10 | 11 | 12 | 13 | 14 | 20 | 21 | 22 |
| Binary | 0 | 1 | 10 | 11 | 100 | 101 | 110 | 111 | 1000 | 1001 | 1010 | 1011 | 1100 |
| Decimal | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| |
| Quinary | 23 | 24 | 30 | 31 | 32 | 33 | 34 | 40 | 41 | 42 | 43 | 44 | 100 |
| Binary | 1101 | 1110 | 1111 | 10000 | 10001 | 10010 | 10011 | 10100 | 10101 | 10110 | 10111 | 11000 | 11001 |
| Decimal | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 |
Fractions in quinary
| Decimal (periodic part) | Quinary (periodic part) | Binary (periodic part) |
| 1/2 = 0.5 | 1/2 = 0.2 | 1/10 = 0.1 |
| 1/3 = 0.3 | 1/3 = 0.13 | 1/11 = 0.01 |
| 1/4 = 0.25 | 1/4 = 0.1 | 1/100 = 0.01 |
| 1/5 = 0.2 | 1/10 = 0.1 | 1/101 = 0.0011 |
| 1/6 = 0.16 | 1/11 = 0.04 | 1/110 = 0.010 |
| 1/7 = 0.142857 | 1/12 = 0.032412 | 1/111 = 0.001 |
| 1/8 = 0.125 | 1/13 = 0.03 | 1/1000 = 0.001 |
| 1/9 = 0.1 | 1/14 = 0.023421 | 1/1001 = 0.000111 |
| 1/10 = 0.1 | 1/20 = 0.02 | 1/1010 = 0.00011 |
| 1/11 = 0.09 | 1/21 = 0.02114 | 1/1011 = 0.0001011101 |
| 1/12 = 0.083 | 1/22 = 0.02 | 1/1100 = 0.0001 |
| 1/13 = 0.076923 | 1/23 = 0.0143 | 1/1101 = 0.000100111011 |
| 1/14 = 0.0714285 | 1/24 = 0.013431 | 1/1110 = 0.0001 |
| 1/15 = 0.06 | 1/30 = 0.013 | 1/1111 = 0.0001 |
| 1/16 = 0.0625 | 1/31 = 0.0124 | 1/10000 = 0.0001 |
| 1/17 = 0.0588235294117647 | 1/32 = 0.0121340243231042 | 1/10001 = 0.00001111 |
| 1/18 = 0.05 | 1/33 = 0.011433 | 1/10010 = 0.0000111 |
| 1/19 = 0.052631578947368421 | 1/34 = 0.011242141 | 1/10011 = 0.000011010111100101 |
| 1/20 = 0.05 | 1/40 = 0.01 | 1/10100 = 0.000011 |
| 1/21 = 0.047619 | 1/41 = 0.010434 | 1/10101 = 0.000011 |
| 1/22 = 0.045 | 1/42 = 0.01032 | 1/10110 = 0.00001011101 |
| 1/23 = 0.0434782608695652173913 | 1/43 = 0.0102041332143424031123 | 1/10111 = 0.00001011001 |
| 1/24 = 0.0416 | 1/44 = 0.01 | 1/11000 = 0.00001 |
| 1/25 = 0.04 | 1/100 = 0.01 | 1/11001 = 0.00001010001111010111 |
Usage
Many languages use quinary number systems, including Gumatj, Nunggubuyu, Kuurn Kopan Noot, Luiseño and Saraveca. Gumatj is a true "5–25" language, in which 25 is the higher group of 5. The Gumatj numerals are shown below:
| Number | Base 5 | Numeral |
---
| 1 | 1 | wanggany |
| 2 | 2 | marrma |
| 3 | 3 | lurrkun |
| 4 | 4 | dambumiriw |
| 5 | 10 | wanggany rulu |
| 10 | 20 | marrma rulu |
| 15 | 30 | lurrkun rulu |
| 20 | 40 | dambumiriw rulu |
| 25 | 100 | dambumirri rulu |
| 50 | 200 | marrma dambumirri rulu |
| 75 | 300 | lurrkun dambumirri rulu |
| 100 | 400 | dambumiriw dambumirri rulu |
| 125 | 1000 | dambumirri dambumirri rulu |
| 625 | 10000 | dambumirri dambumirri dambumirri rulu |
Biquinary
Chinese Abacus or suanpan
A decimal system with 2 and 5 as a sub-bases is called biquinary, and is found in Wolof and Khmer. Roman numerals are an early biquinary system. The numbers 1, 5, 10, and 50 are written as I, V, X, and L respectively. Seven is VII and seventy is LXX. The full list is:
| | | | | | | |
--- --- ---
| I | V | X | L | C | D | M |
| 1 | 5 | 10 | 50 | 100 | 500 | 1000 |
Note that these are not positional number systems. In theory a number such as 73 could be written as IIIXXL without ambiguity as well as LXXIII and it is still not possible to extend it beyond thousands. There is also no sign for zero. But with the introduction of inversions such as IV and IX, it was necessary to keep the order from most to least significant.
Many versions of the abacus, such as the suanpan and soroban, use a biquinary system to simulate a decimal system for ease of calculation. Urnfield culture numerals and some tally mark systems are also biquinary. Units of currencies are commonly partially or wholly biquinary.
Bi-quinary coded decimal is a variant of biquinary that was used on a number of early computers including Colossus and the IBM 650 to represent decimal numbers.
Base 10
The decimal numeral system (also called the base-ten positional numeral system, and occasionally called denary /ˈdiːnəri/ or decanary) is the standard system for denoting integer and non-integer numbers. It is the extension to non-integer numbers of the Hindu–Arabic numeral system. The way of denoting numbers in the decimal system is often referred to as decimal notation.
A decimal numeral (also often just decimal or, less correctly, decimal number), refers generally to the notation of a number in the decimal numeral system. Decimals may sometimes be identified by a decimal separator (usually "." or "," as in 25.9703 or 3,1415). Decimal may also refer specifically to the digits after the decimal separator, such as in "3.14 is the approximation of to two decimals". Zero-digits after a decimal separator serve the purpose of signifying the precision of a value.
The numbers that may be represented in the decimal system are the decimal fractions. That is, fractions of the form a/10n, where a is an integer, and n is a non-negative integer.
The decimal system has been extended to infinite decimals for representing any real number, by using an infinite sequence of digits after the decimal separator (see decimal representation). In this context, the decimal numerals with a finite number of non-zero digits after the decimal separator are sometimes called terminating decimals. A repeating decimal is an infinite decimal that, after some place, repeats indefinitely the same sequence of digits (e.g., 5.123144144144144... = ). An infinite decimal represents a rational number, the quotient of two integers, if and only if it is a repeating decimal or has a finite number of non-zero digits.
Origin
Ten fingers on two hands, the possible origin of decimal counting
Many numeral systems of ancient civilizations use ten and its powers for representing numbers, possibly because there are ten fingers on two hands and people started counting by using their fingers. Examples are firstly the Egyptian numerals, then the Brahmi numerals, Greek numerals, Hebrew numerals, Roman numerals, and Chinese numerals. Very large numbers were difficult to represent in these old numeral systems, and only the best mathematicians were able to multiply or divide large numbers. These difficulties were completely solved with the introduction of the Hindu–Arabic numeral system for representing integers. This system has been extended to represent some non-integer numbers, called decimal fractions or decimal numbers, for forming the decimal numeral system.
Decimal notation
For writing numbers, the decimal system uses ten decimal digits, a decimal mark, and, for negative numbers, a minus sign "−". The decimal digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; and a comma "," in other countries.
For representing a non-negative number, a decimal numeral consists of
either a (finite) sequence of digits (such as "2017"), where the entire sequence represents an integer,
or a decimal mark separating two sequences of digits (such as "20.70828")
: : .
If m > 0, that is, if the first sequence contains at least two digits, it is generally assumed that the first digit am is not zero. In some circumstances it may be useful to have one or more 0's on the left; this does not change the value represented by the decimal: for example, 3.14 = 03.14 = 003.14. Similarly, if the final digit on the right of the decimal mark is zero—that is, if bn = 0—it may be removed; conversely, trailing zeros may be added after the decimal mark without changing the represented number; for example, 15 = 15.0 = 15.00 and 5.2 = 5.20 = 5.200.
For representing a negative number, a minus sign is placed before am.
The numeral represents the number
: .
The integer part or integral part of a decimal numeral is the integer written to the left of the decimal separator (see also truncation). For a non-negative decimal numeral, it is the largest integer that is not greater than the decimal. The part from the decimal separator to the right is the fractional part, which equals the difference between the numeral and its integer part.
When the integral part of a numeral is zero, it may occur, typically in computing, that the integer part is not written (for example .1234, instead of 0.1234). In normal writing, this is generally avoided, because of the risk of confusion between the decimal mark and other punctuation.
In brief, the contribution of each digit to the value of a number depends on its position in the numeral. That is, the decimal system is a positional numeral system.
Decimal fractions
Decimal fractions (sometimes called decimal numbers, especially in contexts involving explicit fractions) are the rational numbers that may be expressed as a fraction whose denominator is a power of ten. For example, the decimals represent the fractions , , , and , and are therefore decimal numbers.
More generally, a decimal with n digits after the separator represents the fraction with denominator 10n, whose numerator is the integer obtained by removing the separator.
It follows that a number is a decimal fraction if and only if it has a finite decimal representation.
Expressed as a fully reduced fraction, the decimal numbers are those whose denominator is a product of a power of 2 and a power of 5. Thus the smallest denominators of decimal numbers are
Real number approximation
Decimal numerals do not allow an exact representation for all real numbers, e.g. for the real number . Nevertheless, they allow approximating every real number with any desired accuracy, e.g., the decimal 3.14159 approximates the real , being less than 10−5 off; so decimals are widely used in science, engineering and everyday life.
More precisely, for every real number x and every positive integer n, there are two decimals L and u with at most n digits after the decimal mark such that L ≤ x ≤ u and (u − L) = 10−n.
Numbers are very often obtained as the result of measurement. As measurements are subject to measurement uncertainty with a known upper bound, the result of a measurement is well-represented by a decimal with n digits after the decimal mark, as soon as the absolute measurement error is bounded from above by 10−n. In practice, measurement results are often given with a certain number of digits after the decimal point, which indicate the error bounds. For example, although 0.080 and 0.08 denote the same number, the decimal numeral 0.080 suggests a measurement with an error less than 0.001, while the numeral 0.08 indicates an absolute error bounded by 0.01. In both cases, the true value of the measured quantity could be, for example, 0.0803 or 0.0796 (see also significant figures).
Infinite decimal expansion
For a real number x and an integer n ≥ 0, let [x]n denote the (finite) decimal expansion of the greatest number that is not greater than x that has exactly n digits after the decimal mark. Let di denote the last digit of [x]i. It is straightforward to see that [x]n may be obtained by appending dn to the right of [x]n−1. This way one has
: [x]n = [x]0.d1d2...dn−1dn,
and the difference of [x]n−1 and [x]n amounts to
: ,
which is either 0, if dn = 0, or gets arbitrarily small as n tends to infinity. According to the definition of a limit, x is the limit of [x]n when n tends to infinity. This is written asor
: x = [x]0.d1d2...dn...,
which is called an infinite decimal expansion of x.
Conversely, for any integer [x]0 and any sequence of digits the (infinite) expression [x]0.d1d2...dn... is an infinite decimal expansion of a real number x. This expansion is unique if neither all dn are equal to 9 nor all dn are equal to 0 for n large enough (for all n greater than some natural number N).
If all dn for n > N equal to 9 and [x]n = [x]0.d1d2...dn, the limit of the sequence is the decimal fraction obtained by replacing the last digit that is not a 9, i.e.: dN, by dN + 1, and replacing all subsequent 9s by 0s (see 0.999...).
Any such decimal fraction, i.e.: dn = 0 for n > N, may be converted to its equivalent infinite decimal expansion by replacing dN by dN − 1 and replacing all subsequent 0s by 9s (see 0.999...).
In summary, every real number that is not a decimal fraction has a unique infinite decimal expansion. Each decimal fraction has exactly two infinite decimal expansions, one containing only 0s after some place, which is obtained by the above definition of [x]n, and the other containing only 9s after some place, which is obtained by defining [x]n as the greatest number that is less than x, having exactly n digits after the decimal mark.
Rational numbers
Long division allows computing the infinite decimal expansion of a rational number. If the rational number is a decimal fraction, the division stops eventually, producing a decimal numeral, which may be prolongated into an infinite expansion by adding infinitely many zeros. If the rational number is not a decimal fraction, the division may continue indefinitely. However, as all successive remainders are less than the divisor, there are only a finite number of possible remainders, and after some place, the same sequence of digits must be repeated indefinitely in the quotient. That is, one has a repeating decimal. For example,
: = 0. 012345679 012... (with the group 012345679 indefinitely repeating).
The converse is also true: if, at some point in the decimal representation of a number, the same string of digits starts repeating indefinitely, the number is rational.
| | |
--- |
| For example, if x is | 0.4156156156... |
| then 10,000x is | 4156.156156156... |
| and 10x is | 4.156156156... |
| so 10,000x − 10x, i.e. 9,990x, is | 4152.000000000... |
| and x is | |
or, dividing both numerator and denominator by 6, .
Decimal computation
Diagram of the world's earliest known multiplication table (c. 305 BCE) from the Warring States period
Most modern computer hardware and software systems commonly use a binary representation internally (although many early computers, such as the ENIAC or the IBM 650, used decimal representation internally).
For external use by computer specialists, this binary representation is sometimes presented in the related octal or hexadecimal systems.
For most purposes, however, binary values are converted to or from the equivalent decimal values for presentation to or input from humans; computer programs express literals in decimal by default. (123.1, for example, is written as such in a computer program, even though many computer languages are unable to encode that number precisely.)
Both computer hardware and software also use internal representations which are effectively decimal for storing decimal values and doing arithmetic. Often this arithmetic is done on data which are encoded using some variant of binary-coded decimal, especially in database implementations, but there are other decimal representations in use (including decimal floating point such as in newer revisions of the IEEE 754|IEEE 754 Standard for Floating-Point Arithmetic).
Decimal arithmetic is used in computers so that decimal fractional results of adding (or subtracting) values with a fixed length of their fractional part always are computed to this same length of precision. This is especially important for financial calculations, e.g., requiring in their results integer multiples of the smallest currency unit for book keeping purposes. This is not possible in binary, because the negative powers of have no finite binary fractional representation; and is generally impossible for multiplication (or division). See Arbitrary-precision arithmetic for exact calculations.
Licensing
Content obtained and/or adapted from:
Radix, Wikipedia under a CC BY-SA license
Binary number, Wikipedia under a CC BY-SA license
Quinary, Wikipedia under a CC BY-SA license
Decimal, Wikipedia under a CC BY-SA license
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2: Solving Linear Equations and Inequalities
Intermediate Algebra
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2.7: Solve Linear Inequalities
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Contents
1. Home
2. Campus Bookshelves
3. Long Beach City College
4. Intermediate Algebra
5. 2: Solving Linear Equations and Inequalities
6. 2.7: Solve Linear Inequalities
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Intermediate Algebra
Front Matter
1: Foundations
2: Solving Linear Equations and Inequalities
3: Math Models
4: Graphs
5: Systems of Linear Equations
6: Polynomials
7: Factoring
8: Rational Expressions and Equations
9: Roots and Radicals
10: Quadratic Equations
Back Matter
2.7: Solve Linear Inequalities
Last updated May 3, 2019
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2.6: Solve a Formula for a Specific Variable
3: Math Models
Page ID 18938
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Note
3. Graph Inequalities on the Number Line
1. Exercise 2.7.1
2. Exercise 2.7.2
3. Exercise 2.7.3
4. INEQUALITIES, NUMBER LINES, AND INTERVAL NOTATION
5. Exercise 2.7.4
6. Exercise 2.7.5
7. Exercise 2.7.6
Solve Inequalities using the Subtraction and Addition Properties of Inequality
PROPERTIES OF EQUALITY
PROPERTIES OF INEQUALITY
Exercise 2.7.7
Exercise 2.7.8
Exercise 2.7.9
Solve Inequalities using the Division and Multiplication Properties of Inequality
PROPERTIES OF EQUALITY
DIVISION AND MULTIPLICATION PROPERTIES OF INEQUALITY
Exercise 2.7.10
Exercise 2.7.11
Exercise 2.7.12
Exercise 2.7.13
Exercise 2.7.14
Exercise 2.7.15
SOLVING INEQUALITIES
Exercise 2.7.16
Exercise 2.7.17
Exercise 2.7.18
Exercise 2.7.19
Exercise 2.7.20
Exercise 2.7.21
Solve Inequalities That Require Simplification
Exercise 2.7.22
Exercise 2.7.23
Exercise 2.7.24
Exercise 2.7.25
Exercise 2.7.26
Exercise 2.7.27
Exercise 2.7.28
Exercise 2.7.29
Exercise 2.7.30
Exercise 2.7.31
Exercise 2.7.32
Exercise 2.7.33
Translate to an Inequality and Solve
Exercise 2.7.34
Exercise 2.7.35
Exercise 2.7.36
Exercise 2.7.37
Exercise 2.7.38
Exercise 2.7.39
Key Concepts
Learning Objectives
By the end of this section, you will be able to:
Graph inequalities on the number line
Solve inequalities using the Subtraction and Addition Properties of inequality
Solve inequalities using the Division and Multiplication Properties of inequality
Solve inequalities that require simplification
Translate to an inequality and solve
Note
Before you get started, take this readiness quiz.
Translate from algebra to English: 15>x.
If you missed this problem, review Exercise 1.3.1.
2. Solve: n−9=−42.
If you missed this problem, review Exercise 2.1.7.
3. Solve: −5p=−23.
If you missed this problem, review Exercise 2.2.1.
4. Solve: 3a−12=7a−20.
If you missed this problem, review Exercise 2.3.22.
Graph Inequalities on the Number Line
Do you remember what it means for a number to be a solution to an equation? A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.
What about the solution of an inequality? What number would make the inequality x>3 true? Are you thinking, ‘x could be 4’? That’s correct, but x could be 5 too, or 20, or even 3.001. Any number greater than 3 is a solution to the inequality x>3.
We show the solutions to the inequality x>3 on the number line by shading in all the numbers to the right of 3, to show that all numbers greater than 3 are solutions. Because the number 3 itself is not a solution, we put an open parenthesis at 3. The graph of x>3 is shown in Figure 2.7.1. Please note that the following convention is used: light blue arrows point in the positive direction and dark blue arrows point in the negative direction.
Figure 2.7.1: The inequality x>3 is graphed on this number line.
The graph of the inequality x≥3 is very much like the graph of x>3, but now we need to show that 3 is a solution, too. We do that by putting a bracket at x=3, as shown in Figure 2.7.2.
Figure 2.7.2: The inequality x≥3 is graphed on this number line.
Notice that the open parentheses symbol, (, shows that the endpoint of the inequality is not included. The open bracket symbol, [, shows that the endpoint is included.
Exercise 2.7.1
Graph on the number line:
x≤1
x<5
x>−1
Answer
1. x≤1 This means all numbers less than or equal to 1. We shade in all the numbers on the number line to the left of 1 and put a bracket at x=1 to show that it is included.
x<5 This means all numbers less than 5, but not including 5. We shade in all the numbers on the number line to the left of 5 and put a parenthesis at x=5 to show it is not included.
x>−1 This means all numbers greater than −1, but not including −1. We shade in all the numbers on the number line to the right of −1, then put a parenthesis at x=−1 to show it is not included.
Exercise 2.7.2
Graph on the number line:
x≤−1
x>2
x<3
Answer
1.
2.
3.
Exercise 2.7.3
Graph on the number line:
x>−2
x<−3
x≥−1
Answer
1.
2.
3.
We can also represent inequalities using interval notation. As we saw above, the inequality x>3 means all numbers greater than 3. There is no upper end to the solution to this inequality. In interval notation, we express x>3 as (3,∞). The symbol ∞ is read as ‘infinity’. It is not an actual number. Figure 2.7.3 shows both the number line and the interval notation.
Figure 2.7.3: The inequality x>3 is graphed on this number line and written in interval notation.
The inequality x≤1 means all numbers less than or equal to 1. There is no lower end to those numbers. We write x≤1 in interval notation as (−∞,1]. The symbol −∞ is read as ‘negative infinity’. Figure 2.7.4 shows both the number line and interval notation.
Figure 2.7.4: The inequality x≤1 is graphed on this number line and written in interval notation.
INEQUALITIES, NUMBER LINES, AND INTERVAL NOTATION
Did you notice how the parenthesis or bracket in the interval notation matches the symbol at the endpoint of the arrow? These relationships are shown in Figure 2.7.5.
Figure 2.7.5: The notation for inequalities on a number line and in interval notation use similar symbols to express the endpoints of intervals.
Exercise 2.7.4
Graph on the number line and write in interval notation.
x≥−3
x<2.5
x≤3 5
Answer
1.
Shade to the right of −3, and put a bracket at −3.
Write in interval notation.
2.
Shade to the left of 2.5, and put a parenthesis at 2.5.
Write in interval notation.
3.
Shade to the left of −3 5, and put a bracket at −3 5.
Write in interval notation.
Exercise 2.7.5
Graph on the number line and write in interval notation:
x>2
x≤−1.5
x≥3 4
Answer
1.
2.
3.
Exercise 2.7.6
Graph on the number line and write in interval notation:
x≤−4
x≥0.5
x<−2 3
Answer
1.
2.
3.
Solve Inequalities using the Subtraction and Addition Properties of Inequality
The Subtraction and Addition Properties of Equality state that if two quantities are equal, when we add or subtract the same amount from both quantities, the results will be equal.
PROPERTIES OF EQUALITY
(2.7.1)Subtraction Property of Equality Addition Property of Equality For any numbers a,b,and c,For any numbers a,b,and c if a=b,if a=b then a−c=b−c.then a+c=b+c
Similar properties hold true for inequalities.
For example, we know that −4 is less than 2.
If we subtract 5 from both quantities, is the left side still less than the right side?
We get −9 on the left and −3 on the right.
And we know −9 is less than −3.
The inequality sign stayed the same.
Table 2.7.1
Similarly we could show that the inequality also stays the same for addition.
This leads us to the Subtraction and Addition Properties of Inequality.
PROPERTIES OF INEQUALITY
(2.7.2)Subtraction Property of Inequality Addition Property of Inequality For any numbers a,b,and c,For any numbers a,b,and c if ab if a>b then a−c>b−c.then a+c>b+c
We use these properties to solve inequalities, taking the same steps we used to solve equations. Solving the inequality x+5>9, the steps would look like this:
(2.7.3)x+5>9 Subtract 5 from both sides to isolate x.x+5−5>9−5 x>4
Any number greater than 4 is a solution to this inequality.
Exercise 2.7.7
Solve the inequality n−1 2≤5 8, graph the solution on the number line, and write the solution in interval notation.
Answer
Add 1 2 to both sides of the inequality.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.
Exercise 2.7.8
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
p−3 4≥1 6
Answer
Exercise 2.7.9
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
r−1 3≤7 12
Answer
Solve Inequalities using the Division and Multiplication Properties of Inequality
The Division and Multiplication Properties of Equality state that if two quantities are equal, when we divide or multiply both quantities by the same amount, the results will also be equal (provided we don’t divide by 0).
PROPERTIES OF EQUALITY
(2.7.4)Division Property of Equality MUltiplication Property of Equality For any numbers a, b, c, and c≠0 For any numbers a, b, c if a=b if a=b then a c=b c then ac=bc
Are there similar properties for inequalities? What happens to an inequality when we divide or multiply both sides by a constant?
Consider some numerical examples.
Divide both sides by 5.Multiply both sides by 5.
Simplify.
Fill in the inequality signs.
Table 2.7.2
The inequality signs stayed the same.
Does the inequality stay the same when we divide or multiply by a negative number?
Divide both sides by -5.Multiply both sides by -5.
Simplify.
Fill in the inequality signs.
Table 2.7.3
The inequality signs reversed their direction.
When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses.
Here are the Division and Multiplication Properties of Inequality for easy reference.
DIVISION AND MULTIPLICATION PROPERTIES OF INEQUALITY
For any real numbers a,b,c
(2.7.5)if a0,then a cb and c>0,then a c>b c and ac>bc if a<b and c<0,then a c>b c and ac>bc if a>b and c<0,then a c<b c and ac<bc
When we divide or multiply an inequality by a:
positive number, the inequality stays the same.
negative number, the inequality reverses.
Exercise 2.7.10
Solve the inequality 7y<42, graph the solution on the number line, and write the solution in interval notation.
Answer
Divide both sides of the inequality by 7.
Since 7>0, the inequality stays the same.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.
Exercise 2.7.11
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
9c>72
Answer
c>8
(8,∞)
Exercise 2.7.12
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
12d≤60
Answer
d≤5
(−∞,5]
Exercise 2.7.13
Solve the inequality −10a≥50, graph the solution on the number line, and write the solution in interval notation.
Answer
Divide both sides of the inequality by −10.
Since −10<0, the inequality reverses.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.
Exercise 2.7.14
Solve each inequality, graph the solution on the number line, and write the solution in interval notation.
−8q<32
Answer
q>−4
Exercise 2.7.15
Solve each inequality, graph the solution on the number line, and write the solution in interval notation.
−7r≤−70
Answer
SOLVING INEQUALITIES
Sometimes when solving an inequality, the variable ends up on the right. We can rewrite the inequality in reverse to get the variable to the left.
(2.7.6)x>a has the same meaning as a<x
Think about it as “If Xavier is taller than Alex, then Alex is shorter than Xavier.”
Exercise 2.7.16
Solve the inequality −20<4 5u, graph the solution on the number line, and write the solution in interval notation.
Answer
Multiply both sides of the inequality by 5 4.
Since 5 4>0, the inequality stays the same.
Simplify.
Rewrite the variable on the left.
Graph the solution on the number line.
Write the solution in interval notation.
Exercise 2.7.17
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
24≤3 8m
Answer
Exercise 2.7.18
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
−24<4 3n
Answer
Exercise 2.7.19
Solve the inequality t−2≥8, graph the solution on the number line, and write the solution in interval notation.
Answer
Multiply both sides of the inequality by −2.
Since −2<0, the inequality reverses.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.
Exercise 2.7.20
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
k−12≤15
Answer
Exercise 2.7.21
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
u−4≥−16
Answer
Solve Inequalities That Require Simplification
Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but be sure to pay close attention during multiplication or division.
Exercise 2.7.22
Solve the inequality 4m≤9m+17, graph the solution on the number line, and write the solution in interval notation.
Answer
Subtract 9m from both sides to collect the variables on the left.
Simplify.
Divide both sides of the inequality by −5, and reverse the inequality.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.
Exercise 2.7.23
Solve the inequality 3q≥7q−23, graph the solution on the number line, and write the solution in interval notation.
Answer
Exercise 2.7.24
Solve the inequality 6x<10x+19, graph the solution on the number line, and write the solution in interval notation.
Answer
Exercise 2.7.25
Solve the inequality 8p+3(p−12)>7p−28 graph the solution on the number line, and write the solution in interval notation.
Answer
Simplify each side as much as possible.8p+3(p−12)>7p−28
Distribute.8p+3p−36>7p−28
Combine like terms.11p−36>7p−28
Subtract 7p from both sides to collect the variables on the left.11p−36−7p>7p−28−7p
Simplify.4p−36>−28
Add 36 to both sides to collect the constants on the right.4p−36+36>−28+36
Simplify.4p>8
Divide both sides of the inequality by 4; the inequality stays the same.4p 4>84
Simplify.p>2
Graph the solution on the number line.
Write the solution in interval notation.(2,∞)
Exercise 2.7.26
Solve the inequality 9y+2(y+6)>5y−24, graph the solution on the number line, and write the solution in interval notation.
Answer
Exercise 2.7.27
Solve the inequality 6u+8(u−1)>10u+32, graph the solution on the number line, and write the solution in interval notation.
Answer
Just like some equations are identities and some are contradictions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.
Exercise 2.7.28
Solve the inequality 8x−2(5−x)<4(x+9)+6x, graph the solution on the number line, and write the solution in interval notation.
Answer
Simplify each side as much as possible.8x−2(5−x)<4(x+9)+6x
Distribute.8x−10+2x<4x+36+6x
Combine like terms.10x−10<10x+36
Subtract 10x from both sides to collect the variables on the left.10x−10−10x<10x+36−10x
Simplify.−10<36
The xx’s are gone, and we have a true statement.The inequality is an identity.
The solution is all real numbers.
Graph the solution on the number line.
Write the solution in interval notation.(−∞,∞)
Exercise 2.7.29
Solve the inequality 4b−3(3−b)>5(b−6)+2b, graph the solution on the number line, and write the solution in interval notation.
Answer
Exercise 2.7.30
Solve the inequality 9h−7(2−h)<8(h+11)+8h, graph the solution on the number line, and write the solution in interval notation.
Answer
Exercise 2.7.31
Solve the inequality 1 3a−1 8a>5 24a+3 4, graph the solution on the number line, and write the solution in interval notation.
Answer
Multiply both sides by the LCD, 24, to clear the fractions.
Simplify.
Combine like terms.
Subtract 5a from both sides to collect the variables on the left.
Simplify.
The statement is false!The inequality is a contradiction.
There is no solution.
Graph the solution on the number line.
Write the solution in interval notation.There is no solution.
Exercise 2.7.32
Solve the inequality 1 4x−1 12x>1 6x+7 8, graph the solution on the number line, and write the solution in interval notation.
Answer
Exercise 2.7.33
Solve the inequality 2 5z−1 3z<1 15z−3 5, graph the solution on the number line, and write the solution in interval notation.
Answer
Translate to an Inequality and Solve
To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like ‘more than’ and ‘less than’. But others are not as obvious.
Think about the phrase ‘at least’ – what does it mean to be ‘at least 21 years old’? It means 21 or more. The phrase ‘at least’ is the same as ‘greater than or equal to’.
Table 2.7.4 shows some common phrases that indicate inequalities.
| > | ≥ | < | ≤ |
--- --- |
| " data-valign="middle" class="lt-math-15134">is greater than | is greater than or equal to | is less than | is less than or equal to |
| " data-valign="middle" class="lt-math-15134">is more than | is at least | is smaller than | is at most |
| " data-valign="middle" class="lt-math-15134">is larger than | is no less than | has fewer than | is no more than |
| " data-valign="middle" class="lt-math-15134">exceeds | is the minimum | is lower than | is the maximum |
Table 2.7.4
Exercise 2.7.34
Translate and solve. Then write the solution in interval notation and graph on the number line.
Twelve times c is no more than 96.
Answer
Translate.
Solve—divide both sides by 12.
Simplify.
Write in interval notation.
Graph on the number line.
Exercise 2.7.35
Translate and solve. Then write the solution in interval notation and graph on the number line.
Twenty times y is at most 100
Answer
Exercise 2.7.36
Translate and solve. Then write the solution in interval notation and graph on the number line.
Nine times z is no less than 135
Answer
Exercise 2.7.37
Translate and solve. Then write the solution in interval notation and graph on the number line.
Thirty less than x is at least 45.
Answer
Translate.
Solve—add 30 to both sides.
Simplify.
Write in interval notation.
Graph on the number line.
Exercise 2.7.38
Translate and solve. Then write the solution in interval notation and graph on the number line.
Nineteen less than p is no less than 47
Answer
Exercise 2.7.39
Translate and solve. Then write the solution in interval notation and graph on the number line.
Four more than a is at most 15.
Answer
Key Concepts
Subtraction Property of Inequality
For any numbers a, b, and c,
if a<b then a−c<b−c and
if a>b then a−c>b−c.
Addition Property of Inequality
For any numbers a, b, and c,
if a<b then a+c<b+c and
if a>b then a+c>b+c.
Division and Multiplication Properties of Inequality
For any numbers a, b, and c,
if a0, then acbc.
if a>b and c>0, then ac>bc and ac>bc.
if a<b and c<0, then ac>bc and ac>bc.
if a>b and c<0, then ac<bc and ac<bc.
When we divide or multiply an inequality by a:
positive number, the inequality stays the same.
negative number, the inequality reverses.
This page titled 2.7: Solve Linear Inequalities is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by OpenStax.
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3: Math Models
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2.1: Solve Equations Using the Subtraction and Addition Properties of Equality
2.2: Solve Equations using the Division and Multiplication Properties of Equality
2.3: Solve Equations with Variables and Constants on Both Sides
2.4: Use a General Strategy to Solve Linear Equations
2.5: Solve Equations with Fractions or Decimals
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3518 | https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/A_Cool_Brisk_Walk_Through_Discrete_Mathematics_(Davies)/02%3A_Sets/2.11%3A_Power_sets | Skip to main content
2.11: Power sets
Last updated
: Jan 21, 2022
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2.10: Subsets
2.12: Partitions
Page ID
: 95473
Stephen Davies
University of Mary Washington via allthemath.org
( \newcommand{\kernel}{\mathrm{null}\,})
Power set is a curious name for a simple concept. We talk about the power set “of" another set, which is the set of all subsets of that other set. Example: suppose A = { Dad, Lizzy }. Then the power set of A, which is written as “P(A)" is: { { Dad, Lizzy }, { Dad }, { Lizzy }, ∅ }. Take a good look at all those curly braces, and don’t lose any. There are four elements to the power set of A, each of which is one of the possible subsets. It might seem strange to talk about “all of the possible subsets" — when I first learned this stuff, I remember thinking at first that there would be no limit to the number of subsets you could make from a set. But of course there is. To create a subset, you can either include, or exclude, each one of the original set’s members. In A’s case, you can either (1) include both Dad and Lizzy, or (2) include Dad but not Lizzy, or (3) include Lizzy but not Dad, or (4) exclude both, in which case your subset is ∅. Therefore, P(A) includes all four of those subsets.
Now what’s the cardinality of P(X) for some set X? That’s an interesting question, and one well worth pondering. The answer ripples through the heart of a lot of combinatorics and the binary number system, topics we’ll cover later. And the answer is right at our fingertips, if we just extrapolate from the previous example. To form a subset of X, we have a choice to either include, or else exclude, each of its elements. So there’s two choices for the first element1, and then whether we choose to include or exclude that first element, there are two choices for the second. Regardless of what we choose for those first two, there are two choices for the third, etc. So if |X|=2 (recall that this notation means “X has two elements" or “X has a cardinality of 2"), then its power set has 2×2 members. If |X|=3, then its power set has 2×2×2 members. In general:
|P(X)|=2|X|.(2.11.1)
As a limiting case (and a brain-bender) notice that if X is the empty set, then P(X) has one (not zero) members, because there is in fact one subset of the empty set: namely, the empty set itself. So |X|=0, and |P(X)|=1. And that jives with the above formula.
I know there’s really no “first” element, but work with me here.
2.10: Subsets
2.12: Partitions |
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Jensen Inequality
In subject area:Mathematics
Jensen's inequality is defined as a mathematical principle stating that for a concave function ϕ and a random variable X, the value of the function at the expected value of X is greater than or equal to the expected value of the function evaluated at X, expressed as ϕ(E(X)) ≥ E(ϕ(X)). It provides a lower bound of the function of the expectation.
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Convex Functions, Partial Orderings, and Statistical Applications
1992, Mathematics in Science and Engineering
2.1Jensen's Inequality
In this section we treat Jensen's inequality in the form of a convex combination of the values of a convex function. Its applications in probability and statistics concerning the expectation of a convex function of a random variable deserve separate attention, and will be treated in Section 13.1.
Jensen's inequality for convex functions is one of the most important inequalities in mathematics and statistics. Many other inequalities can be obtained from it. Thus this inequality and many of its useful consequences are discussed here.
2.1Theorem
(Jensen's Inequality).if is an interval in and f: I → is convex, x = (x1,…,xn) ∈ In (n ≥ 2), p = (p1,…,pn) is a positive n-tuple (i.e., pi, > 0), and Pk = Σi=1k (k = 1,…,n), then
(2.1)
if f is strictly convex, then (2.1) is strict unless x1 = … = xn.
Proof
The proof of (2.1) is by induction. The case n = 2 follows from the definition of convex functions. Suppose that the result is valid for all k, 2 ≤ k ≤ n − 1. Then
holds by the result for n = 2 and the induction hypothesis. The proof for the strict inequality when f is strictly convex is easy and is omitted.
2.2Remarks
(a) The condition that p is a positive n-tuple can be replaced by: “p is a nonnegative n-tuple and Pn > 0”. Similar remarks can be made for other results in this chapter.
(b) Jensen's inequality in (2.1) can be used as an alternative definition of convexity.
(c) Jensen's original papers (1905 and 1906) are related to J-convex functions (functions which1.16)). However, inequality (1.16) appeared much earlier under different assumptions. For example, as Jensen (1905) mentioned in the Appendix of his paper, Hölder (1889) proved inequality (1.16) in 1889 by assuming that/is twice differentiable on [a, b] and that f” (x) ≥: 0 on the interval. If f is twice differentiable, then f″(x) ≥ 0 for x ∈ [a, b] is equivalent to f being convex on [a, b]. Later Henderson (1895/96) proved (2.1) under the same assumptions imposed by Hölder (1889). As far back as 1875, however, a special case of (2.1) in the form
(2.2)
(i.e., p1 = … = pn) was proved by Grolous (1875) by an application of the centroid method. To our knowledge, (2.2) is the first inequality for convex functions in mathematical literature (see Mitrinović and Vasić, 1975). Grolous (1875) also introduced the assumption that f″(x) > 0, but it can be seen from the text that it suffices to assume that f is a convex function in the geometric sense.
(d) By replacing the interval I with a convex set U from a real vector space M, and xi(i = 1,…,n) for points in U, Theorem 2.1 remains valid for convex functions defined on a real vector space.
Integral inequalities analogous to (2.1) can be proved by using Theorem 2.1 or by a direct argument. For example, the following integral analogue of Theorem 2.1 is given in Beesack and Pečarić (1984):
2.3Theorem
(a) Let v be a nonnegative measure on a σ-algebra of subsets of a set D and let q, f be v-measurable functions on D such that q(x) ≥ 0 and − ∞ ≤ a ≤ f(x) ≤ b ≤ ∞ for all x ∈ D and ∫D q dv = 1. If ϕ is continuous and convex on (a, b) and if q ϕ(f) ∈ Lv(D), then either qf + or qf− (or both) is in Lv(D), so that ∫D qf dv = ∫D qf+ dv + ∫D qf− dv is well defined (possibly ∞ or − ∞); moreover, we have
(2.3)
In the case ∫D qf dv = ∞ we have b = supx∈D f(x) = ∞, and the left-hand side of (2.3) means ϕ(∞) = limx→∞ ϕ(x); while if ∫D qf dv = −∞, then a = infx∈D f(x) = −∞ and the left-hand side of (2.3) is ϕ(−∞). Finally, if both integrals in (2.3) are known to exist (finite or infinite), then (2.3) still holds. Moreover, if ∫D qf dv is finite, then ∫D q ϕ(f) dv does exist, either finite or ∞.
(b) If ϕ is continuous and concave on (a, b), then all of the above statements hold true with the direction of inequality in (2.3) reversed. In that case the finiteness of ∫D qf dv implies the existence of ∫D q ϕ(f) dv, either finite or −∞.
Proof
(a) If either ϕ or f is constant for almost all x, then (2.3) holds trivially with equality; thus we assume that this is not the case. The proof of all but the last two sentences in (a) is essentially given in Hardy, Littlewood, and Pólya (1934, 1952, Theorem 202) for ordinary Lebesgue integrals on the line, and that proof still applies in our case. To prove these last sentences we assume that both integrals
exist, either finite or infinite. (The case in which s is finite has already been discussed.) If r is finite, then we proceed as in Hardy, Littlewood, and Pólya (1934, 1952), and the convexity of ϕ implies
(2.4)
where λ is any real number between the right-hand and left-hand derivatives ϕ′R(r), ϕ′L(r). From (2.4) we see that
so that ∫D q[ϕ(f)]− dv is finite. Hence ∫D q dv exists but may have the value ∞. Multiplying (2.4) by q(x) and integrating over D, we obtain (2.3) (and we do not need the existence of s.) If ϕ is concave, then the reverse inequality in (2.4) holds, and similar argument using [ϕ(f)]+ = max{0, ϕ(f)} shows that the finiteness of r implies the existence of s, either finite or − ∞.
Next, let us consider the case r = ∞. Here we have b = ∞, ∫D qf+ dv = ∞, but ∫D qf− dv is finite. Let fn(x) = min{n, f(x)}. Then fn(x) → f(x) as n → ∞ for x ∈ D. Also 0 ≤ fn+(x) ≤ n and 0 fn−(x) ≤ f−(x) hold. Thus both ∫D qfn+ dv and ∫D− qf dv are finite. Consequently, ∫D qf dv is finite, and we may apply (2.3) to fn to obtain
(2.3′)
Now ∫D qfn dv → ∫D qf dv = ∞ as n → ∞. Since ϕ(u) is continuous and convex on (a, ∞), it is also monotonic for large u. Moreover, fn(x) ≤ fn+1(x) for x ∈ D, so that the right-hand side of (2.3′) approaches ϕ(r) = ϕ(∞) as n→∞. To obtain (2.3) it only remains to prove that
(2.5)
We may assume that ϕ(u) is monotonic on (u0, ∞). Let
Then fn+1(x) ≥ fn(x) > u0 for x ∈ D2 and n ≥ n0, where n0 > u0. Thus {q(x)ϕ(fn(x))}n0∞ is a monotonic sequence for each x ∈ D2. By the monotone convergence theorem it follows that
where the limiting integral may have the value ∞ or −∞. For x ∈ D1 we have a ≤ f(x) ≤ u0, so fn(x) ≡ f(x) for all n ≥ n0; hence ∫D1 qϕ(fn) dv ≡ ∫D1 qϕ(f) dv where this integral may also have the value ∞ or − ∞. However, since s = ∫D qϕ(f) dv exists, by assumption it follows that if both ∫D1 qϕ(f) dv and ∫D2 qϕ(f) dv are finite, then they are both ∞ or both −∞, and (2.5) follows. We note that it is possible to prove that ∫D1 qϕ(f) dv is either finite or ∞.
Finally, the case r = −∞ follows in much the same way. In this case, a = −∞, ∫D qf− dv = −∞, and ∫D qf+ dv is finite. We apply (2.3) to the function fn = max{-n, f} for which 0 ≥ fn−(x) ≥ −n, 0 ≤ fn+(x),≤+ (x), and fn+1(x) ≤(x) for all x ∈ D. The proof proceeds as before, and we again obtain (2.3) for f.
(b) This follows by applying (a) to the convex function ϕ1 = −ϕ
Now, let E be a nonempty set and L be a linear class of real-valued functions f: E → having the properties:
L1:
: f, g ∈ L ⇒ (af + bg) ∈ L for all a, b ∈ ;
L2:
: if l ∈ L, i.e., if f(t) = 1 for t ∈ E, then f ∈ L.
We also consider isotonic positive linear functionals A: L→ . That is, we assume that
Al:
: A(af + bg) = aA(f) + bA(g) for f, g ∈ L, a, b ∈ ;
A2:
: f ∈ L, f(t) ≥ 0 on E ⇒ A(f) ≥ 0 (A is isotonic).
If
A3:
: A(1) = 1
also holds, then we say that A(f) is a linear mean defined on L.
Jessen (1931) gave the following generalization of the Jensen's inequality for convex functions (see also Popoviciu, 1944, p. 33):
2.4Theorem
Let L satisfy properties L1, L2 on a nonempty set E, and assume that ϕ is a continuous convex function on an interval I ⊂ . If A is a linear positive functional with A(1) = 1, then for all g ∈ L such that ϕ(g) ∈ L we have A(g) ∈ I and
(2.6)
Proof
(Rasa, 1988a). Let I = [a, b]. From a ≤ g(t) ≤ b for all t ∈ E we obtain a ≤ A(g) ≤ b. For arbitrary but fixed ∈ > 0 there exist real numbers u, v ∈ such that for p = up0 + vp1 (pi(t) = ti for i = 0, 1) we have (i) p ≤ ϕ and (ii) p(A(g)) ≥ ϕ(A(g)) − ∈. (If a < A(g) < b or if f has finite derivative in [a, b], we can replace (ii) by p(A(g)) = ϕ(A(g)).) Now (i) implies p ∘ g ≤ ϕ ° g; hence
Since ∈ is arbitrary, the proof is complete.
Note that, as Rasa (1988) pointed out, if ϕ is not continuous on I, then (2.6) may fail to hold. For example, let
A: L → , A(g) = limx→1 g(x), and ϕ: [0, 1] → such that ϕ(x) = 0 for x ∈ [0, 1) and ϕ(1) = 1. Then ϕ(A(p1)) = 1 > 0 = A(ϕ ° p1).
Additional generalizations of Jessen's inequality are given in McShane (1937). The following Theorems 2.5–2.6 and 2.8–2.10 can be found in his paper. Let n be the n-dimensional Euclidean space, and a point in will be denoted by z = (z1,…,zn). Linear functions Σ aizi on n will be denoted by ℓ(z). If f(x) is an n-tuple of functions (f1(x),…,fn(x)) of L, we denote by A(F) the n-tuple (A(f1),…,A(fn)). From A1 we obtain
A1′: A(ℓ(f)) = ℓ(A(f)) for every function ℓ(z) linear on n.
The first result of McShane (1937) concerns the geometric formulation of Jensen's inequality:
2.5Theorem
Let L1, L2, and A1–A3 be satisfied. Let K be a closed convex point set in n, and f1(x),…,fn(x) be functions of the class L such that f = (f1,…,fn) is in K for all x ∈ E. Then A(f) is in K.
Proof
Let ℓ(z) + c = 0 be a hyperplane in such that K is entirely to one side of the hyperplane, say, ℓ(z) + c ≥ 0 for z in K. Then ℓ(f) + c ≥ 0 for all x, and 0 ≤ A(ℓ(f) + c) = A(ℓ(f)) + A(c) = ℓ(A(f)) + c, so that A(f) lies on the same side of the hyperplane as K does. That is, no hyperplane separates A(f) from K. This is possible only if A(f) is in K.
2.6Theorem
Let L1, L2 and A1–A3 be satisfied. Let K be a closed convex point set in n and ϕ(z) be continuous and convex on K. Let f1(x),…,fn(x) be functions of the class L such that f(x) = (f1(x),…,fn(x)) is in K for all x ∈ E and ϕ(f(x)) is in the class L. Then ϕ(A(f)) is defined and
(2.7)
Proof
Proof Denote by K1 the epigraph of the function ϕ:
Then the set K1 is closed and convex and the point (f(x), ϕ(f(x)) is in K1 for all x ∈ E. Hence, by Theorem 2.4, the point (A(f), A(ϕ(f))) is in K1; that is, A(f) is in K and A(ϕ(f)) ≥ ϕ(A(f)).
2.7Remarks
The following are some examples from McShane's (1937) paper. In the first example the conditions on the convergence or integrability are too obvious to be formally stated.
(a) The range of x is (1,…,m) or (1, 2,…), so that f(x) is a (finite or infinite) sequence {a1, a2,…}, and A(f) = Σ ciai/Σ, ci, where ci ≥ 0 and 0 < Σ ci < ∞.
(b) E is in the interval (0, 1), L is the class of all bounded functions on E, A(f) is the Banach integral of f over (0, 1).
(c) E is the set of all real numbers, L the class of all uniformly almost periodic functions, A(f) is the mean value of f.
(d) More generally, E is any group, L is the class of all functions almost periodic on E, and A(f) is von Neumann's (1934) mean value of f.
In Theorems 2.5 and 2.6 we have not mentioned conditions for strict inequality. To investigate this problem it is convenient to define negligible sets. A set S ⊂ E is negligible (with respect to L and A) if there exists a function f in the class L such that
It follows readily that every subset of a negligible set is negligible and so is every set which is the sum of a finite number of negligible sets.
2.8Theorem
In Theorem 2.5, A(f) is a boundary point of K only if all points f(x) not in a negligible set of x belong to the intersection of K with one of its hyperplanes of support.
Proof
If A(f) is a boundary point of K, then passing through it there exists a hyperplane of support π: ℓ(z) + c = 0 of K; say ℓ(z) + c ≥ 0 for z in K. Let S be the set of x such that f(x) is not in π ∩ K, then ℓ(f(x)) + c > 0 on S. Since ℓ(f(x)) + c ≥ 0 for all x and A(ℓ(f(x)) + c) = ℓ(A(f)) + c = 0, it follows that S is negligible.
The following result is also valid:
2.9Theorem
If the set K is strictly convex and ϕ is strictly convex on K, then in Theorem 2.6 equality holds only if fi(x) = A(fi) = const. (i = 1,…,n) except possibly on a negligible set.
We are unable to state whether the condition “fi = A(fi) except possibly on a negligible set” is sufficient as well as necessary for the equality to hold in Theorem 2.9. However, by adding a condition concerning L and A we can establish such a result even when K is not strictly convex. Thus we restrict our attention to systems of L, A such that the following condition holds:
A4: If S is any negligible subset of E, and f(x) is any (real) function which vanishes on E − S, then f(x) is in the class L and A(f) = 0.
Note that in Remarks 2.7(a) a set S of integers is negligible if Σi∈S ci = 0, and in Remarks 2.7(c) and (d) only the empty set is negligible.
An immediate consequence of conditions L1, L2, A1–A4 is that if f(x) is any function in the class L and g(x) = f(x) except on a negligible set, then g(x) is in L and A(f) = A(g).
The following theorem is proved in McShane (1937):
2.10Theorem
If Condition A4 is satisfied, then in inequality (2.7) equality holds iff the following condition is satisfied: For all x not in a negligible set S, the point (f1(x),…,fn(x)) belongs to a convex subset K′ of K on which ϕ(z) is linear. In particular, if ϕ(z) is strictly convex, then the equality holds iff fi(x) = const, except on a negligible set.
2.11Remarks
(a) (McShane, 1937). It is possible to extend Theorems 2.5, 2.6, 2.8, and 2.9 to functions f(x) assuming values in a Banach space . Suppose that L1, L2, and A1–A3 are satisfied, and that L is a class of functions f(x) defined on E and assuming values in . We shall assume that for every linear function ℓ(z) on and every f(x) in L the function ℓ(f(x)) is in L. Further, we shall assume that there is a linear mean M defined on L such that for every linear function ℓ(z) on and every f in L we have ℓ(M(f)) = M(ℓ(f)). We then find that Theorems 2.5, 2.6, 2.8, and 2.9 can be extended with only one change. The only properties of convex sets needed there are these: through each boundary point of a convex set there exists a hyperplane of support, and every point that does not belong to a convex set can be separated from it by a hyperplane. These properties have been established for convex bodies (closed convex sets having interior points). Hence our theorems can be immediately extended provided that we replace the words “convex set” by “convex body.”
(b) Further generalizations of McShane's (1937) results are given in Guse nov (1987). He considered functions f: E → ∪ {− ∞, ∞}, and assumed that Condition A4 is also valid in Theorems 2.5 and 2.6. In this case Theorems 2.5 and 2.6 are also valid for arbitrary convex sets K (not only closed convex sets). Of course, in this case Guse nov provided an extension of the following example in McShane (1937): Let (Ω, Σ, μ) be a space with positive finite measure. Let L = L1(Ω, Σ, μ) and for f ∈ L define
It is obvious that the Conditions A1–A4 are satisfied. Let K ⊂ n be a convex set, and ϕ be an arbitrary convex function on K (it may have infinite value). If f1,…,fn are functions in the class L1 (i.e., μ- measurable functions), then the following inequality is valid:
(2.8)
Of course, Guse nov (1987) has also given generalizations of McShane's results in Banach space.
(c) To yield results in noncommutative information theory, including the quantum statistics of measurements, several authors (Davis, 1961, 1963; and Nakamura and Umegaki, 1961, 1962) discussed the extension of Jensen's inequality for conditional expectations of finite von Neumann algebras introduced by Umegaki (1954). Since it is obvious in Nakamura, Takesaki, and Umegaki (1960) that a conditional expectation of a von Neumann algebra is completely positive, the following inequality, established by Davis (1957), represents Jensen's inequality in operator algebras:
(2.9)
for a completely positive linear map K, a positive operator x, and an operator convex function f on (λ, v). In Davis (1963) and Ando (1978) it is noted that the following inequality for a contraction a is more general than (2.9) by the Stinespring dilation theorem for completely positive maps:
(2.10)
Hansen (1980) gave an ingenious direct proof on an inequality of Jensen's type: If g is an operator monotone function on [0, ∞), x is a positive operator, and a is a contraction, then
(2.11)
where a (continuous real-valued) function g is operator monotone on [0, ∞) if g(a) ≤ g(b) for 0 ≤ a ≤ b. By the Stinespring theorem, (2.11) implies
(2.12)
for a completely positive map K. Fujii (1979) gave a one-line proof of (2.11) based on the Kubo-Ando theory of means of positive operators (Kubo and Ando, 1980), and he pointed out a bijective correspondence between operator concave functions and means similar to the theory of Kubo-Ando (1980). This result is used in Kainuma and Nakamura (1980) for a short proof of (2.11). Choi (1979) pointed out that (2.9) is also valid for a unital positive linear map K. He also pointed out that the following two inequalities (one of which is the celebrated Schwarz inequality of Kadison) follow from Jensen's inequality (2.3): (i) K(x)2 ≤ K(x2) for a self-adjoint x, and (ii) K(x)−1 ≤ K(x−1) for a positive invertible x. Ando (1978) gave computational proofs of (i) and (ii), by which he proved analytically Jensen's inequality (2.9) based on the integral representation of operator convex functions:
where dm is a regular positive Radom measure on [− 1, 1]. Therefore, Jensen's inequality is essentially equivalent to its special cases (i) and (ii) for a unital positive linear map. An analogous proof for Hansen's inequality (2.12) is given in Kainuma and Nakamura (1980) and based on the following integral representation of an operator monotone function g on [0, ∞):
where dm is a positive Radon measure. They showed that Hansen's inequality is essentially equivalent to its special case (ii) for unital positive maps. For some additional results, see Fujii and Kubo (1982).
In addition to defining J-convex functions in his papers, Jensen (1905, 1906) also applied the famous inductive method (which was used by Cauchy in the proof of the AG inequality) to the proof of the following theorem:
2.12Theorem
If f is J-convex on I, then for all points x1,…,xn ∈ I and all rational nonnegative numbers w1,…,wn such that Wn = Σi=1n wi = 1, we have
(2.13)
Proof
Case 1: wi = 1/n (i = 1,…,n). In this special case (2.13) becomes (2.2). Jensen's (i.e., Cauchy's) inductive proof can be found in many reference books (e.g., Mitrinović, 1970, pp. 14-15; Roberts and Varberg, 1973, pp. 212-13; and Kuczma, 1985, p. 125). However, two simpler inductive proofs in Aczél (1961a, 1961b) are not so well-known. Thus we present these two proofs.
First Proof. For n = 2,
(2.14)
follows by definition of J-convex functions. Now suppose that the result is valid for every k, 2 ≤ k ≤ n. Denoting x = (1/(n + 1)) Σi=1n+1 xi, we have
which is just
(2.15)
Second Proof.
which also yields (2.15).
Case 2: The general case. Since w1,…,wn are nonnegative rational numbers, there exist a natural number m and nonnegative integers P1,…,pn such that m = Pn and wi = pi/m (i = 1,…,n). Now, by Case 1, we have
(2.16)
where x(p) = px. Thus (2.16) yields
2.13Remarks
(a) By replacing the interval I with a convex set U in a real vector space M, and x,(i = 1,…,n) with points in U, Theorem 2.12 remains valid for J-convex functions (mid-convex functions) defined on a real vector space.
(b) Let f: I → [− ∞, ∞) be a function defined on an interval I ⊂ . For t ∈ [0, 1], the function f is called t-convex if f(tu + (1 − t)v) ≤ tf(u) + (1 − t)f(v) holds for all u, v ∈ I, with 0. (− ∞) = 0. Furthermore, we define K(f): = {t ∈ [0, 1]: f is t-convex}. Then Theorem 2.12 shows that the following implication is valid:
The following result is given in Kuhn (1984): Let K(f) ≠ [0, 1]. Then [K(f)] ∩ [0, 1] ⊂ K(f); hence [K(f)] ∩ [0, 1] = K(f). In particular, we have Q ∩ [0, 1] ⊂ K(f). Here for a given subset M ⊂ , [M] defines the subfield generated by M. As a consequence, Kuhn noted that (2.13) is valid for all x1,…,xn ∈ and w1,…,wn ∈ K(f) with Wn = 1.
In Remark 2.2(b) we noted that (2.1) may be used as an alternative definition of convexity. The same is true for Jensen's inequality, i.e., the following result is given in Kocić and Lacković (1986):
2.14Theorem
Let p and q be integrable on [u, v] and p(t) > 0. Let g be monotone and bounded (a ≤ g(t) ≤ b) on [u, v]. Then f ∈ C[a, b] is convex iff for all tt < t2 in [u, v], the inequality
(2.17)
holds.
It is reasonable to ask whether the J-convexity condition in Jensen's inequality can be relaxed. First we observe an answer to this question due to Mitrinovi and Pečarić (1987a):
2.15Theorem
Let f: I → (I = [a, b]) be a real-valued function such that f is Wright-convex on [a, (a + b)/2] and f(x) = −f(b + a − x). If xi ∈ I and (xi + xn+1-i)/2 ∈ [a, (a + b)/2] for i = 1,…,n, then (2.2) is valid.
Proof
First we consider the special case n = 2, i.e., (2.14). If x, y ∈ [a, (a + b)/2], then we have the known case of Jensen's inequality for convex functions. Thus we shall assume that x ∈ [a, (a + b)/2], y ∈((a + b)/2, b] and x + y = d, and we have
If we let δ = a + b − d in the definition of Wright-convex functions by assuming
we obtain that g is a decreasing function on [a, (a + b)/2]. Therefore g(x) ≥ g((a + b)/2), i.e.,
Note that (2.14) can be applied for Wright-convex function f on [a, (a + b)/2] because (d − (a + b)/2) ∈ [a, (a + b)/2]. Now using this result and (2.2), we have
2.16Remark
The conditions in Theorem 2.15 are satisfied if I = [0, 2a], xi ∈ I (i = 1,…,n) and Σi=1n xi ≤ 2a. A special case in which f is a convex function on [0, a] and Σi=1n = 2a is given in Sirotkina (1972).
Another answer to this question is provided in Pečarić and Beesack (1986), where they proved the following result:
2.17Theorem
Let L satisfy properties L1, L2 on a nonempty set E, and let ϕ: I be a function such that there exists a constant k for which
(2.18)
where y0 is a fixed point in I. If A: L → is any isotonic linear functional wtih A(1) = 1, then for all g ∈ L such that ϕ(g) ∈ L and A(g) = y0, the inequality (2.6) holds.
2.18Remarks
(a) If ϕ is convex on I, then for every y0 ∈ I there exists a constant k = k(y0) such that (2.18) holds. Clearly not all ϕ satisfying (2.18) for some y0 ∈ I and k ∈ are convex.
(b) Jessen's inequality for sequences was considered by Gh. Toader (1987b).
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2.23Remarks
(a) For n = 1, we obtain the Jensen-Steffensen Inequality from the Jensen-Boas Inequality and, in the limit as n → ∞, λ is increasing and f needs to be continuous, so that Jensen's inequality is a special case of the Jensen-Boas inequality when passing to the limit.
(b) Theorem 2.22 is given in Boas (1970a).
The following generalization of Jensen-Steffensen's inequality is also valid (Brunk, 1956):
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8.8Remark
In the previous result we assume that all sums are finite. Of course, we can use other generalizations of Jensen's inequality (e.g., a generalization of Theorem 8.1) and related inequalities to obtain similar results. For example, in Vasić and Pečarić (1982c) the Jensen—Petrović Inequality is used, and Imoru (1977) contains a generalized Hardy's inequality.
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Proof
Consider Jensen's inequality for multivariate continuous convex functions on k, (ϕ1, …, ϕk) ∈ C( k), and let dμ be a probability measure on k. Then
(5.34)
and the equality in (5.34) holds iff f ∈ π1 ( k) (the space of all k-variate polynomials of total degree ≤1). Let dμ(x) = M(x | x0,…,xn) dx. Then dμ is a probability measure on k. To establish (5.33) we let fi(x) = xi; i = 1, 2,…,k in (5.34). hence the inequality follows from Theorem 1.6(e).
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Proof
This is a simple consequence of Jensen's inequality after we make the substitutions
in Theorem. 2.1.
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Proof
Using Jensen's inequality for the convex function ϕ we have
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2.2Jensen-Steffensen's Inequality
In Jensen's inequality
(2.19)
it is reasonable to ask whether the condition “p = (p1,…,pn) is a nonnegative n-tuple” can be relaxed at the expense of restricting x = (x1,…,xn) more severely. An answer to this question was given by Steffensen (1919). In a slightly more general form, his result states:
2.19Theorem
If f: I → is a convex function, x is a real monotonic n-tuple such that xi, ∈ I (i = 1,…,n), and p is a real n-tuple such that
(2.20)
is satisfied, then (2.19) holds. If f is strictly convex, then inequality (2.19) is strict unless x1 = x2 = ··· = xn.
Proof
(Pečarić, 1984a). By Theorem 1.6, a convex function f has a line of support at each c ∈ I, i.e., for every z and c we have
(2.21)
holds for some λ. Consequently we have
(2.22)
where λ is the constant satisfying (2.21). Let x and p satisfy the conditions of Theorem 2.19. If we let = (1 /Pn) Σi=1n Pi xi, j = Pn − Pj-1 (j = 2,…,n), and 1 = Pn, then for decreasing x we have
thus ≤ x1. Similarly, we have
and xn ≤ x1. We can easily show that in the case = x1 (or xn) the equality in (2.19) is valid. Thus we need to consider only the case xn < < x1. Let there exist an m such that ∈ [xm+1, xm], and c = . It follows that
(2.23)
Using (2.22) and (2.23) we obtain (2.19).
2.20Remarks
(a) Note that the conditions in (2.20) are necessary and sufficient for the inequality (2.19) to hold for every convex function f: I → and every monotonic n-tuple x. For f(x) = x2, xi, = 0(i = 1,…,k − 1) and xi = 1 (i = k,…,n), (2.19) becomes ( k/Pn)2 ≤ k/Pn, which is equivalent to (2.20) under the assumption Pn > 0.
(b) In the mathematical literature, inequality (2.19), i.e., Theorem 2.19, is known as the Jensen-Steffensen's inequality.
An integral analogue of (2.19) was also given by Steffensen. Here we consider the integral analogue of Jensen-Steffensen's inequality given by Boas (1970a). Let ϕ: I be a continuous convex function where I is the range of the continuous function f: [a, b] → . Note that Jensen's inequality states
(2.24)
provided that λ is increasing, bounded, and λ(a) ≠ λ(b).
2.21Theorem
(Jensen-Steffensen's Inequality). If f is continuous and monotonic {either increasing or decreasing) and λ is either continuous or of bounded variation satisfying
(2.25)
then (2.24) holds.
2.22Theorem
(Jensen-Boas Inequality). If λ is continuous or of bounded variation satisfying
for all xk in (yk-1, yk) (y0 = a, yn = b), and λ(b) > λ(a), and if f is continuous and monotonic (either increasing or decreasing) in each of the n − 1 intervals (yk-1, yk), then inequality (2.24) holds.
Proof
(Pečarić, 1982a). We shall use only Jensen's inequality for sums, i.e., (2.19) for nonnegative n-tuple p, and Theorem 2.21. (Inequality (2.19) can be easily obtained from (2.24).) If λ(a) < λ(y1) < λ(y2) < ··· < λ(yn−1) < λ(b), then from the Jensen-Steffensen Inequality we have
i.e.,
with the notation
Since pk > 0 and tk ∈ I (k = 1,…,n), by (2.19) we have
If λ(yj−1) = λ(yj) for some j, then dλ(x) = 0 on [yj-1, yj] and
thus by (2.19) we can also easily prove that the Jensen-Boas Inequality is valid.
2.23Remarks
(a) For n = 1, we obtain the Jensen-Steffensen Inequality from the Jensen-Boas Inequality and, in the limit as n → ∞, λ is increasing and f needs to be continuous, so that Jensen's inequality is a special case of the Jensen-Boas inequality when passing to the limit.
(b) Theorem 2.22 is given in Boas (1970a).
The following generalization of Jensen-Steffensen's inequality is also valid (Brunk, 1956):
2.24Theorem
(Jensen-Brunk Inequality). Let f be a continuous and increasing function, λ be continuous or of bounded variation, and λ(b) > λ(a). Then (2.24) holds for every convex function ϕ iff
(2.26)
Proof
This is a simple consequence of Theorem 1.7, i.e., it suffices to prove that (2.24) is valid for the functions δ1(x) = vx + uand δ2(x) = (x − c)+ for every c ∈ [a, b]. For arbitrary u, v ∈ the function δ1 satisfies inequality (2.24) (with the inequality in (2.24) being an equality), and there exist real numbers α, β ∈ [a, b] such that f(α) = c and f(β) = = (∫abad f dλ). Let f(α) ≥ f(β), i.e., α ≥ β. Then for the function δ2, (2.24) becomes
(2.27)
and, for f(α) < f(β), i.e., for α < β, (2.24) becomes
(2.28)
It is obvious that the conditions (2.27) and (2.28) are necessary and sufficient for (2.24) to hold. Moreover, these conditions are equivalent to (2.26). Indeed, the implications (2.26) ⇒ (2.27) and (2.26) ⇒ (2.28) are obvious, but the reverse implications are also valid. For example, for or α ≥ β we have
Thus (2.28) is valid for every α ∈ [a, b]. The same is true for (2.27).
2.25Remarks
(a) A similar result is valid for decreasing f. Thus if we assume only that the function f is monotonic, then (2.26) becomes
(2.26′)
(b) By using the identities
(2.29)
Condition (2.26) can be written as in Brunk (1956). By using these two identities we also directly obtain Jensen-Steffensen's inequality from Theorem 2.24.
(c) Simple proofs of Jensen-Steffensen's inequality are also given in Boas (1970a), Magnus (1980), and Pečarić (1985c).
A similar consequence of Theorem 1.7 is the following:
2.26Theorem
Let f be continuous and λ be either continuous or of bounded variation with λ(b) > λ(a). Then (2.24) holds for every continuous convex function ϕ iff
(2.30)
2.27Remark
The result in Theorem 2.26 is given in Pečarić (1981b). As a simple consequence of this theorem, Jensen's, Jensen-Steffensen's, and Jensen-Boas′ inequalities were obtained there.
Let f(t) = (f1(t),…,fk(t)) denote the map from the real interval [a, b] into an interval I in k-dimensional Euclidean space k such that the components fi of f are continuous and monotonic in the same direction. By ∫J f dλ we mean the vector (∫J f1 dλ,…,∫J fk dλ). Brunk (1964) gave a generalization of Theorem 2.24 for functions with increasing increments. As a special case we observe
2.28Theorem
With the above notation, Theorem 2.21 is valid if ϕ is a continuous function with increasing increments.
In a special case we can obtain the discrete version of Theorem 2.28 as follows:
2.29Theorem
Let f: I → (I ⊂ k) be a continuous function with increasing increments, p be a real n-tuple such that (2.20) is satisfied, and xi ∈ I (i = 1,…,n) with
(2.31)
Then (2.19) holds.
In Pečarić (1984g) it is shown that the condition in (2.31) can be relaxed at the expense of restricting p to be a positive n-tuple:
2.30Theorem
Let f: I → (I ⊂ k) be a continuous function with increasing increments and p be a positive n-tuple. If xi ∈ I (i = 1,…,n) and
(2.32)
or if the reverse inequalities in (2.32) hold, then (2.19) is valid.
2.31Remarks
(a) The previous three theorems can also be proved for P-convex functions.
(b) In the previous results we used Riemann-Stieltjes' integral (as in Boas, 1970a). A result concerning Jensen-Steffensen's inequality for Lebesgue's integral is given in Fink and Jodeit (1984).
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Mathematical Inequalities
2005, North-Holland Mathematical Library
Inequality (2) is now known in the literature as Jensen's inequality. It is one of the most important inequalities for convex functions and has been extended and refined in several different directions using different principles or devices. The fundamental work of Jensen was the starting point for the foundation work in convex functions and can be cited as anticipation what was to come. The general theory of convex functions is the origin of powerful tools for the study of problems in analysis. Inequalities involving convex functions are the most efficient tools in the development of several branches of mathematics and has been given considerable attention in the literature.
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Eigenvalues in Riemannian Geometry
1984, Pure and Applied Mathematics
Proof
First one establishes, with standard arguments, on (0, π), and
Next we apply the following version of Jensen's inequality: If F = F(B) is a strictly convex function defined on the convex set of positive definite, self-adjoint linear transformations of V, and v is any positive measure on R, then
(4)
with equality in (4) if and only if B(s) is a constant function.
To apply the Jensen inequality, one sets The Jensen inequality then implies that which, in turn, implies
(5)
One checks that equality is attained in (5) if and only if A = ϕ I on (0, π).
The next step is to apply Hölder's inequality to and measure ε, given by Then the Hölder inequality when 1/p + 1/q = 1 (with equality if and only if there exist constants α, β not both 0, such that α|f|p = β|h|q a.e. [dɛ]), and (5), combine to imply
(6)
where Equality is achieved in (6) if and only if that is, if and only if
(7)
on [0, π].
Thus we are led to study G(ϕ). Note that if ϕ(t) = sin t, then which implies Therefore, inequality (3) is a consequence of the inequality
(8)
We shall prove (8) under the hypothesis that ϕ(t) has the form where 0 ⩽ α, β ⩽ 1, and h(t) is positive and continuous on all of [0, π].
Let Ω be the subset of ℕ3 consisting of those (τ, t, s) for which and, on Ω, define the measure σ by Note that Now write where u is continuous on (0, π). Then the usual form of Jensen's inequality implies
(9)
So (8) will be a consequence of
(10)
for all u. Note that we have yet to use the symmetry hypothesis (cf. (2)) for m(t). We shall show that (10) is valid for all u under consideration if and only if m(t) = m(π - t) for all t ∈ [0, π].
First one employs some manipulation to rewrite (10) as
(11)
for all u, where f(t) is a C2 function on [0, π], satisfying f(0) = 0, and{(sin t)−2 f′(t)}′ = (sin t)−2 {(sin3 t)[m(t) − m(π − t)]}′.If m(t) = m(π − t) for all t ∈ [0, π], then f = 0, and (11) is valid for all u. Conversely, if (11) is valid for all u, then f = 0 and m(t) − m(π - t) = const. To evaluate the constant, set t = π/2. The constant is 0, and inequality (3) is proved.
Equality in (3) implies equality in (6), which implies (7); the theorem is proved.
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Convex Functions, Partial Orderings, and Statistical Applications
1992, Mathematics in Science and Engineering
Proof
By Jensen's inequality we have
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Related terms:
Probability Theory
Poincaré Inequality
Random Variable
Covariate
Lim Inf
Lim Sup
Measurable Function
Steffensen
Convex Function
Conditional Expectation
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3520 | https://pubmed.ncbi.nlm.nih.gov/2986275/ | Sucralfate and alginate/antacid in reflux esophagitis - PubMed
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Scand J Gastroenterol
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. 1985 Mar;20(2):229-32.
doi: 10.3109/00365528509089662.
Sucralfate and alginate/antacid in reflux esophagitis
S Laitinen,M Ståhlberg,M I Kairaluoma,H Kiviniemi,M Pääkkönen,J Lahtinen,E Poikolainen,S Aukee
PMID: 2986275
DOI: 10.3109/00365528509089662
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Clinical Trial
Sucralfate and alginate/antacid in reflux esophagitis
S Laitinen et al. Scand J Gastroenterol.1985 Mar.
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. 1985 Mar;20(2):229-32.
doi: 10.3109/00365528509089662.
Authors
S Laitinen,M Ståhlberg,M I Kairaluoma,H Kiviniemi,M Pääkkönen,J Lahtinen,E Poikolainen,S Aukee
PMID: 2986275
DOI: 10.3109/00365528509089662
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Abstract
The efficacy of sucralfate and of an alginate/antacid compound was compared in a randomized, double-blind 6-week trial in patients with symptomatic, endoscopically confirmed macroscopic reflux esophagitis. Of the 68 patients who completed the study, 36 received sucralfate and 32 alginate/antacid. Significant symptomatic improvement occurred in both treatment groups: almost 70% of the patients became symptom-free or improved. Esophagitis healed completely in 53% of the patients receiving sucralfate and in 34% of the alginate/antacid patients, as measured with endoscopic criteria (p greater than 0.05). Our results indicate that sucralfate seems to be at least as effective as alginate/antacid in relieving symptoms and in healing macroscopic lesions. As a safe, locally active mucosal protecting agent, sucralfate is a promising new drug for the treatment of reflux esophagitis and deserves further trials over longer periods.
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[Treatment of reflux esophagitis with sucralfat].Weiss W, Brunner H, Büttner GR, Gabor M, Miederer S, Mittelstaedt A, Olbermann M, Schwamberger K, Witzel L.Weiss W, et al.Dtsch Med Wochenschr. 1983 Nov 11;108(45):1706-11. doi: 10.1055/s-2008-1069811.Dtsch Med Wochenschr. 1983.PMID: 6354663 Clinical Trial.German.
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3521 | https://courses.lumenlearning.com/odessa-collegealgebra/chapter/using-the-quotient-rule-of-exponents/ | Using the Quotient Rule of Exponents
The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as ymyn, where m>n. Consider the example y9y5. Perform the division by canceling common factors.
y9y5=y⋅y⋅y⋅y⋅y⋅y⋅yy⋅y⋅y⋅y⋅y=y⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅yy⋅y⋅y⋅y⋅y=y⋅y⋅y⋅y1=y4
Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend.
aman=am−n
In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.
y9y5=y9−5=y4
For the time being, we must be aware of the condition m>n. Otherwise, the difference m−n could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.
A General Note: The Quotient Rule of Exponents
For any real number a and natural numbers m and n, such that m>n, the quotient rule of exponents states that
aman=am−n
Example 2: Using the Quotient Rule
Write each of the following products with a single base. Do not simplify further.
(−2)14(−2)9
t23t15
(z√2)5z√2
Solution
Use the quotient rule to simplify each expression.
(−2)14(−2)9=(−2)14−9=(−2)5
t23t15=t23−15=t8
(z√2)5z√2=(z√2)5−1=(z√2)4
Try It 2
Write each of the following products with a single base. Do not simplify further.
s75s68
(−3)6−3
(ef2)5(ef2)3
Solution
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3522 | https://www.academia.edu/41206222/Complex_Variables | (PDF) Complex Variables
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The paper discusses the development and implications of complex numbers within the real number system, outlining key concepts, definitions, and theorems related to complex variables and functions. It aims to provide a comprehensive examination of complex analysis while addressing fundamental principles, including analytic functions, their properties, and various applications.
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Eero Kasanen
European Journal of Psychology of Education, 1997
From an educational point of view, mathematics is supposed to have a completely hierarchical structure in which all new concepts logically follow from prior ones. In this article we try to show that there are also concepts in mathematics which are difJicult to learn because of problematic continuity from prior knowledge to new concepts. Wefocus on the problems of conceptual change connected with the learning of calculus and the shift from rational to real numbers. We demonstrate the difficulty oj this conceptual change with the help of historical and psychological evidence. In the empirical study 65 students of higher secondary school were tested after a 40 hour calculus course. In addition, 11 students participated in individual interview. According to the results the conceptual change from a discrete to a continuous idea of numbers seems to be difficult for students. None of the subjects had developed an adequate understanding of real numbers although they had learned to carry out algorithmic procedures belonging to calculus. fVe discuss how appropriate recent theoretical ideas on conceptual change are for explaining learning problems in this domain. Also some educational implications are presented.
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Basic properties of real numbers
Krzysztof Hryniewiecki
Formalized Mathematics, 1990
FORMALIZED MATHEMATICS Number 1, January 1990 Université Catholique de Louvain Basic Properties of Real Numbers Krzysztof Hryniewiecki1 Warsaw University Summary. Basic facts of arithmetics of real numbers are presented: definitions and properties of the ...
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Evaluate using theorems on limits. In each case, state precisely which theorems are used. (a) lim z!2i (iz 4 þ 3z 2 À 10i), (c) lim z!i=2
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3523 | https://www.wyzant.com/resources/answers/1981/how_do_you_determine_the_axis_of_symmetry_vertex_and_two_points | WYZANT TUTORING
Ruth B.
how do you determine the axis of symmetry, vertex, and two points?
I'm trying to learn about quadratic functions in standard form: conceptual enrichment.
Determine the axis of symmetry, vertex, and at least two points (x, y). The graph the quadratic equation.
2 Answers By Expert Tutors
Robert B. answered • 11/20/12
Experienced (15+ years), Patient, Effective.
If you begin with a quadratic equation written as
y = ax2 + bx + c,
it will be best to rewrite it another way.
By "completing the square," you should be able to write it as
y = a(x - h)2 + k.
Once in this form, everything is a little easier to find.
The vertex is located at (h,k). The axis of symmetry is the vertical line through the vertex, or x = h.
You can find two other points by choosing two values for x (other than h), and find the corresponding y-values. This will give you two more ordered pairs to assist you in graphing the equation. Choosing one value below h, and one above it will give you points on both sides of the vertex.
Hope this helps. If you need further assistance, I or another tutor will be happy to help.
Charles S.
Ruth,
You hail from my old "stomping grounds" ! I'm a former Washingtonian...hail Redskins! Robert B's answer is a great one! I would add only one thing more...
Your vertex will have the abscissa (x-coordinate) of -b/2a , which comes from the formula he listed at the top. Be careful to use your signs wisely, as the "negative" will change the sign of that value. If you substitute that result back into the equation,
then you will find the corresponding y-value to determine the ordered pair for the vertex. If you follow "Cramer" on CNBC and "Mad Money", he alludes to the parabola quite often to denote when a stock will "take off" based on the curve of the parabola. Any
point to the right of the vertex will have a reflective y-value on the left. For example, if your vertex is at zero...on the y-axis, then the result from x=1 will be the same as for x= -1 on the other side. This also works for x=2 and -2. That makes
YOUR work that much easier. By the way, GREAT job Robert. I will give you a thumbs up review. Contact either of us if you have any more questions.
Charles S.
04/11/13
Roman C. answered • 11/20/12
Masters of Education Graduate with Mathematics Expertise
Factor out a and complete the square first:
y = ax2 + bx + c
y = a(x2+(b/a)x+c/a) Factored out a.
y = a(x2 + 2(b/(2a))x + c/a) Prepared for completing square.
y = a(x2 + 2(b/(2a))x + (b/(2a))2 - (b/(2a))2 + c/a) Completed square.
y = a( (x + b/(2a))2 + (4ac - b2 )/(4a2) ) Folded the square + combined terms.
y = a(x + b/(2a))2 + (4ac - b2)/(4a) distributed a.
So the equation of the axis of symmetry is x = -b/(2a), where the square expression vanishes.
The vertex is on this axis and has y-coordinate y = (4ac - b2)/(4a)
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3524 | https://www.cut-the-knot.org/Generalization/PolyStar.shtml | Polygons: formality and intuition. Polygonal metamorphosis.
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Polygons: formality and intuition
A polygon is a closed geometric figure that consists of points - vertices - connected by straight line segments - the sides (or edges) of the polygon. Vertices are sequenced in a cyclic order, and sides only connect pairs of adjacent vertices. The terminology varies. In some sources (The Harper Collins Dictionary of Mathematics, Harper Perennial, 1991), the term polygon only applies to the cases where the sides do not intersect. Elsewhere, especially when star polygons form an object of study, sides are allowed to intersect. In the latter case, side intersections are not considered as vertices of the polygon.
Assume a polygon has n vertices and m sides. Each side connects two vertices, and each vertex belongs to two edges. (In the cyclic order of vertices, one of the edges may be called incoming while the other is naturally outgoing.) We thus have n=2m/2. In other words, n=m. This argument shows that a polygon can be unambiguously referred to as an n-gon. There is no need to indicate whether n is the number of vertices or the sides.
The above argument is not, however, flawless. Imagine an 8-shaped polygon with a vertex shared by the two loops. Is it a polygon at all? Why not? Unless the definition explicitly precludes overlapping vertices, an 8-shaped figure fits the definition perfectly. Should the definition be restricted or not?
If it is, we can, as an argument above shows, freely talk of n-gons. Which is a short and a desirable notation. If overlapping vertices are allowed, the notational convenience seems to be lost; for we no longer may expect that the identity #vertices=#sides will necessarily hold. The situation can be saved, however, with the following device. The key here is the word overlapping. Let's agree that two vertices of a polygon may overlap without being one and the same. In other words, the points that physically represent the vertices may coincide, yet we agree to look at the vertices as different. Does this seem counterintuitive or farfetched? Will this threaten our geometric intuition?
Not necessarily. Once you warm up to the idea, it becomes quite natural. It even adds to our intuition somewhat. Here's an example.
In the discussion on the Isoperimetric Theorem I mention the fact that among all n-gons with the same perimeter, the regular n-gon has the largest area. According to the Isoperimetric Theorem, the largest area of all is enclosed by the circle. The circle can be approximated by regular n-gons. The larger is n, the closer is the approximation. Does it follow then that the area enclosed by regular n-gons grows along with n?
Yes, it does. And the proof becomes quite trivial and well-nigh intuitive if we stick with the definition of n-gon that permits vertices to overlap. Assume we know how to prove that indeed, among all n-gons, the regular n-gon encloses the largest area. Can we now show that a regular n-gon encloses a larger area than a regular (n-1)-gon with the same perimeter? What is here to show? An (n-1)-gon is, by definition, an n-gon with some two adjacent vertices overlapping. Looked at as an n-gon, the regular (n-1)-gon ceases to be regular, and, therefore, by our assumption, has an area less than that enclosed by the regular n-gon. Q.E.D.
The argument seems to push our logic even farther. Not only the vertices may overlap, but the sides should be permitted to have zero length. This is weird. What kind of line segments are those that have zero length? However, the argument is fully supported by the geometric intuition. It would be silly to consider only those polygons whose sides have length that is bounded below by a fixed quantity. So, if sides of a polygon may have arbitrary small lengths, why not to apply the "by continuity" argument, and let them vanish altogether? The idea, as we just saw, is quite fruitful.
{n} is the customary notation for a regular n-gon. To include the star-shaped polygons n is permitted to be a positive rational number above 2, although n=2 is often included. As we know, rational numbers have infinitely many representations as common fractions (5/3=10/6=15/9=...) In general, to construct a star-shaped regular polygon of type {n/d} the fraction is reduced to lowest terms such that n and d become mutually prime. n equidistant points on a circle split the circle into n equals arcs. To obtain an {n/d} polygon, connect n points one after another skipping exactly d arcs at a time.
The renown geometer Branco Grünbaum argues that to carry out this construction, n and d must not be required to be coprime. If they are not, the polygon thus obtained will have its vertices and sides overlap and will appear as a polygon with a smaller n. It is usually said that the construction, in this case, leads to gcd(n,d) overlapping but distinct n/gcd(n,d)-gons. Against the prevailing norm, Grünbaum suggests that the construction yields a single polygon with overlapping vertices. To prove his point, Grünbaum continuously transforms vertices of the polygon by the same amount. Say, the even-numbered vertices slide by an angle t, whereas the odd-numbered vertices slide in the opposite direction, i.e. by the angle -t. Letting t change continuously, it is possible to transform a {14/2}-gon into a polygon of type {14/5}. Which points to the legitimacy of the {14/2}-gon as a single entity as opposed to its being a combination of two distinct 7-gons.
A generalization of Napoleon's Theorem appears to lend support to Grünbaum's point of view as it leads in the same breath to the "conventional" as well as "unconventional" star-shaped polygons.
The applet below illustrates Grünbaum's transformation. It works best when n equals twice a prime. It does not make a lot of sense for odd n because of the loss of symmetry. For n divisible by 4, the transformation may seem to have the effect opposite from the intended one. One consistently observes a polygon with a smaller number of sides. However, uncheck the "Auto update" box and drag the vertices manually. Even in this case there is a single polygon with multiple vertex overlaps.
This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at download and install Java VM and enjoy the applet.
What if applet does not run?
Another problem is pertinent to the discussion. Consider a billiard table in the shape of an equilateral triangle. From a point on the base of the triangle, shoot the ball at 60° angle to the base. After covering a symmetric polygonal itinerary, the ball will return to the starting point. It's interesting that the length of the itinerary - the perimeter of the polygon - does not depend on the starting point. (Prove this.) There is one exception though. If the starting point is located in the middle of the base, the itinerary is less intricate and, in fact, reduces to a plain equilateral triangle whose perimeter is twice as small as the perimeter of all other polygons. Grünbaum's continuity argument suggests that we should talk of a family of 6-gons of the same perimeter of which one (corresponding to the middle point of the base) has vertices overlapping in pairs.
31 August 2015, Created with GeoGebra
This might be good from a geometric perspective. But it does not make sense from the point of view of the billiard ball. If shot from the midpoint, the ball travels along an equilateral triangle and, if not stopped upon its return, will do this indefinitely: going over and over the same triangle. What reason is there to combine 2 triangles into 1 hexagon? Why not to combine 3 or more of them as well? (For the names of (regular) n-gons see a page at the Math Forum.)
So, who is right? And what is the right answer? Are vertex overlaps allowed or not? I do not know about you, but, personally, I do not care. Mathematics is flexible. In one context one definition is more appropriate and/or propitious, in another context one may be better off leaning on another definition. I am often asked whether 0 is a natural number. While the existing ambiguity may be deplorable, I always feel uneasy having to answer this question. What in the world do you care? Pick the answer that best suits your goals. It's all the same to me.
In passing, I want to make one more remark. In popular Calculus texts, there is a tendency to restrict function definitions to functions given by algebraic formulas. Until Weierstrass came up with his example of a nowhere differentiable function some 150 years ago, this perception of function was quite common among mathematicians as well. Nowadays, elementary Calculus texts avoid the subject by insisting that functions that appear in practice (whatever this may mean) are all defined by algebraic expressions; all others belong to a higher mathematics.
I do not know. If we measure the distance traversed by the billiard ball in the above problem, how do we represent this distance as a function of a point on the base of the triangle? If we are only concerned with the distance travelled up to the first return to the starting point, what algebraic formula gives you a quantity constant everywhere except for one point, where it is twice as small? (The question is rhetorical. No such expression exists. The function is an example of a "naturally" occurring discontinuity.)
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| Amount of Bill | 5% | 10% | 15% | 20% | 25% |
--- --- --- |
| $5.00 | $0.25 | $0.50 | $0.75 | $1.00 | $1.25 |
| $10.00 | $0.50 | $1.00 | $1.50 | $2.00 | $2.50 |
| $15.00 | $0.75 | $1.50 | $2.25 | $3.00 | $3.75 |
| $20.00 | $1.00 | $2.00 | $3.00 | $4.00 | $5.00 |
| $25.00 | $1.25 | $2.50 | $3.75 | $5.00 | $6.25 |
| Amount of Bill | 5% | 10% | 15% | 20% | 25% |
--- --- --- |
| $5.00 | $0.25 | $0.50 | $0.75 | $1.00 | $1.25 |
| $10.00 | $0.50 | $1.00 | $1.50 | $2.00 | $2.50 |
| $15.00 | $0.75 | $1.50 | $2.25 | $3.00 | $3.75 |
| $20.00 | $1.00 | $2.00 | $3.00 | $4.00 | $5.00 |
| $25.00 | $1.25 | $2.50 | $3.75 | $5.00 | $6.25 |
A tip is a gift or a sum of money given for a service performed or anticipated. Also known as a gratuity, tips are often given to many types of workers above and beyond the amount of the total bill. While tips are commonplace here in America, in some parts of the world, such as Japan, tips can potentially be seen as insulting and perhaps interpreted as a bribe. We recommend searching for tipping customs in a particular country that you may be visiting to familiarize yourself with local expectations.
In America, the standard (unwritten) rule for calculating a tip is generally 15-20%, but those figures can vary depending on the type of service performed. For example, 15-20% pre-tax is standard for how much to tip at a restaurant, but for bellhops and door attendants, $1-$2 per bag is reasonable.
If you have access to a calculator (on your phone, for example), here is how to calculate tips. First, determine the percentage of your tip. Then, choose one of two ways to do the math. If you want to know exactly how much the tip is going to be, multiply the decimal version of your tip (if you chose 20%, use 0.20) by your total bill. For example, if your total bill is $50, you’d enter 50 x 0.20 on your calculator, giving you 10 – a $10 tip. Or if you want to know the total that you’ll need to spend, add a 1 before the decimal version of your tip. In this instance, if your total bill is $50, you’d enter 50 x 1.20 on your calculator, giving you 60 -- $60 for the total bill, including your tip.
While cash is usually appreciated by most service workers, if you’re paying with a card or personal checks, you can usually add the tip into your total cost. Not paying with a check or card? See if your worker accepts digital currency from an app like Zelle, Venmo or CashApp. Use the handy table at the top of this page to help get you started.
Frequently Asked Questions about Tipping
How much should you tip your hairdresser?
For hair stylists and barbers, a tip of 15% to 20% is common, but consider giving more if you’re particularly happy with how your hair turns out or if you had a great conversation with them during the styling.
How much should you tip movers?
Tipping your moving company isn’t generally expected, but tipping your individual movers is greatly appreciated by them. Depending on the amount of your move, $10 to $20 for each individual mover is a great way to show that you’re thankful for their hard work.
How much should you tip your nail tech?
Like hair stylists, you should expect to tip techs at a nail salon 15% to $20% of the cost of the service before tax. Consider a little more if it’s a lower cost service or you’re incredibly satisfied with their work.
How much should you tip delivery drivers?
A tip of at least 15% for your delivery drivers is usually standard. Consider the distance and total price in your calculation.
How much should you tip at a restaurant?
In sit-down restaurants, expect to tip at least 15% to 20% pretax for good service, and tip more if you feel your server went above and beyond. If you’re simply picking up an order to go, 10% to 15% is more standard and is usually given to the workers that prepared and packed your order.
How much should you tip at a valet?
Tipping your valet $2 to $5 is standard and is usually given to valet that retrieves your car and brings it back to you. Giving the valet that parks your car for you another $2 to $5 is also appreciated and might get you a spot in the shade.
How much should you tip drivers?
Standard tip for your rideshare driver should be around 15% to 20% of the total cost. Consider giving a little more if your driver was particularly helpful with advice or recommendations for the area you’re in, or if they’ve gone out of their way to offer bottled water or phone chargers to their passengers.
What countries don’t tip?
We recommend doing further research on local tipping customs before planning a trip to a different country, but these countries typically do not expect a tip (and in some cases don’t accept them): Australia, China, Malaysia, New Zealand, South Korea, Thailand and Vietnam. There are many European countries that don’t expect tips but do appreciate them, so some research on your destination before you travel can help you prepare.
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3526 | https://www.quora.com/Why-does-lim-limits_-x-to-0-frac-sin-x-x-1 | Why does [math]\lim\limits_{x \to 0} \frac{\sin x}{x} = 1[/math]? - Quora
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Why does lim x→0 sin x x=1 lim x→0 sinx x=1?
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·Updated 8y
Originally Answered: Why does \displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1? ·
Unfortunately, neither Taylor series nor l’Hopital rule-based answers can be qualified as rigorous proofs because they introduce a circular argument: both of these methods require the computation of a derivative of the function f(x)=sin(x)f(x)=sin(x), to compute which we must know what the limit in question is equal to. In other words, while searching for A, we introduce B, but to find B we must know what A is.
It is not that difficult to construct a proof rigorous enough that will pass as acceptable in a “Mathematical Analysis” course. Here is one version: in the drawing below △A O C△A O C is an isos
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Unfortunately, neither Taylor series nor l’Hopital rule-based answers can be qualified as rigorous proofs because they introduce a circular argument: both of these methods require the computation of a derivative of the function f(x)=sin(x)f(x)=sin(x), to compute which we must know what the limit in question is equal to. In other words, while searching for A, we introduce B, but to find B we must know what A is.
It is not that difficult to construct a proof rigorous enough that will pass as acceptable in a “Mathematical Analysis” course. Here is one version: in the drawing below △A O C△A O C is an isosceles triangle contained within the circular sector O A p C O A p C which, in turn, is contained within the right triangle O A B O A B. The line segment A B A B is perpendicular to the ray O A O A:
From Euclid's "Elements" Book 3 3 Proposition 16 16 it follows that the square areas of the above objects are sorted by size as follows:
A△O A C<A O A p C<A△O A B A△O A C<A O A p C<A△O A B
In that proposition Euclid (basically) proves that it is impossible to squeeze another straight line between A B A B and the circumference of the circle q q at the point A A in such a way that that new straight line is positioned between A B A B and p p. Conversely it means that any straight line that cuts the right angle O A B O A B necessarily falls inside the circle - as the line sigment A C A C does above. Next, using the formulas for the areas of a triangle and of a circular sector and the fact that the angle A O C A O C is measured in radians, we have:
O A×C H 2<α n×r 2 2<O A×A B 2 O A×C H 2<α n×r 2 2<O A×A B 2
r 2 sin(α n)<α n×r 2<r 2×tan(α n)r 2 sin(α n)<α n×r 2<r 2×tan(α n)
sin(α n)<α n<tan(α n)sin(α n)<α n<tan(α n)
Observe the leftmost inequality:
sin(α n)<α n sin(α n)<α n
as we will use it later on. Next, we take it that 0<α n<π 2 0<α n<π 2 and that gives us the right to divide the last double inequality by sin(α n)sin(α n):
1<α n sin(α n)<1 cos(α n)1<α n sin(α n)<1 cos(α n)
Since cos(x)cos(x) is an even function and f(x)=x f(x)=x and sin(x)sin(x) are both odd, the reciprocal values of the above inequality are:
1>sin(α n)α n>cos(α n)1>sin(α n)α n>cos(α n)
Multiply the above by −1−1 and flip the inequality signs:
−1<−sin(α n)α n<−cos(α n)−1<−sin(α n)α n<−cos(α n)
Add 1 1 to the above:
0<1−sin(α n)α n<1−cos(α n)0<1−sin(α n)α n<1−cos(α n)
But:
1−cos(α n)=2 sin 2(α n 2)<2 sin(α n 2)<2 α n 2=α n 1−cos(α n)=2 sin 2(α n 2)<2 sin(α n 2)<2 α n 2=α n
because of the “leftmost” inequality that we have proved earlier (see above). Now:
1−cos(α n)<α n 1−cos(α n)<α n
and that means that:
0<1−sin(α n)α n<α n 0<1−sin(α n)α n<α n
Because we assumed earlier that 0<α n<π 2 0<α n<π 2, we can use the absolute values in the above inequalities:
|sin(α n)α n−1|<|α n||sin(α n)α n−1|<|α n|
which conforms to the ϵ,δ ϵ,δ definition of a limit: for any ϵ>0 ϵ>0 we choose δ=m i n(ϵ,π 2)δ=m i n(ϵ,π 2):
|sin(α n)α n−1|<|α n|=|α n−0|<δ|sin(α n)α n−1|<|α n|=|α n−0|<δ
Now, if we were studying not a “continuous” variant but a discrete sequence, then we would set α n=π 2 n α n=π 2 n and have:
α n=π 2 n<δ≤ϵ α n=π 2 n<δ≤ϵ
from where:
n>π 2 ϵ n>π 2 ϵ
and finally:
∀ϵ>0∃N=π 2 ϵ:∀n>N|sin(α n)α n−1|<ϵ∀ϵ>0∃N=π 2 ϵ:∀n>N|sin(α n)α n−1|<ϵ
In either case it means that:
lim x→0 sin(x)x=1 lim x→0 sin(x)x=1
Observe that as an added bonus in this line of reasoning we automatically proved that:
lim x→0 cos(x)=1 lim x→0 cos(x)=1
And from the earlier deduced inequality sin(α n)<α n sin(α n)<α n it follows that as soon as |α n−0|<δ|α n−0|<δ we have |sin(α n)−0|<ϵ|sin(α n)−0|<ϵ which means that:
lim x→0 sin(x)=0 lim x→0 sin(x)=0
From where it immediately follows that we can compute the following limit:
lim x→0 tan(x)=0 lim x→0 tan(x)=0
(left as an exercise to the reader), etc.
To celebrate our victory, let us compute the following limit that has everything to do with the working expression of this question:
lim n→+∞∏n k=1 cos(ϕ 2 k)lim n→+∞∏k=1 n cos(ϕ 2 k)
where ϕ ϕ is an arbitrary non-zero (real) number. Write out first few terms of the product:
cos(ϕ 2)cos(ϕ 2 2)cos(ϕ 2 3)…cos(ϕ 2 n)cos(ϕ 2)cos(ϕ 2 2)cos(ϕ 2 3)…cos(ϕ 2 n)
Start out with the half angle identity:
sin(ϕ)=2 cos(ϕ 2)sin(ϕ 2)sin(ϕ)=2 cos(ϕ 2)sin(ϕ 2)
Apply it again to sin(ϕ 2)sin(ϕ 2):
sin(ϕ)=2 2 cos(ϕ 2)cos(ϕ 2 2)sin(ϕ 2 2)sin(ϕ)=2 2 cos(ϕ 2)cos(ϕ 2 2)sin(ϕ 2 2)
And again - to sin(ϕ 2 2)sin(ϕ 2 2):
sin(ϕ)=2 3 cos(ϕ 2)cos(ϕ 2 2)cos(ϕ 2 3)sin(ϕ 2 3)sin(ϕ)=2 3 cos(ϕ 2)cos(ϕ 2 2)cos(ϕ 2 3)sin(ϕ 2 3)
And so on. We see that after n n such substitutions we will have:
sin(ϕ)=2 n cos(ϕ 2)cos(ϕ 2 2)cos(ϕ 2 3)…cos(ϕ 2 n)sin(ϕ 2 n)sin(ϕ)=2 n cos(ϕ 2)cos(ϕ 2 2)cos(ϕ 2 3)…cos(ϕ 2 n)sin(ϕ 2 n)
From where it follows that our lengthy product of cosines can be represented as:
sin(ϕ)2 n sin(ϕ 2 n)=sin(ϕ)ϕ ϕ 2 n sin(ϕ 2 n)sin(ϕ)2 n sin(ϕ 2 n)=sin(ϕ)ϕ ϕ 2 n sin(ϕ 2 n)
But we already know what the above limit is, and hence:
lim n→+∞∏n k=1 cos(ϕ 2 k)=sin(ϕ)ϕ lim n→+∞∏k=1 n cos(ϕ 2 k)=sin(ϕ)ϕ
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How come sinx/x = 1 when x approaches 0?
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Awnon Bhowmik
Studied at University of Dhaka · Author has 3.7K answers and 11.2M answer views
·Updated 8y
Originally Answered: Why does \lim_{x \to 0} \frac{\sin x}{x} = 1? ·
Method 1: Taylor Expansion
We know
sin x=∞∑n=0(−1)n x 2 n+1(2 n+1)!=x−x 3 3!+x 5 5!−x 7 7!+⋯sin x x=1−x 2 3!+x 4 5!−x 6 7!+⋯sinx=∑n=0∞(−1)n x 2 n+1(2 n+1)!=x−x 3 3!+x 5 5!−x 7 7!+⋯sinx x=1−x 2 3!+x 4 5!−x 6 7!+⋯
Take the limit as x→0 x→0, direct substitution will work. And you have your answer.
Method 2: L’ Hospital’s Rule
Direct substitution yields 0 0 0 0, an indeterminate form. Applying L’ Hospital’s Rule….
lim x→0 sin x x=lim x→0 cos x 1=1 lim x→0 sinx x=lim x→0 cosx 1=1
As Frank Wei pointed out, this limit should not be ev
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Method 1: Taylor Expansion
We know
sin x=∞∑n=0(−1)n x 2 n+1(2 n+1)!=x−x 3 3!+x 5 5!−x 7 7!+⋯sin x x=1−x 2 3!+x 4 5!−x 6 7!+⋯sinx=∑n=0∞(−1)n x 2 n+1(2 n+1)!=x−x 3 3!+x 5 5!−x 7 7!+⋯sinx x=1−x 2 3!+x 4 5!−x 6 7!+⋯
Take the limit as x→0 x→0, direct substitution will work. And you have your answer.
Method 2: L’ Hospital’s Rule
Direct substitution yields 0 0 0 0, an indeterminate form. Applying L’ Hospital’s Rule….
lim x→0 sin x x=lim x→0 cos x 1=1 lim x→0 sinx x=lim x→0 cosx 1=1
As Frank Wei pointed out, this limit should not be evaluated using L’ Hospital’s rule, since the result itself is required to calculate the limit. I know that too, but this is how I was taught in the books I came across, along with the remark that “Using L’ Hospital’s Rule for this limit actually means going in circles, so it is not a favorable method at least for this problem.”
My point being, if limits still have limitations during evaluation, then there should be a slight change in the way we are taught about them. Maybe list some of the limits like these and say yes you can use L’ Hospital’s rule but remember that you are still accepting the result of the proof in order to prove the given problem.
Method 3: Using a well known approximation
Recall that for x≈0 x≈0 we have sin x≈tan x≈x sinx≈tanx≈x
lim x→0 sin x x=lim x→0 x x=1 lim x→0 sinx x=lim x→0 x x=1
Method 4: Geometric Proof with Squeeze Theorem
Consider the unit circle
tan θ=y 1 y=tan θ tanθ=y 1 y=tanθ
Area of Δ O C B≤Area of sector O C B≤Area of Δ O A B 1 2 a b sin θ≤1 2 r 2 θ≤1 2 b h 1 2×1×1×sin θ≤1 2×1 2×θ≤1 2×1×y 1 2 sin θ≤1 2 θ≤1 2 tan θ 1 2 sin θ≤1 2 θ≤1 2⋅sin θ cos θ sin θ≤θ≤sin θ cos θ 1≤θ sin θ≤1 cos θ Area of Δ O C B≤Area of sector O C B≤Area of Δ O A B 1 2 a b sinθ≤1 2 r 2 θ≤1 2 b h 1 2×1×1×sinθ≤1 2×1 2×θ≤1 2×1×y 1 2 sinθ≤1 2 θ≤1 2 tanθ 1 2 sinθ≤1 2 θ≤1 2⋅sinθ cosθ sinθ≤θ≤sinθ cosθ 1≤θ sinθ≤1 cosθ
Since, lim θ→0 1=1=lim θ→0 1 cos θ lim θ→0 1=1=lim θ→0 1 cosθ
Hence lim θ→0 sin θ θ=1 lim θ→0 sinθ θ=1
via the Squeeze Theorem.
Since θ θ is a dummy variable, we can also write
lim x→0 sin x x=1 lim x→0 sinx x=1
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Sridhar Ramesh
Mathematician/Logician/All-Around Great Guy · Upvoted by
José Ilhano Silva
, M. S. Mathematics & Differential Geometry, Federal University of Ceará (2017) · Author has 954 answers and 6.7M answer views
·8y
Originally Answered: Why does \displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1? ·
Essentially, we define radians as the units that make this come out true.
We can think of it this way: What happens when you rotate the unit vector ⟨1,0⟩⟨1,0⟩ by θ θ radians, to yield ⟨cos(θ),sin(θ)⟩⟨cos(θ),sin(θ)⟩? Well, our original vector swings circularly, its tip tracing out an arc of length θ θ (this is what it means to measure in radians; the arclength traced out is the angle times the radius). Also, the tip begins by moving perpendicularly to the original vector (since the tangents of movement along a circle are perpendicular to the radii of the circle).
Thus,
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Essentially, we define radians as the units that make this come out true.
We can think of it this way: What happens when you rotate the unit vector ⟨1,0⟩⟨1,0⟩ by θ θ radians, to yield ⟨cos(θ),sin(θ)⟩⟨cos(θ),sin(θ)⟩? Well, our original vector swings circularly, its tip tracing out an arc of length θ θ (this is what it means to measure in radians; the arclength traced out is the angle times the radius). Also, the tip begins by moving perpendicularly to the original vector (since the tangents of movement along a circle are perpendicular to the radii of the circle).
Thus, if θ θ is “infinitesimally small”, the x-coordinate of the tip stays basically the same while the y-coordinate increases by basically the entire arclength θ θ. This means ⟨cos θ,sin θ⟩≈⟨1,θ⟩⟨cosθ,sinθ⟩≈⟨1,θ⟩ in suitable sense, and from this we find that lim θ→0 sin(θ)θ=1 lim θ→0 sin(θ)θ=1, and also, lim θ→0 cos(θ)−1 θ=0 lim θ→0 cos(θ)−1 θ=0. More generally, we get the derivatives of sine and cosine by this same argument.
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The limit lim x→0 sin x x=1 lim x→0 sinx x=1 can be understood through several approaches: the Squeeze Theorem, Taylor series, or geometric interpretation. Here’s a concise explanation using both the Squeeze Theorem and the Taylor series.
Geometric Interpretation
Consider a unit circle with an angle x x (in radians). We can compare the lengths of different segments:
The length of the arc corresponding to angle x x is x x.
The length of the vertical line segment from the point on the circumference straight down to the x-axis is sin x sinx.
The length of the tangent line segment at the point where the angl
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The limit lim x→0 sin x x=1 lim x→0 sinx x=1 can be understood through several approaches: the Squeeze Theorem, Taylor series, or geometric interpretation. Here’s a concise explanation using both the Squeeze Theorem and the Taylor series.
Geometric Interpretation
Consider a unit circle with an angle x x (in radians). We can compare the lengths of different segments:
The length of the arc corresponding to angle x x is x x.
The length of the vertical line segment from the point on the circumference straight down to the x-axis is sin x sinx.
The length of the tangent line segment at the point where the angle x x meets the circle is tan x tanx.
For 0<x<π 2 0<x<π 2, we can establish the following inequalities based on these segments:
sin x<x<tan x sinx<x<tanx
Dividing all parts of the inequality by sin x sinx (which is positive for x x near 0):
1<x sin x<1 cos x 1<x sinx<1 cosx
Taking the reciprocal (and reversing the inequalities):
cos x<sin x x<1 cosx<sinx x<1
As x→0 x→0, cos x→1 cosx→1. By the Squeeze Theorem:
lim x→0 sin x x=1.lim x→0 sinx x=1.
Taylor Series
The Taylor series expansion for sin x sinx around x=0 x=0 is:
sin x=x−x 3 3!+x 5 5!−x 7 7!+⋯sinx=x−x 3 3!+x 5 5!−x 7 7!+⋯
Now, if we consider the ratio sin x x sinx x:
sin x x=x−x 3 6+x 5 120−⋯x=1−x 2 6+x 4 120−⋯sinx x=x−x 3 6+x 5 120−⋯x=1−x 2 6+x 4 120−⋯
As x→0 x→0, all higher-order terms vanish, leading to:
lim x→0 sin x x=1.lim x→0 sinx x=1.
Conclusion
Both approaches confirm that:
lim x→0 sin x x=1.lim x→0 sinx x=1.
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Douglas Magowan
Private Pilot · Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) and
Stephen Avsec
, Ph.D. Mathematics, University of Illinois at Urbana-Champaign (2012) · Author has 1.2K answers and 1.3M answer views
·8y
Originally Answered: Why does \lim_{x \to 0} \frac{\sin x}{x} = 1? ·
Looking at this figure, we have two triangles and a section of a circle. The radius of the circle is 1. And, all share the same angle that has measure x.
The areas are 1 2|sin x|,1 2|x|,1 2|tan x|1 2|sinx|,1 2|x|,1 2|tanx|
Hopefully it is clear that, when −π 2<x<π 2−π 2<x<π 2
1 2|sin x|≤1 2|x|≤1 2|tan x|1 2|sinx|≤1 2|x|≤1 2|tanx|
And with a little bit of algebra.
1≤x sin x≤sec x 1≥sin x x≥cos x 1≤x sinx≤secx 1≥sinx x≥cosx
Now we look at the limit as x goes to 0.
1\ge\lim_\limits{x\to 0}\frac{\sin x}{x}\ge\lim_\limits{x\to 0}\cos x\1\ge\lim_\limits{x\to 0}\frac{\sin 1\ge\lim_\limits{x\to 0}\frac{\sin x}{x}\ge\lim_\limits{x\to 0}\cos x\1\ge\lim_\limits{x\to 0}\frac{\sin
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Looking at this figure, we have two triangles and a section of a circle. The radius of the circle is 1. And, all share the same angle that has measure x.
The areas are 1 2|sin x|,1 2|x|,1 2|tan x|1 2|sinx|,1 2|x|,1 2|tanx|
Hopefully it is clear that, when −π 2<x<π 2−π 2<x<π 2
1 2|sin x|≤1 2|x|≤1 2|tan x|1 2|sinx|≤1 2|x|≤1 2|tanx|
And with a little bit of algebra.
1≤x sin x≤sec x 1≥sin x x≥cos x 1≤x sinx≤secx 1≥sinx x≥cosx
Now we look at the limit as x goes to 0.
1≥lim x→0 sin x x≥lim x→0 cos x 1≥lim x→0 sin x x≥1 1≥lim x→0 sinx x≥lim x→0 cosx 1≥lim x→0 sinx x≥1
And the limit we seek gets squeezed 1.
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Karel Křesťan
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·8y
Originally Answered: Why does \displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1? ·
Respect to all answers proving this limit in a correct way. Here is my way of showing this limit to a 6 years old child, with no prior knowledge required. Please note that this is not a proof!
Here are our two functions:
Noticed something? Let’s zoom a little bit:
And one more zoom:
Now, we can say that after crossing some point, sin(x) and x are almost identical. For negative numbers, situation is the same:
Nice, right? So we see that in some interval really close to 0, sin(x) and x behaving the same. And what do we get when we divide two identical numbers?
Yes, it’s 1.
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Respect to all answers proving this limit in a correct way. Here is my way of showing this limit to a 6 years old child, with no prior knowledge required. Please note that this is not a proof!
Here are our two functions:
Noticed something? Let’s zoom a little bit:
And one more zoom:
Now, we can say that after crossing some point, sin(x) and x are almost identical. For negative numbers, situation is the same:
Nice, right? So we see that in some interval really close to 0, sin(x) and x behaving the same. And what do we get when we divide two identical numbers?
Yes, it’s 1.
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Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968) · Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) and
Robby Goetschalckx
, Computer scientist for 11+ years and passionate about math since childhood. · Author has 16.6K answers and 29.4M answer views
·8y
Originally Answered: How does one prove that \lim_{x\to 0} \frac{\sin(x)}{x} =1? ·
Several answers have used d’Hopital’s rule. That’s fine if you already know how to prove that the derivative of sin x is cos x. But that proof is usually based on the limit you are asked to prove. This can be proved geometrically. Draw a unit circle and mark an angle x at the centre. Draw lines so that one has length sin x and one, a tangent line, has length tan x. The angle x is the length of an arc.
Based on lengths it seems reasonable that x < tan x. But this is not completely obvious. It’s better to use areas. Then sin x < x < tan x because the each area is nested inside the next. Then 0 <
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Several answers have used d’Hopital’s rule. That’s fine if you already know how to prove that the derivative of sin x is cos x. But that proof is usually based on the limit you are asked to prove. This can be proved geometrically. Draw a unit circle and mark an angle x at the centre. Draw lines so that one has length sin x and one, a tangent line, has length tan x. The angle x is the length of an arc.
Based on lengths it seems reasonable that x < tan x. But this is not completely obvious. It’s better to use areas. Then sin x < x < tan x because the each area is nested inside the next. Then 0 < sin x < x < sin x / cos x.
Then cos x < sin x / x < 1. If you can show that cos x tends to 1 (which should be easy geometrically, then you are almost finished.
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Yim Dongkyun
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·4y
following this image we can express the line segment HC to sinx, and we can also express area of triangle OBC it is sinx/2
we can express sector form OBC’s area to x/2
x is written by radian so 2π:360=x:360x/2π circle O’s area is π so sector form obc’s area is π×(360x/360×2π)=x/2
we can express triangle OBT’s area with tanx/2 because the line segment BC is tan x
know we will use sandwich proof
we can know that area of each triangle and sector form’s can be organized
triangle OBC<sector form OBC < triangle OBT
sinx/2 < x/2 < tanx/2
(tanx=sinx/cosx if) x is not π/2 + Nπ (N is integer)
if you divide all o
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following this image we can express the line segment HC to sinx, and we can also express area of triangle OBC it is sinx/2
we can express sector form OBC’s area to x/2
x is written by radian so 2π:360=x:360x/2π circle O’s area is π so sector form obc’s area is π×(360x/360×2π)=x/2
we can express triangle OBT’s area with tanx/2 because the line segment BC is tan x
know we will use sandwich proof
we can know that area of each triangle and sector form’s can be organized
triangle OBC<sector form OBC < triangle OBT
sinx/2 < x/2 < tanx/2
(tanx=sinx/cosx if) x is not π/2 + Nπ (N is integer)
if you divide all of the expression with sin x
1/2 < x/2sinx < 1/2cosx
if you multiply 2 to all expression
1<x/sinx<1/cosx
if limx→0 all of expression is bigger than 0
you can reverse the expression
cosx<sinx/x<1
know you can find answer
limx→0 cos x = 1
1<limx→0 sinx/x < 1
so you can find limx→0 sinx/x is 1
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Reuven Harmelin
Studied Mathematics at טכניון (Graduated 1978) · Author has 2.3K answers and 1.9M answer views
·1y
Originally Answered: Why does the limit \lim_{x \to 0} \frac{\sin(x)}{x}=1 exist in mathematics? ·
The existence of the limit
and the fact that that limit is equal to 1 , can be deduced from the following chain of inequalities
for every angle x (measured in radians) satisfying
The following drawing explain where these inequalities are coming from
First, when the central angle x=<DOC in a circle of radius 1 centered at point O is measured in radians, then the length of the arc of that circle connecting the points BC is equal x then the length of the chord BC is smaller then x and on the other hand, as the hypotenuse in the right triangle ABC, the length of BC is bigger than the AB=sin(x), which
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The existence of the limit
and the fact that that limit is equal to 1 , can be deduced from the following chain of inequalities
for every angle x (measured in radians) satisfying
The following drawing explain where these inequalities are coming from
First, when the central angle x=<DOC in a circle of radius 1 centered at point O is measured in radians, then the length of the arc of that circle connecting the points BC is equal x then the length of the chord BC is smaller then x and on the other hand, as the hypotenuse in the right triangle ABC, the length of BC is bigger than the AB=sin(x), which implies the first inequality sin(x)<x for every positive angle x.
On the other side, clearly the area of the triangle OCD is bigger than the area of the circular sector OCB, since the sector OCB is included in the triangle OCD. Hence since the area of the sector OCB is
and the area of the triangle OCD is
the inequality x<tan(x) follows immediately.
Now since tan(x)=sin(x)/cos(x), the inequalities
imply, after dividing by x>0 and multiplying by cos(x)>0 ,
We observe that because both cos(x), sin(x)/x are even functions and therefore the last inequalities hold also for negative values of x. now, since
we deduce by the squeeze rule that cos(x) tends to 1 as x→ 0 , and once again by the squeeze rule regarding the previously proven inequality the existence of the limit of sin(x)/x as x→0 and the fact that this limit is 1 are established
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Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987) · Author has 6.3K answers and 1.7M answer views
·1y
Originally Answered: Why does the limit \lim_{x \to 0} \frac{\sin(x)}{x}=1 exist in mathematics? ·
Here a plot of the function sin(x) / x
As you can see at x = 0 there seems to exist a limit of 1.
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Here a plot of the function sin(x) / x
As you can see at x = 0 there seems to exist a limit of 1.
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Alf Salte
Works at Schlumberger (company) · Author has 3.8K answers and 2.6M answer views
·9y
Originally Answered: Why is \displaystyle\lim_{x \to 0}\dfrac{\sin x}{x}=1? ·
First convince yourself that when x is positive small number (less than pi/2) then if you picture a pizza slice - part of a circle or pizza with radius 1 and it has three points. One is the centre of the pizza O, one point is along the x axis following a straight edge of the pizza and we can call that point B and the third point is A being on the straight edge that is not along the X axis. The lines OA and OB are straight lines while AB is the curved edge along the pizza slice. Now. If you put a straight line from A down to the X axis you will hit a new point C which divides the line OB and AC
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First convince yourself that when x is positive small number (less than pi/2) then if you picture a pizza slice - part of a circle or pizza with radius 1 and it has three points. One is the centre of the pizza O, one point is along the x axis following a straight edge of the pizza and we can call that point B and the third point is A being on the straight edge that is not along the X axis. The lines OA and OB are straight lines while AB is the curved edge along the pizza slice. Now. If you put a straight line from A down to the X axis you will hit a new point C which divides the line OB and AC will have length sin(x). Alternatively you can make a line from A tangent to the circle that will hit the X axis at point D somewhere outside of OB. Now. AD will be tan(x) the curved line AB has length x and AC is sin(x). Clearly, sin(x) < x < tan(x). Dividing by sin(x) you then get 1 < x/sin(x) < 1/cos(x). When inverting these expressions you also change the less than to greater than so you get: 1 > sin(x)/x > cos(x). As x goes towards zero cos(x) will go to 1 and so sin(x)/x will be squeezed between those two so it too must have a limit of 1. So Lim x -> 0 [ sin(x)/x ] = 1.
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Jered M.
mathematics educator · Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 3K answers and 5.5M answer views
·9y
Originally Answered: Why is \displaystyle\lim_{x \to 0}\dfrac{\sin x}{x}=1? ·
For a solid geometrical proof, see the first answer at How to prove that lim sin(x)/x=1?
The basic idea is to compare sin x sinx to tan x tanx, per the following diagram:
Continue Reading
For a solid geometrical proof, see the first answer at How to prove that lim sin(x)/x=1?
The basic idea is to compare sin x sinx to tan x tanx, per the following diagram:
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; Data structures using C and C++
; Langsam, Yedidyah, 1952-; ; Augenstein, Moshe, 1947-; Tenenbaum, Aaron M;
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Data structures using C and C++ Available at Hilton C. Buley Library Main Circulating Stacks(QA76.73.C15T46 1996)
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Data structures using C and C++
Data structures using C and C++
Data structures using C and C++
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Tenenbaum, Aaron M. Data structures using C.
Tenenbaum, Aaron M. Data structures using C.
Tenenbaum, Aaron M. Data structures using C.
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Author/Creator
Langsam, Yedidyah, 1952-
Langsam, Yedidyah, 1952-
Langsam, Yedidyah, 1952-
Augenstein, Moshe, 1947-
Augenstein, Moshe, 1947-
Augenstein, Moshe, 1947-
Tenenbaum, Aaron M
Tenenbaum, Aaron M
Tenenbaum, Aaron M
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Upper Saddle River, N.J. : Prentice Hall
Upper Saddle River, N.J. : Prentice Hall
Upper Saddle River, N.J. : Prentice Hall
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c1996
c1996
c1996
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2nd ed.
2nd ed.
2nd ed.
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English
English
English
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xvi, 672 p. : ill. ; 24 cm.
xvi, 672 p. : ill. ; 24 cm.
xvi, 672 p. : ill. ; 24 cm.
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Bibliography
Includes bibliographical references (p. 647-661) and index.
Includes bibliographical references (p. 647-661) and index.
Includes bibliographical references (p. 647-661) and index.
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C (Computer program language)
C (Computer program language)
C (Computer program language)
C++ (Computer program language)
C++ (Computer program language)
C++ (Computer program language)
Data structures (Computer science)
Data structures (Computer science)
Data structures (Computer science)
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Notes
Rev. ed. of: Data structures using C / Aaron M. Tenenbaum, Yedidyah Langsam, Moshe J. Augenstein. c1990.
Rev. ed. of: Data structures using C / Aaron M. Tenenbaum, Yedidyah Langsam, Moshe J. Augenstein. c1990.
Rev. ed. of: Data structures using C / Aaron M. Tenenbaum, Yedidyah Langsam, Moshe J. Augenstein. c1990.
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3528 | https://aviation.stackexchange.com/questions/51588/how-to-realistically-model-propeller-static-thrust | Stack Exchange Network
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How to realistically model propeller static thrust?
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I am graphing static thrust vs prop diameter for a 600 kW motor and a prop diameter up to 5 m. I have seen the post 'What is the equation for calculating static thrust?' and using its formula and fixed prop efficiency I get a pretty linear graph. At 5 m the thrust is 19 kN.
Structural factors become increasingly important at larger diameters, but what other aerodynamic factors (perhaps prop efficiency changes with diameter) need to be considered to make the graph more realistic?
Update - I assume the prop pitch can adjust to maximize static thrust, but the twist is optimized for high speed (200mph) flight. The engines full power is also available at the best speed for the size of prop being tested. This Equation to bind velocity, thrust and power simplified by assuming unitary engine efficiency is the same as what I have been using.
$$T_0 = \sqrt[\LARGE{3\:}]{P^2\cdot\eta_{Prop}^2\cdot\pi\cdot \frac{d_P^2}{2}\cdot\rho}$$
aerodynamics
propeller
performance-calculation
thrust
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edited Jun 16, 2018 at 2:22
PilotheadPilothead
asked May 16, 2018 at 19:31
PilotheadPilothead
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This basically depends on the prop loading and RPM. If you are not satisfied with the momentum disk theory based results then the next step would be to use blade element theory based calculation.
If I were you I would certainly run few test cases with a known propeller geometry by scaling the propeller. One of the easiest way I can think of doing this is to use QPROP from Prof. Mark Drela at MIT. This is a free software and on top of this it is a command line program which is very well suited for this kind of an activity because all the parameters could be changed on the go, preferably via a bash or python script.
HTH
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answered Jun 15, 2018 at 8:21
ABCDABCD
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$\begingroup$ Am experimenting with Qprop and will respond when I have learned something. $\endgroup$
Pilothead
– Pilothead
2018-06-16 02:05:03 +00:00
Commented Jun 16, 2018 at 2:05
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$\begingroup$
A paper to estimate static thrust with a C-172 example is here. Also, a NASA report 447 on calculating static thrust may be of interest. An Excel spreadsheet to calculate static thrust is here. There is a SE discussion here and here.
The formula for thrust is...
Where: F = thrust (Newton) d = prop dia (inch) rpm = rotation (per minute) pitch = pitch (inch) Vo = aircraft speed (m/s)
If you want thrust in other units: to convert newtons to grams, multiply newtons by 1000/9.81. To then convert grams to ounces, multiply grams by 0.035274. To convert ounces to pounds, divide ounces by 16.
Note: the equation is hard-coded for a "standard day" density at sea level of 1.225kg/m^3.
The thrust of a propeller is not constant for different flight speeds. Reducing the inflow velocity generally increases the thrust. A reduction of the aircraft speed down to zero tends to increase the thrust even further, but often a rapid loss of thrust can be observed in this regime. That is why the static thrust of a propeller is not such a terribly important number for a propeller - the picture of a propeller, working under static conditions can be distorted and blurred.
As long as an aircraft does not move, its propeller operates under static conditions. There is no air moving towards the propeller due to the flight speed, the propeller creates its own inflow instead. A propeller, with its chord and twist distribution designed for the operating point under flight conditions, does not perform very well under static conditions. As opposed to a larger helicopter rotor, the flow around the relatively small propeller is heavily distorted and even may be partially separated. From the momentum theory of propellers we learn, that the efficiency at lower speeds is strongly dependent on the power loading (power per disk area), and this ratio for a propeller is much higher than that for a helicopter rotor. We are able to achieve about 80-90% of the thrust, as predicted by momentum theory for the design point, but we can reach only 50% or less of the predicted ideal thrust under static conditions.
Static thrust depends also on the inflow, influenced by the environment of the propeller (fuselage, crosswind, ground clearance). Measurements of static thrust can be easily done, but the theoretical treatment is very complicated and only possible with a lower degree of confidence than calculations in the vicinity of the design point. Due to local flow separation, the behavior of propellers under static conditions can be very sensitive with respect to blade angle settings and airfoil shape.
To get a picture of the bandwidth of static thrust, several older NACA reports and some publications from model magazines have been examined. The results are combined in the following graph.
Source
The Source provided a model airplane example calculation that is applicable to full size aircraft.
We have got two different propellers with a blade angle of 10° and 25° respectively. The first one has a diameter of D = 200 mm, the size of the second one is D = 300 mm. Which one would be better suited to build a VTOL aircraft model? How much thrust can we expect using an .60 engine of 2000 W (assuming a suitable gearbox)?
From the diagram above we read a static thrust parameter of 0.32 [kg^(1/3)/m], respectively 0.1 [kg^(1/3)/m] around the center of the blue band. To calculate the thrust we have to multiply these values with the power P [W] and the diameter D [m] to the power of 2/3. Performing the calculation for the first propeller (10° blade angle) yields T = 0.3254.288 [N] and thus a static thrust of 17.4 N, whereas the second, larger propeller delivers 0.171.138 = 7.1 N only. Using the same engine in a helicopter with its large rotor of 1 m diameter and low pitch angles, would give us a lifting force of more than 55 N !
This example shows, that the diameter of a propeller is as important for static thrust, as it is under flight conditions. But, for static thrust the blade angle is also very important - probably even more important than for the design point, where a gearbox can match almost any propeller pitch and flight speed quite well.
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edited Jun 15, 2018 at 14:39
answered Jun 15, 2018 at 13:37
jwzumwaltjwzumwalt
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1
$\begingroup$ Your first thrust equation uses rpm instead of power which makes it hard to apply to this problem. I have updated the question to include an adjustable pitch prop so outright stalling is not a factor. The "Source" article caption for your diagram states that it shows thrust parameters for props with 2-8 blades, but I could find no legend that tells which datapoints belong to which prop. Are you familiar enough with the material to know? $\endgroup$
Pilothead
– Pilothead
2018-06-16 02:52:27 +00:00
Commented Jun 16, 2018 at 2:52
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28 Is there any equation to bind velocity, thrust and power?
3 What is the equation for calculating static thrust?
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6 Is this a good or a bad propeller?
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2 Can someone please tell me all the units for the propeller thrust equations?
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3529 | https://www.boddlelearning.com/2nd-grade/strategies-for-adding-subtracting-within-100 | Teaching Resources
2nd Grade
Strategies for Adding & Subtracting within 100
Learning different strategies to add and subtract numbers within 100 is a second grade, Common Core math skill: 2.NBT.5. Below we show two videos that demonstrate this standard. Then, we provide a breakdown of the specific steps in the videos to help you teach your class.
Prior Learnings
Your students should be familiar with the first grade skill of counting up to 120 starting from any number below 120 (1.NBT.1), this skill helps them understand greater or less than values. The second grade skill is also closely linked to the first grade skill of understanding place values (ones and tens) in two-digit numbers (1.NBT.2).
Future Learnings
Comparing large numbers as greater than, less than, and equal to will help your students understand future concepts in third grade. In third grade, your students will learn how to interpret the products of whole numbers (i.e. 8 x 3 is the same as 8 groups of 3 objects each) (3.OA.1). They will also learn to use multiplication and division within 100 while solving word problems in situations that involve “equal groups, arrays, and measurement quantities” (3.OA.3).
Common Core Standard: 2.NBT.5 - Fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the relationship between addition and subtraction
Students who understand this principle can:
Successfully apply different strategies to add and subtract within 100.
Explain how addition and subtraction are related to solve problems.
Explain and understand the different strategies used to add and subtract.
Use place value to add and subtract.
Create concrete models or drawings, using strategies to add within 100 (strategies = place value, properties of operations, and how addition & subtraction are related).
Utilize number lines, base-ten blocks, or items to find unknown numbers.
2 Videos to Help You Teach Common Core Standard: 2.NBT.5
Below we provide and breakdown two videos to help you teach your students this standard.
Video 1: Using Rods and Cubes to Subtract
This video shows students how to subtract using drawings. Though it only covers subtraction problems, the method can be adapted for addition problems as well.
The first problem is 92 - 34 = ?. How the video solves this problem is related below.
Start by representing 92 with 10s rods and 1s cubes.
Since there are 9 tens, draw 9 rods.
Since there are 2 ones, draw 2 cubes.
Now, we will subtract (or take away) 34.
The video starts by removing 3 rods.
Now we must subtract 4, but removing another rod will remove 10.
There aren’t 4 ones to take away, only 2.
So take 1 ten rod and break into 10 cubes.
Then subtract 4 from the cubes.
Now count the remaining pieces to solve the problem.a. So, 92 - 34 = 58.
The second problem is 43 - 12 = ?. The video follows the same pattern as the first problem to solve the equation.
Represent 43 with 4 rods and 3 cubes.
Subtract 12 by removing 1 rod and 2 cubes.
Count the remaining pieces to find the answer.a. So, 43 - 12 = 31.
Video 2: Finding the Keys to Treasure Chests
This video covers some strategies for adding and subtracting up to 100. Boddle presents 5 treasure chests that need to be unlocked. However, they can only be opened with the correct key, and to select the correct key, one must solve the math problem.
Boddle asks students if they can help her unlock the chests.
Chest 1 has the problem 82 + 0.a. Remember, any number added to 0 is equal to itself.b. 82 + 0 = 82.c. So use the key that says 82.
Chest 2 has the problem 50 - 10.a. Since the values in the ones place are both 0’s, we can simply subtract those in the tens place.b. 5 - 1 = 4; add the 0 after for the ones place.c. 50 - 10 = 40; so use the key that says 40.
Chest 3 has the problem 41 + __ = 67.a. What do we need to add to 41 to make it 67.b. Subtract 41 from 67; position the numbers in a column to make it easier.c. 67 - 41 = 26d. 41 + 26 = 67
Chest 4 has the problem 62 - 8.a. Align the numbers in a column and subtract ones place first.b. Here we get 2 - 8, which we cannot do, so we borrow one from the 6.c. Now we have 12 - 8 = 4.d. Since we borrowed one from 6, it becomes a 5.e. Nothing is subtracted from the new 5, so the answer is 54.
Chest 5 has the problem 72 + 15.a. There is only one key left, so we are sure it’ll open the chest.b. But let’s see if it's true.c. Place the numbers in columns and add the place values.d. 2 + 5 = 7 and 7 + 1 = 8 e. So, 72 + 15 = 87.
Boddle congratulates the students on a job well done, hoping to see them next time.
Want more practice?
Give your students additional standards-aligned practice with Boddle Learning. Boddle includes questions related to Comparing and Measuring Lengths plus rewarding coins and games for your students to keep them engaged. Click here to sign up for Boddle Learning and create your first assignment today.
Information on standards is gathered from The New Mexico Public Education Department's New Mexico Instructional Scope for Mathematics and the Common Core website.
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3530 | https://onesearch.library.northeastern.edu/discovery/fulldisplay?docid=alma9952490149101401&context=L&vid=01NEU_INST:NU&lang=en&search_scope=MyInst_and_CI&adaptor=Local%20Search%20Engine&tab=Everything&query=sub%2Ccontains%2CInternal%20medicine%2CAND&mode=advanced&offset=30 | Harrison's principles of internal medicine - Northeastern University
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Harrison's principles of internal medicine
; Fauci, Anthony S., 1940- ed.; Hauser, Stephen L., ed.; Jameson, J. Larry., ed.; Kasper, Dennis L., ed.; Longo, Dan L. (Dan Louis), 1949- ed.; Loscalzo, Joseph., ed.;
New York, N.Y. : McGraw-Hill Education LLC.c2018
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Details
Title
Harrison's principles of internal medicine
Harrison's principles of internal medicine
Harrison's principles of internal medicine
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Alternative Title
Principles of internal medicine
Principles of internal medicine
Principles of internal medicine
Harrison's principles of internal medicine, twentieth edition
Harrison's principles of internal medicine, twentieth edition
Harrison's principles of internal medicine, twentieth edition
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Author
Fauci, Anthony S., 1940- ed.
Fauci, Anthony S., 1940- ed.
Fauci, Anthony S., 1940- ed.
Hauser, Stephen L., ed.
Hauser, Stephen L., ed.
Hauser, Stephen L., ed.
Jameson, J. Larry., ed.
Jameson, J. Larry., ed.
Jameson, J. Larry., ed.
Kasper, Dennis L., ed.
Kasper, Dennis L., ed.
Kasper, Dennis L., ed.
Longo, Dan L. (Dan Louis), 1949- ed.
Longo, Dan L. (Dan Louis), 1949- ed.
Longo, Dan L. (Dan Louis), 1949- ed.
Loscalzo, Joseph., ed.
Loscalzo, Joseph., ed.
Loscalzo, Joseph., ed.
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Edition
20th ed.
20th ed.
20th ed.
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Rights
Licensed for use by Northeastern University faculty, staff, and currently enrolled students.
Licensed for use by Northeastern University faculty, staff, and currently enrolled students.
Licensed for use by Northeastern University faculty, staff, and currently enrolled students.
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Subjects
Internal medicine
Internal medicine
Internal medicine
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Publisher
New York, N.Y. : McGraw-Hill Education LLC.
New York, N.Y. : McGraw-Hill Education LLC.
New York, N.Y. : McGraw-Hill Education LLC.
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Creation Date
c2018
c2018
c2018
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Bibliographical Notes
Includes bibliographical references and indexes.
Includes bibliographical references and indexes.
Includes bibliographical references and indexes.
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Contents
Part 1: The Profession of Medicine -- Chapter 1: The Practice of Medicine -- Part 2: Cardinal Manifestations and Presentation of Diseases -- Section 8: Alterations in the Skin -- Chapter 56: Cutaneous Drug Reactions -- Part 3: Pharmacology -- Chapter 63: Principles of Clinical Pharmacology -- Chapter 64: Pharmacogenomics -- Part 5: Infectious Diseases -- Section 1: Basic Considerations in Infectious Diseases -- Chapter 116: Molecular Mechanisms of Microbial Pathogenesis -- Chapter 117: Approach to the Acutely Ill Infected Febrile Patient -- Chapter 118: Immunization Principles and Vaccine Use -- Section 2: Clinical Syndromes: Community-Acquired Infections -- Chapter 121: Pneumonia -- Chapter 122: Lung Abscess -- Chapter 124: Infections of the Skin, Muscles, and Soft Tissues -- Chapter 125: Infectious Arthritis -- Chapter 126: Osteomyelitis -- Chapter 127: Intraabdominal Infections and Abscesses -- Chapter 129: Clostridium difficile Infection, Including Pseudomembranous Colitis -- Chapter 130: Urinary Tract Infections, Pyelonephritis, and Prostatitis -- Part 6: Disorders of the Cardiovascular System -- Section 1: Introduction to Cardiovascular Disorders -- Chapter 231: Approach to the Patient with Possible Cardiovascular Disease -- Chapter 232: Basic Biology of the Cardiovascular System -- Chapter 233: Epidemiology of Cardiovascular Disease -- Section 2: Diagnosis of Cardiovascular Disorders -- Chapter 234: Physical Examination of the Cardiovascular System -- Chapter 235: Electrocardiography -- Chapter 237: Diagnostic Cardiac Catheterization and Coronary Angiography -- Section 3: Disorders of Rhythm -- Chapter 238: Principles of Electrophysiology -- Chapter 239: The Bradyarrhythmias: Disorders of the Sinoatrial Node -- Chapter 240: The Bradyarrhythmias: Disorders of the Atrioventricular Node -- Chapter 241: Approach to Supraventricular Tachyarrhythmias -- Chapter 242: Physiologic and Non-Physiologic Sinus Tachycardia -- Chapter 243: Focal Atrial Tachycardia -- Section 4: Disorders of the Heart -- Chapter 263: Multiple and Mixed Valvular Heart Disease -- Chapter 264: Congenital Heart Disease in the Adult -- Part 7: Disorders of the Respiratory System -- Section 1: Diagnosis of Respiratory Disorders -- Chapter 278: Approach to the Patient with Disease of the Respiratory System.
Part 1: The Profession of Medicine -- Chapter 1: The Practice of Medicine -- Part 2: Cardinal Manifestations and Presentation of Diseases -- Section 8: Alterations in the Skin -- Chapter 56: Cutaneous Drug Reactions -- Part 3: Pharmacology -- Chapter 63: Principles of Clinical Pharmacology -- Chapter 64: Pharmacogenomics -- Part 5: Infectious Diseases -- Section 1: Basic Considerations in Infectious Diseases -- Chapter 116: Molecular Mechanisms of Microbial Pathogenesis -- Chapter 117: Approach to the Acutely Ill Infected Febrile Patient -- Chapter 118: Immunization Principles and Vaccine Use -- Section 2: Clinical Syndromes: Community-Acquired Infections -- Chapter 121: Pneumonia -- Chapter 122: Lung Abscess -- Chapter 124: Infections of the Skin, Muscles, and Soft Tissues -- Chapter 125: Infectious Arthritis -- Chapter 126: Osteomyelitis -- Chapter 127: Intraabdominal Infections and Abscesses -- Chapter 129: Clostridium difficile Infection, Including Pseudomembranous Colitis -- Chapter 130: Urinary Tract Infections, Pyelonephritis, and Prostatitis -- Part 6: Disorders of the Cardiovascular System -- Section 1: Introduction to Cardiovascular Disorders -- Chapter 231: Approach to the Patient with Possible Cardiovascular Disease -- Chapter 232: Basic Biology of the Cardiovascular System -- Chapter 233: Epidemiology of Cardiovascular Disease -- Section 2: Diagnosis of Cardiovascular Disorders -- Chapter 234: Physical Examination of the Cardiovascular System -- Chapter 235: Electrocardiography -- Chapter 237: Diagnostic Cardiac Catheterization and Coronary Angiography -- Section 3: Disorders of Rhythm -- Chapter 238: Principles of Electrophysiology -- Chapter 239: The Bradyarrhythmias: Disorders of the Sinoatrial Node -- Chapter 240: The Bradyarrhythmias: Disorders of the Atrioventricular Node -- Chapter 241: Approach to Supraventricular Tachyarrhythmias -- Chapter 242: Physiologic and Non-Physiologic Sinus Tachycardia -- Chapter 243: Focal Atrial Tachycardia -- Section 4: Disorders of the Heart -- Chapter 263: Multiple and Mixed Valvular Heart Disease -- Chapter 264: Congenital Heart Disease in the Adult -- Part 7: Disorders of the Respiratory System -- Section 1: Diagnosis of Respiratory Disorders -- Chapter 278: Approach to the Patient with Disease of the Respiratory System.
Part 1: The Profession of Medicine -- Chapter 1: The Practice of Medicine -- Part 2: Cardinal Manifestations and Presentation of Diseases -- Section 8: Alterations in the Skin -- Chapter 56: Cutaneous Drug Reactions -- Part 3: Pharmacology -- Chapter 63: Principles of Clinical Pharmacology -- Chapter 64: Pharmacogenomics -- Part 5: Infectious Diseases -- Section 1: Basic Considerations in Infectious Diseases -- Chapter 116: Molecular Mechanisms of Microbial Pathogenesis -- Chapter 117: Approach to the Acutely Ill Infected Febrile Patient -- Chapter 118: Immunization Principles and Vaccine Use -- Section 2: Clinical Syndromes: Community-Acquired Infections -- Chapter 121: Pneumonia -- Chapter 122: Lung Abscess -- Chapter 124: Infections of the Skin, Muscles, and Soft Tissues -- Chapter 125: Infectious Arthritis -- Chapter 126: Osteomyelitis -- Chapter 127: Intraabdominal Infections and Abscesses -- Chapter 129: Clostridium difficile Infection, Including Pseudomembranous Colitis -- Chapter 130: Urinary Tract Infections, Pyelonephritis, and Prostatitis -- Part 6: Disorders of the Cardiovascular System -- Section 1: Introduction to Cardiovascular Disorders -- Chapter 231: Approach to the Patient with Possible Cardiovascular Disease -- Chapter 232: Basic Biology of the Cardiovascular System -- Chapter 233: Epidemiology of Cardiovascular Disease -- Section 2: Diagnosis of Cardiovascular Disorders -- Chapter 234: Physical Examination of the Cardiovascular System -- Chapter 235: Electrocardiography -- Chapter 237: Diagnostic Cardiac Catheterization and Coronary Angiography -- Section 3: Disorders of Rhythm -- Chapter 238: Principles of Electrophysiology -- Chapter 239: The Bradyarrhythmias: Disorders of the Sinoatrial Node -- Chapter 240: The Bradyarrhythmias: Disorders of the Atrioventricular Node -- Chapter 241: Approach to Supraventricular Tachyarrhythmias -- Chapter 242: Physiologic and Non-Physiologic Sinus Tachycardia -- Chapter 243: Focal Atrial Tachycardia -- Section 4: Disorders of the Heart -- Chapter 263: Multiple and Mixed Valvular Heart Disease -- Chapter 264: Congenital Heart Disease in the Adult -- Part 7: Disorders of the Respiratory System -- Section 1: Diagnosis of Respiratory Disorders -- Chapter 278: Approach to the Patient with Disease of the Respiratory System.
Chapter 279: Disturbances of Respiratory Function -- Chapter 280: Diagnostic Procedures in Respiratory Disease -- Section 2: Diseases of the Respiratory System -- Chapter 281: Asthma -- Chapter 282: Hypersensitivity Pneumonitis and Pulmonary Infiltrates with Eosinophilia -- Chapter 291: Sleep Apnea -- Part 8: Critical Care Medicine -- Section 1: Respiratory Critical Care -- Chapter 293: Approach to the Patient with Critical Illness -- Part 9: Disorders of the Kidney and Urinary Tract -- Chapter 303: Cellular and Molecular Biology of the Kidney -- Chapter 304: Acute Kidney Injury -- Chapter 305: Chronic Kidney Disease -- Chapter 306: Dialysis in the Treatment of Renal Failure -- Part 10: Disorders of the Gastrointestinal System -- Section 3: Liver and Biliary Tract Disease -- Chapter 332: Acute Viral Hepatitis -- Chapter 334: Chronic Hepatitis -- Part 11: Immune-Mediated, Inflammatory, and Rheumatologic Disorders -- Section 1: The Immune System in Health and Disease -- Chapter 343: The Major Histocompatibility Complex -- Chapter 344: Primary Immune Deficiency Diseases -- Section 2: Disorders of Immune-Mediated Injury -- Chapter 345: Urticaria, Angioedema, and Allergic Rhinitis -- Chapter 346: Anaphylaxis -- Chapter 347: Mastocytosis -- Chapter 348: Autoimmunity and Autoimmune Diseases -- Chapter 353: Systemic Sclerosis (Scleroderma) and Related Disorders -- Part 12: Endocrinology and Metabolism -- Section 1: Endocrinology -- Chapter 369: Approach to the Patient with Endocrine Disorders -- Chapter 370: Mechanisms of Hormone Action -- Chapter 371: Physiology of Anterior Pituitary Hormones -- Chapter 372: Hypopituitarism -- Chapter 373: Pituitary Tumor Syndromes -- Chapter 374: Disorders of the Neurohypophysis -- Chapter 375: Thyroid Gland Physiology and Testing -- Chapter 376: Hypothyroidism -- Chapter 377: Hyperthyroidism -- Chapter 378: Thyroid Nodular Disease and Thyroid Cancer -- Part 14: Poisoning, Drug Overdose, and Envenomation -- Chapter 449: Heavy Metal Poisoning -- Chapter 450: Poisoning and Drug Overdose -- Part 23: Atlases -- Chapter A1: Atlas of Rashes Associated with Fever -- Chapter A2: Atlas of Oral Manifestations of Disease -- Chapter A3: Atlas of Urinary Sediments and Renal Biopsies -- Chapter A4: Atlas of Skin Manifestations of Internal Disease -- Chapter A5: Atlas of Hematology -- Chapter A7: Atlas of Electrocardiography -- Chapter A9: Atlas of Cardiac Arrhythmias -- Chapter A13: Atlas of Liver Biopsies -- Chapter A14: Atlas of the Vasculitic Syndromes.
Chapter 279: Disturbances of Respiratory Function -- Chapter 280: Diagnostic Procedures in Respiratory Disease -- Section 2: Diseases of the Respiratory System -- Chapter 281: Asthma -- Chapter 282: Hypersensitivity Pneumonitis and Pulmonary Infiltrates with Eosinophilia -- Chapter 291: Sleep Apnea -- Part 8: Critical Care Medicine -- Section 1: Respiratory Critical Care -- Chapter 293: Approach to the Patient with Critical Illness -- Part 9: Disorders of the Kidney and Urinary Tract -- Chapter 303: Cellular and Molecular Biology of the Kidney -- Chapter 304: Acute Kidney Injury -- Chapter 305: Chronic Kidney Disease -- Chapter 306: Dialysis in the Treatment of Renal Failure -- Part 10: Disorders of the Gastrointestinal System -- Section 3: Liver and Biliary Tract Disease -- Chapter 332: Acute Viral Hepatitis -- Chapter 334: Chronic Hepatitis -- Part 11: Immune-Mediated, Inflammatory, and Rheumatologic Disorders -- Section 1: The Immune System in Health and Disease -- Chapter 343: The Major Histocompatibility Complex -- Chapter 344: Primary Immune Deficiency Diseases -- Section 2: Disorders of Immune-Mediated Injury -- Chapter 345: Urticaria, Angioedema, and Allergic Rhinitis -- Chapter 346: Anaphylaxis -- Chapter 347: Mastocytosis -- Chapter 348: Autoimmunity and Autoimmune Diseases -- Chapter 353: Systemic Sclerosis (Scleroderma) and Related Disorders -- Part 12: Endocrinology and Metabolism -- Section 1: Endocrinology -- Chapter 369: Approach to the Patient with Endocrine Disorders -- Chapter 370: Mechanisms of Hormone Action -- Chapter 371: Physiology of Anterior Pituitary Hormones -- Chapter 372: Hypopituitarism -- Chapter 373: Pituitary Tumor Syndromes -- Chapter 374: Disorders of the Neurohypophysis -- Chapter 375: Thyroid Gland Physiology and Testing -- Chapter 376: Hypothyroidism -- Chapter 377: Hyperthyroidism -- Chapter 378: Thyroid Nodular Disease and Thyroid Cancer -- Part 14: Poisoning, Drug Overdose, and Envenomation -- Chapter 449: Heavy Metal Poisoning -- Chapter 450: Poisoning and Drug Overdose -- Part 23: Atlases -- Chapter A1: Atlas of Rashes Associated with Fever -- Chapter A2: Atlas of Oral Manifestations of Disease -- Chapter A3: Atlas of Urinary Sediments and Renal Biopsies -- Chapter A4: Atlas of Skin Manifestations of Internal Disease -- Chapter A5: Atlas of Hematology -- Chapter A7: Atlas of Electrocardiography -- Chapter A9: Atlas of Cardiac Arrhythmias -- Chapter A13: Atlas of Liver Biopsies -- Chapter A14: Atlas of the Vasculitic Syndromes.
Chapter 279: Disturbances of Respiratory Function -- Chapter 280: Diagnostic Procedures in Respiratory Disease -- Section 2: Diseases of the Respiratory System -- Chapter 281: Asthma -- Chapter 282: Hypersensitivity Pneumonitis and Pulmonary Infiltrates with Eosinophilia -- Chapter 291: Sleep Apnea -- Part 8: Critical Care Medicine -- Section 1: Respiratory Critical Care -- Chapter 293: Approach to the Patient with Critical Illness -- Part 9: Disorders of the Kidney and Urinary Tract -- Chapter 303: Cellular and Molecular Biology of the Kidney -- Chapter 304: Acute Kidney Injury -- Chapter 305: Chronic Kidney Disease -- Chapter 306: Dialysis in the Treatment of Renal Failure -- Part 10: Disorders of the Gastrointestinal System -- Section 3: Liver and Biliary Tract Disease -- Chapter 332: Acute Viral Hepatitis -- Chapter 334: Chronic Hepatitis -- Part 11: Immune-Mediated, Inflammatory, and Rheumatologic Disorders -- Section 1: The Immune System in Health and Disease -- Chapter 343: The Major Histocompatibility Complex -- Chapter 344: Primary Immune Deficiency Diseases -- Section 2: Disorders of Immune-Mediated Injury -- Chapter 345: Urticaria, Angioedema, and Allergic Rhinitis -- Chapter 346: Anaphylaxis -- Chapter 347: Mastocytosis -- Chapter 348: Autoimmunity and Autoimmune Diseases -- Chapter 353: Systemic Sclerosis (Scleroderma) and Related Disorders -- Part 12: Endocrinology and Metabolism -- Section 1: Endocrinology -- Chapter 369: Approach to the Patient with Endocrine Disorders -- Chapter 370: Mechanisms of Hormone Action -- Chapter 371: Physiology of Anterior Pituitary Hormones -- Chapter 372: Hypopituitarism -- Chapter 373: Pituitary Tumor Syndromes -- Chapter 374: Disorders of the Neurohypophysis -- Chapter 375: Thyroid Gland Physiology and Testing -- Chapter 376: Hypothyroidism -- Chapter 377: Hyperthyroidism -- Chapter 378: Thyroid Nodular Disease and Thyroid Cancer -- Part 14: Poisoning, Drug Overdose, and Envenomation -- Chapter 449: Heavy Metal Poisoning -- Chapter 450: Poisoning and Drug Overdose -- Part 23: Atlases -- Chapter A1: Atlas of Rashes Associated with Fever -- Chapter A2: Atlas of Oral Manifestations of Disease -- Chapter A3: Atlas of Urinary Sediments and Renal Biopsies -- Chapter A4: Atlas of Skin Manifestations of Internal Disease -- Chapter A5: Atlas of Hematology -- Chapter A7: Atlas of Electrocardiography -- Chapter A9: Atlas of Cardiac Arrhythmias -- Chapter A13: Atlas of Liver Biopsies -- Chapter A14: Atlas of the Vasculitic Syndromes.
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Available in other form: Online version: Harrison's principles of internal medicine. twentieth edition. New York, N.Y. : McGraw-Hill Education LLC., 2018
Available in other form: Online version: Harrison's principles of internal medicine. twentieth edition. New York, N.Y. : McGraw-Hill Education LLC., 2018
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3531 | https://discrete.openmathbooks.org/dmoi2/sec_planar.html | Skip to main content
Section4.2Planar Graphs
¶
Investigate!30
When a connected graph can be drawn without any edges crossing, it is called planar. When a planar graph is drawn in this way, it divides the plane into regions called faces.
Draw, if possible, two different planar graphs with the same number of vertices, edges, and faces.
Draw, if possible, two different planar graphs with the same number of vertices and edges, but a different number of faces.
When is it possible to draw a graph so that none of the edges cross? If this is possible, we say the graph is planar (since you can draw it on the plane).
Notice that the definition of planar includes the phrase “it is possible to.” This means that even if a graph does not look like it is planar, it still might be. Perhaps you can redraw it in a way in which no edges cross. For example, this is a planar graph:
That is because we can redraw it like this:
The graphs are the same, so if one is planar, the other must be too. However, the original drawing of the graph was not a planar representation of the graph.
When a planar graph is drawn without edges crossing, the edges and vertices of the graph divide the plane into regions. We will call each region a face. The graph above has 3 faces (yes, we do include the “outside” region as a face). The number of faces does not change no matter how you draw the graph (as long as you do so without the edges crossing), so it makes sense to ascribe the number of faces as a property of the planar graph.
WARNING: you can only count faces when the graph is drawn in a planar way. For example, consider these two representations of the same graph:
If you try to count faces using the graph on the left, you might say there are 5 faces (including the outside). But drawing the graph with a planar representation shows that in fact there are only 4 faces.
There is a connection between the number of vertices (v), the number of edges (e) and the number of faces (f) in any connected planar graph. This relationship is called Euler's formula.
Euler's Formula for Planar Graphs
For any (connected) planar graph with v vertices, e edges and f faces, we have
v−e+f=2
Why is Euler's formula true? One way to convince yourself of its validity is to draw a planar graph step by step. Start with the graph P2:
Any connected graph (besides just a single isolated vertex) must contain this subgraph. Now build up to your graph by adding edges and vertices. Each step will consist of either adding a new vertex connected by a new edge to part of your graph (so creating a new “spike”) or by connecting two vertices already in the graph with a new edge (completing a circuit).
What do these “moves” do? When adding the spike, the number of edges increases by 1, the number of vertices increases by one, and the number of faces remains the same. But this means that v−e+f does not change. Completing a circuit adds one edge, adds one face, and keeps the number of vertices the same. So again, v−e+f does not change.
Since we can build any graph using a combination of these two moves, and doing so never changes the quantity v−e+f, that quantity will be the same for all graphs. But notice that our starting graph P2 has v=2, e=1 and f=1, so v−e+f=2. This argument is essentially a proof by induction. A good exercise would be to rewrite it as a formal induction proof.
SubsectionNon-planar Graphs
¶
Investigate!31
For the complete graphs Kn, we would like to be able to say something about the number of vertices, edges, and (if the graph is planar) faces. Let's first consider K3:
How many vertices does K3 have? How many edges?
If K3 is planar, how many faces should it have?
Repeat parts (1) and (2) for K4, K5, and K23.
What about complete bipartite graphs? How many vertices, edges, and faces (if it were planar) does K7,4 have? For which values of m and n are Kn and Km,n planar?
Not all graphs are planar. If there are too many edges and too few vertices, then some of the edges will need to intersect. The first time this happens is in K5.
If you try to redraw this without edges crossing, you quickly get into trouble. There seems to be one edge too many. In fact, we can prove that no matter how you draw it, K5 will always have edges crossing.
Theorem4.2.1
K5 is not planar.
Proof
The proof is by contradiction. So assume that K5 is planar. Then the graph must satisfy Euler's formula for planar graphs. K5 has 5 vertices and 10 edges, so we get
5−10+f=2
which says that if the graph is drawn without any edges crossing, there would be f=7 faces.
Now consider how many edges surround each face. Each face must be surrounded by at least 3 edges. Let B be the total number of boundaries around all the faces in the graph. Thus we have that B≥3f. But also B=2e, since each edge is used as a boundary exactly twice. Putting this together we get
3f≤2e
But this is impossible, since we have already determined that f=7 and e=10, and 21≰20. This is a contradiction so in fact K5 is not planar.
The other simplest graph which is not planar is K3,3
Proving that K3,3 is not planar answers the houses and utilities puzzle: it is not possible to connect each of three houses to each of three utilities without the lines crossing.
Theorem4.2.2
K3,3 is not planar.
Proof
Again, we proceed by contradiction. Suppose K3,3 were planar. Then by Euler's formula there will be 5 faces, since v=6, e=9, and 6−9+f=2.
How many boundaries surround these 5 faces? Let B be this number. Since each edge is used as a boundary twice, we have B=2e. Also, B≥4f since each face is surrounded by 4 or more boundaries. We know this is true because K3,3 is bipartite, so does not contain any 3-edge cycles. Thus
4f≤2e.
But this would say that 20≤18, which is clearly false. Thus K3,3 is not planar.
Note the similarities and differences in these proofs. Both are proofs by contradiction, and both start with using Euler's formula to derive the (supposed) number of faces in the graph. Then we find a relationship between the number of faces and the number of edges based on how many edges surround each face. This is the only difference. In the proof for K5, we got 3f≤2e and for K3,3 we go 4f≤2e. The coefficient of f is the key. It is the smallest number of edges which could surround any face. If some number of edges surround a face, then these edges form a cycle. So that number is the size of the smallest cycle in the graph.
In general, if we let g be the size of the smallest cycle in a graph (g stands for girth, which is the technical term for this) then for any planar graph we have gf≤2e. When this disagrees with Euler's formula, we know for sure that the graph cannot be planar.
SubsectionPolyhedra
¶
Investigate!32
A cube is an example of a convex polyhedron. It contains 6 identical squares for its faces, 8 vertices, and 12 edges. The cube is a regular polyhedron (also known as a Platonic solid) because each face is an identical regular polygon and each vertex joins an equal number of faces.
There are exactly four other regular polyhedra: the tetrahedron, octahedron, dodecahedron, and icosahedron with 4, 8, 12 and 20 faces respectively. How many vertices and edges do each of these have?
Another area of mathematics where you might have heard the terms “vertex,” “edge,” and “face” is geometry. A polyhedron is a geometric solid made up of flat polygonal faces joined at edges and vertices. We are especially interested in convex polyhedra, which means that any line segment connecting two points on the interior of the polyhedron must be entirely contained inside the polyhedron.2An alternative definition for convex is that the internal angle formed by any two faces must be less than (180\deg\text{.})
Notice that since 8−12+6=2, the vertices, edges and faces of a cube satisfy Euler's formula for planar graphs. This is not a coincidence. We can represent a cube as a planar graph by projecting the vertices and edges onto the plane. One such projection looks like this:
In fact, every convex polyhedron can be projected onto the plane without edges crossing. Think of placing the polyhedron inside a sphere, with a light at the center of the sphere. The edges and vertices of the polyhedron cast a shadow onto the interior of the sphere. You can then cut a hole in the sphere in the middle of one of the projected faces and “stretch” the sphere to lay down flat on the plane. The face that was punctured becomes the “outside” face of the planar graph.
The point is, we can apply what we know about graphs (in particular planar graphs) to convex polyhedra. Since every convex polyhedron can be represented as a planar graph, we see that Euler's formula for planar graphs holds for all convex polyhedra as well. We also can apply the same sort of reasoning we use for graphs in other contexts to convex polyhedra. For example, we know that there is no convex polyhedron with 11 vertices all of degree 3, as this would make 33/2 edges.
Example4.2.3
Is there a convex polyhedron consisting of three triangles and six pentagons? What about three triangles, six pentagons and five heptagons (7-sided polygons)?
Solution
How many edges would such polyhedra have? For the first proposed polyhedron, the triangles would contribute a total of 9 edges, and the pentagons would contribute 30. However, this counts each edge twice (as each edge borders exactly two faces), giving 39/2 edges, an impossibility. There is no such polyhedron.
The second polyhedron does not have this obstacle. The extra 35 edges contributed by the heptagons give a total of 74/2 = 37 edges. So far so good. Now how many vertices does this supposed polyhedron have? We can use Euler's formula. There are 14 faces, so we have (v - 37 + 14 = 2) or equivalently (v = 25\text{.}) But now use the vertices to count the edges again. Each vertex must have degree at least three (that is, each vertex joins at least three faces since the interior angle of all the polygons must be less that (180^\circ)), so the sum of the degrees of vertices is at least 75. Since the sum of the degrees must be exactly twice the number of edges, this says that there are strictly more than 37 edges. Again, there is no such polyhedron.
To conclude this application of planar graphs, consider the regular polyhedra. Above we claimed there are only five. How do we know this is true? We can prove it using graph theory.
Theorem4.2.4
There are exactly five regular polyhedra.
Proof
Recall that a regular polyhedron has all of its faces identical regular polygons, and that each vertex has the same degree. Consider the cases, broken up by what the regular polygon might be.
Case 1: Each face is a triangle. Let f be the number of faces. There are then 3f/2 edges. Using Euler's formula we have v−3f/2+f=2 so v=2+f/2. Now each vertex has the same degree, say k. So the number of edges is also kv/2. Putting this together gives
e=3f2=k(2+f/2)2
which says
k=6f4+f
We need k and f to both be positive integers. Note that 6f4+f is an increasing function for positive f, and has a horizontal asymptote at 6. Thus the only possible values for k are 3, 4, and 5. Each of these are possible. To get k=3, we need f=4 (this is the tetrahedron). For k=4 we take f=8 (the octahedron). For k=5 take f=20 (the icosahedron). Thus there are exactly three regular polyhedra with triangles for faces.
Case 2: Each face is a square. Now we have e=4f/2=2f. Using Euler's formula we get v=2+f, and counting edges using the degree k of each vertex gives us
e=2f=k(2+f)2
Solving for k gives
k=4f2+f=8f4+2f
This is again an increasing function, but this time the horizontal asymptote is at k=4, so the only possible value that k could take is 3. This produces 6 faces, and we have a cube. There is only one regular polyhedron with square faces.
Case 3: Each face is a pentagon. We perform the same calculation as above, this time getting e=5f/2 so v=2+3f/2. Then
e=5f2=k(2+3f/2)2
so
k=10f4+3f
Now the horizontal asymptote is at 103. This is less than 4, so we can only hope of making k=3. We can do so by using 12 pentagons, getting the dodecahedron. This is the only regular polyhedron with pentagons as faces.
Case 4: Each face is an n-gon with n≥6. Following the same procedure as above, we deduce that
k=2nf4+(n−2)f
which will be increasing to a horizontal asymptote of 2nn−2. When n=6, this asymptote is at k=3. Any larger value of n will give an even smaller asymptote. Therefore no regular polyhedra exist with faces larger than pentagons.3Notice that you can tile the plane with hexagons. This is an infinite planar graph; each vertex has degree 3. These infinitely many hexagons correspond to the limit as (f \to \infty) to make (k = 3\text{.})
SubsectionExercises
¶
1
Is it possible for a planar graph to have 6 vertices, 10 edges and 5 faces? Explain.
Solution
No. A (connected) planar graph must satisfy Euler's formula: (v - e + f = 2\text{.}) Here (v - e + f = 6 - 10 + 5 = 1\text{.})
2
The graph G has 6 vertices with degrees 2,2,3,4,4,5. How many edges does G have? Could G be planar? If so, how many faces would it have. If not, explain.
Solution
(G) has 10 edges, since (10 = \frac{2+2+3+4+4+5}{2}\text{.}) It could be planar, and then it would have 6 faces, using Euler's formula: (6-10+f = 2) means (f = 6\text{.}) To make sure that it is actually planar though, we would need to draw a graph with those vertex degrees without edges crossing. This can be done by trial and error (and is possible).
3
I'm thinking of a polyhedron containing 12 faces. Seven are triangles and four are quadralaterals. The polyhedron has 11 vertices including those around the mystery face. How many sides does the last face have?
Solution
Say the last polyhedron has (n) edges, and also (n) vertices. The total number of edges the polyhedron has then is ((7 \cdot 3 + 4 \cdot 4 + n)/2 = (37 + n)/2\text{.}) In particular, we know the last face must have an odd number of edges. We also have that (v = 11 \text{.}) By Euler's formula, we have (11 - (37+n)/2 + 12 = 2\text{,}) and solving for (n) we get (n = 5\text{,}) so the last face is a pentagon.
4
Consider some classic polyhedrons.
An octahedron is a regular polyhedron made up of 8 equilateral triangles (it sort of looks like two pyramids with their bases glued together). Draw a planar graph representation of an octahedron. How many vertices, edges and faces does an octahedron (and your graph) have?
The traditional design of a soccer ball is in fact a (spherical projection of a) truncated icosahedron. This consists of 12 regular pentagons and 20 regular hexagons. No two pentagons are adjacent (so the edges of each pentagon are shared only by hexagons). How many vertices, edges, and faces does a truncated icosahedron have? Explain how you arrived at your answers. Bonus: draw the planar graph representation of the truncated icosahedron.
Your “friend” claims that he has constructed a convex polyhedron out of 2 triangles, 2 squares, 6 pentagons and 5 octagons. Prove that your friend is lying. Hint: each vertex of a convex polyhedron must border at least three faces.
5
Prove Euler's formula using induction on the number of edges in the graph.
Solution
Proof
Let (P(n)) be the statement, “every planar graph containing (n) edges satisfies (v - n + f = 2\text{.})” We will show (P(n)) is true for all (n \ge 0\text{.}) Base case: there is only one graph with zero edges, namely a single isolated vertex. In this case (v = 1\text{,}) (f = 1) and (e = 0\text{,}) so Euler's formula holds. Inductive case: Suppose (P(k)) is true for some arbitrary (k \ge 0\text{.}) Now consider an arbitrary graph containing (k+1) edges (and (v) vertices and (f) faces). No matter what this graph looks like, we can remove a single edge to get a graph with (k) edges which we can apply the inductive hypothesis to. There are two possibilities. First, the edge we remove might be incident to a degree 1 vertex. In this case, also remove that vertex. The smaller graph will now satisfy (v-1 - k + f = 2) by the induction hypothesis (removing the edge and vertex did not reduce the number of faces). Adding the edge and vertex back gives (v - (k+1) + f = 2\text{,}) as required. The second case is that the edge we remove is incident to vertices of degree greater than one. In this case, removing the edge will keep the number of vertices the same but reduce the number of faces by one. So by the inductive hypothesis we will have (v - k + f-1 = 2\text{.}) Adding the edge back will give (v - (k+1) + f = 2) as needed. Therefore, by the principle of mathematical induction, Euler's formula holds for all planar graphs.
6
Prove Euler's formula using induction on the number of vertices in the graph.
7
Euler's formula (v−e+f=2) holds for all connected planar graphs. What if a graph is not connected? Suppose a planar graph has two components. What is the value of v−e+f now? What if it has k components?
8
Prove that the Petersen graph (below) is not planar.
Hint
What is the length of the shortest cycle? (This quantity is usually called the girth of the graph.)
9
Prove that any planar graph with v vertices and e edges satisfies e≤3v−6.
Solution
Proof
We know in any planar graph the number of faces (f) satisfies (3f \le 2e) since each face is bounded by at least three edges, but each edge borders two faces. Combine this with Euler's formula:
\begin{equation}
v - e + f = 2
\end{equation}
\begin{equation}
v - e + \frac{2e}{3} \ge 2
\end{equation}
\begin{equation}
3v - e \ge 6
\end{equation}
\begin{equation}
3v - 6 \ge e.
\end{equation}
10
Prove that any planar graph must have a vertex of degree 5 or less.
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Prove trigonometric identity cos(2x) with rotation matrix
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How do one prove the following trigonometric identity with the standard rotation matrix T θ T θ
c o s(2 θ)=c o s 2(θ)−s i n 2(θ)c o s(2 θ)=c o s 2(θ)−s i n 2(θ)?
The hint given is to compare T 2 θ T 2 θ and T θ◦T θ T θ◦T θ. We have not learned about complex numbers yet.
I have come to that:
T θ◦T θ T θ◦T θ= [c o s 2 θ−s i n 2 θ 2 s i n θ c o s θ−2 c o s θ s i n θ c o s 2 θ−s i n 2 θ][c o s 2 θ−s i n 2 θ−2 c o s θ s i n θ 2 s i n θ c o s θ c o s 2 θ−s i n 2 θ]
T 2 θ T 2 θ= [c o s 2 θ s i n 2 θ−s i n 2 θ c o s 2 θ][c o s 2 θ−s i n 2 θ s i n 2 θ c o s 2 θ]
Is it enough to show this and say that one element in the matrix equals the other element in the other matrix?
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edited Oct 9, 2018 at 10:25
SamSam
asked Oct 8, 2018 at 22:00
SamSam
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1 You have incorrectly computed T θ∘T θ T θ∘T θ Ben Grossmann –Ben Grossmann 2018-10-08 22:13:25 +00:00 Commented Oct 8, 2018 at 22:13
Why do you think you might need complex number for this?amd –amd 2018-10-08 23:21:03 +00:00 Commented Oct 8, 2018 at 23:21
@Omnomnomnom Sorry, fixed it now.Sam –Sam 2018-10-09 10:30:25 +00:00 Commented Oct 9, 2018 at 10:30
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Your mistake is in computing T θ∘T θ T θ∘T θ. Specifically, the top left and bottom-right entries are NOT 0.
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answered Oct 8, 2018 at 22:13
Alex R.Alex R.
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Oh, we have not learned much about that notation either. Would you please show me how you compute it?Sam –Sam 2018-10-08 22:18:27 +00:00 Commented Oct 8, 2018 at 22:18
@Sam: It's just matrix multiplication. You're multiplying T θ T θ by itself. Are you familiar with matrix multiplication?Alex R. –Alex R. 2018-10-08 23:04:09 +00:00 Commented Oct 8, 2018 at 23:04
Sorry, my bad. I corrected it now. Is it enough to say one element of Tθ◦Tθ is equal to T2θ?Sam –Sam 2018-10-09 10:29:48 +00:00 Commented Oct 9, 2018 at 10:29
@Sam: Yes, as with a 2x2 matrix, all four components must satisfy their respective equation Alex R. –Alex R. 2018-10-11 21:36:42 +00:00 Commented Oct 11, 2018 at 21:36
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Generalized Shared Birthday
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Suppose a year has d d days. How many people should be in a room so that there are at least 2 k 2 k people in the room with birthdays shared with each other (all could be same day or there could be k k distinct birthdays (there is a wide range) but I want in total 2 k 2 k people) with probability ≥1 2≥1 2?
I am looking for asymptotics.
I just want to know how the probability scales.
For traditional birthday problem the odds scale as 1−exp(−c n 2)1−exp(−c n 2) where n n is number of people.
What if I ask 2 k 2 k people who have birthdays shared so that any shared birthday occurs with at least t t people?
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edited Dec 31, 2016 at 8:49
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We have (n 2)(n 2) pairs of people, for each pair the probability to share the birthday being d−1 d−1. Hence, denoting by K K the number of birthday-sharing pairs, the expectation of K K is d−1(n 2)d−1(n 2), and by Markov's inequality, the probability to have K≥k K≥k is less than n 2/(2 k d)n 2/(2 k d). Thus, in order to have at least k k pairs with probability 0.5 0.5, one needs n>k d−−√n>k d.
On the other hand, denoting for each j∈[1,d]j∈[1,d] by n j n j the number of people having their birthday at the j j th day of the year, by Jensen's inequality we have
K=(n 1 2)+⋯+(n d 2)≥d(n/d 2);K=(n 1 2)+⋯+(n d 2)≥d(n/d 2);
therefore, if n≥2 k d−−−√+d n≥2 k d+d, then K≥n(n/d−1)/2>k K≥n(n/d−1)/2>k, meaning that in this case we have at least k k pairs of people with the same birthday with probability 1 1.
To summarize,
The smallest number of people needed to ensure at least k k birthday-sharing pairs with probability 0.5 0.5 satisfies
k d−−√<n<2 k d−−−√+d.k d<n<2 k d+d.
In the regime where d=o(k 1/3)d=o(k 1/3), a precise asymptotic can be given showing that the number of people needed is
n=(1+o(1))2 k d−−−√.n=(1+o(1))2 k d.
To see this, observe that each n j n j (introduced above) is distributed as B(n,d−1)B(n,d−1). Hence, by the Hoeffding's inequality, for any ϵ>0 ϵ>0, we have n(d−1−ϵ)<n j<n(d−1+ϵ)n(d−1−ϵ)<n j<n(d−1+ϵ) with probability a least 1−2 e−2 ϵ 2 n 1−2 e−2 ϵ 2 n. By the union bound, with probability at least 1−2 d e−2 ϵ 2 n 1−2 d e−2 ϵ 2 n, we then have n(d−1−ϵ)<n j<n(d−1+ϵ)n(d−1−ϵ)<n j<n(d−1+ϵ) for each j∈[1,d]j∈[1,d], resulting in d(n(d−1−ϵ)2)≤K≤d(n(d−1+ϵ)2)d(n(d−1−ϵ)2)≤K≤d(n(d−1+ϵ)2). The assertion now follows by letting, say, ϵ:=d−1/2 k−1/6 ϵ:=d−1/2 k−1/6 and observing that for n=(1+o(1))2 k d−−−√n=(1+o(1))2 k d we have then e−ϵ 2 n=o(1)e−ϵ 2 n=o(1) and ϵ=o(d−1)ϵ=o(d−1).
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edited Jan 2, 2017 at 19:12
answered Dec 31, 2016 at 8:47
SevaSeva
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may I ask a fine modification as well. What if I ask 2 k 2 k people who have birthdays shared so that any shared birthday occurs with at least t t people?user94040 –user94040 2016-12-31 08:49:18 +00:00 Commented Dec 31, 2016 at 8:49
@AJ: the same approach may work as well: check the contribution of January 1, say. (Happy New Year!)Seva –Seva 2016-12-31 09:15:21 +00:00 Commented Dec 31, 2016 at 9:15
Happy New year.user94040 –user94040 2016-12-31 09:16:07 +00:00 Commented Dec 31, 2016 at 9:16
Similar considerations show that for t>2 t>2 one needs roughly n≈(t!⋅k⋅d t−1)1/t n≈(t!⋅k⋅d t−1)1/t to get the probability of k k t t-fold birthdays close to 1/2 1/2 esg –esg 2017-01-02 19:30:46 +00:00 Commented Jan 2, 2017 at 19:30
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3535 | https://zhu.physics.ucdavis.edu/Physics9C-C_2021/Spherical%20Coordinates.pdf | Algebra Applied Mathematics Calculus and Analysis Discrete Mathematics Foundations of Mathematics Geometry History and Terminology Number Theory Probability and Statistics Recreational Mathematics Topology Alphabetical Index Interactive Entries Random Entry New in MathWorld MathWorld Classroom About MathWorld Contribute to MathWorld Send a Message to the Team MathWorld Book 12,976 entries Last updated: Sun Jan 17 2010 Geometry > Coordinate Geometry > Interactive Entries > Interactive Demonstrations > Spherical Coordinates Spherical coordinates, also called spherical polar coordinates (Walton 1967, Arfken 1985), are a system of curvilinear coordinates that are natural for describing positions on a sphere or spheroid. Define to be the azimuthal angle in the -plane from the x-axis with (denoted when referred to as the longitude), to be the polar angle (also known as the zenith angle and colatitude, with where is the latitude) from the positive z-axis with , and to be distance (radius) from a point to the origin. This is the convention commonly used in mathematics.
In this work, following the mathematics convention, the symbols for the radial, azimuth, and zenith angle coordinates are taken as , , and , respectively. Note that this definition provides a logical extension of the usual polar coordinates notation, with remaining the angle in the -plane and becoming the angle out of that plane. The sole exception to this convention in this work is in spherical harmonics, where the convention used in the physics literature is retained (resulting, it is hoped, in a bit less confusion than a foolish rigorous consistency might engender).
Unfortunately, the convention in which the symbols and are reversed is also frequently used, especially in physics. The symbol is sometimes also used in place of , and and instead of . The following table summarizes a number of conventions used by various authors; be very careful when consulting the literature.
(radial, azimuthal, polar) reference this work, Zwillinger (1985, pp. 297-298) Beyer (1987, p. 212) Korn and Korn (1968, p. 60) Misner et al. (1973, p. 205) (Rr, Pphi, Ttheta) SetCoordinates[Spherical[r, Ttheta, Pphi]] in the Mathematica package VectorAnalysis`) Arfken (1985, p. 102) Moon and Spencer (1988, p. 24) The spherical coordinates are related to the Cartesian coordinates by (1) (2) (3) where , , and , and the inverse tangent must be suitably defined to take the correct quadrant of into account.
In terms of Cartesian coordinates, (4) (5) (6) The scale factors are (7) (8) (9) so the metric coefficients are (10) (11) (12) The line element is (13) the area element Sp SEARCH MATHWORLD Oth Spherical Coordinates -- from Wolfram MathWorld 1 of 6 1/18/2010 12:05 PM (14) and the volume element (15) The Jacobian is (16) The position vector is (17) so the unit vectors are (18) (19) (20) (21) (22) (23) Derivatives of the unit vectors are (24) (25) (26) (27) (28) (29) (30) (31) (32) The gradient is (33) and its components are (34) (35) (36) (37) (38) (39) (40) (41) (42) (Misner et al. 1973, p. 213, who however use the notation convention ).
Spherical Coordinates -- from Wolfram MathWorld 2 of 6 1/18/2010 12:05 PM The Christoffel symbols of the second kind in the definition of Misner et al. (1973, p. 209) are given by (43) (44) (45) (Misner et al. 1973, p. 213, who however use the notation convention ). The Christoffel symbols of the second kind in the definition of Arfken (1985) are given by (46) (47) (48) (Walton 1967; Moon and Spencer 1988, p. 25a; both of whom however use the notation convention ).
The divergence is (49) or, in vector notation, (50) (51) The covariant derivatives are given by (52) so (53) (54) (55) (56) (57) (58) (59) (60) (61) The commutation coefficients are given by (62) (63) so , where .
Spherical Coordinates -- from Wolfram MathWorld 3 of 6 1/18/2010 12:05 PM (64) so , .
(65) so .
(66) so (67) Summarizing, (68) (69) (70) Time derivatives of the position vector are (71) (72) (73) The speed is therefore given by (74) The acceleration is (75) (76) (77) (78) (79) (80) Plugging these in gives (81) but (82) (83) so (84) (85) Spherical Coordinates -- from Wolfram MathWorld 4 of 6 1/18/2010 12:05 PM Time derivatives of the unit vectors are (86) (87) (88) The curl is (89) The Laplacian is (90) (91) (92) The vector Laplacian in spherical coordinates is given by (93) To express partial derivatives with respect to Cartesian axes in terms of partial derivatives of the spherical coordinates, (94) (95) (96) Upon inversion, the result is (97) The Cartesian partial derivatives in spherical coordinates are therefore (98) (99) (100) (Gasiorowicz 1974, pp. 167-168; Arfken 1985, p. 108).
The Helmholtz differential equation is separable in spherical coordinates.
SEE ALSO: Azimuth, Colatitude, Great Circle, Helmholtz Differential Equation--Spherical Coordinates, Latitude, Longitude, Oblate Spheroidal Coordinates, Polar Angle, Prolate Spheroidal Coordinates, Zenith Angle REFERENCES: Arfken, G. "Spherical Polar Coordinates." §2.5 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 102-111, 1985.
Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, 1987.
Gasiorowicz, S. Quantum Physics. New York: Wiley, 1974.
Korn, G. A. and Korn, T. M. Mathematical Handbook for Scientists and Engineers. New York: McGraw-Hill, 1968.
Misner, C. W.; Thorne, K. S.; and Wheeler, J. A. Gravitation. San Francisco, CA: W. H. Freeman, 1973.
Moon, P. and Spencer, D. E. "Spherical Coordinates ." Table 1.05 in Field Theory Handbook, Including Coordinate Systems, Differential Equations, and Their Solutions, 2nd ed. New York: Springer-Verlag, pp. 24-27, 1988.
Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, p. 658, 1953.
Walton, J. J. "Tensor Calculations on Computer: Appendix." Comm. ACM 10, 183-186, 1967.
Zwillinger, D. (Ed.). "Spherical Coordinates in Space." §4.9.3 in CRC Standard Mathematical Tables and Formulae. Boca Raton, FL: CRC Press, pp. 297-298, 1995.
CITE THIS AS: Weisstein, Eric W. "Spherical Coordinates." From MathWorld--A Wolfram Web Resource. Spherical Coordinates -- from Wolfram MathWorld 5 of 6 1/18/2010 12:05 PM Contact the MathWorld Team © 1999-2010 Wolfram Research, Inc. | Terms of Use Spherical Coordinates -- from Wolfram MathWorld 6 of 6 1/18/2010 12:05 PM |
3536 | https://www.thelancet.com/journals/laninf/article/PIIS1473-3099(06)70412-1/abstract | The role of BCG in prevention of leprosy: a meta-analysis - The Lancet Infectious Diseases
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ReviewVolume 6, Issue 3p162-170 March 2006
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The role of BCG in prevention of leprosy: a meta-analysis
Maninder Singh Setia
Maninder Singh Setia
Affiliations
Division of Epidemiology, School of Public Health, University of California, Berkeley, CA, USA
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a ∙ Craig Steinmaus
Craig Steinmaus
Affiliations
Division of Occupational and Environmental Medicine, School of Public Health, University of California, Berkeley, CA, USA
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b ∙ Christine S Ho
Christine S Ho
Affiliations
Division of Epidemiology, School of Public Health, University of California, Berkeley, CA, USA
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a ∙ Dr George W Rutherford
Dr George W Rutherford
Correspondence
Correspondence to: Dr George W Rutherford, Institute for Global Health, Department of Epidemiology and Biostatistics, University of California, San Francisco, 50 Beale Street, Suite 1200, San Francisco, CA 94105, USA. Tel +1 415 597 9108; fax +1 415 597 8299
GRutherford@psg.ucsf.edu
Affiliations
Institute for Global Health, Department of Epidemiology and Biostatistics, University of California, San Francisco, San Francisco, CA, USA
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cGRutherford@psg.ucsf.edu
Affiliations & Notes Article Info
a Division of Epidemiology, School of Public Health, University of California, Berkeley, CA, USA
b Division of Occupational and Environmental Medicine, School of Public Health, University of California, Berkeley, CA, USA
c Institute for Global Health, Department of Epidemiology and Biostatistics, University of California, San Francisco, San Francisco, CA, USA
Publication History:
Published March 2006
DOI: 10.1016/S1473-3099(06)70412-1 External LinkAlso available on ScienceDirect External Link
Copyright: © 2006 Elsevier Ltd. Published by Elsevier Ltd.
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The present meta-analysis investigates the role of BCG—a widely used yet controversial vaccine—in the prevention of leprosy. The electronic databases Medline, Embase, the Cochrane Library, and LILACS were searched to identify studies assessing the protective effect of BCG against leprosy. We included seven experimental studies and 19 observational studies. The experimental studies demonstrated an overall protective effect of 26% (95% CI 14–37%). At 61% (95% CI 51–70%), the observational studies overestimated the protective effect. The age at vaccination did not predict the protective effect of BCG. An additional dose of BCG was more protective in the prevention of leprosy compared with a single dose. An additional dose of BCG may be warranted for contacts of leprosy patients in areas where leprosy continues to be a public-health problem.
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Figure 2 Forest plots of the studies included in meta-analysis
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(A) Forest plot of the experimental studies. (B) Forest plot of the observational studies. Horizontal lines=95% CI. The rectangles represent the point estimates of the study and the size of the rectangle represents the weight given to each study in the meta-analysis. The diamond and vertical broken line represent the summary estimate; the size of the diamond represents the CIs of the summary estimate. The solid vertical line is the null value.
Figure 3 Funnel plots to assess publication bias in the meta-analysis
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(A) Funnel plot for experimental studies. (B) Funnel plot for observational studies.
RR=relative risk.
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3537 | https://my.clevelandclinic.org/health/body/21872-larynx | Abu Dhabi|Canada|Florida|London|Nevada|Ohio|
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Larynx (Voice Box)
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Larynx (Voice Box)
Your larynx is a hollow tube in the middle of your neck, just above your trachea (windpipe) and esophagus. It makes it possible for you to make sounds, which is why it’s also called your voice box. It also lets air pass from your throat to your trachea and on to your lungs.
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ContentsOverviewFunctionAnatomyConditions and DisordersCareAdditional Common Questions
Overview
What is the larynx (voice box)?
Your larynx is part of your respiratory system. It’s a hollow tube that’s about 4 to 5 centimeters (cm) in length and width. It lets air pass from your throat (pharynx) to your trachea on the way to your lungs. Your larynx is also the reason you’re able to make sounds, so it’s often called your voice box.
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Function
What is the function of the larynx?
Your larynx helps you to:
Breathe. When you take in air through your nose and mouth, your larynx funnels it down to your trachea and to your lungs.
Talk, shout and make other vocal sounds. Your larynx contains your vocal cords, which create sound.
Anatomy
Where is my larynx located?
Your larynx is in the middle of your neck, just above your trachea (windpipe) and your esophagus. But its exact location changes throughout your life.
From birth up until age 2, your larynx was higher in your neck. Over time, your larynx moves down to the middle of your neck.
What are the parts of my larynx?
Your larynx is divided into three parts:
The upper part (supraglottis).
The middle part (glottis).
The lower part (subglottis).
Your supraglottis, glottis and subglottis have different types of cartilage, muscle, ligaments and membranes, but only your glottis contains your vocal folds (vocal cords):
Cartilages in your larynx help to give it structure, the same way interior walls frame a house.
Muscles in your larynx move your larynx while swallowing, help with breathing and produce vocal sounds.
Ligaments in your larynx connect cartilages and link your larynx to nearby structures, like your hyoid bone and trachea.
Membranes help to hold cartilage in place.
Conditions and Disorders
What are conditions and disorders that affect my larynx?
Many things can affect your larynx, from diseases like cancer to simply using your voice too much. Conditions that affect your larynx include:
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Acute laryngitis: You may develop acute laryngitis if you have an infection or if you strain your vocal cords by overusing your voice.
Chronic laryngitis: Long-term laryngitis lasts longer than three weeks. Causes include smoking, allergies and reflux.
Laryngeal cancer: You can develop cancer in any part of your larynx.
Trauma or injury: You can hurt your larynx by overusing your voice when speaking, shouting or singing for a long time. Likewise, being hit in your throat can hurt your larynx.
Vocal cord dysfunction: This occurs when the vocal cords don’t act or work normally.
Vocal cord lesions: This happens when you develop noncancerous lesions, nodules, polyps or cysts, especially with overuse of your voice.
Vocal cord paralysis: This happens when one or both vocal cords don’t move properly.
What are common symptoms of conditions that affect the larynx?
Some common symptoms are:
Sore throat or cough.
Voice changes, including hoarseness.
Pain or other difficulties when you swallow.
Talk to your healthcare provider if you’ve had these symptoms for more than two weeks.
What are common tests to diagnose larynx issues?
Tests will vary depending on the suspected cause, but may include:
Throat culture or blood tests to check for infections.
Imaging tests like magnetic resonance imaging (MRI), computed tomography (CT) scan or positron emission tomography (PET) scan.
Videostroboscopy to examine your vocal cords while they’re vibrating.
Laryngoscopy to examine your larynx.
Biopsy to remove tissue for examination under a microscope.
What are common treatments for larynx issues?
Acute and chronic laryngitis
Antibiotics.
Antifungals.
Laryngeal cancer
Radiation therapy.
Chemotherapy.
Immunotherapy.
Targeted therapy.
Laryngectomy.
Vocal cord dysfunction
Treatment depends on your situation but could include:
Breathing exercises.
Speech therapy.
Heliox (a combination of oxygen and helium).
Tracheostomy. A provider may do this procedure if vocal cord dysfunction keeps air from getting to your lungs.
Vocal cord lesions
Treatment depends on the underlying cause but may include:
Voice therapy.
Help with lifestyle changes that may reduce your risk of vocal cord lesions.
Surgery.
Vocal cord paralysis
Voice therapy.
Vocal cord injection to place filler in vocal cord gaps.
Laryngeal framework surgery to insert a voice box implant.
Tracheostomy if vocal fold paralysis affects breathing.
Care
How do I take care of my larynx?
There are lots of ways to take care of your larynx and your voice, including:
Reduce your risk of laryngeal cancer: Avoid tobacco products and secondhand smoke. Limit the number of beverages containing alcohol that you drink.
Reduce your risk of laryngitis: Protect yourself against respiratory infections by washing your hands and avoiding people who are sick. Drink lots of water to keep your throat from feeling dry.
Protect your vocal cords: Don’t strain your voice. Limit chemicals and medications that can dry your vocal cords, like some over-the-counter (OTC) cold and allergy medications.
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When should I call my healthcare provider?
Several conditions can affect your larynx. You should contact your provider if you have symptoms that don’t go away after treatment or get worse.
Additional Common Questions
Can someone talk without a larynx?
They can, but they’d have to learn new ways to speak and communicate. For example, some people who have surgery to remove their larynx may use an electrolarynx. An electrolarynx is an artificial larynx that you hold against your throat to make your speech clearer.
A note from Cleveland Clinic
Your larynx is a hardworking part of your respiratory system. You have a healthy larynx to thank anytime you make sounds, from speaking up to singing and shouting. Likewise, your larynx helps you to breathe and keeps food and drink from getting into your lungs. You can keep your larynx healthy by avoiding activities that increase your risk of infections, laryngeal cancer or strained vocal cords.
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Last reviewed on 08/23/2023.
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3538 | https://www.mometrix.com/academy/graphing-linear-equations/ | How to Graph Linear Equations (Video & Practice Questions)
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How to Graph Linear Equations
How to Graph Linear Equations (Video & Practice Questions)
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The Importance of Graphing Linear Equations
Review of Linear Equation Forms
Graphing from Standard Form
Graphing from Slope-Intercept Form
Graphing from Point-Slope Form
Graphing Linear Equations Practice Questions
[x] Transcript - [x] Practice
Hi, and welcome to this video on graphing linear equations!
In this video, we’re going to quickly review the various ways that linear equations can be written, and discover the best methods of graphing equations by hand based on the form that you are given.
Let’s get started!
The Importance of Graphing Linear Equations
When graphed, linear equations form a straight line on the x−y x−y coordinate plane. The graph is a visual representation, and it is often useful to identify key features that are not immediately recognizable from the equation.
For example, when we look at a graph, we can see exactly where a line crosses the x x-axis. This is useful because that x x-value is the solution to the equation. Similarly, if we are given a system of equations, meaning one or more equations, we can find the solution of the system by graphing each line and finding the point of intersection, if it exists.
Knowing how to graph a linear equation by hand is an important skill. The graphing method that you will use will depend on the structure that is provided.
Review of Linear Equation Forms
Let’s quickly review the three forms:
An equation in standard form is written as A x+B y=C A x+B y=C, where A A, B B and C C are constants. A A and B B are known as coefficients of the variables x x and y y, respectively.
An equation in slope-intercept form is written as y=m x+b y=m x+b, where m m represents the slope of the line, and b b represents the y y-intercept, which is the point on the y y-axis where the line crosses.
An equation in point-slope form is written as (y–y 1)=m(x–x 1)(y–y 1)=m(x–x 1). This form provides the ordered pair of a point, (x 1,y 1)(x 1,y 1), that the line passes through and the slope of the line.
While the equations may look different, they all represent a line. When you are asked to graph an equation, you will want to choose the method that is most suitable based on what is provided. Let’s now take a look at each form and determine the graphing method that is the best for each.
Graphing from Standard Form
Let’s start with an example that uses standard form: A x+B y=C A x+B y=C.
A few algebraic steps are necessary in order to graph a line in this form. Keep in mind that you only need two points to graph a line, and using a straight edge to carefully draw the line through those points will provide more accurate results.
The two points that can be quickly calculated from this form are the x x-intercept and the y y-intercept. The graph here illustrates this process visually.
Let’s go through the steps to find the intercepts of the equation, 2 x+3 y=12 2 x+3 y=12:
The y y-intercept can be determined by substituting x=0 x=0 into the equation and solving for y y.
2(0)+3 y=12 2(0)+3 y=12
3 y=12 3 y=12
Divide by 3 on both sides because the 0 doesn’t matter; it won’t change anything.
3 y 3=12 3 3 y 3=12 3
y=4 y=4
The line crosses the y y-axis at the y y-intercept, (0,4)(0,4).
Likewise, the x x-intercept can be determined by substituting y=0 y=0 into the equation and solving for x x, as follows:
2 x+3(0)=12 2 x+3(0)=12
2 x=12 2 x=12
Divide by 2 on both sides, because again, the 0 won’t change anything.
2 x 2=12 2 2 x 2=12 2
x=6 x=6
So the x x-intercept is at the point (6,0)(6,0).
Once the intercepts are plotted at the points (0,4)(0,4) and (6,0)(6,0), you are ready to graph the line. The line passes through these points, and arrows at the ends should be drawn to indicate that the line extends forever in both directions.
This is an easy method of graphing an equation in standard form if the constant, C, is a multiple of the coefficients, A and B. Accordingly, this method of graphing is called the “intercept method.”
Let’s try another example problem:
Graph the equation 3 x–4 y=12 3 x–4 y=12.
This equation is in standard form and the constant, 12, is a multiple of both 3 and 4. Therefore, using the intercept method to graph is a good method.
As before, determine both the y y– and x x-intercepts:
To find the y y-intercept, let x=0 x=0 in the standard form equation and solve for y y, as shown:
3(0)–4 y=12 3(0)–4 y=12
−4 y=12−4 y=12
−4 y−4=12−4−4 y−4=12−4
y=−3 y=−3
So our y y-intercept is at the point (0,−3)(0,−3).
The x x-intercept can be determined by substituting y=0 y=0 into the equation and solving for x x:
3 x–4(0)=12 3 x–4(0)=12
3 x=12 3 x=12
3 x 3=12 3 3 x 3=12 3
x=4 x=4
This means our x x-intercept, or the point where our graph crosses the x x-axis, is (4,0)(4,0).
So our 2 points we’re gonna use to graph are (0,−3)(0,−3) and (4,0)(4,0).
Graph this line by plotting the intercepts and drawing a straight line extending through the points, like so: (see image at 6:00)
Graphing from Slope-Intercept Form
Okay, now let’s try slope-intercept form: y=m x+b y=m x+b.
If you are given a standard form equation but the constant, C C, is not a multiple of the coefficients A and B, then converting the equation into slope-intercept form will provide the information needed for a more accurate graph.
To convert to slope-intercept form, simply solve the standard form equation for y y.
Consider the standard form equation, 3 x+2 y=4 3 x+2 y=4. You notice immediately that while 4 is a multiple of 2, it is not a multiple of 3. A better way to graph this line is to first convert it to slope-intercept form by solving for the variable y y.
To do this, we take our equation 3 x+2 y=4 3 x+2 y=4 and subtract 3 x 3 x from both sides.
This gives us 2 y=4−3 x 2 y=4−3 x.
Now we want to divide by 2. When we divide the whole equation by 2, it’s also the same as dividing each part by 2.
So we’re going to do y=2−3 2 x y=2−3 2 x.
Now, this equation looks similar to what we want but it’s not quite right. We want our x x to be in front.
Since we’re adding, we can flip this without worrying about changing anything.
The resulting equation, y=−3 2 x+2 y=−3 2 x+2, is in slope-intercept form, where the slope of this line is shown, m=−3 2 m=−3 2. The constant in this equation, b=2 b=2, represents the point where the line crosses the y y-axis, the point (0,2)(0,2).
To graph a line in this form, plot the y y-intercept at (0,2)(0,2). A second point on the line is plotted by moving in a vertical direction, often referred to as the “rise,” and then a horizontal direction, referred to as the “run.” These directions are known by looking at the slope, m=−3 2 m=−3 2.
Because the numerator of the slope is -3, move 3 units down from the y y-intercept. From there, move 2 units to the right, as indicated by the denominator of the slope, 2.
Note: A positive “rise” will be up, and a negative “rise” is down; a positive “run” is right, a negative “run” is left.
This movement from the y y-intercept brings you to another point that is on the line, (2,−1)(2,−1). With two points plotted, you can graph the straight line that extends through them.
Now let’s try this out with another example problem:
Graph the equation y=2 3 x−1 y=2 3 x−1. This equation is already in slope-intercept form.
The slope is identified as m=2 3 m=2 3, and the y y-intercept is at the ordered pair, (0,−1)(0,−1). That is where the graphing begins. Plot the first point at the y y-intercept, then move from that point according to the slope. Both numerator and denominator are positive, so the “rise” is up 2 and the “run” is right 3.
This brings you to the second point that is needed to graph the line, (3,1)(3,1). Carefully draw a line that extends through these two points.
Graphing from Point-Slope Form
Okay, now to our third and final form, point-slope form: (y–y 1)=m(x–x 1)(y–y 1)=m(x–x 1).
As noted, the point-slope form of an equation names a point, (x 1,y 1)(x 1,y 1), that the line travels through. It also provides the slope. So, graphing a line in this form is similar to the process we detailed before. The exception is that rather than starting at the y y-intercept, the first point of the graph is at the point (x 1,y 1)(x 1,y 1). Identifying that point sometimes causes some confusion.
Note that the point-slope form subtracts the y-coordinate of the point, y 1 y 1, from the variable y y, and the x x-coordinate of the point x 1 x 1 is subtracted from the variable x x. Notice how the point is identified with the following examples:
In this first example, the line passes through the point (2,3)(2,3). The line in the second example runs through points (1,2)(1,2). In the third example, the line passes through the points (7,−5)(7,−5).
Example 3 requires some explanation because the y y-coordinate of the point is added to the y y-variable in the equation, rather than being subtracted. As a result, an adjustment must be made to the equation to identify the point.
Remember, when you subtract a negative value, it has the same effect as addition. As a result, rewriting the equation to reflect subtraction is allowable.
The adjusted equation looks like this: y−(−5)=3(x–7)y−(−5)=3(x–7). Now the point, (x 1,y 1)(x 1,y 1), is recognizable as (7,−5)(7,−5).
Here is another example of a point-slope equation that deserves some attention: y=12(x−4)y=12(x−4).
This equation is unique because there is no y-coordinate being subtracted from the y y-variable. To identify the point, rewrite the equation as: y−0=12(x−4)y−0=12(x−4). Now the point can be seen as the x x-intercept, (4,0)(4,0).
Now that the first point can be identified from point-slope form, use the slope to move from that point to another point on the line using slope.
Okay, let’s go over two more examples.
Graph the following equation: y–3=4(x–2)y–3=4(x–2).
Identify and plot the first point, (2,3)(2,3). Slope is m=4 m=4. This value can be represented by the equivalent fraction, 4 1 4 1.This slope translates to a “rise” of up, 4 and a “run” of right, 1.
This brings you to another point on the graph, (3,7)(3,7). The line can then be graphed extending through these two points.
Graph the following equation: y−2=−1 3(x−1)y−2=−1 3(x−1).
Identify and plot the first point, (1, 2). Slope in this equation is a negative. At this stage, you can decide to move the rise or the run in the negative direction. For this example, let’s consider the rise, or the numerator, to be the negative value: m=r i s e r u n=−1 3 m=r i s e r u n=−1 3.
This translates to a “rise” of down, 1; and a “run” of right, 3. The position of the second point on the line is at (4,1)(4,1).
The line is graphed by drawing a line extending through these two points.
Now that you have had some practice identifying the features of linear equations in various forms, you can recognize why there are different graphing methods. A quick analysis of the equation should give you direction on how to proceed with the graphing process.
I hope this review was helpful! Thanks for watching, and happy studying!
Graphing Linear Equations Practice Questions
Question #1:
Convert the linear equation from standard form to slope-intercept form:
5 x+4 y=8 5 x+4 y=8
4 y=5 x−8 4 y=5 x−8
4 y=5 x+8 4 y=5 x+8
y=5 4 x−2 y=5 4 x−2
y=−5 4 x+2 y=−5 4 x+2
[x] Show Answer
Answer:
To convert to slope-intercept form, solve the standard form equation for y.
5 x+4 y=8 5 x+4 y=8
First, move 5x to the right side of the equation by using inverse operations. Since the opposite of addition is subtraction, subtract 5x from both sides of the equation.
5 x+4 y−5 x=8−5 x 5 x+4 y−5 x=8−5 x
5 x−5 x=0 5 x−5 x=0, so we are left with 4 y 4 y on the left side of the equation.
4 y=8−5 x 4 y=8−5 x
To isolate the variable y y, use inverse operations. Since the opposite of multiplication is division, divide both sides of the equation by 4.
4 y 4=8−5 x 4 4 y 4=8−5 x 4
4÷4=1 4÷4=1, so we have 1y (which can also be written as y). 8÷4=2 8÷4=2 and 5 x÷4=54 x 5 x÷4=54 x.
y=2−5 4 x y=2−5 4 x
Last, switch the order of the two terms so the x-term comes first. Since 5 4 x 5 4 x is subtracted, it becomes a negative term when switched. Since 2 is a positive number, it is added to −5 4 x−5 4 x. Therefore, the correct answer is D.
y=−5 4 x+2 y=−5 4 x+2
[x] Hide Answer
Question #2:
Consider the following linear equation written in standard form: 6 x−4 y=24 6 x−4 y=24
What are the coordinates for the x-intercept and the y-intercept?
x-intercept: (0,−6)(0,−6); y-intercept: (4,0)(4,0)
x-intercept: (4,0)(4,0); y-intercept: (0,−6)(0,−6)
x-intercept: (−6,0)(−6,0); y-intercept: (0,4)(0,4)
x-intercept: (0,4)(0,4); y-intercept: (−6,0)(−6,0)
[x] Show Answer
Answer:
Solve for y when x=0 to identify the x-intercept, and solve for x when y=0 y=0 to identify the y-intercept. See work shown below:
Starting with the equation in standard form, solve for x when y=0 y=0 to identify the x-intercept.
6 x−4 y=24 6 x−4 y=24
Substitute y with 0 to solve for x.
6 x−4(0)=24 6 x−4(0)=24
Since 4×0=0 4×0=0, replace 4 y 4 y with 0 in the equation.
6 x−0=24 6 x−0=24
Next, isolate the variable x by using its inverse operation. Since the opposite of multiplication is division, divide both sides of the equation by 6.
6 x÷6=24÷6 6 x÷6=24÷6
6÷6=1 6÷6=1, so we have 1 x 1 x (which can also be written as x). 24÷6=4 24÷6=4, so x=4 x=4. Therefore, the x-intercept is 4, or (4,0)(4,0).
x=4 x=4
Going back to the original equation in standard form, solve for y when x=0 x=0 to identify the y-intercept.
6 x−4 y=24 6 x−4 y=24
Substitute x with 0 to solve for y.
6(0)−4 y=24 6(0)−4 y=24
Since 6×0=0 6×0=0, replace 6 x 6 x with 0 in the equation.
0−4 y=24 0−4 y=24
Next, isolate the variable y by using its inverse operation. Since the opposite of multiplication is division, divide both sides of the equation by -4.
−4 y÷−4=24÷−4−4 y÷−4=24÷−4
−4÷−4=1−4÷−4=1, so we have 1 y 1 y (which can also be written as y). 24÷−4=−6 24÷−4=−6, so y=−6 y=−6. The y-intercept is -6, or (0,−6)(0,−6).
[x] Hide Answer
Question #3:
Carly pays 5 dollars to rent a bicycle to go for a ride along the river trail. For each hour she rents the bike, Carly pays an extra 2 dollars. This scenario is illustrated in the graph below. Which equation represents the scenario and corresponding graph?
y=2 x−5 y=2 x−5
y=2 x+5 y=2 x+5
y=−2 x+5 y=−2 x+5
y=−2 x−5 y=−2 x−5
[x] Show Answer
Answer:
Since the graph of the line intercepts the y-axis at (0,5)(0,5), the value for b in slope-intercept form y=m x+b y=m x+b must equal 5. To determine the slope (m), calculate the rise over run of the two points shown on the graph (0,5)(0,5) and (−1,3)(−1,3). Calculate the change in y-coordinates and divide by the change in x-coordinates. 5−3=2 5−3=2 and 0−(−1)=1 0−(−1)=1. 2÷1=2 2÷1=2, so the slope (m) is 2. Therefore, the correct answer is B.
[x] Hide Answer
Question #4:
Given the following linear equation in point-slope form, what point does it pass through when graphed?
y−8=1 4(x+3)y−8=1 4(x+3)
(−8,3)(−8,3)
(3,−8)(3,−8)
(8,−3)(8,−3)
(−3,8)(−3,8)
[x] Show Answer
Answer:
When a linear equation is written in point-slope form, the y-coordinate of the point, y 1 y 1, is subtracted from the variable y. Likewise, the x-coordinate of the point, x 1 x 1, is subtracted from the variable x.
y−y 1=m(x−x 1)y−y 1=m(x−x 1)
y−8=1 4(x+3)y−8=1 4(x+3)
Notice that in this equation, the x-coordinate is added to x instead of subtracted. Remember that subtracting a negative number has the same result as adding. We can rewrite the formula as follows:
y−8=14(x−(−3))y−8=14(x−(−3))
The x-coordinate is -3, and the y-coordinate is 8. Therefore, D is the correct answer.
[x] Hide Answer
Question #5:
Karen pays 224 dollars in advance for her gym membership, which is kept in an account. Each time she visits the gym, 7 dollars is deducted from her account. Which equation represents the value of Karen’s account after x visits to the gym?
y=7 x−224 y=7 x−224
y=−7 x−224 y=−7 x−224
y=−7 x+224 y=−7 x+224
y=7 x+224 y=7 x+224
[x] Show Answer
Answer:
Karen starts with $224 in her account. This is shown as a positive number in the equation. Each time Karen visits the gym, $7 is deducted. This is shown as a negative number in the equation. -7 is multiplied by x because $7 is deducted for each visit to the gym, represented by x. C is correct because this equation contains the terms -7x and +224.
[x] Hide Answer
Return to Algebra I Videos
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by Mometrix Test Preparation | Last Updated: August 12, 2025
On this page
The Importance of Graphing Linear Equations
Review of Linear Equation Forms
Graphing from Standard Form
Graphing from Slope-Intercept Form
Graphing from Point-Slope Form
Graphing Linear Equations Practice Questions
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What You Need to Know About Osteomalacia
Medically reviewed by Angela M. Bell, MD, FACP — Written by Christine Case-Lo — Updated on April 17, 2024
Osteomalacia is a condition that softens the bones. It can result in symptoms such as muscle weakness and bone fractures. Vitamin D supplements can help treat the condition.
Osteomalacia, also known as soft bone disease, is a weakening of the bones. The condition stops your bones from hardening, which can cause them to become weak and more vulnerable to breaks and fractures.
Osteomalacia can occur due to a lack of calcium in the bones. Vitamin D deficiency commonly causes it. Oral supplements typically treat it.
Here’s what you need to know about osteomalacia symptoms, causes, treatment, and more.
What are the symptoms of osteomalacia?
Common symptoms of osteomalacia includeTrusted Source:
fracturing bones easily
muscle weakness
difficulty walking
a waddling gait
Bone pain, especially in your hips, is also a common symptom of osteomalacia. You may develop a dull, aching pain, which can spread from your hips to the following places:
lower back
pelvis
legs
ribs
What are the causes of osteomalacia?
A lack of vitamin D is the most common cause of osteomalacia. Vitamin D is an important nutrient that helps your body effectively absorb calcium, which is neededTrusted Source to build and maintain strong bones.
Vitamin D is made within the skin from exposure to the UV rays in sunlight. It can also be absorbed from foods like dairy products and fish. Your body can’t process the calcium your bones need to stay strong if you have low levels of vitamin D.
A vitamin D deficiency can result from:
a lack of sun exposure
taking medication that interferes with the absorption of vitamin D, such as antiseizure drugs
not getting enough vitamin D from your diet
Your body may also have a problem absorbing vitamin D or breaking down food to release it if you’ve had surgery to remove parts of your stomach or small intestine.
Certain health conditions can also interfere with the absorption of vitamin D, including:
celiac disease, which can damage the lining of your intestines and prevent the absorption of key nutrients like vitamin D
certain types of cancer, such as breast and lung cancer, which can interfere with vitamin D processing
certain types of liver disease, which can affect the metabolism of vitamin D
chronic kidney disease
A diet that doesn’t include phosphates can cause phosphate depletion, which can also lead toTrusted Source osteomalacia.
What are the treatments for osteomalacia?
When osteomalacia is detected early, you may only need to take oral vitamin D, calcium, or phosphate supplements. This may be the first line of treatment if you have absorption problems due to intestinal injury or surgery, or a diet low in key nutrients.
It can also be beneficial to spend some time outdoors in sunlight so your body can make enough vitamin D.
If an underlying condition is causing your osteomalacia, a doctor will typically recommend treating it first to alleviate osteomalacia symptoms.
Children with severe cases of osteomalacia or rickets may have to wear braces or get surgery to correct bone deformation.
What are the potential complications of osteomalacia?
If you don’t treat the cause of your osteomalacia, it can result in several complications. Adults can fracture bones easily, such as rib, leg, and spine bones.
In children, osteomalacia and rickets often occur together, which can lead to bowing (outward curve) of the legs or premature tooth loss.
It’s important to keep in mind that symptoms may return if you stop taking vitamin D supplements or if you don’t address underlying conditions like kidney failure.
You may see improvements in a few weeks by increasing your vitamin D, calcium, and phosphorus intake. Complete healing of the bones can take anywhere from several monthsTrusted Source to a year.
How is osteomalacia diagnosed?
To diagnose osteomalacia, a healthcare professional typically orders a blood test. If it shows any of the following, you may have osteomalacia or another bone disorder:
low levels of vitamin D
low levels of calcium
low levels of phosphorus
While a general blood test is the most common method of diagnosing osteomalacia, a healthcare professional may also order the following tests:
ALP bone isoenzyme test: High levels can indicate that you have osteomalacia.
Parathyroid hormone (PTH) test: High levels of this hormone suggest insufficient vitamin D.
X-rays and other imaging tests: These tests can highlight if there are small cracks in your bones. These cracks are called Looser transformation zones. Fractures can begin in these zones, even with small injuries.
Bone biopsy: This method is rarely used. It involves a healthcare professional inserting a needle through your skin and muscle and into your bone to get a small sample. The sample is then placed on a slide and examined under a microscope for signs of osteomalacia.
Usually, an X-ray and blood test are enough to make a diagnosis, and a bone biopsy isn’t necessary.
What should I eat if I have osteomalacia?
Because osteomalacia can occur from vitamin D deficiency, consuming more foods and drinks rich in this essential nutrient can be beneficial.
Foods high in vitamin D includeTrusted Source:
cod liver oil
white mushrooms
rainbow trout
salmon
vitamin D fortified milk
If osteomalacia is interfering with your daily life, consider speaking with a healthcare professional, such as a dietitian. They can recommend a personalized dietary plan to help alleviate your symptoms.
Frequently asked questions
Is osteomalacia the same as osteoporosis?
Osteomalacia is often confused with osteoporosis, but they are not the same. Both conditions cause the bones to weaken. However, osteoporosis is a weakening of living bone that’s already formed, while osteomalacia occurs when new bone cannot harden.
Are there different types of osteomalacia?
Because osteomalacia has several different causes, each with its own treatment options, some expertsTrusted Source suggest there are various types of the condition. In this case, the type of osteomalacia can be characterized by the nutritional deficiency causing the condition, such as vitamin D or phosphate deficiency.
Is osteomalacia the same as rickets?
Rickets and osteomalacia are differentTrusted Source conditions. Rickets impairs new bone growth in children, whereas osteomalacia can develop in adults and children. Both conditions can result in soft and weak bones due to nutrient deficiencies.
The bottom line
Osteomalacia, also known as soft bone disease, is a condition that weakens the bones. It’s commonly caused by vitamin D deficiency and a lack of calcium in the bones.
Osteomalacia can result in symptoms such as muscle weakness and make you vulnerable to breaks and fractures. If left untreated, it can lead to broken bones and severe deformity.
If you think you may be experiencing symptoms of osteomalacia, consider speaking with a healthcare professional. Various treatment options are available to help manage the condition. When it’s found early, you may only need to take oral supplements to treat symptoms.
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How we reviewed this article:
Healthline has strict sourcing guidelines and relies on peer-reviewed studies, academic research institutions, and medical journals and associations. We only use quality, credible sources to ensure content accuracy and integrity. You can learn more about how we ensure our content is accurate and current by reading our editorial policy.
Cianferotti L. (2022). Osteomalacia is not a single disease.
Serna J, et al. (2020). Importance of dietary phosphorus for bone metabolism and healthy aging.
Uday S, et al. (2020). Nutritional rickets & osteomalacia: A practical approach to management.
Vitamin D deficiency. (n.d.).
Vitamin D: Fact sheet for health professionals. (2023).
Zimmerman L, et al. (2023). Osteomalacia.
Our experts continually monitor the health and wellness space, and we update our articles when new information becomes available.
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Apr 17, 2024
Written By
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Medically Reviewed By
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Written By
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University Physics Volume 1
12.2 Examples of Static Equilibrium
University Physics Volume 1
12.2
Examples of Static Equilibrium
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Learning Objectives
By the end of this section, you will be able to:
Identify and analyze static equilibrium situations
Set up a free-body diagram for an extended object in static equilibrium
Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations
All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Equation 12.7 to Equation 12.9. We introduced a problem-solving strategy in Example 12.1 to illustrate the physical meaning of the equilibrium conditions. Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. We proceed in five practical steps.
Problem-Solving Strategy
Static Equilibrium
Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
Set up a free-body diagram for the object. (a) Choose the xy-reference frame for the problem. Draw a free-body diagram for the object, including only the forces that act on it. When suitable, represent the forces in terms of their components in the chosen reference frame. As you do this for each force, cross out the original force so that you do not erroneously include the same force twice in equations. Label all forces—you will need this for correct computations of net forces in the x- and y-directions. For an unknown force, the direction must be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The correct direction is determined by the sign that you obtain in the final solution. A plus sign (+) means that the working direction is the actual direction. A minus sign (−) means that the actual direction is opposite to the assumed working direction. (b) Choose the location of the rotation axis; in other words, choose the pivot point with respect to which you will compute torques of acting forces. On the free-body diagram, indicate the location of the pivot and the lever arms of acting forces—you will need this for correct computations of torques. In the selection of the pivot, keep in mind that the pivot can be placed anywhere you wish, but the guiding principle is that the best choice will simplify as much as possible the calculation of the net torque along the rotation axis.
Set up the equations of equilibrium for the object. (a) Use the free-body diagram to write a correct equilibrium condition Equation 12.7 for force components in the x-direction. (b) Use the free-body diagram to write a correct equilibrium condition Equation 12.11 for force components in the y-direction. (c) Use the free-body diagram to write a correct equilibrium condition Equation 12.9 for torques along the axis of rotation. Use Equation 12.10 to evaluate torque magnitudes and senses.
Simplify and solve the system of equations for equilibrium to obtain unknown quantities. At this point, your work involves algebra only. Keep in mind that the number of equations must be the same as the number of unknowns. If the number of unknowns is larger than the number of equations, the problem cannot be solved.
Evaluate the expressions for the unknown quantities that you obtained in your solution. Your final answers should have correct numerical values and correct physical units. If they do not, then use the previous steps to track back a mistake to its origin and correct it. Also, you may independently check for your numerical answers by shifting the pivot to a different location and solving the problem again, which is what we did in Example 12.1.
Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium. Also note that a free-body diagram for an extended rigid body that may undergo rotational motion is different from a free-body diagram for a body that experiences only translational motion (as you saw in the chapters on Newton’s laws of motion). In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. The reason for this is that in analyzing rotation, we must identify torques acting on the body, and torque depends both on the acting force and on its lever arm. Here, the free-body diagram for an extended rigid body helps us identify external torques.
Example 12.3
The Torque Balance
Three masses are attached to a uniform meter stick, as shown in Figure 12.9. The mass of the meter stick is 150.0 g and the masses to the left of the fulcrum are m1=50.0g and m2=75.0g. Find the mass m3 that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced.
Figure
12.9
In a torque balance, a horizontal beam is supported at a fulcrum (indicated by S) and masses are attached to both sides of the fulcrum. The system is in static equilibrium when the beam does not rotate. It is balanced when the beam remains level.
Strategy
For the arrangement shown in the figure, we identify the following five forces acting on the meter stick:
w1=m1g is the weight of mass m1; w2=m2g is the weight of mass m2;
w=mg is the weight of the entire meter stick; w3=m3g is the weight of unknown mass m3;
FS is the normal reaction force at the support point S.
We choose a frame of reference where the direction of the y-axis is the direction of gravity, the direction of the x-axis is along the meter stick, and the axis of rotation (the z-axis) is perpendicular to the x-axis and passes through the support point S. In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates. Now we are ready to set up the free-body diagram for the meter stick. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot Figure 12.10. At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w, at the 50-cm mark.
Figure
12.10
Free-body diagram for the meter stick. The pivot is chosen at the support point S.
Solution
With Figure 12.9 and Figure 12.10 for reference, we begin by finding the lever arms of the five forces acting on the stick:
r1r2rrSr3=====30.0cm+40.0cm=70.0cm40.0cm50.0cm−30.0cm=20.0cm0.0cm(becauseFSis attached at the pivot)30.0cm.
Now we can find the five torques with respect to the chosen pivot:
τ1τ2ττSτ3=====+r1w1sin90°=+r1m1g+r2w2sin90°=+r2m2g+rwsin90°=+rmgrSFSsinθS=0−r3w3sin90°=−r3m3g(counterclockwise rotation, positive sense)(counterclockwise rotation, positive sense)(gravitational torque)(becauserS=0cm)(clockwise rotation, negative sense)
The second equilibrium condition (equation for the torques) for the meter stick is
τ1+τ2+τ+τS+τ3=0.
When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is
+r1m1g+r2m2g+rmg−r3m3g=0.
12.17
Selecting the +y-direction to be parallel to F⃗ S, the first equilibrium condition for the stick is
−w1−w2−w+FS−w3=0.
Substituting the forces, the first equilibrium condition becomes
−m1g−m2g−mg+FS−m3g=0.
12.18
We solve these equations simultaneously for the unknown values m3 and FS. In Equation 12.17, we cancel the g factor and rearrange the terms to obtain
r3m3=r1m1+r2m2+rm.
To obtain m3 we divide both sides by r3, so we have
m3=r1r3m1+r2r3m2+rr3m=7030(50.0g)+4030(75.0g)+2030(150.0g)=316.023g≃317g.
12.19
To find the normal reaction force, we rearrange the terms in Equation 12.18, converting grams to kilograms:
FS=(m1+m2+m+m3)g=(50.0+75.0+150.0+316.7)×10−3kg×9.8ms2=5.8N.
12.20
Significance
Notice that Equation 12.17 is independent of the value of g. The torque balance may therefore be used to measure mass, since variations in g-values on Earth’s surface do not affect these measurements. This is not the case for a spring balance because it measures the force.
Check Your Understanding 12.3
Repeat Example 12.3 using the left end of the meter stick to calculate the torques; that is, by placing the pivot at the left end of the meter stick.
In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by Equation 12.7 and Equation 12.8. We present this solution to illustrate the importance of a suitable choice of reference frame. Although all inertial reference frames are equivalent and numerical solutions obtained in one frame are the same as in any other, an unsuitable choice of reference frame can make the solution quite lengthy and convoluted, whereas a wise choice of reference frame makes the solution straightforward. We show this in the equivalent solution to the same problem. This particular example illustrates an application of static equilibrium to biomechanics.
Example 12.4
Forces in the Forearm
A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in Figure 12.11. His forearm is positioned at β=60° with respect to his upper arm. The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give your final answers in SI units.
Figure
12.11
The forearm is rotated around the elbow (E) by a contraction of the biceps muscle, which causes tension T⃗ M.
Strategy
We identify three forces acting on the forearm: the unknown force F⃗ at the elbow; the unknown tension T⃗ M in the muscle; and the weight w⃗ with magnitude w=50lb. We adopt the frame of reference with the x-axis along the forearm and the pivot at the elbow. The vertical direction is the direction of the weight, which is the same as the direction of the upper arm. The x-axis makes an angle β=60° with the vertical. The y-axis is perpendicular to the x-axis. Now we set up the free-body diagram for the forearm. First, we draw the axes, the pivot, and the three vectors representing the three identified forces. Then we locate the angle β and represent each force by its x- and y-components, remembering to cross out the original force vector to avoid double counting. Finally, we label the forces and their lever arms. The free-body diagram for the forearm is shown in Figure 12.12. At this point, we are ready to set up equilibrium conditions for the forearm. Each force has x- and y-components; therefore, we have two equations for the first equilibrium condition, one equation for each component of the net force acting on the forearm.
Figure
12.12
Free-body diagram for the forearm: The pivot is located at point E (elbow).
Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y-components of the forces because the x-components of the forces are all parallel to their lever arms, so that for any of them we have sinθ=0 in Equation 12.10. For the y-components we have θ=±90° in Equation 12.10. Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of Ty and of wy.
Solution
We see from the free-body diagram that the x-component of the net force satisfies the equation
+Fx+Tx−wx=0
12.21
and the y-component of the net force satisfies
+Fy+Ty−wy=0.
12.22
Equation 12.21 and Equation 12.22 are two equations of the first equilibrium condition (for forces). Next, we read from the free-body diagram that the net torque along the axis of rotation is
+rTTy−rwwy=0.
12.23
Equation 12.23 is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are rT=1.5in. and rw=13.0in. At this point, we do not need to convert inches into SI units, because as long as these units are consistent in Equation 12.23, they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:
FxTxwxFyTywy======Fcosβ=Fcos60°=F/2Tcosβ=Tcos60°=T/2wcosβ=wcos60°=w/2Fsinβ=Fsin60°=F3–√/2Tsinβ=Tsin60°=T3–√/2wsinβ=wsin60°=w3–√/2.
We substitute these magnitudes into Equation 12.21, Equation 12.22, and Equation 12.23 to obtain, respectively,
F/2+T/2−w/2F3–√/2+T3–√/2−w3–√/2rTT3–√/2−rww3–√/2===000.
When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T, because Equation 12.21 for the x-component is equivalent to Equation 12.22 for the y-component. In this way, we obtain the first equilibrium condition for forces
F+T−w=0
12.24
and the second equilibrium condition for torques
rTT−rww=0.
12.25
The magnitude of tension in the muscle is obtained by solving Equation 12.25:
T=rwrTw=13.01.5(50 lb)=43313lb≃433.3lb.
The force at the elbow is obtained by solving Equation 12.24:
F=w−T=50.0lb−433.3lb=−383.3lb.
The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram. In the final answer, we convert the forces into SI units of force. The answer is
F=383.3lb=383.3(4.448N)=1705N downwardT=433.3lb=433.3(4.448N)=1927N upward.
Significance
Two important issues here are worth noting. The first concerns conversion into SI units, which can be done at the very end of the solution as long as we keep consistency in units. The second important issue concerns the hinge joints such as the elbow. In the initial analysis of a problem, hinge joints should always be assumed to exert a force in an arbitrary direction, and then you must solve for all components of a hinge force independently. In this example, the elbow force happens to be vertical because the problem assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a general rule.
Solution
Suppose we adopt a reference frame with the direction of the y-axis along the 50-lb weight and the pivot placed at the elbow. In this frame, all three forces have only y-components, so we have only one equation for the first equilibrium condition (for forces). We draw the free-body diagram for the forearm as shown in Figure 12.13, indicating the pivot, the acting forces and their lever arms with respect to the pivot, and the angles θT and θw that the forces T⃗ M and w⃗ (respectively) make with their lever arms. In the definition of torque given by Equation 12.10, the angle θT is the direction angle of the vector T⃗ M, counted counterclockwise from the radial direction of the lever arm that always points away from the pivot. By the same convention, the angle θw is measured counterclockwise from the radial direction of the lever arm to the vector w⃗ . Done this way, the non-zero torques are most easily computed by directly substituting into Equation 12.10 as follows:
τT=rTTsinθT=rTTsinβ=rTTsin60°=+rTT3–√/2τw=rwwsinθw=rwwsin(β+180°)=−rwwsinβ=−rww3–√/2.
Figure
12.13
Free-body diagram for the forearm for the equivalent solution. The pivot is located at point E (elbow).
The second equilibrium condition, τT+τw=0, can be now written as
rTT3–√/2−rww3–√/2=0.
12.26
From the free-body diagram, the first equilibrium condition (for forces) is
−F+T−w=0.
12.27
Equation 12.26 is identical to Equation 12.25 and gives the result T=433.3lb. Equation 12.27 gives
F=T−w=433.3lb−50.0lb=383.3lb.
We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components.
Check Your Understanding 12.4
Repeat Example 12.4 assuming that the forearm is an object of uniform density that weighs 8.896 N.
Example 12.5
A Ladder Resting Against a Wall
A uniform ladder is L=5.0m long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in Figure 12.14. The minimum inclination angle between the ladder and the rough floor below which the ladder slips is β=53°. Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction μs at the interface of the ladder with the floor that prevents the ladder from slipping.
Figure
12.14
A 5.0-m-long ladder rests against a frictionless wall.
Strategy
We can identify four forces acting on the ladder. The first force is the normal reaction force N from the floor in the upward vertical direction. The second force is the static friction force f=μsN directed horizontally along the floor toward the wall—this force prevents the ladder from slipping. These two forces act on the ladder at its contact point with the floor. The third force is the weight w of the ladder, attached at its CM located midway between its ends. The fourth force is the normal reaction force F from the wall in the horizontal direction away from the wall, attached at the contact point with the wall. There are no other forces because the wall is slippery, which means there is no friction between the wall and the ladder. Based on this analysis, we adopt the frame of reference with the y-axis in the vertical direction (parallel to the wall) and the x-axis in the horizontal direction (parallel to the floor). In this frame, each force has either a horizontal component or a vertical component but not both, which simplifies the solution. We select the pivot at the contact point with the floor. In the free-body diagram for the ladder, we indicate the pivot, all four forces and their lever arms, and the angles between lever arms and the forces, as shown in Figure 12.15. With our choice of the pivot location, there is no torque either from the normal reaction force N or from the static friction f because they both act at the pivot.
Figure
12.15
Free-body diagram for a ladder resting against a frictionless wall.
Solution
From the free-body diagram, the net force in the x-direction is
+f−F=0
12.28
the net force in the y-direction is
+N−w=0
12.29
and the net torque along the rotation axis at the pivot point is
τw+τF=0.
where τw is the torque of the weight w and τF is the torque of the reaction F. From the free-body diagram, we identify that the lever arm of the reaction at the wall is rF=L=5.0m and the lever arm of the weight is rw=L/2=2.5m. With the help of the free-body diagram, we identify the angles to be used in Equation 12.10 for torques: θF=180°−β for the torque from the reaction force with the wall, and θw=180°+(90°−β) for the torque due to the weight. Now we are ready to use Equation 12.10 to compute torques:
τw=rwwsinθw=rwwsin(180°+90°−β)=−L2wsin(90°−β)=−L2wcosβτF=rFFsinθF=rFFsin(180°−β)=LFsinβ.
We substitute the torques into Equation 12.30 and solve for F:
−L2wcosβ+LFsinβF=w2cotβ=400.0N2cot53°==0150.7N
We obtain the normal reaction force with the floor by solving Equation 12.29: N=w=400.0N. The magnitude of friction is obtained by solving Equation 12.28: f=F=150.7N. Since the ladder will slip if the angle is any smaller, the static friction force is at its maximum angle and the coefficient of static friction is μs=f/N=150.7/400.0=0.377.
The net force on the ladder at the contact point with the floor is the vector sum of the normal reaction from the floor and the static friction forces:
F⃗ floor=f⃗ +N⃗ =(150.7 N)(−iˆ)+(400.0N)(+jˆ)=(−150.7iˆ+400.0jˆ)N.
Its magnitude is
Ffloor=f2+N2−−−−−−−√=150.72+400.02−−−−−−−−−−−−√N=427.4N
and its direction is
φ=tan−1(N/f)=tan−1(400.0/150.7)=69.3°above the floor.
We should emphasize here two general observations of practical use. First, notice that when we choose a pivot point, there is no expectation that the system will actually pivot around the chosen point. The ladder in this example is not rotating at all but firmly stands on the floor; nonetheless, its contact point with the floor is a good choice for the pivot. Second, notice when we use Equation 12.10 for the computation of individual torques, we do not need to resolve the forces into their normal and parallel components with respect to the direction of the lever arm, and we do not need to consider a sense of the torque. As long as the angle in Equation 12.10 is correctly identified—with the help of a free-body diagram—as the angle measured counterclockwise from the direction of the lever arm to the direction of the force vector, Equation 12.10 gives both the magnitude and the sense of the torque. This is because torque is the vector product of the lever-arm vector crossed with the force vector, and Equation 12.10 expresses the rectangular component of this vector product along the axis of rotation.
Significance
This result is independent of the length of the ladder because L is cancelled in the second equilibrium condition, Equation 12.31. No matter how long or short the ladder is, as long as its weight is 400 N and the angle with the floor is 53°, our results hold. But the ladder will slip if the net torque becomes negative in Equation 12.31. This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping.
Check Your Understanding 12.5
For the situation described in Example 12.5, determine the values of the coefficient μs of static friction for which the ladder starts slipping, given that β is the angle that the ladder makes with the floor.
Example 12.6
Forces on Door Hinges
A swinging door that weighs w=400.0N is supported by hinges A and B so that the door can swing about a vertical axis passing through the hinges Figure 12.16. The door has a width of b=1.00m, and the door slab has a uniform mass density. The hinges are placed symmetrically at the door’s edge in such a way that the door’s weight is evenly distributed between them. The hinges are separated by distance a=2.00m. Find the forces on the hinges when the door rests half-open.
Figure
12.16
A 400-N swinging vertical door is supported by two hinges attached at points A and B.
Strategy
The forces that the door exerts on its hinges can be found by simply reversing the directions of the forces that the hinges exert on the door. Hence, our task is to find the forces from the hinges on the door. Three forces act on the door slab: an unknown force A⃗ from hinge A, an unknown force B⃗ from hinge B, and the known weight w⃗ attached at the center of mass of the door slab. The CM is located at the geometrical center of the door because the slab has a uniform mass density. We adopt a rectangular frame of reference with the y-axis along the direction of gravity and the x-axis in the plane of the slab, as shown in panel (a) of Figure 12.17, and resolve all forces into their rectangular components. In this way, we have four unknown component forces: two components of force A⃗ (Ax and Ay), and two components of force B⃗ (Bx and By). In the free-body diagram, we represent the two forces at the hinges by their vector components, whose assumed orientations are arbitrary. Because there are four unknowns (Ax, Bx, Ay, and By), we must set up four independent equations. One equation is the equilibrium condition for forces in the x-direction. The second equation is the equilibrium condition for forces in the y-direction. The third equation is the equilibrium condition for torques in rotation about a hinge. Because the weight is evenly distributed between the hinges, we have the fourth equation, Ay=By. To set up the equilibrium conditions, we draw a free-body diagram and choose the pivot point at the upper hinge, as shown in panel (b) of Figure 12.17. Finally, we solve the equations for the unknown force components and find the forces.
Figure
12.17
(a) Geometry and (b) free-body diagram for the door.
Solution
From the free-body diagram for the door we have the first equilibrium condition for forces:
inx-direction:iny-direction:−Ax+Bx=0⇒Ax=Bx+Ay+By−w=0⇒Ay=By=w2=400.0N2=200.0N.
We select the pivot at point P (upper hinge, per the free-body diagram) and write the second equilibrium condition for torques in rotation about point P:
pivot atP:τw+τBx+τBy=0.
We use the free-body diagram to find all the terms in this equation:
τwτBxτBy===dwsin(−β)=−dwsinβ=−dwb/2d=−wb2aBxsin90°=+aBxaBysin180°=0.
In evaluating sinβ, we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into Equation 12.32 and compute Bx:
pivot atP:−wb2+aBx=0⇒Bx=wb2a=(400.0N)12⋅2=100.0N.
Therefore the magnitudes of the horizontal component forces are Ax=Bx=100.0N. The forces on the door are
at the upper hinge:F⃗ Aon door=−100.0Niˆ+200.0Njˆat the lower hinge:F⃗ Bon door=+100.0Niˆ+200.0Njˆ.
The forces on the hinges are found from Newton’s third law as
on the upper hinge:F⃗ door onA=100.0Niˆ−200.0Njˆon the lower hinge:F⃗ door onB=−100.0Niˆ−200.0Njˆ.
Significance
Note that if the problem were formulated without the assumption of the weight being equally distributed between the two hinges, we wouldn’t be able to solve it because the number of the unknowns would be greater than the number of equations expressing equilibrium conditions.
Check Your Understanding 12.6
Solve the problem in Example 12.6 by taking the pivot position at the center of mass.
Check Your Understanding 12.7
A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg. Find the tensions in the two vertical ropes supporting the scaffold.
Check Your Understanding 12.8
A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut.
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Authors: William Moebs, Samuel J. Ling, Jeff Sanny
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Book title: University Physics Volume 1
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3541 | https://dcomcme.lmunet.edu/sites/default/files/304%20113%2014%20Edwards%20UA%20Casts.pdf | MLS Continuing Education Conference November 2014 PACE Session # 304 – 113 - 14 Urinary Casts: The Importance of Laboratory Identification 1 Urinalysis – The Beginning • The field of laboratory medicine started with the analysis of urine – Records of urine study are found in cave drawings and Egyptian hieroglyphics – In the Middle Ages, urine examination was a major focus of physicians • Observations were basic, including: – Color – Clarity – Odor – Volume – Sweetness 2 Urinalysis - Progress • 17th century – Microscope was invented • This led to examination of urine sediment – Methods were developed for the quantitation of urine sediment 3 Sediment Constituents • RBCs • WBCs • Epithelial cells (3 types) • Oval fat bodies • Bacteria • Yeast • Parasites • Spermatozoa • Mucus • Various casts • Various crystals 4 Urinary Casts • Among the various constituents of urinary sediment, casts are unique in that: – They provide us with a microscopic view of the conditions within the nephrons • Nephrons are functional units of the kidney – Each kidney contains between 1 and 1.5 million 5 Cast Composition • All urinary casts are composed of the renal glycoprotein known as Uromodulin – More commonly known as Tamm-Horsfall protein • Uromodulin is produced exclusively in the kidney by the epithelial cells of the thick ascending loop of Henle – Also by the beginning segment of the distal convoluted tubule – Under normal conditions uromodulin is the most abundant protein found in urine 6 Uromodulin Characteristics • Characteristics: – It is a mucoprotein • Approximately 70% protein & 30% carbohydrate • Secreted by RTE cells of the thick ascending loop and the distal convoluted tubule • Secreted into the filtrate as a soluble monomer – It has a high tendency to form polymers • There is a high gel-tendency – Tamm-Horsfall protein is not detected by the reagent strip protein reaction 7 Uromodulin Secretion 8 Uromodulin Function • Biological function is still not fully understood • What we do know: – Uromodulin is linked to water and electrolyte balance (polymeric form) – Soluble form helps to protect against urinary tract infection by fimbriated bacteria – Research has also suggested that uromodulin helps to prevent kidney stones by inhibiting the growth of the monohydrate form of calcium oxalate crystals 9 Factors Contributing to Cast Formation • Urinary casts are formed when there is an increase in uromodulin polymerization which increases gel-formation • Factors which contribute to increased polymerization: – Urinary stasis which results in higher osmolality – Increasing concentrations of sodium and calcium – Decreasing pH – Increasing levels of uromodulin • Secretion increases with stress and physical exertion 10 Cast Formation • Step-by-step formation of a cast matrix as studied by electron microscopy: – Uromodulin aggregates into individual protein fibrils which are attached to tubular epithelial cells – Protein fibrils interweave to form a loose network • At this point any constituents present in the filtrate may become enmeshed – Protein fibrils continue to interweave, resulting in a solid matrix • Blockage of the lumen decreases urine flow & increases pressure – Protein fibrils become detached from the epithelial cells – Cast is excreted 11 Cast Formation 12 Specimen Concerns • For proper cast identification: – Specimens should be as fresh as possible • Casts and cells disintegrate quickly especially in alkaline urine – Centrifuge at least 10 mL of urine – Scan for casts whenever the protein is positive – Scan for casts with 10x magnification • If using slide and coverslip, be sure to scan outer edges • Identify with 40x magnification 13 Hyaline Casts • Most common & basic cast seen • Almost pure uromodulin • Colorless and difficult to see • Significance: – Up to 2 casts/lpf is a normal finding – Large numbers may be seen in cases of dehydration or in times of physical or emotional stress 14 Hyaline Casts – Unstained & Stained 15 RBC Casts • RBC casts indicate bleeding within the nephron – Primarily associated with Glomerulonephritis – Protein and free-standing RBCs should also be present – Ability to detect a visible matrix is important for ID 16 RBC Casts – Stained & Unstained 17 Importance of Identification • Glomerulonephritis accounts for 10%-15% of end stage renal failure cases in the USA – Begins with inflammation within the glomerulus mediated by autoimmune processes – Secondary impairment of renal function occurs over days to weeks • Early diagnosis is extremely important – Even patients with mild renal impairment may quickly lose kidney function – Patients must be treated urgently 18 Types of Glomerulonephritis • Post-infectious Glomerulonephritis – A form of acute glomerulonephritis seen after certain bacterial, viral, fungal, or parasitic infections – Becoming an increasingly seen complication of endocarditis after IV drug abuse – Post-streptococcal glomerulonephritis is the most common • Occurs within 12 weeks of initial infection with certain strains of Streptococcus pyogenes • Immune complexes composed of specific antibodies and bacterial M-protein are deposited on the basement membrane • Anti-streptolysin O titer is diagnostically useful 19 Types of Glomerulonephritis • IgA nephropathy – Most common form of glomerulonephritis – Immune complexes containing IgA are deposited primarily on the mesangium of the glomerulus – Patients have increased levels of IgA – Disease is relatively silent for up to 20 years except for periodic episodes of hematuria – Gradually progresses to chronic glomerulonephritis and end stage renal disease 20 Types of Glomerulonephritis • Henoch-Schonlein Purpura – Small vessel vasculitis – Primarily in children following respiratory infections – Usually begins with a rash of raised, red patches – Respiratory symptoms may include bloody sputum – Arthritis may be present – Abdominal pain with bloody stools – Renal involvement is most serious complication • Up to 50% of patients progress to a more severe form of glomerulonephritis and possibly renal failure 21 Types of Glomerulonephritis • Goodpasture Syndrome – Appearance of a specific, cytotoxic autoantibody against the basement membranes of the alveoli and the glomeruli following a viral respiratory infection • Antiglomerular basement membrane antibody (anti-GBM) – Initial symptoms are respiratory including bloody sputum and difficulty breathing – Chronic glomerulonephritis and end stage renal disease are common • Disease progression can be rapid 22 Types of Glomerulonephritis • Wegener Granulomatosis – Granuloma-producing, small vessel vasculitis • Primarily affecting lungs and kidneys • Pulmonary involvement is first presentation of disease – Key to diagnosis is the production of an autoantibody known as anti-neutrophilic cytoplasmic antibody (ANCA) – Commonly progresses to chronic glomerulonephritis and end stage renal disease 23 Types of Glomerulonephritis • Rapidly Progressive Glomerulonephritis – Initiated by the depositing of immune complexes in the glomerulus • Usually a complication of a systemic immune disorder such as systemic lupus erythematosus – Also known as crescentic glomerulonephritis – Poor prognosis • Often ends in renal failure 24 Renal Tubular Epithelial Cell Casts • Presence represents advanced destruction of the renal tubules – Acute tubular necrosis • Potential causes: – Heavy metal toxicity – Chemical toxicity – Drug-induced toxicity – Viral infections – Transplant rejection – Interruption of renal blood flow 25 Acute Tubular Necrosis • Nephrotoxic agents include: – Aminoglycoside antibiotics – Amphotericin B – Ethylene glycol – Heavy metal exposure – Mushroom poisoning – Hemoglobin & myoglobin • May also see granular casts, waxy casts, & broad casts 26 RTE Casts 27 • Staining can help in identifying RTE casts • This RTE cast has been stained with Sternheimer-Malbin stain RTE Casts • This RTE cast came from a patient with active Hepatitis B • The cells are stained due to the absorption of bilirubin from the filtrate 28 Fatty Casts • Presence indicates lipiduria which is most often associated with Nephrotic syndrome • Usually seen along with oval fat bodies and free fat droplets • Identity may be confirmed with polarized light microscopy or staining with lipid stains such as Sudan III or Oil Red O 29 Fatty Casts - Stained & Polarized 30 The Nephrotic Syndrome • Acute onset due to systemic shock or the progression of other glomerular disorders, such as glomerulonephritis – Glomerular membrane is damaged, resulting in a less tightly connected barrier – Net Result: • High molecular weight proteins and lipids are passed into the urine • Loss of albumin stimulates the hepatic production of lipids • Loss of protein leads to lower oncotic pressure which leads to edema • Depletion of immunoglobulins and coagulation factors places the patient at an increased risk of infection & coagulation disorders 31 The Nephrotic Syndrome • Both glomerular and tubular damage may occur • Condition may progress to chronic renal failure • Additional sediment that may be found: – Oval fat bodies – Free fat droplets – RTE cells & casts – Waxy casts 32 WBC Casts • WBC casts with the presence of bacteria indicate pyelonephritis – May be acute or chronic – Results from the ascending movement of bacteria – Possible complications: • Renal abscess • Renal impairment • Septic shock 33 Chronic Pyelonephritis • Chronic cases of pyelonephritis are usually due to structural defects which allow urinary reflux – Usually diagnosed in children – Leads to tubular damage and renal failure • May also see waxy casts & broad casts 34 WBC Casts • WBC casts without the presence of bacteria indicate acute interstitial nephritis – Primarily associated with an allergic reaction to a medication that occurs within the renal interstitium, leading to inflammation of renal interstitial tissues and renal tubules • penicillin, ampicillin, cephalosporins, NSAIDs, thiazide diuretics 35 Acute Interstitial Nephritis • High percentage of eosinophils present • Hansel stain may be used for identification of urinary eosinophils 36 Granular Casts Fine Granular Coarse Granular 37 Granular Casts • May occur as coarse or fine granular casts • Origin of granules can be: – RTE lysosomes (same significance as hyaline casts) • Excreted during normal metabolism • More are seen after exercise and activity – Disintegration of cellular casts • Pay extra attention to specimens with cells and granular casts 38 Degenerating RBC & WBC Casts 39 Waxy Casts • Presence indicates extreme urine stasis – Seen in cases of renal failure • Degenerated hyaline and granular casts 40 Waxy Casts • Brittle appearance • Highly refractile • Often fragmented with jagged ends and notches • Well visualized with stain 41 Broad Casts • Also significant of extreme renal stasis – Also referred to as “renal failure casts” • Formed in: – Collecting duct, or – Distal tubule that is widened due to destruction 42 Granular & Waxy Broad Casts 43 Renal Failure • May be chronic or acute • Chronic renal failure is seen as a progression of original renal diseases to end stage renal disease • Casts that may be present: – Granular – Waxy – Broad 44 Acute Renal Failure • Onset is sudden • Usually reversible by correcting the cause • Causes may be: – Prerenal – decreased blood flow – Renal – acute disease – Postrenal – obstruction 45 Acute Renal Failure • Expected sediment related to cause: – RTE cells & casts = decreased blood flow – RBCs & casts = glomerular damage – WBCs & casts = infection/inflammation 46 Causes of Acute Renal Failure • Prerenal – Decreased blood pressure/cardiac output – Hemorrhage – Burns – Surgery – Septicemia • Renal – Acute glomerulonephritis – Acute tubular necrosis – Acute pyelonephritis – Acute interstitial nephritis • Postrenal – Renal calculi – Tumors – Crystallization of ingested substances 47 What is the most likely condition? 48 What is the most likely condition? 49 What is the most likely condition? 50 References • Chung, V., C. Tai, C. Fan, and C Tang. 2014. Severe Acute Pyelonephritis: A Review of Clinical Outcome and Risk Factors for Mortality. Hong Kong Med J. 2014;20(4):285-89. • Neumann, I, and P. Moore. 2014. Pyelonephritis (Acute) in Non-Pregnant Women. Clin Evid (Online). 2014:25373019. • Perazella, M. 2010. Drug-induced Acute Interstitial Nephritis. Nat Rev Nephrol. 2010;6(8):461-70. • Strasinger, S.K. 2014. Urinalysis and Body Fluids, 6th ed. F.A. Davis, Philadelphia. • Vinen, C., and D. Oliveira. 2003. Acute Glomerulonephritis. Postgrad Med J. 2003;79:206-213. Contact Information Steven Edwards, MS(MLS) Assistant Professor of Medical Lab Science Lincoln Memorial University 6965 Cumberland Gap Pkwy Harrogate, TN 37752 Office: 423-869-6232 Email: steven.edwards@LMUnet.edu |
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Binary Trees
A Binary Tree is a type of tree data structure where each node can have a maximum of two child nodes, a left child node and a right child node.
This restriction, that a node can have a maximum of two child nodes, gives us many benefits:
Algorithms like traversing, searching, insertion and deletion become easier to understand, to implement, and run faster.
Keeping data sorted in a Binary Search Tree (BST) makes searching very efficient.
Balancing trees is easier to do with a limited number of child nodes, using an AVL Binary Tree for example.
Binary Trees can be represented as arrays, making the tree more memory efficient.
Use the animation below to see how a Binary Tree looks, and what words we use to describe it.
A parent node, or internal node, in a Binary Tree is a node with one or two child nodes.
The left child node is the child node to the left.
The right child node is the child node to the right.
The tree height is the maximum number of edges from the root node to a leaf node.
Binary Trees vs Arrays and Linked Lists
Benefits of Binary Trees over Arrays and Linked Lists:
Arrays are fast when you want to access an element directly, like element number 700 in an array of 1000 elements for example. But inserting and deleting elements require other elements to shift in memory to make place for the new element, or to take the deleted elements place, and that is time consuming.
Linked Lists are fast when inserting or deleting nodes, no memory shifting needed, but to access an element inside the list, the list must be traversed, and that takes time.
Binary Trees, such as Binary Search Trees and AVL Trees, are great compared to Arrays and Linked Lists because they are BOTH fast at accessing a node, AND fast when it comes to deleting or inserting a node, with no shifts in memory needed.
We will take a closer look at how Binary Search Trees (BSTs) and AVL Trees work on the next two pages, but first let's look at how a Binary Tree can be implemented, and how it can be traversed.
Types of Binary Trees
There are different variants, or types, of Binary Trees worth discussing to get a better understanding of how Binary Trees can be structured.
The different kinds of Binary Trees are also worth mentioning now as these words and concepts will be used later in the tutorial.
Below are short explanations of different types of Binary Tree structures, and below the explanations are drawings of these kinds of structures to make it as easy to understand as possible.
A balanced Binary Tree has at most 1 in difference between its left and right subtree heights, for each node in the tree.
A complete Binary Tree has all levels full of nodes, except the last level, which is can also be full, or filled from left to right. The properties of a complete Binary Tree means it is also balanced.
A full Binary Tree is a kind of tree where each node has either 0 or 2 child nodes.
A perfect Binary Tree has all leaf nodes on the same level, which means that all levels are full of nodes, and all internal nodes have two child nodes.The properties of a perfect Binary Tree means it is also full, balanced, and complete.
Binary Tree Implementation
Let's implement this Binary Tree:
The Binary Tree above can be implemented much like we implemented a Singly Linked List, except that instead of linking each node to one next node, we create a structure where each node can be linked to both its left and right child nodes.
This is how a Binary Tree can be implemented:
Example
class TreeNode: def __init__(self, data): self.data = data self.left = None self.right = None root = TreeNode('R') nodeA = TreeNode('A') nodeB = TreeNode('B') nodeC = TreeNode('C') nodeD = TreeNode('D') nodeE = TreeNode('E') nodeF = TreeNode('F') nodeG = TreeNode('G') root.left = nodeA root.right = nodeB nodeA.left = nodeC nodeA.right = nodeD nodeB.left = nodeE nodeB.right = nodeF nodeF.left = nodeG # Test print("root.right.left.data:", root.right.left.data)
Run Example »
Binary Tree Traversal
Going through a Tree by visiting every node, one node at a time, is called traversal.
Since Arrays and Linked Lists are linear data structures, there is only one obvious way to traverse these: start at the first element, or node, and continue to visit the next until you have visited them all.
But since a Tree can branch out in different directions (non-linear), there are different ways of traversing Trees.
There are two main categories of Tree traversal methods:
Breadth First Search (BFS) is when the nodes on the same level are visited before going to the next level in the tree. This means that the tree is explored in a more sideways direction.
Depth First Search (DFS) is when the traversal moves down the tree all the way to the leaf nodes, exploring the tree branch by branch in a downwards direction.
There are three different types of DFS traversals:
pre-order
in-order
post-order
These three Depth First Search traversals are described in detail on the next pages.
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3544 | https://www.khanacademy.org/science/organic-chemistry/gen-chem-review/hybrid-orbitals-jay/v/sp3-hybridized-orbitals-and-sigma-bonds | sp³ hybridized orbitals and sigma bonds (video) | Khan Academy
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Organic chemistry
Course: Organic chemistry>Unit 1
Lesson 2: Hybridization
sp³ hybridized orbitals and sigma bonds
sp² hybridized orbitals and pi bonds
sp³ hybridization
Steric number
sp² hybridization
sp hybridization
Worked examples: Finding the hybridization of atoms in organic molecules
Tetrahedral bond angle proof
Science>
Organic chemistry>
Structure and bonding>
Hybridization
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Shaun Smith 12 years ago Posted 12 years ago. Direct link to Shaun Smith's post “As the carbon needs energ...” more As the carbon needs energy to make the electron jump from "s" to "p", would the nature of this hybridization remain the same at absolute zero? I know electrons still orbit at 0 K but what would happen?
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Haecio a year ago Posted a year ago. Direct link to Haecio's post “Why is this introduced in...” more Why is this introduced in the Organic Chemistry course without a previous introduction to quantum and electron configuration?
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Samiksha Sahu a year ago Posted a year ago. Direct link to Samiksha Sahu's post “Why does the hybridized o...” more Why does the hybridized orbital have onle lope bigger than the other one??!
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Richard a year ago Posted a year ago. Direct link to Richard's post “It has to do with the int...” more It has to do with the interference between the atomic s and p orbitals. Atomic orbitals are wavefunctions, or essentially three-dimensional waves. And like waves they have amplitudes, which we refer to as phases here. Orbitals can have negative or positive phases. When orbitals interfere, they can do so constructively or destructively, just like waves. A constructive interference is when the phases of two waves are the same their amplitudes combine to yield a larger amplitude than either individual wave. Destructive interference is when two waves with different phases combine yielding a smaller amplitude than either wave.
For an s orbital, it can only have a single phase over its entire self, either positive or negative. A p orbital though has two lobes separated by a node at the center of the lobes. Each lobe has a different phase from the other. The node, in addition to being a region without any electron density, also acts as a boundary where the phase changes. An s orbital lacks such a node so it doesn’t have these multiple phases.
When s and p orbitals combine, they do so both constructively and destructively. The s orbital combines constructively with the p orbital lobe with the same phase, and simultaneously destructively with the lobe with the opposite phase. This yields a hybridized orbital with two lobes as well. The larger lobe being the result of the constructive interference, and the smaller lobe from the destructive interference.
Hope that helps.
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shreyashfs 13 years ago Posted 13 years ago. Direct link to shreyashfs's post “While drawing the structu...” more While drawing the structure of the CH4 molecule ,why doesn`t Sal draw the 1s2 orbital?
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ivanantipov86 a year ago Posted a year ago. Direct link to ivanantipov86's post “I found online that the 2...” more I found online that the 2p orbitals are actually slightly smaller than the 2s orbital (not as drawn in the video), is that right? Why then electrons in 2p orbitals hold more potential energy than those in the 2s one?
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Richard a year ago Posted a year ago. Direct link to Richard's post “It’s good to remember the...” more It’s good to remember the wave-like nature of electrons and therefore orbitals. The electrons exist in a disperse volume around the nucleus, so their distances from the nucleus vary. So we never truly know where the electron will be at any one point in time, just where it will most likely be. Which is what the orbitals are; most probably places of finding electrons. So it isn’t really a matter of an orbital being smaller or larger, rather just a greater likelihood of finding the electron closer or farther from the nucleus.
To assess this likelihood, we can use a radial distribution function which plots where an electron is most probable to be a certain distance from the nucleus. This plot can be done for each of the subshells which yield their own curve. Again because of the wave-like nature, these curves are a distribution of likely distances and are not single points. Therefore, these orbitals can have several maxima, or likely distances. For the 2p subshell, it has a maximum of about 200 pm, while the 2s subshell has a maximum of about 300 pm. This is essentially their distances within the second electron shell, which initially shows that the 2s subshell is indeed farther from the nucleus than the 2p.
However, we also have to keep in mind shielding and penetration. The 2s and 2p electrons are attracted to the protons in the nucleus, but they also feel a repulsive force from the 1s electrons shielding the 2s and 2p electrons from the full charge of the nucleus. Outer orbitals are able to penetrate into these lower orbitals and get closer to the nucleus to experience more of the positive charge. For the 2s subshell, it is able to penetrate successfully through the 1s subshell, but the 2p subshell is not able to penetrate the same 1s subshell. This is reflected on the radial distribution plot which shows that the 2s subshell has a lesser, but significant, maximum about 50 pm from the nucleus (essentially in the first shell). So because the 2s subshell can penetrate into the lower subshell, it feels more of the positive charge from the protons which reduces its potential energy. The 2p does not penetrate the 1s and is instead more effectively shielded from the positive charge giving it relatively higher potential energy.
So within the second electron shell, the 2p electrons are generally closer to the nucleus than the 2s electrons. But because of the 2s’s penetration capability, is has a significant presence closer to the nucleus too in the first shell. These two affects combine to make the 2s subshell lower in energy than the 2p. This same trend continues with other subshells (d and f) in other shells. It’s the degree of penetration which largely determines the subshell’s energy.
Hope that helps.
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Abhik Pal 12 years ago Posted 12 years ago. Direct link to Abhik Pal's post “Is there any special reas...” more Is there any special reason why the orbitals looks the way they do? also, how is the spin of an electron determined?
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Pulkit Batra 12 years ago Posted 12 years ago. Direct link to Pulkit Batra's post “Do electrons collide duri...” more Do electrons collide during performing spins in opposite directions?
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Patrick Bernardi de Freitas a year ago Posted a year ago. Direct link to Patrick Bernardi de Freitas's post “Does the sp3 hybrid lobes...” more Does the sp3 hybrid lobes stay in a pattern, like big lobe at -x, the other x?
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Samuel 2 months ago Posted 2 months ago. Direct link to Samuel's post “The sp³ hybrid orbitals f...” more The sp³ hybrid orbitals form a tetrahedral geometry, with the lobes pointing towards the corners of a tetrahedron. This arrangement minimizes electron repulsion between the orbitals. The orientation of the lobes does not strictly align with the Cartesian axes (like -x or +x) but rather forms angles of approximately 109.5° between each lobe.
Each lobe consists of one larger and one smaller portion due to the unequal contribution of the s and p orbitals in hybridization. The larger lobe is typically involved in bonding, as it overlaps more effectively with other orbitals. However, the specific orientation of the lobes depends on the molecule's geometry and bonding context.
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sylviaokafor44 4 months ago Posted 4 months ago. Direct link to sylviaokafor44's post “What are stigma bond” more What are stigma bond
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Richard 4 months ago Posted 4 months ago. Direct link to Richard's post “Do you mean sigma bonds?” more Do you mean sigma bonds?
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Natalie Price 8 years ago Posted 8 years ago. Direct link to Natalie Price's post “At 1:37, Sal says that th...” more At 1:37 , Sal says that the two arrows are opposite because the spins are opposite. There's a correction that says "Sal said 'the first electron that goes in the 1s orbital has one spin and the next electron to go in the 1s orbital has the opposite spin' but mean[t] 'the first electron that goes in the 1st orbital has 1/2 spin up and the second electron has 1/2 spin down.'" Why is Sal's statement wrong?
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Video transcript
Let's remind ourselves a little bit of what we already know about orbitals and I've gone over this early on in the regular chemistry playlist. Let's say that this is the nucleus of our atom, super small, and around that we have our first orbital, the 1s orbital. The 1s orbital, you can kind of just view it as a cloud around the nucleus. So you have your 1s orbital and it can fit two electrons, so the first electron will go into the 1s orbital and then the second electron will also go into the 1s orbital. For example, hydrogen has only one electron, so it would go into 1s. Helium has one more, so that will also go into the 1s orbital. After that is filled, then you move onto the 2s orbital. The 2s orbital, you can view it as a shell around the 1s orbital, and all of these, you can't really view it in our conventional way of thinking. You can kind of view it as a probability cloud of where you might find the electrons. But for visualization purposes, just imagine it's kind of a shell cloud around the 1s orbital. So imagine that it's kind of a fuzzy shell around the 1s orbital, so it's around the 1s orbital, and your next electron will go there. Then the fourth electron will also go there, and I drew these arrows upward and downward because the first electron that goes into the 1s orbital has one spin and then the next electron to go into 1s orbital will have the opposite spin, and so they keep pairing up in that way. They have opposite spins. Now, if we keep adding electrons, now we move to the 2p orbitals. Actually, you can view it as there are three 2p orbitals and each of them can hold two electrons, so it can hold a total of six electrons in the 2p orbitals. Let me draw them for you just so you can visualize it. So if we were to label our axis here, so think in three dimensions. So imagine that that right there is the x-axis. Let me do this in different colors. Let's say that this right here is our y-axis and then we have a z-axis. I'll do that in blue. Let's say we have a z-axis just like that. You actually have a p orbital that goes along each of those axes. So you could have your two-- let me do it in the same color. So you have your 2p sub x orbital, and so what that'll look like is a dumbbell shape that's going in the x-direction. So let me try my best attempt at drawing this. It's a dumbbell shape that goes in the x-direction, in kind of both directions, and it's actually symmetric. I'm drawing this end bigger than that end so it looks like it's coming out at you a little bit, but let me draw it a little bit better than that. I can do a better job. And maybe it comes out like that. Remember, these are really just probability clouds, but it's helpful to kind of visualize them as maybe a little bit more things that we would see in our world, but I think probability cloud is the best way to think about it. So that is the 2px orbital, and then I haven't talked about how they get filled yet, but then you also have your 2py orbital, which'll go in this axis, but same idea, kind of a dumbbell shape in the y-direction, going in both along the y-axis, going in that direction and in that direction. Then, of course, so let me do this 2py, and then you also have your 2pz, and that goes in the z-direction up like that and then downwards like that. So when you keep adding electrons, the first-- so far, we've added four electrons. If you add a fifth electron, you would expect it to go into the 2px orbital right there. So even though this 2px orbital can fit two electrons, the first one goes there. The very next one won't go into that one. It actually wants to separate itself within the p orbital, so the very next electron that you add won't go into 2px, it'll go into 2py. And then the one after that won't go into 2py or 2px, it'll go into 2pz. They try to separate themselves. Then if you add another electron, if you add-- let's see, we've added one, two, three, four, five, six, seven. If you add an eighth electron, that will then go into the 2px orbital, so the eighth electron would go there, but it would have the opposite spin. So this is just a little bit of review with a little bit of visualization. Now, given what we just reviewed, let's think about what's happening with carbon. Carbon has six electrons. Its electron configuration, it is 1s2, two electrons in the 1s orbital. Then 2s2, then 2p2, right? It only has two left, because it has a total of six electrons. Two go here, then there, then two are left to fill the p orbitals. If you go based on what we just drew and what we just talked about here, what you would expect for carbon-- let me just draw it out the way I did this. So you have your 1s orbital, your 2s orbital, and then you have your 2px orbital, your 2py orbital, and then you have your 2pz orbital. If you just go straight from the electron configuration, you would expect carbon, so the 1s orbital fills first, so that's our first electron, our second electron, our third electron. Then we go to our 2s orbital, That fills next, third electron, then fourth electron. Then you would expect maybe your fifth electron to go in the 2px. We could have said 2py or 2z. It just depends on how you label the axis. But you would have your fifth electron go into one of the p orbitals, and then you would expect your sixth to go into another. So you would expect that to be kind of the configuration for carbon. And if we were to draw it-- let me draw our axes. That is our y-axis and then this is our x-axis. Let me draw it a little bit better than that. So that is the x-axis and, of course, you have your z-axis. You have to think in three dimensions a little bit. Then you have your z-axis, just like that. So first we fill the 1s orbital, so if our nucleus is sitting here, our 1s orbital gets filled with two electrons. You can imagine that as a little cloud around the nucleus. Then we fill the 2s orbital and that would be a cloud around that, kind of a shell around that. Then we would put one electron in the 2px orbital, so one electron would start kind of jumping around or moving around, depending how you want to think about it, in that orbital over there, 2px. Then you'd have the next electron jumping around or moving around in the 2py orbital, so it would be moving around like this. If you went just off of this, you would say, you know what? These guys, this guy over here and that guy over there is lonely. He's looking for a opposite spin partner. This would be the only places that bonds would form. You would expect some type of bonding to form with the x-orbitals or the y-orbitals. Now, that's what you would expect if you just straight-up kind of stayed with this model of how things fill and how orbitals look. The reality of carbon, and I guess the simplest reality of carbon, is if you look at a methane molecule, is very different than what you would expect here. First of all, what you would expect here is that carbon would probably-- maybe it would form two bonds. But we know carbon forms four bonds and it wants to pretend like it has eight electrons. Frankly, almost every atom wants to pretend like it has eight electrons. So in order for that to happen, you have to think about a different reality. This isn't really what's happening when carbon bonds, so not what happens when carbon bonds. What's really happening when carbon bonds, and this will kind of go into the discussion of sp3 hybridization, but what you're going to see is it's not that complicated of a topic. It sounds very daunting, but it's actually pretty straightforward. What really happens when carbon bonds, because it wants to form four bonds with things, is its configuration, you could imagine, looks more like this. So you have 1s. We have two electrons there. Then you have your 2s, 2px, 2py and 2pz. Now what you can imagine is it wants to form four bonds. It has four electrons that are willing to pair up with electrons from other molecules. In the case of methane, that other molecule is a hydrogen. So what you could imagine is that the electrons actually-- maybe the hydrogen brings this electron right here into a higher energy state and puts it into 2z. That's one way to visualize it. So this other guy here maybe ends up over there, and then these two guys are over there and over there. Now, all of a sudden, it looks like you have four lonely guys and they are ready to bond, and that's actually more accurate of how carbon bonds. It likes to bond with four other people. Now, it's a little bit arbitrary which electron ends up in each of these things, and even if you had this type of bonding, you would expect things to bond along the x, y, and z axis. The reality is, the reality of carbon, is that these four electrons in its second shell don't look like they're in just-- the first one doesn't look like it's just in the s orbital and then the p and y and z for the other three. They all look like they're a little bit in the s and a little bit in the p orbitals. Let me make that clear. So instead of this being a 2s, what it really looks like for carbon is that this looks like a 2sp3 orbital. This looks like a 2sp3 orbital, that looks like a 2sp3 orbital, that looks like a 2sp3 orbital. They all look like they're kind of in the same orbital. This special type of-- it sounds very fancy. This sp3 hybridized orbital, what it actually looks like is something that's in between an s and a p orbital. It has a 25% s nature and a 75% p nature. You can imagine it as being a mixture of these four things. That's the behavior that carbon has. So when you mix them all, instead of having an s orbital, so if this is a nucleus and we do a cross-section, an s orbital looks like that and the p orbital looks something like that in cross-section. So this is a an s and that is a p. When they get mixed up, the orbital looks like this. An sp3 orbital looks something like this. This is a hybridized sp3 orbital. Hybrid just means a combination of two things. A hybrid car is a combination of gas and electric. A hybridized orbital is a combination of s and p. Hybridized sp3 orbitals are the orbitals when carbon bonds with things like hydrogen or really when it bonds with anything. So if you looked at a molecule of methane, and people talk about sp3 hybridized orbitals, all they're saying is that you have a carbon in the center. Let's say that's the carbon nucleus right there. And instead of having one s and three p orbitals, it has four sp3 orbitals. So let me try my best at drawing the four sp3 orbitals. Let's say this is the big lobe that is kind of pointing near us, and then it has a small lobe in the back. Then you have another one that has a big lobe like that and a small lobe in the back. Then you have one that's going back behind the page, so let me draw that. You can kind of imagine a three-legged stool, and then its small lobe will come out like that. And then you have one where the big lobe is pointing straight up, and it has a small lobe going down. You can imagine it as kind of a three-legged stool. One of them is behind like that and it's pointing straight up, So a three-legged stool with something-- it's kind of like a tripod, I guess is the best way to think about it. So that's the carbon nucleus in the center and then you have the hydrogens, so that's our carbon right there. Then you have your hydrogens. You have a hydrogen here. A hydrogen just has one electron in the 1s orbital, so the hydrogen has a 1s orbital. You have a hydrogen here that just has a 1s orbital. It has a hydrogen here, 1s orbital, hydrogen here, 1s orbital. So this is how the hydrogen orbital and the carbon orbitals get mixed. The hydrogens 1s orbital bonds with-- well, each of the hydrogen's 1s orbital bonds with each of the carbon's sp3 orbitals. Just so you get a little bit more notation, so when people talk about hybridized sp3 orbitals, all they're saying is, look, carbon doesn't bond. Once carbon-- this right here is a molecule of methane, right? This is CH4, or methane, and it doesn't bond like you would expect if you just want with straight vanilla s and p orbitals. If you just went with straight vanilla s and p orbitals, the bonds would form. Maybe the hydrogen might be there and there, and if it had four hydrogens, maybe there and there, depending on how you want to think about it. But the reality is it doesn't look like that. It looks more like a tripod. It has a tetrahedral shape. The best way that that can be explained, I guess the shape of the structure, is if you have four equally-- four of the same types of orbital shapes, and those four types of orbital shapes are hybrids between s's and p's. One other piece of notation to know, sometimes people think it's a very fancy term, but when you have a bond between two molecules, where the orbitals are kind of pointing at each other, so you can imagine right here, this hydrogen orbital is pointing in that direction. This sp3 orbital is pointing that direction, and they're overlapping right around here. This is called a sigma bond, where the overlap is along the same axis as if you connected the two molecules. Over here, you connect the two molecules, the overlap is on that same axis. This is the strongest form of covalent bonds, and this'll be a good basis for discussion maybe in the next video when we talk a little bit about pi bonds. The big takeaway of this video is to just understand what does it mean? What is an sp3 hybridized orbital? Nothing fancy, just a combination of s and p orbitals. It has 25% s character, 75% p character, which makes sense. It's what exists when carbon forms bonds, especially in the case of methane. That's what describes it's tetrahedral structure. That's why we have an angle between the various branches of a 109.5 degrees, which some teachers might want you know, so it's useful to know. If you take this angle right here, 109.5, that's the same thing as that angle, or if you were to go behind it, that angle right there, 109.5 degrees, explained by sp3 hybridization. The bonds themselves are sigma bonds. The overlap is along the axis connecting the hydrogen.
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3545 | https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators | Skip to main content
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Web
JavaScript
Reference
Expressions and operators
Expressions and operators
This chapter documents all the JavaScript language operators, expressions and keywords.
Expressions and operators by category
For an alphabetical listing see the sidebar on the left.
Primary expressions
Basic keywords and general expressions in JavaScript. These expressions have the highest precedence (higher than operators).
this
: The this keyword refers to a special property of an execution context.
Literals
: Basic null, boolean, number, and string literals.
[]
: Array initializer/literal syntax.
{}
: Object initializer/literal syntax.
function
: The function keyword defines a function expression.
class
: The class keyword defines a class expression.
function
: The function keyword defines a generator function expression.
async function
: The async function defines an async function expression.
async function
: The async function keywords define an async generator function expression.
/ab+c/i
: Regular expression literal syntax.
string
: Template literal syntax.
( )
: Grouping operator.
Left-hand-side expressions
Left values are the destination of an assignment.
Property accessors
: Member operators provide access to a property or method of an object (object.property and object["property"]).
?.
: The optional chaining operator returns undefined instead of causing an error if a reference is nullish (null or undefined).
new
: The new operator creates an instance of a constructor.
new.target
: In constructors, new.target refers to the constructor that was invoked by new.
import.meta
: An object exposing context-specific metadata to a JavaScript module.
super
: The super keyword calls the parent constructor or allows accessing properties of the parent object.
import()
: The import() syntax allows loading a module asynchronously and dynamically into a potentially non-module environment.
Increment and decrement
Postfix/prefix increment and postfix/prefix decrement operators.
A++
: Postfix increment operator.
A--
: Postfix decrement operator.
++A
: Prefix increment operator.
--A
: Prefix decrement operator.
Unary operators
A unary operation is an operation with only one operand.
delete
: The delete operator deletes a property from an object.
void
: The void operator evaluates an expression and discards its return value.
typeof
: The typeof operator determines the type of a given object.
+
: The unary plus operator converts its operand to Number type.
-
: The unary negation operator converts its operand to Number type and then negates it.
~
: Bitwise NOT operator.
!
: Logical NOT operator.
await
: Pause and resume an async function and wait for the promise's fulfillment/rejection.
Arithmetic operators
Arithmetic operators take numerical values (either literals or variables) as their operands and return a single numerical value.
: Exponentiation operator.
: Multiplication operator.
/
: Division operator.
%
: Remainder operator.
+ (Plus)
: Addition operator.
-
: Subtraction operator.
Relational operators
A comparison operator compares its operands and returns a boolean value based on whether the comparison is true.
< (Less than)
: Less than operator.
> (Greater than)
: Greater than operator.
<=
: Less than or equal operator.
>=
: Greater than or equal operator.
instanceof
: The instanceof operator determines whether an object is an instance of another object.
in
: The in operator determines whether an object has a given property.
Note:
=> is not an operator, but the notation for Arrow functions.
Equality operators
The result of evaluating an equality operator is always of type boolean based on whether the comparison is true.
==
: Equality operator.
!=
: Inequality operator.
: Strict equality operator.
!==
: Strict inequality operator.
Bitwise shift operators
Operations to shift all bits of the operand.
<<
: Bitwise left shift operator.
>>
: Bitwise right shift operator.
>>>
: Bitwise unsigned right shift operator.
Binary bitwise operators
Bitwise operators treat their operands as a set of 32 bits (zeros and ones) and return standard JavaScript numerical values.
&
: Bitwise AND.
|
: Bitwise OR.
^
: Bitwise XOR.
Binary logical operators
Logical operators implement boolean (logical) values and have short-circuiting behavior.
&&
: Logical AND.
||
: Logical OR.
??
: Nullish Coalescing Operator.
Conditional (ternary) operator
(condition ? ifTrue : ifFalse)
: The conditional operator returns one of two values based on the logical value of the condition.
Assignment operators
An assignment operator assigns a value to its left operand based on the value of its right operand.
=
: Assignment operator.
=
: Multiplication assignment.
/=
: Division assignment.
%=
: Remainder assignment.
+=
: Addition assignment.
-=
: Subtraction assignment
<<=
: Left shift assignment.
>>=
: Right shift assignment.
>>>=
: Unsigned right shift assignment.
&=
: Bitwise AND assignment.
^=
: Bitwise XOR assignment.
|=
: Bitwise OR assignment.
=
: Exponentiation assignment.
&&=
: Logical AND assignment.
||=
: Logical OR assignment.
??=
: Nullish coalescing assignment.
[a, b] = arr, { a, b } = obj
: Destructuring allows you to assign the properties of an array or object to variables using syntax that looks similar to array or object literals.
Yield operators
yield
: Pause and resume a generator function.
yield
: Delegate to another generator function or iterable object.
Spread syntax
...obj
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3546 | https://wtcs.pressbooks.pub/healthalts/chapter/11-12-irritable-bowel-syndrome/ | 11.12 Irritable Bowel Syndrome – Health Alterations
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Contents
Introduction
Preface
Standards and Conceptual Approach
I. Chapter 1 Leadership Principles
1.1 Leadership Introduction
1.2 Leadership Skills
1.3 Leadership Styles
1.4 Characteristics of Effective Leadership
1.5 Conflict Resolution
1.6 Constructive Feedback
1.7 Managing the Nursing Team
1.8 Delegation and Supervision
1.9 Prioritization Strategies
1.10 Time Management
1.11 Learning Activities
I Glossary
II. Chapter 2 Perioperative Care
2.1 Perioperative Introduction
2.2 Basic Concepts Related to Surgery
2.3 Preoperative Nursing Care
2.4 Intraoperative Nursing Care
2.5 Postoperative Nursing Care
2.6 Learning Activities
II Glossary
III. Chapter 3 Hematological Alterations
3.1 Hematology Introduction
3.2 Review of Hematology Anatomy & Physiology
3.3 General Hematological System Assessment
3.4 Common Hematological Disorders
3.5 Anemia
3.6 Iron-Deficiency Anemia
3.7 Vitamin B12 and Folate Deficiency Anemia
3.8 Sickle Cell Disease
3.9 Polycythemia
3.10 Thrombocytopenia
3.11 Leukopenia
3.12 Aplastic Anemia
3.13 Learning Activities
Glossary
IV. Chapter 4 Malignancy and Autoimmune Alterations
4.1 Autoimmune and Malignancy Introduction
4.2 Review of Anatomy and Physiology of the Immune System
4.3 Cancer
4.4 Applying the Nursing Process to Cancer Treatment
4.5 Autoimmune and Hypersensitivity Reactions
4.6 Systemic Lupus Erythematosus
4.7 Learning Activities
IV Glossary
V. Chapter 5 Cardiovascular Alterations
5.1 Cardiovascular Introduction
5.2 Review of Anatomy & Physiology of the Cardiovascular System
5.3 General Cardiovascular System Assessment
5.4 Common Cardiovascular Disorders
5.5 Hypertension
5.6 Arteriosclerosis & Atherosclerosis
5.7 Coronary Artery Disease
5.8 Heart Failure
5.9 Peripheral Arterial Disease
5.10 Venous Insufficiency
5.11 Deep Vein Thrombosis
5.12 Aneurysm
5.13 Infective Endocarditis
5.14 Learning Activities
V Glossary
VI. Chapter 6 Respiratory Alterations
6.1 Respiratory Introduction
6.2 Review of Anatomy and Physiology of the Respiratory System
6.3 General Respiratory System Assessment and Interventions
6.4 Common Respiratory Disorders
6.5 Asthma
6.6 COPD
6.7 Pneumonia
6.8 Lung Cancer
6.9 Tuberculosis
6.10 Respiratory Viral Infections
6.11 Other Disorders of Head and Neck
6.12 Learning Activities
VI Glossary
VII. Chapter 7 Endocrine Alterations
7.1 Endocrine Introduction
7.2 Review of Anatomy and Physiology of the Endocrine System
7.3 General Endocrine System Assessment
7.4 Endocrine Disorders
7.5 Diabetes Mellitus
7.6 Thyroid Disorders
7.7 Adrenal Disorders
7.8 Parathyroid Disorders
7.9 Learning Activities
VII Glossary
VIII. Chapter 8 Renal and Urinary System Alterations
8.1 Renal Introduction
8.2 Review of Anatomy & Physiology of the Urinary System
8.3 General Urinary System Assessment
8.4 Common Urinary System Disorders
8.5 Acute Renal Failure
8.6 Chronic Kidney Disease
8.7 Cystitis
8.8 Pyelonephritis
8.9 Glomerulonephritis
8.10 Urinary Incontinence
8.11 Urolithiasis
8.12 Benign Prostate Hypertrophy
8.13 Bladder and Prostate Cancer
8.14 Learning Activities
VIII Glossary
IX. Chapter 9 Nervous System Alterations
9.1 Neurological Introduction
9.2 Review of Anatomy & Physiology of the Nervous System
9.3 General Nervous System Assessment
9.4 Common Neurological Conditions
9.5 Alzheimer's Disease
9.6 Parkinson’s Disease
9.7 Seizures and Epilepsy
9.8 Meningitis
9.9 Cerebrovascular Accident
9.10 Multiple Sclerosis
9.11 Myasthenia Gravis
9.12 Other Nervous System Disorders
9.13 Learning Activities
IX Glossary
X. Chapter 10 Musculoskeletal System Alterations
10.1 Musculoskeletal Introduction
10.2 Review of Anatomy & Physiology of the Musculoskeletal System
10.3 General Musculoskeletal System Assessment
10.4 Common Musculoskeletal Disorders
10.5 Amputation
10.6 Fracture
10.7 Osteoarthritis
10.8 Osteoporosis
10.9 Rheumatoid Arthritis
10.10 Other Musculoskeletal Disorders
10.11 Learning Activities
X Glossary
XI. Chapter 11 Gastrointestinal System Alterations
11.1 Gastrointestinal Introduction
11.2 Review of Anatomy & Physiology of the Gastrointestinal System
11.3 General Assessment of the Gastrointestinal System
11.4 Gastrointestinal Disorders
11.5 Appendicitis
11.6 Cholecystitis
11.7 Gastroenteritis
11.8 Diverticulitis
11.9 Gastroesophageal Reflux Disease
11.10 Peptic Ulcer Disease
11.11 Inflammatory Bowel Disease
11.12 Irritable Bowel Syndrome
11.13 Bowel Obstruction
11.14 Hernia
11.15 Hepatitis
11.16 Other Gastrointestinal Disorders
11.17 Learning Activities
XI Glossary
XII. Chapter 12 Transition to Practice
12.1 Transition to Practice Introduction
12.2 Preparing for the NCLEX-PN
12.3 Obtaining Your Nursing License
12.4 Applying for a Nursing Position
12.5 Transitioning to the PN Role
12.6 Lifelong Learners
12.7 Learning Activities
XII Glossary
XIII. Answer Key
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Appendix A - Normal Reference Ranges
Appendix B - Overall Glossary
Health Alterations
11.12 Irritable Bowel Syndrome
Irritable bowel syndrome (IBS) is a chronic gastrointestinal disorder in which there is abdominal discomfort, along with a change in bowel patterns. There are three different subclasses of IBS: IBS-C, IBS-D, and IBS-M. In IBS-C, constipation is the predominant stool pattern. In IBS-D, diarrhea is the predominant stool pattern. With IBD-M, there is a mix of bowel movement types, alternating between diarrhea and constipation.
IBS is most common in women younger than 50 years old. Additionally, clients with anxiety, depression, or increased stress are more likely to experience IBS. Family history/genetics is a risk factor, and IBS is most common in clients from South America and least common in clients from Southeast Asia. Certain foods can also trigger IBS symptoms, such as dairy products; gas-producing foods such as cabbage and beans; carbonated beverages; and citrus fruits.
Pathophysiology
IBS is a functional disorder, meaning there are symptoms present but there is no correlating gastrointestinal disease. The cause of IBS is unknown, but there are many theories regarding the development of this disorder such as the following,:
Alterations in intestinal muscle contractions, with increased muscle contractions causing diarrhea and decreased or weak contractions causing constipation.
Inappropriate nervous system signaling in the GI tract.
Alterations in the normal intestinal flora when compared to clients without IBS.
Changes in the immune response of the gastrointestinal tract.
Environmental factors such as significant stress early in life, antibiotic use, infections such as gastroenteritis, and the inability to tolerate certain foods.
Assessment
Physical Exam
Common signs and symptoms of IBS include the following:
Abdominal pain
A change in bowel habits such as diarrhea, constipation, or alternating diarrhea and constipation
Abdominal distention or bloating
Correlation of symptoms with intake of certain foods that are relieved by bowel movements
There is no specific diagnostic test for IBS, but it is typically diagnosed by a health care provider using Rome IV criteria.Criteria include recurrent abdominal pain at least one day per week associated with two or more of the following symptoms over a three-month time span:
Abdominal pain that improves after having a bowel movement
Alterations in stool frequency
Alterations in the appearance of stool
Common Laboratory and Diagnostic Tests
There are no diagnostic tests for IBS. However, if other alarming symptoms are present, such as blood in stool or unexpected weight loss, blood tests such as a complete blood count or metabolic panel may be done to rule out other causative factors. If diarrhea is an associated symptom, stool tests may be ordered to rule out an infectious cause. A colonoscopy may also be ordered, especially if there is a family history of gastrointestinal cancer or IBD.
Nursing Diagnosis
Nursing priorities for those suffering from irritable bowel syndrome include managing symptoms, promoting good nutrition, and preventing dehydration.
Nursing diagnoses for clients with irritable bowel syndrome are created based on the specific needs of the client, their signs and symptoms, and the etiology of the disorder. These nursing diagnoses guide the creation of client specific care plans that encompass client outcomes and nursing interventions, as well the evaluation of those outcomes. These individualized care plans then serve as a guide for client treatment.
Possible nursing diagnoses for those with irritable bowel syndrome are as follows,:
Acute Pain
Constipation
Diarrhea
Imbalanced Nutrition: Less than Body Requirements
Risk for Fluid Volume Deficit
Outcome Identification
Outcome identification encompasses the creation of short- and long-term goals for the client. These goals are used to create expected outcome statements that are based on the specific needs of the client. Expected outcomes should be specific, measurable, and realistic. These outcomes should be achievable within a set time frame based on the application of appropriate nursing interventions.
Sample expected outcomes include the following:
The client will rate their pain at 3 or less on a scale of 0 to 10 on a daily basis.
The client will pass bowel movements without straining until their next scheduled appointment.
The client will exhibit a formed stool that occurs at the client’s normal frequency until their next scheduled appointment.
The client will maintain a weight within a healthy range that is appropriate for their height over the next six months.
The client will exhibit blood pressure and heart rate within normal limits for age, moist mucous membranes, and urine output appropriate for age until their next scheduled appointment.
Interventions
Medical Interventions
The goal of IBS treatment is to control symptoms that are experienced by the client. This can be accomplished with lifestyle changes and the use of medications.
Lifestyle Changes
Increased physical activity can decrease symptoms because it increases the amount of time it takes for food to move through the colon.
Avoid foods that are poorly absorbed that increase symptoms such as wheat products, onions, dairy, and some fruits and vegetables.
Medications
For clients with constipation as the predominant symptom, fiber supplements and/or laxatives may be prescribed.
For clients with diarrhea as the predominant symptom, antidiarrheals and/or probiotics may be prescribed.
Low doses of antidepressants such as tricyclic antidepressants (TCAs) or selective serotonin reuptake inhibitors (SSRIs) can be helpful for clients with severe symptoms.
Alosetron has shown symptom improvement in women with IBS-D.
Rifaximin, an antibiotic, has also reduced the occurrence of diarrhea and abdominal pain in those with IBS.
Nursing Interventions
When providing nursing care to a client diagnosed with IBS, nursing interventions can be divided into nursing assessments, nursing actions, and client teaching,:
Nursing Assessments
Assess the client’s daily intake of food to establish a baseline of their dietary status.
Assess the client’s weight daily to monitor their progress towards weight-related goals.
Assess the frequency, consistency, and amount of stool.
Monitor electrolytes levels and replenish if necessary if diarrhea is present.
Assess for signs of dehydration due to fluid loss: dry skin and mucous membranes, increased heart rate, low blood pressure, and decreased urine output.
Nursing Actions
Administer IBS medications as prescribed by the provider to help manage symptoms.
Encourage referral to gastroenterologist because they can individualize the client’s IBS treatment plan.
Encourage referral to a dietician to help create a dietary plan that is client specific.
Encourage oral fluid intake/administer intravenous fluids to prevent/treat dehydration when diarrhea is present. Cold liquids should be avoided as they can increase gastrointestinal motility.
Client Teaching
Avoid food triggers, such as gas-producing foods, gluten, and carbohydrates that contain lactose and fructose. A diet high in fiber and increased fluid intake should be encouraged if constipation is an issue.
Use stress management strategies (such as deep breathing and meditation) to help decrease symptoms.
Keep a food diary to help identify personal food triggers.
For clients with IBS-C, monitor stool pattern and frequency to identify when additional medications are needed to treat constipation.
Evaluation
Evaluation of client outcomes refers to the process of determining whether or not client outcomes were met by the indicated time frame. This is done by reevaluating the client as a whole and determining if their outcomes have been met, partially met, or not met. If the client outcomes were not met in their entirety, the care plan should be revised and reimplemented. Evaluation of outcomes should occur each time the nurse assesses the client, examines new laboratory or diagnostic data, or interacts with another member of the client’s interdisciplinary team.
View a supplementary YouTube video on IBS: Irritable bowel syndrome (IBS) – causes, symptoms, risk factors, treatment, pathology,
RN Recap: Irritable Bowel Syndrome
View a brief YouTube video overview of irritable bowel syndrome:
Patel, N., & Shackelford, K. B. (2022).Irritable Bowel Syndrome. StatPearls [Internet].
Curran, A. (2023). Irritable bowel syndrome (IBS) nursing diagnosis and care plan.
Patel, N., & Shackelford, K. B. (2022).Irritable Bowel Syndrome. StatPearls [Internet].
Patel, N., & Shackelford, K. B. (2022).Irritable Bowel Syndrome. StatPearls [Internet].
Patel, N., & Shackelford, K. B. (2022).Irritable Bowel Syndrome. StatPearls [Internet].
Herdman, T. H., Kamitsuru, S., & Lopes, C. T. (Eds.). (2020). Nursing diagnoses: Definitions and classification, 2021-2023 (12th ed.). Thieme. ↵
Curran, A. (2023). Irritable bowel syndrome (IBS) nursing diagnosis and care plan.
Patel, N., & Shackelford, K. B. (2022).Irritable Bowel Syndrome. StatPearls [Internet].
Patel, N., & Shackelford, K. B. (2022).Irritable Bowel Syndrome. StatPearls [Internet].
Curran, A. (2023). Irritable bowel syndrome (IBS) nursing diagnosis and care plan.
Osmosis from Elsevier. (2022, September 13). Irritable bowel syndrome (IBS) - causes, symptoms, risk factors, treatment, pathology[Video]. YouTube. All rights reserved.
Open RN Project. (2024, June 23). Health Alterations - Chapter 11 - Irritable bowel syndrome[Video]. You Tube. CC BY-NC 4.0
definition
A chronic gastrointestinal disorder in which there is abdominal discomfort, along with a change in bowel patterns.
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A set of diagnostic criteria utilized to help diagnose IBS.
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3547 | https://brainly.com/question/28140758 | [FREE] Which formula gives the zeros of y = \sin(x) ? A. k\pi for any positive integer k B. k\pi for any - brainly.com
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Which formula gives the zeros of y=sin(x)?
A. kπ for any positive integer k
B. kπ for any integer k
C. 2 kπ for any positive integer k
D. 2 kπ for any integer k
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The correct formula that gives the zeros of y=sin(x) is kπ for any integer k.
To understand why this is the correct formula, let's analyze the sine function and its properties. The sine function, y=sin(x), has zeros where the value of the sine is zero. The sine function equals zero at integer multiples of π along the x-axis. This is because the sine function represents the y-coordinate of the unit circle, and it equals zero when the angle x corresponds to the horizontal axis of the circle, which occurs at 0,π,2 π,3 π,… radians, and also at −π,−2 π,−3 π,… radians.
Expressing these angles as multiples of π, we get x=kπ, where k is any integer (positive or negative). This ensures that all possible angles where the sine function equals zero are included.
The other options provided in the question are incorrect for the following reasons:
kπ for any positive integer kπ: This option excludes the negative multiples of π where the sine function also equals zero.
2 kπ for any positive integer k: This formula actually gives the points where the sine function has local maxima and minima, not where it equals zero.
2 kπ for any integer : Similar to the previous point, this formula gives the points where the sine function has local maxima, minima, and zeros. However, it includes too many points, as it also includes the points where the cosine function equals zero, which is not what we are looking for when finding the zeros of the sine function alone.
Therefore, the correct formula that gives the zeros of y=sin(x) is kπ for any integer k.
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The formula that gives the zeros of y=sin(x) is kπ for any integer k. This is because the sine function is zero at integer multiples of π. Therefore, the correct choice is option B.
Explanation
The correct formula that gives the zeros of y=sin(x) is kπ for any integer k.
To understand why this formula works, let's look at the sine function and its behavior. The sine function, y=sin(x), is zero at specific angles, which correspond to the points where the y-coordinate of the unit circle is zero. This happens at all integer multiples of π radian angles along the x-axis, where x \ can be values like 0,π,2 π,−π,−2 π, and so on.
When we indicate this relationship mathematically, we express the angles where the sine function equals zero as x=kπ, where k can be any integer (this includes both positive and negative integers, as well as zero).
The other multiple-choice options are incorrect for the following reasons:
A. kπ for any positive integer k: This excludes the negative values and zero, which are also zeros of the sine function.
C. 2 kπ for any positive integer k: This formula indicates points where the sine function reaches its maximum and minimum values (not where it is zero).
D. 2 kπ for any integer k: While this does include some zeros, it also includes points that correspond to the maximum and minimum values of both sine and cosine functions, leading to inaccuracies in identifying just the zeros of y=sin(x).
Thus, the correct answer is B.kπ for any integer k.
Examples & Evidence
Consider values such as k=0 (which gives x=0), k=1 (which gives x=π), and k=−1 (which gives x=−π). All these angles are where sin(x)=0.
The behavior of the sine function can be confirmed through its periodic nature and symmetry on the unit circle, where it crosses the x-axis precisely at integer multiples of π.
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Community Answer 4.6 53 Which formula gives the x-coordinates of the maximum values for y = cos(x)? k pi for any integer k k pi for k = 0, plus-or-minus 2, plus-or-minus 4, ellipsis StartFraction k pi Over 2 EndFraction for any positive integer k StartFraction k pi Over 2 EndFraction for k = 0, plus-or-minus 2, plus-or-minus 4, ellipsis
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
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish?
Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit?
Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25
Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points.
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Complex Numbers, The Fundamental Theorem of Algebra, & Euler's Formula
I had originally intended to write a blog post encompassing all the fundamental theorems in the fields of mathematics that I've studied. But part way through, specifically when I got to the section about the Fundamental Theorem of Algebra (surprise surprise), I realized I just had so much to say about complex numbers that I couldn't refrain from writing a whole blog post about that alone. So here it is.
A complex number is a number that has both a Real and an Imaginary part. That is it has 'length' residing along the Real number line (the usual numbers we're all familiar with), what we call the Real axis, and a 'height' residing along an axis perpendicular to that Real number line, which we call the Imaginary axis.
In fact, all the numbers you usually deal with are complex numbers, just with a vanishing Imaginary component. It is truly unfortunate that Imaginary numbers are termed "Imaginary" - they are, in fact, very real. Carl Friedrich Gauss himself considered the terminology horrendous and suggested we call Imaginary numbers "lateral numbers" to indicate that they reside laterally to the Real number line.
Here's a little spatial exercise of the mind that may assist you in building the intuition for complex numbers. Consider yourself moving to the right at, say, 1 m/s. We will now denote the right orientation to be the positive velocities and the moving left to be negative velocities. So, if instead, you were moving to the left you would be moving at -1 m/s. So, algebraically what operation would we have to perform to turn from right to left, that is, what do I have to do algebraically to 1 m/s to make it equal to -1 m/s (caveat, it has to work regardless of my speed in the right direction - so if I moved at 2 m/s, what needs to be performed algebraically to have me move -2 m/s).
Hopefully you all realized the answer is to multiply 1 m/s by -1 to get -1 m/s. All fine and dandy, now what about the reverse. I am now moving at -1 m/s, that is to the left at a speed of 1 m/s. What do I have to do algebraically when I am moving -1 m/s to move at 1 m/s - that is how do go from moving left at 1 m/s to moving right at 1 m/s.
Well, it's the same as before, multiply -1 m/s by -1 to get 1 m/s. That is, multiplying by -1 can be seen as an operation that causes a 180 degree rotation in the direction of my movement (180 degree rotation is essentially a reflection along the left-to-right line).
Alright, well now, what if I only wanted to turn 90 degrees? That is, I'm moving 1 m/s to the right, what if I wanted to move 1 m/s to the north? Say you must perform this algebraically, but just how the heck are you supposed to do this with an algebraic operation?
Well, let's be clever about this. What do we know? We know that multiplying by -1 will rotate us 180 degrees. And multiplying by -1 again will rotate us back to our original position so multiplying by -1 twice is the same as adding two 180 degree rotations together. So what we want to do is multiply to 1 m/s by some number to move 1 m/s northwards, that is, a 90 degree rotation. Well, if multiplying by this number is equivalent to a 90 degree rotation, then if we were to multiply 1 m/s northwards by this number again, then we would've turned 180 degrees since multiplying by these rotation numbers is the same as adding the degrees of rotation. So the square of this number is -1! That is, a 90 degree rotation is equivalent to multiplication by the square root of -1, which we've come to call i for Imaginary (unfortunately).
Now, if you've been staying sharp, you would've noticed there's another way I could've achieved a 180 degree rotation in two equivalent multiplications. That is, I've only now went clockwise to 90 degrees, but I could've just as well gone 270 degrees first (or -90 degrees) then to 180 degrees. That is, I could've traversed the plane clockwise rather than counterclockwise. I would've then gone from 1 m/s to the right to 1 m/s to the south. Now we've already defined 1 m/s to the north to be i m/s and so 1 m/s to the south must be equal to i multiplied by -1 (it's respective 180 degree rotation), or -i m/s. However, this choice was completely arbitrary, I could've just as well said moving south is positive i and moving north is negative i. And so it's true, that ANY equation written with complex numbers, is EQUALLY TRUE if every single i is replaced by -i and vice versa - this is what's called taking the complex conjugate and it's used intimately in quantum mechanics as well as in the study of other cyclical phenomena.
[maybe another graphic...]
Now that you've abstracted to this point, it's only a little more to be able to figure out the numbers that must be multiplied to achieve rotation to an arbitrary direction in the plane (the complex plane). Lets say you wanted to make a 45 degree rotation from 1 m/s (moving right with speed 1 m/s). Well we know i is a 90 degree rotation, so the number corresponding to the 45 degree rotation squared is 90 degrees since multiplying by this number twice would add up to a 90 degree rotation. That is, a 45 degree rotation is the square root of i m/s...But wait, it looks like we are again stuck, we seemingly have to invent another number - the square root of i. Fear not, it turns out this is not the case, the square root of a complex number, is still a complex number, we do not have to invent any new numbers.
I will showcase two methods of finding out this number that produces a 45 degree rotation. First, the ugly method. We will straight-up ASSUME that the square root of i is itself a complex number, that is, it can be written in the form a + ib for some real numbers a and b. So,
i = (a + ib)2
Now simply FOIL the right side out following the definitions we've made for i - namely, i 2 = -1. So,
i = a2 + i 2ab - b2 i = ( a2 - b2 ) + i 2ab
Now, the Real component must be equal to zero because i has no Real component, hence
0 = a2 - b2 a2 = b2 or (+/-) a = (+/-) b
Similarly,
i = i 2ab 1 = 2ab
So with,
1 = 2ab a = b
(let's just use a only now)
we get, 1 = 2a2 a = +/- Sqrt(1/2)
So, the
Sqrt(i) = +/- ( 1/Sqrt(2) + i (1/Sqrt(2)) ).
Take some time to FOIL this out to convince yourself it is correct. Now, here's a more visual method to solve for the 45 degree rotation, that is the Sqrt( i ).
Notice that we are still within the plane! So if we believe that this visual representation is correct, then we can simply follow along the Real axis and Imaginary axis to find the lengths of the components that lay along each. And this is just simple trigonometry. I will spare you the details, a vector at 45 degrees with a length of one, will have a horizontal (Real) component of length 1/Sqrt(2) and the same 1/Sqrt(2) for the vertical (Imaginary) component. And we are done! Simple isn't it?
The same methods can be applied to find the complex number corresponding to any rotation within the complex plane. And this leads us to the Fundamental Theorem of Algebra.
The Fundamental Theorem of Algebra states that "every non-constant single-variable polynomial with complex coefficients has at least one complex root" OR "every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n roots" (via wikipedia). Equivalently, what this theorem says is that the field of complex numbers is algebraically closed. Perhaps the most compelling viewpoint is the visual one - that what this theorem really tells us is that algebraic operations on the complex can be seen as merely translations between points in the complex plane - hence, closed. You cannot leave the field of the complex plane with algebraic operations - the complex number is the most generic number.
and
Not only that, this theorem was the first historical theorem that truly linked the otherwise disparate fields of algebra and geometry.
In fact, the allure doesn't stop there. If you can recall from elementary geometry we define the equation of a circle to be
r2 = x2 + y2
where r is the radius of the circle and the two arguments are the length and height of the point on the circle (x and y, respectively). However, there is another way the equation of a circle can be represented, which you should also have learned from elementary geometry - and that is parametrically where the argument is the radius and the angle (r and t) with the trigonometric functions sine and cosine (often called polar coordinate representation).
x = r cos(t) y = r sin(t)
And this is where the mind-boggling Euler Formula comes into play. Those who've seen or heard of the Euler Formula will know where I'm going with this. I will first write out a version of the Euler Formula:
e i τ = 1 where τ = 2 π
The standard Euler Equation is written
e i π = -1 or via the variable τ e i (τ/2) = -1
Recall that π radians is equal to 180 degrees (a half rotation) and 2 π radians is equal to 360 degrees (a full rotation). And look what we have! e to i times 180 degrees is equal to -1, a half rotation! And e to the i times 360 degrees is equal to 1, a full rotation! Just as how we've been describing earlier! We've now come full circle (pun intended).
Unfortunately, diving deep into how this is actually proven is beyond the scope of this article. But what I want you to understand out of this is that what is truly fascinating about the number e (besides it's rather boring, but usual introduction as a way of calculating compound interest) is that it is THE NATURAL (UNIT) ROTATION OPERATOR IN THE COMPLEX PLANE!
Just as how in elementary geometry, any point along the unit circle (r = 1) can be written parametrically as
x = cos(t) y = sin(t)
The same can be achieved in only one expression in the complex plane:
ei t = cos(t) + i sin(t)
In fact, every complex number in the complex plane can be written equivalently in the form
r ei t
where r is the distance of the point from the origin and t is the angle the line to the point makes with the Real axis.
So any complex number a + i b can be written as r ei t which, for those with a stronger mathematical background, is essentially the polar coordinate representation in the complex plane.
I will once again re-write Euler's Formula but in a more all-encompassing form:
ei τ = 1 + i 0
This equation relates the five most important constants in all of mathematics: 0, 1, e, i, and τ (actually 2π, but more on that shortly). Richard Feynman, one of the most brilliant physicists in all of the 21st century called this formula our "our jewel" and the "most important formula in all of mathematics". It is used extensively in all of theoretical physics, and not only there, it is pervasive in almost all of engineering because of its intimate connection with cyclical phenomena (which encompasses practically everything).
Enough about that, I will now segue briefly into my usage of the variable τ = 2π. You may be wondering why I have introduced this arbitrary new constant τ into my explanation of the Euler Formula. The point of the matter is, we have been brought up learning the wrong circle constant all along.
π is the wrong circle constant. There is nothing inherently wrong with the number π, but a circle is defined not by it's diameter, but by it's radius. The original derivation of the number π is that π times the diameter of a circle is equal to the circumference of the circle. However, what really defines a circle is its radius. So π d = C should really be written as 2 π r = C - and this leaves us with an arbitrary factor of 2. The true circle constant should be 2 π which has come to be written as τ, that is
τ r = C
There has, slowly, but surely, been a silent movement in the mathematics community at large to implement this change from π to τ as it greatly simplifies MANY formulas and trigonometric identities which otherwise have an extra factor of 2 simply because we've been using half the correct circle constant all along. Switching to τ not only simplifies all equations involving π, but it also makes the concept or radians as angles in trigonometry easier to understand.
Hopefully it's obvious the image on the left is much more intuitive...
Unfortunately, this usage of π is so deeply ingrained in our historical mathematics literature, it's unsure if this understanding could ever truly come to light. Also, this post is starting to get quite lengthy so I can only link you the article that started this whole debacle/debate here and an explanatory video(s) by the lovely Vi Hart here [and here, and here...].
Once again, if you've made it all the way here, I hope I have enlightened you in some way or another. If not, I apologize for boring you, but realize that this is probably one of the most ubiquitous concepts in all of mathematics - more so than any other topic I've covered thus far. Being able to understand and appreciate this concept is a stepping stone to a greater interpretation of the beauty and universality of mathematics as a whole. And as always - stay positive, stay strong, stay passionate, and have a great 2016! |
3549 | https://www.nayuki.io/page/smallest-enclosing-circle | Smallest enclosing circle
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Smallest enclosing circle
Demo (JavaScript)
Random static points Random moving points Manual positioning
The above program computes the smallest circle that encloses an arbitrary set of points in the plane. Points defining the circumference are colored red; points in the interior are colored gray. You can manually add/move/remove points with the mouse:
Left-click in a blank space to add a new point
Left-click an existing point and drag to move it
Right-click an existing point to delete it
Description
Although the problem looks easy to solve visually, the algorithm is in fact moderately long and quite tricky. (This deceptive difficulty is a common theme in computational geometry.) I don’t fully understand how to solve this problem from scratch, and my program’s implementation relies heavily on the mathematics and reasoning covered in the linked PDF file.
The code implements a variant of Emo Welzl’s algorithm. With randomization, it runs in expected Θ(n) (linear) time, unlike the brute-force Θ(n 4) algorithm. Thus it can handle thousands of points with ease.
Source code
Java (SE 7+)
SmallestEnclosingCircle.java (computation functions)
SmallestEnclosingCircleTest.java (JUnit test suite)
TypeScript / JavaScript
smallest-enclosing-circle.ts / smallest-enclosing-circle.js (computation functions)
smallest-enclosing-circle-demo.ts / smallest-enclosing-circle-demo.js (the demo on this page)
Python
smallestenclosingcircle.py (computation functions)
smallestenclosingcircle-test.py (test suite)
C# (.NET 4.0+)
SmallestEnclosingCircle.cs (computation functions)
SmallestEnclosingCircleTest.cs (test suite)
C++ (C++11 and above)
SmallestEnclosingCircle.cpp (computation functions)
SmallestEnclosingCircle.hpp (header file)
SmallestEnclosingCircleTest.cpp (test suite)
The Java version is considered the reference version because it has object types and the code is structured more clearly. License: GNU LesserGeneral Public License v3.0+
More info
Wikipedia: Smallest-circle problem
Computational Geometry: Smallest enclosing circles and more (PDF) by Van Kreveld
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3550 | https://brilliant.org/landing/pre-algebra/consequences-15/ | Solving Equations
Start your algebra journey here with an introduction to variables and equations.
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Level 2
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Substitution
Level 3
Distributing and Factoring
Groups in Equations
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Level 4
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Level 5
Inequalities
Unbalanced Scales
Inequalities Both Ways
Balancing Scales
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Level 6
Solving Inequalities
Finding the Boundary
Solving Inequalities
Including the Boundary
Writing Inequalities
Negations
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Level 7
Solving Systems
Systems with Scales
Balancing with Substitution
Writing Systems of Equations
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Level 8
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Level 9
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More Reasoning about Weights
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Reasoning about Two Scales
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Level 10
Reasoning about Groups of Variables
Unpacking Boxes
Packing Boxes
Reasoning about Unpacking
Reasoning about Packing Boxes
Reasoning about Multiples
Level 11
Reasoning about Variables
Constraints
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Reasoning with Balanced Scales
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Reasoning with Scale Systems
Constraints in Scale Systems
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Solving Equations
Start your algebra journey here with an introduction to variables and equations. |
3551 | https://www.albert.io/blog/polynomial-functions-and-rates-of-change/ | Skip to content
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AP® Precalculus
Polynomial Functions and Rates of Change: AP® Precalculus Review
The Albert Team
Last Updated On:
What We Review
Introduction
Polynomial functions are a critical topic in AP® Precalculus, forming the foundation for understanding more advanced mathematical concepts. These functions show up everywhere—from predicting the trajectory of a ball to modeling population growth over time.
In this guide, we’ll break down everything you need to know about polynomial functions and their rate of change. You’ll learn about their components, how to find zeros and extrema, and how to analyze their behavior with confidence. We’ll also include a handy quick reference chart to summarize key ideas.
By the end of this article, you’ll have the tools you need to tackle problems from section 1.4 Polynomial Functions and Rates of Change from the AP® Precalculus CED like a pro. Let’s get started!
Start practicing AP® Precalculus on Albert now!
What Is a Polynomial Function?
Polynomial functions are a key part of AP® Precalculus and appear in many real-world scenarios, like physics and economics. A polynomial function is defined as any equation that can be written in the form:
p(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \dots + a_1x + a_0
Here’s what each part means:
n: The highest power of x, called the degree of the polynomial.
a_n: The coefficient of the highest degree term, known as the leading coefficient.
a_i: Coefficients of the terms, where i ranges from 0 to n. These are real numbers.
a_0: The constant term, which does not have x.
Example
Let’s take this polynomial:p(x) = 3x^4 - 2x^3 + 5x^2 - 7x + 9
The degree is 4 (the highest power of x).
The leading term is 3x^4, and the leading coefficient is 3.
The constant term is 9.
Special Case: Constant Polynomial Functions
A constant function, like p(x) = 7, is also a polynomial. Its degree is 0 because it contains no variable x.
Key Characteristics of Polynomial Functions
Smooth and Continuous: Polynomial graphs have no sharp corners or breaks.
Defined Everywhere: Polynomial functions are valid for all real numbers.
Why Polynomial Functions Matter
Polynomials model countless real-world problems, from predicting population trends to calculating the height of a ball in motion. By mastering them, you’ll build a strong foundation for AP® Precalculus and beyond!
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Local and Global Extrema in Polynomial Functions
Polynomial functions often have high and low points on their graphs, known as extrema. These extrema can help us analyze the behavior of a function, which is critical for solving real-world problems or graphing accurately.
Local vs. Global Extrema
A local maximum is a point where the function’s value is greater than the values of the function at nearby points.
A local minimum is a point where the function’s value is less than the values of the function at nearby points.
Among all the local maxima or minima, the greatest local maximum is the global maximum, and the least local minimum is the global minimum.
For example:
p(x) = x^3 - 6x^2 + 9x + 2
This polynomial has a local maximum and a local minimum.
If the domain is unrestricted, it has no global maximum or minimum because the cubic function increases or decreases indefinitely.
Even-Degree Polynomial Functions Always Have Global Extrema
Polynomials with even degrees (like x^2 or x^4) always have either a global maximum or a global minimum. Their graphs either open upwards or downwards:
If the leading coefficient is negative, the graph opens downwards, and the function has a global maximum.
If the leading coefficient is positive, the graph opens upwards, and the function has a global minimum.
Zeros and Local Extrema
The zeros of a polynomial function are the points where the graph crosses or touches the x-axis. These zeros play a crucial role in understanding the graph’s behavior, especially when paired with local extrema.
What Are Zeros?
Zeros (or roots) of a polynomial function are the input values x where the function equals zero:p(x) = 0.
For example, consider:p(x) = x^3 - 4x^2 + x + 6
To find the zeros, solve:x^3 - 4x^2 + x + 6 = 0
By factoring or using other techniques, the zeros are found to be x = -1, x = 2, x = 3.
Relationship Between Zeros and Local Extrema
Between every two distinct real zeros of a polynomial function, there must be at least one local maximum or minimum. This is because the graph of the polynomial changes direction as it crosses the x-axis at its zeros.
For example, in the cubic polynomial p(x) = x^3 - 6x^2 + 9x:
The zeros are x = 0, x = 3.
A local maximum occurs between these zeros at x = 1.
Visualizing This Concept
Imagine a roller coaster that rises and falls as it travels through hills (local maxima) and valleys (local minima). These hills and valleys must appear between points where the coaster crosses the ground (the zeros).
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Points of Inflection and Concavity in Polynomial Functions
Polynomial graphs often change their “shape,” curving upward in some sections and downward in others. These changes in curvature are described using concavity and points of inflection.
What Is Concavity?
Concavity refers to the direction the graph curves:
A graph is concave up if it looks like a “U” (the slope is increasing).
A graph is concave down if it looks like an upside-down “U” (the slope is decreasing).
What Are Points of Inflection?
A point of inflection is where the graph changes concavity—from concave up to concave down, or vice versa. These points occur when the rate of change of the function changes from increasing to decreasing or from decreasing to increasing.
Example:
Analyze p(x) = x^3 - 3x^2 + 2x to find the points of inflection and understand the graph’s concavity.
By graphing the function p(x) , we can see where the points of inflection will occur.
Locate Points of Inflection
At x = 1, the graph changes from concave up to concave down .
x = 1 is a point of inflection.
At x = 1, the graph changes concavity, making it a point of inflection.
Conclusion
Polynomial functions are a cornerstone of AP® Precalculus, and mastering them is essential for tackling challenging problems and understanding real-world applications. Here’s what you’ve learned:
How to break down a polynomial into its key components, like degree, leading term, and zeros.
The difference between local and global extrema and how to identify them.
The relationship between zeros and local extrema, as well as the role of points of inflection and concavity.
By combining these concepts, you can analyze and graph any polynomial function with confidence!
Quick Reference Chart for Polynomial Functions
| Feature | Definition | How to Identify |
---
| Degree | The highest power of x in the polynomial. | Look at the largest exponent. |
| Leading Term | The term with the highest degree. | Use the term with the largest exponent of x. |
| Zeros | The points where p(x) = 0. | Factor the polynomial or use the quadratic formula or numerical methods. |
| Local Extrema | Turning points where the graph changes direction. | Occur between zeros; check intervals where the graph rises or falls. |
| Global Extrema | The highest (maximum) or lowest (minimum) value of the function over its entire domain. | Only exist for bounded graphs, like those of even-degree polynomials. |
| Concavity | The direction the graph curves: concave up or concave down. | Test intervals or use graphing tools. |
| Points of Inflection | Points where the graph changes concavity. | Identify where the graph changes from curving up to curving down, or vice versa. |
Keep practicing these concepts by analyzing different polynomial functions. The more you practice, the more intuitive polynomial behavior will become!
With this solid understanding, you’re well on your way to conquering AP® Precalculus and applying these skills to real-world problems. Good luck, and keep exploring math with curiosity!
Need help preparing for your AP® Precalculus exam?
Albert has hundreds of AP® Precalculus practice questions, free response, and an ap precalculus practice test to try out.
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3552 | https://www.doubtnut.com/qna/645259069 | IfA+B+C=0 , Prove :cos2A+cos2B+cos2C=1+2cosAcosBcosC.
More from this Exercise
(a) LHS=cos2A+cos2A+cos2B+cos2C ltbr> =3+(cos2A+cos2B+cos2C)2 now given A+B=C or A+B+(π−C)=π or A+B+D=π, where D=π−C ⇒LHS=3+(cos2A+cos2B+cos2C)2 =3+(Cos2A+Cos2B+cos2D)2 =1−2cosAcosBcosD =1−2cosAcosBcos(π−C) =1+2cosAcosBcosC=RHS. (b) Given α+β=60∘ cos2α+cos2β−cosαcosβ =1−sin2α+cos2β−cosαcosβ =1+cos(α+β)cos(α−β)−cosαcosβ =1+cos60∘cos(α−β)−cosαcosβ =1+cosαcosβ+sinαsinβ2−cosαcosβ =1−cos(α+β)2 =1−14=34
Similar Questions
If A+B+C=180∘, then prove thatcos2A+cos2B+cos2C=1−2cosAcosBcosC.
If A+B+C=π, prove that : cos2A+cos2B+cos2C=−1−4cosAcosBcosC
If A+B+C = π, prove that : cos2A+cos2B+cos2C=−1−4cosAcosBcosC.
If A+B+C=2S, prove that : cos2S+cos2(S−A)+cos2(S−B)+cos2(S−C)=2+2cosAcosBcosC.
If A+B+C=1800, prove that : sin2A+sin2B+sin2C=2(1+cosAcosBcosC)
If A+B+C=2π, prove that : cos2B+cos2C−sin2A−2cosAcosBcosC=0.
In a triangle ABC, prove that: cos4A+cos4B+cos4C=32+2cosAcosBcosC+12cos2Acos2Bcos2C
If A+B+C=π2, prove that:
sin2A+sin2B+sin2C=4cosAcosBcosC
If A+B+C=π, prove that : cosA+cosB−cosC=4cos(A2)cos(B2)sin(C2)−1
If A+B+C=1800, prove that : cos2(A2)+cos2(B2)−cos2(C2)=2cos(A2)cos(B2)sin(C2)
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3553 | https://users.math.msu.edu/users/zhan/Notes2.pdf | Theorem 1 (Squeeze Lemma, Exercise 8.5). Suppose three sequences (an), (bn), (sn) satisfy that there is N0 ∈N such that for all n > N0, an ≤sn ≤bn. If (an) and (bn) both converge to the same number s, then (sn) also converges to s.
Discussion We want to arrive at the inequality |sn −s| < ε, which is equivalent to s −ε < sn < s + ε. In order to have sn < s + ε, we want to use sn ≤bn and bn < s + ε. In order to have sn > s −ε, we want to use sn ≥an and an > s −ε. The inequalities sn ≤bn and sn ≥an are satisfied if n > N0. The inequalities bn < s + ε and an > s −ε can be obtained for big n using the convergence of (an) and (bn).
Proof. Let ε > 0. Since (an) converges to s, there is N1 ∈N such that for n > N1, |an −s| < ε, which implies that an > s −ε. Since (bn) converges to s, there is N2 ∈N such that for n > N2, |bn −s| < ε, which implies that bn < s + ε. Let N = max{N0, N1, N2}. If n > N, then an ≤sn ≤bn, an > s −ε, and bn < s + ε, which together imply that s −ε < sn < s + ε. So we get |sn −s| < ε for n > N.
For x ∈[0, ∞) and r = p/q ∈Q with p, q ∈N, we define xr = (x1/q)p. We have (xr1)r2 = xr1r2, and for x, y ≥0, x < y if and only if xr < yr.
Theorem 2 (Theorem 9.7). We have the following basic examples.
(a) limn→∞( 1 n)r = 0 for any r ∈Q and r > 0.
(b) limn→∞an = 0 if |a| < 1.
(c) limn→∞n1/n = 1.
(d) limn→∞a1/n = 1 if a > 0.
Proof. (a) Let ε > 0. Then ε1/r > 0. Since 1/n →0, there is N ∈N such that for n > N, |1/n −0| < ε1/r, i.e., 1/n < ε1/r, which then implies that |(1/n)r −0| = (1/n)r < (ε1/r)r = ε.
(b) If a = 0, then obviously 0n →0. Suppose a ̸= 0. Then |a| = 1 1+b for some b > 0. We use the binomial theorem (1 + b)n = n X k=0 n k bk, where n k = n!
k!(n −k)! = n(n −1) · · · (n −k + 1) k!
.
Especially, n 0 = 1 and n 1 = n. Also, n k ≥0 for all 0 ≤k ≤n. So (1 + b)n ≥1 + nb > nb.
Thus, |an| = |a|n < 1 nb. Since 1/n →0, we get 1 nb →0. By the corollary of the squeeze lemma, we get an →0.
(c) Let sn = n1/n −1. Then sn ≥0 and n = (1 + sn)n. We use the binomial theorem again and the fact that n 2 = 1 2n(n −1). So if n ≥2, then n = (1 + sn)n ≥1 + nsn + 1 2n(n −1)s2 n > 1 2n(n −1)s2 n.
1 So s2 n < 2 n−1 for n ≥2. Since 0 ≤s2 n < 2 n−1 for n ≥2 and 2 n−1 →0, by squeeze lemma, s2 n →0.
So sn = p s2 n →0. Thus, n1/n = 1 + sn →1.
(d) If a ≥1, then 1 ≤a1/n ≤n1/n for n ≥a. By (iii) and squeeze lemma, we get a1/n →1.
If 0 < a ≤1, then 1/a ≥1, and so (1/a)1/n →1. Thus, a1/n = 1/(1/a)1/n →1/1 = 1.
Infinite Limits Definition 1. Let (sn) be a sequence of real numbers. We write limn→∞sn = +∞or sn →+∞ if ∀M > 0, ∃N ∈N, such that n > N implies sn > M.
(1) When this holds, we say that (sn) diverges to +∞. We write limn→∞sn = −∞or sn →−∞if ∀M < 0, ∃N ∈N, such that n > N implies sn < M.
(2) When this holds, we say that (sn) diverges to −∞.
Since for any x ∈R, there are M1 > 0 and M2 < 0 such that M2 < x < M1, the assumption on the M in (1) and (2) can all be replaced by M ∈R. We may write ∞for +∞.
Example 1. If sn = n for each n ∈N, then sn →+∞. To see this is true, for any M > 0, by Archimedean property, there is N ∈N such that N > M. Then for any n > N, sn = n > N > M.
Recall that we say that limn→∞sn = s or sn →s for some s ∈R if ∀ε > 0, ∃N ∈N, such that n > N implies |sn −s| < ε.
(3) If any of (1,2,3) hold, we say that limn→∞sn exists. But only when (3) holds, we say that (sn) converges.
When sn →s ∈R, (sn) is bounded. This is not the case if sn →+∞or sn →−∞. If sn →+∞, then (sn) is not bounded above since for any M ∈R there is n such that sn > M.
But (sn) is bounded below. To see this, taking M = 1, we get an N ∈N such that sn > 1 for all n > N. Then min{s1, . . . , sN, 1} is a lower bound of {sn : n ∈N}. Similarly, if sn →−∞, then (sn) is bounded above and not bounded below.
Example 2. The sequence sn = (−1)n, n ∈N, has no limit. We have seen that it is divergent.
It can not diverge to +∞or −∞since it is bounded.
Remark 1. The limit property of a sequence is not affected by a change of finitely many elements. This means that, if we have two sequences (sn) and (tn) and a number N0 ∈N such that sn = tn for n > N0, then if lim sn exists, then lim tn also exists and equals lim sn. To see this, suppose sn →s ∈R. Let ε > 0. There is Ns ∈N such that |sn −s| < ε if n > Ns. Let Nt = max{Ns, N0}. If n > Nt, then n > Ns and n > N0. From n > N0 we know tn = sn; from n > N we know |sn −s| < ε, which together imply that |tn −s| < ε. If sn →+∞or −∞, 2 the argument is similar. Thus, when we talk about lim sn, we may allow that the sn to be not defined for finitely many n.
A related fact is that when lim sn exists, then lim sn+1 also exists, and two limits are equal.
The converse is also true.
Remark 2. We have another way to understand the statements (3,1,2).
Consider a game played by Player A and Player B. To describe the limit sn →s ∈R, let the sequence (sn) and the number s be fixed. Then the two players choose the following numbers in order: 1. Player A chooses ε > 0.
2. Player B chooses N ∈N.
3. Player A chooses n ∈N with n > N.
The rule is: if |sn −s| < ε, then Player B wins the game; otherwise, Player A wins the game.
If sn →s, then Player B has a strategy to always win the game. Otherwise Player A can always win the game.
To describe lim sn = +∞, we modify the game such that in the first step Player A chooses M > 0, and the final rule is changed such that B wins the game if sn > M. If sn →+∞, then Player B has a strategy to win the game; otherwise Player A can always win the game. One may similarly describe a game for lim sn = −∞.
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3554 | https://mccme.ru/circles/oim/mmks/works/zaslavski.pdf | / / 0 1 2 0 3 4 , 5 # // "!( ! 2 O
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Compute volume of parallelepiped using triple vector
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Before anyone claims that I am not studying, my university recently changed their coursework that does not require linear algebra as a pre-requisite to multi-variable calculus.
My professor gave us these questions even though we have no experience with matrix, he did not even teach us about it and I had to study them online by myself.
I only got as far as I could and I only want one question help so I can work on the other similar questions by myself. I would not be here if I can't find any other help.
linear-algebra
multivariable-calculus
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edited Jan 11, 2019 at 5:33
ZealotoryZealotory
asked Jan 11, 2019 at 5:28
ZealotoryZealotory
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$\begingroup$ Ok, so much have you figured out so far? $\endgroup$
Jonathan Dunay
– Jonathan Dunay
2019-01-11 05:35:25 +00:00
Commented Jan 11, 2019 at 5:35
2
$\begingroup$ What are "the results of problem 2"? $\endgroup$
Angina Seng
– Angina Seng
2019-01-11 05:44:02 +00:00
Commented Jan 11, 2019 at 5:44
2
$\begingroup$ Two very basic operations from multivariable calculus that you should know are the cross product of two vectors, and the dot product of two vectors. To calculate the volume of the parallelepiped, you need to compute $(\vec{\bf a} \times \vec{\bf b})\cdot{\bf c}$ (known as the scalar triple product). $\endgroup$
Decaf-Math
– Decaf-Math
2019-01-11 05:44:04 +00:00
Commented Jan 11, 2019 at 5:44
$\begingroup$ @LordSharktheUnknown the area of the parallelogram of vector a and b. Which I got to as /sqrt(20). but I think the more relevant thing is the cross product of vector a and b, which is 4j -2k. (can someone tell me where I can format?) $\endgroup$
Zealotory
– Zealotory
2019-01-11 06:42:01 +00:00
Commented Jan 11, 2019 at 6:42
$\begingroup$ Reference for Mathjax $\endgroup$
Shubham Johri
– Shubham Johri
2019-01-11 06:48:55 +00:00
Commented Jan 11, 2019 at 6:48
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$\begingroup$
The volume of the parallelepiped is the base area times the height. The base area, as you know, is the magnitude of $\vec a\times\vec b$, that is $\|\vec a\times\vec b\|$.
You also know that $\vec a\times\vec b$ is a vector perpendicular to the plane containing $\vec a,\vec b$. Thus, the height of the parallelepiped is the absolute value of the projection of $\vec c$ along $\vec a\times\vec b$, that is $\Big|\vec c\cdot\frac{\vec a\times\vec b}{\|\vec a\times\vec b\|}\Big|$ (for the derivation of this quantity, see this answer).
Multiply them together to get$$V=\Big|\vec c\cdot\frac{\vec a\times\vec b}{\|\vec a\times\vec b\|}\Big|\|\vec a\times\vec b\|=|\vec c\cdot\vec a\times\vec b|$$
$\vec c\cdot\vec a\times\vec b$ is called the scalar triple product of vectors $\vec a,\vec b,\vec c$ and denoted as $[\vec a\ \vec b\ \vec c]$. The volume of the parallelepiped formed by three vectors is the geometrical interpretation of the absolute value of their scalar triple product.
In your case, note that $\vec a,\vec b,\vec c$ are coplanar, as $\vec b-\vec c=\vec a$. The volume of the parallelepiped formed by three vectors lying in the same plane is zero. You can verify the same using scalar triple product.
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edited Jul 7, 2020 at 14:35
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Monomial – Definition, Degree, Parts, Examples, Facts, FAQs
Home » Math Vocabulary » Monomial – Definition, Degree, Parts, Examples, Facts, FAQs
What Is a Monomial in Math?
How to Find the Degree of a Monomial
How to Factor a Monomial
Solved Examples on Monomials
Practice Problems on Monomials
Frequently Asked Questions about Monomials
What Is a Monomial in Math?
A monomial is a polynomial that contains only a single non-zero term.
“Mono” means one. When polynomials are classified on the basis of the number of terms, the polynomials with only a single term are called monomials. This single term can be a number (a constant), a variable, or a product of multiple variables with a coefficient.
The coefficient can be any real number. There won’t be any operator present in a monomial.
Note that the degree of a monomial need not be 1. The exponents of variables cannot be negative or fractional. The exponent of all variables must be a whole number (non-negative integer).
Monomial examples:
5 a 2
−3 p q
−1 2 x y 2 z p q
4 x 2 y
x y z
−7 x 4 y 8 z 3
Monomial: Definition
A monomial can be defined as a polynomial with a single non-zero term.
Parts of a Monomial
Different parts of a monomial are as follows:
Variable: the letters present in the expression
Coefficient: the number attached with the variable
Degree: the sum of exponents of all the variables present
Literal part: the alphabets and their exponents
| Monomial | Variable | Degree | Literal part |
--- --- |
| 5 x 3 | x | 3 | x 3 |
| −6 x 2 y z 3 | x,y,z | 6 | x 2 y z 3 |
| 9 x y 7 | x,y | 8 | x y 7 |
| 8 | No variable | 0 | – |
How to Find Monomials
You can identify a monomial easily using the following points:
The expression must have a single term, which implies that the combination of alphabets (variables) must not be separated by any operator.
The exponents must always be positive (a whole number). The exponents cannot be fractions.
For instance, (2 x+y),(3 p−t),x 1 2 are not monomials.
How to Find the Degree of a Monomial
The degree of a polynomial will always be a positive integer owing to the rule of monomials to contain positive exponents. The power of variables is summed up to calculate the degree of the polynomial as depicted in the diagram. Given below are the steps to finding the degree of a monomial:
Step 1:Identify the variables and their exponents.
Step 2:Add all the exponents.
Step 3:The sum represents the degree.
Example:3 p 7 q 2 s can be written as 4 p 7 q 2 s 1.
Factoring Monomials
Factoring monomials simply means writing the coefficients and the variables in the factored form. In other words, monomial factorization is a way of expressing a given monomial as a product of two or more monomials.
How to Factor a Monomial
To factor monomials, we factor the coefficients and the literal part (variables) separately. Factorization refers to separating the terms into constituents whose product is the original term.
Example 1:4 p 2 q 3 s=2×2×p×p×q×q×q×s
Example 2:15 x y 2 z 3=3×5×x×y×y×z×z×z
Monomials, Binomials, Trinomials
The polynomial can have only a finite number of terms. We can classify polynomials based on the number of terms. A single-term polynomial is called a monomial. Polynomials with two terms are called binomials. Polynomials with three terms are called trinomials.
| Monomial | Binomial | Trinomial |
---
| A polynomial with only one term. | A polynomial with two terms. | A polynomial with three terms. |
| 4 p 5 q 3 s | 4 p 5 q 3+s | 4 p 5−q 3+s |
Operations on Monomials
Addition of two monomials having the same literal part will result in a monomial.
Example:4 p 5 q 3 s+10 p 5 q 3 s=14 p 5 q 3 s
Subtraction of two monomials having the same literal part will result in a monomial.
Example:4 p 5 q 3 s−10 p 5 q 3 s=−6 p 5 q 3 s
Multiplication of two monomials is always a monomial.
Example:4 p 5 q 3 s×10 p 5 q 3 s=40 p 10 q 6 s 2
To divide two monomials having the same variables, we use the laws of indices (rules of exponents).
Example:40 p 10 q 6 s 2 4 p 5 q 3 s=10 p 5 q 3 s
Facts about Monomials
The product of two monomials will always be a monomial.
If the coefficient of a monomial is not present, it is 1. Examples: xyz, x, pq
A constant polynomial is a monomial with degree zero.
The coefficient of a monomial can be any real number. It can be negative or fractional, but not its exponent.
Conclusion
Monomials are polynomials categorized based on terms. They contain one term with varying degrees calculated by summing up the exponents. They can comprise single or multiple variables. The factorization and other operations of monomials are performed similarly to general mathematics.
Solved Examples on Monomials
1. State the parts of a monomial10 a 2 b 2.
Solution:
The parts of a monomial are coefficient, variable, degree and literal part.
Coefficient: 10
Literal part: a 2 b 2
Variable: a, b
Degree: 2+2=4
2. Identify the monomials, binomials, and trinomials from the following polynomials.
x 2 y
a 2+b 2
p 3 q 2−s 2 t 2
a 2+b 2+2 a b
Solution:
Monomial: x 2 y
Binomials: a 2+b 2,p 3 q 2−s 2 t 2
Trinomial:a 2+b 2+2 a b
3. What is the degree of the given monomials?
i)5 x 3
ii)x 2 y
Solution:
i) Polynomial =5 x 3
Degree =3
ii) Polynomial =x 2 y
Degree =2+1=3
Practice Problems on Monomials
Monomial - Definition, Degree, Parts, Examples, Facts, FAQs
Attend this quiz & Test your knowledge.
1
Which is the correct factorization of monomial 3 x 3?
3×x×x×x
1×3
x×x×x
3×x
Correct Incorrect
Correct answer is: 3×x×x×x
3 x 3=3×x×x×x
2
What is the degree of monomial 39 x 3 y z?
3
39
5
2
Correct Incorrect
Correct answer is: 5
Degree of monomial 39 x 3 y z=3+1+1=5
3
Identify a monomial.
19 x 4 y z
3xyz
6
All of the above
Correct Incorrect
Correct answer is: All of the above
Monomials have a single term, which can be a number or a variable (or multiple variables) with coefficients.
4
A monomial can have
a variable with negative coefficient
a variable with negative exponent
a variable with fractional exponent
All of the above
Correct Incorrect
Correct answer is: a variable with negative coefficient
A variable in a monomial cannot have a fractional or negative exponent.
Frequently Asked Questions about Monomials
What are the rules of a monomial?
There is a single rule for monomials: they must be raised only to positive integers.
Can a monomial be negative?
The coefficient of a monomial can be negative, but its exponent cannot be negative.
What happens to exponents while dividing and multiplying the monomial?
When multiplying monomials with a single variable, we add the exponents.
When dividing monomials with a single variable, we subtract the exponent in the denominator from the exponent value of the numerator.
Is a constant polynomial a monomial?
Yes, every constant polynomial is a monomial.
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3558 | https://www.doubtnut.com/qna/647802800 | Potassium permanganate gives the following reactions in neutral medium M n O 4 - + 2 H 2 O + 3 e - → M n O 2 + 4 O H - . The equivalent weight of K M n O 4 is (atomic mass of Mn= 55u )
158
79
52.66
31.6
The correct Answer is:C
Equivalent weight of K M n O 2 in neutral medium M n O 4 - + 2 H 2 O + 3 e - → M n O 2 + 4 O H - Total oxidation state of Mn In K M n O 4 = 14 units Total oxidation state of Mn in product M n O 2 = + 8 units Change in oxidation state of Mn = 14 - 8 =6 units ∴ Equivalent weight of K M n O 4 = 2 M 6 = M 3 = 158 3 = 52.66
Topper's Solved these Questions
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Similar Questions
Consider the following reaction
MnO−4+8H+5e−→Mn2++4H2O,E∘=1.51V.
The quantity of electricity required in Faraday to reduce five moles of MnO−4 is _____________ .
Equivalent weight of MnO−4 in the following reaction will be
2MnO−4+l−+H2O→2MnO2+lO−3+2OH−
Knowledge Check
Potassium permanganate gives the following reactions in neutral medium MnO4)−+2H2O+3e−→MnO2+4OH−. The equivalent weight of KMnO4 is (atomic mass of Mn=55u)
In strongly alkaline medium, MnO−4 MnO−4+e−→MnO2−4 In this medium, equivalent weight of KMnO4 is reduced as:
If the molecular weight of Ba(MnO4)2 is M , then the equivalent weight of Ba(MnO4)2 in acidic medium is
In acidic medium,potassium manganate undergoes disproportionation according to equation.
3MnO2−4+4H+long→2MnO−4+MnO2+2H2O
Equivalent weight of K2MnO4 will be (Mol.wt.of K2MnO4=M)
If the molecular weight of Ba(MnO4)2 is M , then the equivalent weight of Ba(MnO4)2 in acidic medium is
How many of the following are obtained on heating Potassium permanganate?
K2MnO4,MnO2,O2,Mn2O3
In acidic medium, potassium manganate undergoes disproportionation according to equation.3MnO2−4+4H+→2MnO−4+MnO2+2H2O
Equivalent weight of K2MnO4 will be (Mol.wt.of K2MnO4=M)
The equivalent mass of KMnO4in the following reaction is ________________.
(M = Molecular mass)
MnO−+45Fe2++8H+→Mn2++5Fe3++4H2O
ARIHANT PUBLICATION JHARKHAND-CONCEPT OF ATOMIC MOLECULAR AND EQUIVALENT MASSES-EXAM BOOSTER FOR CRACKING EXAM
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3559 | https://bpb-us-w2.wpmucdn.com/people.smu.edu/dist/8/1647/files/2024/10/Optimization_Part_I.pdf | ECON 5100 Lecture Notes 6 Mathematics for Economics and Business Bo Chen Economics Fall 2024 Bo Chen (HKUST) Optimization-Part I Fall, 2024 1 / 29 Optimization Introduction Unconstrained Optimization Constrained Optimization Envelope Theorem and Applications The Maximum Theorem Elements of Dynamic Optimization Bo Chen (HKUST) Optimization-Part I Fall, 2024 2 / 29 Mathematical Optimization A mathematical optimization problem has the form: (⋆) max f (x) subject to fi(x) ≤bi, i = 1, . . . , m.
Variable x = (x1, . . . , xn) is the choice variable Function f : Rn →R is the objective function Functions fi : Rn →R, i = 1, . . . , m, are the constraint functions, and constants b1, . . . , bm are bounds for the constraints Vector x∗is optimal, or a solution of the optimization problem (⋆), if for all z with f1(z) ≤b1, . . . , fm(z) ≤bm, we have f (x∗) ≥f (z) We also write a optimization problem as “max f (x) sub. to x ∈D”, or max{f (x)|x ∈D}, where D is the constraint set. The set of all solutions or maximizers x∗is denoted as arg max{f (x)|x ∈D}: arg max{f (x)|x ∈D} = {x ∈D|f (x) ≥f (z) for all z ∈D}.
The optimization problem (⋆) is called a linear program if the objective function and the constraint functions f , f1, ..., fm are linear.
If the problem (⋆) is not linear, it is called a nonlinear program.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 3 / 29 Mathematical Optimization For a given optimization problem, (1) a solution may fail to exist (i.e., problem (⋆) may have no solution); (2) even if a solution does exist, it need not necessarily be unique (i.e., there are multiple solutions).
Example 1 Let D = R+ and f (x) = 2x3 for x ∈D. We have f (D) = R+ and sup f (D) = ∞, or max{f (x)|x ∈D} has no solution.
2 Let D = [0, 1] and f (x) = x(1 −x) for x ∈D. We have that max{f (x)|x ∈D} has exactly one solution, at x∗= 1/2.
3 Let D = [−1, 1], f (x) = x2 for x ∈D. Problem max{f (x)|x ∈D} has two solutions: arg max{f (x)|x ∈D} = {−1, 1}.
In economics and business, optimization problems are typically presented in a parametric form, or the objective function and/or the feasible set D depend on the value of a parameter θ ∈Θ: max{f (x, θ)|x ∈D(θ)}.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 4 / 29 Optimization–Examples Utility Maximization. A consumer’s utility from consuming xi ≥0 units of good i ∈{1, . . . , n} is u(x1, . . . , xn). She has a budget I ≥0 and faces prices (p1, . . . , pn) with pi ≥0 being the price of good i.
Her budget set is B(p, I) = {x ∈Rn + | p · x ≤I}. Her objective is to maximize her utility over the set of affordable bundles, i.e., to solve: max{u(x) | x ∈B(p, I)}.
Expenditure Minimization. This is the flip side of utility maximization. Given a price vector p ∈Rn + and a fixed utility level u, the consumer minimizes the income required to achieve u by solving: min{p · x | x ∈X(u)}, where X(u) = {x ∈Rn + | u(x) ≥u}.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 5 / 29 Optimization–Examples Consumption-Leisure Choice. Consider the labor-supply decision of households where an agent, endowed with H ≥0 units of “time”, faces wage w ≥0. The agent also enjoys time not working (leisure) ℓ with H −ℓ= L being her working time. The agent’s utility from xi ≥0 units of good i with price pi and leisure ℓis u(x1, . . . , xn; ℓ), with budget set F(p, w) = {(x, ℓ) ∈Rn+1 + | p · x ≤w(H −ℓ)}. Her objective is to maximize her utility: max{u(x, ℓ) | (x, ℓ) ∈F(p, w)}.
Optimal Provision of Public Goods. A central issue in public finance. There are n agents, a public good (initially 0) and a private good, money (initially ω). ω is used for both private consumption and public good via h : R+ →R+. The set of feasible allocations is Φ(ω) = {(x1, ..., xn, y) ∈Rn+1 + |y = h(x), x + P i xi ≤ω}. The social planner maximizes a social welfare function W (u1, ..., un): max{W [u1(x1, y), ..., un(xn, y)] | (x1, ..., xn; y) ∈Φ(ω)}.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 6 / 29 Optimization–Examples Pareto Optima. Consider allocating resource ω ∈Rn + to 2 agents.
Agent i’s utility from allocation xi is U : Rn + →R. Allocation (x1, x2) is feasible if (x1, x2) ∈F(ω) = {(x1, x2) ∈Rn + × Rn +|x1 + x2 ≤ω}. A feasible allocation (x1, x2) is Pareto optimal if ∄feasible (x′ 1, x′ 2) s.t.
ui(x′ i ) ≥ui(xi) for i = 1, 2 with at least one strict inequality. Pick α ∈[0, 1], let U(x1, x2; α) = αu1(x1) + (1 −α)u2(x2), and solve max{U(x1, x2; α) | (x1, x2) ∈F(ω}, where every solution (x∗ 1, x∗ 2) identifies a Pareto optimal allocation.
Optimal Commodity Taxation. Optimal tax on commodities to raise revenue R. Given x = (x1, ..., xn), p = (p1, ..., pn), excise taxes τ = (τ1, ..., τn), income I, utility u, consumers solve max u(x) s.t.
x ∈B(p + τ, I) = {x ∈Rn +|(p + τ) · x ≤I} with demand x∗(p + τ, I), indirect utility v(p + τ, I) = u(x∗(p + τ, I)). The government solves max{v(p + τ, I) | τ ∈T (p, I)}, where T (p, I) = {τ ∈Rn +|τ · x∗(p + τ, I) ≥R}.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 7 / 29 Unconstrained Optimization-FOC Definitions Let f (x) be a real-valued function with x ∈A ⊆Rn. (1) A point x∗∈A is a global maximum point of f on A if f (x∗) ≥f (x) ∀x ∈A, where f (x∗) is the maximum value, and a global minimum point of f if f (x∗) ≤f (x) ∀x ∈S, where f (x∗) is the minimum value. (2) A point x∗∈A is an (unconstrained) local maximum point if there is an open ball Br(x∗) s.t. f (x∗) ≥f (x) for all x ∈Br(x∗) ∩A.
We use extreme points and extreme values to denote both maxima and minima. We denote a point where all the first-order partial derivatives are 0 as a stationary point or a critical point.
Theorem (Necessary First-Order Condition) Given a real-valued f ∈C1 defined on A ⊂Rn, a necessary condition for x∗∈A to be an extremal point of f is that f ′ i (x∗) = 0 for all i = 1, ..., n.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 8 / 29 Local and Global Extrema Figure: Local and Global Extrema of cos(3πx)/x, 0.1 ≤x ≤1.1.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 9 / 29 Unconstrained Optimization-Sufficiency of SOC Why require f ′ i (x∗) = 0 for all i? Let Br(x∗) ⊂A satisfy f (x∗) ≥f (x) ∀x ∈Br(x∗). Then x∗ i maximizes xi 7→f (x∗ 1, ..., x∗ i−1, xi, x∗ i+1, ..., x∗ n).
Sufficient Second-Order Condition (SOC): Given the necessary FOC’s f ′ i (x∗) = 0 for all i, we use SOC’s to determine which of the stationary points satisfying f ′ i (x∗) = 0 are local max/min.
Hessian of f : f ′′ (x∗) = f11(x∗) · · · f1n(x∗) .
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...
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fn1(x∗) · · · fnn(x∗) .
Theorem (Sufficient Second-Order Conditions) Let f : A →R be C2 with A ⊂Rn open, and x∗satisfies f ′ i (x∗) = 0 ∀i.
1 If f ′′(x∗) is negative definite, then x∗is a strict local max.
2 If f ′′(x∗) is positive definite, then x∗is a strict local min.
3 If f ′′(x∗) is indefinite, then x∗is neither a local max or a local min.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 10 / 29 Unconstrainted Optimization-Saddle Point Definition (Saddle Point) A stationary point x∗of f for which f ′′(x∗) is indefinite is called a saddle point (sometimes also called a minmax point) of f . A saddle point x∗is a minimum of f in some directions, and a maximum of f in other directions.
Example Consider f (x1, x2) = x2 1 −x2 2 and x∗= (0, 0).
f ′′ (x∗) = 1 0 0 −1 .
The leading principal minors are D1 = 1 and D2 = −1. Hence, the Hessian f ′′(x∗) is indefinite and x∗= (0, 0) is a saddle point of f . Graphically, the graph of f curves up along the x1 direction and curves down along the x2 direction at x∗= (0, 0), resembling a riding saddle.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 11 / 29 Unconstrainted Optimization-Necessity of SOC We now consider necessary second-order conditions. For motivation, let f : R →R, the second-order necessary condition (weaker than second-order sufficient condition) says that if an interior point x∗is a local max (min), then f ′(x∗) = 0 and f ′′(x∗) ≤0 (f ′′(x∗) ≥0).
Example Consider f (x) = x4 and g(x) = −x4. It can be verified that f (x) is strictly convex with stationary point x∗= 0 and g(x) is strictly concave with x∗= 0. However, f ′(0) = f ′′(0) = g′(0) = g′′(0) = 0.
Theorem (Necessity of SOC for Multivariate Functions) Let f : A ⊂Rn →R with A open and f ∈C2. Suppose x∗∈int(A) is a local max (min) of f . Then f ′(x∗) = 0 and f ′′(x∗) is negative semi-definite (positive semi-definite).
Bo Chen (HKUST) Optimization-Part I Fall, 2024 12 / 29 Necessity of SOC-An Example Example Consider f (x, y) = x3 −y3 + 9xy. We have that f ′ x = 3x2 + 9y = 0 and f ′ y = −3y2 + 9x = 0. There are two stationary points (x∗, y∗) ∈{(0, 0), (3, −3)}. The Hessian matrix is: f ′′ (x, y) = 6x 9 9 −6y , with the leading principal minors being D1 = 6x and D2 = −36xy −81.
First, consider (x∗, y∗) = (0, 0). We have D1 = 0 and D2 = −81, or the Hessian f ′′(x∗, y∗) is indefinite. This implies that (x∗, y∗) = (0, 0) is a saddle point.
Next, consider (x∗, y∗) = (3, −3). We have D1 = 18, D2 = 243, or the Hessian f ′′(x∗, y∗) is positive definite. Hence, (x∗, y∗) = (3, −3) is a strictly local min point.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 13 / 29 Unconstrained Optimization–Global Maxima/Minima The previous FOC’s and SOC’s are only for local extrema of a C2 function. What about sufficient conditions for global extrema?
Theorem (Globa Extrema for Univariate Functions) Consider f : A →R where f ∈C2 and A is an open interval. If x∗is the only local maximum (minimum) point, and f ′′(x) ≤0 (f ′′(x) ≥0) ∀x ∈A, i.e., f is concave (convex), then x∗is a global max (min) of f .
Theorem (Globa Extrema for Multivariate Functions) Consider f : A →R where f ∈C2 and A ⊂Rn is open and convex. If f is a concave function (or f (y) −f (x) ≤f ′(x) · (y −x), or f ′′(x) is negative semi-definite for all x ∈A), and f ′(x∗) = 0 for x∗∈A, then x∗is a global max of f . (Global min is analogous) Proof Sketch for n = 1. Concavity of f implies that ∀x, y ∈A, λ ∈[0, 1], f (y) −f (x) ≤f (λy+(1−λ)x)−f (x) λ ⇒ f (y) −f (x) ≤f (x+h)−f (x) h (y −x) with h = λ(y −x).
Bo Chen (HKUST) Optimization-Part I Fall, 2024 14 / 29 Unconstrained Optimization–Global Maxima/Minima Take limit h →0 and we have f (y) −f (x) ≤f ′(x)(y −x).
Since f is concave and f ′(x∗) = 0, we have f (y) −f (x∗) ≤f ′(x∗)(y −x∗) = 0, or f (y) ≤f (x∗) for all y ∈A.
Example Let f (x, y, z) = x2 + 2y2 + 3z2 + 2xy + 2xz + 3, (x, y, z) ∈R3.
Determine concavity/convexity and extreme points of f (x, y, z).
As f ′ x = 2x + 2y + 2z, f ′ y = 4y + 2x, f ′ z = 6z + 2x. The Hessian is f ′′ (x) = 2 2 2 2 4 0 2 0 6 .
Leading principal minors: D1 = 2, D2 = 4, D3 = 8. Hence, f ′′(x) is positive definite, implying that f is strictly convex.
Unique stationary point (x∗, y∗, z∗) = (0, 0, 0) from f ′ x = f ′ y = f ′ z = 0.
Strict convexity ⇒(x∗, y∗, z∗) unique global minimum.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 15 / 29 Application–Profit Maximization Suppose a firm uses n inputs to produce a single output. An input bundle x = (x1, ..., xn) ∈Rn. Production function Q = F(x) is C2 and market price is P. The cost function is c(x) = w · x = Pn i=1 wixi.
The firm’s profit function is: Π(x) = PF(x) −c(x) with FOC’s being ∂Π(x∗) ∂xi = P ∂F(x∗) ∂xi −∂c(x∗) ∂xi = 0.
Marginal revenue balances marginal cost: P ∂F(x∗) ∂xi = wi, ∀i ∈{1, ..., n}.
SOC (necessary): Π′′(x∗) is negative semi-definite ⇔F ′′(x∗) is negative semi-definite, which implies that ∂2F(x∗) ∂x2 i ≤0 for all i, or we have diminishing marginal productivity for all i.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 16 / 29 Application–Method of Least Squares Suppose we are given n observations/points (xi, yi) ∈R2, i = 1, . . . , n. We want to find a linear graph that best fits the observations (hence n ≥3), i.e., find f (x) = ax + b s.t.
Pn i=1[f (xi) −yi]2 is minimized.
Define F : R2 →R and solve: max a,b F(a, b) = max a,b − Xn i=1(axi + b −yi)2 F is in quadratic form and is hence C2: ∂F ∂a = −2 n X i=1 (axi + b −yi)xi = −2 Xn i=1(ax2 i + bxi −xiyi), ∂F ∂b = −2 Xn i=1(axi + b −yi), ∂2F ∂a2 = −2 Xn i=1 x2 i , ∂2F ∂b2 = −2n, ∂2F ∂a∂b = −2 Xn i=1 xi We can calculate |D2F(a, b)| = 4n Pn i=1 x2 i −4(Pn i=1 xi)2.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 17 / 29 Application–Method of Least Squares By Cauchy-Schwarz inequality ((P uivi)2 ≤(P u2 i )(P v2 i )), |D2F(a, b)| ≥4n Xn i=1 x2 i −4n Xn i=1 x2 i = 0.
Since ∂2F/∂a2 ≤0, ∂2F/∂b2 ≤0, |D2F(a, b)| ≥0, the Hessian D2F(a, b) is negative semi-definite. Hence, if (a∗, b∗) satisfies ∂F/∂a = ∂F/∂b = 0, then (a∗, b∗) is a global maximum: a∗Xn i=1 x2 i + b∗Xn i=1 xi = Xn i=1 xiyi, a∗Xn i=1 xi + b∗n = Xn i=1 yi.
Let x = Pn i=1 xi/n and y = Pn i=1 yi/n. Then: a∗= 1 n Pn i=1 xiyi −xy 1 n Pn i=1 x2 i −x2 , b∗= y −a∗x, which are the least squares estimates of a and b.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 18 / 29 Constrained Optimization–Equality Economics is the study of optimal allocation of scarce resources.
“Scarce” implies that resources are somewhat constrained. As a result, constrained optimization lies @ the heart of economic theory: 1 a household’s consumption is constrained by its income 2 a firm’s production is constrained by cost and availability of its inputs 3 a monopolist’s optimal pricing is constrained by the market info Canonical Form of Constrained Optimization: max {x1,...,xn}∈Rn f (x1, ..., xn) s.t.
g1(x1, ..., xn) ≤b1; h1(x1, ..., xn) = c1; .
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gk(x1, ..., xn) ≤bk; hm(x1, ..., xn) = cm.
with “k” inequality constraints and “m” equality constraints.
Example–Leisure Choice and Labor Supply: max U(x1, ..., xn; ℓ) s.t.
Pn i=1 pixi ≤w0 + w(H −ℓ), xi ≥0 for all i, and 0 ≤ℓ≤H.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 19 / 29 Constrained Optimization–Equality Figure: Utility Maximization max{u(x1, x2) | p1x1 + p2x2 = y}.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 20 / 29 Constrained Optimization–Equality Consider the following simple constrained max problem in R2: (⋆) max{f (x1, x2) | h(x1, x2) = c}.
1 One possibility is solving (⋆) by substitution: derive x1 as a function of x2, say x1 = g(x2), from h(x1, x2) = c and then solve the unconstrained problem maxx2 f (g(x2), x2) 2 We can also solve (⋆) geometrically with a tangency condition and h(x1, x2) = c as in the utility max problem: geometrically, the max of f (x1, x2) is attained when f ’s highest level curve is tangent to h(x1, x2) = c: −f1(x∗) f2(x∗) = −h1(x∗) h2(x∗) ⇒f1(x∗) h1(x∗) = f2(x∗) h2(x∗) = µ, i.e., slope of level curve is the same as slope of constraint. This leads to f1(x∗) −µh1(x∗) = 0 1 ○ f2(x∗) −µh2(x∗) = 0 2 ○ h(x∗ 1 , x∗ 2 ) −c = 0 3 ○ 3 Lagrangian Method. It is convenient to set up the Lagrangian: L(x1, x2; µ) = f (x1, x2) −µ[h(x1, x2) −c].
We also have ∂L ∂x1 = 0 ⇔1 ○; ∂L ∂x2 = 0 ⇔2 ○; ∂L ∂µ = 0 ⇔3 ○.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 21 / 29 Constrained Optimization–Equality The Lagrangian method effectively transforms a two-variable constrained problem into a three-variable unconstrained problem.
The new variable µ, which multiplies the constraint is called the Lagrangian multiplier, which is loaded with economic interpretation (a measure of value of the constraint or the “scarce resources”).
The Lagrangian was due to Italian (French later, a creator of calculus of variations) mathematician Joseph-Louis Lagrange (1936-1813).
Constraint Qualification. For the above methods to work, it’s important that h1(x∗) and/or h2(x∗) is not 0. This restriction is called the Constraint Qualification (under equality constraints).
An alternative interpretation of the tangency condition is that the gradient vector ∇f (x) and ∇h(x) must “line up” at x∗, i.e., ∇f (x) and ∇h(x) point in the same or opposite directions: ∇f (x∗) = (f1(x∗), f2(x∗))′ = µ∗∇h(x∗) = µ∗(h1(x∗), h2(x∗))′.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 22 / 29 Constrained Optimization–Equality Figure: Problem max{f (x, y) = 3x + y | g(x, y) = x2 2 + y 2 2 = c} with c = 2.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 23 / 29 Constrained Optimization–Equality Theorem (Maximization with One Equality Constraint) Let f , h : R2 →R be C 1 and (x∗ 1, x∗ 2) a solution to max f (x1, x2) subject to h(x1, x2) = c. Suppose that (x∗ 1, x∗ 2) is not a stationary point of h(x1, x2). Then there is µ∗∈R s.t. (x∗ 1, x∗ 2, µ∗) is a stationary point of the Lagrangian function L(x1, x2; µ) = f (x1, x2) −µ[h(x1, x2) −c], i.e., ∂L ∂x1 = 0, ∂L ∂x2 = 0, ∂L ∂µ = 0 at (x∗ 1, x∗ 2, µ∗).
Example. Consider the problem max{f (x1, x2) = x2 1x2 | (x1, x2) ∈C} with C = {(x1, x2) : 2x2 1 + x2 2 = 3}. We solve this problem in 3 steps.
Step 1. Check Constraint Qualification. The unique stationary point for h(x1, x2) = 2x2 1 + x2 2 is (0, 0), but (0, 0) / ∈C.
Step 2. Set up the Lagrangian: L(x1, x2; µ) = x2 1x2 −µ(2x2 1 + x2 2 −3) Bo Chen (HKUST) Optimization-Part I Fall, 2024 24 / 29 Constrained Optimization–Equality Step 2 (Cont). The first-order conditions for L are ∂L ∂x1 = 2x1(x2−2µ) = 0; ∂L ∂x2 = x2 1−2µx2 = 0; ∂L ∂µ = 2x2 1+x2 2−3 = 0.
Solving the non-linear system of equations with 3 unknowns leads to 6 candidates: (x1, x2; µ) ∈{(0, ± √ 3, 0), (±1, 1, 0.5), (−1, −1, 0.5), (1, −1, −0.5)}.
Step 3. Since f (x1, x2) is C 1 and C is compact, Weierstrass Theorem implies that a max and a min exist. By Theorem of Max with One Equality Constraint, the constrained max is one of the 6 candidates: max: f (1, 1) = f (−1, 1) = 1; min: f (1, −1) = f (−1, −1) = −1, while f (0, √ 3) = f (0, − √ 3) = 0 is neither max nor min.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 25 / 29 Optimization with Multiple Equality Constraints Consider the maximization problem with x = (x1, ..., xn): max f (x) s.t.
x ∈Ch = {x : h1(x) = a1, ..., hm(x) = am}.
Constraint Qualification (CQ): ( ∂hi ∂x1 (x∗), ..., ∂hi ∂xn (x∗)) ̸= 0 for all i or Dh(x∗) = ∂h1 ∂x1 (x∗) · · · ∂hi ∂xn (x∗) .
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∂hm ∂x1 (x∗) · · · ∂hm ∂xn (x∗) .
Non-degenerate CQ (NDCQ): rank(Dh(x∗)) = m, i.e., the constraint set has a well-defined m–dimensional tangent plane everywhere.
Theorem (Maximization with Multiple Equality Constraints) Consider max{f (x) | x ∈Ch} with f , h1, ..., hm being C 1. Suppose x∗∈Ch is a (local) max of f on Ch and NDCQ is satisfied. Then ∃µ∗ 1, ..., µ∗ m s.t.
(x∗, µ∗) = (x∗ 1, ..., x∗ n; µ∗ 1, ..., µ∗ m) is a stationary point of Lagrangian L(x, µ) = f (x) −µ1[h1(x) −a1] −· · · −µm[hm(x) −am], i.e., ∂L ∂x1 = 0,..., ∂L ∂xn = 0; ∂L ∂µ1 = 0,..., ∂L ∂µm = 0 at (x∗ 1, ..., x∗ n; µ∗ 1, ..., µ∗ m).
Bo Chen (HKUST) Optimization-Part I Fall, 2024 26 / 29 Optimization with Multiple Equality Constraints Example Consider the maximization problem max f (x, y, z) = xyz s.t.
( h1(x, y, z) = x2 + y2 = 1 (a cylinder parallel to the z–axis) h2(x, y, z) = x + z = 1 (a plane parallel to the y–axis) We first calculate the Jacobian matrix: Dh(x, y, z) = 2x 2y 0 1 0 1 .
rank(Dh) < 2 ⇔x = y = 0, violating h1(x, y, z) = 1. NDCQ ✓.
Set up the Lagrangian L(x, y, z; µ1, µ2) = xyz −µ1(x2 + y2 −1) −µ2(x + z −1).
The FOC’s ∂L ∂x = 0, ∂L ∂y = 0, ∂L ∂z = 0, ∂L ∂µ1 = 0, ∂L ∂µ2 = 0 lead to 4 solution candidates. By the previous Theorem, evaluating the objective function at the 4 candidates yields the solution.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 27 / 29 Optimization with Multiple Equality Constraints Example (The Lagrangian Method Can Fail) Consider the maximization problem max f (x, y) = −y s.t.
h(x, y) = y3 −x2 = 0.
We first calculate the Jacobian matrix: Dh(x, y, z) = −2x 3y2 .
rank(Dh) < 1 ⇔x = y = 0. NDCQ fails at (0, 0).
Set up the Lagrangian L(x, y; µ) = −y −µ(y3 −x2).
The FOC’s are −2µx = 0, −1 + 3µy2 = 0, −x2 + y3 = 0. There is however no solution: if µ = 0, the second equation fails, while if x = 0, then y = 0 and the second equation again fails.
The problem has a unique global max at (0, 0), but NDCQ is not met at the optimum, leading to zero stationary points for the Lagrangian.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 28 / 29 Optimization with Multiple Equality Constraints Example (The Lagrangian Method Can Fail) Consider the maximization problem max f (x, y) = x2 −y2 s.t.
h(x, y) = 1 −x −y = 0.
We first calculate the Jacobian matrix: Dh(x, y, z) = −1 −1 .
rank(Dh) = 1 for all (x, y). NDCQ always holds.
Set up the Lagrangian L(x, y; µ) = x2 −y2 −µ(1 −x −y).
FOC’s: 2x + µ = 0, −2y + µ = 0, 1 −x −y = 0. Again no solution exists: if µ ̸= 0, then x = −y, violating the third equation, while if µ = 0, then x = y = 0 and the third equation again fails.
The issue of this example is that global maxima and minima fail to exist and it is unrelated to NDCQ, which is always satisfied.
Bo Chen (HKUST) Optimization-Part I Fall, 2024 29 / 29 |
3560 | https://www.math.cmu.edu/~lohp/docs/math/b3-func-eqn.pdf | B-III. Functional Equations Po-Shen Loh 18 June 2004 Warm-Ups 1. (Russia 2000/9) Find all functions f : R →R which satisfy f(x+y)+f(y+z)+f(z+x) ≥3f(x+2y+3z) for all x, y, z.
Solution: Answer: f constant. Solution: Put x = a, y = z = 0, then 2f(a) + f(0) ≥3f(a), so f(0) ≥f(a). Put x = a/2, y = a/2, z = −a/2.
Then f(a) + f(0) + f(0) ≥3f(0), so f(a) ≥f(0). Hence f(a) = f(0) for all a. But any constant function obviously satisfies the given relation.
2. (MOP97/2/1) Let f be a real-valued function which satisfies (a) for all real x, y, f(x + y) + f(x −y) = 2f(x)f(y).
(b) there exists a real number x0 such that f(x0) = −1.
Prove that f is periodic.
Solution: Swapping x and y yields that function is even. Yet plugging in x = y = 0 we get f(0) = 0 or 1. If it is 0, then plugging in y = 0 yields f ≡0, done.
Otherwise, f(0) = 1, and plug in x = y = x0/2 to get f(x0)+1 = 2f(x0/2)2, implying that f(x0/2) = 0.
Now plugging in y = x0/2, we get that f(x + x0/2) = −f(x −x0/2), so function inverts sign every x0.
Hence periodic with period 2x0.
3. (Balkan 1987/1) f is a real valued function on the reals satisfying (1) f(0) = 1/2, (2) for some real a we have f(x + y) = f(x)f(a −y) + f(y)f(a −x) for all x, y. Prove that f is constant.
Solution: Put x = y = 0.
We get f(0) = 2f(0)f(a), so f(a) = 1/2.
Put y = 0, we get f(x) = f(x)f(a) + f(0)f(a −x), so f(x) = f(a −x). Put y = a −x, we get f(a) = f(x)2 + f(a −x)2, so f(x) = 1/2 or −1/2.
Now take any x. We have f(x/2) = 1/2 or −1/2 and f(a−x/2) = f(x/2). Hence f(x) = f(x/2+x/2) = 2f(x)f(a −x/2) = 1/2.
Problems 1. (IMO 2002/5) Find all real-valued functions f on the reals such that [f(x) + f(y)][f(u) + f(v)] = f(xu −yv) + f(xv + yu) for all x, y, u, v.
Solution: Plug in all 0; then f(0) = 0 or 1/2. If 1/2, then plug in x = y = 0 and get f(u)+f(v) = 1 implying that constant at 1/2. Now suppose f(0) = 0.
Plug in x = v = 0. Then f(y)f(u) = f(yu). So f(1)2 = f(1) implying that f(1) = 0 or 1. If 0, then multiplicativity implies that constant at 0. Else: Plug in x = y = 1. Now 2[f(u) + f(v)] = f(u −v) + f(u + v). Using u = 0, v = 1, get f(−1) = 1.
Multiplicativity implies f is even.
1 Plug in x = y, u = v.
Then 4f(x)f(u) = f(2xu).
Multiplicativity implies that f(2) = 4.
More multiplicativity gives that f(x) = x2 for all powers of 2. Inductively using 2[f(u) + f(v)] = f(u −v) + f(u + v), get that f(z) = z2 for all integers. Reverse multiplicativity implies that f(q) = q2 for all rationals.
Multiplicativity implies f(x2) = f(x)2 so f ≥0. Yet plug in x = v, y = u and get f(x2 + y2) = [f(x) + f(y)]2 ≥f(x)2 by nonnegativity, so increasing function.
2. (IMO 1999/6) Determine all functions f : R →R such that f(x −f(y)) = f(f(y)) + xf(y) + f(x) −1 for all x, y ∈R.
Solution: Let c = f(0) and A be the image f(R). If a is in A, then it is straightforward to find f(a): putting a = f(y) and x = a, we get f(a −a) = f(a) + a2 + f(a) −1, so f(a) = (1 + c)/2 −a2/2 ().
The next step is to show that A −A = R. Note first that c cannot be zero, for if it were, then putting y = 0, we get: f(x −c) = f(c) + xc + f(x) −1 () and hence f(0) = f(c) = 1. Contradiction. But () also shows that f(x −c) −f(x) = xc + (f(c) −1). Here x is free to vary over R, so xc + (f(c) −1) can take any value in R.
Thus given any x in R, we may find a, b ∈A such that x = a −b.
Hence f(x) = f(a −b) = f(b) + ab + f(a) −1. So, using (): f(x) = c −b2/2 + ab −a2/2 = c −x2/2.
In particular, this is true for x ∈A. Comparing with () we deduce that c = 1. So for all x ∈R we must have f(x) = 1−x2/2. Finally, it is easy to check that this satisfies the original relation and hence is the unique solution.
3. Let f(x) be a continuous function with f(0) = 1. Suppose that for every n ∈Z+ and any t ∈R: (f(t))n = f √nt .
Prove that there exists a constant c such that on R+, f(t) = ect2.
Solution: Suppose there exists t > 0 such that f(t) ≤0. Then there exists a minimal s0 > 0 such that f(s0) = 0. But then f(s0) = f(s0/ √ 2)2, contradicting minimality. Same holds for t < 0.
Therefore, this function exists: L(t) = log f(t). But then for any m, n ∈Z+, we have L((n/m)t) = (n/m)2L(t). Continuity tells us that L(r) = r2L(1) for any r ∈R+.
4. (IMO 1996/3) Let S be the set of non-negative integers. Find all functions f : S →S such that f(m + f(n)) = f(f(m)) + f(n) for all m, n.
Solution: Setting m = n = 0, the given relation becomes: f(f(0)) = f(f(0)) + f(0). Hence f(0) = 0. Hence also f(f(0)) = 0. Setting m = 0, now gives f(f(n)) = f(n), so we may write the original relation as f(m + f(n)) = f(m) + f(n).
So f(n) is a fixed point. Let k be the smallest non-zero fixed point. If k does not exist, then f(n) is zero for all n, which is a possible solution. If k does exist, then an easy induction shows that f(qk) = qk for all non-negative integers q. Now if n is another fixed point, write n = kq +r, with 0 ≤r < k. Then f(n) = f(r + f(kq)) = f(r) + f(kq) = kq + f(r). Hence f(r) = r, so r must be zero. Hence the fixed points are precisely the multiples of k.
But f(n) is a fixed point for any n, so f(n) is a multiple of k for any n. Let us take n1, n2, . . . , nk−1 to be arbitrary non-negative integers and set n0 = 0. Then the most general function satisfying the conditions we have established so far is: f(qk + r) = qk + nrk for 0 ≤r < k.
We can check that this satisfies the functional equation. Let m = ak + r, n = bk + s, with 0 ≤r, s < k.
Then f(f(m)) = f(m) = ak + nrk, and f(n) = bk + nsk, so f(m + f(n)) = ak + bk + nrk + nsk, and f(f(m)) + f(n) = ak + bk + nrk + nsk. So this is a solution and hence the most general solution.
2 5. (IMO 1994/2) Let S be the set of all real numbers greater than −1. Find all functions f : S →S such that f(x + f(y) + xf(y)) = y + f(x) + yf(x) for all x and y, and f(x)/x is strictly increasing on each of the intervals −1 < x < 0 and 0 < x.
Solution: Suppose f(a) = a. Then putting x = y = a in the relation given, we get f(b) = b, where b = 2a + a2. If −1 < a < 0, then −1 < b < a. But f(a)/a = f(b)/b. Contradiction. Similarly, if a > 0, then b > a, but f(a)/a = f(b)/b. Contradiction. So we must have a = 0.
But putting x = y in the relation given we get f(k) = k for k = x + f(x) + xf(x). Hence for any x we have x + f(x) + xf(x) = 0 and hence f(x) = −x/(x + 1).
Finally, it is straightforward to check that f(x) = −x/(x + 1) satisfies the two conditions.
6. (IMO 1992/2) Find all functions f defined on the set of all real numbers with real values, such that f(x2 + f(y)) = y + f(x)2 for all x, y.
Solution: The first step is to establish that f(0) = 0.
Putting x = y = 0, and f(0) = t, we get f(t) = t2. Also, f(x2 + t) = f(x)2, and f(f(x)) = x + t2. We now evaluate f(t2 + f(1)2) two ways. First, it is f(f(1)2 + f(t)) = t + f(f(1))2 = t + (1 + t2)2 = 1 + t + 2t2 + t4. Second, it is f(t2 + f(1 + t)) = 1 + t + f(t)2 = 1 + t + t4. So t = 0, as required.
It follows immediately that f(f(x)) = x, and f(x2) = f(x)2.
Given any y, let z = f(y).
Then y = f(z), so f(x2 + y) = z + f(x)2 = f(y) + f(x)2. Now given any positive x, take z so that x = z2.
Then f(x + y) = f(z2 + y) = f(y) + f(z)2 = f(y) + f(z2) = f(x) + f(y). Putting y = −x, we get 0 = f(0) = f(x + −x) = f(x) + f(−x). Hence f(−x) = −f(x). It follows that f(x + y) = f(x) + f(y) and f(x −y) = f(x) −f(y) hold for all x, y.
Take any x. Let f(x) = y. If y > x, then let z = y −x. f(z) = f(y −x) = f(y)−f(x) = x−y = −z. If y < x, then let z = x −y and f(z) = f(x −y) = f(x) −f(y) = y −x. In either case we get some z > 0 with f(z) = −z < 0. But now take w so that w2 = z, then f(z) = f(w2) = f(w)2 ≥0. Contradiction.
So we must have f(x) = x.
7. (Balkan 2000/1) Find all real-valued functions on the reals which satisfy f(xf(x) + f(y)) = f(x)2 + y for all x, y.
Solution: Answer: (1) f(x) = x for all x; (2) f(x) = −x for all x.
Put x = 0, then f(f(y)) = f(0)2 + y. Put y = −f(0)2 and k = f(y). Then f(k) = 0. Now put x = y = k. Then f(0) = 0 + k, so k = f(0). Put y = k, x = 0, then f(0) = f(0)2 + k, so k = 0. Hence f(0) = 0.
Put x = 0, f(f(y)) = y (). Put y = 0, f(xf(x)) = f(x)2 (). Put x = f(z) in (), then using f(z) = x, we have f(zf(z)) = z2. Hence z2 = f(z)2 for all z (). In particular, f(1) = 1 or −1.
Suppose f(1) = 1. Then putting x = 1 in the original relation we get f(1 + f(y)) = 1 + y. Hence (1 + f(y))2 = (1 + y)2. Hence f(y) = y for all y.
Similarly if f(1) = −1, then putting x = 1 in the original relation we get f(−1 + f(y)) = 1 + y. Hence (−1 + f(y))2 = (1 + y)2, so f(y) = −y for all y.
Finally, it is easy to check that f(x) = x does indeed satisfy the original relation, as does f(x) = −x.
8. (IMO 1990/1) Construct a function from the set of positive rational numbers into itself such that f(xf(y)) = f(x)/y for all x, y.
Solution: We show first that f(1) = 1. Taking x = y = 1, we have f(f(1)) = f(1). Hence f(1) = f(f(1)) = f(1f(f(1))) = f(1)/f(1) = 1.
Next we show that f(xy) = f(x)f(y). For any y we have 1 = f(1) = f(1/f(y)f(y)) = f(1/f(y))/y, so if z = 1/f(y) then f(z) = y. Hence f(xy) = f(xf(z)) = f(x)/z = f(x)f(y).
Finally, f(f(x)) = f(1f(x)) = f(1)/x = 1/x.
We are not required to find all functions, just one. So divide the primes into two infinite sets S = {p1, p2, . . .} and T = {q1, q2, . . .}. Define f(pn) = qn, and f(qn) = 1/pn. We extend this definition to all rationals using f(xy) = f(x)f(y): f(pi1pi2 · · · qj1qj2 · · · /(pk1 · · · qm1 · · ·)) = pm1 · · · qi1 · · · /(pj1 · · · qk1 · · ·).
It is now trivial to verify that f(xf(y)) = f(x)/y.
3 9. (IMO Shortlist 1995/A5) Does there exist a real-valued function f on the reals such that f(x) is bounded, f(1) = 1 and f(x + 1/x2) = f(x) + f(1/x)2 for all non-zero x?
Solution: Answer: no.
Suppose there is such a function.
Let c be the least upper bound of the set of values f(x).
We have f(2) = f(1 + 1/12) = f(1) + f(1/1)2 = 2. So c ≥2. But definition we can find y such that f(y) > c −1/4. So c ≥f(y + 1/y2) = f(y) + f(1/y)2 > c −1/4 + f(1/y)2. So f(1/y)2 < 1/4 and hence f(1/y) > −1/2.
We also have c ≥f(1/y+y2) = f(1/y)+f(y)2 > −1/2+(c−1/4)2 = c2−c/2−7/16. So c2−3c/2−7/16 < 0, or (c −3/4)2 < 1. But c ≥2, so that is false. Contradiction. So there cannot be any such function.
4 |
3561 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5864449/ | Lyme disease: clinical diagnosis and treatment - PMC
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Can Commun Dis Rep
. 2014 May 29;40(11):194–208. doi: 10.14745/ccdr.v40i11a01
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Lyme disease: clinical diagnosis and treatment
TF Hatchette
TF Hatchette
1 Department of Pathology and Laboratory Medicine, Capital District Health Authority (CDHA), Halifax, Nova Scotia
2 Department of Medicine, CDHA and Dalhousie University, Halifax, Nova Scotia
3 Department of Pathology, Dalhousie University, Halifax, Nova Scotia
Find articles by TF Hatchette
1,2,3,, I Davis
I Davis
1 Department of Pathology and Laboratory Medicine, Capital District Health Authority (CDHA), Halifax, Nova Scotia
2 Department of Medicine, CDHA and Dalhousie University, Halifax, Nova Scotia
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1,2, BL Johnston
BL Johnston
2 Department of Medicine, CDHA and Dalhousie University, Halifax, Nova Scotia
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2
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1 Department of Pathology and Laboratory Medicine, Capital District Health Authority (CDHA), Halifax, Nova Scotia
2 Department of Medicine, CDHA and Dalhousie University, Halifax, Nova Scotia
3 Department of Pathology, Dalhousie University, Halifax, Nova Scotia
Corresponding author: todd.hatchette@cdha.nshealth.ca
Series information
Clinical aspects of Lyme disease
Collection date 2014 May 29.
PMC Copyright notice
PMCID: PMC5864449 PMID: 29769842
Abstract
Background
Lyme disease is an emerging zoonotic infection in Canada. As the Ixodes tick expands its range, more Canadians will be exposed to Borrelia burgdorferi, the bacterium that causes Lyme disease.
Objective
To review the clinical diagnosis and treatment of Lyme disease for front-line clinicians.
Methods
A literature search using PubMed and restricted to articles published in English between 1977 and 2014.
Results
Individuals in Lyme-endemic areas are at greatest risk, but not all tick bites transmit Lyme disease. The diagnosis is predominantly clinical. Patients with Lyme disease may present with early disease that is characterized by a “bull’s eye rash”, fever and myalgias or with early disseminated disease that can manifest with arthralgias, cardiac conduction abnormalities or neurologic symptoms. Late Lyme disease in North America typically manifests with oligoarticular arthritis but can present with a subacute encephalopathy. Antibiotic treatment is effective against Lyme disease and works best when given early in the infection. Prophylaxis with doxycyline may be indicated in certain circumstances. While a minority of patients may have persistent symptoms, evidence does not demonstrate that prolonged courses of antibiotics improve outcome.
Conclusion
Clinicians need to be aware of the signs and symptoms of Lyme disease. Knowing the regions where Borrelia infection is endemic in North America is important for recognizing patients at risk and informing the need for treatment.
Introduction
Lyme disease is the most common vector-borne zoonosis in North America, where it is estimated that each year more than 300,000 individuals in the United States (US) are infected by Borrelia burgdorferi, the bacterium that causes Lyme disease (1). Data from Canada are sparse, as Lyme disease only became nationally reportable in 2009. The number of reported infections in Canada in 2012 was 315, which is likely an underestimate and includes cases that are the result of infections acquired during travel to other endemic regions, such as Europe (2-4).
The range of Ixodes ticks, the vector for Borrelia, has expanded greatly over the last 20 years, making Lyme disease an emerging infection in Canada. Ixodes ticks have been identified in British Columbia, southeastern Manitoba, southern and eastern Ontario, and specific areas in New Brunswick, Nova Scotia and Quebec (4,5). As the Ixodes tick expands its range, more Canadians will be exposed to B. burgdorferi, and clinicians will need the expertise to manage tick bites and diagnose and treat Lyme disease.
Awareness of the extent of local Lyme disease, early detection and treatment are important for managing this emerging infection. The objective of this article is to review the clinical diagnosis and treatment of B. burgdorferi infection for the front-line clinician.
Methods
A literature search was done using PubMed and restricted to articles published in English between 1977 and 2014. The following search terms were used: Lyme disease and treatment, Lyme disease and arthritis, Lyme disease and clinical presentation, Lyme disease ticks and risks, Lyme disease and pregnancy, neuroborreliosis, chronic Lyme OR post-Lyme syndrome AND treatment (clinical trials). Abstracts were reviewed for clinically relevant articles, and additional references were obtained from the articles identified in the initial search.
Although there are different strains of Borrelia in North America and Europe, which are associated with overlapping but distinct clinical presentations, this review will focus on the clinical manifestations that are common to Canada. It does not describe the various aspects of diagnostic testing, which is reviewed in a companion paper (6).
Findings
Who is at risk of Lyme disease and why?
An individual must first be bitten by a tick that is infected with B. burgdorferi for infection to occur. While infected ticks can be sporadically deposited in different regions in Canada by migratory birds, the risk of getting Lyme disease is greatest in an area where the ticks are persistently present or endemic, and the risk depends on the proportion of ticks that are infected. Even in areas where the ticks have become endemic, the proportion infected by this spirochete varies and is evolving. Clinicians may want to check with their local public health departments to understand where the current Lyme endemic areas are within their jurisdiction. It is important to obtain a thorough travel history from the patient, as Lyme disease is endemic in other areas of North America and Europe, and travel-related Lyme disease accounts for a proportion of the documented cases in Canada. It is also important to relay this travel history or suspicion of European Borrelia infection to the diagnostic laboratory, as the confirmatory testing for North American and European Borrelia strains differs.
What is the risk of Lyme after a tick bite?
While studies are limited in numbers and size, data from the US and Europe suggest that the risk of acquiring Lyme disease, even after a bite from an infected tick, is small – in the range of < 1% to6% (7). In studies that have examined the efficacy of antibiotic prophylaxis in persons presenting with a tick bite in Lyme-endemic areas (where the prevalence of B. burgdorferi in ticks ranged from 15% to 50%), the rate of infection among those who received placebo ranged from 1.1% to 3.4% (8-11). These rates are consistent with a prospective study of patients bitten by ticks in a Lyme-endemic area of New York State, where infection developed in 3.7% of those bitten by a tick (12).
The risk of infection following a tick bite is related to how long the tick has been attached and is explained by the Borrelia life cycle in the tick. The spirochete first has to migrate from the tick’s gut to its salivary glands before it can be injected into the human through the bite. In order to do this, the bacterium must undergo antigenic change to its outer coat that favours human infection (13-16). This process takes about 36 hours (14). If the tick is removed during this time frame, infection is almost always prevented. In an antibiotic prophylaxis trial in the US, no study participants receiving placebo or prophylaxis were infected if the tick fed for less than 72 hours (11). This is further supported by data from another prospective study in an endemic region, in which participants were at greatest risk of infection if the tick had been attached for more than 72 hours (12).
Who should receive prophylaxis after a tick bite?
There are a number of factors that require consideration when deciding whether to offer prophylaxis after a tick bite (Table 1). All of the prophylaxis trials have been conducted in regions that would be considered hyperendemic for Lyme disease, where the burden of infection in ticks is > 20% and where the antibiotics were administered within 72 hours of removing the tick (8-11). An early meta-analysis of antibiotics for prophylaxis of Lyme disease reported a pooled infection rate in those who received placebo of 1.4%, compared with 0% for those receiving an antibiotic (18). Each of the included trials was small and failed to show any significant effect of prophylaxis, as the incidence of infection in those taking placebo was low (8-10). A subsequent randomized placebo controlled trial showed that a single dose of doxycycline could reduce the rate of Lyme disease (11).
Table 1. Criteria and recommended prophylaxis after a tick bite (17).
Criteria for prophylaxis1. The attached tick can be reliably identified as an I. scapularis (deer/blacklegged) tick that is estimated to have been attached for > 36 h on the basis of the degree of engorgement or by certainty about the time of tick acquisition.
2. Prophylaxis can be started within 72 h of tick removal.
3. The local rate of B. burgdorferi infection in ticks is > 20% (check with local public health).
4. Doxycycline is not contraindicated.
Recommended prophylaxisAdults and children > 8 years of age Single 200 mg dose of doxycycline given orally (4.4 mg/kg for patients < 45 kg)
Children < 8 years of age and pregnancy Not recommended
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Consider antibiotic prophylaxis after a tick bite when the individual meets all of the criteria.
An updated systematic review and meta-analysis suggested that, to prevent one case of Lyme disease, prophylaxis of 49 cases would be needed (19). Patients in whom infection developed were successfully treated without any long-term sequelae. Given that there were relatively few infections in each of the studies and that these were successfully treated, the use of antimicrobial prophylaxis must be considered in the context of potential side effects, which were reported but were not severe.
How do patients with Lyme disease present and how are they treated?
Although patients with B burgdorferi infection can be asymptomatic, most cases of Lyme disease present as one of three stages, which may occur sequentially if an earlier stage was untreated: early localized disease (usually < 30 days from exposure), early disseminated disease (< 3 months after exposure) and late disseminated disease (> 3 months after exposure) (Table 2).
Table 2. Clinical manifestations of Lyme disease (17,20,21).
| Stage | System | Manifestation |
:---
| Early localized disease (< 30 days) | Skin | Erythema migrans (note: must be > 5 cm in diameter, painless and slowly expanding) |
| Systemic | Fever Arthralgias Headache |
| Early disseminated disease (< 3 months) | Skin | Multiple erythema migrans |
| Systemic | Fever Arthralgias Headache Lymphadenophathy |
| Heart | Atrioventricular block Tachyarrhythmias Myopericarditis Myocardial dysfunction |
| CNS | Aseptic meningitis Cranial neuropathy (especially facial nerve palsy) Motor or sensory radiculopathy |
| Ocular | Conjunctivitis (rare) |
| Late disseminated disease (> 3 months) | MSK | Oligoarticular arthritis |
| CNS | Encephalopathy Axonal polyradiculoneuropathy Chronic encephalomyelitis |
| Ocular | Retinitis (rare) |
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Note: While there may be exceptions, these time frames provide clinicians with a general guide on when the different manifestations tend to occur. CNS = central nervous system. MSK = musculoskeletal.
The treatment of Lyme disease depends on the stage and organ system involved and is summarized in a clinical practice guideline that has been developed by the Infectious Diseases Society of America (IDSA) (Table 3) (17). The clinical presentation and some of the literature that supports those treatment recommendations are summarized in the sections that follow.
Table3. Infectious Disease Society of America guidelines for the treatment of Lyme disease (17).
Treatment of adults and children older than 8 years with Lyme diseaseErythema migrans or early disseminated disease, including Bell’s palsy, but without other CNS involvement Doxycycline 100 mg po bid x 14–21 days (contraindicated in pregnancy)
Amoxicillin 500 mg po tid x 14–21 days
Cefuroxime 500 mg po bid x 14–21 days
Early Lyme with CNS involvement Ceftriaxone 2 g IV once daily x 14–28 days
Pen G 4 x10 6 units IV q 4 h x 14–28 days
Doxycycline 100–200 mg po bid x 28 days (alternative if others not possible)
Early Lyme with carditis Same treatment as early Lyme with CNS involvement, but use IV initially with high grade heart block or if admission to hospital is necessary.
Late Lyme without CNS involvement Doxycycline 100 mg po bid x 8 days
Amoxicillin 500 mg po id x 28 days
Cefuroxime 500 mg po bid x 28 days
Late Lyme with CNS involvement (late neuroborreliosis) Ceftriaxone 2 g IV once daily x 14–28 days
Pen G 4 x10 6 units IV q 4 h x 14–28 days
Treatment of children 8 years or younger with Lyme diseaseEarly localized disease Amoxicillin 30 mg/kg per day, orally, divided into three doses (max 1.5 g/day) for 14–21 days
For children allergic to penicillin, cefuroxime 30 mg/kg per day, orally, in two divided doses (maximum 1 g/day) for 14–21 days
Early disseminated and late disease: multiple erythema migrans Oral treatment as the above for 21 days
Early disseminated and late disease: isolated facial palsy and first episodes of arthritis Oral treatment as the above for 21 days
Early disseminated and late disease: persistent/recurrent arthritis, carditis and meningitis/encephalitis Ceftriaxone or penicillin IV at pediatric dosing
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These recommendations are off-label, evidence-based best practices. Recurrent or persistent joint swelling – repeat 4 week course of oral antibiotic as above. Use of IV ceftriaxone should be reserved for relapse or persistent joint swelling without improvement with oral treatment. CNS = central nervous system.
Asymptomatic infection
Approximately 1.6%–7% of infected individuals may have asymptomatic infection (7,22). The prognosis for patients with asymptomatic infection is generally good. Within a large vaccine study conducted in 10 US states where Lyme is endemic, asymptomatic infections were documented in 6% (15/269) of study participants. While the majority of study participants received treatment when the seroconversion had been documented, only 1 of 8 who did not receive treatment went on to show arthritis over the 12 months after seroconversion (22). Although the follow-up period was relatively short, these data support the notion that asymptomatic infection does not require treatment, and disease will ultimately declare itself if infection persists.
Early localized disease (< 30 days)
Early disease usually presents as an acute illness characterized by fever, headache and myalgia with the presence of a single, localized skin lesion known as erythema migrans (EM). This characteristic skin eruption is present in approximately 80% of patients with early disease and will resolve without antibiotic treatment over a median of 28 days (23,24). While most patients will present with erythema migrans within seven days of the initial tick bite, the incubation period can vary between 3 and 30 days (25). The skin lesion is characteristically an annular erythematous lesion > 5 cm in diameter that slowly increases in size and is usually painless and non-pruritic. The lesion sometimes develops central clearing, but it can be more homogenously erythematous (17,20,25). It is important to note that many people can have a reaction to the tick saliva and show a localized reaction resembling erythema migrans. However, unlike erythema migrans this reaction often develops within the first three days of the bite, is 5 cm or less in diameter and does not expand.
Erythema migrans can be subtle or even go unnoticed, particularly if the bite was in an area that is difficult for the patient to see, such as behind the knee. Diagnostic testing is insensitive at this stage and not recommended in a patient who has links to an endemic area (6).
Treatment: Several clinical trials conducted in the US (25-29) and Europe (30-32) has informed the antimicrobial treatment of early Lyme disease. In these trials, individuals with a clinical diagnosis of erythema migrans were randomly assigned to one of several treatment options. The agents used included amoxicillin, penicillin, ceftriaxone, a macrolide (erythromycin or azithromycin) and a tetracycline or doxycycline in somewhat varying doses and for varying lengths of times. Azithromycin therapy was given for 5 or 10 days, beta-lactams were given for 10–30 days, and tetracycline/doxycycline was given for 10, 14 or 20–21 days. Follow-up frequently lasted for several months after completion of therapy, and patients were monitored for complete and partial resolution of symptoms, treatment failure and late manifestations of illness that were considered either minor or major.
The studies found similar efficacy for amoxicillin, azithromycin, cefuroxime and doxycycline independent of the dose and duration of therapy. The one trial that included erythromycin showed it to be less effective than penicillin and tetracycline. A large retrospective cohort study of 607 patients confirmed these findings, 99% of patients with early local or disseminated Lyme remaining treatment-failure free at two years whether treated for ≤ 10 days, 11–15 days or > 15 days (33). One study that looked specifically at early disseminated Lyme disease compared ceftriaxone 2 g daily for 2 weeks with doxycycline 100 mg twice daily for 3 weeks (34). Clinical cure was similar for the two (85% for ceftriaxone and 88% for the doxycycline), with 14% (doxycycline) and 27% (ceftriaxone) of patients having residual symptoms, usually mild arthralgias and fatigue, at nine months after completing treatment.
Prognosis: Both clinical trials (25-32) and observational studies (33,35-37) provide information about the prognosis for treated early Lyme disease. With the currently recommended therapies, greater than 80% of patients will have complete resolution of symptoms at long-term follow-up. While it is not unusual for individuals to have symptoms such as fatigue and arthralgias after treatment for Lyme, most studies reported either no or few (< 5%) late manifestations of disseminated Lyme.
To determine whether patients who had early Lyme disease are more likely to have persistent symptoms after their acute infection, investigators in Slovenia compared a cohort of patients treated for erythema migrans with a control group of individuals who did not have Lyme disease (37). They found that the frequency of new or increased symptoms in patients with this condition did not exceed the frequency of such symptoms in the control group. At 12 months, only 2.2% of Lyme disease patients reported new or increased symptoms, and in none of the patients were the symptoms disabling. These findings are similar to those of a study done in the US involving patients with and without Lyme disease from 1984 to 1991 (38). The frequency of new symptoms and increased difficulties with daily activities were similar in the two groups, those with Lyme disease and age-matched controls without Lyme disease.
In summary, the evidence from these studies suggests that although patients with Lyme may have symptoms that persist or appear after treatment is complete, the frequency of those symptoms is comparable with what one would see in individuals who do not have Lyme disease.
Early disseminated Lyme disease (< 3 months)
Early disease may be followed by disseminated disease, with the development of multiple secondary annular lesions and multisystem and intermittent cardiac, neurological, ocular or articular manifestations. The development of multiple secondary annular lesions and systemic symptoms, including fever, arthralgias, headache and lymphadenopathy, usually occurs within several weeks of the localized erythema migrans (39). In one early study, half of the patients presenting with this condition went on to show multiple erythema migrans lesions (39).
The other manifestations of early disseminated infection, such as Lyme carditis and neurologic infection (known as neuroborreliosis), tend to occur weeks to months after initial infection (39). Patients with Lyme carditis can present with conduction abnormalities, tachyarrhythmias, myopericarditis or myocardial dysfunction (21). The most common presentation is atrioventricular block ranging from first to third degree heart block, often requiring temporary pacing. Carditis usual occurs within two months of the presenting erythema migrans, but a history of the condition is not always present.
Treatment is effective, and carditis is rarely fatal (40). While earlier data showed that carditis developed in up to 10% of patients with untreated Lyme (21), recent data suggest that rates are much lower; this may be due to improved recognition of Borrelia infection and effective treatment (17). US data from 1996 to 2006 suggest that only 0.8% of patients reported to have Lyme disease have conduction abnormalities (41). Physicians should think of Lyme as a cause of unexplained conduction abnormalities in patients who have recently been to a Lyme-endemic area. However, in the absence of a history of erythema migrans, the presentation is too non-specific to warrant empiric treatment without serological confirmation (17).
Neuroborreliosis has a clinical spectrum that can include meningitis, facial paralysis, motor or sensory radiculopathy, and cognitive symptoms. In North America, neuroborreliosis commonly presents with cranial nerve involvement (especially Bell’s palsy), with or without aseptic meningitis. Symptoms can appear anywhere from two to eight weeks after erythema migrans (42). Lyme disease can be responsible for Bell’s palsy in 10%–50% of children and adults in Lyme-endemic areas and can be bilateral in up to 23% of patients (43-47). Clinicians should suspect Lyme disease as the cause of facial palsy when it occurs in Lyme-endemic areas during the tick season, is associated with headache and other non-cranial nerve neurologic findings, including papilledema, or is bilateral (44,47-50).
In regions where Lyme is endemic, it can be challenging to differentiate early neuroborreliosis from other causes of central nervous system infection. The cerebrospinal fluid is usually abnormal, with a white cell count that is elevated to levels similar to those of patients presenting with viral meningitis (48,51). However, compared with other causes of aseptic meningitis, patients with Lyme meningitis are less likely to have fever and more likely to have a higher preponderance of mononuclear cells (total percentage of monocytes and lymphocytes) in the cerebrospinal fluid, a longer duration of headache, and cranial nerve involvement (48,52,53).
The rule of sevens: A clinical prediction rule has been developed to help stratify the likelihood of Lyme-related meningitis in children. Children are unlikely to have Lyme-related meningitis if they have all of the following: headache for fewer than seven days, less than 70% mononuclear cells in their cerebrospinal fluid and absence of a seventh nerve palsy. This “rule of 7s” has 96% sensitivity and 95% specificity for distinguishing aseptic meningitis from Lyme neuroborreliosis. On the basis of the rule of 7s, patients who lack these features can be managed conservatively while Lyme serological results are pending (53,54).
Treatment: A study done in the US in the early 1990s of patients with disseminated Lyme disease(without meningitis) compared ceftriaxone 2 g daily for 14 days with doxycycline 100 mg twice daily for 21 days (34). The cure rate was similar for the two regimens (88% for ceftriaxone and 85% for doxycycline), 27% of ceftriaxone-treated patients and 14% of doxycycline-treated patients having residual symptoms at the end of treatment.
In a large series of 101 adults with Lyme-related Bell’s palsy, the palsy resolved completely in more than 86% at a median time of 26 days (range 1–270 days) despite the fact that only 36% had received antibiotic treatment. Fifteen percent had mild residual weakness, and only one person had a severe deficit. Patients with residual dysfunction were more likely to have bilateral disease (43).
The prognosis in children is equally good. In a cohort study, 94% of children treated with antibiotics were asymptomatic four weeks after treatment. All but one patient had resolution of symptoms by six months, and no children had evidence of chronic disease two years later (35). In 43 children who presented with Bell’s palsy attributable to Lyme disease, most felt that they were cured, and there was no difference in the daily activities of infected children compared with those of a non-infected group (55). These studies in patients with Lyme neuroborreliosis highlight the importance of early recognition and treatment in order to improve response to treatment. At the same time, it was demonstrated that a prolonged course of amoxicillin given orally after a three week course of intravenous ceftriaxone did not alter outcome as compared with placebo (56).
Late Lyme disease (> 3 months)
If left untreated, Lyme disease can evolve into late disease, which presents with persistent arthritis and/or persistent neuroborreliosis. The different strains of Borrelia in North America and Europe are associated with overlapping but distinct clinical presentations. In Europe, late Lyme disease presents with dermatologic syndromes that are rare in North America, including acrodermatitis chronica atrophicans and lymphocytoma cutis (20). In North America, arthritis is a much more common manifestation of late Lyme disease, whereas in Europe neuroborrelia is more commonly seen and presents as a chronic encephalitis or painful radiculopathy called Bannwarth’s syndrome in up to 86% of cases (57).
If European Lyme disease is suspected on the basis of the clinical presentation and the patient’s travel history, the laboratory should be notified to ensure that the appropriate serological test can be used, as the confirmation tests for North American and European Borrelia are different. A more detailed comparison of the differences between North American and European Lyme disease has been provided by Hengge and colleagues (20).
Late Lyme arthritis
In North America, up to 60% of untreated Lyme disease cases show monoarticular or oligoarticular arthritis, usually involving the knees, with other joints such as the ankle, elbow and wrist less commonly affected (58-60). In one study, all patients with late Lyme arthritis had symptoms appear within two years of infection. In 14 of 16 patients with untreated Lyme-related facial nerve palsies arthritis ultimately developed (61). Remembering a tick bite has not been a helpful feature in diagnosing Lyme arthritis, as up to 84% of cases did not recall exposure to a tick (33,62).
Lyme disease serology is very sensitive and specific for the diagnosis of Lyme arthritis, and a negative test essentially rules out infection (6). However, the results of Lyme disease testing can take time, and because the presentation can mimic septic arthritis, which requires immediate treatment, it is important to try and differentiate the two.
Three studies have attempted to develop a clinical prediction model to distinguish Lyme arthritis from septic arthritis in Lyme-endemic areas. Although there is an overlap between the two entities, Lyme arthritis tends to involve the knees preferentially, is less frequently accompanied by fever and local signs of inflammation, and in general has lower inflammatory markers (62-64). One study found refusal to weight bear as the strongest predictor septic arthritis compared with Lyme disease (63). Another study in children found that an erythrocyte sedimentation rate of < 40 mm/hour and a peripheral white blood cell count < 10 x 10 3 cells/ mm 3 could effectively rule out septic arthritis (64). While the synovial fluid is inflammatory with elevated leukocyte counts, there is overlap with septic arthritis, and the cell count can range from 1,700 x 10 3 cells/µL to > 100,000 x 10 3 cells/µL, predominantly polymorphonuclear cells (58,60,63). In one review, children with Lyme disease had higher inflammatory markers than adults (60).
Up to 10% of patients will have evidence of synovitis six months after treatment for Lyme arthritis, termed antibiotic refractory Lyme arthritis (65). Currently, the only recommendation for the use of the polymerase chain reaction (PCR) to detect Borrelia DNA in the diagnosis of Lyme disease has been in patients with persistent synovial swelling, to determine whether retreatment is necessary (17). However, the significance of a positive test result after primary treatment is unclear. There are data to suggest that any Borrelia detected are not viable, calling into question whether there really is persistent infection in these patients (66).
Late Lyme arthritis treatment: There are few studies that have examined the treatment of late Lyme arthritis. Two studies (67,68) comparing penicillin or ceftriaxone with placebo in late Lyme arthritis in adults found benefit with antibiotic therapy, but the cure rate was still < 50%, although better with ceftriaxone than penicillin. A European randomized trial comparing penicillin and ceftriaxone, both intravenously for 10 days, in patients 13 years of age and older demonstrated a symptom remission rate of 87.9% for ceftriaxone and 61.3% for penicillin. Finally, a study involving children and adults with late Lyme in the US in the mid-2000s (69) found that 28 days (70% cure) of ceftriaxone had no advantage over 14 days (76% cure) in terms of clinical cure and was associated with more treatment discontinuations due to adverse events. In some series up to 25% of children and 50% of adults required a second course of antibiotics because of refractory arthritis symptoms. Those who went on to show refractory arthritis were successfully treated with nonsteroidal anti-inflammatories, steroid injections or disease-modifying drugs. None developed long-term sequelae (59,60).
Late neuroborreliosis
The most common manifestation of neuroborreliosis in North America is subacute encephalopathy with subtle cognitive changes, whereas a chronic encephalomyelitis characterized by spastic paraparesis and cognitive impairment is more common in Europe (20,70). In one case series of 37 American patients, neuroborreliosis manifested approximately two years after erythema migrans as subacute encephalopathy; axonal polyradiculoneuropathy with objective sensory and electromyography abnormalities; or leukoencephalitis; or a combination of the three (42). It is important to note that the patients had objective findings and serological evidence of infection.
Late neuroborreliosis treatment: All of the original studies examining treatment of neuroborreliosis that were identified in our search were undertaken in Europe (56,71-77). While the clinical manifestations of Lyme disease are somewhat different in Europe and North America, presumably the treatment principles are similar. These different studies compared a variety of different drugs, treatment doses and treatment durations. Generally, however, the comparators were intravenous ceftriaxone/cefotaxime, intravenous penicillin or oral doxycycline (200 mg–400 mg daily) for anywhere between 10 days and 3 weeks. Patients had a mix of early and late neurological disease, and follow-up was often for several months after treatment had stopped. The treatment success was comparable for the different treatment arms, but the dose of doxycycline was higher in some of these studies than that used for early Lyme disease. Treatment success was generally higher than 80% but as low as 33% in one group of patients with late Lyme neuroborreliosis who received ceftriaxone for two weeks.
What is the prognosis after delayed treatment?
There are relatively few data examining the effects of delayed treatment. Two retrospective studies showed that patients who had a longer duration of symptoms before treatment were more likely to have persistent subjective musculoskeletal and cognitive symptoms. However, there were no objective physical findings or neurocognitive testing abnormalities compared with uninfected controls (78,79). Similar findings were reported in patients involved in studies of early Lyme disease, in which it was found that those who did not receive treatment on initial presentation with facial palsies were more likely to have body pain, physical limitations and a lower physical composite score on the SF36 (36 item Short-Form General Health Survey) compared with controls who received antibiotics at the time of presentation. They were also more likely to have mild neurocognitive symptoms, but their overall “mental composite” score on the SF-36 standardized form was not different from that of the control group (61).
Can patients with Lyme disease have persistent symptoms after treatment?
It is clear that a proportion of patients with confirmed evidence of previous Lyme disease continue to have symptoms after standard antimicrobial treatment. The terms “chronic Lyme disease” and “post-Lyme disease syndrome” have been applied by some clinicians to patients with symptoms persisting more than six months after treatment with the recommended agents. Persistent symptoms include fatigue, generalized musculoskeletal pain and cognitive impairment without objective findings or microbiological evidence of active infection.
Given that some patients have persistent symptoms after treatment of their Lyme disease, the question has been raised of whether there is benefit to prolonged antimicrobial therapy. Four randomized placebo controlled trials, all conducted in the US, have addressed this issue. The larger of the studies (80) included two separate studies (i.e. seropositive and seronegative studies) of patients clinically diagnosed with acute Lyme disease. Both studies enrolled patients who had previously been treated with a recommended antibiotic regimen and reported ongoing symptoms. The subjects were placed into the seropositive or seronegative study according to their serological status at the time of enrollment. There were 129 patients enrolled in the two studies, and participants were randomly assigned to receive either ceftriaxone 2 g daily for 30 days followed by doxycycline for 60 days, or placebo. Outcome measures included cognition, memory, pain and activities of daily living.
There were no differences between the treatment vs placebo groups in outcomes in either the seronegative or seropositive studies, and they were terminated early after an interim analysis by the Data and Safety Monitoring Board, on the basis that the studies would not be able to identify a difference between treatment vs placebo. A sub-study using data from this trial also found no differences between the treatment and control groups in mood and cognitive-related quality of life measures (81).
Krupp and colleagues compared 2 g daily of ceftriaxone for 28 days with placebo in 55 patients whose Lyme disease had been previously treated with a standard regimen (82). While the ceftriaxone group showed improvement in fatigue, there was no difference in cognition on formal testing. Three patients had intravenous line infections and one had anaphylaxis in reaction to ceftriaxone. Finally, 37 adults were randomly assigned to receive 10 weeks of ceftriaxone or placebo after completing treatment for Lyme disease (83). The patients who received ceftriaxone had non-sustained, mild improvements in cognition and a 26.1% adverse event rate versus 7.1% in the placebo group.
On the basis of these findings, the Infectious Diseases Society of America guideline does not recommend additional antimicrobial or prolonged treatment for patients who have completed a standard course of therapy for their Lyme disease (17). Any benefit that might be derived is small and not sustained, and has been associated with an excess risk of adverse events that may be life-threatening.
How should Lyme disease in pregnancy be managed?
While there have been reports of Lyme disease in pregnant women and sporadic case reports of transplacental transmission of B burgdorferi, there is not a clear link between fetal infections and adverse outcomes (84,85). Clinical, serological and epidemiological studies of Borrelia infection in pregnancy have failed to demonstrate an association between infection and adverse outcomes (84-86). Because doxycycline is contraindicated in pregnancy, treatment and prophylactic options are different than in the non-pregnant adult.
Some data suggest that a 10 day course of ampicillin may be an effective prophylactic strategy, although the risk of a rash developing in reaction to the beta lactam is greater than the risk of Lyme disease, and therefore the Infectious Diseases Society of America(IDSA) guideline does not recommend prophylaxis in the setting of pregnancy (17,19,86). If an Ixodes tick bites a pregnant woman, she should be monitored for 30 days for signs and symptoms of Lyme disease and treated with amoxicillin or cefuroxime if it develops (17).
Can individuals be re-infected with Borrelia burgdorferi?
Relapses are recurrent symptoms that are the result of failure to cure the original infection, whereas re-infection is the recurrence of symptoms as a result of a new exposure to an infected tick, leading to a new infection. Although erythema migrans lesions can relapse if not treated with antibiotics (25), their recurrence after successful treatment is more likely to be re-infection than relapse (87).
Re-infection can occur in as many as 2%–21% of patients living in endemic areas who have had Lyme disease (78,88-90). On examination, re-infection typically presents with an erythema migrans lesion at a different site than the original lesion more than 1–2.5 years after the original infection and not within 11 months of the first infection (87,89,91). In one series, 79% of patients with re-infection presented with erythema migrans at a different site than the previous infection, and 21% presented with a febrile illness with myalgias. Re-infection after late Lyme disease characterized by arthritis or neuroborreliosis is very rare (90,91).
Laboratory diagnosis of re-infection becomes a challenge, given that serology, including IgM, can remain positive for many years (92). Diagnosis is reliant on the clinical presentation of a new erythema migrans lesion at a different site. Recent data suggest that patients treated for re-infection have excellent outcomes. Patients with re-infection reported less fatigue than both the non-infected control group and patients with their first infection, suggesting that repeated infections present a lower risk of persistent symptoms (91).
Conclusions
Lyme disease is an emerging infection in Canada, and it is important that clinicians be aware of its epidemiology, clinical presentations and management. While sporadic cases of Lyme disease are possible from infected ticks that are imported on migrating birds, individuals living in or traveling to Lyme-endemic regions will be at greatest risk of infection. A thorough travel history is essential in a person presenting with symptoms suggestive of Lyme disease, particularly if symptoms compatible with one of the neurologic syndromes that are more common in Europe are present. It is important to inform the laboratory of a positive travel history, as the serologic testing for North American and European Borrelia strains is different.
Given the constantly expanding tick population it can be challenging to clearly define regions constituting the highest risk for Lyme disease, and it is important to remain abreast of local Lyme epidemiology through local public health.
Footnotes
Conflict of interest: The authors have no conflicts of interest to declare.
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Percent
A percent is a type of ratio where something is compared to a hundred. Probabilities, scores, and success rates are commonly written in percents.
Contents
1 Conversion to Fractions and Decimals
2 Percent of a Number
3 Percent Change
4 Percent Increase and Decrease
5 Problems
5.1 Introductory Problems
5.2 Intermediate Problems
Conversion to Fractions and Decimals
By definition, means , which means hundredths. In other words, converting percents to decimals means moving two decimal places to the left, and converting decimals to percents means moving two decimal places to the right. For instance, , , and .
Percent of a Number
Let be the percentage, be the number we want to take the percentage of, and be the wanted quantity. By definition, , so . In other words, to find the percent of the number, we convert the percent to the fraction (or decimal) and then multiply it with the number we want to take the percentage of. For example, of equals .
Percent Change
The percent change equals the ratio of the positive difference between the original quantity and the new quantity to the original quantity when written as a percent. For example, the percent change from 10 to 15 equals .
Percent Increase and Decrease
If the new quantity is greater than the original quantity, then the percent change is called a percent increase. If the new quantity is smaller than the original quantity, then the percent change is called a percent decrease.
Let be the percentage, be the new quantity, and be the old quantity. From the definition of percent change, a percent increase is written in the form , and solving for results in . Additionally, a percent decrease isn’t written in the form , and solving for results in .
For a written example, a increase from is , and a decrease from is .
Problems
Introductory Problems
Practice Problems on Alcumus
Simple Percents (Prealgebra)
Combining Percents (Prealgebra)
2006 AMC 10B Problems/Problem 4
Intermediate Problems
1990 AIME Problems/Problem 6
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3564 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9813436/ | Prevalence of pneumonia and its determinant factors among under-five children in Gamo Zone, southern Ethiopia, 2021 - PMC
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Front Pediatr
. 2022 Dec 22;10:1017386. doi: 10.3389/fped.2022.1017386
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Prevalence of pneumonia and its determinant factors among under-five children in Gamo Zone, southern Ethiopia, 2021
Yerukneh Solomon
Yerukneh Solomon
1 Department of Biomedical Sciences, College of Medicine and Health Sciences, Debre Berhan University, Debre Berhan, Ethiopia
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1,, Zelalem Kofole
Zelalem Kofole
2 Department of Biomedical Sciences, School of Medicine, College of Health Science, Arba Minch University, Arba Minch, Ethiopia
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2, Tewodros Fantaye
Tewodros Fantaye
2 Department of Biomedical Sciences, School of Medicine, College of Health Science, Arba Minch University, Arba Minch, Ethiopia
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2, Solomon Ejigu
Solomon Ejigu
3 Department of Biomedical Sciences, College of Medicine and Health Science, Mizan-Tepi University, Mizan-Aman, Ethiopia
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3
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1 Department of Biomedical Sciences, College of Medicine and Health Sciences, Debre Berhan University, Debre Berhan, Ethiopia
2 Department of Biomedical Sciences, School of Medicine, College of Health Science, Arba Minch University, Arba Minch, Ethiopia
3 Department of Biomedical Sciences, College of Medicine and Health Science, Mizan-Tepi University, Mizan-Aman, Ethiopia
Edited by: Gönül Tanır Dr Sami Ulus Child Health and Diseases Training and Research Hospital, Turkey
Reviewed by: Mohammed Ahmed Haramaya University, Ethiopia Furqan Khurshid Hashmi Punjab University, Pakistan
Correspondence: Yerukneh Solomon yeruknesolomon@gmail.com
Specialty Section: This article was submitted to Pediatric Infectious Diseases, a section of the journal Frontiers in Pediatrics
Received 2022 Aug 11; Accepted 2022 Nov 17; Collection date 2022.
© 2022 Solomon, Kofole, Fantaye and Ejigu.
This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
PMC Copyright notice
PMCID: PMC9813436 PMID: 36619517
Abstract
Background
Pneumonia, which is a form of acute lower respiratory tract infection, affects the lung parenchyma and destructs alveolar air space. Pneumonia is the leading cause of morbidity and mortality in under-five children. It was estimated that pneumonia kills 900,000 under-five children each year worldwide. Approximately 172 deaths per 1,000 live births occur in sub-Saharan African countries, with pneumonia being the major cause. This study aimed to assess the prevalence and determinant factors of pneumonia inunder-five children in southern Ethiopia.
Methodology
An institutional cross-sectional study was employed. A total of 239 child–caregiver pairs were included. Data were collected by trained nurses using a semi-structured questionnaire. The collected data were checked for completeness, coded and entered into EPI data version 4.6, and exported to SPSS version 25 for analysis. Results were reported as the mean, frequency, and percentile. Logistic regression was employed to assess statistically significant predictors of pneumonia. Variables with a p-value <0.05 were considered statistically significant factors of pneumonia.
Result
The prevalence of pneumonia in the study area was 30%. Among the factors assessed, place of food cooking—inside the living room [adjusted odd ratio (AOR) = 5.79, 95% confidence interval (CI): 2.47–13.58], nonexclusive breastfeeding (AOR = 3.26, 95% CI: 1.42–7.52), vitamin A supplementation status (AOR = 5.62, 95% CI: 2.65–11.94), and vaccination status (AOR = 3.59, 95% CI: 1.49–8.66) were significantly associated with the occurrence of pneumonia in under-five children.
Conclusion
This study showed that the prevalence of pneumonia was relatively higher in Arba Minch town than other parts of the country. Place of food cooking, nonexclusive breastfeeding, vitamin A supplementation status, and vaccination status of children were significant factors of pneumonia among under-five children. Enhancing caregivers’/mothers’ awareness of predicted factors was needed to reduce the incidence of childhood pneumonia and to enhance children's quality of health.
Keywords: pneumonia, childhood illness, Arba Minch, Gamo, under-five
Introduction
Pneumonia is a acute lower respiratory tract infection that affects the lung parenchyma and destructs alveolar air space (1). Pneumonia is the leading cause of morbidity and mortality in under-five children. In 2016, it was estimated that 900,000 under-five children died of pneumonia worldwide (2). From these, sub-Saharan African countries took the lion's share with 50% of the burden of worldwide under-five children mortality rate due to pneumonia. Approximately 172 deaths per 1,000 live births occur in sub-Saharan African countries, mainly due to pneumonia (3, 4). Between 2000 and 2019, the global under-five children mortality rate decreased by about 50%, but progress is still slower, and 65 (32%) of 204 countries, primarily in sub-Saharan Africa and south Asia, are not on course to fulfil either sustainable development goal (SDG) 3.2 targeted by 2030 (5). Nearly half of all under-five deaths in 2019 occurred in just five countries: Nigeria, India, Pakistan, the Democratic Republic of Congo and Ethiopia. Globally, infectious diseases, including pneumonia, diarrhea, and malaria, remain a leading cause of under-five deaths (6). Children with compromised immune systems like malnutrition, especially those not exclusively fed under-five children, are at a higher risk of developing pneumonia (7).
In developing countries like Ethiopia, the risk factors for pneumonia in under-five children are numerous. Among factors, nonexclusive breastfeeding, lack of/incomplete immunization, environmental conditions, outdoor/indoor air pollution, micronutrients, and vitamin deficiencies are predominantly reported (8–10).
The World Health Organization's Integrated Management of Childhood Illness (IMNCI) recommendations have been used in Ethiopia, which has done so since 2001 (11). Additionally, Ethiopia made a policy breakthrough by introducing community-based treatment of pneumonia through health extension workers in 2010. This was done to increase access to lifesaving interventions. Since then, integrated community case management has been used to treat pneumonia in the community by health extension workers over the whole health posts in the country. Even though child mortality due to pneumonia decreased, progress among under-five children is still slow (12).
In Ethiopia, pneumonia is one the leading causes of death in under-five children, with an estimated contribution of more than 40,000 deaths annually (13, 14). These deaths could be prevented by cost-effective interventions like immunization, health education, good nutrition, exclusive breastfeeding, and sanitary management (9). Reports showed that the prevalence of pneumonia among under-five children in Ethiopia reached 20.68% (14). According to a study conducted in Wondo Genet, southern Ethiopia, the magnitude of pneumonia in under-five children visiting health centers was reported to be 33.5% (15). Globally, Ethiopia ranks sixth among the top 15 countries in morbidity and mortality from pneumonia (16). Predictive factors of pneumonia need to be studied to better inform healthcare providers and other stakeholders/policymakers to consider additional pneumonia prevalence reduction strategies. Although studies are conducted in Ethiopia, the determinant factors for pneumonia differ in different societies and study populations (17). Taking the high burden of pneumonia and the variability of risk factors into account, this study aimed to identify the predicted factors of pneumonia among under-five children. Furthermore, there is no previous study in the area that could show the current status of pneumonia and its determinant factors.
Method and materials
Source population
All under-five children visiting the Pediatrics Outpatient Department of Arba Minch General Hospital.
Study population
All under-five children visiting the Pediatrics Outpatient Department of Arba Minch General Hospital during the data collection period.
Inclusion and exclusion criteria
The study included children who were under 5 years of age, those who were residents of Arba Minch town for a minimum of 6 months, and those who visited the pediatric unit of Arba Minch referral Hospital during the study period. Children with the following conditions were excluded from the study: cough that lasted for more than 15 days, cough because of the recent history of aspiration of liquid or foreign body, cardiac disease, and caregiver who did not have any information about the child during the time of data collection.
Sample size determination
The study utilized the single population proportion formula technique to determine the study population sample size
where n is the desired sample size,
P is the population of under-five pneumonia children, which is 28.1% (18), have taken from previously published study from jimma town
Z α/2 is the critical value at the 95% confidence interval level of certainty (1.96), and
d is the margin of error between the sample and population (5%)
where source population (N)= 728, which was a monthly plan on the proportion of under-five children visiting the outpatient department in Arba Minch General Hospital. Because the source population was less than 10,000, we can use the correction formula to get the actual sample size of the study population:
where
nf = final sample size,
n = first calculated sample size,
N = source population.
. By adding 10% of the nonresponse rate, the total sample size of this study was 217 + 22 = 239.
Sampling procedure
A total of 239 participants were enrolled in this study. The study participants were selected by using the simple random sampling technique from under-five children outpatient registration book of Arba Minch General Hospital. The first study participant was selected by using K-value. That is
The first was selected from 1 − k (that is, from 1 to 3) by using the lottery method on the first day of data collection.
Study variables
Dependent variables: Pneumonia among under-five children.
Independent variables:
Sociodemographic factors: Age, sex, and residence.
Health facility and childcare factors: Low birth weight, prematurity, exclusive breastfeeding, breastfeeding duration, vaccination status, and vitamin “A” supplementation.
Environmental risk factors
Cook food inside the living room, home aeration, hand washing practice, domestic smoking, and carrying baby during cooking.
Data collection procedures and tools
The data collection tool was developed after reviewing previously published relevant literature (14, 18, 19). The adopted structured questionnaire included sociodemographic factors, environmental factors, and health facility and childcare factors. Before data collection was started, the questionnaire was pretested on a 5% sample size at Chencha Primary Hospital to ensure the validity and reliability of the study tool. After collecting the pretest data, it was checked for potential problems related to the tool, such as any difficult question that was not easily understandable and unclear to reply, and corrective measures were taken. The data were collected through face-to-face interviews and medical chart reviews. Charts were reviewed to collect information about the diagnosis, pre-existing or comorbidities, and anthropometric data. Data were collected by four diploma nurses and one first-degree nurse who received 2 days of training on the objective and contents of the study.
Data quality assurance and analysis
The collected data were checked for completeness, coded and entered into Epi Info version 6, and exported to SPSS version 25 for analysis. Then, data were cleaned for consistency and the extent of outliers. Different statistical assumptions and appropriate corrections were made prior to analysis. Descriptive analyses were made for each independent variable. Bivariate and multivariate binary logistic regression analyses were used to test the association between the independent and dependent variables. Bivariate analysis was performed for each of the independent variables with the outcome variable. Variables with a p-value of <0.25 on bivariate analysis were taken as candidates for multivariate binary logistic regression model analysis to identify predictors of the outcome variable. Variables with a p-value of <0.05 on multivariate binary logistic regression analysis were considered predictive factors for pneumonia among under-five children. The strength of the association between the outcome variable and independent variables was expressed using an adjusted odd ratio (AOR) with 95% confidence intervals (CIs).
Results
Sociodemographic characteristics of study participants
A total of 239 children's mothers/caregivers have participated in the current study, which made a response rate of 100%. About 125 (52.3%) of the children were boys. The highest proportions (41%) of participating children were in the age group of 12–36 months. In the current study, 75% of the respondents (children's mothers/caregivers) had formal education. Regarding the marital status of children's mothers/caregivers, 230 (96.2%) of the participants were married. The occupational profile of children's mothers/caregivers showed that the majority (189, 79%) were self-employed and 30 (12.5%) were civil servants. More than half (54.8%) of the study participants were residing in rural parts of the study area (shown in Table 1).
Table 1.
Sociodemographic characteristics of children and mothers/caregivers studied at the Arba Minch General Hospital, southern Ethiopia (2021).
| Variables | Category | Frequency | Percent |
:---: :---: |
| Sex | Male | 125 | 52.3 |
| Female | 114 | 47.7 |
| Age of child (months) | 2–12 | 71 | 29.7 |
| 13–36 | 98 | 41 |
| 37–59 | 70 | 29.3 |
| Residence of child | Urban | 131 | 54.8 |
| Rural | 108 | 45.2 |
| Family’s monthly income Ethiopian Birr | <2,000 | 31 | 13 |
| 2,001–3,999 | 122 | 51 |
| 4,000–5,999 | 62 | 25.9 |
| ≥6,000 | 24 | 10 |
| Mother’s marital status | Married | 230 | 96.2 |
| Unmarried | 9 | 3.8 |
| Mother’s educational status | Illiterate | 58 | 24.3 |
| Literate | 181 | 75.7 |
| Mother’s occupation | Self-employed | 189 | 79.1 |
| Unemployed | 11 | 4.6 |
| Civil servant | 30 | 12.5 |
| Other | 9 | 3.7 |
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Prevalence of pneumonia and its signs and symptoms among participants
From the total participating children, 38% had a history of cough, 18% had difficulty breathing, and 34% had fast breathing at the time of the study. In addition to these signs and symptoms, 169 (70.7%) had a fever, and 35 (14.6%) had chest in-drawing. Our study showed that the overall prevalence of pneumonia among under-five children during the study period in Arba Minch General Hospital was 30% (73) (as shown in Figure 1). Among the children diagnosed with pneumonia, 21 (28.8%) had severe pneumonia and 15 (71.4%) had moderate to severe respiratory distress (as shown in Figure 2).
Figure 1.
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Pneumonia among under-five children visiting the Arba Minch General Hospital in 2021.
Figure 2.
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Severe pneumonia among under-five children visiting the Arba Minch General Hospital in 2021.
Factors affecting the occurrence of pneumonia among under-five children
Bivariate analysis showed that cooking food inside the living room, nonexclusive breastfeeding in the first 6 months, vitamin A supplementation status, caring for child on mothers’/caregivers’ back or beside the mother during food cooking, kitchen without a window, and children's contact history of a family with acute respiratory tract infection were significantly associated with pneumonia among under-five children (shown in Table 2). Variables with a p-value of <0.25 in the bivariate analysis were candidates for multivariate analysis. Multivariate analysis showed that the kitchen not being separated from the main house for food cooking, nonexclusive breastfeeding, vitamin A supplementation status, and vaccination status of children were statistically significant factors to be considered as independent causes of pneumonia in under-five children (shown in Table 3).
Table 2.
Bivariate logistic regression analysis of independent variables and occurrence of pneumonia in under-five children in the Arba Minch General Hospital, 2021.
| Independent variable | Category | Pneumonia | COR (95% CI) | p-value |
:---: :---:
| Yes | No |
| Place of cooking | Living room | 64 | 96 | 6.757 (3.329–13.715) | 0.001 |
| Kitchen | 9 | 70 | 1 |
| Breastfeeding status | Nonexclusive | 47 | 70 | 2.922 (1.527–5.592) | 0.001 |
| Exclusive | 26 | 96 | 1 |
| Vitamin A supplementation | No | 64 | 86 | 5.621 (2.645–11.94) | 0.001 |
| Yes | 19 | 70 | 1 |
| Contact history of the family with acute RTI | Yes | 24 | 96 | 2.80 (1.752–4.988) | 0.002 |
| No | 49 | 70 | 1 |
| Vaccination status | Fully | 37 | 114 | 1 | |
| Up to date | 14 | 20 | 3.000 (1.994–9.051) | 0.051 |
| Partial | 7 | 14 | 1.146 (0.559–2.350) | 0.710 |
| Unvaccinated | 15 | 18 | 2.270 (1.265–4.075) | 0.006 |
| Location of the child while cooking | On mother back | 25 | 61 | 1.901 (1.992–3.644) | 0.053 |
| Outside cooking room | 70 | 83 | 1 |
| Kitchen with a window | Yes | 24 | 96 | 1.812 (1.883–3.721) | 0.105 |
| No | 49 | 70 | 1 |
Open in a new tab
CI, confidence interval; COR, crude odd ration; RTI, respiratory tract infection.
Table 3.
Logistic regression analysis of independent variables and occurrence of pneumonia in under-five children in the Arba Minch General Hospital, 2021.
| Independent variable | Category | Child diagnosis of pneumonia | AOR (95% CI) | p-value |
:---: :---:
| Yes | No |
| Place of cooking | Living room | 64 | 96 | 5.791 (2.469–13.584) | 0.001 |
| Kitchen | 9 | 70 | 1 |
| Breastfeeding status | Mixed | 47 | 70 | 3.262 (1.415–7.522) | 0.002 |
| Exclusive | 26 | 96 | 1 |
| Vitamin A supplementation | No | 64 | 86 | 4.803 (2.260–10.205) | 0.001 |
| Yes | 19 | 70 | 1 |
| Vaccination status | Fully | 37 | 114 | 1 |
| Up to date | 14 | 20 | 1.416 (0.659–3.043) | 0.372 |
| Partial | 7 | 14 | 1.046 (0.679–2.420) | 0.610 |
| Unvaccinated | 15 | 18 | 3.59 (1.491–8.662) | 0.004 |
Open in a new tab
AOR, adjusted odds ratio; CI, confidence interval.
Discussion
In this study, the kitchen not being separated from the main house for food cooking, nonexclusive breastfeeding, vitamin A supplementation status, and vaccination status of children were found to be significant risk factors associated with the occurrence of pneumonia among under-five children. This study also showed a high prevalence of pneumonia, with 30% among under-five children (as shown in Figure 1). Our finding was in agreement with a study done in other parts of Ethiopia (15, 18). However, the result of this study was greater than that of the study conducted by Merkeb and Fentahun(20.68%) (14) and lower than that of a study conducted in Uganda (53.7%) (20). This difference may be due to the study design (Merkeb and Fentahun conducted a systematic review and meta-analysis) and socioeconomic status variations and health service facility accessibility in the study area for the study in Uganda. Although the magnitude of pneumonia is reduced from time to time due to different health policies and strategies, pneumonia remains the leading cause of death in under-five children in sub-Saharan and developing countries. Utilization of effective strategies for the sustainable reduction of pneumonia in under-five children was hindered by many different factors (4). Our current study showed that under-five children from the household who used the main house for a place of cooking were 5.8 times (AOR = 5.791, 95% CI: 2.469–13.584, p = 0.001) more likely to develop pneumonia than those from the family having a separated kitchen (as shown in Table 3). Our study result was in line with other studies conducted in Wondo Genet, southern Ethiopia (15), a study conducted in Estie town Northwest Ethiopia (21), a study by Markos et al. (19), and a study conducted in central India and Dominican Republic (22, 23). The association between pneumonia and cooking environment might be due to the use of wood as a fuel produces irritants in the form of smoke. Inhalation of this smoke results in the impairment of alveolar cells and macrophages of the lung. This study indicated that unvaccinated under-five children were 3.5 times [AOR = 3.59, 95% CI: 1.491–8.662, p = 0.004] more vulnerable to suffering from pneumonia as compared to those children who completed their vaccination (as shown in Table 3). This result was supported by different studies conducted on predictors of pneumonia (14, 18, 24). This association could be due to the loss of strong enough immunity for causative agents. Concerning breastfeeding in the first 6 months of the children's age, the odds of being vulnerable to developing pneumonia was 3.3 times (AOR = 3.262, 95% CI: 1.415, 7.522, P = 0.002) more likely among children who were not exclusively breastfed as compared to those under-five children who were exclusively breastfed (Table 3). The finding was supported by Ramezani et al.’s report on the predictors of pneumonia (10) and was nearly similar to the UNICEF 2012 report on pneumonia predictors (24). This study also found that under-five children who were not supplemented by vitamin A were nearly five times (AOR = 4.803, 95% CI: 2.260–10.205, p = 0.001) more likely to suffer from pneumonia as compared to children who had vitamin A supplementation (Table 3). Studies in Rwanda also reported similar results (25). This association could be explained by the fact that vitamin A is an essential micronutrient that governs many biological processes and a deficiency of vitamin A will cause an imbalance between pro- and anti-inflammatory factors and excessive immune response (26).
Limitations of the study
Because this study was institution-based research with data gathered from a single hospital in Ethiopia, conclusions may not be easily extrapolated to patients admitted to other areas. Second, even though the cases of pneumonia were classified by physicians using 2014 WHO standard clinical and integrated management of newborn child illness, this study did not use chest x-ray, blood cultures, and other cultures to confirm pneumonia. We merely used physician's diagnosis from patient cards. Hence, this may not be as reliable as pneumonia confirmation using laboratory diagnostic methods.
Conclusion
This study showed that the prevalence of pneumonia was higher in Arba Minch town compared to study reports from other parts of the country. Place of food cooking, nonexclusive breastfeeding, vitamin A supplementation status, and vaccination status of children were significant factors of pneumonia among under-five children. Enhancing caregivers’/mothers’ awareness of predicted factors needed to reduce the incidence of childhood pneumonia and to increase enhanced children's quality of health.
Acknowledgments
The authors thank their study participants and data collectors for their priceless cooperation.
Funding
All of the funding for the research was covered by the authors.
Data availability statement
The original contributions presented in the study are included in the article/Supplementary Material; further inquiries can be directed to the corresponding author.
Ethics statement
The studies involving human participants were reviewed and approved by Ethics Review Committee of Paramed Health Sciences and Business College. Written informed consent to participate in this study was provided by the participants’ legal guardian/next of kin.
Author contributions
YS: data curation, formal analysis, investigation, methodology, and writing—review and editing. ZK: conceptualization, resources, software, supervision, methodology, and investigation. TF: data curation, methodology, resource, software, supervision, and writing. SE: formal analysis, resource, software, supervision, and data curation. All authors contributed to the article and approved the submitted version.
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher's note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Data Availability Statement
The original contributions presented in the study are included in the article/Supplementary Material; further inquiries can be directed to the corresponding author.
Articles from Frontiers in Pediatrics are provided here courtesy of Frontiers Media SA
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Subjects
Chemistry
A-level Chemistry
A-level Chemistry 7405
SpecificationSpecificationPlanning resourcesTeaching resourcesAssessment resourcesKey dates
A Chemistry alkyl groups
See AS and A-level Chemistry guidance about alkyl groups.
Show Index
1.0 Introduction
2.0 Specification at a glance
3.0 Subject content
3.1 Physical chemistry
3.2 Inorganic chemistry
3.3 Organic chemistry
4.0 Scheme of assessment
5.0 General administration
6.0 Mathematical requirements and exemplifications
7.0 Practical assessment
AS and A-level Chemistry Specification Specifications for first teaching in 2015 12 Jan 2015 PDF | 1.21 MB
3.2 Inorganic chemistry
3.2.1 Periodicity
The Periodic Table provides chemists with a structured organisation of the known chemical elements from which they can make sense of their physical and chemical properties. The historical development of the Periodic Table and models of atomic structure provide good examples of how scientific ideas and explanations develop over time.
3.2.1.1 Classification
| Content | Opportunities for skills development |
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| An element is classified as s, p, d or f block according to its position in the Periodic Table, which is determined by its proton number. | |
3.2.1.2 Physical properties of Period 3 elements
| Content | Opportunities for skills development |
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| The trends in atomic radius, first ionisation energy and melting point of the elements Na–Ar The reasons for these trends in terms of the structure of and bonding in the elements. Students should be able to: explain the trends in atomic radius and first ionisation energy explain the melting point of the elements in terms of their structure and bonding. | |
3.2.2 Group 2, the alkaline earth metals
The elements in Group 2 are called the alkaline earth metals. The trends in the solubilities of the hydroxides and the sulfates of these elements are linked to their use. Barium sulfate, magnesium hydroxide and magnesium sulfate have applications in medicines whilst calcium hydroxide is used in agriculture to change soil pH, which is essential for good crop production and maintaining the food supply.
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| The trends in atomic radius, first ionisation energy and melting point of the elements Mg–Ba Students should be able to: explain the trends in atomic radius and first ionisation energy explain the melting point of the elements in terms of their structure and bonding. The reactions of the elements Mg–Ba with water. The use of magnesium in the extraction of titanium from TiCl 4 The relative solubilities of the hydroxides of the elements Mg–Ba in water. Mg(OH)2 is sparingly soluble. The use of Mg(OH)2 in medicine and of Ca(OH)2 in agriculture. The use of CaO or CaCO 3 to remove SO 2 from flue gases. The relative solubilities of the sulfates of the elements Mg–Ba in water. BaSO 4 is insoluble. The use of acidified BaCl 2 solution to test for sulfate ions. The use of BaSO 4 in medicine. Students should be able to: explain why BaCl 2 solution is used to test for sulfate ions and why it is acidified. | AT c and k PS 2.2 Students could test the reactions of Mg–Ba with water and Mg with steam and record their results. AT d and k PS 2.2 Students could test the solubility of Group 2 hydroxides by mixing solutions of soluble Group 2 salts with sodium hydroxide and record their results. Students could test the solubility of Group 2 sulfates by mixing solutions of soluble Group 2 salts with sulfuric acid and record their results. Students could test for sulfate ions using acidified barium chloride and record their results. Research opportunity Students could investigate the use of BaSO 4 in medicine. |
3.2.3 Group 7(17), the halogens
The halogens in Group 7 are very reactive non-metals. Trends in their physical properties are examined and explained. Fluorine is too dangerous to be used in a school laboratory but the reactions of chlorine are studied. Challenges in studying the properties of elements in this group include explaining the trends in ability of the halogens to behave as oxidising agents and the halide ions to behave as reducing agents.
3.2.3.1 Trends in properties
| Content | Opportunities for skills development |
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| The trends in electronegativity and boiling point of the halogens. Students should be able to: explain the trend in electronegativity explain the trend in the boiling point of the elements in terms of their structure and bonding. The trend in oxidising ability of the halogens down the group, including displacement reactions of halide ions in aqueous solution. The trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid. The use of acidified silver nitrate solution to identify and distinguish between halide ions. The trend in solubility of the silver halides in ammonia. Students should be able to explain why: silver nitrate solution is used to identify halide ions the silver nitrate solution is acidified ammonia solution is added. | AT d and k PS 2.2 Students could carry out test-tube reactions of solutions of the halogens (Cl 2, Br 2, I 2) with solutions containing their halide ions (eg KCl, KBr, KI). Students could record observations from reactions of NaCl, NaBr and NaI with concentrated sulfuric acid. Students could carry out tests for halide ions using acidified silver nitrate, including the use of ammonia to distinguish the silver halides formed. |
3.2.3.2 Uses of chlorine and chlorate(I)
| Content | Opportunities for skills development |
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| The reaction of chlorine with water to form chloride ions and chlorate(I) ions. The reaction of chlorine with water to form chloride ions and oxygen. Appreciate that society assesses the advantages and disadvantages when deciding if chemicals should be added to water supplies. The use of chlorine in water treatment. Appreciate that the benefits to health of water treatment by chlorine outweigh its toxic effects. The reaction of chlorine with cold, dilute, aqueous NaOH and uses of the solution formed. | Research opportunity Students could investigate the treatment of drinking water with chlorine. Students could investigate the addition of sodium fluoride to water supplies. |
| Required practical 4 Carry out simple test-tube reactions to identify: cations – Group 2, NH 4+ anions – Group 7 (halide ions), OH–, CO 3 2–, SO 4 2– | |
3.2.4 Properties of Period 3 elements and their oxides (A-level only)
The reactions of the Period 3 elements with oxygen are considered. The pH of the solutions formed when the oxides react with water illustrates further trends in properties across this period. Explanations of these reactions offer opportunities to develop an in-depth understanding of how and why these reactions occur.
| Content | Opportunities for skills development |
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| The reactions of Na and Mg with water. The trends in the reactions of the elements Na, Mg, Al, Si, P and S with oxygen, limited to the formation of Na 2 O, MgO, Al 2 O 3, SiO 2, P 4 O 10, SO 2 and SO 3 The trend in the melting point of the highest oxides of the elements Na–S The reactions of the oxides of the elements Na–S with water, limited to Na 2 O, MgO, Al 2 O 3, SiO 2, P 4 O 10, SO 2 and SO 3, and the pH of the solutions formed. The structures of the acids and the anions formed when P 4 O 10, SO 2 and SO 3 react with water. Students should be able to: explain the trend in the melting point of the oxides of the elements Na–S in terms of their structure and bonding explain the trends in the reactions of the oxides with water in terms of the type of bonding present in each oxide write equations for the reactions that occur between the oxides of the elements Na–S and given acids and bases. | AT a, c and k PS 2.2 Students could carry out reactions of elements with oxygen and test the pH of the resulting oxides. |
3.2.5 Transition metals (A-level only)
The 3d block contains 10 elements, all of which are metals. Unlike the metals in Groups 1 and 2, the transition metals Ti to Cu form coloured compounds and compounds where the transition metal exists in different oxidation states. Some of these metals are familiar as catalysts. The properties of these elements are studied in this section with opportunities for a wide range of practical investigations.
3.2.5.1 General properties of transition metals (A-level only)
| Content | Opportunities for skills development |
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| Transition metal characteristics of elements Ti–Cu arise from an incomplete d sub-level in atoms or ions. The characteristic properties include: complex formation formation of coloured ions variable oxidation state catalytic activity. A ligand is a molecule or ion that forms a co-ordinate bond with a transition metal by donating a pair of electrons. A complex is a central metal atom or ion surrounded by ligands. Co-ordination number is number of co-ordinate bonds to the central metal atom or ion. | |
3.2.5.2 Substitution reactions (A-level only)
| Content | Opportunities for skills development |
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| H 2 O, NH 3 and Cl−can act as monodentate ligands. The ligands NH 3 and H 2 O are similar in size and are uncharged. Exchange of the ligands NH 3 and H 2 O occurs without change of co-ordination number (eg Co 2+and Cu 2+). Substitution may be incomplete (eg the formation of [Cu(NH 3)4(H 2 O)2]2+). The Cl−ligand is larger than the uncharged ligands NH 3 and H 2 O Exchange of the ligand H 2 O by Cl–can involve a change of co-ordination number (eg Co 2+, Cu 2+and Fe 3+). Ligands can be bidentate (eg H 2 NCH 2 CH 2 NH 2 and C 2 O 4 2–). Ligands can be multidentate (eg EDTA 4–). Haem is an iron(II) complex with a multidentate ligand. Oxygen forms a co-ordinate bond to Fe(II) in haemoglobin, enabling oxygen to be transported in the blood. Carbon monoxide is toxic because it replaces oxygen co-ordinately bonded to Fe(II) in haemoglobin. Bidentate and multidentate ligands replace monodentate ligands from complexes. This is called the chelate effect. Students should be able to: explain the chelate effect, in terms of the balance between the entropy and enthalpy change in these reactions. | AT d and k PS 1.2 Students could carry out test-tube reactions of complexes with monodentate, bidentate and multidentate ligands to compare ease of substitution. AT d and k PS 2.2 Students could carry out test-tube reactions of solutions of metal aqua ions with ammonia or concentrated hydrochloric acid. |
3.2.5.3 Shapes of complex ions (A-level only)
| Content | Opportunities for skills development |
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| Transition metal ions commonly form octahedral complexes with small ligands (eg H 2 O and NH 3). Octahedral complexes can display cis–trans isomerism (a special case of E–Z isomerism) with monodentate ligands and optical isomerism with bidentate ligands. Transition metal ions commonly form tetrahedral complexes with larger ligands (eg Cl–). Square planar complexes are also formed and can display cis–trans isomerism. Cisplatin is the cis isomer. Ag+forms the linear complex [Ag(NH 3)2]+as used in Tollens’ reagent. | MS 4.1 and 4.2 Students understand and draw the shape of complex ions. MS 4.3 Students understand the origin of cis–trans and optical isomerism. Students draw cis–trans and optical isomers. Students describe the types of stereoisomerism shown by molecules/complexes. |
3.2.5.4 Formation of coloured ions (A-level only)
| Content | Opportunities for skills development |
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| Transition metal ions can be identified by their colour. Colour arises when some of the wavelengths of visible light are absorbed and the remaining wavelengths of light are transmitted or reflected. d electrons move from the ground state to an excited state when light is absorbed. The energy difference between the ground state and the excited state of the d electrons is given by: ∆E=h ν =hc/λ Changes in oxidation state, co-ordination number and ligand alter ∆E and this leads to a change in colour. The absorption of visible light is used in spectroscopy. A simple colorimeter can be used to determine the concentration of coloured ions in solution. | PS 3.1 and 3.2 Students could determine the concentration of a solution of copper(II) ions by colorimetry. MS 3.1 and 3.2 Students determine the concentration of a solution from a graph of absorption versus concentration. AT a, e and k Students could determine the concentration of a coloured complex ion by colorimetry. |
3.2.5.5 Variable oxidation states (A-level only)
| Content | Opportunities for skills development |
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| Transition elements show variable oxidation states. Vanadium species in oxidation states IV, III and II are formed by the reduction of vanadate(V) ions by zinc in acidic solution. The redox potential for a transition metal ion changing from a higher to a lower oxidation state is influenced by pH and by the ligand. The reduction of [Ag(NH 3)2]+(Tollens’ reagent) to metallic silver is used to distinguish between aldehydes and ketones. The redox titrations of Fe 2+and C 2 O 4 2–with MnO 4– Students should be able to: perform calculations for these titrations and similar redox reactions. | AT d and k PS 1.2 Students could reduce vanadate(V) with zinc in acidic solution. AT b, d and k PS 4.1 Students could carry out test-tube reactions of Tollens' reagent to distinguish aldehydes and ketones. AT a, d, e and k PS 2.3, 3.2 and 3.3 Students could carry out redox titrations. Examples include, finding: the mass of iron in an iron tablet the percentage of iron in steel the M r of hydrated ammonium iron(II) sulfate the M r of ethanedioic acid the concentration of H 2 O 2 in hair bleach. |
3.2.5.6 Catalysts (A-level only)
| Content | Opportunities for skills development |
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| Transition metals and their compounds can act as heterogeneous and homogeneous catalysts. A heterogeneous catalyst is in a different phase from the reactants and the reaction occurs at active sites on the surface. The use of a support medium to maximise the surface area of a heterogeneous catalyst and minimise the cost. V 2 O 5 acts as a heterogeneous catalyst in the Contact process. Fe is used as a heterogeneous catalyst in the Haber process. Heterogeneous catalysts can become poisoned by impurities that block the active sites and consequently have reduced efficiency; this has a cost implication. A homogeneous catalyst is in the same phase as the reactants. When catalysts and reactants are in the same phase, the reaction proceeds through an intermediate species. Students should be able to: explain the importance of variable oxidation states in catalysis explain, with the aid of equations, how V 2 O 5 acts as a catalyst in the Contact process explain, with the aid of equations, how Fe 2+ions catalyse the reaction between I−and S 2 O 8 2– explain, with the aid of equations, how Mn 2+ions autocatalyse the reaction between C 2 O 4 2–and MnO 4– | AT d and k PS 4.1 Students could investigate Mn 2+as the autocatalyst in the reaction between ethanedioic acid and acidified potassium manganate(VII). |
3.2.6 Reactions of ions in aqueous solution (A-level only)
The reactions of transition metal ions in aqueous solution provide a practical opportunity for students to show and to understand how transition metal ions can be identified by test-tube reactions in the laboratory.
| Content | Opportunities for skills development |
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| In aqueous solution, the following metal-aqua ions are formed: [M(H 2 O)6]2+, limited to M = Fe and Cu [M(H 2 O)6]3+, limited to M = Al and Fe The acidity of [M(H 2 O)6]3+is greater than that of [M(H 2 O)6]2+ Some metal hydroxides show amphoteric character by dissolving in both acids and bases (eg hydroxides of Al 3+). Students should be able to: explain, in terms of the charge/size ratio of the metal ion, why the acidity of [M(H 2 O)6]3+is greater than that of [M(H 2 O)6]2+ describe and explain the simple test-tube reactions of: M 2+(aq) ions, limited to M = Fe and Cu, and of M 3+(aq) ions, limited to M = Al and Fe, with the bases OH–, NH 3 and CO 3 2– | AT d and K PS 1.2 Students could carry out test-tube reactions of metal-aqua ions with NaOH, NH 3 and Na 2 CO 3 AT d and k PS 2.2 Students could carry out test-tube reactions to identify the positive and negative ions in this specification. PS 1.1 Students could identify unknown substances using reagents. |
| Required practical 11 Carry out simple test-tube reactions to identify transition metal ions in aqueous solution. | |
3.1 Physical chemistry3.3 Organic chemistry
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3566 | https://www.youtube.com/watch?v=spkim5Ns304 | How to interpret N(d1) and N(d2) in Black Scholes Merton (FRM T4-12)
Bionic Turtle
103000 subscribers
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Description
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Posted: 12 Feb 2019
(my xls is here N(d1) is the option's delta and N(d2) is the probability that a call option will be exercised; that is, N(d2) is the probability that S(T) will be greater than K.
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42 comments
Transcript:
Introduction my previous video walked through the details of my black Scholes Merton option pricing model in this video I just want to specifically address a question that I think we get every year from candidates about the black Scholes Merton option pricing model and that is you can see I've colored them here can we interpret n of D 1 and n of D 2 and the answer is yes we can especially n of D 2 and we'll start by noticing that these are cumulative normal distribution functions that means these are probabilities they range from 0 to 100% in all cases so my yellow Inputs cells in the worksheet as usual these are inputs into the black Scholes Merton BSM as black Scholes Merton there are 6 inputs I'm including the form of the model that incorporates dividend yields sometimes we first see the the more basic form that does not include dividends I'm also de-emphasizing the put-call parity on this sheet you may recall previous video I've mentioned that I'd like to include put-call parity just as a reality check on the number after all black Scholes Merton spits out for us in sort of a black box fashion with the price of the call in the put R and put-call parity just a nice reality check to make sure that they're correct right it says that the call plus discounted cash needs to equal the price the price of a protective put and you can see that's true here I'm also de-emphasizing d1 and d2 which is are tedious although we can access an intuition on the d2 that's discussed in my forum I think I've written a post in my forum where I show that it's actually not difficult to access the intuition of d2 in particular so here for the price of the call option I've unpacked some of the variables here that go into the input you can see here for these values which I think match John whole 10th edition problem or example 15-point not exactly sure but I think they match and you can see the call price here for these assumptions is four dollars and 28 cents you can always do a gut check right we know the call price needs to be less than the stock price that stock price is an upper bound but it's going to be significantly less as a and we can look at it as a percentage right you can see I have about 10% the price of the call is about 10% on the stock and strike price and for a six month option that's an important input here six months it's about right which sounds about right and the most important determinant here is going to be volatility okay so what I have here is for the formula black Scholes n of D - n of D 1 and n of D - and I've colored them to match and I'm addressing here a question that we get frequently which is can we interpret the n of D 1 and n of D 2 in black Scholes Merton and I think you sort of can the n of D 2 is easier than the end of D 1 but I've said before that my style just in memorizing the formula if you're going to sit for an exam frm or CFA I like to start I like to think of this as the lower bound and then we wrap in these probability functions right the lower bound of the call option is this current stock price - the discounted strike price some have that memorized some do not why is that well we're in a risk neutral world you want to us keep that in mind a lot of theory behind that but we're in a risk neutral whirl which means that we expect the stock to grow at the risk-free rate among other things and if the stock were to grow at the risk-free rate right we compound continuously at the risk-free rate over the maturity that's our continuous compounding and then we would pay this Detroit price the strike price does not grow it's $40 here it's constant or fixed right we would pay that and you can hopefully this makes sense is the future gain on this call option if the grows at the risk-free rate but we're in the risk-neutral world so that's the reasonable assumption that's the future gain we would we're computing a present price which is a present value right price her value usually means present value so we take that future gain and discount it at the risk-free rate as you might expect you see if you distribute that how these cancel and we're back to right here if I distribute the minimum value so that's Probability functions the minimum value or that was math wanted to give you intuition on of the minimum value as stock price might too - discounted strike price I sometimes call this discounted cash because that's what this is fixed so we can think of it as cash that we're gonna pay the minimum value and then we wrap in these probability functions why did I call them probability functions well these this n notation is we could just say generically that signifies the cumulative normal distribution function or specifically the standard normal cumulative distribution function so right here you can see on the upper left I'm using excels standard normal cumulative distribution function by standard normal I mean a normal with zero mean and variance and therefore also of all at standard deviation of one and what that means is we're getting a probability by definition this is a probability so I I've taken to calling it a probability function it needs to lie between 0 and 1 as these values always will when the options get vary in or out of the money they will these will tend asymptotically to 0 and 1 but they are probabilities so the less intuitive one is the N of d1 and it turns out that if we take this stock price and grow it at the risk-free rate in the risk-neutral world this is the expected future stock price is it not however this is a call option such that underwater outcomes will be worth zero so it turns out that if we multiply and of this by n of D 1 what we get is an expected future stock price if the outcomes that are underwater are counted as zero so it's a kind of average and so that's my first term here well again that's future so it'd be discounted back and so this drops out that's my first term here stock price times n of d1 to adjust for that probability but I've also inserted the dividend haircut right the dividend haircut you wouldn't see it in all forms but as I've also covered in previous videos in general in option pricing a dividend or dividend yield has the effect of reducing the current stock price that's because an option holder for goes the dividends they miss out on those and for a given assumption about total shareholder return if there's more dividend we would expect less capital a capital or price appreciation so you can see here that's that's the more difficult one but we can view this here as a function of the expected future stock price we're underwater outcomes are counted as 0 because this n of D one introduces a probability multiplier the more intuitive one is n FD 2 remember I mentioned that the d2 itself has a very intuitive way we can access it but I'm not going to go into that now just to say that I'm just gonna say at the end of d2 itself is the probability that this option will be exercised it's that straightforward if we think about the stock price starting here there is a future probability distribution and then here is the strike price let's say NFD - right well stock prices up here for this call option will be exercised if it expires in-the-money and will not be exercised if it expires worthless Lior out of the money end of d2 here literally is the probability that this dot price will end up in the money so it's the area under this curve as a percentage of the entire probability which is a hundred percent so you can see here and again again this is easy to miss or forget or just gloss over when studying all the numbers around the black Scholes again I remind NF D 1 and n FD 2 are probabilities so when we see here n FD 2 is well very close to 70% there's a very intuitive interpretation of this it is that again here's the caveat in the risk-neutral world where we've made this assumption about the expected growth of risk-free rate among other things given the caveat a risk neutral world n FD - as is 0.7 is 70% probability that this stock price will finish above the strike price the OP that the option will be in the money that the option will be exercised very intuitive right so that means this K is the fixed price that we that's fixed it doesn't change we either pay that or we do not we're multiplying that by the probability that we pay it so this is the probability adjusted expected strike price payment in fact very intuitive so you'd seen this way we've taken the minimum value which I started with a sub zero subtract the discounted cash and we've just wrapped in the probability functions to account for the fact we don't know what's going to happen both of these terms are probability adjusted to account for what's the probability the stock price will be in the money in the case of this call so obviously the higher the stock price starts the greater that probability as the stock price goes up and of d2 which is directly as the probability they'll be in the money will go up as well so I Put options hope that is a helpful I have on the second page in the in the sheet that all make up make available for download I add the put option here and logic is actually just very similar if we consider starting at the stock price here we do have positive drift and then not a very good distribution and but then I'll just assume a fixed strike price here right in the case of a put now higher outcomes to the stock price that implies that the options out of the money right in the case of the put it's only going to be in the money or exercised for these outcomes where the future stock price is lower that's S sub t right these are out of the money these are in the money for the put and so I have the same parameters here no recall we said n fd2 was a 70% probability that this call option will be exercised meaning a 70% probability that will be up here these are probabilities it's one of the other above or below that means what we have here for n of negative d2 is 100% minus one minus 0.7 or 100% minus the 70% you can see we do in fact have here 30% right if all the other assumptions the same if we had a 70% chance of the call option being exercised we have a 30% chance right here of the put option being exercised so an of negative d2 going to the symmetry the normal here has also an intuitive explanation the N of negative d1 is 0.25 and you can see it's here a function of the end of d1 don't think I mentioned the previous video that NF d1 is also the options Delta so we say Delta right that's change in the call price as a function of changing the stock price well the put options Delta is n of D 1 minus 1 and in this case it equal that means it equals negative 0.25 3 6 so the put options Delta here is just this value that goes in the black Scholes but with a negative so a little more little more difficult on the in a the end of negative d1 input but this is these are still probabilities so I hope that's helpful if you did like this video subscribe to the channel and we'll update you because I'm pretty much recording 2 videos a week thank you |
3567 | https://home.mathematik.uni-freiburg.de/arithgeom/lehre/ss20/algzt/Keith%20Conrad%20-%20Unit%20theorem.pdf | DIRICHLET’S UNIT THEOREM KEITH CONRAD 1. Introduction Dirichlet’s unit theorem describes the structure of the unit group of orders in a number field.
Theorem 1.1 (Dirichlet, 1846). Let K be a number field with r1 real embeddings and 2r2 pairs of complex conjugate embeddings. The unit group of an order in K is finitely generated with r1 + r2 −1 independent generators of infinite order.
More precisely, letting r = r1 + r2 −1, each order O in K contains multiplicatively inde-pendent units ε1, . . . , εr of infinite order such that every unit in O can be written uniquely in the form ζεm1 1 · · · εmr r , where ζ is a root of unity in O and the mi’s are in Z. Abstractly, O× ∼ = µ(O) × Zr1+r2−1, where µ(O) is the finite cyclic group of roots of unity in O.
Units u1, . . . , uk are called multiplicatively independent, or just independent, when they satisfy no multiplicative relations except the trivial one: um1 1 · · · umk k = 1 ⇒mi = 0 for all i. It then follows that exponents in such a product are unique: if um1 1 · · · umk k = un1 1 · · · unk k then mi = ni for all i. This looks like linear independence, and that is exactly what it is: when we view O× as a Z-module using its group law, multiplicative independence means Z-linear independence.
If r1 > 0 then µ(O) = {±1} since ±1 are the only roots of unity in R. If r1 = 0 we might also have µ(O) = {±1}, e.g., O = Z[ √ d] for d < −1.
It is important that the unit groups of all orders in K have the same number of inde-pendent generators of infinite order: r1 + r2 −1. Therefore [O× K : O×] is finite. A choice of generators ε1, . . . , εr for O× (really, for the quotient group O×/µ(O)) is called a system of fundamental units. We call r1 + r2 −1 the rank of the unit group.
The unit groups of orders in number fields were, historically, the first important examples of finitely generated abelian groups. Finding algorithms to produce explicit generators for unit groups is one of the tasks of computational number theory.
In Section 2 we will look at some examples of the unit theorem. The theorem will be proved in Section 3 and some more examples are described in Sections 4 and 5.
2. Examples Example 2.1. For Q( √ 2) we have r1 + r2 −1 = 1, so the unit group of each order in Q( √ 2) has the form ±εZ for some unit ε. In particular, Z[ √ 2]× = ±(1 + √ 2)Z and Z[3 √ 2]× = ±(17 + 12 √ 2)Z.
1 2 KEITH CONRAD Table 1 describes unit groups in the full ring of integers in several number fields. The unit ε in the row for Q( 3 √ 2, ζ3) is ε = −1 + 2 3 √ 2 + 3 √ 4 3 + 1 − 3 √ 2 + 3 √ 4 3 ζ3.
K r1 r2 r1 + r2 −1 µ(OK) O× K Q( √ 3) 2 0 1 ±1 ±(2 + √ 3)Z Q( √ 5) 2 0 1 ±1 ±( 1+ √ 5 2 )Z Q(ζ5) 0 2 1 µ10 µ10( 1+ √ 5 2 )Z Q( 3 √ 2) 1 1 1 ±1 ±(1 + 3 √ 2 + 3 √ 4)Z Q( 3 √ 6) 1 1 1 ±1 ±(1 −6 3 √ 6 + 3 3 √ 36)Z Q( 4 √ 2) 2 1 2 ±1 ±(1 + 4 √ 2)Z(1 + √ 2)Z Q( 3 √ 2, ζ3) 0 3 2 µ6 µ6 · εZεZ Q( √ 2, √ 3) 4 0 3 ±1 ±(1 + √ 2)Z( √ 2 + √ 3)Z( √ 2+ √ 6 2 )Z Table 1. Unit Group of OK Example 2.2. The unit group of an order is finite if and only if r1 +r2 −1 = 0. This means (r1, r2) is (1, 0) or (0, 1), so K is Q or an imaginary quadratic field. Moreover, the unit group of each order in an imaginary quadratic field is {±1} except for the maximal orders Z[i] and Z[ζ3], whose units groups have size 4 and 6, respectively. There are a number of important results in algebraic number theory that have a simpler form for Q and imaginary quadratic fields than for other number fields, precisely because in these (and only these) cases the unit group is finite.
Example 2.3. We have r1 + r2 −1 = 1 if and only if (r1, r2) = (2, 0), (1, 1), or (0, 2), i.e., K is real quadratic (e.g., Q( √ 2)), a cubic field with only one real embedding (e.g., Q( 3 √ 2)), or a totally complex quartic field (e.g., Q(ζ5)).
Example 2.4. If K is a totally real cubic field then r1 + r2 −1 = 2, so each order in K has unit group of the form ±εZ 1 εZ 2 .
Example 2.5. We always have r1 + r2 −1 ≤n −1, where n = [K : Q] = r1 + 2r2. Easily r1 + r2 −1 = n −1 if and only if r2 = 0, i.e., K is a totally real number field.
Example 2.6. Let’s look at a unit group with rank greater than 1 and see how to find multiplicative relations between units numerically, by using logarithms to discover them as linear relations. Set K = Q(α) where α3 −3α −1 = 0. The polynomial f(T) = T 3 −3T −1 has 3 real roots, so O× K has rank r1 + r2 −1 = 3 −1 = 2.
Before looking at O× K, let’s show OK = Z[α]. Since disc(Z[α]) = −4(−3)3 −27(−1)2 = 81 = 34, [OK : Z[α]] divides 9. Therefore elements of OK when written in the basis {1, α, α2} have coefficients with denominator dividing 9. Since f(T + 1) = T 3 + 3T 2 −3 is Eisenstein at 3 with α−1 as a root, elements of OK when written in the basis {1, α−1, (α−1)2} have coefficients with denominator prime to 3. This carries over to {1, α, α2}, so OK = Z[α].
(The Minkowski bound is exactly 2, and there is no prime ideal with norm 2 since T 3−3T −1 is irreducible modulo 2, so h(K) = 1: Z[α] is a PID.) DIRICHLET’S UNIT THEOREM 3 We now write down several units in Z[α]. For a, b ∈Q, NK/Q(aα + b) = −a3f(−b/a).
Check with this formula that α, α + 1, α −2, and 2α + 3 all have norm ±1, so they are all in Z[α]×. The three roots of f(T) are α, 2 −α2, and α2 −α −2, so 2 −α2 and α2 −α −2 are in Z[α]×. The product of all three roots of f(T) is −f(0) = 1.
Since Z[α]× has rank 2, the nontrivial units we just wrote down must admit some nontriv-ial multiplicative relations. How can we find such relations? We will use the three different embeddings K →R. Call them σ1, σ2, and σ3. The real roots of f(T) are σ1(α), σ2(α), and σ3(α). Arranging the roots in increasing order, σ1(α) = −1.532 . . . , σ2(α) = −.347 . . . , σ3(α) = 1.879 . . . .
For γ ∈K, NK/Q(γ) = σ1(γ)σ2(γ)σ3(γ). For u ∈O× K, |σ1(u)σ2(u)σ3(u)| = 1. Taking logarithms, (2.1) u ∈O× K = ⇒log |σ1(u)| + log |σ2(u)| + log |σ3(u)| = 0.
Define the logarithmic mapping L: K× →R3 by L(γ) = (log |σ1(γ)|, log |σ2(γ)|, log |σ3(γ)|).
We will use such a map L in the proof of the general unit theorem; here we will see how this map is useful computationally. Easily L is a group homomorphism and by (2.1), L(O× K) is in the hyperplane {(x, y, z) ∈R3 : x+y+z = 0}. The kernel of L on O× K is {±1} (why?). Table 2 gives numerical approximations to the images of units under the logarithmic mapping.
γ L(γ) (approx.) α (.4266, −1.0575, .6309) α + 1 (−.6309, −.4266, 1.0575) α −2 (1.2618, .8532, −2.1151) 2α + 3 (−2.7460, .8352, 1.9108) 2 −α2 (−1.0575, .6309, .4266) α2 −α −2 (.6309, .4266, −1.0575) Table 2. Log Images of Units From the table, it appears that L(α −2) = −2L(α + 1) = L(1/(α + 1)2), so α −2 = ±1/(α + 1)2. You can check that the minus sign is needed. Using a computer algebra package, the 3 × 3 matrix (L(α) L(α + 1) L(2α + 3)) has (2, −3, 1) in its kernel, so α2(α + 1)−3(2α + 3) has L-value 0. Therefore 2α + 3 = ±α−2(α + 1)3. Check that the plus sign holds. Since it looks like L(2 −α2) = L(α + 1) −L(α), 2 −α2 = ±(α + 1)/α and the minus sign is needed. Then, since the three roots of f(T) multiply to 1, α2 −α −2 = 1/(α(2 −α2)) = (1/α)(−α/(α + 1)) = −1/(α + 1).
This evidence suggests that α and α + 1 are a system of fundamental units for Z[α]×.
3. Proof of the unit theorem Our proof of the unit theorem is based on [3, Sect. 1.5] and [4, pp. 214–215] (see also [5, p. 5]), and is deduced from a compactness theorem: the unit theorem is a consequence of a certain group being compact.
We will use Minkowski’s convex-body theorem in our proof. This is a standard tool for proofs of the unit theorem, although by comparison with typical applications of Minkowski’s 4 KEITH CONRAD theorem we will be able to get by with a crudely chosen convex body: a sufficiently large ball will work.
Dirichlet did not use Minkowski’s theorem; he proved the unit theorem in 1846 while Minkowski’s theorem appeared in 1889. Dirichlet’s substitute for the convex-body theorem was the pigeonhole principle. (An account of Dirichlet’s proof in German is in [2, Sect. 183] and in English is in [6, Sect. 2.8–2.10].) Dirichlet did not state the unit theorem for all orders, but only those of the form Z[α], since at the time these were the kinds of rings that were considered. According to an oft-repeated story, the main idea for the proof of the unit theorem came to Dirichlet while he attended a concert in the Sistine Chapel.1 We set some notation. As in the statement of the unit theorem, K is a number field of degree n, r1 is the number of real embeddings of K and 2r2 is the number of complex embeddings of K (that is, embeddings K →C whose image is not in R), so n = r1 + 2r2.
Set V = Rr1 × Cr2, so dimR(V ) = n. The Euclidean embedding θK : K →V is defined using the real and complex embeddings of K, as follows. Let the real embeddings of K be σ1, . . . , σr1 and let the complex embeddings of K be σr1+1, σr1+1, . . . , σr1+r2, σr1+r2, where we collect the complex embeddings into conjugate pairs σj, σj. For α ∈K, we set2 θK(α) = (σ1(α), · · · , σr1(α), σr1+1(α), · · · , σr1+r2(α)) ∈V.
Algebraically, V is a commutative ring using component wise operations. Give V its natural topology as a Euclidean space and all subsets of V will be given the subspace topology. A particular subset we will care about is V × = (R×)r1 × (C×)r2.
Let N: V →R by N(x1, . . . , xr1, z1, . . . , zr2) = x1 · · · xr1|z1|2 · · · |zr2|2 = x1 · · · xr1z1z1 · · · zr2zr2.
On the image of K in V , N looks like the norm: N(θK(α)) = NK/Q(α) for all α ∈K. Set G = {v ∈V × : |N(v)| = 1}.
This is a subgroup of V ×, and it is closed in V since G is the inverse image of {1} under the continuous map V →R given by v 7→|N(v)|. Thus G is a closed subgroup of V ×.
Let O be an order in K and set U = θK(O×) = G ∩θK(O).
(Think “U = units”.) We have U ⊂G since O× = {α ∈O : |NK/Q(α)| = 1}. Since we give G the subspace topology from V and the image of O in V under the Euclidean embedding is discrete, U is discrete in G. We will be interested in the quotient group G/U.
1For instance, in 1905 Minkowski [9, pp. 156–7] wrote “Es wird erz¨ ahlt, da nach langj¨ ahrigen vergeblichen Bem¨ uhungen um das schwierige Problem Dirichlet die L¨ osung in Rom in der Sixtinischen Kapelle w¨ ahrend des Anh¨ orens der Ostermusik ergr¨ undet hat.
Inwieweit dieses Faktum f¨ ur die von manchen behauptete Wahlverwandtschaft zwischen Mathematik und Musik spricht, wage ich nicht zu er¨ ortern.” (translation: “People say that, after many years of unsuccessful efforts in trying to solve this difficult problem, Dirichlet found the solution in Rome in the Sistine Chapel while listening to Easter music. I do not dare to discuss to what extent this fact confirms the conjectured (by some people) relationship between mathematics and music.”) 2The Euclidean embedding of K, as defined here, depends on the ordering of the different real and complex embeddings as well as on the choice of one complex embedding from each conjugate pair. A coordinate-free way of defining the Euclidean embedding uses tensor products: the natural mapping K →R ⊗Q K where α 7→1 ⊗α is a ring homomorphism into a finite-dimensional real vector space of dimension n, just like V .
DIRICHLET’S UNIT THEOREM 5 Example 3.1. Let K = Q( √ 2) and O = OK = Z[ √ 2]. Then V = R2 and N: V →R by N(x, y) = xy. The Euclidean embedding θ: Q( √ 2) →R2 places Z[ √ 2]× on the curve G = {(x, y) ∈R2 : |xy| = 1}, a union of two hyperbolas. We know Z[ √ 2]× = ±(1 + √ 2)Z and U = θK(Z[ √ 2]×) is a discrete subset of G (“equally spaced” in a multiplicative sense).
See Figure 1.
θ((1 + √ 2)2) θ(1) θ −1 1+ √ 2 θ(−(1 + √ 2)) θ(−(1 + √ 2)2) θ(−1) θ 1 1+ √ 2 θ(1 + √ 2) s s s s s s s s Figure 1. Units in Z[ √ 2] on G = {(x, y) ∈R2 : |xy| = 1}.
Let’s see how the subgroup U moves G around by multiplication in Figure 1. Multiplying G by some u ∈U moves the arcs between consecutive points of U in Figure 1 to other arcs between consecutive point, and it exchanges the hyperbolas y = 1/x and y = −1/x if N(u) = −1. Multiplication by θ(−1) = (−1, −1) on G exchanges the two branches on each hyperbola.
Modulo U each (x, y) ∈G is congruent to a point on the arc between θ(1) and θ((1+ √ 2)2), so the map [1, (1+ √ 2)2] →G/U given by x 7→(x, 1/x)U is surjective and continuous, which implies G/U is compact.
In Example 3.1 we used knowledge of the unit group of Z[ √ 2] to see G/U is compact.
The key to proving the unit theorem is showing the compactness of G/U without knowing the structure of the unit group in advance.
Lemma 3.2. For nonzero a in O, [O : (a)] = |NK/Q(a)|.
Proof. This follows from O being a free Z-module of rank [K : Q].
□ Lemma 3.3. For each positive integer N, finitely many a ∈O satisfy |NK/Q(a)| = N up to multiplication by O×. That is, there are a1, . . . , ak ∈O, where k depends on N, such that |NK/Q(ai)| = N and each a ∈O satisfying |NK/Q(a)| = N is a unit multiple of an ai.
Proof. If |NK/Q(a)| = N then [O : (a)] = N by Lemma 3.2, so NO ⊂(a) ⊂O. Since O/NO is finite, there are only finitely many principal ideals between NO and O. Let (a1), . . . , (ak) be those ideals. Then (a) = (ai) for some i, so a and ai are unit multiples.
□ 6 KEITH CONRAD Theorem 3.4. The group G/U is compact in the quotient topology.
Proof. We will find a compact subset S of G that represents all cosets in G/U. The con-tinuous map S →G/U is onto and thus G/U is compact. (Usually G itself is not compact.
See Figure 1.) We begin with a remark about volumes. For v ∈V ×, multiplication of V = Rr1 × Cr2 by v is an R-linear map (hence continuous) given by a matrix with determinant N(v), so for a region R ⊂V with finite volume, the volume of vR is |N(v)| times the volume of R.
In particular, if v ∈G then vol(vR) = vol(R) because |N(v)| = 1. When R is compact so is vR, by continuity of multiplication.
Pick a compact, convex, centrally symmetric region C ⊂V with vol(C) > 2n vol(θK(O)), where the “volume” of the lattice θK(O) means the volume of a fundamental domain for this lattice as a subset of V . For instance, C could be a large ball in V centered at the origin. For each g ∈G, gC is also compact and centrally symmetric. It is convex too, since multiplication by g on V is an invertible linear transformation, and invertible linear transformations send convex sets to convex sets. Using g−1 instead of g, Minkowski’s convex body theorem applies to g−1C and the lattice O ⊂V (we identify O with θK(O)): g−1C ∩(O −{0}) ̸= ∅.
Let a be a nonzero element of O lying in g−1C. Then |NK/Q(a)| = |N(a)| ∈|N(g−1C)| = |N(C)|, which is a bounded subset of R since C is compact. Note |N(C)| is independent of g. The number |NK/Q(a)| is also an integer, so |NK/Q(a)| lies in a finite set (a bounded set of integers is finite). Combining that with Lemma 3.3, there is a finite set {a1, . . . , am} of nonzero elements of O such that every g−1C meets some aiO× = aiU, which implies every gU meets some a−1 i C.
We have shown the quotient group G/U is represented by G ∩Sm i=1 a−1 i C. The union Sm i=1 a−1 i C is a compact subset of V , since each a−1 i C is compact, and since G is closed in V the intersection G ∩Sm i=1 a−1 i C is compact in G. Hence G/U has a compact set of representatives in G, so G/U is compact in the quotient topology.
□ Now we prove the unit theorem. Recall that, by definition, G = {v ∈V : |N(v)| = 1}.
Each element of V = Rr1 × Cr2 can be written in the form (x1, . . . , xr1, zr1+1, . . . , zr1+r2).
Define the logarithmic mapping L: V × →Rr1+r2 by L(x1, . . . , zr1+r2) := (. . . , log |xi|, . . . , 2 log |zj|, . . . ), where the coefficients 2 in this formula are related to the exponents 2 in the definition of N. The function L is a continuous group homomorphism and, for each g ∈G, L(g) lies in the hyperplane H = {(y1, . . . , yr1+r2) ∈Rr1+r2 : X i yi = 0}.
It is easy to see that L(G) = H, so L(G) has dimension r1 + r2 −1 over R. What we really care about is L(U), which provides a linearized geometric picture for U (once we determine the kernel of L|U). The basic plan is to show L(U) is a “full” lattice in the hyperplane L(G) and the kernel of L restricted to U is finite cyclic (coming from roots of unity in U).
First we treat the kernel of L|U. As a map out on V ×, L has compact kernel: ker L = {±1}r1 × (S1)r2.
Every root of unity in U gets sent to 0 by L. Let’s check these are the only elements of U = O× in ker L. Since U is closed in V × (all discrete subsets are closed), the kernel of DIRICHLET’S UNIT THEOREM 7 L|U is closed and thus (as a subset of {±1}r1 × (S1)r2) is compact. Since O is discrete in V (it’s a lattice), U is discrete in V ×, so the kernel of L|U is also discrete (a subset of a discrete set is discrete), so ker(L|U) is compact and discrete: it is finite! A subgroup of U with finite order can only contain roots of unity. Therefore the elements of ker(L|U) are the roots of unity in U = O×, which form a finite cyclic group since every finite subgroup of K× is a cyclic group. (Warning: it is false that the kernel of L as a map out of K× is only the roots of unity in K. An element of K× that has all of its Q-conjugates lying on the unit circle must be in the kernel of L. An example is 3/5 + (4/5)i, or more generally a/c + (b/c)i where (a, b, c) is a Pythagorean triple. But these are not algebraic integers, so they don’t belong to U.) Now we look at the image L(U) in the hyperplane L(G) ⊂Rr1+r2. We have already seen (and used) that the group U is discrete in V ×, so also in G. The image of a discrete set under a continuous map need not be discrete (consider Z2 →R by (m, n) 7→m + n √ 2), but L(U) is discrete in L(G) since there are only finitely many elements in L(U) that lie in a bounded region of Rr1+r2. Indeed, consider the box {(y1, . . . , yr1+r2) ∈Rr1+r2 : |yi| ≤b}.
Suppose L(u) is in this box for some u ∈U. The real embeddings3 of u have absolute value at most eb and the complex embeddings of u have absolute value at most eb/2. That puts an upper bound in terms of b (and n = [K : Q]) on the coefficients of the polynomial Q σ(T −σ(u)) ∈Z[T]. The coefficients have only finitely many possibilities, since there are finitely many integers with absolute value below a given bound, so there are finitely many such polynomials. As u is a root of such a polynomial, there are finitely many choices for u. This shows L(U) is discrete.
Since L(U) is a discrete subgroup of L(G) ∼ = Rr1+r2−1, L(U) ∼ = Zr′ where r′ ≤r1+r2−1.
Since L: G →L(G) is a continuous and surjective group homomorphism, the induced map G/U →L(G)/L(U) is also continuous and surjective where both quotient groups get the quotient topology. From Theorem 3.4, G/U is compact so L(G)/L(U) is compact. Since L(G) is (r1+r2−1)-dimensional over R and L(U) has Z-rank r′ ≤r1+r2−1, compactness of L(G)/L(U) forces r′ = r1 + r2 −1: Euclidean space modulo a discrete subgroup is compact only when the subgroup has rank equal to the dimension of the space (e.g., R2/(Z × {0}) is a non-compact infinite cylinder). That proves L(U) ∼ = Zr1+r2−1 and L(U) is a lattice in the hyperplane H.
We’re now basically done. Let ε1, . . . , εr (r = r1 + r2 −1) be elements of O× whose Euclidean embeddings in U provide a Z-basis of L(U). The εi’s are multiplicatively inde-pendent, since their L-images are Z-linearly independent. For ε ∈O×, L(ε) = m1L(ε1) + · · ·+mrL(εr) for some integers mi, so L(ε) = L(εm1 1 · · · εmr r ). Since ker(L|U) is the Euclidean image of the roots of unity in O×, ε = ζεm1 1 · · · εmr r for some ζ ∈µ(O). This concludes the proof of the unit theorem.
The most difficult part of the proof of the unit theorem is showing there are r1 + r2 −1 independent units of infinite order. For instance, using the logarithmic map it was not hard for us to show L(U) is a discrete subgroup of L(G) ∼ = Rr1+r2−1, so O× ∼ = U ∼ = W × Zr′ where r′ ≤r1 + r2 −1 and W is the group of roots of unity in O×. Thus O× has at most r1 + r2 −1 independent units of infinite order, but this alone doesn’t tell us there are units 3We are identifying U = θK(O×) with O× when we speak of real embeddings of u. If we did not make that identification, and wrote u = θK(α), then we would speak instead of real embeddings of α, which are the initial coordinates of u.
8 KEITH CONRAD of infinite order in O×. The place in the proof where saw there are units of infinite order (if r1 + r2 −1 > 0) is when we went from r′ ≤r1 + r2 −1 to r′ = r1 + r2 −1. This happened two paragraphs up and relied on G/U being compact.
4. Fundamental Unit in the Rank 1 Case As noted already in Example 2.3, an order O in number field K has a rank 1 unit group precisely when K is real quadratic, cubic with 1 real embedding (that is, a cubic field that is not totally real), or a totally complex quartic field. In the first two cases, the only roots of unity in K are ±1, which are always in O, so O× = ±εZ.4 Viewing K in R, the choice of ε > 1 is called the fundamental unit of O.
Example 4.1. Since Z[ √ 2]× = ±(1 + √ 2)Z, the fundamental unit of Z[ √ 2] is 1 + √ 2.
Example 4.2. Since Z[3 √ 2]× = ±(17 + 12 √ 2)Z, Z[3 √ 2] has fundamental unit 17 + 12 √ 2.
Example 4.3. In Example 4.9 we will show Z[ 3 √ 6]× = ±(109 + 60 3 √ 6 + 33 3 √ 36)Z, so 109 + 60 3 √ 6 + 33 3 √ 36 ≈326.990 is the fundamental unit of Z[ 3 √ 6].
In a real quadratic field, one way to find the fundamental unit in an order is by brute force: if we write a unit greater than 1 as a+b √ d or a+b(1+ √ d)/2 with a, b ∈Z, necessarily a ≥0 and b ≥1 (check!). This allows one to systematically search for the smallest unit greater than 1 by sifting through pairs of integers in the first quadrant by increasing values of a and b. (There is a more efficient method, using continued fractions.) To give examples of fundamental unit computations in the cubic case, we will use an inequality due to Artin [1, pp. 169–170]. Mordell , near the end of his review of in 1962, described Artin’s inequality as a “surprise” since “one would have thought that there was not much opportunity for new results on cubic units”.
Theorem 4.4 (Artin). Let O be an order in a cubic field K with r1 = 1. Viewing K in R, if v > 1 is a unit of O× then | disc(O)| < 4v3 + 24.
Proof. This argument is similar to Artin’s in and may look like a messy calculation.
Consider reading the corollary and its applications first, and then return to this proof.
Since v is a unit and is not ±1, v ̸∈Q. Thus Q(v) = K, so Z[v] is an order inside O.
From Z[v] ⊂O, | disc(O)| ≤| disc(Z[v])|. We will show | disc(Z[v])| < 4v3 + 24.
Let σ: K →C be one of the non-real embeddings of K. Then NK/Q(v) = vσ(v)σ(v) = v|σ(v)|2 > 0, so v has norm 1. Let x = √v (as a positive real number), so 1 = x2|σ(v)|2.
Therefore |σ(v)| = 1/x, so in polar form σ(v) = x−1eit for some real number t. Then disc(Z[v]) = ((σ(v) −v)(σ(v) −v)(σ(v) −σ(v)))2 = ((x−1eit −x2)(x−1e−it −x2)(x−1eit −x−1e−it))2 = ((x−2 + x4 −2x cos t)(−2ix−1 sin t))2 = −4(sin2 t)(x3 + x−3 −2 cos t)2.
Since x > 1, x3 + x−3 > 2. Set a = (x3 + x−3)/2, so a > 1 and by taking absolute values, | disc(Z[v])| = 4(sin2 t)(2a −2 cos t)2 = 16(1 −cos2 t)(a −cos t)2.
(4.1) 4Don’t confuse ±εZ with ε±Z; the latter is just εZ.
DIRICHLET’S UNIT THEOREM 9 Set y = cos t, so y ∈[−1, 1]. Then (4.1) is f(y) = 16(1 −y2)(a −y)2 and we want to maximize this on [−1, 1]. Let a maximum occur at y0. Since f(y) ≥0 on [−1, 1] with f(1) = f(−1) = 0 and f(0) = 16a2 > 0, we have y0 ∈(−1, 1) and f′(y0) = 0.
By the product rule, f′(y) = 32(a −y)(2y2 −ay −1) and the root of the linear factor is a > 1, so 2y2 0 −ay0 −1 = 0. Rewrite this as (4.2) ay0 = 2y2 0 −1.
Thus (4.3) | disc(Z[v])| = f(cos t) ≤f(y0) = 16(1 −y2 0)(a −y0)2.
Expanding (a −y0)2 and using the relation (4.2) a couple of times, we get 16(1 −y2 0)(a −y0)2 = 16(a2 + 1 −y4 0 −y2 0).
Substituting a = (x3 + x−3)/2 into this, f(y0) = 16 x6 4 + 3 2 + x−6 4 −y4 0 −y2 0 .
We will show x−6/4 < y2 0, so the right side is less than 16(x6/4+3/2) = 4x6+24 = 4v3+24.
Then by (4.3), | disc(Z[v])| < 4v3 + 24, as desired.
Let h(y) = 2y2−ay−1, the quadratic factor of f′(y), so h′(y0) = 0. The roots of h(y) have product −1/2, so they have opposite sign. Since h(1) = 1 −a < 0 and h(y) > 0 for large y, h(y) has a root in (1, ∞). Thus −1 < y0 < 0, and recall x > 1, so the inequality x−6/4 < y2 0 is the same as y0 < −1/(2x3). The graph of h(y) is a concave up parabola and y0 is the smaller root of h(y), so to prove y0 < −1/(2x3) it is enough to show h(−1/(2x3)) < 0: h −1 2x3 = 2 4x6 + a 2x3 −1 = 1 2x6 + 1 2x3 x3 + x−3 2 −1 = 3 4x6 −3 4 = 3 4 1 x6 −1 < 0 since x > 1.
□ Remark 4.5. The condition on v in Theorem 4.4 is v > 1, not v > 0. If we could use 0 < v < 1 in Artin’s inequality, then replacing v with a high power of itself would imply | disc(O)| < 24, which is false for all cubic orders with one exception. See Footnote 6 below.
Corollary 4.6. Let O be an order in a cubic field K with r1 = 1. Viewing K inside R, let ε > 1 be the fundamental unit of O. For each unit u > 1 in O×, if 4u3/m + 24 ≤| disc(O)| for an integer m ≥2 then u = εk where 1 ≤k < m. In particular, if 4u3/2 + 24 ≤| disc(O)| then u = ε.
Proof. The group O× is ±εZ, so u = εk for some positive integer k. Artin’s inequality using v = ε says | disc(O)| < 4ε3 + 24 = 4u3/k + 24.
If k ≥m then | disc(O)| < 4u3/k +24 ≤4u3/m +24 ≤| disc(O)|, so we have a contradiction.
Thus k < m.
□ Example 4.7. Let K = Q( 3 √ 2), so OK = Z[ 3 √ 2] and disc(OK) = disc(T 3 −2) = −108.
Since 1 = 3 √ 2 3 −1 = ( 3 √ 2 −1)( 3 √ 4 + 3 √ 2 + 1), we have a unit u = 1 + 3 √ 2 + 3 √ 4 ≈3.847. Since 4u3/2 + 24 ≈54.185 < 108, u is the fundamental unit of OK.
10 KEITH CONRAD Example 4.8. Let K = Q(α), where α3+2α+1 = 0. The polynomial is irreducible modulo 3, so K/Q is cubic. It has one real root, approximately −.45. Since disc(T 3+2T +1) = −59, OK = Z[α]. Clearly α is a unit. We view K in R. Since α ≈−.45, we get a unit greater than 1 using u = −1 α ≈2.205.
Since 4u3/2 + 24 ≈37.10 < 59, u is the fundamental unit of OK.
Example 4.9. Let K = Q( 3 √ 6). This will be an example where the unit u we find will satisfy 4u3/2 + 24 > | disc(OK)|, so we will have to be more creative to prove u is the fundamental unit.
First we show OK = Z[ 3 √ 6]. Since disc(Z[ 3 √ 6]) = disc(T 3 −6) = −2235 = −972 = [OK : Z[ 3 √ 6]]2 disc(OK), the index [OK : Z[ 3 √ 6]] is a factor of 2 · 32. At the same time, since T 3 −6 is Eisenstein at 2 and 3 the index [OK : Z[ 3 √ 6]] is not divisible by 2 or 3. Therefore the index is 1, so OK = Z[ 3 √ 6].
To find units in OK, we seek two different descriptions of a principal ideal in OK: if (α) = (β) then α = βu where u is a unit. Here is a table of how the first few primes p decompose in OK, based on how T 3 −6 mod p decomposes.
p T 3 −6 mod p (p) 2 T 3 p3 2 3 T 3 p3 3 5 (T −1)(T 2 + T + 1) p5p25 7 (T −3)(T −5)(T −6) p7p′ 7p′′ 7 The only ideal of norm 2 is p2. We will prove p2 is principal by finding an element of absolute norm 2. For c ∈Z, NK/Q( 3 √ 6 + c) = c3 + 6. Therefore NK/Q( 3 √ 6 −2) = −2, so the ideal ( 3 √ 6 −2) has norm 2 and must be p2. We have the equality of principal ideals (2) = p3 2 = ( 3 √ 6 −2)3 = (( 3 √ 6 −2)3), so the numbers 2 and ( 3 √ 6 −2)3 are equal up to a unit multiple in OK. Since ( 3 √ 6 −2)3 ≈ −.0061, to get a unit greater than 1 we use the ratio5 u = − 2 ( 3 √ 6 −2)3 ≈326.9908.
Let ε > 1 be the fundamental unit of Z[ 3 √ 6]. Does u = ε? By the unit theorem u = εk for some k ≥1 and we want to show k = 1. Artin’s inequality with v = ε says that in R, | disc(OK)| < 4ε3 + 24 = ⇒972 < 4u3/k + 24.
For large k this inequality must fail, since the right side tends to 4 + 24 = 28 as k →∞.
In fact 4u3/4 + 24 ≈331.5 < 972, so k is either 1, 2, or 3. How do we rule out u = ε2 and u = ε3?
Here’s a great idea: to prove an algebraic integer is not a square or cube, prove it is not a square or cube modulo p for some prime ideal p. Looking at the above table of prime ideal factorizations, we will use the ideals p5 and p7.
5Explicitly, u = 109 + 60 3 √ 6 + 33 3 √ 36, but this representation will not be needed.
DIRICHLET’S UNIT THEOREM 11 In OK/p5 ∼ = Z/(5) we have 3 √ 6 ≡1 mod p5, so u ≡2/(2 −1)3 ≡2 mod p5. The nonzero squares in Z/(5) are 1 and 4, so u is not a square in OK/p5 and thus is not a square in OK.
In OK/p7 ∼ = Z/(7) we have 3 √ 6 ≡3 mod p7, so u ≡2/(2 −3)3 ≡−2 ≡5 mod p7. The nonzero cubes in Z/(7) are 1 and 6, so u is not a cube in OK. (Using p′ 7 or p′′ 7 would have led to the same conclusion.) We have shown k = 1, so u = ε is the fundamental unit of Z[ 3 √ 6].
Example 4.10. Let K = Q(α) where α3 −α −1 = 0. The polynomial T 3 −T −1 is irreducible mod 5, so K/Q is cubic. The polynomial has one real root α ≈1.324, so r1 = 1.
Since disc(T 3 −T −1) = −23 is squarefree, OK = Z[α]. Clearly α is a unit in OK. It is natural to wonder if α is the fundamental unit of OK since it is so close to 1 in the real embedding. We can’t use Artin’s inequality because | disc(OK)| < 24,6 so for every unit u > 1 and positive integer m, | disc(OK)| < 4u3/m + 24.
Since we know the unit group of OK (modulo ±1) is infinite cyclic, to show α is the fundamental unit we show α is the smallest unit greater than 1: no unit u ∈Z[α]× satisfies 1 < u < α.
Let σ: K →C be one of the complex embeddings of K, so NK/Q(u) = uσ(u)σ(u) = u|σ(u)|2 > 0. Therefore NK/Q(u) = 1. Since u ̸∈Q, the minimal polynomial of u over Q is T 3 + aT 2 + bT −1 for some integers a and b. The roots are u, σ(u), and σ(u), so a = −(u + σ(u) + σ(u)), b = uσ(u) + uσ(u) + σ(u)σ(u).
Then |a| ≤u + 2|σ(u)|, |b| ≤2u|σ(u)| + |σ(u)|2.
Since 1 = u|σ(u)|2, the bound 1 ≤u implies |σ(u)| ≤1, so from 1 < u < α we get |a| < α + 2 ≈3.3, |b| ≤2α + 1 ≈3.6.
Thus a and b both lie in {0, ±1, ±2, ±3}. Among the 49 polynomials T 3 + aT 2 + bT −1 with a and b in this set, every such polynomial that has a unit u > 1 of OK as a root must have discriminant equal to a nonzero square multiple of disc(OK) = −23 (because disc(Z[u]) = [OK : Z[u]]2 disc(OK)). There are four such polynomials (see table below), including T 3 −T −1 itself, and the real root of each polynomial other than T 3 −T −1 is larger than α.
Polynomial Discriminant Real Root T 3−T −1 −23 α ≈1.324 T 3 −2T 2 + T −1 −23 α2 ≈1.754 T 3 −3T 2 + 2T −1 −23 α3 ≈2.324 T 3 −2T 2 −3T −1 −23 α4 ≈3.079 Thus α is the fundamental unit of OK.
5. Units in a multiquadratic field A real quadratic field has unit rank 1. A biquadratic field Q(√m, √n), where m, n, and mn are positive integers and not squares, has unit rank 4−1 = 3. There are three quadratic subfields, Q(√m), Q(√n), and Q(√mn), and each has a fundamental unit. A choice of one unit from each quadratic subfield need not be a set of 3 fundamental units for the biquadratic field.
6 This field K is the only cubic field, up to isomorphism, with absolute discriminant less than 24, so OK is the only cubic order with absolute discriminant less than 24.
12 KEITH CONRAD Example 5.1. In the field Q( √ 2, √ 3), a system of fundamental units is 1 + √ 2, √ 2 + √ 3, and √ 2+ √ 6 2 (see Table 1).
Q( √ 2, √ 3) Q( √ 2) Q( √ 3) Q( √ 6) Q Fundamental units for the three quadratic subfields are u1 = 1 + √ 2, u2 = 2 + √ 3, and u3 = 5 + 2 √ 6. In terms of the fundamental units of Q( √ 2, √ 3), u2 = √ 2+ √ 6 2 2 and u3 = ( √ 2 + √ 3)2. In terms of the fundamental units of the quadratic subfields, the fundamental units we listed for Q( √ 2, √ 3) are u1, √u3, √u2.
Example 5.2. In the field Q( √ 3, √ 5), a system of fundamental units is 1+ √ 5 2 , 4 + √ 15, and 3+ √ 3+ √ 5+ √ 15 2 .
Q( √ 3, √ 5) Q( √ 3) Q( √ 5) Q( √ 15) Q Fundamental units of the quadratic subfields are u1 = 2+ √ 3, u2 = 1+ √ 5 2 , and u3 = 4+ √ 15.
In terms of the fundamental units of Q( √ 3, √ 5), u1 = 1 4+ √ 15 3+ √ 3+ √ 5+ √ 15 2 2 by PARI. In terms of the fundamental units of the quadratic subfields, the fundamental units we listed for Q( √ 3, √ 5) are u2, u3, √u1u3.
Kuroda showed that for every real biquadratic field Q(√m, √n), with fundamental units u1, u2, u3 for its 3 quadratic subfields, a set of fundamental units for the biquadratic field is one of the following 7 lists up to relabeling the ui’s: {u1, u2, u3}, {√u1, u2, u3}, {√u1u2, u2, u3}, {√u1u2u3, u2, u3}, {√u1, √u2, u3}, {√u1u2, √u3, u2}, {√u1u2, √u2u3, √u3u1}.
Examples 5.1 and 5.2 illustrate two of these possibilities and the list shows ⟨−1, u1, u2, u3⟩ has index 1, 2, 4, or 8 in the unit group of the biquadratic field.
Consider now a general multiquadratic field K = Q( p d1, . . . , p dk), where the di’s are nonsquare positive integers that are multiplicatively independent modulo squares (that is, they are independent in Q×/(Q×)2). By Galois theory and induction, [K : Q] = 2k and Gal(K/Q) ∼ = (Z/2Z)k by making sign changes on every √di. The unit rank of K is r1−1 = 2k−1, and this is also the number of quadratic subfields: such subfields DIRICHLET’S UNIT THEOREM 13 are of the form Q(√dI), where I = {i1, . . . , im} is a nonempty subset of {1, 2, . . . , k} and dI = di1 · · · dim. Since each Q(√dI) has unit rank 1, it is natural to suspect that choosing one unit (besides ±1) from each quadratic subfield of K should give us a multiplicatively independent set of units in K.
Theorem 5.3. With notation as above, let uI be a unit in Q(√dI) other than ±1. These units are multiplicatively independent: if Q I uaI I = 1, where the exponents aI are in Z, then each aI is 0.
Proof. Our argument is taken from [8, Lemma 2] (which includes some extraneous hypothe-ses on the di’s). The special feature of a unit in a real quadratic field is that its Q-conjugate is, up to sign, its inverse: u′ = ±u−1. This fact will interact well with multiplication rela-tions.
One Q-basis of K is all the square roots √dI together with 1 (we could set d∅= 1 and 1 = p d∅). For each nonempty subset J of {1, 2, . . . , k}, there is a σ ∈Gal(K/Q) such that σ(√dJ) = −√dJ and σ(√dI) = √dI for all I ̸= J. Since σ is the identity on Q(√dI) and is nontrivial on Q(√dJ), σ(uI) = uI while σ(uJ) = ±u−1 J .
Applying σ to Q I uaI I = 1 turns it into Q I̸=J uaI I · (±u−1 J )aJ = 1. Dividing one multi-plicative relation by the other, (±u2 J)aJ = 1. Since uJ has infinite order, aJ = 0.
□ Corollary 5.4. The units uI generate a subgroup of O× K with finite index.
Proof. By their multiplicative independence, the uI’s generate a group of rank 2k −1, which is the rank of O× K.
□ References E. Artin, Theory of Algebraic Numbers, George Striker, Schildweg 12, G¨ ottingen, 1959.
P. G. L. Dirichlet, Vorlesungen ¨ uber Zahlentheorie, 4th ed., Vieweg und Sohn, Braunschweig, 1894. Online at
R. Godement, Introduction ` a la Th´ eorie des Groups de Lie, Springer–Verlag, Berlin, 2003.
E. Kleinert, “Units of Classical Orders: A Survey,” L’Enseignement Math. 40 (1994), 205–248.
E. Kleinert, Units in Skew Fields, Birkha¨ user, Basel, 2000.
H. Koch, Number Theory: Algebraic Numbers and Functions, Amer. Math. Society, Providence, 2000.
S. Kuroda, “¨ Uber den Dirichletschen K¨ orper,” J. Fac. Sci. Imp. Univ. Tokyo Sect. I 4 (1943), 383–406.
F. Luca and I. E. Shparlinski, “On the Square-free Parts of ⌊en!⌋,” Glasgow Math. J. 49 (2007), 391–403.
H. Minkowski, “Peter Gustav Lejeune Dirichlet und seine Bedeutung f¨ ur die heutige Mathematik,” Jahresbericht der Deutschen Mathematiker-Vereinigung 14 (1905), 149–163.
L. J. Mordell, Review of E. Artin’s “Theory of Algebraic Numbers”, Bull. Amer. Math. Soc. 3 (1962), 162–166. |
3568 | https://www.spanishlearninglab.com/daily-routine-in-spanish-pdf-worksheet/ | My Daily Routine in Spanish - PDF Worksheet - Spanish Learning Lab
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My Daily Routine in Spanish – PDF Worksheet
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¡Hola! We have designed this worksheet to help you practice the key vocabulary and grammar to talk about your daily routine in Spanish with some simple and interesting exercises. You will see both reflexive and non-reflexive verbs in use to write and talk about the activities someone does every day.
Directions:
Students could work by themselves or in pairs to solve the exercises on this worksheet.
In order to solve the exercises in this worksheet, students have to read the paragraphs about someone’s daily routine in Spanish in the worksheet. They can take some time to underline all of the activities that this guy does every day. For the first exercise, they must write down seven more daily activities in Spanish that he does every day. For the second exercise, students have to reply to the questions about the daily routine in Spanish with the information provided in the text.
Worksheet information:
Level: Intermediate
Skill: Reading
Related lessons:
Sentences with Daily Routine Activities in Spanish
La Rutina – Describing your Daily Routine in Spanish
Conjugating and Using Spanish Reflexive Verbs
My Daily Routine in Spanish – PDF Worksheet (with answers)Download PDF
Solve it online!
Exercise No. 1
Write the name of each of the activities in Fernando’s daily routine so that they match the flashcards below.
Exercise No. 2
Choose the correct answer for each question about Fernando’s daily activities in Spanish based on the content of the reading.
Extra Practice:
Homework:
Students write a paragraph about their daily routine in Spanish similar to the one in the worksheet. Then, they share their routine with a partner and ask questions about their daily activities.
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3569 | https://byjus.com/maths/dihedral-angle/ | Dihedral angle is defined as the angle formed when two planes intersect each other. The two intersecting planes here are the cartesian planes. The cartesian geometry is defined for two-dimensional and three-dimensional planes, which determines the shapes of different objects.
In general, we usually a combination of two lines or line segments to represent the angles. But in Geometry, we deal with the behaviour of two or more planes together. The problems related to two or more planes can be seen very often. The planes can also intersect one another. When we talk about the three-dimensional system, given two distinct planes either intersect each other or are parallel to each other. When two planes intersect, they are inclined at some angle which is known as the dihedral angle. In this article, we will discuss its definition, formula, method of its calculation and some solved examples.
Dihedral Angle Definition
In three dimensional geometry, two planes can directly or indirectly (by extension) intersect each other. The angle formed between the intersection of two different planes is called a dihedral angle. In other words, we can say that the interior angle between the two planes is known as a dihedral angle.
From the above figure, we can see, the two planes A and B are intersecting each other with an inclination of angle θ.
Properties of Dihedral Angle
As we can see that usually there are two angles formed between two planes – one is acute, and another is obtuse. Generally, the dihedral angle refers to the acute angle between the two planes. The obtuse angle can be found by subtracting an acute angle from 180°.
Instead of intersecting two planes, sometimes they are parallel to each other. In this case, the dihedral angle between the two planes is zero. Hence, to prove that two planes are parallel, calculating the dihedral angle can be very useful.
The dihedral angle can also be referred to as the interior angle made by the two planes in any three-dimensional shape, such as polyhedron. In the diagram, below β represents the dihedral angle between the two faces of a tetrahedron.
Dihedral Angle Formula
The dihedral angle is the angle between the normal vectors of two planes. Let us recall that the normal vector is defined as a vector that is perpendicular or normal or perpendicular to the given plane. Let us suppose that a plane is denoted by the equation :
px + qy + rz + s = 0 and its normal vector is represented by
, then=(p, q, r)
Assume that the equations of two planes are given as:
p1x + q1y + r1z + s1 = 0 and p2x + q2y + r2z + s2 = 0
Let the corresponding normal vector to these planes be
andrespectively. Then:= (p1, q1, r1) and= (p2, q2, r2)
If the acute angle between these planes be θ, then the formula for α would be given as below :
It can also be written as:
How to Find the Dihedral Angle?
To calculate the value of the dihedral angle, follow the below steps:
Step 1: Determine the equations of the plane and rearrange the terms in the following format: px + qy + rz + s = 0.
Step 2: Now, write the normal vector for each given plane, i.e. (p, q, r).
Step 3: Plug above values in the formula of dihedral angle which is:
Step 4: Solve and calculate the value of θ.
Related Articles
Angles
Angle Between Two Lines
Angle Between Two Planes
Alternate Angles
Alternate Interior Angles
Dihedral Angle Solved Example
Question: Calculate the angle between two planes: x + 4y + z = 0 and 3 x + y + 4z = 0
Solution:
Given two planes: x + 4y + z = 0 and 3 x + y + 4z = 0
Compare the given plane equation with the standard form:
p1x + q1y + r1z + s1 = 0 and p2x + q2y + r2z + s2 = 0
So, we get :
p1 = 1, q1 = 4, r1 = 1
p2 = 3, q2 = 1, r2 = 4
The formula to find the dihedral angle is
Stay tuned with BYJU’S – The Learning App and also download the app to learn all the important Maths-related articles and explore videos.
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3570 | https://it.wikipedia.org/wiki/Cianuro_di_potassio | Cianuro di potassio - Wikipedia
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Cianuro di potassio
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| Cianuro di potassio |
| Cristalli di cianuro accanto a un dischetto grande quanto una moneta da 1 centesimo di euro |
| Caratteristiche generali |
| Formula bruta o molecolare | KCN |
| Massa molecolare (u) | 65,12 g/mol |
| Aspetto | solido bianco |
| Numero CAS | 151-50-8 |
| Numero EINECS | 205-792-3 |
| PubChem | 9032 |
| SMILES | [C-]#N.[K+] |
| Proprietà chimico-fisiche |
| Densità (g/cm 3, in c.s.) | 1,55 (20°C) |
| Solubilità in acqua | 716 g/l (25°C) |
| Temperatura di fusione | 634°C (907,15 K) |
| Temperatura di ebollizione | 1.625°C (1.898,15 K) (1013 hPa) |
| Proprietà termochimiche |
| Δ f H 0 (kJ·mol−1) | −131.5 |
| Proprietà tossicologiche |
| DL 50 (mg/kg) | 2,857 |
| Indicazioni di sicurezza |
| TLV (ppm) | 5 mg/m³ |
| Simboli di rischio chimico |
| |
| pericolo |
| Frasi H | 300 - 310 - 330 - 410 |
| Consigli P | 260 - 264 - 273 - 280 - 284 - 301+310 |
| Modifica dati su Wikidata·Manuale |
Il cianuro di potassio è il sale di potassio dell'acido cianidrico.
Aspetto
[modifica | modifica wikitesto]
A temperatura e pressione standard si presenta come un solido cristallino bianco e igroscopico; inodore se perfettamente anidro, ma esalante il tipico odore di mandorle amare se umido o a contatto con l'aria.
Utilizzo
[modifica | modifica wikitesto]
Il cianuro di potassio, e lo ione cianuro in generale, è molto importante in chimica perché capace di creare complessi solubili con molti metalli (oro, ferro, argento, nichel) e viene quindi sfruttato nelle tecniche di analisi per solubilizzare i vari cationi metallici, come per la cianurazione dell'acciaio. Viene usato anche in fotografia per i bagni di fissaggio, per rimuovere gli alogenuri d'argento.
In chimica organica è un tipico reagente di sintesi che può essere addizionato mediante sostituzioni nucleofile, allungando la catena carboniosa. Per la sua grande capacità di complessare i metalli di transizione, è usato nella preparazione di sali di potassio e di sodio, tra questi spiccano quelli di oro e argento: KAu(CN)2 , KAu(CN)4 , KAg(CN)2. Questi sono impiegati in galvanica per placcare (doratura o argentatura), sia gioielli sia componenti elettronici.
Tossicologia
[modifica | modifica wikitesto]
Lo stesso argomento in dettaglio: Fosforilazione ossidativa e Respirazione cellulare.
Il tipico odore di mandorle amare è dovuto al rilascio nell'aria di acido cianidrico (infatti le mandorle amare contengono acido cianidrico), il quale, essendo un acido debole, viene spostato dai suoi sali anche a causa dell'anidride carbonica (contenuta nell'aria) in presenza di umidità secondo la reazione:
2 KCN+CO 2+H 2 O⟶K 2 CO 3+2 HCN{\displaystyle {\ce {2KCN\ +\ CO2\ +\ H2O->K2CO3\ +\ 2HCN}}}
Il cianuro di potassio è il sale dell'acido cianidrico che più comunemente viene utilizzato nella ricerca chimica, nelle industrie di estrazione di metalli preziosi e in alcuni processi di sintesi.
Il cianuro di potassio è un potentissimo veleno: lo ione CN- infatti, una volta nel corpo umano, si lega all'atomo di ferro (III) contenuto nell'enzimacitocromo ossidasi, inattivandolo irreversibilmente. L'enzima citocromo ossidasi è uno degli enzimi più importanti nel corpo umano: fa in modo che le cellule possano utilizzare l'ossigeno assunto respirando, e senza di esso le cellule morirebbero per mancanza di ossigeno.
Il cianuro di potassio ha sapore acre e, anche se ingerito in piccole dosi, causa rapidamente morte per anossia istotossica: irritazione-intorpidimento della gola, ansia e confusione, vertigini e dispnea/iperpnea (l'aria sembra mancare, sensazione di soffocamento, respirazione convulsa), la vittima perde velocemente conoscenza, convulsioni e infine morte per arresto respiratorio e cardiocircolatorio.
La dose letale del cianuro di potassio varia da 150 a 300 mg, mentre per l'acido cianidrico (liquido o gassoso) la dose letale oscilla tra i 50 ed i 100 mg.
Antidoti
[modifica | modifica wikitesto]
Gli antidoti contro il cianuro si basano sulla trasformazione, nel corpo della vittima, di parte dell'emoglobina in metaemoglobina, la quale ha la proprietà di "catturare" gli ioni cianuro prima che essi a loro volta inattivino l'enzima citocromo-ossidasi. Poiché il cianuro "catturato" dalla metaemoglobina dovrà essere eliminato, viene somministrato quindi tiosolfato di sodio nel corpo; l'enzima rodanasi (presente nell'organismo umano) ha così sufficiente zolfo per trasformare gli ioni cianuro in ioni tiocianato, molto meno pericolosi, che vengono eliminati successivamente dal corpo.
Per produrre metaemoglobina, negli avvelenamenti da cianuro, vengono somministrati nitrito di amile, nitrito di sodio e altri vasodilatatori (nitriti alchilici e organici) in dosi controllate (sono anch'essi tossici) e quindi viene somministrato il tiosolfato di sodio.
Attualmente si stanno cercando altri antidoti, meno tossici e più maneggevoli dei nitriti, tra i quali l'idrossicobalamina (vitamina B 12) che sarebbe capace di catturare essa stessa il cianuro, a dosi da 20 a 50 mg per volta.
Note
[modifica | modifica wikitesto]
^MSDS di Sigma-Aldrich, rev. 10.01.2011
Voci correlate
[modifica | modifica wikitesto]
Cianuro
Ferrocianuro di potassio
Altri progetti
[modifica | modifica wikitesto]
Altri progetti
Wikimedia Commons
Wikimedia Commons contiene immagini o altri file su cianuro di potassio
Collegamenti esterni
[modifica | modifica wikitesto]
(EN) potassium cyanide, su Enciclopedia Britannica, Encyclopædia Britannica, Inc.
| mostra V·D·M Sali di potassio |
| Composti inorganici | K 2 O·KO 2·KAl(SO 4)2·KBr·KBrO 3·KCN·KCNO·KCl·KClO 3·KClO 4·KF·KCr(SO 4)2·KH·KHCO 3·KHF 2·KHSO 3·KHSO 4·KI·KIO 3·KH(IO 3)2·KIO 4·KMnO 4·K 2 MnO 4·KNO 2·KNO 3·KOCN·KOH·KSCN·K 2 CO 3·K 2 CrO 4·K 2 Cr 2 O 7·K 2 S·K 2 SO 3·K 2 SO 4·K 2 S 2 O 5·K 2 S 2 O 7·K 2 S 2 O 8·K 2 S 4 O 6·K[Sb(OH)6]·K 2 SiO 3·K 3[Fe(CN)6]·K 4[Fe(CN)6]·KH 2 PO 4·K 2 HPO 4·K 3 PO 4·K 2 TeO 3 |
| Composti organici | Acetato di potassio·Ascorbato di potassio·Biftalato di potassio·Bitartrato di potassio·Citrato di potassio·Formiato di potassio·Ossalato di potassio·Sorbato di potassio·Tartrato di potassio·Tartrato di sodio e potassio |
| Controllo di autorità | LCCN(EN)sh85105605·GND(DE)4163067-1·J9U(EN,HE)987007529535605171 |
Portale Chimica: il portale della scienza della composizione, delle proprietà e delle trasformazioni della materia
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3571 | https://www.youtube.com/watch?v=1QRq9qPmBBs | What Are Common Square Coordinate Geometry Mistakes? - All About Geometry
All About Geometry
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Posted: 19 Sep 2025
What Are Common Square Coordinate Geometry Mistakes? Are you interested in mastering coordinate geometry and avoiding common pitfalls? In this detailed video, we'll explore the typical mistakes students make when working with squares and other quadrilaterals on the coordinate plane. We'll start by explaining how to correctly interpret and plot points using ordered pairs. You'll learn the importance of maintaining the right order of coordinates to ensure accurate calculations.
Next, we’ll cover the proper steps for using the distance formula to find side lengths and how skipping or mixing these steps can lead to errors. We’ll also discuss how to verify whether a shape is a square by checking side lengths, diagonals, and angles, and why overlooking some properties might cause incorrect conclusions.
Additionally, we’ll highlight common errors in calculating perimeter and area, emphasizing the importance of using the correct formulas and units. The video will also guide you through applying transformations like rotations and reflections correctly, so your figures stay accurate after each move.
Finally, we’ll explain how to accurately measure angles and use slopes to confirm right angles, as well as how to correctly find midpoints and dividing points on sides. Paying attention to these details will help you improve your problem-solving skills and confidence in coordinate geometry. Join us to avoid these mistakes and sharpen your understanding of squares and quadrilaterals!
⬇️ Subscribe to our channel for more valuable insights.
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CoordinateGeometry #GeometryTips #Squares #MathMistakes #GeometryHelp #MathTutorial #CoordinatePlane #DistanceFormula #ShapeProperties #Angles #Slopes #Transformations #Midpoints #MathLearning #GeometryForStudents
About Us: Welcome to All About Geometry! Our channel is dedicated to breaking down the world of geometry, making it easy to grasp key concepts like shapes, angles, triangles, circles, and polygons. Whether you're tackling area, perimeter, volume, or surface area, our engaging content will help you master essential geometry theorems and topics, from the Pythagorean theorem to congruence and similarity. We also cover coordinate and Euclidean geometry, along with helpful geometry tips for high school students.
Transcript:
What are common square coordinate geometry mistakes? Imagine trying to solve a puzzle with pieces that fit perfectly but missing a few small details. That's what working with squares in coordinate geometry can feel like if you make simple mistakes. These errors happen more often than you think and can throw off your entire problem solving process. Understanding what these mistakes are can help you avoid them and get accurate results every time. One common mistake is mixing up the order of coordinates. When you see an ordered pair like 3 5, the first number is always the x coordinate which tells you how far left or right you are. The second number is the y-coordinate which shows how high or low you are. Reversing these can lead to wrong calculations of distances or slopes. Always double check that you are using the correct order when plotting points or doing calculations. Another mistake is incorrect use of the distance formula. To find the length of a side of a square, you need to subtract the x coordinates, subtract the y-coordinates, square those differences, add them together, and then take the square root. Skipping steps or mixing operations can give you wrong side lengths. Following each step carefully helps ensure your calculations are correct. Failing to verify the properties of a square is also a common error. To confirm a shape is a square using coordinates, you need to check that alpha sides are equal, the two diagonals are equal, and that adjacent sides meet at right angles. Sometimes students only check side lengths and forget to verify the diagonals or the right angles which can lead to mistaken conclusions about the shape. Many students confuse perimeter and area calculations. The perimeter is the total length around the square which is for times the side length. The area is the side length squared. Using the wrong formula or units can cause confusion. Remember perimeter is measured in linear units while area is in square units. When applying transformations like rotations or reflections, mistakes happen if the rules are not followed precisely. For example, rotating a square clockwise instead of counterclockwise or reflecting over the wrong axis can change the coordinates incorrectly. Always follow the specific rules for each transformation to get accurate results. Another mistake is miscalculating angles. All angles in a square are right angles which measure 90°. Sometimes students confuse degrees and radians or try to measure angles without proper methods. Using slopes or the dot product can help verify that the angles are correct. Using slopes to check for right angles is important. The slope of one side multiplied by the slope of the adjacent side should be -1 if they are perpendicular. Forgetting to calculate slopes correctly or misinterpreting the results can lead to incorrect assumptions about the shape being a square. Finally, errors often happen when finding midpoints or dividing points on sides. The midpoint formula involves adding the xcoordinates and dividing by two and doing the same for y-coordinates. Mistakes in applying these formulas or mixing up the coordinates can lead to wrong points which affects the entire problem. Knowing these common mistakes helps you doublech checkck your work and avoid losing points. Whether you are verifying if points from a square, calculating the area or perimeter, or performing transformations, paying attention to these details makes a big difference. Practice these steps carefully, and you will become more confident in solving coordinate geometry problems involving squares and other quadrilaterals. |
3572 | https://ell.stackexchange.com/questions/106642/difference-between-amazing-at-and-amazing-in | prepositions - Difference between 'amazing at' and 'amazing in' - English Language Learners Stack Exchange
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Difference between 'amazing at' and 'amazing in'
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Asked 8 years, 11 months ago
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She is amazing ---------- painting.
What's the difference between amazing at and amazing in?
What would be used, at or in? And also tell me what is the grammatical role of painting in the sentence.
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edited Oct 16, 2016 at 7:32
Cardinal
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asked Oct 16, 2016 at 6:27
Samra SalmanSamra Salman
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Referring to your comment about the grammatical role of painting: at and in are both prepositions: they require a noun. Words derived from verbs and ending in -ing are either participles (which act like adjectives) or gerunds (which act like nouns). In this case a noun is required, so you know that it's a gerund. Many activities are described by gerunds: riding, skiing, swimming, etc.JavaLatte –JavaLatte 2016-10-16 18:29:10 +00:00 Commented Oct 16, 2016 at 18:29
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I feel like amazing at/in will (more or less) follow the same pattern as good at/in. This is likely because the two words reflect something positive about the subject.
In and at are sometimes interchangeable. In this case, I would use at:
She is good at painting. → She is amazing at painting.
According to this post, good at is generally used with activities. It provides other examples:
He’s good at football.
She’s good at product design.
Her mother is good at Trivial Pursuit.
When Fatima was only six, she was good at drawing.
In each case, you can swap good with amazing and each case remains idiomatic.
The post also makes a note about academics. It says
When it comes to school subjects, both “good at” and “good in” are used.
For example,
Max is good at math.
Max is good in math.
Of course, these are general guidelines. There are exceptions.
She's good in bed
This means she's good at sex. You cannot replace in with at in 8 because bed is not an activity.
This can sometimes be extended to other places where activities are done (as this post points out), for example
He's good in the kitchen. (He's good at cooking; He cooks well.)
You can swap good with amazing in the previous examples and they should sound fine. Other positive descriptions, like skillful, should work too. Again, this is a rough guideline and there probably more exceptions.
You can rewrite this as "She is an amazing painter." I prefer this over the original, although the original is ok.
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edited Apr 13, 2017 at 12:38
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answered Oct 16, 2016 at 7:35
Em.Em.
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Your "exceptions" both refer to physical places: you would also say "he is in the kitchen" and not "at the kitchen". The trick here is that in examples 1 - 5, you directly mention the activity they are good at. In examples 8 and 9, you don't mean literally that he is a good person in the kitchen but not in the living room; rather "bed" and "kitchen" are locations that stand for the activities "sex" and "cooking", respectively. Similarly, you would say "he is good with a ball" because that is what you hold when you are good at (for example) football.CompuChip –CompuChip 2016-10-16 16:32:38 +00:00 Commented Oct 16, 2016 at 16:32
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The main distinction, though it is somewhat blurred, is that at is used mainly with activities requiring our physical attendance and participation, and in is used mainly with domains or areas or circumstances or environments.
We can be good at ping pong, guessing games, or calming a frightened child.
Someone (or something) can be good in the kitchen, in delicate situations, in the board room, in chemistry, in times of political unrest.
The camel is good in the desert.
Stainless steel is good in the salt-air.
The virtues of the camel and the stainless steel are manifest in those particular environments.
Some activities can be understood in either manner. When we say
She is good at math
we are presenting her as someone who can solve number problems, or who offers answers in algebra class, that is, "math" is presented as an activity in which she participates.
And when we say
She is good in math
we are presenting math as a domain of knowledge, an area of intellectual pursuit.
P.S. Good means "possessing some skill or virtue", not "showing mastery" or "possessing unparalleled excellence". So it would be comical to say
Albert Einstein was good at physics.
Beethoven was good at music.
Pelé was good at soccer.
We can be "amazing","terrible", "fantastic", "horrible", "so-so", "excellent", "very good" or "not very good", or even "a genius" at or in something.
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edited Oct 16, 2016 at 11:21
answered Oct 16, 2016 at 11:00
TimR-gone from hereTimR-gone from here
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1 What about computers? "She is amazing (or good) at / in computers,"Mari-Lou A –Mari-Lou A♦ 2016-10-16 11:26:15 +00:00 Commented Oct 16, 2016 at 11:26
Would you recommend a learner to say The camel is good in the desert?Mari-Lou A –Mari-Lou A♦ 2016-10-16 11:29:05 +00:00 Commented Oct 16, 2016 at 11:29
1 You see, I'd be inclined to say that someone is good with computers.Mari-Lou A –Mari-Lou A♦ 2016-10-16 11:31:59 +00:00 Commented Oct 16, 2016 at 11:31
1 with is a viable option. With emphasizes instrumentality. at suggests "knows how to work these gadgets"; "in" suggests "computers as domain of knowledge".TimR-gone from here –TimR-gone from here 2016-10-16 11:32:57 +00:00 Commented Oct 16, 2016 at 11:32
1 "Good" there would be understood by native speakers to refer to virtues or capabilities which are beneficial in the desert. No native speaker would understand it to mean "behaves well" in the sense of "doesn't misbehave like an unruly child". Compare A Jeep is good in rough terrain.TimR-gone from here –TimR-gone from here 2016-10-16 11:43:03 +00:00 Commented Oct 16, 2016 at 11:43
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February 21, 2017, 02:33Dynamic similarity for full-scale model#1
siw
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Not sure if this is the correct forum for a non-CFD but aerodynamics technical question.
For dynamic similarity between real conditions and wind tunnel tests means that the important dimensionless parameters must be the same. In my case I am wind tunnel (open working section) testing a full scale flight vehicle and I want the tunnel set-up to match as closely as possible the flight condition (Mach number and altitude). So the main difference is the pressure at altitude and sea-level (where the wind tunnel is). I can match the Mach number by adjusting the tunnel airspeed but cannot match the Reynolds number. However, why cannot I just match the dynamic pressure so the loads on the vehicle are the same, even though this is a dimensional parameter? All the texts on dynamic similarity assume a scaled-down model so that the Reynolds number can be matched, as well as the Mach number.
Thanks
February 21, 2017, 06:51#2
HectorRedal
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You are stating that you are trying to test a full scale model. Maybe I am misunderstanding you, but as far as I know, a full scale model has a 1:1 scale relation.
This mean that if you want to achieve the same Reynolds number, keeping the same L, then you have to change the other properties in the same proprotion.
Re = rho U L / nu = rho' U' L / nu'
So, rho U / nu = rho' U' / nu'
What are the fluid are you using in your wind testing?
Is the same as in real flight (that is, air)?
February 21, 2017, 07:10#3
siw
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Yes, I am using a full scale in the wind tunnel so the length ratio will be 1:1.
This is all air.
But the wind tunnel is at sea-level and the real flight condition is at an altitude so the test air pressure will be different. So although the Mach number can be matched between the two scenarios by the wind tunnel airspeed, the Reynolds number cannot be matched as the air properties (pressure, temperature) in the wind tunnel will be the local sea-level values. The wind tunnel is not a closed loop facility in which the air properties can be varied.
So part of me thinks should I just match the dynamic pressure, even though that is not a non-dimensional parameter.
February 21, 2017, 18:01#4
HectorRedal
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Quote: Originally Posted by siw Yes, I am using a full scale in the wind tunnel so the length ratio will be 1:1. This is all air. But the wind tunnel is at sea-level and the real flight condition is at an altitude so the test air pressure will be different. So although the Mach number can be matched between the two scenarios by the wind tunnel airspeed, the Reynolds number cannot be matched as the air properties (pressure, temperature) in the wind tunnel will be the local sea-level values. The wind tunnel is not a closed loop facility in which the air properties can be varied. So part of me thinks should I just match the dynamic pressure, even though that is not a non-dimensional parameter. In my humble opinion, I think that dynamic similarity is needed. If not, I don't see how you can compare the experiment with the real behavoiur. For the compressible fluid flow, dynamic similary implies, equal Reynolds Number, equal Mach number and equal specific heat relation. But, maybe, someone with more experience on this can impart wisdom on the subject.
February 21, 2017, 19:29#5
LuckyTran
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Even if you pretend you can satisfy dynamic similarity for all other parameters, it's very hard to match Reynolds number and Mach number in an open wind tunnel. You simply don't have enough knobs, the best knob would be pressure and then temperature.
Matching dynamic pressure doesn't solve your problem, you still don't meet dynamic similitude for Reynolds number, Mach number, or the load coefficient. I'm not saying it is a bad idea, it's actually a great idea if what you're trying to measure is the load. But that doesn't answer the question of how to match Reynolds & Mach number.
February 22, 2017, 04:30#6
siw
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Thanks for the replies.
I know that not all of the non-dimensional parameters ( etc.) can be matched between the flight and wind tunnel so is it better to adjust the wind tunnel speed to match the flight and wind tunnel Mach numbers or adjust the wind tunnel speed to match the flight and wind tunnel dynamic pressures?
February 22, 2017, 08:43#7
LuckyTran
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Quote: Originally Posted by siw is it better to adjust the wind tunnel speed to match the flight and wind tunnel Mach numbers or adjust the wind tunnel speed to match the flight and wind tunnel dynamic pressures? First you match whatever it is you're trying to test. Doing so establishes a reference, a baseline. Then you vary the other parameters (turn your knobs) to see how the thing you're trying to test varies.
February 22, 2017, 09:43#8
siw
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LuckyTran, ultimately I have a rotating device (driven by the airflow) on the flight vehicle and it is that I want to have rotate at the same speed in the wind tunnel due to the aerodynamic loads on it.
March 1, 2017, 03:43#9
siw
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It states in the following "...the aircraft's aerodynamic structure responds to dynamic pressure alone, and the aircraft will perform the same when at the same dynamic pressure...". So why ensure dynamic similarity when it is only the dynamic pressure that must be matched for the aerodynamic loads to match between free flight and wind tunnel? This seems both intuitive and counter intuitive at the same time
March 1, 2017, 08:15#10
agd
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Because historically most wind tunnel testing doesn't care about the air frame response. If you are testing for lift and drag, then in theory matching the non-dimensional parameters should guarantee the same (non-dimensional) solution regardless of scale. If you also want to look at buffet or some other fluid-structure interaction the issue of scaling becomes much more complex, as you are finding.
March 1, 2017, 09:48#11
siw
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Yes, this is proving very tricky in my case and I have not reached a conclusion for the wind tunnel set-up parameters: i.e. 1) match Mach Number, 2) match Reynolds number or 3) match dynamic pressure, as cannot match them all. All this so that the exposed turbine device on the flight vehicle in the wind tunnel rotates at the same speed that it would in the free flight condition at altitude. The wind tunnel model and the free flight vehicle are the same size.
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3575 | https://water.mecc.edu/courses/Env211/lesson7_print.htm | Lesson 7:
Acids and Bases
In this lesson we will answer the
following questions:
What are acids and bases?
How do acids and bases participate in chemical reactions?
How are acids and bases measured?
Reading Assignment
In addition to the online lecture, read chapter 6 in Basic Chemistry for Water and Wastewater Operators and Chapter 2 and Chapter 21 in Simplified Procedures for Water Examination.
Lecture
Acids and Bases
Throughout history, chemists have created different definitions of acids and bases. Today, many people use the Brønsted-Lowry version. It describes an acid as a molecule that will give away a proton from one of its hydrogen atoms. At a minimum, that tells us that all Brønsted-Lowry acids must contain hydrogen as one of their building blocks. Hydrogen, the simplest atom, is made up of one proton and one electron. When an acid gives away its proton, it hangs on to the hydrogen atom's electron. This is why scientists sometimes call acids "proton donors".
Brønsted-Lowry bases, in contrast, are good at stealing protons, and they'll gladly take them from acids.
Not to confuse you, but scientists sometimes use another scheme, the Lewis system, to define acids and bases. Instead of protons, this Lewis definition describes what molecules do with their electrons. In fact, a Lewis acid doesn't need to contain any hydrogen atoms at all. Lewis acids only need to be able to accept electron pairs.
Water is chemically neutral. That means it is neither an acid nor a base. But mix an acid with water and the water molecules will act as bases. They'll snag hydrogen protons from the acid. The altered water molecules are now called hydronium. Mix water with a base and that water will play the part of the acid. Now the water molecules give up their own protons to the base and become what are known as hydroxide molecules.
Acid-base Chemistry
Acid-base chemistry is based on one simple reaction - the ionization
of water:
H2O H+ + OH-
As we saw in the last lesson, this equation means that water can break apart into a hydrogen ion and a hydroxide ion. A hydrogen and a hydroxide ion can also join together to form a water molecule.
In pure water, hydrogen and hydroxide ions are present in a 1:1 ratio since the only source of these ions is the ionization of water. Acids and bases are substances which change the balance of hydrogen and hydroxide ions in water. In this lesson, we will first explore the chemistry of acids and bases, then we will show how to calculate their concentration mathematically. In the next lesson, we will find out how to calculate the concentration of acids and bases in the lab.
Vinegar contains acetic acid, which makes it taste sour.
Onions release a gas which turns into sulfuric acid when
it reaches your eyes, making them burn.
What do you think of when you think of acids? You might think of sour-tasting acidic foods such as lemons. Or you might think of strong acids, such as battery acid, which can burn your skin or corrode metal. In this lesson, we will be concerned with the less visible properties of acids - their chemical properties.
Chemically, acids are substances which increase the concentration of hydrogen ions when they are placed in water. In this section we will be concerned only with strong acids. Weak acids act slightly differently and will be discussed in a later section.
You can recognize strong acids because they are ionic compounds which contain hydrogen ions. For example, consider the strong acids listed in the table below. Notice that each compound consists of a hydrogen ion bonded to some sort of anion.
| | | | |
--- --- |
| | | | |
| Nitric acid | HNO3 | H+ | NO3- |
| Perchloric acid | HClO4 | H+ | ClO4- |
| Sulfuric acid | H2SO4 | H+ | SO42- |
| Hydrochloric acid | HCl | H+ | Cl- |
| Hydrobromic acid | HBr | H+ | Br- |
| Hydriodic acid | HI | H+ | I- |
When a strong acid is placed in water, it will ionize completely, breaking down into its constituent ions. For example, hydrochloric acid reacts as shown below:
HCl H+ + Cl-
The ionization of a molecule of hydrochloric acid introduces a hydrogen ion and a chloride ion to the solution. The chloride ion has no effect on the acidity of the water, but the hydrogen ion makes the solution more acidic.
The diagrams above show what happens when hydrochloric acid is added to water. The top diagram merely contains water. Notice that most of the water is present in its un-ionized state, but that two water molecules have ionized into hydroxide and hydrogen ions. Despite the presence of hydrogen ions, this solution is neutral (meaning that it is neither acidic nor basic) because the number of hydroxide ions equals the number of hydrogen ions.
The bottom diagram shows what happens to the water after the addition of hydrochloric acid. Notice that all of the hydrochloric acid has ionized, so now there are five hydrogen ions and only two hydroxide ions in the solution. Since there are more hydrogen ions than hydroxide ions present, the solution has become acidic.
You probably have less familiarity with bases than with acids, so you may be surprised to learn how many bases you deal with in your everyday life. Soap, baking soda, milk of magnesia, and ammonia all contain bases. These substances exhibit some of the physical properties of bases, such as feeling slippery, tasting bitter, and dissolving greases. Bases also irritate the skin and eyes just as acids do.
Chemically, a base is a substance which decreases the concentration of hydrogen ions when it is placed in water. Strong bases decrease the hydrogen ion concentration by increasing the hydroxide ion concentration.
Let's consider what happens when we put sodium hydroxide (a strong base) in water. Like strong acids, strong bases ionize completely, so the sodium hydroxide breaks down into a sodium ion and a hydroxide ion:
NaOH Na+ + OH-
The hydroxide ion released by the sodium hydroxide then reacts with a hydrogen ion from the water, forming a water molecule:
OH- + H+ H2O
The reaction between the hydroxide ion and the hydrogen ion removes the hydrogen ion from the solution, making the solution less acidic and more basic. The two diagrams below show the net reaction of sodium hydroxide with water schematically:
Notice that in the top diagram, three water molecules have ionized to produce hydrogen and hydroxide ions. When three sodium hydroxide molecules are added (in the bottom diagram) the hydroxide ions from the base's dissociation combine with the three hydrogen ions in the solution, forming water. So the net result of the addition of sodium hydroxide to the water is that the concentration of hydrogen ions becomes lower and the acidity of the solution is decreased.
In the table below, I have listed some of the strong bases which you may run into in the lab:
| | |
--- |
| Name | Formula |
| Sodium hydroxide | NaOH |
| Lithium hydroxide | LiOH |
| Potassium hydroxide | KOH |
| Rubidium hydroxide | RbOH |
| Cesium hydroxide | CsOH |
| Calcium hydroxide | Ca(OH)2 |
| Strontium hydroxide | Sr(OH)2 |
| Barium hydroxide | Ba(OH)2 |
Weak Acids and Bases
So far we have talked about strong acids which contain a hydrogen ion and strong bases which contain a hydroxide ion. However, some acids and bases work slightly differently. Technically, an acid is any substance which donates a hydrogen ion to a solution and a base is any substance which accepts a hydrogen ion.
The table below lists some weak acids and bases:
| | |
--- |
| Weak Acids | Weak Bases |
| | | | --- | | Name | Formula | | Acetic acid | HC2H3O2 | | Phosphoric acid | H3PO4 | | Carbonic acid | H2CO3 | | Hydrofluoric acid | HF | | Fluosilicic acid | H2SiF6 | | | | | --- | | Name | Formula | | Ammonia | NH4OH | | Magnesium hydroxide | Mg(OH)2 | | Aluminum hydroxide | Al(OH)3 | | Lime | CaO | | Sodium silicate | Na2SiO2 | | HTH | NaOCl and Ca(OCl)2 | | Soda ash | Na2CO3 | |
Some of these weak acids and bases ionize just like their strong counterparts, donating hydrogen or hydroxide ions to the solution. For example, ammonia will ionize as follows:
NH4OH NH4+ + OH-
The difference between strong and weak bases (and between strong and weak acids) is that weak bases do not ionize completely when placed in solution. So, while every molecule of sodium hydroxide will break down into a sodium ion and a hydroxide ion when placed in solution, only some of the ammonia molecules will ionize under the same conditions. The illustration below gives a schematic representation of what happens under these two circumstances.
The top rectangle is an example of a solution containing sodium hydroxide. All of the molecules have ionized into their constituent ions. The bottom rectangle is an example of a solution containing ammonia. Most of the ammonia stayed in its current state, with only a small percentage breaking apart into ammonium and hydroxide ions.
Some weak bases not only do not ionize fully, they also do not release a hydroxide ion. For example, ammonia in its gaseous form reacts as follows:
H2O + NH3 NH4+ + OH-
Notice that the ammonia accepted a hydrogen ion from the water, decreasing the acidity of the solution. So it acted as a base even though it didn't produce hydroxide by ionization.
pH
Introduction
You should already be familiar with pH, which is the scale we use to measure the acidity or alkalinity of water. You will remember that the pH scale runs from 0 to 14, with numbers less than 7 being acidic and with numbers more than 7 being basic (also known as alkaline.) A pH of 7 means that the substance is neutral.
Although this elementary understanding of the pH scale is enough for many water and wastewater treatment processes, you will need to know more in order to fully understand the chemistry of acids and bases. pH is actually a measure of the concentration of hydrogen ions in the solution. Since acids donate hydrogen ions to a solution, they tend to make the pH lower. Bases, by accepting hydrogen ions, make the pH higher. A neutral pH of 7, which is the pH of distilled water, contains the same number of hydrogen ions as hydroxide ions.
The math used to calculate a solution's pH is relatively complicated, and we will not present it here. Instead, you just need to remember that the lower the pH value, the higher the concentration of hydrogen ions. For example, stomach acid with a pH of 1 has a much higher concentration of hydrogen ions than tomatoes do with a pH of 4. Baking soda with a pH of 8 has more hydrogen ions than household ammonia with a pH of 11.
You shouldn't find it surprising that pH is a measure of the hydrogen ion concentration of a solution, since we have already explained that acids and bases change that concentration. What you may not realize is that the pH scale only covers a small range of acidity and alkalinity. In fact, the pH scale is meant to mimic nature by covering the acidity and alkalinity values which might be found in natural waters. Strong acids and bases, like those discussed on the last page, have pH values at the far ends of the scale or even off the scale. Concentrated hydrochloric acid has a pH of 0, and one drop of 33% hydrochloric acid in a liter of distilled water can lower the pH from neutral to about 3. We will discuss ways to measure the concentration of strong acids and bases in a later section.
Neutralization
A neutral pH of 7 may mean that you are dealing with distilled water containing no acids and bases. In this case, the amount of hydrogen ions and hydroxide ions will be equivalent because they will both be due to the ionization of water. However, a neutral pH can also be achieved in a solution containing acids and bases as long as the acids and bases have neutralized each other, meaning that the acids have donated as many hydrogen ions as have been accepted by the bases.
Neutralization reactions occur whenever acids and bases are placed in proximity. An acid combines with a base to create water and a salt, as shown below:
HCl + NaOH H2O + NaCl
Titration, which we will introduce in a later lesson, is based on this neutralization reaction. You will also need to understand neutralization if you spill an acid or base in lab and want to clean it up safely. Neutralizing an acid with a base (or vice versa) can aid in cleanup, but you should also be aware that strong acids and bases can react explosively. Always use weak or low concentration acids or bases for neutralization reactions.
Neutralization occurs in nature as well. For example, organisms living in very acidic environments tend to excrete basic wastes which bring the environment back into equilibrium. Organisms living in basic environments excrete acids instead. This is one of the reasons that most natural waters, including septic tanks and wastewater treatment ponds which have been allowed to work for some time, tend to have a pH near 7. In addition, buffers (which we will explain in the next section) neutralize acids and bases both in natural waters and in the laboratory.
Buffers
A buffer is a solution containing a weak acid and one of its salts or a weak base and one of its salts. This solution is able to neutralize acids and bases without allowing the pH of the solution to change greatly. In lab, buffers are used when the pH of a solution must remain stable.
Some examples of the pairs which make up buffer solutions are shown in the table below.
| | |
--- |
| Acid or Base | Salt |
| Acetic acid | Sodium acetate |
| Phosphoric acid | Potassium phosphate |
| Oxalic acid | Lithium oxalate |
| Carbonic acid | Sodium carbonate |
| Ammonium hydroxide | Ammonium nitrate |
In order for a buffer to "resist" the effect of adding strong acids or bases, it must have both an acidic and a basic component. However, you cannot mix any two acid/base combination together and get a buffer. If you mix HCl and NaOH, for example, you will simply neutralize the acid with the base and obtain a neutral salt, not a buffer. For a buffer to work, both the acid and the base component must be part of the same equilibrium system - that way, neutralizing one or the other component (by adding strong acid or base) will transform it into the other component, and maintain the buffer mixture. Therefore, a buffer must consist of a mixture of a weak conjugate acid-base pair.
Of course, a buffer will not continue to neutralize the solution indefinitely. Eventually, the acid or salt will be used up, and the pH of the solution will begin to change. The amount of acid or base which a buffer solution is able to neutralize is known as the buffer capacity.
Buffer solutions are not limited to the lab. In natural water systems, carbon dioxide from the air often enters the water, forming carbonic acid. A salt of carbonic acid, such as calcium carbonate (limestone), may become dissolved in the water from the surrounding rocks and soil. Thus, a natural buffer solution is formed.
Normality
While pH is used to record the acidity or alkalinity of
natural waters, we use a measurement known as normality to show the concentration
of the much stronger acid and base solutions we use in the lab. Normality
is based on molarity, but also takes into account a characteristic of acids
and bases which we will call "equivalents" and will describe in the next section.
Normality, equivalents and equivalent weight are all related terms typically used in titrations when the titration reaction is unknown or just not used. Consequently, definitions for these terms vary depending on the type of chemical reaction that is being used for the titration. The two most common types of reactions for which normality is used are acid-base reactions and redox (reduction-oxidation) reactions.
The basic unit for normality related conventions is the equivalent. Equivalents are comparable to moles and used to relate one substance to another. Normality is a measure of concentration equal to the gram equivalent weight per liter of solution. Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.
We have already determined in a previous lesson that molarity of a solution refers to its concentration (the solute dissolved in the solution). The normality of a solutionr refers to the number of equivalents of solute per Liter of solution. The definition of chemical equivalent depends on the substance or type of chemical reaction under consideration. Because the concept of equivalents is based on the reacting power of an element or compound, it follows that a specific number of equivalents of one substance will react with the same number of equivalents of another substance. When the concept of equivalents is taken into consideration, it is less likely that chemicals will be wasted as excess amounts. Keeping in mind that normality is a measure of the reacting power of a solution, we use the following equation to determine normality:
Example 1:
If 2.0 equivalents of a chemical are dissolved in 1.5 L of solution, what is the normality of the solution?
Example 2:
A 800-mL solution contains 1.6 equivalents of a chemical. What is the normality of the solution?
First convert 800 mL to Liters:
800 mL / 1000 mL = 0.8 L
Now calculate the normality of the solution:
Equivalents
The reactive capacity of a chemical species, the ions or electrons, depends on what is being transferred in a chemical reaction. In acid-base reactions, an equivalent is the amount of a substance that will react with one mole of hydrogen ions. In oxidation-reduction (redox) reactions, where electrons are either gained or lost in a chemical reaction, it is one mole of electrons. Finding equivalents depends on the chemical species under consideration.
The oxidation state of an element describes the number of electrons transferred in reactions. For example, the oxidation, or valence states, of the following elements are equal to the number of equivalents:
Calcium: (Ca+2) ion: valence of 2; number of equivalents = 2
Aluminum: (Al+3) ion: valence of 3; number of equivalents = 3
For acids, an equivalent is the number of hydrogen ions a molecule transfers. In acids it is straightforward to find equivalent units. Look at the number directly after the hydrogen (H) in the chemical formulas below. The number provides the number of equivalents per mole of that acid:
Hydrochloric acid (HCl): equivalents = 1
Sulfuric acid (H2SO4): equivalents = 2
Phosphoric acid (H3PO4): equivalents = 3
Nitric acid (HNO3): equivalents = 1
For bases, it is the number of hydroxide ions (OH-) provided for a reaction, such as:
Sodium hydroxide (NaOH): equivalents = 1
Barium hydroxide (Ba(OH)2): equivalent = 2
One equivalent of an acid reacts with one equivalent of a base. For the acid HCl and base NaOH, both with one equivalent, they have the same reactivity. For H2SO4, with two equivalents, and NaOH, it will take twice the amount of the NaOH to react with the sulfuric acid. Mixing equal equivalents of acidic and basic solutions will result in a neutral solution.
Altough molarity can be used to measure the concentration
of acids, it is a relatively unuseful measurement for understanding neutralization
reactions. Why? Because not every acid or base can add (or remove)
the same number of hydrogen ions from solution.
Equivalent Weight
The equivalent weight can be thought of as the weight (or mass, to be precise) of a substance that will contain a single reactive proton (or hydrogen ion) or a single reactive hydroxide ion. The former case applies to acids, which are proton donors, while the second applies to bases, which are proton acceptors. The reason the concept of equivalent weight is needed is that some compounds can donate or accept more than one proton, meaning that for every mole present, the substance is in effect doubly reactive. The equivalent weight can be determined by:
Equivalent weight is defined as the ratio of molar mass of a substance to the valence of the substance. Valence is also denoted as equivalence factor. The valence is the number of hydrogen atoms in an acid, or hydroxide atoms in a base, and for salt, charge present in ionic forms. For reduction/oxidation (redox) reactions, it is the number of electrons than an oxidizing or reducing agent can accept or dontate that are counted as valence or equivalence factor. When the equivalent weight is expressed in grams using molar mass in grams, it is called gram equivalent weight.
Let's get some practice determining the equivalent weight for the following formulas:
H2SO4:
Molar mass of H2SO4 = 98 g/mol
Looking at the formula, there are 2 hydrogen atoms, so "n" will be 2 when determine the equivalent weight:
NaCl:
Molar mass of NaCl = 58.5 g/mol
Looking at the formula, because there are no hydrogen or hydroxide atoms, the number of equivalents is 1, because there will always be at least one equivalent before a rection can occur.
NaOH:
Molar mass of NaOH = 40 g/mol
In looking at the formula, there is 1 hydroxide (OH) atom, so the equivalent is 1.
Now let's look at a salt (a salt determines its equivalents differently because there are no hydrogen or hydroxide atoms involved, so we look at the charge):
Na2CO3:
Molar mass of Na2CO3 = 106 g/mol
The salt Na2CO3 ionizes to form 2Na+ and CO3-2, so the charge present on both is 2.
Making Normal Solutions (N)
Normality is the most
common measurement used for showing the concentration of acids and bases.
Normality takes into account both the molarity of the solution and the equivalent
content of the acid or base. It is defined as the number of gram equivalent present in per liter solution and can be determined through the following formula:
To calculate normality of the solution, follow these steps:
Find the equivalent weight of the solute based on the chemical reaction it is going to be using
Calculate the number of gram equivalents of the solute
Calculate the volume in liters
Calculate the normality using the formula given above
Making normal solutions can be a bit confusing. Aqueous solutions of acids and bases are often described in terms of their normality rather than their molarity. In order to properly make a Normal solution, the student must understand the difference between a pure reagent and a diluted reagent.
A "1 Normal" solution (1 N) contains 1 "gram equivalent weight" of solute, topped-off to one liter of solution. The gram equivalent weight is equal to the solute's molecular weight (molar mass), expressed as grams, divided by the valence (n) of the solute:
After the equivalent weight (or milliequivalent weight) has been calculated, then the following equation is used:
or
The equivalent weight of a substance depends upon the type of reaction in which the substance is taking part. Some different types of chemical reactions, along with how to determine a solute's equivalent weight for each reaction, are given below.
Making a Normal Solution With Salts
Example:
Calculate the normality of a sodium chloride (NaCl) solution prepared by dissolved 2.9216 grams of NaCl in water and then topping it off with more water to a total volume of 500.0 mL.
First, check the periodic table to determine the molar mass, or molecular weight, of NaCl, which is 58.44.
In looking at the formula (NaCl), the equivalent is 1 because there is room in the molecule for only one replaceable (H+) ion. In other words, one hydrogen atom can replace the sodium atom in NaCl.
Now determine the equivalent weight of NaCl:
You will need to know the milliequivalent weight of NaCl (since the solute volume is in mL):
Next calculate the normality of the sodium chloride solution:
Making Normal Solutions With Pure (Non-Aqueous) Acids
The equivalent weight of an acid is its molecular weight, divided by the number of replaceable hydrogen atoms in the reaction. To clarify this concept, consider the following acids:
Hydrochloric acid (HCl) has one replaceable hydrogen ion (H+). Sulfuric acid (H2SO4) has two replaceable hydrogen ions (2H+). The valences of these acids are determined by their respective replaceable hydrogen ions, displayed below.
| | |
--- |
| Acid | Valence (Replaceable hydrogen ions) |
| HCl | n = 1 |
| HNO3 | n = 1 |
| H2SO4 | n = 2 |
| HF | n = 1 |
So, for pure HCl, its molecular weight is 36.46, its equivalent weight is 36.46 and therefore a 1N solution would be 36.46 grams of the pure chemical per liter. Note that, in the case of HCl, a 1N solution has the same concentration as a 1M solution.
To make a 1N H2SO4 solution from pure H2SO4, its molecular weight is 98.08, and its equivalent weight is (98.08/2 = 49.04 grams/Liter or 49.04 grams per 1000 mL). So a 1N solution would be 49.04 grams of the pure chemical per liter.
Making Normal Solutions From Pure Alkalis (Bases)
The equivalent weight of a base is defined as "its molecular weight divided by the number of hydrogen ions that are required to neutralize the base". To understand the valences of alkalis, consider the following examples:
The (OH-) ion in Sodium hydroxide (NaOH) can be neutralized by one hydrogen ion. the (OH)2-- ions in Calcium hydroxide (Ca(OH)2) can be neutralized by two hydrogen ions. As was the case with acids, the valences (n) of these bases are determined by their respective replaceable hydrogen ions, displayed below.
| | |
--- |
| Base | Valence (Replaceable hydrogen ions) |
| NaOH | n = 1 |
| Ca(OH)2 | n = 2 |
So, for NaOH, its molecular weight is 40, its equivalent weight is 40, and therefore a 1N solution would be 40 grams of the pure chemical per Liter of water. You will also note, in the case of NaOH, a 1N solution is the same concentration as a 1M solution.
For Ca(OH)2, its molecular weight is 74, its equivalent weight is (74/2 = 37, because n = 2). Therefore, a 1N aqueous solution of Ca(OH)2 is 37 grams of the pure chemical per Liter of water. Of course, many acidic reagents and basic reagents come from the factory in a diluted aqueous form, which is a form of preparation we will not cover in this lesson.
Example:
Calculate the normality of a NaOH solution formed by dissolving 0.2 g NaOH to make a 250 mL solution.
First we need to determine the molar mass, which is 40 g/mol.
Then determine the milliequivalent weight (since the solution is in mL):
Now, calculate the normality of the NaOH solution:
Relationship Between Normality and Molarity
Here is Normality in terms of molarity:
Normality = n x Molarity
Where n = number of Hydrogen in acids, or Hydroxides in bases and for salt, charge present in ionic forms.
Calculating Dilutions (Acid-Base Titration)
Once you have calculated the normality of an acid or base solution,
you can easily calculate the concentration of any dilutions of that solution.
The formula used is essentially the same as that used for any other dilution
calculation:
N1V1 = N2V2
Where:
N1 = normality of the first solution
V1 = volume of the first solution
N2 = normality of the second solution
V2 = volume of the second solution
Example 1:
After producing the 0.5 L of a 0.37N solution of calcium
hydroxide in the last section, how would you dilute it to form a 0.25N solution?
(0.37N) (0.50 L) = (0.25N) V2
(0.37N)(0.50 L) / 0.25N = V2
0.74 L = V2
Based on the calculations above, we know that we have to add enough
water to the 0.37N solution so that the total volume reaches 0.74 L.
Then we would have a 0.25N solution.
Example 2:
How many milliliters of 2N NaOH are needed to prepare 300 mL of 1.2N NaOH?
N1V1 = N2V2
(2N)(V1) = (1.2N)(300 mL)
V1 = (1.2N)(300 mL) / 2N
V1 = 180 mL
So, to prepare the 1.2N NaOH solution, you pour 180 mL of 2N NaOH into your container and add water to get 300 mL total volume.
Safety Tips
Always use a fume hood when handling highly concentrated acids and bases. Wear a platic apron, plastic gloves, and eye goggles. This is particular important when working with hydrofluoric acid.
If you need a reagent that can produce highly precise results in quantitative analysis, then you should store all newly prepared strong basic solutions in labeled plastic containers.
Highly acidic reagents can be stored in either glass containers or in plastic containers. The exception is hydrofluoric acid, which etches glass and weakens it. Store hydrofluoric acid in a labeled plastic container.
It is advisable to store concentrated reagents that come straight from the supplier in their original containers, with their labels intact.
When preparing solutions, students sometimes make the mistake of adding the solute to a set volume of solvent. Doing this will produce a solution of the wrong concentration. Example: when making one liter of a 1N solution of NaCl, do not add 58.5 grams of NaCl to 1 Liter of water. Instead, the correct way to make the solution is to add 58.5 grams of NaCl into a container and then top it off with water to a total volume of 1 Liter.
Never add water into a large volume of concentrated acid! You risk creating an explosion! The rule is:
"Acid into water = you're doing what ya oughta."
"Water into acid = you might get blasted!"
Alkalinity
The final topic we will introduce in this lesson is alkalinity, which is the capacity of a solution to neutralize a strong acid. Despite the name, you should not think of alkalinity as the amount of base found in the solution; instead, alkalinity is a measurement of the buffering ability of the solution. Water high in alkalinity will be able to maintain its pH, despite the addition of acids.
Alkalinity in natural waters is caused primarily by carbonate (CO3-2) and bicarbonate (HCO3-) ions. A few other causes of alkalinity may also be present in water, but usually at much lower concentrations. These other types of alkalinity can include hydroxide, borate, silicate, phosphate, ammonium, and sulfide. Bicarbonates are the major components because of carbon dioxide action on basic materials of the soil. The alkalinity of raw water may also contain salts formed from organic acids such as humic acids.
Alkalinity in water acts as a buffer that tends to stabilize and prevent fluctuations in pH. In fact, alkalinity is closely related to pH, but the two must not be confused. Total alkalinity is a measure of the amount of alkaline materials in the water. The alkaline materials act as buffers to changes in the pH. If the alkalinity is too low (below 80 ppm or mg/L), the pH can fluctuate rapidly because of insufficient buffer. High alkalinity (above 200 ppm) results in the water being too buffered. Thus having significant alkalinity in water is usually beneficial because it tends to prevent quick changes in pH that interfere with the effectiveness of common water treatment processes. Low alkalinity also contributes to the corrosive tendencies of water.
Acidity in water is usually due to carbon dioxide, mineral acids, or hydrolysis of some heavy metal salts, such as aluminum sulfate. Hydrolysis is decomposition of a substance by reacting with water.
How are hardness and alkalinity related? They are both expressed as mg/L CaCO3. In tap water, when the concentration of hardness and alkalinity are the same, both are probably due to dissolved calcium carbonate. Calcium carbonate does not dissolve well above pH 7.0, so other chemicals like calcium chloride and sodium bicarbonate can be used to adjust hardness and alkalinity independently.
Alkalinity usually enters the water as salts, such as calcium carbonate (also known as limestone). Carbonate and bicarbonate can also be formed when carbon dioxide from the air is dissolved in the water. Hydrolized metal salts, like aluminum sulfate and various iron salt coagulants, produce acidity and, thus, consume alkalinity. Although the exact amount may vary, 1 mg/L of alum requires about 0.5 mg/L of alkalinity. It is essential that operators understand this reaction because coagulation may be affected if enough alkalinity is not present.
In any case, all forms of alkalinity operate similarly to neutralize acids. Let's see what happens when we add sulfuric acid to water containing bicarbonate alkalinity:
2HCO3- + H2SO4 → 2H2CO3 + SO42-
First, the sulfuric acid loses its hydrogen ions, which would usually increase the acidity of the water. However, the bicarbonate accepts the hydrogen ions, turning into carbonic acid. Since carbonic acid is a weak acid, it tends to stay in its un-ionized state, so the pH of the water remains neutral.
The general formula for alkalinity derived from a titration is:
Example:
A 100 mL sample of water is tested for alkalinity. The normality of the sulfuric acid used for titrating is 0.02N. If 7.6 mL of titrant is used, what is the alkalinity of the sample?
Alkalinity in Natural Waters
Alkalinity is a measure of a river's buffering capacity, or its ability to neutralize acids. Alkaline compounds in the water, such as bicarbonates (baking soda is one type), carbonates, and hydroxides remove hydrogen ions and lower the acidity of the water, which means increased pH. They do this usually be combining with the hydrogen ions to make new compounds. Without this acid neutralizing capacity, any acid added to a river would cause an immediate change in the pH. Measuring alkalinity is important to determining a river's ability to neutralize acidic pollution from rainfall or snowmelt. It's one of the best measures of the sensitivity of the river to acid inputs. Alkalinity comes from rocks and soils, salts, certain plant activities, and certain industrial wastewater discharges.
The Carbonate System
The relationship between alkalinity and pH is relatively complex. As mentioned in the last section, higher alkalinity tends to prevent water from becoming acidic. In addition, pH influences the type of alkalinity found in water. In this section, we will consider the carbonate system, which is the relationship between pH and the different forms of carbonate - carbonic acid, bicarbonate, and carbonate.
The graph above summarizes the carbonate system. Notice that the form of carbonate in the water is very dependent upon pH. At a low pH, the carbonate is present as carbonic acid. Bicarbonate can be found in water with a pH between 4.3 and 12.3. Above a pH of 8.3, carbonate is also present.
What causes this relationship between carbonate species and pH? Carbon dioxide, carbonic acid, bicarbonate, and carbonate can transform back and forth by gaining or losing hydrogen ions in the reactions shown below:
CO2 + H2O H2CO3
H2CO3 HCO3- + H+
HCO3- CO32- + H+
Based on the concentration of hydrogen ions in the solution (the acidity of the solution), various of these reactions will be more or less likely to take place. For example, when the water is very acidic (containing a high concentration of hydrogen ions), the hydrogen ions tend to attach to carbonate or bicarbonate, forming carbonic acid. However, if the water is very basic (containing a lower concentration of hydrogen ions), then carbonic acid tends to break apart, adding hydrogen ions to the solution. Do not concern yourself with the reactions above, it is just to show you how gaining and losing hydrogen ions can transform the carbonate species.
In the Treatment Plant
Alkalinity is of little sanitary significance, although extremely high levels of alkalinity in water can be problematic. It has been reported that a carbonate concentration of 350 ppm is unhealthy. In addition, free caustic (hydroxide) alkalinity may impart a bitter taste to the water and can even cause a burning sensation at high concentrations. However, in most water treatment plants, operators are more concerned with ensuring that there is enough alkalinity in the water to promote coagulation and prevent corrosion than they are with removing alkalinity from water.
Alkalinity is essential for proper coagulation. When alum is added to raw water, the alum reacts with the alkalinity present to form floc. As a rule of thumb, coagulation usually requires an alkalinity equal to half of the amount of alum used. So, if your plant treats raw water with 25 ppm of alum, then a minimum of 12.5 ppm of alkalinity must be present in the water for floc to form. If insufficient alkalinity is present in the water during coagulation, then dense floc with not form and soluble alum will be left in the water.
Since coagulation uses up alkalinity in the water, treated water often tends to be corrosive (acidic). The operator must make sure that an adequate amount of alkalinity is present in the treated water as well as in the raw water to prevent corrosion of pipes in the distribution system. We will discuss the math used to determine the appropriate alkalinity of treated water in the alkalinity lab.
Water treatment may require the addition of artificial alkalinity to enhance coagulation or to prevent corrosion. If the source water does not contain an adequate amount of natural alkalinity, treatment will usually include the addition of lime or soda ash to artificially increase the amount of alkalinity in the water. Water with high pH is generally depositing and water with low pH is corroding.
Acids
and bases are substances which change the balance of hydrogen and hydroxide
ions in water. Both are divided into two categories - strong acids or
bases which ionize completely in water, and weak acids or bases which ionize
only partially in water.
Acids and bases undergo a variety of chemical reactions.
Acids donate hydrogen ions to water while bases remove hydrogen ions from
water. When acids and bases are brought together, they neutralize each
other. Buffers can neutralize both acids and bases.
pH can be used to measure the balance of acidity and alkalinity
in water. Normality, in contrast, is used to measure the concentration
of a specific acid or base in a relatively strong solution.
Alkalinity is the buffer capacity of water, usually caused by carbonate and bicarbonate ions. The type of alkalinity found in water will depend on the pH of the water. Regardless of the type of alkalinity, a certain concentration of alkalinity is required in the water treatment plant to promote good coagulation and to prevent corrosive water.
New Formulas
Used
Normality1Volume1 = Normality2Volume2 or (N1V1 = N2V2)
Complete the worksheet for this lesson. You must be logged into Canvas to submit this assignment. Make sure you choose the appropriate semester.
Read the pH and alkalinity labs.
Complete the Acids and Bases Labster Simulation. You must be logged into Canvas to submit this assignment. Make sure you choose the appropriate semester.
Answer the questions in the lesson quiz. You must be logged into Canvas to take this quiz. You may take the quiz up to three times; an average will be taken for final grade calculation. Make sure you choose the appropriate semester. |
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How can the generalization error be estimated?
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How would you estimate the generalization error? What are the methods of achieving this?
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Generalization error is the error obtained by applying a model to data it has not seen before. So, if you want to measure generalization error, you need to remove a subset from your data and don't train your model on it. After training, you verify your model accuracy (or other performance measures) on the subset you have removed since your model hasn't seen it before. Hence, this subset is called a test set.
Additionally, another subset can also be used for parameter selection, which we call a validation set. We can't use the training set for parameter tuning, since it does not measure generalization error, but we can't use the test set too since our parameter tuning would overfit test data. That's why we need a third subset.
Finally, in order to obtain more predictive performance measures, we can use many different train/test partitions and average the results. This is called cross-validation.
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edited Apr 11, 2018 at 23:42
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This def of gen error is wrong. "Generalization error" is defnd as abs(empirical error calculated (using the hypothesis from a hypothesis class) on a set of datapoints known as a sample, drawn from data distriution MINUS the error calculated (again using the same hypothesis in the hypothesis class) on the entire "true" data distribution), i.e. : \Delta_{gen} = |\hat{L_S}(h) - L_D(h)| where \hat{L_S}(h) is the "empirical error" on sample S drawn from D, and L_D(h) is the true error on the actual data distribution. Reference simonshaoleidu.com/teaching/cs599tdl/DLbook.pdf - pg 32, chap 4 Megh –Megh 2025-08-27 20:00:37 +00:00 Commented Aug 27 at 20:00
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Error Estimation is a subject with a long history. The test-set method is only one way to estimate generalization error. Others include resubstitution, cross-validation, bootstrap, posterior-probability estimators, and bolstered estimators. These and more are reviewed, for instance, in the book: Braga-Neto and Dougherty, "Error Estimation for Pattern Recognition," IEEE-Wiley, 2015.
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answered Aug 16, 2020 at 23:41
Ulisses Braga-NetoUlisses Braga-Neto
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It's basically not possible to test besides some empirical experiments. All the generalization bounds only apply if your process actually follows the model assumptions which you don't actually know to be true.
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edited Nov 8, 2020 at 13:07
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answered Nov 7, 2020 at 17:29
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1 This is wrong. The generalization error can be estimated. This answer provides a way to estimate the generalization error. Now, if you're talking about generalization bounds, that's something different. However, the question was about estimating the generalization error.nbro –nbro 2020-11-07 17:39:03 +00:00 Commented Nov 7, 2020 at 17:39
You don't know if your validation data actually comes from the same distribution or.not basically it's all guess, you never know much of anything until you run the system real time.FourierFlux –FourierFlux 2020-11-07 20:37:14 +00:00 Commented Nov 7, 2020 at 20:37
The point being made by nbro is that you are estimating something or approximating something. It may not be correct but it is the best you can do, hence the word 'estimate'. What you are talking about is the actual generalization error on the true distribution compared to the error on the sampled distribution (which gives you the estimated error of the true distribution).user9947 –user9947 2020-11-08 13:06:48 +00:00 Commented Nov 8, 2020 at 13:06
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3577 | https://lieferscience.weebly.com/uploads/1/0/9/9/109985151/wkst_4f_titrations_mccc2-ws3f.pdf | Steps for problem 3 on website Slightly simplified version of problem 3 on website Slightly simplified version of problem 3 with steps on website AP Worksheet 4f (Titration) 1. Calculate the concentration of a phosphoric acid solution if 45.0 mL of it were required to neutralize 20.8 mL of 0.532 M sodium hydroxide. Assume complete ionization of the acid. 2. If 0.664 g of an unknown acid were required to neutralize 10.0 mL of 0.800 M NaOH, calculate the molar mass of the acid given it reacts with sodium hydroxide in a 1:2 ratio, i.e., 1 acid : 2 NaOH. 3. 3.364 g of hydrated barium chloride, BaCl2xH2O, was dissolved in water and made up to a total volume of 250.0 mL. 10.00 mL of this solution required 46.92 mL of 2.530 x 10-2M silver nitrate for complete reaction. Calculate the value of x in the formula of hydrated barium chloride, given the net ionic equation for precipitation below. Cl-(aq) + Ag+(aq) AgCl(s) Steps for problem 3 on website Slightly simplified version of problem 3 on website Slightly simplified version of problem 3 with steps on website AP Worksheet 4f (Titration) 1. Calculate the concentration of a phosphoric acid solution if 45.0 mL of it were required to neutralize 20.8 mL of 0.532 M sodium hydroxide. Assume complete ionization of the acid. 2. If 0.664 g of an unknown acid were required to neutralize 10.0 mL of 0.800 M NaOH, calculate the molar mass of the acid given it reacts with sodium hydroxide in a 1:2 ratio, i.e., 1 acid : 2 NaOH. 3. 3.364 g of hydrated barium chloride, BaCl2xH2O, was dissolved in water and made up to a total volume of 250.0 mL. 10.00 mL of this solution required 46.92 mL of 2.530 x 10-2M silver nitrate for complete reaction. Calculate the value of x in the formula of hydrated barium chloride, given the net ionic equation for precipitation below. Cl-(aq) + Ag+(aq) AgCl(s) Step 1: Write a molecular equation for the reaction of barium chloride with silver nitrate Step 2: Perform a titration calculation to determine the number of moles of barium chloride present in the reaction of 10.0 mL of the barium chloride solution Step 3: Create a ratio comparing the number of moles of barium chloride in 10.0 mL of the solution with the number of moles that would be present in the 250.0 mL solution (THIS STEP IS NOT PART OF VERSION) Step 4: Calculate the mass of barium chloride from the moles in the 250.0 mL solution Step 5: Determine the mass of water in the hydrated sample Steps for problem 3 on website Slightly simplified version of problem 3 on website Slightly simplified version of problem 3 with steps on website Step 6: Use the mass of anhydrous barium chloride and the mass of water to determine the formula of the hydrate. AP Worksheet 4f (Titration) 1. Calculate the concentration of a phosphoric acid solution if 45.0 mL of it were required to neutralize 20.8 mL of 0.532 M sodium hydroxide. Assume complete ionization of the acid. 2. If 0.664 g of an unknown acid were required to neutralize 10.0 mL of 0.800 M NaOH, calculate the molar mass of the acid given it reacts with sodium hydroxide in a 1:2 ratio, i.e., 1 acid : 2 NaOH. 3. 0.1346 g of hydrated barium chloride, BaCl2xH2O, was dissolved in water and made up to a total volume of 10.00 mL. This solution required 46.92 mL of 2.530 x 10-2M silver nitrate for a complete reaction. Calculate the value of x in the formula of hydrated barium chloride, given the net ionic equation for precipitation below. Cl-(aq) + Ag+(aq) AgCl(s) Steps for problem 3 on website Slightly simplified version of problem 3 on website Slightly simplified version of problem 3 with steps on website AP Worksheet 4f (Titration) 4. Calculate the concentration of a phosphoric acid solution if 45.0 mL of it were required to neutralize 20.8 mL of 0.532 M sodium hydroxide. Assume complete ionization of the acid. 5. If 0.664 g of an unknown acid were required to neutralize 10.0 mL of 0.800 M NaOH, calculate the molar mass of the acid given it reacts with sodium hydroxide in a 1:2 ratio, i.e., 1 acid : 2 NaOH. 6. 3.364 g of hydrated barium chloride, BaCl2xH2O, was dissolved in water and made up to a total volume of 250.0 mL. 10.00 mL of this solution required 46.92 mL of 2.530 x 10-2M silver nitrate for complete reaction. Calculate the value of x in the formula of hydrated barium chloride, given the net ionic equation for precipitation below. Cl-(aq) + Ag+(aq) AgCl(s) Step 1: Write a molecular equation for the reaction of barium chloride with silver nitrate Step 2: Perform a titration calculation to determine the number of moles of barium chloride present in the reaction of 10.0 mL of the barium chloride solution Step 4: Calculate the mass of barium chloride in the solution Step 5: Determine the mass of water in the hydrated sample Step 6: Use the mass of anhydrous barium chloride and the mass of water to determine the formula of the hydrate. |
3578 | https://au.indeed.com/career-advice/career-development/social-loafing | What Is Social Loafing? (Definition, Causes and Prevention) | Indeed.com Australia
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What Is Social Loafing? (Definition, Causes and Prevention)
What Is Social Loafing? (Definition, Causes and Prevention)
Written by
Indeed Editorial Team
Updated 4 March 2025
Social loafing, also known as the Ringelmann effect, is a term in social psychology that refers to an individual putting in less effort when working as part of a group than they may when working on a project alone. This behaviour can have a significant impact on projects, teams and businesses. Understanding what the Ringelmann effect is and how to prevent it can help you to be a successful team leader. In this article, we define social loafing, explain what usually causes it, discuss ways to prevent it as a team leader and share the benefits of reducing this social behavioural phenomenon.Key takeaways:
Social loafing, also known as the Ringelmann effect, is when people exert less effort while working in a group compared to working alone, and understanding this phenomenon can aid in successful teamwork and group performance management.
The Ringelmann effect can be caused by several factors including group size, inequitable contributions, absence of evaluation, poor communication, unclear goals, and detached group settings.
To prevent social loafing as a team leader, consider having smaller groups, clearly define tasks and objectives, manage communication effectively, conduct team-building activities, acknowledge individual contributions and develop team contracts.
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What is social loafing?
Social loafing is a concept in psychology that suggests that individuals working in groups have an inclination to expend less effort than when undertaking a task on their own. Maximilian Ringelmann, a French agricultural engineer, discovered this behaviour in 1913 by observing a group taking part in a rope-pulling experiment. He noticed their pulling power as a group was inferior to that of the sum of their individual pulling powers. This group dynamics phenomenon is typically evident in collective effort model scenarios where individual effort and contribution can be challenging to evaluate.The Ringelmann effect is often a subconscious decrease in social awareness that individuals may experience when engaging in group work. It indicates that individuals commonly feel demotivated and unindividuated when there's a lack of judgement of their accountability, like in a large group project scenario. This social behaviour can have a significant impact on teams and individuals in the workplace and affect organisation-wide productivity levels. The presence of others may either increase or reduce effort depending on whether social facilitation or social loafing takes place.
Causes of the Ringelmann effect
Many theories can logically explain why the Ringelmann effect occurs. Below, you can find some of them:
Group size
The fundamental cause of social loafing is group size or the number of people. The larger a group, the less inclined members of the group can feel to perform. This is usually because of issues surrounding visibility. Because of this, the Ringelmann effect may be more pronounced in larger groups than in smaller ones.Related:5 Steps to Become a Better Ally at Work
Inequitable contribution
Members of a group or team may believe that other people in their group or on their team aren't putting in as much effort as they are. Because of this, they may decide to reduce their contribution to be more equal to the contribution of other participants. This usually creates a downward spiral in the group or team's productivity levels, where the endpoint is minimal output. This can be one of the consequences of social loafing.Related:Employee Control: Empowering Employees in the Workplace
Absence of evaluation
It's typical for there to be a lack of individual evaluation present in group environments. Without being judged on individual effort, social loafing can occur, as people believe their contribution or lack thereof can easily go unnoticed. Consider a sales team's performance that's measurable by their group sales figures instead of individual tasks. Here, there can be a tendency for individual salespeople to underperform because of the lack of evaluation.Related:Performance Review: What It Is and Why It Matters
Poor communication
When team members feel like their contribution is likely to not matter or make a difference to the group or team's outcome, they can be more likely to experience the Ringelmann effect. Team members can often feel this way because of a lack of communication. Without clarity surrounding group task objectives and an environment in which people feel like they can share their thoughts and ideas, social loafing can occur.Related:Why Interpersonal Communication Is So Important at Work
Unclear goals
When the desired result of a team or group project is unclear, social loafing can occur. The absence of an important objective and a plan for reaching it can cause group participants to feel demotivated. Because of this, they can be less likely to commit to contributing to the common goal to the best of their ability.Related:Understanding Objectives vs. Goals (Including Examples)
Detached groups
Individuals in groups that are disjointed and lacking cohesion can be more prone to experiencing the Ringelmann effect. When individuals lack respect for or have poor expectations of those on their team, they may reduce their efforts because of demotivation and diffusion of responsibility. They may also feel less concerned about letting their co-worker down because of their free-riding or bystander tendencies. In these scenarios, social loafers may emerge, further impacting team dynamics.
How to prevent social loafing as a team leader
Below, you can find several ways of reducing social loafing in team environments:
1. Consider smaller groups
Where possible, it can be a good idea to minimise group numbers. With fewer people to share responsibility amongst, individuals within the group may be less prone to social loafing. Smaller groups may also allow for enhanced relationships and communications, which can reduce the likelihood of individuals experiencing feelings associated with this behaviour.
2. Clearly define tasks and objectives
To reduce social loafing, it can be important that group participants have clearly defined objectives to work with. Clearly defined goals and an understanding of the tasks required to achieve them can provide individuals with a sense of purpose, accountability and motivation to work toward a common and important goal. It's usually a good idea that team goals and their associated tasks have well-established expectations and noticeable and quantifiable outcomes.
3. Manage communication
Effective communication can contribute significantly to a reduction in social loafing in group environments. Effective communication refers to the clear definition of group goals and tasks, and also to how group participants communicate with each other. Developing a framework for group communication channels and ensuring frequent group discussions during which all participants can express their ideas can be two effective ways of managing communication to reduce social loafing.
4. Conduct team building
Conducting team-building activities to enhance workplace relationships can be effective in reducing the Ringelmann effect within groups. When team members have good relationships with each other, built on mutual understanding, trust and respect, they're more likely to work cohesively together. A positive team environment can also foster a sense of responsibility and motivation amongst individuals to perform for the benefit of the group's objectives.Related:10 Team Effectiveness Models for Building Effective Teams
5. Acknowledge individual achievement
As social loafing is commonly a result of the lack of individual evaluation in group settings, evaluating individual progress and acknowledging team members' achievements can be effective ways of boosting intrinsic involvement. This might include, for example, meeting with group members individually to discuss their contributions, accomplishments and any areas in which they might improve. It might also involve highlighting team members' successes alongside the group's successes in team meetings and progress updates.
6. Develop team contracts
As unclear goals and poor communication can contribute to social loafing, a key prevention method is ensuring that team members have a solid understanding of their roles, objectives and how to work effectively together. An effective way of doing this can be to develop team contracts or terms for collaborative tasks. These contracts can outline the expectations of the group, what each individual is responsible for and the framework for group communications, for example.
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Benefits of preventing the Ringelmann effect
Teams and organisations can experience several benefits from reducing social loafing in group settings. Some of these benefits may include:
Enhanced productivity: When team members feel accountable for their individual performance in a group setting, they're often likely to feel motivated to contribute more to the group's common goal. A group comprising motivated and committed individuals can experience higher levels of output.
Improved employee relations: Reducing social loafing in group settings can create enhanced employee relationships and reduce workplace conflicts. When individuals feel that the people on their team are contributing to their common goal equitably, they're more likely to have positive feelings about each other and care more about the group's success.
Better results: Teams that work effectively together to achieve their common goals generally produce better results. When team members are less prone to social loafing in a group setting, they're more likely to perform at their best and achieve results that can motivate them for future tasks.
Lower employee turnover: Reducing social loafing can have a positive impact on how employees perceive their workplace environment, while also preventing burnout amongst team members who may perform more than their required work to compensate where social loafing is present. Employees who perceive their work positively and feel their efforts are sustainable may be more likely to experience job satisfaction and demonstrate loyalty to their employer.
The information on this site is provided as a courtesy and for informational purposes only. Indeed is not a career or legal advisor and does not guarantee job interviews or offers.
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Finding Eccentricity of A Hyperbola
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Modified 6 years, 9 months ago
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1
$\begingroup$
Given an asymptote to an hyperbola and that a line perpendicular to it, intersects it at a single point, we need to find its eccentricity.
Asymptote : $5x-4y+5=0$ and Tangent : $4x+5y-7=0$.
I thought that if we consider asymptote to be limiting tangent at infinity, then the point of intersection $(3/41,55/41)$ should lie on the director circle of the hyperbola as it is the locus of the perpendicular tangents. I am finding too many variables to handle here. And I suspect this question should be solved by an argument for a specific type of hyperbola (Maybe rectangular), but I could use a hint over here
conic-sections
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edited Dec 9, 2018 at 15:51
Sarthak RoutSarthak Rout
asked Dec 7, 2018 at 8:14
Sarthak RoutSarthak Rout
36711 silver badge1212 bronze badges
$\endgroup$
5
$\begingroup$ The orthoptic for an ellipse and a hyperbola is its director circle. Please google 'director circle' . $\endgroup$
Sarthak Rout
– Sarthak Rout
2018-12-07 09:10:14 +00:00
Commented Dec 7, 2018 at 9:10
$\begingroup$ Google orthoptic and director circle yourself. The orthoptic of an ellipse is its director circle, but the orthoptic of a hyperbola is neither of its director circles. $\endgroup$
amd
– amd
2018-12-07 09:20:20 +00:00
Commented Dec 7, 2018 at 9:20
$\begingroup$ Look, I googled and I got this Wiki: In geometry, the director circle of an ellipse or hyperbola is a circle consisting of all points where two perpendicular tangent lines to the ellipse or hyperbola cross each other. And Orthoptic of a hyperbola as per wiki" The orthoptic of a hyperbola x^2/a^2 y^2/b^2 = 1, a > b, is the circle x^2 + y^2 = a^2 b^2 (in case of a ¤ b there are no orthogonal tangents, see below)". And , that is the director circle of the hyperbola. Please elaborate. $\endgroup$
Sarthak Rout
– Sarthak Rout
2018-12-07 09:29:24 +00:00
Commented Dec 7, 2018 at 9:29
$\begingroup$ There appear to be conflicting articles and terminology (not the first time). In en.wikipedia.org/wiki/Orthoptic_(geometry) a hyperbolas orthoptic is pointedly not called its director circle; en.wikipedia.org/wiki/Hyperbola also avoids calling the orthoptic the director circle, but also uses the term circular directrix for the latter. Unlike the orthoptic, these circles exist for all hyperbolas, a useful trait if youre going to use them for construction. A footnote suggests that the term director circle is used differently in German and English. $\endgroup$
amd
– amd
2018-12-07 10:10:39 +00:00
Commented Dec 7, 2018 at 10:10
1
$\begingroup$ At any rate, without further information or assumptions, I think that the best you can do here is to state an upper bound for the eccentricity. $\endgroup$
amd
– amd
2018-12-07 10:11:13 +00:00
Commented Dec 7, 2018 at 10:11
Add a comment |
1 Answer 1
Reset to default
1
$\begingroup$
The given data are not enough to fix the hyperbola eccentricity. You can better understand that with a simpler example: suppose you are given $y=0$ as an asymptote and $x=a$ as tangent, then take $y=mx$ as the second asymptote (with $m$ any non-vanishing real number). The equation of a hyperbola having those two asymptotes and tangent to line $x=a$ can be easily found to be $$y(y-mx)=-{a^2m^2\over4}.$$ The eccentricity of the above hyperbola is $$ e={\sqrt2\over m}\sqrt{1+m^2-\sqrt{1+m^2}}, $$ hence it can take any value between $1$ and $\sqrt{2}$, depending on the value of $m$.
EDIT.
With a rotation and a translation, the above example can be adapted to your data. You may check that the hyperbolas of equation $$ (5x-4y+5)\left((4-5m)y-(5+4m)x-5+{25\over4}m\right)={9\over64}m^2 $$ are all tangent to line $4x+5y-7=0$ and have line $5x-4y+5=0$ as asymptote, for any value of $m$. The eccentricities of these hyperbolas are given by the same formula as before.
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edited Dec 10, 2018 at 13:41
answered Dec 8, 2018 at 17:07
Intelligenti paucaIntelligenti pauca
56.2k44 gold badges5050 silver badges9191 bronze badges
$\endgroup$
2
$\begingroup$ I have edited the question. Please see, if you can arrive at a result. $\endgroup$
Sarthak Rout
– Sarthak Rout
2018-12-09 15:52:41 +00:00
Commented Dec 9, 2018 at 15:52
$\begingroup$ I don't see any relevant change in your question: without some other information you cannot find the eccentricity. I used different lines in the above example just to keep calculations easy, but you could do exactly the same reasoning with your data, and find the same result. $\endgroup$
Intelligenti pauca
– Intelligenti pauca
2018-12-09 16:27:26 +00:00
Commented Dec 9, 2018 at 16:27
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3580 | https://www.johndcook.com/blog/2023/08/27/intersect-circles/ | (832) 422-8646
Contact
Calculating the intersection of two circles
Posted on by John
Given the equations for two circles, how can you tell whether they intersect? And if they do intersect, how do you find the point(s) of intersection?
MathWorld gives a derivation, but I’d like to add the derivation there in two ways. First, I’d like to be more explicit about the number of solutions. Second, I’d like to make the solution more general.
The derivation begins with the simplifying assumption that one circle is centered at the origin and the other circle is centered somewhere along the x-axis. You can always change coordinates so that this is the case, and doing so simplifies the presentation. Undoing this simplification is implicitly left as an exercise to the reader. I will go through this exercise here because I want a solution I can use in software.
Finding the x coordinate
Suppose the first circle, the one centered at the origin, has radius R. The other circle is centered at (d, 0) for some d, and has radius r. The x-coordinate of the intersection is shown to satisfy the following equation.
When I got to this point in the derivation I was wondering what assumption was made that guaranteed there is a solution. Clearly if you increase d enough, moving the second circle to the right, the circles won’t intersect. And yet the derivation for x always succeeds, unless d = 0. If d does equal 0, the two circles are concentric. In that case they’re the same circle if R = r; otherwise they never intersect.
Finding the y coordinate
Why does the derivation for x always succeed even though the intersection might be empty? The resolution depends on the solution for y. What we’ve found is that if the circles intersect, the x coordinate of the point(s) of intersection is given by the equation above.
The y coordinate of the point(s) of intersection satisfies
If the numerator is negative, there is no real solution, no intersection. If the numerator is zero, there is one solution, and the two circles are tangent. if the numerator is positive, there are two solutions.
The distance between the two intersection points, if there are two intersection points, is a = 2|y|. This will be needed below.
General position
Now suppose we’re first circle is centered at (x0, y0) and the second circle is centered at (x1, y1). Again we let d be the distance between the centers of the circles. The circles intersect twice if d < R + r, once if d = R + r, and never if d > R + r.
Imagine for a moment shifting and rotating the plane so that (x0, y0) goes to the origin and goes to (d, 0). The length of the line segment between the two intersection points is still given by a above. And the distance from the center of the first circle to that line segment is given by the equation for x above.
So to find the points of intersection, we first form a unit vector in the direction of the center of the first circle headed toward the center of the second circle. This is the black line at the top of the post. We move a distance x along this line, with x as in the equation above, and then move perpendicularly a distance a/2 in either direction. This is the dashed gray line.
Python code
The following code implements the algorithm described above.
def circle_intersect(x0, y0, r0, x1, y1, r1): c0 = np.array([x0, y0]) c1 = np.array([x1, y1]) v = c1 - c0 d = np.linalg.norm(v) if d > r0 + r1 or d == 0: return None u = v/np.linalg.norm(v) xvec = c0 + (d2 - r12 + r02)u/(2d) uperp = np.array([u, -u]) a = ((-d+r1-r0)(-d-r1+r0)(-d+r1+r0)(d+r1+r0))0.5/d return (xvec + auperp/2, xvec - auperp/2)
Application
This post started out to be part of the next post, but it turned out to be big enough to make its own post. The next post looks carefully at an example that illustrates how you could discover that you’re living on a curved surface just by measuring the distances to points around you. I needed the code in this post to make the image in the next post.
3 thoughts on “Calculating the intersection of two circles”
Kaleberg
I had to compute the intersection of two circles some years back. I think I wound up with a quadratic equation and used the formula to solve it. The discriminant told whether the two intersected at two, one or zero points.
(Now, where is that code?)
2. Oscar Cunningham
I wanted to do the same thing in Python recently, but I was working with complex numbers as points. So I wanted to find the intersections of the circles with centers d and D (complex numbers), and radii r and R (real numbers).The equations for these are thus (z – d)(z – d) = r^2 and (z – D)(z – D) = R^2. Eliminating z gives a quadratic for z whose solutions are the intersection points.I found the coefficients were:a = (d – D)b = r^2 – R^2 + (d+D)(D – d)c = dR^2 – Dr^2 + D|d|^2 – d|D|^2
3. Mohammad Zeidan
The circles intersect twice if |R-r|
Comments are closed. |
3581 | https://scholarworks.lib.csusb.edu/cgi/viewcontent.cgi?article=4457&context=etd-project | Published Time: Tue, 13 Oct 2020 22:07:49 GMT
California State University, San Bernardino California State University, San Bernardino
CSUSB ScholarWorks CSUSB ScholarWorks
Theses Digitization Project John M. Pfau Library
2008
Chinese remainder theorem and its applications Chinese remainder theorem and its applications
Jacquelyn Ha Lac
Follow this and additional works at:
Part of the Algebra Commons
Recommended Citation Recommended Citation
Lac, Jacquelyn Ha, "Chinese remainder theorem and its applications" (2008).
Theses Digitization Project .
This Thesis is brought to you for free and open access by the John M. Pfau Library at CSUSB ScholarWorks. It has
been accepted for inclusion in Theses Digitization Project by an authorized administrator of CSUSB ScholarWorks.
For more information, please contact scholarworks@csusb.edu .Chinese Remainder Theorem and Its Applications
A Thesis
Presented to the
Faculty of
California State University,
San Bernardino
In Partial Fulfillment
of the Requirements for the Degree
Master of Arts
in
Mathematics
by
Jacquelyn Ha Lac
December 2008 Chinese Remainder Theorem and Its Applications
A Thesis
Presented to the
Faculty of
California State University,
San Bernardino
by
Jacquelyn Ha Lac
December 2008
Approved by:
Dr. Laura Wallace, Committee Chair
Joseph Chavez, C^/nittee Member
Date
Dr. Zahid Hasan, Committee Member
Dr. Peter Williams, Chair,
Department of Mathematics
oseph Chavez,
Graduate Coordir^or,
Department of Mathematics iii
Abstract
The Chinese Remainder Problem appeared around the first century AD in Sun
Zie ’s book. Its uses ranged from the computation of calendars and counting soldiers
to building the wall and base of a house. Later on, it became known as the Chinese
Remainder Theorem involving integers and remainders under division. Over a period of
time, people had expanded the theorem into abstract algebra for rings and principal ideal
domains. Furthermore, the application of the Chinese Remainder Theorem can be found
in computing, codes, and cryptography.
In this manuscript, the Chinese Remainder Theorem will be introduced as the
original theorem dealing with integers. Then, its expansion and application into rings,
principal ideal domains, and Dedekind Domains will be discussed. Finally, we will see how
the theorem, as a secret-sharing scheme, takes part in the development of cryptography. Acknowledgements
I would like to express my sincerest gratitude to Dr. Laura Wallace for her
continued patience, support, and guidance in the completion of this project. I would also
like to extend my appreciation to Dr. Chavez and Dr. Hasan for being on the commiittee
reviewing this project.
Special thanks are due to my wonderful husband, Tuan Tran for his love and
encouragement while I pursued my educational goals. Additionally, I would like to thank
my good friends, Dilma Bonzer and Lynn Nguyen for their support. V
Table of Contents
Abstract iii
Acknowledgements iv
1 Introduction 1
2 Foundation and Development of the Chinese Remainder Theorem 4
2.1 Historical Development .............. '.............................................................................. 4
2.2 Basic Properties of Relatively Prime Integers .................................................... 6
2.3 Chinese Remainder Theorem for Integers .......................................................... 9
2.4 Chinese Remainder Algorithm for Integers ............................................................. 10
3 Various Formulations of the Chinese Remainder Theorem 16
3.1 Rings, Ideals, and Homomorphisms ........................................................................ 16
3.2 Chinese Remainder Theorem for Rings and Domains ...................................... 20
3.3 Chinese Remainder Theorem for Polynomial Rings ......................................... 24
4 Applications of the Chinese Remainder Theorem 27
4.1 Finite Sequence of Integers ...................................................................................... 27
4.2 A Characterization of Dedekind Domains .......................................................... 28
4.3 Cryptography Schemes ................................................................................................ 30
5 Conclusion 34
Bibliography 35 1
Chapter 1
Introduction
In the mid thirteenth century, a method for solving systems of linear congruences
was published by the Chinese mathematican Ch ’in Chiu-Shao. He wrote the Mathematical
Treatises in the Nine Sections. This method was then called the Chinese Remainder
Theorem due to the contribution of Ch ’in Chiu-Shao ([Gal06]).
In an old guide book for magicians ([DPS96]), the form of the Chinese Remainder
Theorem is found as a mind-reading trick to impress the audience. A magician would
ask a helper to think of a number less than 60. Then the helper is asked to divide this
number by 3 and tell the remainder. The process continues as the helper divides the
original number by 4 and 5. Upon hearing the remainders, the magician will announce
the number. For example, the number will be 23 corresponding to the remainders 2, 3,
and 3, obtained by dividing by 3, 4, and 5 respectively. By the instruction of the guide
book, the magician divides the numbers 40a + 455 + 36c by 60 where a, b, c are the three
remainders. So, in the above case 40x2+45x3+36x3 = 80+135+108 = 323+60 = 5r23.
The last remainder 23 is the answer. This is a concrete example of using the Chinese
Remainder Theorem with three moduli.
In arithmetic, modulo indicates a congruence relations on the integers. Two
integers a and b are said to be congruent of modulo m if their difference a—b is a multiple of
m. Also if we divide both a and b by m, their remainders will be the same. The magician
problem above starts with simple moduli of the integers; however, the congruence relation
is expanded into abstract algebra with the operations on rings, domains, fields, and so
on. 2
The Chinese Remainder Theorem began with a problem similar to that of the
magician and the Chinese used its algorithm to calculate the calendar, compute the
number of soldiers when marching in lines, or compute the construction of building a
wall. Nowadays, we have found more uses involving the application of this theorem.
In dealing with logic and mathematics, the theorem was used to prove that any finite
sequence of integers can be represented in terms of two integers ([DPS96]). The property
shows part of the power of the Chinse Remainder Theorem which will be proved in chapter
Property:
Let at, 0 < i < t, be a finite sequence of nonnegative integers. Then
there are integers u and v such that (u mod (1 + (i + l)v)) = cq, for every i = 0,1,..., t.
Modern mathematicians also generalized the theorem into rings and integral
domains which is our topic in chapter 3. Other applications that directly involve the
theorem are seen in Dedekind domains and cryptography which will be discussed in
chapter 4. In dealing with cryptography, the theorem itself is already a secret-sharing
scheme which as mentioned above, was employed to compute the number of soldiers
to prevent the enemy from such information as follows. A general asks his soldiers to
stand in ri,r2, rows in turn, and each time he counts the remainders. Finally, he
computes the number of his soldiers using the Chinese Remainder Algorithm, the process
of applying the Chinese Remainder Theorem. This is a secret method to calculate the
number of soldiers.
Even though the theorem, originated as a puzzle, first appeared in China, the
concept was also recognized in other areas of the world. There were several mathemati
cians exploring this idea. The work of Brahmagupta in Indian involved planar geometry,
arithmetic progressions, and quadratic equations. A form of the theorem was also men
tioned in his work. Even though the Chinese Remainder Theorem was just a glimpse in
Fibonacci ’s work, we could see the substantial spread of the theorem. We will take a brief
glance of how the Chinese Remainder Theorem is treated by Fibonacci.
In Fibonacci ’s book Liber Abaci, the Chinese Remainder Theorem was discussed
as follows. “Let a contrived number be divided by 3, also by 5, also by 7; and ask each
time what remains from each division. For each unity that remains from the division
by
3, retain 70; for each unity that remains from the division by 5, retain 21; and for
each unity that remains from the division by 7, retain 15. And as much as the number 3
surpasses 105, subtract from it 105; and what remains to you is the contrived number.
Example: suppose from the division by 3 the remainder is 2; for this you retain twice 70,
or 140; from which you subtract 105, and 35 remains. From the the division by 5, the
remainder is
3; for which you retain three times 21, or 63, which you add to the above 35;
you get 98. From the division by 7, the remainder is 4, for which you retain four times
15, or 60; which you add to the above 98, and you get 158, from which you subtract 105,
and the remainder is 53, which is the contrived number. From this rule comes a pleasant
game, namely if someone has learned this rule with you; if somebody else should say some
number privately to him, then your companion, not interrogated, should silently divide
the
number for himself by 3, by 5, and by 7 according to the above-mentioned rule; the
remainders from each of these divisions he says to you in order; and in this way you can
know the number said to him in private. ” ([DPS96])
Fibonacci ’s presentation is very similar to Sun Zi ’s approach in generating a
method to find the mystery number. Chinese Remainder type of problems as mentioned
above were also considered by other mathematicans such as Euler, Gauss, and Lagrange.
Its popularity took a great part in our modern application to cryptography. 4
Chapter 2
Foundation and Development of
the Chinese Remainder Theorem
2.1 Historical Development
In historical times, problems involving finding the number of objects, such as
the numbers of baskets, blocks of bricks, or numbers of soldiers in a group under certain
conditions were to compute the remainders when dividing the mystery number in different
steps. One of the examples was as follow: “We have a number of things, but do not know
exactly how many. If we count them by threes we have two left over. If we count them
by fives we have three left over. If we count them by sevens we have two left over. How
many things are there? ” ([DPS96])
This problem is presented in the mathematical classic of Sun Zi, a mathematician
in ancient China. Sun Zi Suanjing, Sun ’s Mathematical Manual was dated approximately
to the beginning of Graeco-Roman time, A.D. 100 - A.D. 500. The oldest Chinese math
ematical classic is Chou Pei Suanjing. This book recorded mathematics for astronomical
calculations. It was dated about 1000 B.C. The Pythagorean Theorem was used in the
astronomical calculations in this book. Therefore, Sun Zi ’s book is not the oldest Chinese
mathematical classic; however, the Chinese Remainder Theorem appeared in it for the
very first time.
The calculation of calendars in ancient China was the main source of the re
mainder theorem. Around A.D. 237, the Chinese astronomers defined the starting point 5
of the calendar as “shangyuan ”, which is a moment that occurred simultaneously with
the midnight of the first day of the 60 days cycle, the Winter Solstice and the new moon.
The system of congruences
xN = n mod 60
xN = r2 mod y
indicates the number of years N after shangyuan. So for the above system of congurences,
if the Winter Solstice of a certain year occurred ri days after shangyuan and r2 days after
the
new moon, then that year was N years after shangyuan, where x is the number of
days in a tropical year and y is the number of days in a lunar month.
This example is considered the very first application of the Chinese Remainder
Theorem ([DPS96]. This kind of computation was also used in building a wall or the base
of a house such as the construction of the Great Wall during feudal times 475-221 B.C.
It is described as follows. To construct a rectangular base for a building, there are four
kinds of materials available: big cubic materials with each side 130 units long; small cubic
materials with each side 110 units; city bricks that are 120 units long, 60 units wide, and
25
units deep; and “six-door ” bricks that are 100 units long, 50 units wide, and 20 units
deep. These four materials were used to build the base without breaking any of them into
little pieces. Therefore, we end up with a system of congruences as we calculate different
materials for each time a specific type of material is used. If big cubic materials are used,
then 60 units base length is left, but 60 units more base width is needed. If small cubic
materials are used, then 20 units based length is left, but 30 units more base width is
needed. If the length of the city bricks is used, then 30 units base length is left, but 10
units more base width is needed. If the width of the city brick is used, then 30 units base
length is
left, but 10 units more base width is needed. If the depth of the city bricks is
used, then 5 units base length is left, and 10 units base width is needed. If the length,
width and depth of the six-door bricks are used, the base length has 30, 30 and 10 units
left respectively, and the base width has 10, 10 and 10 units left respectively. The goal is
to determine how large the base length X and base width Y are.
The above example is simplified into the following congruences where X is the
material
used each time and Y is the material needed:
X = 60 mod 130
=30
mod 120 6
= 20 mod 110
= 30 mod 100
= 30 mod 60
= 30 mod 50
= 5 mod 25
= 10 mod 20
and Y = 60 mod 130
= 10 mod 120
= 30 mod 110
= 10 mod 100
= 10 mod 60
= 10 mod 50
= 10 mod 25
= 10 mod 20.
As mentioned in Chapter 1, this algorithm is also used to compute the number
of soldiers that went out for battles. To avoid the enemy of knowing the number of
soldiers he has, a general would count his soldiers in a certain way. For example, first he
asks his soldiers to line up in rows of 11, then in rows of 17, 29, and 31. Respectively,
each time, he is reported with remainder 8, 5, 16, and 24. Then he will calculate his
soldiers in private. Since not many people know of this secret computation, the general
can conceal his number of soldiers. From this example, we compile the following system
of congruences with relatively prime moduli where x would be the number of soldiers:
x = 8 mod 11,
x = 5 mod 17,
x = 16 mod 29,
x = 24 mod 31.
We will solve these three types of problems in the last section of this chapter.
2.2 Basic Properties of Relatively Prime Integers
Before introducing the theorem, there are several terms, lemmas, and theorems
that we need to know concerning relatively prime integers. 7
Definition 2.1.
A set of integers is said to be pairwise relatively prime if every pair of
integers a and b in the set have no common divisor other than 1, in other words (a, 6) = 1
where (a, b) is the greatest common divisor of a and b.
Example 2.2.
The set {10,7,33,13} is pairwise relatively prime because any pair of
numbers has greatest common divisor equal to 1.
(10,7) = (10,33) = (10,13) = (7, 33) = (7,13) = (33,13) = 1.
This concept of pairwise relatively prime is used commonly in the Chinese Re
mainder Theorem. The property in the next theorem and those that follow are useful in
dealing with greatest common divisors.
Theorem 2.3.
The greatest common divisor of the integers a and b, not both 0, is the
least positive integer that is a linear combination of a and b.
Proof: Let m be the least positive integer that is a linear combination of a. and
b. Then we have m = ax + by for some integers x and y. By the Division Algorithm,
there exists integers q and r such that a = mq + r, 0 < r < m. So r = a — mq. Substitute
m by the linear combination above, we get r — a — {ax + by)q = (1 — xq)a — byq. Then
r is a linear combination of a and b. Since 0 < r < m, and m is the least positive linear
combination of a and b, we have r = 0. Therefore, m | a. Similarly m | b. Hence, m is a
common divisor of a and b. Now let n be another divisor of a and b. For m = ax + by, if
n | a and n | b then n | m, so n < m. Consequently, m is the greatest common divisor of
a and b.
□
Proposition 2.4.
For any integers a,b E Z, (a,b) = 1 if and only if ax + by = 1 for
some integers x and y.
Proof: By Theorem 2.3, we have (a, b) = 1 leading to ax+by = 1 for any integers
x and y. Conversely, suppose that ax + by = 1, and let m = {a, b) then m | a and m | b.
So m | (ax + by) . Hence m | 1. Thus m = 1.
□
Lemma 2.5.
For any integers a,b,c E Z, [a, b] | c, where [a,b] is the least common
multiple of a and b, if and only if a | c and b | c. 8
Proof: Suppose [a, b] | c. Then c = [a, b]x for some integer x. Also a | [a, b], so
[a, b] = ay for some integer y. So, we have c = axy. Hence, a | c. Similarly, we obtain
b | c.
Conversely, if a | c and b | c, we try to prove that [a, 5] | c. Let [a, b] = m. By
the Division Algorithm, there exists integers q and r such that c = mq + r, 0 < r < m.
We will show r = 0. Since a | c, a | mq + r. However, [a, b] = m so a | m. Then a | r,
so r = ax for some x. Similarly, r = by for some y. But, r < m and m = [a, b] , hence,
r = 0. Thus, c = mq, and m | c. Therefore [a, 6] | c.
□
Definition 2.6.
Let m be a positive integer. If a and b are integers, we say that a is
congruent to b modulo m if m | (a — b). We write a = b mod m.
Theorem 2.7.
If a = b (mod mi), a = b (mod m2), ■■■, a = b mod (m^), where
a, b, mi, m2, ■■■, m^ G Z and mi,m,2, ..., m^ > 1, then a = b (mod [mi, m2,..., m^.]), where
[mi, m2, m^] is the least common multiple of mi, m2, ...,mk.
Proof: Let a = b (mod mi), a, = b (mod m2),..., a = b (mod m^). Tthen
mi | (a — b), m2 | (a — b), ..., m^ | (a — b). So, by Lemma 2.5, [mi, m2, ...,mA,] | (a — b).
So a = b (mod [mi, m2, ...m/J).
□
Lemma 2.8.
a. For any a,b,c G Z, (a, b) = (b, c) = 1, then (ac, b) = 1.
b. For any a\,a2,---,a n G Z, if (ai,b) = (0,2, b) = ... = (a n, b) = 1 then
(aia 2...a n, b) = 1.
Proof:
a. Suppose (a, b) = 1. Then ax + by = 1 for some integers x, y. Similarly (b, c) = 1 implies
bs+ct = 1 for some integers s, t. Then (ax+by'jlbs+ct) — 1, so abxs+acxt+b 2ys+bcyt = 1
and acxt + b(axs + bys + cyt) = 1. Hence (ac, b) = 1 by Proposition 2.4.
b. We are going to prove this part by induction.
For n = 1, we have (a, b) = 1 => (a, b) = 1.
Suppose (ai,b) = 1, (02, b) = 1,... (a n+i,b) = 1. Then by the induction hyposthesis,
(aia2-..a n, b) = 1 and (a n_|_i, b) = 1. Since ai, 02,..., an G Z, then let ai • a2 • • • an = c. So
by part a, (c, b) = 1 and (a n+ i, b) = 1 implies (ca n+i , b) = 1. Thus, (aia2...a nan+ i, b) = 1.
□9
2.3 Chinese Remainder Theorem for Integers
Now that we have some basic concepts to help us in solving the Chinese Remain
der puzzle given at the beginning of this chapter, we will start with our original theorem
that gives the method to acquire the solution of the puzzle.
Theorem 2.9. Let mi, m2,..., mr be pairwise relatively prime positive integers. Then the
system of congruences:
x = ai (mod mi)
x = a2 (mod m2)
x = ar (mod mr)
has a unique solution modulo M — mim2-..m r.
M
Proof: Let Mk = ---- = mpai ■ ■ ■ mk-imk+i ■ ■ ■ mT. By Lemma 2.8, we know
W
that = 1 because (mj,mk) = 1 whenever j / k. Then m^x + Mkyk = 1 for
some x,yk- So MkPk = 1 (mod mf). Hence, yk is the inverse of Mk (mod m^). We form
the sum:
x = aiMiyi + a2M2y2 + + arMryr
where x would be the solution of the r congruences.
Since mk | Mj whenever j k, we have Mj = 0 (mod mf). So ajMjyj = 0 mod mk for
j k. Hence, from the sum we get x = akMk'yk = o-fc (mod mk) since Mk'tjk = 1 (mod
mk).
To prove that the solution is unique modulo M, we let xi and x2 be two solutions to the
system of r congruences. Then xi = X2 = ak (mod mk) for each k. So mk | (x2 — a?i).
By Theorem 2.7, M | (x 2 — a?i). Hence, xi = x2 (mod M).
□
Now, we are going to use the construction of the solution in the proof of the
Chinese Remainder Theorem to solve the problem first mentioned in section 1. Let x be
the unknown number of objects. Then we have the system of congruences:
x = 2 (mod 3)
x = 3 (mod 5)
x = 2 (mod 7). 10
So M = 3 • 5 • 7 = 105. Hence, Mx = — = 35, M2 = — = 21, and M3 = — = 15.
3 5 I
Then Miyi = 1 mod 3 becomes 35i;i = 1 mod 3. Simplifying this congruence and solving
for yx we get 2y x = 1 mod 3 so that y\ = 2 mod 3. Similarly, M-ryz = 1 mod 5 becomes
21y 2 = 1 mod 5. We get y2 = 1 mod 5. Again, M3y3 = 1 mod 7 becomes 15 t/3 = 1 mod
7 and we get y3 = 1 mod 7.
So x = 2 • 35 • 2 + 3 • 21 • 1 + 2 ■ 15 • 1 = 140 + 63 + 30 = 233 = 23 (mod 105). To check
this, note that 23 = 2 (mod 3), 23 = 3 (mod 5), 23 = 2 (mod 7).
The answer to this problem was explained in a verse in Chen Dawei ’s book
Suanfa Tongzong: “Three people walking together, it is rare that one be seventy. Five
cherry blossom trees, twenty one branches bearing flowers, seven disciples reunite for the
half-month. Taking away one hundred and five you shall know. ” ([DPS96]) To understand
this saying, let x be the unknown number. Divide x by three (people) and multiply the
remainder by 70 = 35 • 2, divide x by five (cherry blossoms) and multiply that remainder
by 21 = 21-1. Finally, divide x by seven (disciples) and multiply the remainder by
15 = 15 • 1. Add all three results and subtract a suitable multiple of 105, i.e. find the
remainder modulo 105, and you shall find x.
2.4 Chinese Remainder Algorithm for Integers
The Chinese Remainder Algorithm was generated based on Sun Zi ’s method to
solve the original problem. By the 13th century, Quin Jiushao gave a more general method
which did not restrict the moduli mi to pairwise relatively prime numbers. His method,
however, also converted the moduli into pairwise relatively prime numbers ([DPS96]). It
is described as follows:
Let mx,m 2,..., mfc be the moduli and I = lcm[mi, ..., m/.], the least common
multiple of mi, ..., We are going to find a set of integers cvi, a2,..., a^ satisfying:
ai divides mi, i = 1,2,..., k-,
gcd aj) = 1 for all j j-
aiaz-.-.ak = lcm[mi, ..., m^].
Then the system of congruences x = ai mod mi for i = 1,..., k is converted into x = ai
mod ai for i = 1,..., k, where the moduli a, are pairwise relatively prime.
To find the set of integers ai for i = l,...,k, we need to complete the following 11
procedure. Consider the first case of k = 2:
a. Let (mi, m2) = d. If (m\/d\,m 2) = 1, then take 01 = mi/di and a2 — m2.
b. If (mi,m2/di) = 1, then take oi = mi and 02 = m2/c?2-
c. If (mi,m.2/di) = d2 > 1, then calculate ds = (mi/cU, m^cU/di) where c/2
divides di and ds divides di/d2. If cfa = 1, then take ai = mi/c/2 and 02 = m2d2/di,
otherwise calculate d^ = (mi/d 2ds,m 2d2ds/dD). Continue this process until there exists
an integer s such that d s+ i = 1. Such an s exists because di > c/2 > T 0. Then take
mi m2d2ds...d s
ai = 3-3 ---- 7- and a2 =-------3--------• So (01, a2) = 1.
For the case of k moduli, apply the above algorithm to m& and m^-i first to obtain a£\
o^-p Then we apply the same algorithm to o^ and m^_2 which will give us o^\ a'k-2-
Continue this procedure and finally apply the algorithm to ak 7 and mi, obtaining
a/;, cq. Then the integers aq, a2,..., o: k_1, satisfy (ofc, a?) = 1 for i = 1, 2,..., k — 1 and
lcm [cui, a2, ctfc] = Icm^, a2, o/fc _1, o/.] = aklcm[a 1, a2, •••, o^-J.
So we have reduced the case of k moduli into k — 1 moduli. If we repeat this procedure,
we will obtain the required 01,..., a^.
Example 2.10.
Let mi = 12 and m2 = 20 which are not pairwise relatively prime. Then
12
lcm[12, 20] = 60. So by the procedure of step (a), (12,20) = 4. Then (—,20) = 1. So
let oi = 3, and 02 = 20. Thus (01,02) = 1 Therefore, we have 0102 = 3.20 = 60 =
lcm[mim 2]. If we have a system of congruences given by x = ai mod 12 and x = a2 mod
20, then it will become x = eg mod 3 and x = a2 mod 20 where 3 and 20 are relatively
prime.
Example 2.11.
Let mi = 312 and m2 = 16 where mi and m2 are not pairwise relatively
prime. Using the algorithm, let di = (mi, m2) = (312,16) = 8. Then d2 = (mi, ^^) =
(312,2) = 2. So ds = = (156,4) = 4. Continuing, d4 = (y^-, m2< ^d3 .) =
a2 &1 a2a3 ai
(39, 16) = 1. Since d4 = 1, we take 01 = 77 = 39 and a2 = m2d2ds = 16. Thus, aq
d2ds ai
and a2 satisfy the three conditions:
ai divides mi and a2 divides m2,
(01,02) = 1,
0102 = 39 • 16 = 624 = lcm [mi, m2].
Hence, the algorithm sets the moduli back to pairwise relatively prime. 12
In the calculation of the calendars application mentioned in section 1 the algo
rithm is used with 2 moduli.
Example 2.12.
Suppose the number of days in a tropical year is 365 and the number
of days in a lunar month is 30. Then x — 365 and y = 30. Let ri be the number of days
that the Winter Solstice occurs after shangyuan and be the number of days that the
Winter Solstice occurs after the new moon, then we have the system of congruences
365AI = n mod 60
3651V = T2 mod 30
which is converted into
51V = ri mod 60
51V = r2 mod 30
where N is the number of years after shangyuan. By the Chinese Remainder Algorithm,
we get di = (60,30) = 30. Then dy = (60,1) = 1. So we take = 60 and O!2 = 1. So
the new system of congruences will be
51V = ri mod 60
51V = r2 mod 1.
We will also use the above algorithm to illustrate the example of building walls
in k moduli with k > 2.
Example 2.13.
Let X be the base length of the wall and Y be the base width. We are
going to use the algorithm to set all the moduli of X and Y into pairwise relatively prime
integers. Since the moduli of X and Y are the same, we apply the agorithm to X and
use the new pairwise relatively prime moduli for Y as well.
Recall that X = 60 mod 130
= 30 mod 120
= 20 mod 110
= 30 mod 100
= 30 mod 60
= 30 mod 50
= 5 mod 25 13
= 10 mod 20.
We start from the bottom up. First of all, let mi, i = 1,2, ...,8 be as follow:
mi = 130, m2 = 120, m3 = 110, m.4 = 100, ms = 60, = 50, my = 25, and ms = 20.
25
So we start with ms = 20 and my = 25. Then, dy — (20,25) = 5. So d2 = (20, —) = 5.
0 20 Consequently, c/3 = (—, —-—) = 1. Thus ctg 1^ = — — 4 and c/ 7 = 25.
000
Now apply the algorithm to Og 1/1 = 4 and m§ = 50. Similarly, we have di = (4, 50) = 2.
Then d2 = (4, ^) = 1. So cig 2^ = 4 and a' & = 25.
(2)
Continuing this process, apply the algorithm to =4 and ms = 60. Then dy =
(4,60) = 4. So d2 = (4, —) = (4,15) = 1. We have ag 3^ = 4 and = 15.
Next, apply the procedure to ag 3^ = 4 and m4 = 100. First, dy = (4,100) = 4. Then
d2 = (4, ^^) = (4,25) = 1. So oig 4) = 4 and c/ 4 = 25.
Again, the next pair is ctg 4^ = 4 and m3 = 110. So dy = (4,110) = 2. Then d2 =
(4, ~x~) = (4, 55) = 1. Hence, = 4 and ctg = 55.
Consequently, the next pair is Og 5^ = 4 and m2 = 120. We get dy — (4,120) = 4. Then
120 4 120•2
d2 = (4, -4-) = (4, 30) = 2. Hence, we continue to ds = (-, —-—) -= (2, 60) = 2. Thus,
A 1 DP) O O
d4 = (^ —^, ----- ------- ) = (1> 120) = 1. Therefore, = 1 and a' 2 = 120.
Lastly, the pair is — 1 and mi = 130. Obviously, the two numbers are relatively
prime, so we obtain eng = 1 and a' 4 = 130. Now that we have just reduced the 8
moduli into 7 with the last one relatively prime to the rest, we have the 7 moduli in
order: 130,120,55,25,15,25,25. Applying this procedure again we have di = (a?, c/ 6) =
(25,25) = 25, then d2 = (25, ||) = 1 So take = 25 and (3g = 1.
1
Next, di = (25,15) = 5, so d2 = (25, —) = 1. Hence, /3? 2) = 25 and /3' 5 — 3.
0
Again, di = (25,25) = 25, then d2 = (25, ^|) = 1. Thus, take (3^ = 25 and /3 4 = 1.
Zu
55 fA\
Similarly, di = (25, 55) = 5, and d2 = (25, —) = 1. Take (3? = 25 and (3% = 11.
o1on
Now we have di = (25,120) = 5, so d2 = (25, —-) = 1. Then take (3^ = 25 and /3 2 = 24.
130
Last, di = (25,130) = 5, then d2 = (25, ——) = 1. Therefore (3y = 25 and = 26.
5
The new list of moduli now has become 26,24,11,1, 3,1, 25. We can see that every ele
ment is pairwise relatively prime with each other except the two moduli 26 and 24. So
we will use the algorithm one more time to convert them to pairwise relatively prime. 14
Let di = (26, 24) = 2, then d2 = (26, —) = 2, so c/3 = (—, —-—) = 1. Thus the two
moduli 26 and 24 now become 13 and 12. This, however, creates another non-pairwise
relatively primes which are 12 and 3. Once again, we apply the algorithm to these two
moduli.
We have di = (12, 3) = 3, then d2 = (12,1) = 1. So the two new moduli are 12 and 1.
Now we have a new system of congruences with pairwise relatively prime integers:
X = 60 mod 13 = 8 mod 13
= 30 mod 12 = 6 mod 12
= 20 mod 11 = 9 mod 11
= 30 mod 1 = 0
= 30 mod 1 = 0
= 30 mod 1 = 0
= 5 mod 25
= 10 mod 1 = 0.
Using the Chinese Remainder Theorem,excluding all the mod 1 congruences, we obtain
42 900 42 900
M = 13 • 12 • 11 • 25 = 42,900. Therefore, Mi = = 3,300, M2 = = 3,575,
M3 = = 3,900, and M4 = = 1716.
11 25
Then 3, 300yi = 1 mod 13 becomes 1 lyi mod 13, so yi = 6 mod 13. Again, 3, 575 t/2 = 1
mod 12, so llj/2 = 1 mod 12, and 1/2 = U mod 12. Similarly, 3, 900 t/3 = 1 mod 11 be
comes 67/3 = 1 mod 11, so 2/3 = 2 mod 11. Finally, 1, 716y 4 = 1 mod 25, so 16y 4 = 1 mod
25, and y4 = 11 mod 25. Hence, x = 8 • 3,300 • 6 + 6 • 3, 575 • 11 + 9 • 3,900 ■ 2 + 5 • 1,716 ■ 11
= 558, 930 mod 42, 900 = 1, 230 mod 42, 900. So the base length of the wall is a multiple
of 1,2300 mod 42,900.
To solve for the base width of the wall, we set up the system of congruences just
as above with the moduli relatively prime.
Y = 60 mod 13 = 8 mod 13
= 10 mod 12
= 30 mod 11 = 8 mod 11
= 10 mod 1 = 0
= 10 mod 1 = 0
= 10 mod 1 = 015
= 10 mod 25
= 10 mod 1 = 0.
So we have Y = 8 • 3, 300 • 6 +10 • 3,575 • 11 + 8 • 3,900 • 2 + 10 ■ 1,716 • 11 = 802,220 mod
49,000 = 18,220 mod 49,000. So the base width of the wall is a multiple of 18,220 mod
42,900.
Lastly, we will solve the problem of counting the number of soldiers.
Example 2.14.
As in section 1, we have a system of congruences for the number of
soldiers going out to battle.
x = 8 mod 11,
x = 5 mod 17,
x = 16 mod 29,
x = 24 mod 31.
Since all the moduli are already relatively prime, we use the Chinese Remainder Theorem
-t ZJ Q 1 1 Q
to solve for x. We have M = 11 • 17 • 29 • 31 = 168,113, so Mi = —L-— = 15,283,
M2 = -168 ’113 = 9,889, M3 = -8l -113 = 5,797, and M4 = 168,113 = 5,423. We
-L i o j.
determine yi by solving the congruence 15, 283r/i = 1 mod 11, or equivalently, 4yi = 1
mod 11. This yields j/i = 3 mod 11. By solving 9, 889?/2 = 1 mod 17, or equivalently,
12?/2 = 1 mod 17, we find y2 = 7 mod 17. Similarly, 5, 7971/3 = 1 mod 29, which is equal
to 26y 3 = 1 mod 29. We get y3 = 10 mod 29. Finally, we solve 5, 423y 4 = 1 mod 31 or
29y 4 = 1 mod 31. This gives y4 = 16 mod 31.
Hence, we calculate the number of soldiers going out to battles by
x = 8 • 15,283 • 3 + 5 • 9, 889 • 7 + 16 • 5,797 -10 + 24-5,423 • 16 = 3,722.859 = 24,373
mod 168,113.
So the number of soldiers would be 24,373 for that specific battle. 16
Chapter 3
Various Formulations of the
Chinese Remainder Theorem
3.1 Rings, Ideals, and Homomorphisms
From the original theorem dealing with integers, the Chinese Remainder Theo
rem is expanded into rings and domains. We now are looking at the Chinese Remainder
Theorem that can be formulated for rings which have pairwise coprime ideals. Before
introducing the expansion of Chinese Remainder Theorem for rings, we are going to get
acquainted with a few definitions.
Definition 3.1. A ring R is a nonempty set with two binary operations, addition (de
noted by a + b) and multiplication (denoted ab), such that for all a, b, c in R:
a + b = b + a.
(a T b) T c = a T (b T c) .
There is an additive identity 0. That is, there is an element 0 in R such that a + 0 = a
for all a in R.
There is an element —a in R such that a + (—a) = 0.
a(bc) = (ab)c.
a(b + c) = ab + ac and (b + c)a — ba + be.
A ring is commutative when multiplication is commutative.
A subset S of a ring R is a subring of R if S itself is a ring with the operations of R.
If R has a multiplicative identity, i.e. an element 1 G R such that x • 1 = 1 • x = x then 17
R is said to be a ring with unity.
Example 3.2.
The set Z of integers under ordinary addition and multiplication is a
commutative ring with unity.
Example 3.3.
The set nZ of integers multiples of n G Z under ordinary addition and
multiplication is a commutative ring without unity and also a subring of Z.
Example 3.4.
The set R[®] of all polynomials in the variable x with real coefficients
under polynomial addition and multiplication is a commutative ring.
Definition 3.5. A
subring Z of a ring R is called an ideal of R if for every r G R and
every i G I both ri and ir are in I.
Theorem 3.6.
Ideal Test
A nonempty subset I of a ring R is an ideal of R if
a — b G I whenever a,b G I.
ra and ar are in I whenever a E I and r G R.
Proposition 3.7.
Let R be a commutative ring with unity and let a G R. Then the set
{a} = {ra|r G R} is an ideal of R called the principal ideal generated by a.
Proof: By the Ideal Test, let ra, sa G (a) where r,s G R. Then ra — sa =
(r — s)a G (a) since r — s G R. Also, let x G R then xar = xra = (xr)a G (a) since R is a
commuative ring. Therefore, {a) is an ideal.
□
Example 3.8.
For any positive integer n, the set of multiples of n, nZ = {0, ±n, ±2n, ...}
= (ri) is an ideal of Z.
Example 3.9.
Let R[rr] be the set of all polynomials with real coefficients and let I be
the subset of all polynomials with constant term 0. Then I is an ideal of R[x] and I = (x).
Definition 3.10.
An integral domain is a commutative ring R with unity and no zero
divisors, i.e. if ab = 0 where a,b G R, then a = 0 or b = 0 for all a, b.
Example 3.11.
The ring of integers Z is an integral domain.
Example 3.12.
The ring Zp of integers modulo a prime p is an integral domain. 18
Definition 3.13.
A principal ideal domain is an integral domain R in which every ideal
has the form (a) = {ra|r G R} for some a in R.
Definition 3.14.
Let R be ring and let I be an ideal of R and s,t E R. The set of cosets
{r + I\r E -R} denoted R/I is a ring under the operations (s + I) + (t + I) = s + t + I
and (s + I)(t + /) = st + I. The ring R/I is called a factor ring.
Example 3.15.
Let Z be the ring of integers. The ring Z/4Z = {0 + 4Z, 1 + 4Z, 2 +
4Z, 3 + 4Z} is a factor ring. For example, we have
(2 + 4Z) + (3 + 4Z) = 5 + 4Z = 1 + 4 + 4Z = 1 + 4Z, and
(2 + 4Z)(3 + 4Z) = 6 + 4Z = 2 + 4 + 4Z = 2 + 4Z.
Example 3.16.
Let R = R[x] and I = (x). We can see that (x) = {r(a?) • rr|r(a;) E R[rr]}
so R[z]/(x) = {f(x) + (x)\f(x) E R[z]} = {a + (rr)|a G R} which is similar to the ring R
as we will verify at the end of this section.
Example 3.17.
Let R = R[x] and I = (a? 2 + l) be the principal ideal generated by a: 2 + l.
Then R[a?]/(ar 2 +1) is a factor ring. We have R[x]/(2: 2 +1) = {y(a;) + (a; 2 + l)|y(a;) G R[x]}
= {ax + b + (x 2 + 1) |a, b E R}.
Definition 3.18.
Let R±, R2,..., Rn be rings and Ii,I 2,..., In be ideals. Construct a new
ring as follows. Let
Rl/Il ® R?./Ii ® ■ ■ ■ ® Rn/In = {( al + A, a2 + Ry •••, an + Ai)| ai + li G Ri/Ii}
and perform component-wise addition and mutiplication, that is
(ai + Ii, a2 + I2,..., an + In) + (bi + Ii, b2 + I2,..., bn + In) = (ai+bi+Ii,a 2+b 2+1 2, ...,a n+
bn + In)
and
(ai + Ii,a 2 + I2,..., an + In)(bi + Ii, b2 + I2,..., bn + In) = (o-i^i + A, 02^2 +^2 ; ■■■, a, nbn + In)-
This ring is called the direct sum of Ri/I\, R2/I 2,Rn/I n-
Example 3.19.
Let Z/(3)®Z/ (5) be a direct sum. Then (l+(3),4+(5))+(2+(3),4+(5))
= (1 + 2 + (3),2 + 4 + (5)) = (0 + (3), 1 + (5)) and (1 + (3),4 + (5))(2 + (3),4 + (5))
= (1 • 2 + (3), 4 • 4 + (5)) = (2 + (3), 1 + (5)).
Definition 3.20.
A ring homomorphism f> from a ring R to a ring S is a mapping from
R to S that preserves the two ring operations; that is, for all a, b in R, 19
f)(a + b) = f)(a) + (b) and 0(ab) = (a)0(b).
A ring homomorphism that is both one-to-one and onto is called a ring isomorphism.
Example 3.21.
Let f) be the mapping from Z4 to ZXo with x —> x. Then f>(x + y)
— 5(x + y) = 5x + 5y = f>(x) + f)(y). Also, f)(xy) = 5(xy) = 5.5(xy) since 5.5 = 5 in Zio-
Then f>(xy) = 5x.5y = f)(x]f)(y). So is a homomorphism.
Theorem 3.22.
Let f> be a homomorphism from a ring R to a ring S. Then kernel of
f), ker f) = {r G #|( r) = 0} an ideal of R.
Proof: To prove this theorem, we are going to apply the ideal test.
Let r,s E kerf) then 0(r) = f)(s) = 0. Hence, f>(r) — (s) = 0 = f>(r — s) since f> is a
ring homomorphism. Therefore, r — s € ker.
Let t G R and r G kerf). Then f)(r) = 0. Since f> is a ring homomorphism f>(tr) =
f)(t)f)(r') = f)(t] -0 = 0. Hence tr G kerf).
From (1) and (2), ker f> is an ideal.
□
Example 3.23.
Let f> be the mapping from Z[ar] onto Z given by f>(f(xf) = /(0) and let
f(x),g(x) G Z[x]. Then f> is a ring homomorphism since f>(f(x)+g(x)) = f>(ff + g)(0)) =
(f + 9)(0) = /(0) + g(0) = 0(/O)) + f>(g(xf), and f)(f(x)g(xf) = f>(fg(O)) = (fg)(O) =
7(0) • 9(0) = (.g(x)\ Hence ker f> = {f(x) E Z[z]|/(0) = 0} = (x), i.e. the
kernel of f> is the set of polynomial with 0 constant term.
Theorem 3.24.
First Isomorphism Theorem for Rings
Let f> be a ring homomorphism from R to S. Then the mapping from R/kerf) to f>(R),
given by r + kerf) —> f>(r) is an isomorphism. In symbols, R/kerf) « f>(R)-
Proof:
Let f) : R —> S be a ring homomorphism.
Let f) : R/kerf) f>(R) be the mapping defined by
f>(r + ker(f>f) = f)(r). We will show this mapping is one-to-one, onto, and that
operations are preserved.
The mapping f> is well-defined.
Let r + kerf> = s + kerf). Then r — s G kerf). So f>(r — s) = 0. It follows that f>(r) = f)(s). 20
Hence, 0(r + kerfy = 0(r) = 0(s) = (s + kerfy.
The mapping 0 is one-to-one.
Let 0(r + kertf) = 0(s + ker). Then 0(r) — (s) which implies 0(r) — 0(s) = 0 since <p
is a ring homomorphism. So (r — s) = 0, then r — s G kertp and r + kerf> = s + ker<p.
The mapping <f is onto.
Let x G 0(7?). Then x = 0(r) = 0(r + ker) for some r G R.
The mapping 0 preserves addition and multiplication.
We have 0((r + kertp) + (s + ker)) = 0((r + s) + ker) — 0(r + s) = 0(r) + 0(s)
= 0(r + kercf} + 0(s + kertf).
Also 0((r + A;er0)(s + A:er0)) = 0((rs) + ker) = (rs) = (r}<!>(s) — 0(r + ker)^(s +
ker<f\
Then by 1-4, 0 is an isomorphism. Therefore, R/ker(f 0(7?).
□
Example 3.25. Let 0 be the mapping from Z to Zn given by 0(x) = Ox mod n. Then
ker 0 = (n) so Z/(n) ~ Zn.
Example 3.26. Let 0 be the mapping from R[x] to R given by 0(/(a;)) = /(0) then ker
0 = (x). So R[a:]/(a:) ~ R where (x) is the ideal of polynomials with zero constant term.
3.2 Chinese Remainder Theorem for Rings and Domains
We now extend the notion of relatively prime integers to coprime ideals in a ring
7?. This will allow us to extend the Chinese Remainder Theorem to rings and integral
domains.
Definition 3.27. In a commutative ring 7?, two ideals A and B are called coprime if
A + B = 7?.
Note that two principal ideals (a) and (6) are coprime in the ring of integers Z
if and only if a and b are relatively prime. Therefore, coprime ideals are analogous to
relatively prime integers.
Proposition 3.28. Let R be a commutative ring with unity
1.I/I + J = R, then IJ = InJ
If Ii,l2, In are coprime in pairs, then I1I2 ■ ■ • In = AiLi h-21
Proof:
Let r 6 I J. Then r = ij where i G I and j G J. So r G J, and r G I since I and J are
ideals. Hence r G I A J. Therefore, IJQIOJ.
Now we have (I + J)(I A J) = 1(1 A J) + J(I A J) = II A IJ + JI A J J C I J. Since
I+J = R then (I + J)(I A J) = (I A J) CIJ.D
We proceed by induction on n. The case for n = 2 is proven by part (1). Assume
I\h ■ ■ ■ In-1 = 0?=? k- Suppose n > 2 and Ii = 0?=? k- Let J = [7?=^ Ii =
A/Tj 1 Ii. Since Ii + In = R for 1 < i < n — 1, then Xj + yi = 1 for some Xi G Ii
and yi G In- Thus nS. 1 _ Vi) = 1 m°d In- So In + J = R. Therefore,
rii=l Ii = IIn = J Fl In = Ai=l Ii-
□
Now it is time to use the First Isomorphism Theorem for Rings to prove the
Chinese Remainder Theorem for rings with two ideals.
Theorem 3.29.
The Chinese Remainder Theorem for Two Ideals
If R is a commutative ring and I and J are proper ideals with I + J = R, then R/ (I A J)
is isomorphic to R/I ffi R/ J.
Proof: We are going to use the First Isomorphism Theorem to prove this theo
rem. Let R/1 © R/ J be the mapping defined by
0(r) = (r + I,r + J).
<p is a well-defined map.
Suppose there exist r, s G R and r = s. Then </>(r) = (r+I, r+J) and p(s') = (s+I, s + J).
Since r — s, we have r+I = s+I andr+J = s+J. Therefore, (r+I, r+J) = (s+I, s+J).
Hence, cf>(r) = p(s).
<p is a homomorphism.
Suppose there exist a, b G R. Then <fr(a + b) = (a + b +I,a + b +J) = (a + I + b +
I,a + J + b + J) = (a + I, a + J) + (b + I, b + J) = 0(a) + </>(b). We also have (ab) =
(ab +1, ab + J) = ((a + I)(b +I), (a + J)(b + J)) = (a +1, a + J)(b +1, b + J) = 0(a) 0(b) .
0 is surjective.
Let (a, b) G R/I © R/ J■ Then (a, b) = (a + I, b + J) for some a, b G R. Since R — I + J,
we have a = x + y and b = s + t for some x,s G I and y,t G J. Consider y + s. We
obtain 0(y + s) = (y + s + I,y + s + J) — (y + I,s + J) since s G I,y G J. However, 22
(a,b) = (a + I,b + J) = (x + y + I,s + t + J) = (y + I,s + J) since x G I and t G J. So
f>(y + s) = (a, b). Therefore is onto.
By the First Isomorphism Theorem, we now have R/kercp « (!?). We know that
= R/I ffi R/ J, so R/kerf> « R/I ffi R/ J.
Kercf = I A J.
Let r G I A J. So r E I and r G J. Then (r) = (r + I, r + J) = (I, J)- So
r G ker I A J C ker<f).
Suppose b G kerf>. Then (6 + I, b + J) = (&) = (Z, J). So b G I and b G J. Hence
b G I A J => ker(j) C (Z A J).
From 1-4, we have R/(I A J) « R/I ffi R/ J.
□
We have to make sure that the ideals I and J are coprime, in other words,
I ® J = R so that is not onto, then the mapping will not
necessarily be an isomorphism. The following example illustrates how the condition of
pairwise coprime is necessary.
Example 3.30.
If R = Z, Ay = (6) and A2 = (4) then the mapping : R/(Ay A A2) —>
R/Ay ffi R/Az is not surjective.
Proof: Since the gcd(6, 4) = 2, 6 and 4 are not relatively prime. So the ideals (6 and
(4) are not coprime. Since (6) + (4) = (2) so Ay + A2 R since 1 (2). Also,
Z/((6) A (4)) « Z/(2) « Z2; but, Z2 Z4 ffi Zg. There are only 2 elements in Z2 but
there are 24 elements in Z4 ffi Zg. Hence, cannot be surjective.
□
Now, we are going to generalize the above theorem to n ideals.
Theorem 3.31.
The Chinese Remainder Theorem for n Ideals
If R is a ring and Iy, ...,I n are ideals of R which are pairwise coprime, i.e. I{ + Ij = R
whenever i 7= j, and I = Q”=1 Ii, then R/I is isomorphic to R/Iy ffi R/I2 ffi ffi R/In-
Proof: By induction, for n = 1, we have R/I « R/I.
Assume that Iy, ..., In, In+ y are ideals of R which are pairwise coprime (Zj + Ij = R
whenever i =4 j) and R/I is isomorphic to R/Iy ffi R/I2 ffi ffi R/In where I — C^yli-23
Then by letting J = Zn+ i and using Theorem 3.29 and Theorem 3.30, we have R/(I D
J) Ri R/I © R/J « R/Ii ffi R/Ii © ••• © R/I n © R/ J- Therefore R/I is isomorphic to
R/Ii © R/h © •••© R/In+i where I = D^Ii-
□
Corollary 3.32.
Ifm 6 Z has the prime decomposition m = p^.-.p^ by the Fundamental
Theorem of Arithmetic, (ki > 0, pi distinct primes), then there is an isomorphism of rings
Zm ~ Z fc x x ... x Z fcj .
Pi Pi
Proof: Since all the pfs are distinct primes, they are pairwise coprime and
Zm ps Z/(m). Therefore, by Theorem 3.32 and Proposition 3.29, we can see that Zm ~
Z fc, x ... x Z l
Pi Pi
□
Now that we have looked at the Chinese Remainder Theorem for rings through
the mapping of isomorphism, we are going to reformulate theorem to be analogous to the
theorem for integers.
Theorem 3.33.
General Chinese Remainder Theorem for Rings
Let Ii, ..., In be ideals in a ring R such that I{ + Ij = R for all i j. If bi, ..., bn G R,
there there exists b G R such that b = b{ (mod Ii) for i = 1,2, ...,n. Furthermore, b is
uniquely determined up to congruence modulo the ideal Zj A Z2 Cl ... A Zn.
Just as Theorem 3.32, the ideals are pairwise coprime; however, instead of having
the factor ring R/I isomorphic to the direct sum of all factor rings, we have the case of
the intersection of all ideals which also combines several moduli to a new, larger modulo.
Proof:
Since Zj + Z2 = R and Zi + Z3 = R, then
R= R2 — (Zi+Z 2)(Zi+Z3) = zf+ZxZs+ZaA+Z.Zg C I1+I2I3 C Zi+(Z 2AZ3) C R.
Therefore, R = Ii + (Z 2 l~l Z3). Assume inductively that
z? = z1 + (z 2nz 3n...nz fc _1).
Then
R = R2 = (Zi + (z 2 a... az&_i))(Zi + z&) c Zi + (z 2 aZ3 a... az&) c r
Therefore, R = Ii + (Z 2 Pl ... A Ik)- Consequently, Z? = Zi + (Ai/iZj). Similarly, for each 24
k = 1, 2, n, R = I/. + Consequently, for each k there exist elements a,k G Ik
and rk G Cgtkli such that bk = fflfc + rfc. Furthermore rk = bk (mod Ik) and rk = 0 (mod
Ii) for i k.
Let b = ri + r2 + ... + rn. Then b = ri mod fi and hence b = bi (mod ZJ for every i.
Finally, if c G R is such that c = bi (mod ZJ for every i, then b = c (mod ZJ for each i,
where b — c E Ii for all i. Therefore, b — c & Ci =1 Ii and b = c(mod Z$) .
□
Example 3.34. Let (2) and (3) be ideals in the ring Z. Then we can see that (2)+
(3) = Z For any ai and a2 € Z, there exists an a G Z such that a = ai mod 2 and a = a2
mod 3. Then a = m mod (2) C (3), or a = m mod (6) which takes us back to the general
theorem for integers, Theorem 2.9.
3.3 Chinese Remainder Theorem for Polynomial Rings
Now let ’s look at how the Chinese Remainder Theorem is applied to polynomial
rings. There are a few terms we need to be familiar with.
Definition 3.35. A unit in a ring R is an invertible element of R, i.e. , an element b such
that there is an a in R with ab = ba = 1^.
Definition 3.36. A field is a commutative ring with unity in which every nonzero
element is a unit.
Example 3.37. For every prime p, the ring of integersmodulo p, denoted Zp is a field.
Definition 3.38. Let D be an integral domain. A polynomial f(x) from Z?[m] that is
neither the zero polynomial nor a unit in D[x] is said to be irreducible over D if, whenever
f(x) is expressed as a product f(x) = g(x)h(x), with g(x) and h(x) from Z?[z], then g(x)
or h(x) is a unit in Z)[x].
Note: Elements a and b of an integral domain D are called associates if a = ub
where u is a unit of D.
Theorem 3.39. Let f(x) and g(x) be irreducible polynomials over a field F. If f(x) and
g(x) are not associates, then F[x\/(f(x)g(x)} is isomorphic to F[x]/{f(x)) ®F[x]/{g(x)). 25
Proof: By Theorem 3.29, and the First Isomorphism Theorem, to show that
F[x]/(/(a:)^(a;)) is isomorphic to F[x]/{f(x)) ®F[a;]/(^(a;)) , 2e only need to check that
if f(x) and g(x) are not associates, then {f(x)g(xf) — {f (x)}{g(x)} = (/(»)) A {g(x)).
Let r(x) G (f(x)g(x)), so r(x) = f(x')g(x')h(x) for some h(x) G F[x]. Then
r(x) — [f(x)h(x)]g(x) since F is a field. Hence, r(x) G (J(x)} Pl (g(x)) => (f(x)g(x)) C
A {g(xf).
Conversely, let s(x) G {f(xf) A {g(xf), so .fix') G {f(xf) and s(x) G {g(xf). Then,
s(z) = and s(x) = g(x)r(x) for some h(x),r(x) G F[a;]. Consequently, we
have f(x)h(x) = g(x)r(x). Because f(x),g(x) are irreducible, (f(x)} + {g(x)} = F[x], so
f(x')u(x) + glx)v(x) = 1. So s(x)f(x)u(x) + s(x)g(x)v(x) = s(x).
s(x) G (f(x)g(x)}, and hence {f(xf) A (g(xf) C {f(x)g(x)).
□
Example 3.40.
For every prime p, the ring of integers modulo p, denoted Zp is a field.
Definition 3.41.
A monic polynomial is a polynomial whose leading coefficient is 1.
Definition 3.42.
Let a(x) and b(x) be polynomials not both zero with coefficients in a
field F. The greatest common divisor of a(x) and b(x) is the monic polynomial d(x) of
highest degree such that d(x) is a divisor of a(x) and b(x).
Example 3.43.
Let a(x) — x2 + 7x + 6 and fix') = x2 — 5z — 6. Then a(x) = (s+l)(x-+6)
and b(x) = (x + l)(z — 6). Hence the gcd(a(x), b(xf) = z + 1.
Theorem 3.44.
General Chinese Remainder Theorem for Polynomial Rings.
Let F be afield and let bi(x), ...,b n(x) be arbitrary polynomials o/F[x], mi(z), ..., mn (x)
and afix), ...,a n(x) be polynomials of F[x] such that
gcd (mfixfim j{x')') = 1 , i / j-
gcd (afix), mfixfif) = 1, i = 1, 2,..., n.
Then the system of congruences
afix)u(x) = bfix) mod mfix), i — 1, 2,..., n
has exactly one solution modulo m(x) = nfifipfix) ■ • -m n(x). 26
Proof: Follow the proof of Theorem 3.32 and the Chinese Remainder Algorithm,
we have gcd(aj(x) , mi(xf) = 1 for 1 < i < n. We can compute a polynomial ci(x) E F[x]
such that Ci(x)ai(x) = 1 mod m-i/x) for all i. Therefore, a,i(x')u(x') = bi(x) mod mi(x),
i = 1,2, ...,n becomes u(x) = Ci(x)bi(x) mod mfa) for 1 < i < n. Using the Chinese
(. t)
Remainder Algorithm, we can find M(x) = 117=1 ■mi(. x)’ then Mi(x) =--- —r, Mz(x) =
1711 (Xj
Mix') „ . Mix') _ , , . , , .....
■---- ..., Mn(x) — ■—jrnr. Then proceeding as the algorithm, the solution is given by:
m2(x) mn(x)
•••)
u(x) = bi^ci^Mi^y^x) + b2(x)c 2(x)M2(x')y2(x') + ... + bn(x)cn(x)M n(x)y n(x) mod
where yt(x) is the inverse of Mi(x) mod mi(x) for i = 1,2,
Example 3.45.
Let mi(j) = x3 + x + 1 and m2(x) = x3 + x2 + 1 in Z/(2). Also, let
a± = x2 + x + 1 and a2 = x + 1. Since mi and a-i cannot be factored, they are relatively
prime for each i. By the Euclidean Algorithm we have (x + l)mi(a;) + o; 2ai(o;) = 1 and
7712(0;) + x2a,2(x) — 1. So the inverses Ci(x) of a.i(x) are ci(o;) = x2 and c2(o:) = x2. Since
17711(2:) + (x + l)-m, 2(:z;) = 1 we have a system of congruences
a(x)u(x) = bi(x) mod 7711(0;)
a2{x)u(x) = b2{x) mod 7702(0;).
The solution is given by u(x) = (x+ I)x 2m2(x)ri(x') + x ■ x2mi(x)r 2(x) mod 771(0;), where
m(x) = mi(x)m2(x) .27
Chapter 4
Applications of the Chinese
Remainder Theorem
4.1 Finite Sequence of Integers
As mentioned in Chapter 1, the first indication of the power of the Chinese
Remainder Theorem applies to finite sequences of integers. For any finite sequence of
integers, we can find another two integers to represent it. Let first look at the theorem
and its proof.
Theorem 4.1. Let ai, 0 < i < t, be a finite sequence of nonnegative integers. Then there
are integers u and v such that (u mod (1 + (i + l)u)) = Oi, for every £ = 0,1,..., t.
Proof: Let a be the largest integer of the sequence ai, 0 < i < t, and define
v = 2a • t! and mi = 1 + v(i + 1), 0 < i < t. We claim that the integers m,, 0 < i < t are
relatively prime in pairs. By contradiction, let p be a prime number that divides both mi
and mj, for some i > j. Then p divides the difference (i + l)mj — (j + l)mj = i — j <t.
Since p let divides my and v is divisible by all integers less than or equal to t, we obtain
that p = 1, which is not a prime. So the integers mi qualify as moduli for the Chinese
Remainder Theorem. Hence, there is a number u such that
u = ai mod mi, i = 0,1,... t.
Then, u mod mi = ai mod mi for all 0 < i < t. However, since ai < v < m,. we can
conclude that mod mi = ai for 0 < i < t. Therefore, u mod mi = ai for 0 < i < t. 28
□
Below is an example using a small finite sequence of integers. For this sequence,
we find two integers representing each term. If we apply this theorem to larger sequence,
we should still be able to find two integers representing each term in the sequence.
Example 4.2. Let {2,3,5,6} be a finite sequence with ao = 2, ai = 3, a2 = 5 and a3 = 6.
Then by the theorem, there are integers u and v such that u mod (1 + (i + l)v) = a;. We
will show how to find u and v. The largest integer is 6, so by the proof of the theorem we
have v = 2 • 6 • 3! = 72, and: mo = 1 + 72 = 73, mi = 1 + 144 = 145, m2 = 1 + 216 = 217,
m3 = 1 + 288 = 289. Then we obtain a system of congruences:
u = 2 mod 73,
u = 3 mod 145,
u = 5 mod 217,
u = 6 mod 289.
This system of congruence has a solution. Using the Chinese Remainder Theorem, we
obtain M = 663,817,105, MQ = 9,093, 385, Mx = 4,578,049, M2 = 3,059,065, and
M3 = 2,296, 945. Next, we get 9,093, 385yo = 1 mod 73 which yields yo = 12 mod 73;
4, 578, 049yi = 1 mod 145 which yields yi = 4 mod 145, next 3, 059,065^2 = 1 mod 217
giving y2 = 95 mod 217, finally 2, 296, 945y 3 = 1 mod 289 giving y3 = 107 mod 289.
Therefore, u = 2 • 9,093, 385 • 12 + 3 • 4, 578,049-4 + 5-3, 059, 065 • 95 + 6 • 2, 296, 945 • 107
= 545,603,973 mod 663,817,105. Hence, we can represent the terms of the sequence
using the two integers 72 and 545, 603, 973:
2 = 545,603,973 mod (1 + 72 • 1),
3 = 545,603,973 mod (1 + 72 ■ 2),
5 = 545,603,973 mod (1 + 72-3),
6 = 545,603,973 mod (1 + 72-4).
4.2 A Characterization of Dedekind Domains
In number theory, the Fundamental Theorey of Arithmetic states that every
natural number greater that 1 can be written as unique product of prime numbers. In
abstract algebra, a Dedekind domain has a similar set up. 29
Definition 4.3. A
Dedekind domain is an integral domain in which each ideal can be
written as a product of a finite number of prime ideals.
Definition 4.4.
Let R be a ring and I is its ideal. If there exists an inverse ideal
I-1 = {r G /C\xi G I?} where K is the quotient field of R then II~^ = R.
Proposition 4.5.
In a Dedekind domain, every nonzero prime ideal is a maximal ideal.
Proof: Let R be a Dedekind domain and p is nonzero prime ideal and p is not
maximal. Let a be another ideal of R such that p C. a. Then C a-1 a = R so
a_1 p is an ideal of R. Since a(a _1 p) — p then a C p or a_1 p C p. If C p then
a-1 C = R which implies that R C a so a = R. On the other hand, if a C p and by
assumption, p C a then p = a. Therefore p is maximal.
□
Example 4.6.
The ring of integers Z is an example of Dedekind domain. The principal
ideals of Z are all generated by each integer such as (2) = {0, ±2, ±4, ±6,... }. If a G Z,
it has a unique prime decomposition; therefore (a) can be written as a product of a finite
number of ideals. That is, if a = p^f ■ ■ -Pn n, then (a) = (p^ 1) ■ ■ ■ (Pn n)-
A principal ideal domain is always a Dedekind domain; however, a Dedekind
domain may not be a principal ideal domain. The following proposition will state the
condition in which a Dedekind domain is a principal domain.
Proposition 4.7.
If a Dedekind domain R has only a finite number of nonzero prime
ideals Pi, ...,P n, then R is a principal ideal domain.
Proof: For each i, choose bi E Pi —Pf. We are going to prove that (bf) = Pi which
implies that every prime ideal is principal, hence, A is a principal ideal domain. Since R is
a Dedekind domain and it has only a finite number of nonzero prime ideals Pj, 1 < i < n,
then these prime ideals are also maximal ideals, and so Pj + Pj = R, i j. Because R is
a Dedekind domain, bi = Iip-.-Im, where It = Pj. By the Chinese Remainder Theorem
3.32, we have bi = ij mod Pj and bi = 1 mod Pj. Then since bj G Pi, (bj) C Pj. So
Iih-.-Im Q Pi- Then Ij = Pi for some j. Then we can rearrange the ideals so that
(bj) = PjZi...I m_i. If Ir = Pi for some r, then (bj) = P/fi...I m-2, but this contradicts
bi Pj 2. So continuing this process, we have (bj) = PiIy...I m-i = PjnZiD...r)/ m_i where 30
Ii, ..., Im-i are distinct primes. Then b{ G Ii and since bi = 1 (mod Ii), bi — 1 G Ii- So
1 G Ii and hence Ii = R. Therefore, (bi) = P,.
□
4.3 Cryptography Schemes
The Chinese Remainder Theorem is applied in secret sharing, which is an im
portant topic of cryptography. The Chinese Remainder Theorem itself is a secret sharing
scheme without any modification. Let mi, m2, ■ •■,mt be t pairwise relatively prime posi
tive integers. Also let m = JlLu mi- Suppose that we have a secret which is an integer s
such that 0 < s < m. Let Pi, P2, ..., Pt be the t parties who are going to share the secret.
Then Pi has the residue Sj = s mod mi as the secret that is only known to Pi. By the
Chinese Remainder Theorem, the t pieces of information s, are sufficient to determine
the original s.
For the t parties, if we give out k shares, then the secret can be computed;
otherwise, k — 1 shares will give a possible range of the secret. A (k, t) secret-sharing
scheme is defined as follows. The t parties Pi share a secret s with the following conditions:
Each party has a share Sj about the secret s which is not known to other parties.
The secret s can be computed from any k shares Sj.
No k — 1 shares Sj give any information about the secret s.
We are going to look at the two secret sharing schemes; one involves the integers and the
other is for polynomials.
Scheme 1
Let mt, i = 1, 2,..., t, be t pairwise relatively prime integers no less than 2. We
define
m,in(k) = min(m.i 1mi 2 ■ ■ -m^jl < ii < ... < < t},
max(k — 1) = ■ ■ ■ mi k__J1 < ?1 < < ■ • • < ik-i < t},
where 1 < k < t. In other words, min(k) is the smallest product of k of the integers mt
and max(k — 1) is the greatest product of k — 1 of the integers mt. Choose w to be the
largest positive integer such that 31
min(k)
max(k — 1)
• gcd (w, m,i) = 1, i = 1,2,..., t.
Let m = min(k).
The secret is the integer s such that 0 < s < w. Therefore, we assume that the secret is
equally likely to be any integer between 0 and w — 1.
We compute the shares for t parties as follows.
Let a G Z such that 0 < s + aw < m, and let s = s + aw. The shares are then
given by $$ = s mod mi , i = 1,2,..., t, where Si is the share of party Pi- We are going to
prove that k — 1 or fewer shares give no information about the secret; but any k or more
shares determine the secret.
Without loss of generality, suppose sy,S2, ■■■,Sh are known and 1 < h < t. Let
M = HiLi mi and Mj = for j = 1,2, ...,h. Then (Mj,mj) = 1. By Euclidean
algorithm, there exist Uj, Vj G Z such that MjUj + mjVj — 1. This can be done by solving
MjUj = 1 mod mj and rrijVj = 1 mod Mj. Then by the Chinese Remainder Algorithm,
the system of congruences x = Si mod mi for 1 < i < h has a unique solution modulo M
given by x = s-yMyu-y + s^M^ uq , + ... + shMhuh.
Let s" = ^j u3sj m°d M where 0 < s" < M, then s" = Sj mod mi. We have two
cases:
Case 1:
If h > k then M > min(k) = m > w. By the Chinese Remainder Theorem s" = s' since
s' = Si mod m,i, s' < m < M and solutions to the system x = Si mod mi are unique
modulo M. Now s' mod w — (s + aw) mod w = s so the secret is given by s = s" mod
w.
Case 2:
w max[k — 1]
If h = k, — 1 then M < max[k — 1] < —. This follows since w < -------j-- implies
for some b where 0 < s' < m since the solutions to the system x = Si mod mi are unique
modulo M. Then 0 < s' < m implies 0 < s" + bM < m, so —s' 1 < bM < m — s" which
. j , —s . , m — s . m — s" . . , m — s m — M
leads to —— < b < ———. Therefore, 0 < b < ——— . Also, — > —
_ M ~ M L M J M M
m m
— — 1 > w — 1 since w < —. So s = s' mod w = ($" + bm) mod w and since b ranges 32
771 —s”777 —s"
from 0 to [———J and w — 1 < [———J, s takes on 0,1, w — 1 equally likely.
Thus any k — 1 or fewer shares give no information about secret s.
Now let ’s see an example to see how the secret-sharing scheme works.
Example 4.8.
Let k = 3, t = 4, my = 5, m2 = 7, m3 = 11, and 7714 = 13. Then m =
min(k) = 5 • 7 • 11 = 385, and
, , min(k)
such that w < -------—------ - =
max(k — 1)
an integer where 0 < s < w.
max (k — 1) = 11 • 13 = 143. So there exists an integer w
385
■^3 and gcd (w,™,) — 1. We get w = 2. The secret s is
So s is either 0 or 1. Since t = 4, we need to compute 4
shares. We choose an a where 0 < s + aw < m. Hence 0 < aw < m — s which implies
0 < a < -------- =------- ---- = 192. Choose a = 30, for example. Let s' = s + aw = s + 60.
w 2
The four shares are given by s; = s' mod m{. Then si = (s + 60) mod 5, s2 = (s + 60)
mod 7, S3 = (s + 60) mod 11, and s4 = (s + 60) mod 13. As the secret keeper, we let
s = 1 then si = 1 and s2 = 5. Then M = miffl2 = 35, Mi = 7, and M2 = 5. By the
Euclidean Algorithm, we get 7 • (—2) + 5-3 = 1. Therefore, u = 2 and v = 3. Then by
the Chinese Remainder Algorithm, we have s" = M1U1S1 + M2u2s2 since h = 2. Then
s" = (7 • (—2) • si + 5 • 3 • s2) mod 35 = (—14 + 75) mod 35 = 61 mod 35 = 26 mod 35.
Q Q K QZJ
Since 6. = 3 — 1 = 2 we have s' = s" + bM = 26 + 356 where 0 < b <---- —---- = 10. It
35
follows that s = s' mod 2 = 6 mod 2. Since 6 G [0,10], s takes on 1 and 0 equally likely.
Therefore, the two shares give no information about the secret.
If we have three shares, we can calculate the secret s. So for the example above, suppose
we have three shares sx = 1, s2 = 5, and s3 = 6. Then M = mim 2rri3 = 5 • 7 • 11 = 385,
so Mi = 77, M2 = 55, and M3 = 35. Therefore, by Euclidean Algorithm we can find u.j
and Vj, for j = 1, 2, 3 such that:
77ui + 5vi = 1;
55 u2 + 7v2 = 1,
35 h3 + lli> 3 = 1
Solving the three equations, we get:
77- (-2)+ 5-31 = 1,
55-(-1)+ 7-8 = 1,
35-6 +11-(-19) = 1.
Hence, s" = 77 • (—2) • sx + 55 • (—1) • s2 + 35 • 6 • s3 = 831 mod 385 = 61 mod 385.
Consequently, the secret s would be s = 61 mod 2 = 1 mod 2. So the secret s is 1. 33
Scheme 2
This secret-sharing scheme is almost similar to scheme 1, but it deals with
polynomials.
Let F be a finite field and mfix), i — 1, ...,t be t pairwise relatively prime polynomials
of FR] with degree greater of equal to 1. Then
min[k] = min^deg^mgm^ ■ ■ ■ mi k)|1 < ii < • • • < < t},
max[k — 1] = max{deg(mi 1mi 2 ■ ■ ■mj fc _1)|l < ii < • • • < ifc-1 < i}>
where 1 < k < t. Let w be the largest positive integer such that
• there is a polynomial IV(z;) of degree w over F with (W(a;), mfix)) = 1 for i =
1,2,. .., t; and
min(k]
• w < -------7-—-.
max[k — 1]
The secret is a polynomial s(x) in Ffrr] of degree less than w. The shares for t parties
are computed as follows.
Choose a(x) G F[x] such that deg(s(a;) + a(x)W(x)) < min[k]. Let s'(x) =
s(x) + a(x)w(x). The shares are then given by sfix) = s'(x) mod mi, i = 1,2,
Therefore, (k, t) is a threshold scheme.
Suppose k shares si(x), s2(x),Sk(x) are given. Let M(x) =
and Mj(x) = for j — 1, Then Mj(x) and mj(x) are relatively prime. By
Euclidean Algorithm, there exist two polynomials Uj(x) and Vj(x) in F[rr] such that
Mj(x)uj(x) + mj(x)vj(x) = 1. By the Chinese Remainder Algorithm, we have s'(x) =
Mj( x)uj(x)sj(x) mod M(x). Then the secret is given by s(x) = s'(x) mod W(z). 34
Chapter 5
Conclusion
In this short manuscript, we have shown some expansion and powerful applica
tions of the Chinese Remainder Theorem. It is amazing to see how the theorem evolved
from the three basic problems of calendar, wall-building, and soldier-counting. Even
though the theorem, first generated as a problem, is taken credit from the Chinese schol
ars, it was quite well-known in other parts of the world. Many other mathematicians
were also trying to solve similar problem
Here we discuss the applications of the theorem to finite sequence of integers,
Dedekind domains, and briefly crytography. However, the Chinese Remainder Theorem
is widely applied in other areas such as computing, and codes. We hope readers find this
topic interesting enough to pursue further research on those areas. 35
Bibliography
[DPS96] C. Ding, D. Pei, and A. Salomaa. Chinese remainder theorem, application in
computing, coding, cryptography. World Scientific Publishing Co., Singapore,
[Gal06] Joseph A. Gallian. Contemporary abstract algebra. Houghton Mifflin, New York,
sixth edition, 2006.
[Hun74] Thomas W. Hungerford. Algebra-graduate texts in mathematics. Springer, New
York, 1974. |
3582 | https://www.popai.pro/resources/homework/how-do-you-use-the-unit-circle-to-determine-the-exact-values-of-sin-cos-and-tan-for-common-angles-such-as-30-45-and-60/ | How do you use the unit circle to determine the exact values of sin, cos, and tan for common angles such as 30°, 45°, and 60°? - PopAi
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Home>Resources>Homework>Math
How do you use the unit circle to determine the exact values of sin, cos, and tan for common angles such as 30°, 45°, and 60°?
Answer 1
Angela Powers
To determine the exact values of sin, cos, and tan for 30°, 45°, and 60° using the unit circle, locate these angles on the circle. For 30° (π/6), sin(30°) = 1/2, cos(30°) = √3/2, tan(30°) = 1/√3. For 45° (π/4), sin(45°) = cos(45°) = √2/2, tan(45°) = 1. For 60° (π/3), sin(60°) = √3/2, cos(60°) = 1/2, tan(60°) = √3.
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3583 | https://mathbitsnotebook.com/Geometry/CoordinateGeometry/CGBoxMethod.html | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- | | | | | --- | | | Box Method - Area on a Coordinate Grid MathBitsNotebook.com Topical Outline | Geometry Outline | MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts | We are going to see that finding the area of polygons drawn on a coordinate axis can be an easy process by using a simple box. Whether the box is needed, depends upon the placement of the polygon on the coordinate grid. Two situations are possible: | | | | | --- --- | | | | | Sides are parallel to the axes: If the polygon is drawn such that its sides (or needed segments) lie ON the grids of the graph paper, COUNT the lengths and use the appropriate area formula. | | | | | Sides are NOT parallel to the axes: If the polygon is drawn such that its sides (or needed segments) do NOT lie ON the grids of the graph paper, draw a "BOX" around the polygon to determine the area. | | | | | | --- | | | | Sides are parallel to the axes: | COUNT to find the needed lengths. The base of the triangle lies ON the grids of the graph paper (horizontal), and the altitude (vertical height) also lies ON the grids of the graph paper, since it is perpendicular to the base. By counting, the base length is 7 units and the altitude is 3 units. Use the area of a triangle formula: The answer is 10½ square units. AC can also be found by subtracting the x-coordinates of the two points (4 - (-3) = 7). Since the base of the altitude is located at (2,1), the altitude can be found by subtracting the y-coordinates (4 - 1 = 3). | To COUNT: stand at A and take one step to the right to the next grid line. Continue stepping and counting until you reach C. | | | | | | | | | --- --- | | | | | Sides are NOT parallel to the axes: | The sides of this ΔABC do NOT lie on the grid lines of the graph paper. Use the "Box Method" to solve for the area of the triangle. • Draw the smallest "box" possible to enclose the polygon (ΔABC). Be sure that the "box" follows the grids of the graph paper. • Number each of the parts of the box with a Roman numeral (ignore the axes when numbering). • "The whole is equal to the sum of its parts." The area of each of the parts of the "box" added together equals the area of the "box". The answer is 14 square units. | | | | Note: Find the area of the box by counting. Represent the triangle you need by x. Find the area of each of the right triangles by counting and using the formula for the area of a triangle. | | | | | | --- | | | | Dealing with odd shaped pieces: | There are times when the "box" will not form nice right triangles in each of the corners, as shown here in the upper left corner. In such a case, it is necessary to further subdivide these sections into one rectangle and two right triangles, so the lengths can be easily counted. The answer is 36 square units. | | | Remember to keep your work simple by always forming shapes around the figure that are easy to count. | | | | | NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". | Topical Outline | Geometry Outline | MathBitsNotebook.com | MathBits' Teacher Resources Terms of UseContact Person: Donna Roberts Copyright © 2012-2025 MathBitsNotebook.com. All Rights Reserved. | | |
3584 | https://en.wikipedia.org/?title=Necessary_and_sufficient_condition&redirect=no | Necessary and sufficient condition - Wikipedia
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3585 | https://www.youtube.com/watch?v=5iOBirDp6ik | Proof of the sine rule
Whiteboard Maths
28100 subscribers
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Description
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Posted: 20 Jan 2021
In this video, we give a nice geometric proof for the sine rule.
29 comments
Transcript:
[Music] okay so in the previous video we saw some examples of using the sine rule now in this one i want to talk about the proof of this so how do we show this is always true for any angles a b and c and links a b and c now the way we do this well we see that we have signs in here so let's uh first remember what the definition of sine is if you have a right angle triangle then sine of the angle x this is the ratio of the opposite side divided by the hypotenuse so if we just have a little triangle like this and this is our angle x this is the opposite side and this is hypotenuse so this is the definition of sine but to use sine we need a right angle triangle so this has to be 90 degrees so we want to find expressions for sine of a and sine of b but to do that we need a right angle triangle so we're going to use a little trick we're going to drop down this line here i'm going to draw this dotted line all the way down so we create two triangles and these are both now right angle triangles and i'm just going to label the length of this line h so now we have a right angle triangle with this angle a which means we can write what sine of a is using this definition so sine of the angle a is the ratio of the opposite side divided by the hypotenuse so here the opposite side is this one which has length h so we have h divided by the length of the hypotenuse and the hypotenuse is the longest side so it's this one it's b so sine of a equals h over b and we don't know what h is but what we can do is we can rearrange again in terms of h so i'm just going to bring up the b and we have h equals b times sine of a and our plan is to find another expression for b sine of b and then we can eliminate h so to find sine of b is kind of a similar strategy we now look at this right angle triangle the one on the right so this is also 90 degrees and we could just use this definition so the ratio of the opposite length to the hypotenuse the opposite is again h but now the hypotenuse isn't b it's actually this length the longest side of this small triangle and this is a so we have h divided by a and again we can rearrange to get in terms of h so h equals a times sine of b and this is really nice because we have two equations for h and so we can equate them because they're going to be equal because they're both equal to h so this tells us we can deduce that b sine a it equals h so it also equals this stuff equals a times sine of b and you can see we're almost there what we do is we divide by a actually we want to divide by sine of a and sine of b and then we're going to be left with b divided by sine of b if i divide by this and then this equals a divided by this if i divide by sine of a a divided by sine of a and this is exactly the sine rule so we just showed this first equality but the second one follows just by relabeling all the points so i'm not going to do it here but if you just draw the vertical line in a different vertex then you can show this equality as well but it's the same strategy essentially so let's just recap what we've done so we're trying to show this equation is true and to do that we looked at the definition of sine which is the length of the opposite length divided by the hypotenuse if we're working in a right angle triangle so we created two or angle triangles and then we wrote the definition of sine of a and sine of b in terms of these new triangles and in terms of h then we rearranged for h and equated these two equations because they're both equal to h and this is how we got the equation involving sine of a and sine of b and if we just did a bit of rearranging this actually gave us our result so it's a bit complicated just make sure go for a couple times make sure you understand it but this is the strategy to proving this so this is our proof |
3586 | https://www.vedantu.com/maths/square | Square in Maths: Properties, Formulas & Real-Life Examples
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Square in Maths: Meaning, Properties & Examples
Square in Maths: Meaning, Properties & Examples
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What are the Properties and Formulas of a Square?
The concept of square in maths plays a key role in mathematics and is widely applicable to both real-life situations and exam scenarios. Every student studies squares in geometry, number theory, and even mental maths tricks for exams like NTSE, Olympiad, and CBSE boards.
What Is Square in Maths?
A square in maths is defined as a quadrilateral (a four-sided polygon) where all the sides are equal, and all the angles are right angles (90°). In other words, a square is a regular quadrilateral. You’ll find this concept applied in geometry, mensuration, and coordinate geometry.
| Property | Value |
--- |
| Number of sides | 4 (all equal) |
| Number of angles | 4 (all 90°) |
| Diagonals | 2 (equal length, bisect at 90°) |
| Lines of symmetry | 4 |
| Rotational symmetry | Order 4 |
| Area | side × side |
| Perimeter | 4 × side |
Key Formula for Square in Maths
Here’s the standard formula:
Area = s×s = s 2
Perimeter = 4×s
Diagonal = s×2
Properties of a Square
All four sides are equal and parallel opposite sides.
All four angles measure 90°.
Diagonals are equal, bisect each other at 90°.
Four lines of symmetry and rotational symmetry of order 4.
Diagonals also bisect the angles of the square.
Difference Between Square and Rectangle
| Feature | Square in Maths | Rectangle |
---
| Sides | All sides equal | Opposite sides equal |
| Angles | All 90° | All 90° |
| Diagonals | Equal, bisect at 90° | Equal, bisect (not always at 90°) |
| Symmetry | 4 lines | 2 lines |
Read more: Difference Between Square and Rectangle
How to Construct a Square
Draw one side AB of a desired length (let’s say 5 cm).
At point A, use your compass and protractor to construct a 90° angle.
Mark point D so that AD = AB.
At point B, construct another 90° angle and mark point C so that BC = AB.
Join points C and D. The figure ABCD is your square in maths!
To see this step visually, visit Construction of Square.
Square in Real Life
Tiles on the floor
Chessboard squares
Window panes
Post-it notes
These everyday objects make it easy to visualize and apply the concept of square in maths to the outside world.
Step-by-Step Illustration
Find the area and perimeter of a square whose side is 6 cm.
Area = 6×6=36 cm²
Perimeter = 4×6=24 cm
Speed Trick or Vedic Shortcut
Here’s a quick shortcut that helps solve problems faster when working with square in maths. Many students use this trick during timed exams to save crucial seconds.
Example Trick: To square numbers ending in 5 (example: 35²):
Multiply the tens digit by its next number:
3 × 4 = 12
2. Write 25 next to it:
1225
3. So, 35² = 1225.
Tricks like this aren’t just cool — they’re practical in competitive exams like NTSE, Olympiads, and even JEE. Vedantu’s live sessions include more such shortcuts to help you build speed and accuracy.
Try These Yourself
Draw a square of side 7 cm and label all its properties.
Find out how many small squares there are on a standard chessboard.
Check if a quadrilateral with equal diagonals that bisect at 90° is always a square.
Calculate the length of the diagonal of a square with a side of 9 cm.
Frequent Errors and Misunderstandings
Mixing up squares with rectangles (remember: all sides must be equal for a square).
Confusing diagonal formula for rectangles and squares.
Forgetting that every angle in a square is always 90°.
Relation to Other Concepts
The idea of square in maths is closely related to rectangle, quadrilaterals, and rhombus. Mastering these properties helps with proofs, Venn diagrams, and problem-solving in higher classes.
Classroom Tip
A quick way to remember square in maths is to visualize a chessboard or tile. All sides must be equal — if even one side is shorter or longer, it’s not a square! Vedantu’s teachers often use model cutouts and digital diagrams to reinforce this during live classes.
We explored square in maths—from definition, formula, examples, mistakes, and connections to other subjects. Continue practicing with Vedantu to become confident in solving questions on this topic. For a deep dive into formulas and more MCQs, checkArea of Square Using Diagonal.
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3587 | https://artofproblemsolving.com/downloads/printable_post_collections/3002465.pdf?srsltid=AfmBOoph8P0apGHQLjJOuq3H9wZSv2Fz0k1QEX19I6f4pCBawLhQs1FX | AoPS Community Math Hour Olympiad, Grades 5-7
University of Washington - Math Hour Olympiad, an individual oral math Olympiad,
www.artofproblemsolving.com/community/c3002465 by parmenides51
2010.67 Round 1
p1. Is it possible to draw some number of diagonals in a convex hexagon so that every diagonal crosses EXACTLY three others in the interior of the hexagon? (Diagonals that touch at one of the corners of the hexagon DO NOT count as crossing.)
p2. A 3 × 3 square grid is filled with positive numbers so that (a) the product of the numbers in every row is 1,(b) the product of the numbers in every column is 1,(c) the product of the numbers in any of the four 2 × 2 squares is 2.What is the middle number in the grid? Find all possible answers and show that there are no others.
p3. Each letter in HAGRID ’s name represents a distinct digit between 0 and 9. Show that
HAGRID × H × A × G × R × I × D
is divisible by 3. (For example, if H = 1 , A = 2 , G = 3 , R = 4 , I = 5 , D = 64 , then HAGRID ×
H × A × G × R × I × D = 123456 × 1 × 2 × 3 × 4 × 5 × 6).
p4. You walk into a room and find five boxes sitting on a table. Each box contains some number of coins, and you can see how many coins are in each box. In the corner of the room, there is a large pile of coins. You can take two coins at a time from the pile and place them in different boxes. If you can add coins to boxes in this way as many times as you like, can you guarantee that each box on the table will eventually contain the same number of coins?
p5. Alex, Bob and Chad are playing a table tennis tournament. During each game, two boys are playing each other and one is resting. In the next game the boy who lost a game goes to rest, and the boy who was resting plays the winner. By the end of tournament, Alex played a total of
10 games, Bob played 15 games, and Chad played 17 games. Who lost the second game? Round 2
p6. After going for a swim in his vault of gold coins, Scrooge McDuck decides he wants to try to arrange some of his gold coins on a table so that every coin he places on the table touches
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exactly three others. Can he possibly do this? You need to justify your answer. (Assume the gold coins are circular, and that they all have the same size. Coins must be laid at on the table, and no two of them can overlap.)
p7. You have a deck of 50 cards, each of which is labeled with a number between 1 and 25 . In the deck, there are exactly two cards with each label. The cards are shuffled and dealt to 25 students who are sitting at a round table, and each student receives two cards. The students will now play a game. On every move of the game, each student takes the card with the smaller number out of his or her hand and passes it to the person on his/her right. Each student makes this move at the same time so that everyone always has exactly two cards. The game continues until some student has a pair of cards with the same number. Show that this game will eventually end. PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2011.67 Round 1
p1. In a chemical lab there are three vials: one that can hold 1 oz of fluid, another that can hold
2 oz, and a third that can hold 3 oz. The first is filled with grape juice, the second with sulfuric acid, and the third with water. There are also 3 empty vials in the cupboard, also of sizes 1 oz, 2
oz, and 3 oz. In order to save the world with grape-flavored acid, James Bond must make three full bottles, one of each size, filled with a mixture of all three liquids so that each bottle has the same ratio of juice to acid to water. How can he do this, considering he was silly enough not to bring any equipment?
p2. Twelve people, some are knights and some are knaves, are sitting around a table. Knaves always lie and knights always tell the truth. At some point they start up a conversation. The first person says, ”There are no knights around this table.” The second says, ”There is at most one knight at this table.” The third – ”There are at most two knights at the table.” And so on until the
12 th says, ”There are at most eleven knights at the table.” How many knights are at the table? Justify your answer.
p3. Aquaman has a barrel divided up into six sections, and he has placed a red herring in each. Aquaman can command any fish of his choice to either ’jump counterclockwise to the next sec-tor’ or ’jump clockwise to the next sector.’ Using a sequence of exactly 30 of these commands, can he relocate all the red herrings to one sector? If yes, show how. If no, explain why not.
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p4. Is it possible to place 13 integers around a circle so that the sum of any 3 adjacent numbers is exactly 13 ?
p5. Two girls are playing a game. The first player writes the letters A or B in a row, left to right, adding one letter on her turn. The second player switches any two letters after each move by the first player (the letters do not have to be adjacent), or does nothing, which also counts as a move. The game is over when each player has made 2011 moves. Can the second player plan her moves so that the resulting letters form a palindrome? (A palindrome is a sequence that reads the same forward and backwards, e.g. AABABAA .) Round 2
p6. Eight students participated in a math competition. There were eight problems to solve. Each problem was solved by exactly five people. Show that there are two students who solved all eight problems between them.
p7. There are 3n checkers of three different colors: n red, n green and n blue. They were used to randomly fill a board with 3 rows and n columns so that each square of the board has one checker on it. Prove that it is possible to reshuffle the checkers within each row so that in each column there are checkers of all three colors. Moving checkers to a different row is not allowed. PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2012.57 Round 1
p1. Tom and Jerry stole a chain of 7 sausages and are now trying to divide the bounty. They take turns biting the sausages at one of the connections. When one of them breaks a connection, he may eat any single sausages that may fall out. Tom takes the first bite. Each of them is trying his best to eat more sausages than his opponent. Who will succeed?
p2. The King of the Mountain Dwarves wants to light his underground throne room by placing several torches so that the whole room is lit. The king, being very miserly, wants to use as few torches as possible. What is the least number of torches he could use? (You should show why he can’t do it with a smaller number of torches.) This is the shape of the throne room:
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Also, the walls in all rooms are lined with velvet and do not reflect the light. For example, the
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picture on the right shows how another room in the castle is partially lit.
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p3. In the Hundred Acre Wood, all the animals are either knights or liars. Knights always tell the truth and liars always lie. One day in the Wood, Winnie-the-Pooh, a knight, decides to visit his friend Rabbit, also a noble knight. Upon arrival, Pooh finds his friend sitting at a round table with
5 other guests. One-by-one, Pooh asks each person at the table how many of his two neighbors are knights. Surprisingly, he gets the same answer from everybody! ”Oh bother!” proclaims Pooh. ”I still don’t have enough information to figure out how many knights are at this table.” ”But it’s my birthday,” adds one of the guests. ”Yes, it’s his birthday!” agrees his neighbor. Now Pooh can tell how many knights are at the table. Can you?
p4. Several girls participate in a tennis tournament in which each player plays each other player exactly once. At the end of the tournament, it turns out that each player has lost at least one of her games. Prove that it is possible to find three players A, B, and C such that A defeated B, B
defeated C, and C defeated A.
p5. There are 40 piles of stones with an equal number of stones in each. Two players, Ann and Bob, can select any two piles of stones and combine them into one bigger pile, as long as this pile would not contain more than half of all the stones on the table. A player who can’t make a move loses. Ann goes first. Who wins? Round 2
p6. In a galaxy far, far away, there is a United Galactic Senate with 100 Senators. Each Senator has no more than three enemies. Tired of their arguments, the Senators want to split into two parties so that each Senator has no more than one enemy in his own party. Prove that they can do this. (Note: If A is an enemy of B, then B is an enemy of A.)
p7. Harry has a 2012 by 2012 chessboard and checkers numbered from 1 to 2012 × 2012 . Can he place all the checkers on the chessboard in such a way that whatever row and column Pro-fessor Snape picks, Harry will be able to choose three checkers from this row and this column such that the product of the numbers on two of the checkers will be equal to the number on the third?
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PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2013.67 Round 1
p1. Goldilocks enters the home of the three bears – Papa Bear, Mama Bear, and Baby Bear. Each bear is wearing a different-colored shirt – red, green, or blue. All the bears look the same to Goldilocks, so she cannot otherwise tell them apart. The bears in the red and blue shirts each make one true statement and one false statement. The bear in the red shirt says: ”I’m Blue’s dad. I’m Green’s daughter.” The bear in the blue shirt says: ”Red and Green are of opposite gender. Red and Green are my parents.” Help Goldilocks find out which bear is wearing which shirt.
p2. The University of Washington is holding a talent competition. The competition has five con-tests: math, physics, chemistry, biology, and ballroom dancing. Any student can enter into any number of the contests but only once for each one. For example, a student may participate in math, biology, and ballroom. It turned out that each student participated in an odd number of contests. Also, each contest had an odd number of participants. Was the total number of contestants odd or even?
p3. The 99 greatest scientists of Mars and Venus are seated evenly around a circular table. If any scientist sees two colleagues from her own planet sitting an equal number of seats to her left and right, she waves to them. For example, if you are from Mars and the scientists sitting two seats to your left and right are also from Mars, you will wave to them. Prove that at least one of the 99 scientists will be waving, no matter how they are seated around the table.
p4. One hundred boys participated in a tennis tournament in which every player played each other player exactly once and there were no ties. Prove that after the tournament, it is possible for the boys to line up for pizza so that each boy defeated the boy standing right behind him in line.
p5. To celebrate space exploration, the Science Fiction Museum is going to read Star Wars and Star Trek stories for 24 hours straight. A different story will be read each hour for a total of 12
Star Wars stories and 12 Star Trek stories. George and Gene want to listen to exactly 6 Star Wars and 6 Star Trek stories. Show that no matter how the readings are scheduled, the friends can find a block of 12 consecutive hours to listen to the stories together. Round 2
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p6. 2013 people attended Cinderella’s ball. Some of the guests were friends with each other. At midnight, the guests started turning into mice. After the first minute, everyone who had no friends at the ball turned into a mouse. After the second minute, everyone who had exactly one friend among the remaining people turned into a mouse. After the third minute, everyone who had two human friends left in the room turned into a mouse, and so on. What is the maximal number of people that could have been left at the ball after 2013 minutes?
p7. Bill and Charlie are playing a game on an infinite strip of graph paper. On Bill’s turn, he marks two empty squares of his choice (not necessarily adjacent) with crosses. Charlie, on his turn, can erase any number of crosses, as long as they are all adjacent to each other. Bill wants to create a line of 2013 crosses in a row. Can Charlie stop him? PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2014.57 Round 1
p1. Three snails – Alice, Bobby, and Cindy – were racing down a road. Whenever one snail passed another, it waved at the snail it passed. During the race, Alice waved 3 times and was waved at twice. Bobby waved 4 times and was waved at 3 times. Cindy waved 5 times. How many times was she waved at?
p2. Sherlock and Mycroft are playing Battleship on a 4 × 4 grid. Mycroft hides a single 3 × 1
cruiser somewhere on the board. Sherlock can pick squares on the grid and fire upon them. What is the smallest number of shots Sherlock has to fire to guarantee at least one hit on the cruiser?
p3. Thirty girls – 13 of them in red dresses and 17 in blue dresses – were dancing in a circle, hand-in-hand. Afterwards, each girl was asked if the girl to her right was in a blue dress. Only the girls who had both neighbors in red dresses or both in blue dresses told the truth. How many girls could have answered ”Yes”?
p4. Herman and Alex play a game on a 5 × 5 board. On his turn, a player can claim any open square as his territory. Once all the squares are claimed, the winner is the player whose territory has the longer border. Herman goes first. If both play their best, who will win, or will the game end in a draw?
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p5. Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them? Round 2
p6. Hermione and Ron play a game that starts with 129 hats arranged in a circle. They take turns magically transforming the hats into animals. On each turn, a player picks a hat and chooses whether to change it into a badger or into a raven. A player loses if after his or her turn there are two animals of the same species right next to each other. Hermione goes first. Who loses?
p7. Three warring states control the corner provinces of the island whose map is shown below.
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As a result of war, each of the remaining 18 provinces was occupied by one of the states. None of the states was able to occupy any province on the coast opposite their corner. The states would like to sign a peace treaty. To do this, they each must send ambassadors to a place where three provinces, one controlled by each state, come together. Prove that they can always find such a place to meet. For example, if the provinces are occupied as shown here, the squares mark possible meeting spots.
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PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2015.57 Round 1
p1. A party is attended by ten people (men and women). Among them is Pat, who always lies to people of the opposite gender and tells the truth to people of the same gender. Pat tells five of the guests: ”There are more men than women at the party.” Pat tells four of the guests: ”There are more women than men at the party.” Is Pat a man or a woman?
p2. Once upon a time in a land far, far away there lived 100 knights, 99 princesses, and 101 drag-ons. Over time, knights beheaded dragons, dragons ate princesses, and princesses poisoned knights. But they always obeyed an ancient law that prohibits killing any creature who has killed an odd number of others. Eventually only one creature remained alive. Could it have been a knight? A dragon? A princess?
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p3. The numbers 1 ◦ 2 ◦ 3 ◦ 4 ◦ 5 ◦ 6 ◦ 7 ◦ 8 ◦ 9 ◦ 10 are written in a line. Alex and Vicky play a game, taking turns inserting either an addition or a multiplication symbol between adjacent numbers. The last player to place a symbol wins if the resulting expression is odd and loses if it is even. Alex moves first. Who wins? (Remember that multiplication is performed before addition.)
p4. A chess tournament had 8 participants. Each participant played each other participant once. The winner of a game got 1 point, the loser 0 points, and in the case of a draw each got 1/2 apoint. Each participant scored a different number of points, and the person who got 2nd prize scored the same number of points as the 5th, 6th, 7th and 8th place participants combined. Can you determine the result of the game between the 3rd place player and the 5th place player?
p5. One hundred gnomes sit in a circle. Each gnome gets a card with a number written on one side and a different number written on the other side. Prove that it is possible for all the gnomes to lay down their cards so that no two neighbors have the same numbers facing up. Round 2
p6. A casino machine accepts tokens of 32 different colors, one at a time. For each color, the player can choose between two fixed rewards. Each reward is up to $10 cash, plus maybe an-other token. For example, a blue token always gives the player a choice of getting either $5 plus a red token or $3 plus a yellow token; a black token can always be exchanged either for $10 (but no token) or for a brown token (but no cash). A player may keep playing as long as he has a token. Rob and Bob each have one white token. Rob watches Bob play and win $500 . Prove that Rob can win at least $1000 .
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p7. Each of the 100 residents of Pleasantville has at least 30 friends in town. A resident votes in the mayoral election only if one of her friends is a candidate. Prove that it is possible to nominate two candidates for mayor so that at least half of the residents will vote. PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2016.67 Round 1
p1. At a fortune-telling exam, 13 witches are sitting in a circle. To pass the exam, a witch must correctly predict, for everybody except herself and her two neighbors, whether they will pass
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or fail. Each witch predicts that each of the 10 witches she is asked about will fail. How many witches could pass?
p2. Out of 152 coins, 7 are counterfeit. All counterfeit coins have the same weight, and all real coins have the same weight, but counterfeit coins are lighter than real coins. How can you find
19 real coins if you are allowed to use a balance scale three times?
p3. The digits of a number N increase from left to right. What could the sum of the digits of
9 × N be?
p4. The sides and diagonals of a pentagon are colored either blue or red. You can choose three vertices and flip the colors of all three lines that join them. Can every possible coloring be turned all blue by a sequence of such moves?
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p5. You have 100 pancakes, one with a single blueberry, one with two blueberries, one with three blueberries, and so on. The pancakes are stacked in a random order. Count the number of blue-berries in the top pancake and call that number N . Pick up the stack of the top N pancakes and flip it upside down. Prove that if you repeat this counting-and-flipping process, the pancake with one blueberry will eventually end up at the top of the stack. Round 2
p6. A circus owner will arrange 100 fleas on a long string of beads, each flea on her own bead. Once arranged, the fleas start jumping using the following rules. Every second, each flea chooses the closest bead occupied by one or more of the other fleas, and then all fleas jump simultane-ously to their chosen beads. If there are two places where a flea could jump, she jumps to the right. At the start, the circus owner arranged the fleas so that, after some time, they all gather on just two beads. What is the shortest amount of time it could take for this to happen?
p7. The faraway land of Noetheria has 2016 cities. There is a nonstop flight between every pair of cities. The price of a nonstop ticket is the same in both directions, but flights between different pairs of cities have different prices. Prove that you can plan a route of 2015 consecutive flights so that each flight is cheaper than the previous one. It is permissible to visit the same city several times along the way. PS. You should use hide for answers. Collected here (
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community/c5h2760506p24143309 ).
2017.67 Round 1
p1. Ten children arrive at a birthday party and leave their shoes by the door. All the children have different shoe sizes. Later, as they leave one at a time, each child randomly grabs a pair of shoes their size or larger. After some kids have left, all of the remaining shoes are too small for any of the remaining children. What is the greatest number of shoes that might remain by the door?
p2. Turans, the king of Saturn, invented a new language for his people. The alphabet has only 6
letters: A, N, R, S, T, U; however, the alphabetic order is different than in English. A word is any sequence of 6 different letters. In the dictionary for this language, the first word is SATURN. Which word follows immediately after TURANS?
p3. Benji chooses five integers. For each pair of these numbers, he writes down the pair’s sum. Can all ten sums end with different digits?
p4. Nine dwarves live in a house with nine rooms arranged in a 3×3 square. On Monday morning, each dwarf rubs noses with the dwarves in the adjacent rooms that share a wall. On Monday night, all the dwarves switch rooms. On Tuesday morning, they again rub noses with their adja-cent neighbors. On Tuesday night, they move again. On Wednesday morning, they rub noses for the last time. Show that there are still two dwarves who haven’t rubbed noses with one another.
p5. Anna and Bobby take turns placing rooks in any empty square of a pyramid-shaped board with 100 rows and 200 columns. If a player places a rook in a square that can be attacked by a previously placed rook, he or she loses. Anna goes first. Can Bobby win no matter how well Anna plays?
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Round 2
p6. Some boys and girls, all of different ages, had a snowball fight. Each girl threw one snowball at every kid who was older than her. Each boy threw one snowball at every kid who was younger than him. Three friends were hit by the same number of snowballs, and everyone else took fewer hits than they did. Prove that at least one of the three is a girl.
p7. Last year, jugglers from around the world travelled to Jakarta to participate in the Jubilant
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Juggling Jamboree. The festival lasted 32 days, with six solo performances scheduled each day. The organizers noticed that for any two days, there was exactly one juggler scheduled to perform on both days. No juggler performed more than once on a single day. Prove there was a juggler who performed every day. PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2018.67 Round 1
p1. Alice and Bob played 25 games of rock-paper-scissors. Alice played rock 12 times, scissors
6 times, and paper 7 times. Bob played rock 13 times, scissors 9 times, and paper 3 times. If there were no ties, who won the most games? (Remember, in each game each player picks one of rock, paper, or scissors. Rock beats scissors, scissors beat paper, and paper beats rock. If they choose the same object, the result is a tie.)
p2. On the planet Vulcan there are eight big volcanoes and six small volcanoes. Big volcanoes erupt every three years and small volcanoes erupt every two years. In the past five years, there were 30 eruptions. How many volcanoes could erupt this year?
p3. A tangle is a sequence of digits constructed by picking a number N ≥ 0 and writing the integers from 0 to N in some order, with no spaces. For example, 010123459876 is a tangle with
N = 10 . A palindromic sequence reads the same forward or backward, such as 878 or 6226 . The shortest palindromic tangle is 0. How long is the second-shortest palindromic tangle?
p4. Balls numbered 1 to N have been randomly arranged in a long input tube that feeds into the upper left square of an 8 × 8 board. An empty exit tube leads out of the lower right square of the board. Your goal is to arrange the balls in order from 1 to N in the exit tube. As a move, you may 1. move the next ball in line from the input tube into the upper left square of the board, 2. move a ball already on the board to an adjacent square to its right or below, or 3. move a ball from the lower right square into the exit tube. No square may ever hold more than one ball. What is the largest number N for which you can achieve your goal, no matter how the balls are initially arranged? You can see the order of the balls in the input tube before you start.
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p5. A 2018 × 2018 board is covered by non-overlapping 2 × 1 dominoes, with each domino cover-ing two squares of the board. From a given square, a robot takes one step to the other square of the domino it is on and then takes one more step in the same direction. Could the robot continue
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moving this way forever without falling off the board?
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Round 2
p6. Seventeen teams participated in a soccer tournament where a win is worth 1 point, a tie is worth 0 points, and a loss is worth −1 point. Each team played each other team exactly once. At least 34 of all games ended in a tie. Show that there must be two teams with the same number of points at the end of the tournament.
p7. The city of Old Haven is known for having a large number of secret societies. Any person may be a member of multiple societies. A secret society is called influential if its membership includes at least half the population of Old Haven. Today, there are 2018 influential secret soci-eties. Show that it is possible to form a council of at most 11 people such that each influential secret society has at least one member on the council. PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2019.67 Round 1
p1. Three two-digit numbers are written on a board. One starts with 5, another with 6, and the last one with 7. Annie added the first and the second numbers; Benny added the second and the third numbers; Denny added the third and the first numbers. Could it be that one of these sums is equal to 148 , and the two other sums are three-digit numbers that both start with 12 ?
p2. Three rocks, three seashells, and one pearl are placed in identical boxes on a circular plate in the order shown. The lids of the boxes are then closed, and the plate is secretly rotated. You can open one box at a time. What is the smallest number of boxes you need to open to know where the pearl is, no matter how the plate was rotated?
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p3. Two detectives, Holmes and Watson, are hunting the thief Raffles in a library, which has the floorplan exactly as shown in the diagram. Holmes and Watson start from the center room marked D. Show that no matter where Raffles is or how he moves, Holmes and Watson can find him. Holmes and Watson do not need to stay together. A detective sees Raffles only if they are in the same room. A detective cannot stand in a doorway to see two rooms at the same time.
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p4. A museum has a 4 × 4 grid of rooms. Every two rooms that share a wall are connected by a door. Each room contains some paintings. The total number of paintings along any path of
7 rooms from the lower left to the upper right room is always the same. Furthermore, the total number of paintings along any path of 7 rooms from the lower right to the upper left room is always the same. The guide states that the museum has exactly 500 paintings. Show that the guide is mistaken.
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p5. The numbers 1–14 are placed around a circle in some order. You can swap two neighbors if they differ by more than 1. Is it always possible to rearrange the numbers using swaps so they are ordered clockwise from 1 to 14 ?Round 2
p6. A triangulation of a regular polygon is a way of drawing line segments between its vertices so that no two segments cross, and the interior of the polygon is divided into triangles. A flip move erases a line segment between two triangles, creating a quadrilateral, and replaces it with the opposite diagonal through that quadrilateral. This results in a new triangulation.
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Given any two triangulations of a polygon, is it always possible to find a sequence of flip moves that transforms the first one into the second one?
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p7. Is it possible to place the numbers from 1 to 121 in an 11 ×11 table so that numbers that differ by 1 are in horizontally or vertically adjacent cells and all the perfect squares (1 , 4, 9, ..., 121) are in one column? PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2022.67 Round 1
p1. Nineteen witches, all of different heights, stand in a circle around a campfire. Each witch
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says whether she is taller than both of her neighbors, shorter than both, or in-between. Exactly three said ”I am taller.” How many said ”I am in-between”?
p2. Alex is writing a sequence of A’s and B’s on a chalkboard. Any 20 consecutive letters must have an equal number of A’s and B’s, but any 22 consecutive letters must have a different number of A’s and B’s. What is the length of the longest sequence Alex can write?.
p3. A police officer patrols a town whose map is shown. The officer must walk down every street segment at least once and return to the starting point, only changing direction at intersections and corners. It takes the officer one minute to walk each segment. What is the fastest the officer can complete a patrol?
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p4. A zebra is a new chess piece that jumps in the shape of an ”L” to a location three squares away in one direction and two squares away in a perpendicular direction. The picture shows all the moves a zebra can make from its given position. Is it possible for a zebra to make a sequence of 64 moves on an 8 × 8 chessboard so that it visits each square exactly once and returns to its starting position?
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p5. Ann places the integers 1, 2, ..., 100 in a 10 × 10 grid, however she wants. In each round, Bob picks a row or column, and Ann sorts it from lowest to highest (left-to-right for rows; top-to-bottom for columns). However, Bob never sees the grid and gets no information from Ann. After eleven rounds, Bob must name a single cell that is guaranteed to contain a number that is at least 30 and no more than 71 . Can he find a strategy to do this, no matter how Ann originally arranged the numbers? Round 2
p6. Evelyn and Odette are playing a game with a deck of 101 cards numbered 1 through 101 . At the start of the game the deck is split, with Evelyn taking all the even cards and Odette taking all the odd cards. Each shuffles her cards. On every move, each player takes the top card from her deck and places it on a table. The player whose number is higher takes both cards from the table and adds them to the bottom of her deck, first the opponent’s card, then her own. The first player to run out of cards loses. Card 101 was played against card 2 on the 10 th move. Prove that this game will never end.
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p7. The Vogon spaceship Tempest is descending on planet Earth. It will land on five adjacent buildings within a 10 ×10 grid, crushing any teacups on roofs of buildings within a 5×1 length of blocks (vertically or horizontally). As Commander of the Space Force, you can place any number of teacups on rooftops in advance. When the ship lands, you will hear how many teacups the spaceship breaks, but not where they were. (In the figure, you would hear 4 cups break.) What is the smallest number of teacups you need to place to ensure you can identify at least one building the spaceship landed on?
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PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2023.67 Round 1
p1. Ash is running around town catching Pok ´ emon. Each day, he may add 3, 4, or 5 Pok ´ emon to his collection, but he can never add the same number of Pok ´ emon on two consecutive days. What is the smallest number of days it could take for him to collect exactly 100 Pok ´ emon?
p2. Jack and Jill have ten buckets. One bucket can hold up to 1 gallon of water, another can hold up to 2 gallons, and so on, with the largest able to hold up to 10 gallons. The ten buckets are arranged in a line as shown below. Jack and Jill can pour some amount of water into each bucket, but no bucket can have less water than the one to its left. Is it possible that together, the ten buckets can hold 36 gallons of water?
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p3. There are 2023 knights and liars standing in a row. Knights always tell the truth and liars always lie. Each of them says, ”the number of liars to the left of me is greater than the number of knights to the right.” How many liars are there?
p4. Camila has a deck of 101 cards numbered 1, 2, ..., 101 . She starts with 50 random cards in her hand and the rest on a table with the numbers visible. In an exchange, she replaces all 50
cards in her hand with her choice of 50 of the 51 cards from the table. Show that Camila can make at most 50 exchanges and end up with cards 1, 2, ..., 50 .
png
p5. There are 101 pirates on a pirate ship: the captain and 100 crew. Each pirate, including the captain, starts with 1 gold coin. The captain makes proposals for redistributing the coins, and the crew vote on these proposals. The captain does not vote. For every proposal, each crew member greedily votes ”yes” if he gains coins as a result of the proposal, ”no” if he loses coins,
©2025 AoPS Incorporated
15 AoPS Community Math Hour Olympiad, Grades 5-7
and passes otherwise. If strictly more crew members vote ”yes” than ”no,” the proposal takes effect. The captain can make any number of proposals, one after the other. What is the largest number of coins the captain can accumulate? Round 2
p6. The town of Lumenville has 100 houses and is preparing for the math festival. The Tesla wiring company will lay lengths of power wire in straight lines between the houses so that power flows between any two houses, possibly by passing through other houses. The Edison lighting company will hang strings of lights in straight lines between pairs of houses so that each house is connected by a string to exactly one other. Show that however the houses are arranged, the Edison company can always hang their strings of lights so that the total length of the strings is no more than the total length of the power wires the Tesla company used.
png
p7. You are given a sequence of 16 digits. Is it always possible to select one or more digits in a row, so that multiplying them results in a square number?
png
PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2024.67 Round 1
p1. There are 2024 parakeets in the Rainbow Forest. Each one either always tells the truth or always lies. Four of the parakeets say: Polly: ”There are no blue parakeets in the forest.” Molly: ”Polly is blue.” Dolly: ”Polly is not blue.” Holly: ”There are more liars than truth-tellers among the four of us.” Are there any blue parakeets in the forest?
p2. Arturo needs to move six paintings by different artists between floors in a museum: • Paul’s painting: Floor 4 to Floor 9 • Yayoi’s painting: Floor 2 to Floor 7 • Frida’s painting: Floor 4 to Floor 1
• Rene’s painting: Floor 2 to Floor 4 • Leonardo’s painting: Floor 7 to Floor 4 • Georgia’s painting: Floor 8 to Floor 2
The paintings can be moved in any order, but the elevator can only hold one painting at a time. Arturo starts and ends on Floor 1.What is the smallest number of floors he can travel and deliver all the paintings to their desired destinations? (For example, if he takes the elevator from Floor
1 to Floor 7,then from Floor 7 to Floor 3, he has traveled (7 − 1) + (7 − 3) = 10 floors.)
p3. There are 100 lights in a circle, all initially turned off. You can turn on lights one at a time. If you turn on a light and both of its neighbors are already on, you must turn off exactly one of
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16 AoPS Community Math Hour Olympiad, Grades 5-7
its neighbors immediately. If only one neighbor is on, then that neighbor must immediately be turned off. What is the maximum number of lights that can be turned on at the same time?
p4. The squares of a 101 ×101 dance floor must be colored red, green, and blue. No two squares that share a side are allowed to have the same color. Cinderella suggests a coloring which al-ternates red, green, and blue diagonals as shown. Drizella suggests a coloring that is different from Cinderella’s at each of the 400 squares on the perimeter. Show that Drizella’s coloring is different from Cinderella’s at every square.
png
p5. Pirate Jim has 10 chests that all contain different numbers of gold coins. In total there are
1000 coins. Every hour, Pirate Jim takes 9 coins from the chest that contains the largest number of coins and puts one coin into each of the other chests. Pirate Jim must stop if any two chests have the same number of coins. Is it possible that Pirate Jim will never stop moving coins? Round 2
p6. A group of children were playing ”king of the tennis court.” To start, one child is named king, and the rest form a line. The first player in line challenges the king to a tennis game. The winner becomes the new king, the loser goes to the back of the line, and the process repeats. The group played a total of 75 games. Naomi won 12 games and lost 8. Prove that Naomi was not the king after the last game.
p7. There are 21 cards, each labeled with a positive integer. The cards are arranged face up in
10 piles of two and 1 pile of one. All 21 card numbers are always visible. Alice and Bob alternate turns picking cards. On their turn, a player must take one card that is currently on top of its pile. Once all 21 cards are taken, the player whose cards have the larger sum wins. Alice takes the first card. Prove that Alice can always win.
png
PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
2025.67 Round 1
p1. There were 11 people standing in a line. The order in the line got mixed up a bit, but each person ended up within one spot of their original position. (For example, the person who started in the 6th position ended up in either 5th, 6th, or 7th position.) Show that at least one person bends up in their original position in line.
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17 AoPS Community Math Hour Olympiad, Grades 5-7
p2. There are 19 people standing in a circle, facing the center. Each person is either a knight or a liar. Knights always tell the truth and liars always tell lies. Each person says, ”Out of the four people to the right of me, at least two are knights.” How many knights could there be in the circle?
p3. A piece of graph paper extends forever in all directions. Each square contains one of the four digits 1, 2, 3, or 4. Each digit is used at least once. Two squares are called neighbors if they share a side. A square is called balanced if its digit is equal to the number of different digits in its four neighbors. (In the example shown, the blue square is balanced but the orange one is not. Both have three different digits as neighbors.) Is it possible that all squares can be balanced?
png
p4. Ten tennis players compete in a tournament that lasts 9 days. Each person plays one game per day. The winner of each game gets 1 point, and the loser gets 0 points. Ties are not allowed. After 5 days, Daphne is in second place: exactly one player has more points than she does, and the rest of the players have fewer points than Daphne. Prove that it is impossible for Daphne to have fewer points than each of the other players at the end of the tournament.
p5. Lily and Mark are playing a game. Each of them has an army of soldiers. They are competing to capture 5 castles. The players secretly send all of their soldiers to the castles. Each soldier goes to exactly one castle. The player who sends more soldiers to a castle captures it. In case of a tie, nobody captures the castle. In the example shown, Mark captures castles A and D, Lily captures castle C, and nobody captures castle B or E. Lily has 10 soldiers in her army. What is the smallest number of soldiers Mark needs in his army to guarantee he can capture at least 3
castles, no matter how Lily places her army?
png
Round 2
p6. A jeweler has six golden rings. He knows they weigh 22, 23, 24, 32, 34, and 36 grams. The rings look identical. The jeweler’s apprentice labeled each ring with one of these weights. By us-ing a balance scale twice, can the jeweler check if his apprentice labeled all the rings correctly?
p7. In the faraway land of Artinia, all roads are one-way roads. According to the law of the land, for any two cities A and B, if there is a one-way road from A to B, then a one-way road from B to
A is not allowed. The Ministry of Transport wants a budget to build one more road. They notice that no matter where the road is built, it will be possible to drive from any city to any other city. Prove that it is already possible to drive between any pair of cities in Artinia without adding the new road.
©2025 AoPS Incorporated
18 AoPS Community Math Hour Olympiad, Grades 5-7
PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ).
©2025 AoPS Incorporated
19
Art of Problem Solving is an ACS WASC Accredited School. |
3588 | https://www.mayocliniclabs.com/test-catalog/download-setup?format=pdf&unit_code=607495 | Test Definition: MEV1
Methemoglobinemia Evaluation, Blood
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Document generated September 28, 2025 at 10:33 PM CT Page 1 of 8
Overview
Useful For
Diagnosis of methemoglobinemia and sulfhemoglobinemia and possible hereditary (congenital) causes
Differentiation of methemoglobinemia and sulfhemoglobinemia from other causes of cyanosis (eg, congenital heart
disease)
Profile Information
Test Id Reporting Name Available Separately Always Performed
MEVI Methemoglobinemia
Interpretation
No Yes
HGBCE Hb Variant, A2 and F
Quantitation,B
Yes Yes
HPLC HPLC Hb Variant, B No Yes
METH Methemoglobin, B Yes, (Order MET) Yes
SULF Sulfhemoglobin, B Yes, (Order MET) Yes
METR1 Cytochrome b5 Reductase,
B
Yes Yes
Reflex Tests
Test Id Reporting Name Available Separately Always Performed
SDEX Sickle Solubility, B Yes No
IEF Isoelectric Focusing, B No No
MASS Hb Variant by Mass Spec, B No No
UNHB Hb Stability, B No No
HPFH Hb F Distribution, B No No
WASQR Alpha Globin Gene
Sequencing, B
Yes, (Order WASEQ) No
WBSQR Beta Globin Gene
Sequencing, B
Yes, (Order WBSEQ) No
WGSQR Gamma Globin Full Gene
Sequencing
Yes, (Order WGSEQ) No
MEV0 Methemoglobin Summary
Interp
No No
WAGDR Alpha Globin Clustr Locus
Del/Dup,B
Yes, (Order AGDD) No
WBGDR Beta Globin Gene Cluster,
Del/Dup,B
Yes, (Order WBGDD) No Test Definition: MEV1
Methemoglobinemia Evaluation, Blood
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Document generated September 28, 2025 at 10:33 PM CT Page 2 of 8
Testing Algorithm
This is a consultative evaluation in which the case will be evaluated at Mayo Clinic Laboratories, the appropriate tests
performed at an additional charge, and the results interpreted. This is an evaluation for methemoglobin and
sulfhemoglobin levels and possible hereditary causes. Methemoglobin, sulfhemoglobin levels, cytochrome-b5 reductase
(methemoglobin reductase) activity, and protein analysis screening for hemoglobin variants (capillary electrophoresis,
cation exchange high performance liquid chromatography and capillary electrophoresis) will always be performed. If
additional hemoglobin variant confirmatory testing is required, appropriate reflex testing will be performed. This will
vary from additional protein analysis methods to molecular testing, as needed.
One or more of the following molecular tests may be reflexed:
-WAGDR / Alpha Globin Cluster Locus Deletion/Duplication, Blood
-WASQR / Alpha-Globin Gene Sequencing, Blood
-WBSQR / Beta-Globin Gene Sequencing, Blood
-WBGDR / Beta-Globin Gene Cluster Deletion/Duplication, Blood
-WGSQR / Gamma-Globin Full Gene Sequencing, Varies
After all test results are finalized, an additional consultative interpretation that summarizes all testing and incorporates
subsequent genetic results will be provided.
For more information see Benign Hematology Evaluation Comparison .
Special Instructions
• Informed Consent for Genetic Testing
• Metabolic Hematology Patient Information
• Benign Hematology Evaluation Comparison
• Informed Consent for Genetic Testing (Spanish)
Method Name
MEVI, MEV0: Medical Interpretation
HGBCE: Capillary Electrophoresis
HPLC: Cation Exchange/High-Performance Liquid Chromatography (HPLC)
METH, SULF: Spectrophotometry (SP)
METR1: Kinetic Spectrophotometry
IEF: Isoelectric Focusing
HPFH: Flow Cytometry
UNHB: Isopropanol and Heat Stability
MASS: Mass Spectrometry (MS)
NY State Available
Yes
Specimen
Specimen Type Test Definition: MEV1
Methemoglobinemia Evaluation, Blood
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Whole Blood ACD-B
Whole Blood EDTA
Shipping Instructions
Specimen must arrive within 3 days (72 hours) of collection.
Necessary Information
Include recent transfusion information.
Include most recent complete blood cell count results.
Metabolic Hematology Patient Information (T810) is strongly recommended . Testing may proceed without this
information, however if the information requested is received, any pertinent reported clinical features and data will
drive the focus of the evaluation and be considered in the interpretation.
The laboratory has extensive experience in hemoglobin variant identification and many cases can be confidently
classified without molecular testing. However, molecular confirmation is always available, subject to sufficient sample
quantity (eg, multiplex ligation-dependent probe amplification testing requires at least 2 mL of sample in addition to
protein testing requirements). If no molecular testing or specific molecular tests are desired, utilize the appropriate
check boxes on the form. If the form or other communication is not received, the reviewing hematopathologist will
select appropriate tests to sufficiently explain the protein findings, which may or may not include molecular testing.
Specimen Required
The following specimens are required for testing:
Whole blood ACD-B specimen
2 Whole blood EDTA specimens
Container/Tube: Lavender top (EDTA) and yellow top (ACD solution B)
Specimen Volume:
EDTA: Two 4 mL tubes
ACD: One 6 mL tube
Collection Instructions: Send whole blood specimen in original tube. Do not aliquot.
Forms
New York Clients-Informed consent is required. Document on the request form or electronic order that a copy is on
file. The following documents are available:
-Informed Consent for Genetic Testing (T576)
-Informed Consent for Genetic Testing-Spanish (T826)
Metabolic Hematology Patient Information (T810)
If not ordering electronically, complete, print, and send a Benign Hematology Test Request (T755) with the specimen
Specimen Minimum Volume
EDTA blood: 3 mL ; ACD blood: 2.7 mL
Reject Due To Test Definition: MEV1
Methemoglobinemia Evaluation, Blood
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Document generated September 28, 2025 at 10:33 PM CT Page 4 of 8
Gross
hemolysis
Reject
Specimen Stability Information
Specimen Type Temperature Time Special Container
Whole Blood ACD-B Refrigerated 72 hours
Whole Blood EDTA Refrigerated 72 hours
Clinical & Interpretive
Clinical Information
Methemoglobin:
Methemoglobin forms when the hemoglobin (Hb) molecule iron is in the ferric (Fe3[+]) form instead of the functional
ferrous (Fe2[+]) form. Methemoglobinemia can be hereditary or acquired and is present by definition when
methemoglobin levels are greater than the normal range. Acquired methemoglobinemia results after toxic exposure to
nitrates and nitrites/nitrates (fertilizer, nitric oxide), topical anesthetics ("caines"), dapsone, naphthalene (moth
balls/toilet deodorant cakes), and industrial use of aromatic compounds (aniline dyes).
Congenital methemoglobinemias are rare. They are due either to:
-A deficiency of cytochrome b5 reductase (methemoglobin reductase) in erythrocytes, an autosomal recessive disorder
resulting from genetic variants in either CYB5R3 or CYB5A .(1,2) Type IV is thought to be extraordinarily rare. Type III is no
longer a category.
-One of several intrinsic structural disorders of Hb, called M-Hbs; all of which are inherited in an autosomal dominant
manner.(3,4) Classically, M-Hbs result from histidine-to tyrosine substitutions at the proximal or distal histidine
important in coordinating the oxygen molecule. These include alpha-, beta- and gamma-chain variants. Rarely, other
substitutions outside the proximal and distal histidine location can cause Hb variants that increase methemoglobin or
sulfhemoglobin levels. Most M-Hb variants are readily identified by h igh performance liquid chromatography (HPLC) or
mass spectrometry methods with characteristic electrophoresis patterns; however, some require more specialized
techniques. Most are associated with increased methemoglobin with or without an increase in sulfhemoglobin. Alpha
chain M-Hb variants can be associated with increased sulfhemoglobin without an increase in methemoglobin.
Sulfhemoglobin:
Sulfhemoglobin cannot combine with oxygen. When acquired, sulfhemoglobinemia can be associated with cyanosis and
often accompanies methemoglobinemia. Sulfhemoglobinemia has been associated with exposure to sumatriptan,
sulfonamides, metoclopramide, paint or varnish vapors, dimethyl sulfoxide, acetanilide, phenacetin, trinitroluene, zinc
ethylene bisdithiocarbamate (a fungicide), and flutamide. It is important to note that some Hb variants are known to
interfere with this test (especially M-Hbs) and sulfhemoglobin absorbance can be increased due to the Hb variant. Hb
evaluation that includes the HPLC method is recommended to exclude this possibility.
In contrast to methemoglobinemia, sulfhemoglobinemia persists until the erythrocytes containing it are destroyed.
Therefore, blood level of sulfhemoglobin declines gradually over a period of weeks. Test Definition: MEV1
Methemoglobinemia Evaluation, Blood
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Document generated September 28, 2025 at 10:33 PM CT Page 5 of 8
Reference Values
Definitive results and an interpretive report will be provided.
Interpretation
This is a consultative evaluation in which the history and previous laboratory values are reviewed by a hematologist who
is an expert on these disorders. Appropriate tests are performed, and an interpretive report is issued.
Cautions
Sulfhemoglobin is exceedingly stable and does not change in stored or shipped specimens.
Methemoglobin is unstable and can degrade at a rate of about 40% per 24 hours.
A normal methemoglobin value obtained with stored or shipped specimens does not exclude prior methemoglobinemia
of minimal degree. However, significant methemoglobinemia will still be demonstrable.
Clinical Reference
OMIM: 250800 Methemoglobinemia due to deficiency of methemoglobin reductase. Updated May 20, 2019.
Accessed January 23, 2024. Available at www.omim.org/entry/250800?search=250800&highlight=250800
OMIM: 250790 Methemoglobinemia and ambiguous genitalia. Updated December 9, 2022. Accessed January 23,
Available at www.omim.org/entry/250790?search=250790&highlight=250790
OMIM: 141800 Hemoglobin alpha locus 1; HBA1. Updated September 15, 2023. Accessed January 23, 2024. Available
at www.omim.org/entry/141800?search=141800&highlight=141800
OMIM: 141900 Hemoglobin beta locus; HBB. Updated September 15, 2023. Accessed January 23, 2024. Available at
www.omim.org/entry/141900?search=141900&highlight=141900
Haymond S, Cariappa R, Eby CS, Scott MG. Laboratory assessment of oxygenation in methemoglobinemia. Clin Chem.
2005;51(2):434-444
Noor M, Beutler E. Acquired sulfhemoglobinemia. An underreported diagnosis? West J Med. 1998;169(6):386-389
Thom CS, Dickson CF, Gell DA, Weiss MJ. Hemoglobin variants: biochemical properties and clinical correlates. Cold
Spring Harb Perspect Med. 2013;3(3):a011858
Percy MJ, McFerran NV, Lappin TR. Disorders of oxidized haemoglobin. Blood Rev. 2005;19(2):61-68
Agarwal AM, Prchal JT. Methemoglobinemia and Other Dyshemoglobinemias. In: Kaushansky K, Lichtman MA, Prchal
JT, Levi MM, Press OW, Burns LJ, Caligiuri M, eds. Williams Hematology. 9th ed. McGraw-Hill; 2016: 789-800
Performance
Method Description
The CAPILLARYS System is an automated system that uses capillary electrophoresis to separate charged molecules by
their electrophoretic mobility in an alkaline buffer. Separation occurs according to the electrolyte pH and
electro-osmotic flow. A sample dilution with hemolyzing solution is injected by aspiration. A high-voltage protein
separation occurs and direct detection of the hemoglobin (Hb) protein fractions is at 415 nm, which is specific to Hbs.
The resulting electrophoregrams peaks are evaluated for pattern abnormalities and are quantified as a percentage of
the total Hb present. Examples of position of commonly found Hb fractions are, from cathode to anode: Hb A2', C,
A2/O-Arab, E, S, D, G-Philadelphia, F, A, Hope, Bart, J, N-Baltimore, and H.(Louahabi A, Philippe M, Lali S, Wallemacq P,
Maisin D. Evaluation of a new Sebia kit for analysis of hemoglobin fractions and variants on the Capillarys system. Clin Test Definition: MEV1
Methemoglobinemia Evaluation, Blood
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Document generated September 28, 2025 at 10:33 PM CT Page 6 of 8
Chem Lab Med. 2006;44:340-345; instruction manual: CAPILLARYS Hemoglobin(E) using the CAPILLARYS 2
flex-piercing instrument. Sebia; 06/2014)
High Performance Liquid Chromatography Hemoglobin Variant:
Hemolysate of whole blood is injected into an analysis stream passing through a cation exchange column using high
performance liquid chromatography. A preprogrammed gradient controls the elution buffer mixture that also passes
through the analytical cartridge. The ionic strength of the elution buffer is raised by increasing the percentage of a
second buffer. As the ionic strength of the buffer increases the more strongly retained Hbs elute from the cartridge.
Absorbance changes are detected by a dual-wavelength filter photometer. Changes in absorbance are displayed as a
chromatogram of absorbance versus time.(Huismann TH, Schroeder WA, Brodie AN, Mayson SM, Jakway J.
Microchromotography of hemoglobins. III. A simplified procedure for the determination of hemoglobin A2. J Lab Clin
Med. 1975;86:700-702; Ou CN, Buffone GJ, Reimer GL, Alpert AJ. High-performance liquid chromatography of human
hemoglobins on a new cation exchanger. J Chromatogr. 1983;266:197-205; instruction manual: Bio-Rad Variant II
Beta-thalassemia Short Program Instructions for Use, L70203705. Bio-Rad Laboratories, Inc; 11/2011)
Methemoglobin:
The normal absorption spectrum of oxyhemoglobin has very little optical density above 600 nm. The absorption
spectrum of methemoglobin exhibits a small, characteristic peak at 630 nm. This peak is abolished as methemoglobin is
converted to cyanmethemoglobin upon addition of potassium cyanide, and the drop in optical density is proportional to
methemoglobin concentration.(Evelyn KA, Malloy HT. Microdetermination of oxyhemoglobin, methemoglobin, and
sulfhemoglobin in a single sample of blood. J Biol Chem. 1938;126:655-662; Fairbanks VF, Klee GG. Biochemical aspects
of hematology. In: Burtis CA, Ashwood ER, eds. Tietz Textbook of Clinical Chemistry. WB Saunders Company;
1999:1676-1678; Robertson LD, Roper D. Laboratory methods used in the investigation of haemolytic anaemias. In: Bain
BJ, Bates I, Laffan MA, eds. Dacie and Lewis Practical Haematology. 12th ed. Elsevier; 2017:214-227)
Sulfhemoglobin:
The normal absorption spectrum of oxyhemoglobin has very little optical density above 600 nm. However, if certain
poorly defined Hb denaturation products are present in a hemolysate, there is a broad elevation of the absorption curve
in the range of 600 to 620 nm. This sulfhemoglobin plateau is not affected by treatment with cyanide. Sulfhemoglobin is
not available, nor can it be prepared, in a pure form for preparation of a sulfhemoglobin standard. In calculating
sulfhemoglobin concentration, the factor for sulfhemoglobin quantitation is based on studies of Carrico et al.(Evelyn KA,
Malloy HT. Microdetermination of oxyhemoglobin, methemoglobin, and sulfhemoglobin in a single sample of blood. J
Biol Chem. 1938;126:655-662; Carrico RJ, Peisach J, Alben JO. The preparation and some physical properties of
sulfhemoglobin. J Analyt Biochem. 1978;253:2386-2391; Fairbanks VF, Klee GG. Biochemical aspects of hematology. In:
Burtis CA, Ashwood ER, eds. Tietz Textbook of Clinical Chemistry. WB Saunders Company; 1999:1676-1678; Robertson
LD, Roper D. Laboratory methods used in the investigation of haemolytic anaemias. In: Bain BJ, Bates I, Laffan MA, eds.
Dacie and Lewis Practical Haematology. 12th ed. Elsevier; 2017:214-227)
Cytochrome b5 Reductase:
Cytochrome B5 reductase (methemoglobin reductase) catalyzes the 1,4-dihydronicotinamide adenine dinucleotide
(NADH)-linked reduction of several substrates, including ferricyanide. The activity at 30 degrees C is followed
spectrophotometrically by measuring the oxidation of NADH at 340 nm.(Fairbank VF, Klee GG. Biochemical aspects of
hematology. In: Burtis CA, Ashwood ER, eds. Tietz Textbook of Clinical Chemistry. 3rd ed. WB Saunders Company;
1999:1647-1648; van Solinge WW, van Wijk. Enzymes of the red blood cell. In: Rifai N, Horvath AR, Wittwer CT: eds.
Tietz Textbook of Clinical Chemistry and Molecular Diagnostics. 6th ed. Elsevier; 2018:chap 30) Test Definition: MEV1
Methemoglobinemia Evaluation, Blood
_________
_________
Document generated September 28, 2025 at 10:33 PM CT Page 7 of 8
PDF Report
No
Day(s) Performed
Monday through Saturday
Report Available
3 to 25 days
Specimen Retention Time
28 days
Performing Laboratory Location
Mayo Clinic Laboratories - Rochester Main Campus
Fees & Codes
Fees
Authorized users can sign in to Test Prices for detailed fee information.
Clients without access to Test Prices can contact Customer Service 24 hours a day, seven days a week.
Prospective clients should contact their account representative. For assistance, contact Customer Service .
Test Classification
This test was developed and its performance characteristics determined by Mayo Clinic in a manner consistent with CLIA
requirements. It has not been cleared or approved by the US Food and Drug Administration.
CPT Code Information
83020-26-Hemoglobinopathy Interpretation
83020-Hb Variant, A2 and F Quantitation
83021-HPLC Hb Variant
82657-Methemoglobin reductase
83050-Methemoglobin, quantitative
83060-Sulfhemoglobin, quantitative
82664 (if appropriate)
83068 (if appropriate)
83789 (if appropriate)
88184 (if appropriate)
LOINC ® Information
Test ID Test Order Name Order LOINC® Value
MEV1 Methemoglobinemia Evaluation In Process
Result ID Test Result Name Result LOINC® Value Test Definition: MEV1
Methemoglobinemia Evaluation, Blood
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Document generated September 28, 2025 at 10:33 PM CT Page 8 of 8
8268 Methemoglobin, B 2614-6
8272 Sulfhemoglobin, B 4685-4
41927 Hb A 20572-4
41928 Hb F 32682-7
41929 Hb A2 4552-6
41930 Variant 1 24469-9
41931 Variant 2 24469-9
41932 Variant 3 24469-9
41933 HGBCE Interpretation 78748-1
65615 HPLC Hb Variant, B No LOINC Needed
METRB Cytochrome b5 Reductase, B 32703-1
608086 Methemoglobinemia Interpretation 59465-5
608108 Reviewed By 18771-6 |
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Carotid Artery Pulse
In subject area:Nursing and Health Professions
The carotid artery pulse is defined as the palpable rhythm of blood flow in the carotid artery, graded from 0 (absence of pulse) to 4+ (forceful pulse), indicating varying strengths of the pulse.
AI generated definition based on: Clinical Procedures in Primary Eye Care (Third Edition), 2007
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Chapters and Articles
You might find these chapters and articles relevant to this topic.
Chapter
Pulse Rate and Contour
2018, Evidence-Based Physical Diagnosis (Fourth Edition)Steven McGee MD
IVPulsus Parvus ET Tardus
AThe Finding and Technique
Pulsus parvus et tardus describes a carotid pulse with a small volume (pulsus parvus) that rises slowly and has a delayed systolic peak (pulsus tardus; see Fig. 15.1).22 It is routinely detected by palpation.
BClinical Significance
Pulsus parvus et tardus is a finding of aortic stenosis. Of its two components, pulsus tardus is the better discriminator, detecting severe aortic stenosis with a sensitivity of 31% to 91%, specificity of 68% to 93%, positive LR of 3.5, and negative LR of 0.4 (see Chapter 44).
CPathogenesis
Pulsus tardus depends on both obstruction to flow and the compliance of the vessel distal to the obstruction. The pulse waveform rises rapidly in stiff vessels but slowly in more compliant vessels that act like low-pass filters and remove the high frequency components of the waveform.68 That the delay in the pulse reflects the severity of obstruction is a principle also used by Doppler sonography to gauge the severity of renal artery stenosis.68
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Read full chapter
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Book2018, Evidence-Based Physical Diagnosis (Fourth Edition)Steven McGee MD
Chapter
Aortic Stenosis
2007, Evidence-Based Physical Diagnosis (Second Edition)
BASSOCIATED CARDIAC SIGNS
Other traditional findings of severe aortic stenosis are (1) a carotid pulse that is abnormally small in volume and delayed (“pulsus parvus et tardus”); (2) a palpable apical impulse that is abnormally sustained (see Chapter 34 for definition of sustained impulse); and (3) reduced intensity of the second heart sound, which occurs because the inflexible aortic leaflets close with less force than normal. Another traditional finding is a prominent A wave in the neck veins (i.e., the “Bernheim phenomenon”), although this wave is more often seen on pressure tracings than at the bedside. Its mechanism is still disputed.4
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Read full chapter
URL:
Book2007, Evidence-Based Physical Diagnosis (Second Edition)
Chapter
History and physical examination
2007, Practice of Geriatrics (Fourth Edition)Edmund H. DuthieJr. M.D.
Cardiovascular examination
Examination of the blood pressure, pulse, neck veins, and carotid pulse has already been reviewed. Frequently, the apical impulse and point of maximal intensity are difficult to locate in a geriatric patient. Palpable thrills, especially over the aortic area, should be sought because of the frequency of systolic murmurs. Splitting of the second heart sound may be difficult to detect in older patients. The presence of a third heart sound is not physiologic in elderly patients, as it is in young adults. Debate exists about whether a fourth heart sound may be accepted as normal in aged patients. Because heart disease is so common in older persons, it is not surprising that fourth heart sounds are frequently reported. This does not mean, however, that a fourth heart sound is the inevitable consequence of aging; rather, it reflects the high prevalence of cardiac disease in the geriatric population.
Systolic heart murmurs have been reported in as many as one third to one half of octogenarians. These murmurs may be due to aortic sclerosis, aortic stenosis, mitral regurgitation from numerous causes, mitral valve prolapse, hypertrophic obstructive cardiomyopathy, tricuspid regurgitation, or atrial septal defect. Clinicians examining geriatric patients should, therefore, expect to hear systolic heart murmurs often and be prepared to assess patients further through maneuvers and associated findings to determine the cause of the murmurs. “Innocent” murmurs, described in children or young adults, are not found in the geriatric age group. Valvular pathology and cardiac dysfunction are the likely explanations of a murmur in a geriatric patient.
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Book2007, Practice of Geriatrics (Fourth Edition)Edmund H. DuthieJr. M.D.
Chapter
A worldwide yearly survey of new data in adverse drug reactions
2015, Side Effects of Drugs AnnualA. Nobili, ... R. Latini
Periportal Edema
A 41-year-old man with a history of diabetes mellitus and hypertension was admitted to Emergency Department with changes in consciousness, with a heart rate of 22 beats/min, a weak carotid pulse, respiration of 8 breaths/min and a Glasgow coma score of 3. Soon after arrival pulseless electrical activity cardiac arrest occurred. Circulation was restored after 2 minutes of cardiopulmonary resuscitation and one bolus dose of 1 mg IV adrenaline. After abdominal CT was evidenced a hypodense area around the portal veins of both lobes of the liver, compatible with periportal edema, dilated inferior vena cava, minimal ascites and marked submucosal edema of gallbladder wall. It was ascertained that the patient attempted suicide and claimed to have ingested 15 sustained-release tablets containing 240 mg of verapamil hydrochloride. The patient attained a complete recovery. Periportal edema is a compensatory increased flow of hepatic lymph due to hepatic lymphatic obstruction or fluid overload. It has rarely been reported after cardiac arrest. In this case, the increase in mediastinal pressure during cardiopulmonary resuscitation, centralization of blood volume secondary to shock and vigorous IV fluid resuscitation may be the pathophysiological mechanism that caused periportal edema [30A].
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Book series2015, Side Effects of Drugs AnnualA. Nobili, ... R. Latini
Chapter
Valvular Heart Disease
2018, Practical CardiologyFeridoun Noohi FACC, FESC, ... Azin Alizadehasl MD, FACC, FASE
Physical Examination
The physical findings of patients with AS depend on the severity of valvular stenosis, LV function, and stroke volume.
The most important features of severe AS are as follows4,7:
1.
: Parvus et tardus arterial pulse, which means a slow-rising, low-amplitude, and late-peaking carotid pulse, is specific for severe AS; however, associated systemic hypertension or aortic regurgitation (AR) can affect the arterial pulse (Fig. 25.3).
2.
: Late-peaking systolic ejection murmur (SEM) is best heard at the base of the heart in the aortic region (upper right sternal border) and can be a musical or seagull sound.
3.
: Radiation of the high-frequency components of SEM to the apex is termed “Gallavardin’s phenomenon” and can be mistaken with the murmur of mitral regurgitation (MR).
4.
: Change in the intensity of the murmur by the Valsalva and standing maneuver is in favor of AS
5.
: Carotid shudder or palpable systolic thrill is a specific, albeit not sensitive, sign of severe AS.
6.
: Single S2 or paradoxical S2 happens because of late A2.
7.
: S4 gallop sound
8.
: In young patients with BAV, A2 can be heard normally or accentuated in association with aortic ejection sound, showing a flexible and noncalcified valve.
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Book2018, Practical CardiologyFeridoun Noohi FACC, FESC, ... Azin Alizadehasl MD, FACC, FASE
Chapter
Field Evaluation of the Injured Athlete
2007, Clinical Sports MedicineKai-Ming Chan, ... Fiona Chui-Yan Wong
Circulation
Cardiogenic and hypovolemic shock are the most common problems encountered in sports medicine. Hypovolemic shock can arise among athletes with severe dehydration or when there is a continuous bleeding source; whereas a cardiogenic shock may arise among athletes with pre-existing cardiovascular diseases. Hence, it is important for the team physician to identify these signs. Clinical features of hypovolemic shock include hypotension, tachycardia, pallor, sweaty cold and clammy extremities, delayed capillary refill (>2 s), low jugular venous pressure (JVP) or confusion. The clinical presentation of cardiogenic shock is similar to hypovolemic shock in its clinical presentation but with high JVP.
The cardiovascular system should initially be assessed by feeling for the carotid pulse. If the carotid pulse is not palpable on both sides along with the signs of shock, then cardiac arrest or arrhythmia is suspected. Hence, immediate cardiopulmonary resuscitation should be done. The rhythm can be determined with a portable defibrillation, and rhythm such as ventricular fibrillation or ventricular tachycardia can be converted to a stable rhythm by defibrillation. Time is precious in these conditions as for every minute of delay, there is a 10% decrease in survival.3
In the presence of shock, establishment of peripheral intravenous access with large bore cannula for volume resuscitation should commence as soon as airway and ventilation is secured. The principle for stabilizing an acutely injured athlete with ‘ABC’ should make no difference between conscious and unconscious athletes. Then the athlete should be transferred to the nearest medical center for further management. In the event of trauma, internal bleeding into third spaces such as the chest, abdomen or from fractures of long bones must be considered as a cause of hypovolemia. Therefore, look for any bruising and abdominal distention, and feel for any abdominal tenderness. Also, apply firm consistent pressure for any visible hemorrhage.
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Book2007, Clinical Sports MedicineKai-Ming Chan, ... Fiona Chui-Yan Wong
Chapter
Assessment of the Patient With a Cardiac Arrhythmia
2018, Cardiac Electrophysiology: From Cell to Bedside (Seventh Edition)Mithilesh K. Das, Douglas P. Zipes
Carotid Sinus Massage and Valsalva
Modulating autonomic tone by carotid sinus massage during the physical examination can be useful to expose the patient with the hypersensitive carotid sinus reflex. The clinician first needs to listen carefully over both carotids to be certain that no bruit is present, palpate lightly to determine that a normal carotid pulse is present, and then gently depress or rub the carotid sinus. Gentle massage for approximately 10 to 15 seconds or less usually is all that is necessary to produce significant periods of sinus arrest or AV block in susceptible patients.
The response to carotid sinus massage or other vagal maneuvers can be helpful in differentiating one tachycardia from another. In the most definitive responses, carotid sinus massage acutely terminates tachycardias such as AVRT, AVNRT, sinus node reentry, adenosine-sensitive atrial tachycardia, and idiopathic right ventricular outflow tract tachycardia. Carotid sinus massage can gradually slow a sinus tachycardia without termination and will decrease the ventricular response to atrial tachycardia, atrial flutter, and atrial fibrillation without termination, thereby exposing atrial activity. Carotid sinus massage transiently terminates the permanent form of AV junctional reciprocating tachycardia, which then restarts when carotid massage ceases. Carotid sinus massage does not affect reentrant ventricular or junctional tachycardias. Unfortunately, not all presentations of these tachycardias behave in such a predictable fashion, and intermediate or overlapping responses can occur.
A variation to the usual Valsalva maneuver has been proposed during which the individual is semirecumbent with passive leg raising immediately after the Valsalva strain to avoid venous pooling. Such an approach tripled the successful SVT termination rate.1
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Book2018, Cardiac Electrophysiology: From Cell to Bedside (Seventh Edition)Mithilesh K. Das, Douglas P. Zipes
Chapter
Sudden Cardiac Death
2010, Decision Making in Medicine (Third Edition)Tomaz Hruczkowski MD, Usha B. Tedrow MD. MS
1.
: Whether witnessing a collapse or being called to an unwitnessed event, assess the environment to allow delivery of care without putting the rescuers at risk. Assess (ask about) the circumstances to help narrow down the diagnostic possibilities. For example, a preceding chest pain versus an interrupted meal may point toward a cardiac or respiratory cause. Establish responsiveness by talking or shouting to the victim. On verification of unresponsiveness, call for help.
2.
: Activate EMS (911), call the code, and request bystander and/or paramedical/medical assistance. Request an AED or a code cart with a defibrillator, airway support equipment, and ACLS drugs. Begin the Primary Survey. Open/clear the airway, and “look, listen, feel” for spontaneous respirations.
3.
: Use one-way valve ventilation barrier device or bag-mask device and oropharyngeal airway if available, assess, and manage airway obstruction. Check carotid artery for pulse 5–10 seconds.
4.
: Full cardiac arrest confirmed, initiate adequate chest compressions and continue artificial respirations. Attach AED/defibrillator. Take charge or ask a more experienced rescuer; delegate tasks to bystanders/code team members.
5.
: Continue chest compressions, and monitor adequacy (carotid pulse). Assess rhythm.
6.
: If VF or pulseless VT.
7.
: Deliver three direct current (DC) shocks as needed (200 J, 300 J, 360 J—monophasic or equivalent biphasic). After three shocks, check pulse; if absent, resume CPR. Initiate secondary survey.
8.
: If asystole, recheck electrodes. Consider duration of the arrest and termination of efforts. If pulseless electrical activity (PEA), DO NOT SHOCK! Ascertain pulselessness versus low output state and consider typical causes of shock (differential diagnosis).
9.
: Consider transcutaneous pacing early, especially in a witnessed arrest. Monitor rhythm by differentiating pacing spikes from true QRS complexes. Monitor pulse by differentiating true pulse (carotid) from skeletal muscle twitches induced by electric current that may be observed in upper extremities.
10.
: Secondary survey and continuation of rescue protocol will generally require paramedical/medical (ACLS-trained) personnel. Delegate or assume predefined roles within the team. Intubate, confirm, and secure endotracheal tube placement and adequacy of ventilations (auscultation, pulse oximetry, end-tidal CO2). Secure IV access. Monitor vital signs. If pulse is palpable, monitor blood pressure (BP). Administer adrenergic agents. Consider rhythm-appropriate antiarrhythmic medications:
a.
: Amiodarone 300 mg IV push for persistent VF/VT; additional 150 mg IV for recurrent VF/VT. Maximum dose 2.2 g in 24 hr.
b.
: Lidocaine 1–1.5 mg/kg IV push for persistent or recurrent VF/VT; may repeat in 3–5 minutes, maximum dose 3 mg/kg.
c.
: Magnesium sulfate 1–2 g IV in torsades de pointes or if hypomagnesemia suspected.
d.
: Procainamide up to 50 mg/min IV, maximum dose 17 mg/kg for recurrent VF/VT.
Consider sodium bicarbonate 1 mEq/kg IV in known, preexisting hyperkalemia, acidosis, or tricyclic or aspirin overdose, after a long arrest interval.
11.
: Search for and treat reversible metabolic causes. Consider/memorize the list of typical causes of PEA and asystole.
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Book2010, Decision Making in Medicine (Third Edition)Tomaz Hruczkowski MD, Usha B. Tedrow MD. MS
Chapter
Evaluation of the Patient with Suspected Arrhythmias
2017, Arrhythmia Essentials (Second Edition)Brian Olshansky MD, ... Nora Goldschlager MD
Carotid sinus massage
Carotid sinus massage (CSM) is performed to terminate some supraventricular tachycardias (SVTs), such as AV node reentry and AV reentry SVTs. It may also be useful for sinoatrial (SA) reentrant SVT. CSM can be helpful to diagnose the atrial rhythm during a tachycardia by producing AV block. For example, CSM-induced AV block can allow atrial flutter (AFL) waves to be seen. CSM may also be helpful to cause ventriculoatrial conduction block when there is a wide QRS complex tachycardia. CSM may also be useful to determine whether carotid sinus syndrome is present in syncopal patients.
The proper way to perform CSM is to make sure that the carotid pulse is felt directly under the fingertips. Two or three fingertips placed on the carotid artery are important to make sure that a vagal reflex is initiated. It can be performed on the right or the left side, but the left side is more likely to act on the AV node, whereas the right side acts on the sinus node. When performing CSM, it is important to realize that there is a small risk of stroke if there is carotid artery plaque; thus, it is important to check for carotid bruits before performing this procedure. The carotid sinus area should be massaged for about 5 seconds, beginning gently and progressing more rapidly to heavier pressure. Monitoring the patient for signs of cerebral hypoperfusion such as weakness, paresthesias, and numbness are important in avoiding transient ischemic attacks.
The head is turned away from the carotid sinus area to be massaged. For example, if right-sided CSM is to be performed, the head is turned to the left. The patient should be supine or even in a Trendelenburg position to maximize intravascular volume. To terminate tachycardias, it may be useful in some cases to combine CSM with Trendelenburg positioning and with a Valsalva or handgrip maneuver.
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Book2017, Arrhythmia Essentials (Second Edition)Brian Olshansky MD, ... Nora Goldschlager MD
Related terms:
Resuscitation
Physical Examination
Blood Pressure
Hypertrophic Cardiomyopathy
Obstruction
Pulse Pressure
Pulse Rate
Aortic Stenosis
Epileptic Absence
Aortic Regurgitation
View all Topics |
3590 | https://www.youtube.com/watch?v=8wBUutF_pZ8 | Perimeter Ratio of Similar Figure
Mathema Teach
9970 subscribers
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Description
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Posted: 8 Jun 2020
This video will show how to solve for wither a missing side or a missing perimeter of similar figure.
Transcript:
[Music] hello everyone today we're gonna have the parameter ratio of similar figures so I put a little note over here if two figures are similar then the ratio of their parameter is equal to the ratio of their corresponding side lengths so these are the corresponding side lengths first side length M is corresponding to side T so these two are corresponding and then the word parameter that we have here which is represented as P parameter in math means it's the distance around a figure so parameter if you add all the lengths of all the sides here that's the parameter so the parameter of one over the parameter of two so the ratio of the two parameters here is equal to the ratio of their corresponding side length so that's how I got the formula over here that P 1 over P 2 P is the parameter is equal to side length M over side length T so let's have an example here to better see how this formula work so let's take this example right here so suppose I have here this two pictures so these two pictures it says here that they are similar what is the parameter of the first figures we're looking for P 1 so the side length for the first picture is 10 centimeters and the side length of the second is 8 centimeters and then the parameter parameter is the distance around this triangle is 32 centimeters and then the the distance around this is missing so how do we find the the distance around this triangle here so again the rule says that the ratio of their parameter so I can go ahead and write p1 over p2 is equal to the ratio of their side lengths so that would be that's 10 centimeters right here over 8 centimeters so I can go ahead and plug in values to this one right here so P one is missing so I can keep that as P 1 P 2 here is 32 centimeters equals 10 centimeters over 8 centimeters from here we want the P 1 or the parameter the first picture so we are going to cross multiply these two here so we are left with 8 centimeters P 1 so 8 centimeters and then P 1 is equal to 32 centimeters times 10 centimeters so then we can go ahead and simplify this so this is 8 centimeters P 1 equals 32 times 10 is 320 centimeters squared because centimeters and times centimeters is centimeters squared so we divide both because we want P 1 by itself so we're gonna divide 8 centimeters from both sides I divide this by 8 centimeters here so I can cross out the 8 centimeters and 8 centimeters centimeters squared means there are 2 centimeters there so going to take one of them and then I'll take the centimeter here so that if you divide of 320 divided by 8 our P 1 is 40 so we are left already with centimeters because centimeter square squared means 2 centimeters and we have 1 centimeters at the bottom so we cross out one of the 2 and then cross out the centimeters at the bottoms were left with 40 and the unit is centimeters so this is 40 centimeters at this time I would encourage you to pause this video and try this problem out on your own and when you're done and pause it and check your answer okay so we go over the problem here so pretty much the same thing these two figures here are similar so what is the length of the corresponding although it's not written here but we always remember that the side should be corresponding in order for their period to be equivalent or to be the ratio of their parameter to be equal so then it means that the corresponding side of 13 is this I can name this as X you can write any letter for that it doesn't matter so going through this same process that we have here so p1 over so our P was I'm just gonna write the formula here first so p1 over p2 is equal to the first side length which is 13 feet over X feet and then we go ahead and write the value for P 1 that would be 65 feet over 85 feet and that is equal to that's gonna be 13 feet over X feet so we cross multiply so we're gonna cross multiply these two here so we have X x times 65 feet is equal to 85 feet times 13 feet so then we go ahead and divide both sides by because we want the X by itself so we divide this by 65 feet and then we divide this by 65 feet so then from here we can cross out 1 feet and 1 feet here so then we are we can cross out the 65 feet and 65 feet no others would want to multiply this 2 it doesn't it doesn't really matter the answer is still the same it's different from the way how it did it over here but it's still gonna get the same answer so this would come out X is equal to that's gonna be 85 times 13 because we already cross out the feet and the feet or you can do it the you can do it the same way over here where you can get a feet squared it doesn't matter the answer is still the same and the math is pretty much the same thing so we multiply 85 times 13 so that would be 1100 and five feet and then we divide that why we still have the 65 so then our answer for X is 17 so the final unit for this by the way is feet did you get the same answer is this good perfect if you find this video helpful keep like and subscribe for more math videos see ya |
3591 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_24?srsltid=AfmBOop8g6AxpiMUR0UX5d2sJJwTc3s3rx8BAqZaIxMpUz4JWSEIY-sx | Art of Problem Solving
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2015 AMC 12A Problems/Problem 24
Contents
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1 Problem
2 Solution
3 Solution 2
3.1 Video Solution by Richard Rusczyk
4 See Also
Problem
Rational numbers and are chosen at random among all rational numbers in the interval ![Image 13: $0,2)$ that can be written as fractions where and are integers with . What is the probability that is a real number?
Solution
Let and . Consider the binomial expansion of the expression:
We notice that the only terms with are the second and the fourth terms. Thus for the expression to be a real number, either or must be , or the second term and the fourth term cancel each other out (because in the fourth term, you have ).
Either or is .
The two satisfying this are and , and the two satisfying this are and . Because and can both be expressed as fractions with a denominator less than or equal to , there are a total of possible values for and :
Calculating the total number of sets of results in sets. Calculating the total number of invalid sets (sets where doesn't equal or and doesn't equal or ), resulting in .
Thus the number of valid sets is .
: The two terms cancel.
We then have:
So:
which means for a given value of or , there are valid values(one in each quadrant).
When either or are equal to , however, there are only two corresponding values. We don't count the sets where either or equals , for we would get repeated sets. We also exclude values where the denominator is an odd number, for we cannot find any corresponding values(for example, if is , then must be , which we don't have). Thus the total number of sets for this case is .
Thus, our final answer is , which is .
Solution 2
Multiplying complex numbers is equivalent to multiplying their magnitudes and summing their angles. In order for to be a real number, then the angle of must be a multiple of , so satisfies , , , or .
There are possible values of and . at and at . The probability of or is (We overcounted the case where .)
We also consider the case where . This only happens when or . The probability is The total probability is
~zeric
Video Solution by Richard Rusczyk
~ dolphin7
See Also
2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23Followed by
Problem 25
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
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3592 | https://en.wiktionary.org/wiki/behind_bars | behind bars - Wiktionary, the free dictionary
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1.1 Pronunciation
1.2 Prepositional phrase
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1.3 See also
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behind bars
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From Wiktionary, the free dictionary
English
[edit]
Pronunciation
[edit]
Audio (General Australian):Duration: 3 seconds.0:03(file)
Prepositional phrase
[edit]
behindbars
(figuratively,idiomatic)In jail, in prison. quotations▼
1905 April–October, Upton Sinclair, chapter XXVII, in The Jungle, New York, N.Y.: Doubleday, Page & Company, published 26 February 1906, →OCLC:There is one kind of prison where the man is behind bars, and everything that he desires is outside; and there is another kind where the things are behind the bars, and the man is outside.
2001 May 7, Margot Roosevelt et al., “The War Against The War On Drugs”, in Time:Some 460,000 Americans are behind bars for drug offenses.
2014 November 27, Ian Black, “Courts kept busy as Jordan works to crush support for Isis”, in The Guardian:Leading Jordanian exponents of the Salafi-jihadi world view, such as Abu Muhammad al-Maqdisi, are now behind bars or silent, fearing arrest by the powerful mukhabarat secret police.
Translations
[edit]
show ▼±in jail or prison
[Select preferred languages] [Clear all]
Arabic: (calques of Engish)خَلْفَ الْقُضْبَان(ḵalfa l-quḍbān), وَرَاءَ الْقُضْبَان(warāʔa l-quḍbān)
Azerbaijani: dəmir barmaqlıq arxasında
Belarusian: за кра́тамі(za krátami)
Bulgarian: зад реше́тките(zad rešétkite)
Catalan: a la garjola
Chinese: Mandarin: 關在監牢裡/ 关在监牢里(guān zài jiānláo lǐ), 身陷囹圄(zh)(shēnxiànlíngyǔ)
Czech: za mřížemi
Danish: bag tremmer
Dutch: achter de tralies
Finnish: telkien takana, kalterientakana, vankilassa
French: derrière les barreaux(fr), au violon(fr)
German: hinter Gitter
Greek: στη φυλακή(sti fylakí)
Hebrew: מֵאֲחוֹרֵי סוֹרָג וּבְרִיחַ(me'akhoréi sorág uvríakh)
Hungarian: rács mögött
Icelandic: bakvið lás og sleggju
Indonesian: di balik jeruji besi
Italian: dietro le sbarre
Japanese: 獄中で(ja)(ごくちゅうで, gokuchū de)
Korean: 감옥에서(ko)(gamogeseo)
Latvian: aiz restēm
Norwegian: Bokmål: bak lås og slå(no)Nynorsk: bak lås og slå
Polish: za kratkami(pl)
Portuguese: atrás das grades
Romanian: după gratii
Russian: за решёткой(za rešótkoj)
Serbo-Croatian: иза решетака
Slovak: za mrežami
Spanish: entre rejas(es), tras las rejas
Swedish: bakom lås och bom(sv)
Tagalog: himas-rehas
Telugu: కటకటాల్లో(kaṭakaṭāllō)
Thai: please add this translation if you can
Ukrainian: за ґра́тами(za grátamy)
Add translation: More
[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
See also
[edit]
See also Thesaurus:jail
References
[edit]
“behind bars”, in Lexico, Dictionary.com; Oxford University Press, 2019–2022.
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3593 | https://math.stackexchange.com/questions/3339212/inequality-with-cardinal-numbers | set theory - Inequality with cardinal numbers - Mathematics Stack Exchange
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Inequality with cardinal numbers
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I want to check if the following statement holds:
If a,b,c a,b,c are cardinal numbers and a<b a<b then a c<b c a c<b c.
If the statement would hold, given that there is a bijective funcion f:a→b f:a→b there will also be a bijective function g:a c→b c g:a c→b c.
Since f f is bijective, we have that f(a)=b f(a)=b.
So there will be a y y such that f(a c)=y f(a c)=y.
Is this right so far? If so, how can we continue?
set-theory
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edited Aug 30, 2019 at 17:23
Arturo Magidin
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asked Aug 30, 2019 at 15:07
Mary StarMary Star
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1 The statement is wrong. You have 2<ℵ 0 2<ℵ 0. However 2 ℵ 0=ℵ ℵ 0 0=c 2 ℵ 0=ℵ 0 ℵ 0=c.mathcounterexamples.net –mathcounterexamples.net 2019-08-30 15:19:29 +00:00 Commented Aug 30, 2019 at 15:19
2 And a<b a<b means that there is an injection f:a→b f:a→b. Not a bijection.mathcounterexamples.net –mathcounterexamples.net 2019-08-30 15:22:32 +00:00 Commented Aug 30, 2019 at 15:22
How could we show that it holds that 2 N 0=N 0 N 0 2 N 0=N 0 N 0 ? "mathcounterexamples.net Mary Star –Mary Star 2019-08-30 16:03:01 +00:00 Commented Aug 30, 2019 at 16:03
You have 2<ℵ 0<2 ℵ 0 2<ℵ 0<2 ℵ 0. Therefore according to cardinal arithmetic2 ℵ 0≤ℵ ℵ 0 0≤2 ℵ 0×ℵ 0=2 ℵ 0 2 ℵ 0≤ℵ 0 ℵ 0≤2 ℵ 0×ℵ 0=2 ℵ 0 as ℵ 0×ℵ 0=ℵ 0 ℵ 0×ℵ 0=ℵ 0.mathcounterexamples.net –mathcounterexamples.net 2019-08-30 16:10:33 +00:00 Commented Aug 30, 2019 at 16:10
2 I'm pretty sure that I wrote at least one counterexample to this question before on this site.Asaf Karagila –Asaf Karagila♦ 2019-08-30 16:30:48 +00:00 Commented Aug 30, 2019 at 16:30
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Your statement is false. For example, N N N N and 2 N 2 N have the same cardinality - put another way,
(ℵ 0)ℵ 0=2 ℵ 0.(ℵ 0)ℵ 0=2 ℵ 0.
We can see this by invoking basic cardinal arithmetic rules, as
2 ℵ 0≤(ℵ 0)ℵ 0≤(2 ℵ 0)ℵ 0=2 ℵ 0×ℵ 0=2 ℵ 0,2 ℵ 0≤(ℵ 0)ℵ 0≤(2 ℵ 0)ℵ 0=2 ℵ 0×ℵ 0=2 ℵ 0,
with the first inequality being trivial, the second following from the weak monotonicity of exponentiation (which is also pretty trivial), and the third and fourth following from the definitions of cardinal multiplication and exponentiation (together with a bit of Currying).
However, we can also see this directly via Cantor-Bernstein (which, as a reminder, does not require Choice). Namely, fix your favorite pairing function⟨⋅,⋅⟩:N 2→N⟨⋅,⋅⟩:N 2→N, and given f:N→N f:N→N let f^:N→2 f^:N→2 be defined by
f^(⟨a,b⟩)=1⟺f(a)=b f^(⟨a,b⟩)=1⟺f(a)=b
(and f^(⟨a,b⟩)=0 f^(⟨a,b⟩)=0 otherwise). It's easy to check that the map
f↦f^f↦f^
is an injection from N N N N to 2 N 2 N. Since the identity provides an injection the other way, Cantor-Bernstein gives a bijection.
(Incidentally, your "proof" also gets cardinal inequality wrong: "a<b a<b" means that there is an injection from a a to b b but there is not an injection from b b to a a.)
It's worth pointing out that increasing the exponent doesn't necessarily increase the whole power either: letting κ=2 ℵ 1 κ=2 ℵ 1, we have
κ ℵ 0=2 ℵ 1×ℵ 0=2 ℵ 1 and κ ℵ 1=2 ℵ 1×ℵ 1=2 ℵ 1.κ ℵ 0=2 ℵ 1×ℵ 0=2 ℵ 1 and κ ℵ 1=2 ℵ 1×ℵ 1=2 ℵ 1.
This is indeed the usual situation:
Operations on cardinals (including addition, multiplication, and exponentiation) are very often weakly monotonic in each of their arguments but not strongly monotonic in any of them.
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edited Aug 30, 2019 at 16:30
answered Aug 30, 2019 at 16:25
Noah SchweberNoah Schweber
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2 It's worth pointing out that increasing the exponent doesn't necessarily increase the whole power either --- Actually, your earlier example also shows this. Denoting 2 ℵ 0 2 ℵ 0 by c,c, you showed that c 1=c ℵ 0.c 1=c ℵ 0. For what it's worth, the fact that c ℵ 0=c c ℵ 0=c has nothing to do with c c being too large to increase through exponentiation by ℵ 0,ℵ 0, because there exist arbitrarily large cardinals b b such that b<b ℵ 0 b<b ℵ 0 and there exist arbitrarily large cardinals b b such that b=b ℵ 0.b=b ℵ 0.(continued)Dave L. Renfro –Dave L. Renfro 2019-08-30 17:34:51 +00:00 Commented Aug 30, 2019 at 17:34
To the proposer: In particular ℵ ℵ 0 1=2 ℵ 0.ℵ 1 ℵ 0=2 ℵ 0.DanielWainfleet –DanielWainfleet 2019-08-30 17:36:01 +00:00 Commented Aug 30, 2019 at 17:36
For strict inequality examples, let {a n}{a n} be any strictly increasing sequence of cardinal numbers (e.g. a 1 a 1 chosen arbitrarily large, a 2=2 a 1,a 2=2 a 1,a 3=2 a 2,a 3=2 a 2, etc.). Then by Konig's strict inequality lemma for infinite products and infinite sums, and letting b b be the sum of the cardinals in the sequence, we have b b less than the product of the cardinals in the sequence, which in turn is less than or equal to b ℵ 0 b ℵ 0(continued)Dave L. Renfro –Dave L. Renfro 2019-08-30 17:46:39 +00:00 Commented Aug 30, 2019 at 17:46
(the last ≤≤ is because a n<b a n<b for each n,n, hence the product of the a n a n's is ≤≤ the product of ℵ 0 ℵ 0 many b b's). For equality examples, let b=a ℵ 0 b=a ℵ 0 for arbitrarily large cardinals a.a.Dave L. Renfro –Dave L. Renfro 2019-08-30 17:46:47 +00:00 Commented Aug 30, 2019 at 17:46
@DaveL.Renfro That's absolutely true - I just thought that the separate example might help the OP.Noah Schweber –Noah Schweber 2019-08-30 17:53:13 +00:00 Commented Aug 30, 2019 at 17:53
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3594 | https://en.wikipedia.org/wiki/Real_coordinate_space | Published Time: 2005-02-08T18:35:03Z
Real coordinate space - Wikipedia
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1 Definition and structures
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3 Vector spaceToggle Vector space subsection
3.1 Matrix notation
3.2 Standard basis
4 Geometric properties and usesToggle Geometric properties and uses subsection
4.1 Orientation
4.2 Affine space
4.3 Convexity
4.4 Euclidean space
4.5 In algebraic and differential geometry
4.6 Other appearances
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5 Topological properties
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6.1 n ≤ 1
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From Wikipedia, the free encyclopedia
Space formed by the n-tuples of real numbers
This article includes a list of references, related reading, or external links, but its sources remain unclear because it lacks inline citations. Please help improve this article by introducing more precise citations.(February 2024) (Learn how and when to remove this message)
Cartesian coordinates identify points of the Euclidean plane with pairs of real numbers
In mathematics, the real coordinate space or real coordinate n-space, of dimensionn, denoted Rn or R n{\displaystyle \mathbb {R} ^{n}}, is the set of all ordered n-tuples of real numbers, that is the set of all sequences of n real numbers, also known as coordinate vectors. Special cases are called the real lineR1, the real coordinate planeR2, and the real coordinate three-dimensional spaceR3. With component-wise addition and scalar multiplication, it is a real vector space.
The coordinates over any basis of the elements of a real vector space form a real coordinate space of the same dimension as that of the vector space. Similarly, the Cartesian coordinates of the points of a Euclidean space of dimension n, En (Euclidean line, E; Euclidean plane, E2; Euclidean three-dimensional space, E3) form a real coordinate space of dimension n.
These one to one correspondences between vectors, points and coordinate vectors explain the names of coordinate space and coordinate vector. It allows using geometric terms and methods for studying real coordinate spaces, and, conversely, to use methods of calculus in geometry. This approach of geometry was introduced by René Descartes in the 17th century. It is widely used, as it allows locating points in Euclidean spaces, and computing with them.
Definition and structures
[edit]
For any natural numbern, the setRn consists of all n-tuples of real numbers (R). It is called the "n-dimensional real space" or the "real n-space".
An element of Rn is thus a n-tuple, and is written (x 1,x 2,…,x n){\displaystyle (x_{1},x_{2},\ldots ,x_{n})} where each x i is a real number. So, in multivariable calculus, the domain of a function of several real variables and the codomain of a real vector valued function are subsets of Rn for some n.
The real n-space has several further properties, notably:
With componentwise addition and scalar multiplication, it is a real vector space. Every n-dimensional real vector space is isomorphic to it.
With the dot product (sum of the term by term product of the components), it is an inner product space. Every n-dimensional real inner product space is isomorphic to it.
As every inner product space, it is a topological space, and a topological vector space.
It is a Euclidean space and a real affine space, and every Euclidean or affine space is isomorphic to it.
It is an analytic manifold, and can be considered as the prototype of all manifolds, as, by definition, a manifold is, near each point, isomorphic to an open subset of Rn.
It is an algebraic variety, and every real algebraic variety is a subset of Rn.
These properties and structures of Rn make it fundamental in almost all areas of mathematics and their application domains, such as statistics, probability theory, and many parts of physics.
The domain of a function of several variables
[edit]
Main articles: Multivariable calculus and Real multivariable function
Any function f(x 1, x 2, ..., x n) of n real variables can be considered as a function on Rn (that is, with Rn as its domain). The use of the real n-space, instead of several variables considered separately, can simplify notation and suggest reasonable definitions. Consider, for n = 2, a function composition of the following form: F(t)=f(g 1(t),g 2(t)),{\displaystyle F(t)=f(g_{1}(t),g_{2}(t)),} where functions g 1 and g 2 are continuous. If
∀x 1 ∈ R : f(x 1, ·) is continuous (by x 2)
∀x 2 ∈ R : f(·, x 2) is continuous (by x 1)
then F is not necessarily continuous. Continuity is a stronger condition: the continuity of f in the natural R2 topology (discussed below), also called multivariable continuity, which is sufficient for continuity of the composition F.
Vector space
[edit]
The coordinate space Rn forms an n-dimensional vector space over the field of real numbers with the addition of the structure of linearity, and is often still denoted Rn. The operations on Rn as a vector space are typically defined by x+y=(x 1+y 1,x 2+y 2,…,x n+y n){\displaystyle \mathbf {x} +\mathbf {y} =(x_{1}+y_{1},x_{2}+y_{2},\ldots ,x_{n}+y_{n})}α x=(α x 1,α x 2,…,α x n).{\displaystyle \alpha \mathbf {x} =(\alpha x_{1},\alpha x_{2},\ldots ,\alpha x_{n}).} The zero vector is given by 0=(0,0,…,0){\displaystyle \mathbf {0} =(0,0,\ldots ,0)} and the additive inverse of the vector x is given by −x=(−x 1,−x 2,…,−x n).{\displaystyle -\mathbf {x} =(-x_{1},-x_{2},\ldots ,-x_{n}).}
This structure is important because any n-dimensional real vector space is isomorphic to the vector space Rn.
Matrix notation
[edit]
Main article: Matrix (mathematics)
In standard matrix notation, each element of Rn is typically written as a column vectorx=[x 1 x 2⋮x n]{\displaystyle \mathbf {x} ={\begin{bmatrix}x_{1}\x_{2}\\vdots \x_{n}\end{bmatrix}}} and sometimes as a row vector: x=[x 1 x 2⋯x n].{\displaystyle \mathbf {x} ={\begin{bmatrix}x_{1}&x_{2}&\cdots &x_{n}\end{bmatrix}}.}
The coordinate space Rn may then be interpreted as the space of all n × 1column vectors, or all 1 × nrow vectors with the ordinary matrix operations of addition and scalar multiplication.
Linear transformations from Rn to Rm may then be written as m × n matrices which act on the elements of Rn via left multiplication (when the elements of Rn are column vectors) and on elements of Rm via right multiplication (when they are row vectors). The formula for left multiplication, a special case of matrix multiplication, is: (A x)k=∑l=1 n A k l x l{\displaystyle (A{\mathbf {x} }){k}=\sum {l=1}^{n}A_{kl}x_{l}}
Any linear transformation is a continuous function (see below). Also, a matrix defines an open map from Rn to Rm if and only if the rank of the matrix equals to m.
Standard basis
[edit]
Main article: Standard basis
The coordinate space Rn comes with a standard basis: e 1=(1,0,…,0)e 2=(0,1,…,0)⋮e n=(0,0,…,1){\displaystyle {\begin{aligned}\mathbf {e} {1}&=(1,0,\ldots ,0)\\mathbf {e} {2}&=(0,1,\ldots ,0)\&{}\;\;\vdots \\mathbf {e} _{n}&=(0,0,\ldots ,1)\end{aligned}}}
To see that this is a basis, note that an arbitrary vector in Rn can be written uniquely in the form x=∑i=1 n x i e i.{\displaystyle \mathbf {x} =\sum {i=1}^{n}x{i}\mathbf {e} _{i}.}
Geometric properties and uses
[edit]
Orientation
[edit]
The fact that real numbers, unlike many other fields, constitute an ordered field yields an orientation structure on Rn. Any full-rank linear map of Rn to itself either preserves or reverses orientation of the space depending on the sign of the determinant of its matrix. If one permutes coordinates (or, in other words, elements of the basis), the resulting orientation will depend on the parity of the permutation.
Diffeomorphisms of Rn or domains in it, by their virtue to avoid zero Jacobian, are also classified to orientation-preserving and orientation-reversing. It has important consequences for the theory of differential forms, whose applications include electrodynamics.
Another manifestation of this structure is that the point reflection in Rn has different properties depending on evenness of n. For even n it preserves orientation, while for odd n it is reversed (see also improper rotation).
Affine space
[edit]
Further information: Affine space
Rn understood as an affine space is the same space, where Rn as a vector space acts by translations. Conversely, a vector has to be understood as a "difference between two points", usually illustrated by a directed line segment connecting two points. The distinction says that there is no canonical choice of where the origin should go in an affine n-space, because it can be translated anywhere.
Convexity
[edit]
The n-simplex (see below) is the standard convex set, that maps to every polytope, and is the intersection of the standard (n + 1) affine hyperplane (standard affine space) and the standard (n + 1) orthant (standard cone).
Further information: Convex analysis
In a real vector space, such as Rn, one can define a convex cone, which contains all non-negative linear combinations of its vectors. Corresponding concept in an affine space is a convex set, which allows only convex combinations (non-negative linear combinations that sum to 1).
In the language of universal algebra, a vector space is an algebra over the universal vector space R∞ of finite sequences of coefficients, corresponding to finite sums of vectors, while an affine space is an algebra over the universal affine hyperplane in this space (of finite sequences summing to 1), a cone is an algebra over the universal orthant (of finite sequences of nonnegative numbers), and a convex set is an algebra over the universal simplex (of finite sequences of nonnegative numbers summing to 1). This geometrizes the axioms in terms of "sums with (possible) restrictions on the coordinates".
Another concept from convex analysis is a convex function from Rn to real numbers, which is defined through an inequality between its value on a convex combination of points and sum of values in those points with the same coefficients.
Euclidean space
[edit]
Main articles: Euclidean space and Cartesian coordinate system
The dot productx⋅y=∑i=1 n x i y i=x 1 y 1+x 2 y 2+⋯+x n y n{\displaystyle \mathbf {x} \cdot \mathbf {y} =\sum {i=1}^{n}x{i}y_{i}=x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n}} defines the norm|x| = √x ⋅ x on the vector space Rn. If every vector has its Euclidean norm, then for any pair of points the distance d(x,y)=‖x−y‖=∑i=1 n(x i−y i)2{\displaystyle d(\mathbf {x} ,\mathbf {y} )=\|\mathbf {x} -\mathbf {y} \|={\sqrt {\sum {i=1}^{n}(x{i}-y_{i})^{2}}}} is defined, providing a metric space structure on Rn in addition to its affine structure.
As for vector space structure, the dot product and Euclidean distance usually are assumed to exist in Rn without special explanations. However, the real n-space and a Euclidean n-space are distinct objects, strictly speaking. Any Euclidean n-space has a coordinate system where the dot product and Euclidean distance have the form shown above, called Cartesian. But there are many Cartesian coordinate systems on a Euclidean space.
Conversely, the above formula for the Euclidean metric defines the standard Euclidean structure on Rn, but it is not the only possible one. Actually, any positive-definite quadratic formq defines its own "distance" √q(x − y), but it is not very different from the Euclidean one in the sense that ∃C 1>0,∃C 2>0,∀x,y∈R n:C 1 d(x,y)≤q(x−y)≤C 2 d(x,y).{\displaystyle \exists C_{1}>0,\ \exists C_{2}>0,\ \forall \mathbf {x} ,\mathbf {y} \in \mathbb {R} ^{n}:C_{1}d(\mathbf {x} ,\mathbf {y} )\leq {\sqrt {q(\mathbf {x} -\mathbf {y} )}}\leq C_{2}d(\mathbf {x} ,\mathbf {y} ).} Such a change of the metric preserves some of its properties, for example the property of being a complete metric space. This also implies that any full-rank linear transformation of Rn, or its affine transformation, does not magnify distances more than by some fixed C 2, and does not make distances smaller than 1 / C 1 times, a fixed finite number times smaller.[clarification needed]
The aforementioned equivalence of metric functions remains valid if √q(x − y) is replaced with M(x − y), where M is any convex positive homogeneous function of degree 1, i.e. a vector norm (see Minkowski distance for useful examples). Because of this fact that any "natural" metric on Rn is not especially different from the Euclidean metric, Rn is not always distinguished from a Euclidean n-space even in professional mathematical works.
In algebraic and differential geometry
[edit]
Although the definition of a manifold does not require that its model space should be Rn, this choice is the most common, and almost exclusive one in differential geometry.
On the other hand, Whitney embedding theorems state that any real differentiable m-dimensional manifold can be embedded into R2 m.
Other appearances
[edit]
Other structures considered on Rn include the one of a pseudo-Euclidean space, symplectic structure (even n), and contact structure (odd n). All these structures, although can be defined in a coordinate-free manner, admit standard (and reasonably simple) forms in coordinates.
Rn is also a real vector subspace of Cn which is invariant to complex conjugation; see also complexification.
Polytopes in R n
[edit]
See also: Linear programming and Convex polytope
There are three families of polytopes which have simple representations in Rn spaces, for any n, and can be used to visualize any affine coordinate system in a real n-space. Vertices of a hypercube have coordinates (x 1, x 2, ..., x n) where each x k takes on one of only two values, typically 0 or 1. However, any two numbers can be chosen instead of 0 and 1, for example −1 and 1. An n-hypercube can be thought of as the Cartesian product of n identical intervals (such as the unit interval[0,1]) on the real line. As an n-dimensional subset it can be described with a system of 2 n inequalities: 0≤x 1≤1⋮0≤x n≤1{\displaystyle {\begin{matrix}0\leq x_{1}\leq 1\\vdots \0\leq x_{n}\leq 1\end{matrix}}} for [0,1], and |x 1|≤1⋮|x n|≤1{\displaystyle {\begin{matrix}|x_{1}|\leq 1\\vdots \|x_{n}|\leq 1\end{matrix}}} for [−1,1].
Each vertex of the cross-polytope has, for some k, the x k coordinate equal to ±1 and all other coordinates equal to 0 (such that it is the k th standard basis vector up to sign). This is a dual polytope of hypercube. As an n-dimensional subset it can be described with a single inequality which uses the absolute value operation: ∑k=1 n|x k|≤1,{\displaystyle \sum {k=1}^{n}|x{k}|\leq 1\,,} but this can be expressed with a system of 2 n linear inequalities as well.
The third polytope with simply enumerable coordinates is the standard simplex, whose vertices are n standard basis vectors and the origin(0, 0, ..., 0). As an n-dimensional subset it is described with a system of n + 1 linear inequalities: 0≤x 1⋮0≤x n∑k=1 n x k≤1{\displaystyle {\begin{matrix}0\leq x_{1}\\vdots \0\leq x_{n}\\sum \limits {k=1}^{n}x{k}\leq 1\end{matrix}}} Replacement of all "≤" with "<" gives interiors of these polytopes.
Topological properties
[edit]
The topological structure of Rn (called standard topology, Euclidean topology, or usual topology) can be obtained not only from Cartesian product. It is also identical to the natural topology induced by Euclidean metric discussed above: a set is open in the Euclidean topology if and only if it contains an open ball around each of its points. Also, Rn is a linear topological space (see continuity of linear maps above), and there is only one possible (non-trivial) topology compatible with its linear structure. As there are many open linear maps from Rn to itself which are not isometries, there can be many Euclidean structures on Rn which correspond to the same topology. Actually, it does not depend much even on the linear structure: there are many non-linear diffeomorphisms (and other homeomorphisms) of Rn onto itself, or its parts such as a Euclidean open ball or the interior of a hypercube).
Rn has the topological dimensionn.
An important result on the topology of Rn, that is far from superficial, is Brouwer's invariance of domain. Any subset of Rn (with its subspace topology) that is homeomorphic to another open subset of Rn is itself open. An immediate consequence of this is that Rm is not homeomorphic to Rn if m ≠ n – an intuitively "obvious" result which is nonetheless difficult to prove.
Despite the difference in topological dimension, and contrary to a naïve perception, it is possible to map a lesser-dimensional[clarification needed] real space continuously and surjectively onto Rn. A continuous (although not smooth) space-filling curve (an image of R1) is possible.[clarification needed]
Examples
[edit]
Empty column vector,
the only element of R0
n ≤ 1
[edit]
Cases of 0 ≤ n ≤ 1 do not offer anything new: R1 is the real line, whereas R0 (the space containing the empty column vector) is a singleton, understood as a zero vector space. However, it is useful to include these as trivial cases of theories that describe different n.
n = 2
[edit]
Both hypercube and cross-polytope in R2 are squares, but coordinates of vertices are arranged differently
Further information: Two-dimensional space
See also: SL2(R)
The case of (x,y) where x and y are real numbers has been developed as the Cartesian planeP. Further structure has been attached with Euclidean vectors representing directed line segments in P. The plane has also been developed as the field extensionC{\displaystyle \mathbf {C} } by appending roots of X 2 + 1 = 0 to the real field R.{\displaystyle \mathbf {R} .} The root i acts on P as a quarter turn with counterclockwise orientation. This root generates the group{i,−1,−i,+1}≡Z/4 Z{\displaystyle {i,-1,-i,+1}\equiv \mathbf {Z} /4\mathbf {Z} }. When (x,y) is written x + y i it is a complex number.
Another group action by Z/2 Z{\displaystyle \mathbf {Z} /2\mathbf {Z} }, where the actor has been expressed as j, uses the line y=x for the involution of flipping the plane (x,y) ↦ (y,x), an exchange of coordinates. In this case points of P are written x + y j and called split-complex numbers. These numbers, with the coordinate-wise addition and multiplication according to jj=+1, form a ring that is not a field.
Another ring structure on P uses a nilpotent e to write x + y e for (x,y). The action of e on P reduces the plane to a line: It can be decomposed into the projection into the x-coordinate, then quarter-turning the result to the y-axis: e (x + y e) = x e since e 2 = 0. A number x + y e is a dual number. The dual numbers form a ring, but, since e has no multiplicative inverse, it does not generate a group so the action is not a group action.
Excluding (0,0) from P makes [x: y] projective coordinates which describe the real projective line, a one-dimensional space. Since the origin is excluded, at least one of the ratios x/y and y/x exists. Then [x: y] = [x/y: 1] or [x: y] = [1: y/x]. The projective line P1(R) is a topological manifold covered by two coordinate charts, [z: 1] → z or [1: z] → z, which form an atlas. For points covered by both charts the transition function is multiplicative inversion on an open neighborhood of the point, which provides a homeomorphism as required in a manifold. One application of the real projective line is found in Cayley–Klein metric geometry.
n = 3
[edit]
Cube (the hypercube) and octahedron (the cross-polytope) of R3. Coordinates are not shown
Main article: Three-dimensional space
n = 4
[edit]
Further information: Four-dimensional space
R4 can be imagined using the fact that 16 points (x 1, x 2, x 3, x 4), where each x k is either 0 or 1, are vertices of a tesseract (pictured), the 4-hypercube (see above).
The first major use of R4 is a spacetime model: three spatial coordinates plus one temporal. This is usually associated with theory of relativity, although four dimensions were used for such models since Galilei. The choice of theory leads to different structure, though: in Galilean relativity the t coordinate is privileged, but in Einsteinian relativity it is not. Special relativity is set in Minkowski space. General relativity uses curved spaces, which may be thought of as R4 with a curved metric for most practical purposes. None of these structures provide a (positive-definite) metric on R4.
Euclidean R4 also attracts the attention of mathematicians, for example due to its relation to quaternions, a 4-dimensional real algebra themselves. See rotations in 4-dimensional Euclidean space for some information.
In differential geometry, n = 4 is the only case where Rn admits a non-standard differential structure: see exotic R 4.
Norms on Rn
[edit]
One could define many norms on the vector spaceRn. Some common examples are
the p-norm, defined by ‖x‖p:=∑i=1 n|x i|p p{\textstyle \|\mathbf {x} \|{p}:={\sqrt[{p}]{\sum {i=1}^{n}|x_{i}|^{p}}}} for all x∈R n{\displaystyle \mathbf {x} \in \mathbf {R} ^{n}} where p{\displaystyle p} is a positive integer. The case p=2{\displaystyle p=2} is very important, because it is exactly the Euclidean norm.
the ∞{\displaystyle \infty }-norm or maximum norm, defined by ‖x‖∞:=max{x 1,…,x n}{\displaystyle \|\mathbf {x} \|{\infty }:=\max{x{1},\dots ,x_{n}}} for all x∈R n{\displaystyle \mathbf {x} \in \mathbf {R} ^{n}}. This is the limit of all the p-norms: ‖x‖∞=lim p→∞∑i=1 n|x i|p p{\textstyle \|\mathbf {x} \|{\infty }=\lim {p\to \infty }{\sqrt[{p}]{\sum {i=1}^{n}|x{i}|^{p}}}}.
A really surprising and helpful result is that every norm defined on Rn is equivalent. This means for two arbitrary norms ‖⋅‖{\displaystyle \|\cdot \|} and ‖⋅‖′{\displaystyle \|\cdot \|'} on Rn you can always find positive real numbers α,β>0{\displaystyle \alpha ,\beta >0}, such that α⋅‖x‖≤‖x‖′≤β⋅‖x‖{\displaystyle \alpha \cdot \|\mathbf {x} \|\leq \|\mathbf {x} \|'\leq \beta \cdot \|\mathbf {x} \|} for all x∈R n{\displaystyle \mathbf {x} \in \mathbb {R} ^{n}}.
This defines an equivalence relation on the set of all norms on Rn. With this result you can check that a sequence of vectors in Rn converges with ‖⋅‖{\displaystyle \|\cdot \|} if and only if it converges with ‖⋅‖′{\displaystyle \|\cdot \|'}.
Here is a sketch of what a proof of this result may look like:
Because of the equivalence relation it is enough to show that every norm on Rn is equivalent to the Euclidean norm‖⋅‖2{\displaystyle \|\cdot \|{2}}. Let ‖⋅‖{\displaystyle \|\cdot \|} be an arbitrary norm on R_n. The proof is divided in two steps:
We show that there exists a β>0{\displaystyle \beta >0}, such that ‖x‖≤β⋅‖x‖2{\displaystyle \|\mathbf {x} \|\leq \beta \cdot \|\mathbf {x} \|{2}} for all x∈R n{\displaystyle \mathbf {x} \in \mathbf {R} ^{n}}. In this step you use the fact that every x=(x 1,…,x n)∈R n{\displaystyle \mathbf {x} =(x{1},\dots ,x_{n})\in \mathbf {R} ^{n}} can be represented as a linear combination of the standard basis: x=∑i=1 n e i⋅x i{\textstyle \mathbf {x} =\sum {i=1}^{n}e{i}\cdot x_{i}}. Then with the Cauchy–Schwarz inequality‖x‖=‖∑i=1 n e i⋅x i‖≤∑i=1 n‖e i‖⋅|x i|≤∑i=1 n‖e i‖2⋅∑i=1 n|x i|2=β⋅‖x‖2,{\displaystyle \|\mathbf {x} \|=\left\|\sum {i=1}^{n}e{i}\cdot x_{i}\right\|\leq \sum {i=1}^{n}\|e{i}\|\cdot |x_{i}|\leq {\sqrt {\sum {i=1}^{n}\|e{i}\|^{2}}}\cdot {\sqrt {\sum {i=1}^{n}|x{i}|^{2}}}=\beta \cdot \|\mathbf {x} \|{2},} where β:=∑i=1 n‖e i‖2{\textstyle \beta :={\sqrt {\sum {i=1}^{n}\|e_{i}\|^{2}}}}.
Now we have to find an α>0{\displaystyle \alpha >0}, such that α⋅‖x‖2≤‖x‖{\displaystyle \alpha \cdot \|\mathbf {x} \|{2}\leq \|\mathbf {x} \|} for all x∈R n{\displaystyle \mathbf {x} \in \mathbf {R} ^{n}}. Assume there is no such α{\displaystyle \alpha }. Then there exists for every k∈N{\displaystyle k\in \mathbf {N} } a x k∈R n{\displaystyle \mathbf {x} {k}\in \mathbf {R} ^{n}}, such that ‖x k‖2>k⋅‖x k‖{\displaystyle \|\mathbf {x} {k}\|{2}>k\cdot \|\mathbf {x} {k}\|}. Define a second sequence (x~k)k∈N{\displaystyle ({\tilde {\mathbf {x} }}{k}){k\in \mathbf {N} }} by x~k:=x k‖x k‖2{\textstyle {\tilde {\mathbf {x} }}{k}:={\frac {\mathbf {x} {k}}{\|\mathbf {x} {k}\|{2}}}}. This sequence is bounded because ‖x~k‖2=1{\displaystyle \|{\tilde {\mathbf {x} }}{k}\|{2}=1}. So because of the Bolzano–Weierstrass theorem there exists a convergent subsequence (x~k j)j∈N{\displaystyle ({\tilde {\mathbf {x} }}{k_{j}}){j\in \mathbf {N} }} with limit a∈{\displaystyle \mathbf {a} \in }R_n. Now we show that ‖a‖2=1{\displaystyle \|\mathbf {a} \|{2}=1} but a=0{\displaystyle \mathbf {a} =\mathbf {0} }, which is a contradiction. It is ‖a‖≤‖a−x~k j‖+‖x~k j‖≤β⋅‖a−x~k j‖2+‖x k j‖‖x k j‖2⟶j→∞0,{\displaystyle \|\mathbf {a} \|\leq \left\|\mathbf {a} -{\tilde {\mathbf {x} }}{k_{j}}\right\|+\left\|{\tilde {\mathbf {x} }}{k{j}}\right\|\leq \beta \cdot \left\|\mathbf {a} -{\tilde {\mathbf {x} }}{k{j}}\right\|{2}+{\frac {\|\mathbf {x} {k_{j}}\|}{\|\mathbf {x} {k{j}}\|{2}}}\ {\overset {j\to \infty }{\longrightarrow }}\ 0,} because ‖a−x~k j‖→0{\displaystyle \|\mathbf {a} -{\tilde {\mathbf {x} }}{k_{j}}\|\to 0} and 0≤‖x k j‖‖x k j‖2<1 k j{\displaystyle 0\leq {\frac {\|\mathbf {x} {k{j}}\|}{\|\mathbf {x} {k{j}}\|{2}}}<{\frac {1}{k{j}}}}, so ‖x k j‖‖x k j‖2→0{\displaystyle {\frac {\|\mathbf {x} {k{j}}\|}{\|\mathbf {x} {k{j}}\|{2}}}\to 0}. This implies ‖a‖=0{\displaystyle \|\mathbf {a} \|=0}, so a=0{\displaystyle \mathbf {a} =\mathbf {0} }. On the other hand ‖a‖2=1{\displaystyle \|\mathbf {a} \|{2}=1}, because ‖a‖2=‖lim j→∞x~k j‖2=lim j→∞‖x~k j‖2=1{\displaystyle \|\mathbf {a} \|{2}=\left\|\lim {j\to \infty }{\tilde {\mathbf {x} }}{k{j}}\right\|{2}=\lim {j\to \infty }\left\|{\tilde {\mathbf {x} }}{k{j}}\right\|_{2}=1}. This can not ever be true, so the assumption was false and there exists such a α>0{\displaystyle \alpha >0}.
See also
[edit]
Exponential object, for theoretical explanation of the superscript notation
Geometric space
Real projective space
Sources
[edit]
Kelley, John L. (1975). General Topology. Springer-Verlag. ISBN0-387-90125-6.
Munkres, James (1999). Topology. Prentice-Hall. ISBN0-13-181629-2.
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3595 | http://homepage.ntu.edu.tw/~luohm/micro2020f/chapter9.pdf | Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Part IV: Production and Supply 9. Production Functions 10. Cost Functions 11. Proft Maximization 1 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Chapter 9 Production Functions Ming-Ching Luoh 2020.12.18.
2 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Marginal Productivity Isoquant Maps and the Rate of Technical Substitution Returns to Scale Te Elasticity of Substitution Four Simple Production Functions Technical Progress Extensions: Many-Input Production Functions 3 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Marginal Productivity ●Production function. Te frm’s production function for a particular good (q), q = f (k, l), shows the maximum amount of the good that can be produced using alternative combinations of capital (k) and labor (l).
Marginal physical product ●Marginal physical product. Te marginal physical product of an input is the additional output that can be produced by using one more unit of that input while holding other inputs constant.
MPk = ∂q ∂k = fk MPl = ∂q ∂l = fl 4 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Diminishing marginal productivity ●Te marginal physical product depends on how much of that input is used.
●Diminishing marginal productivity is about the second-order derivatives ∂MPk ∂k = ∂2 f ∂k2 = fkk < 0, for high enough k, ∂MPl ∂l = ∂2 f ∂l2 = fll < 0, for high enough l.
●Te assumption of diminishing marginal productivity was originally proposed by 19th-century economist Tomas Malthus, who worried that rapid increases in population would result in lower labor productivity.
●Malthus’ gloomy predictions led economics to be called the “dismal science." 5 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Changes in the marginal productivity of labor also depend on changes in other inputs such as capital.
●We need to consider flk, flk = ∂MPl ∂k which is ofen the case flk > 0.
●It appears that labor productivity has risen signifcantly since Malthus’ time, primarily because increases in capital inputs (along with technical improvements) have ofset the impact of decreasing marginal productivity alone.
6 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Average productivity ●In common usage, the term labor productivity ofen means average productivity.
●We defne the average product of labor (APl) to be APl = output labor input = q l = f (k, l) l Notice that APl also depends on the level of capital used.
7 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Example 9.1 A Two-Input Production Function ●Suppose the production function for fyswatters during a particular period can be represented by q = f (k, l) = 600k2l2 −k3l3.
To construct MPl and APl, we must assume a value for k.
Let k = 10, then the production function becomes q = 60, 000l2 −1, 000l3 ●Te marginal productivity function is MPl = ∂q ∂l = 120, 000l −3, 000l2, which diminishes as l increases. q has a maximum value when MPl = 0, 120, 000l −3000l2 = 0, l = 40 8 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●To fnd average productivity, we hold k = 10 and solve APl = q l = 60, 000l −1000l2 APl reaches its maximum where ∂APl ∂l = 60, 000 −2000l = 0 l = 30 When l = 30, APl = MPl = 900, 000. When APl is at its maximum, APl and MPl are equal.
●Tis result is general. Because ∂APl ∂l = ∂( q l ) ∂l = l ⋅MPl −q ⋅1 l2 , at a maximum l ⋅MPl = q or MPl = ALl.
9 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Isoquant Maps and the Rate of Technical Substitution ●To illustrate the possible substitution of one input for another in a production function, we use its isoquant map.
●Isoquant, An isoquant shows those combinations of k and l that can produce a given quantity of output (say, q0).
Mathematically, an isoquant records the set of k and l that satisfes f (k, l) = q0 ●Tere are infnitely many isoquants in the k −l plane. Each isoquant represents a diferent level of output.
10 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Figure 9.1 An Isoquant Map 11 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Marginal rate of technical substitution (RTS) ●Marginal rate of technical substitution. Te marginal rate of technical substitution (RTS) shows the rate at which, having added a unit of labor, capital be decreased while holding output constant along an isoquant.
RTS(l for k) = −dk dl ∣ q=q0 12 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions RTS and marginal productivities ●Te equation for an isoquant is q0 = f (k(l), l), Total diferentiate the equation with q0 constant gives 0 = fk dk dl + fl = MPk dk dl + MPl.
Terefore, RTS(l for k) = −dk dl ∣ q=q0 = MPl MPk RTS is given by the ratio of the inputs’ marginal productivity, and because MPl and MPk are both nonnegative, RTS will be positive.
13 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Reasons for a diminishing RTS ●It is not possible to derive a diminishing RTS from the assumption of diminishing marginal productivity alone.
●To show that isoquants are convex, we would like to show that dRTS dl < 0. Because RTS = fl fk , we have dRTS dl = d(fl/fk) dl ●Using the fact that dk dl = −fl fk along an isoquant and Young’s theorem (fkl = flk), we have dRTS dl = fk(fll + flk ⋅dk/dl) −fl(fkl + fkk ⋅dk/dl) (fk)2 = f 2 k fll −2fk fl fkl + f 2 l fkk (fk)3 < 0 if fkl > 0.
14 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions dRTS dl = f 2 k fll −2fk fl fkl + f 2 l fkk (fk)3 ●Te denominator is positive because we have assumed fk > 0.
●Te ratio will be negative if fkl is positive because fll and fkk are both assumed to be negative.
●Intuitively, it seems reasonable that fkl = flk should be positive. If workers have more capital, they will be more productive.
●But some production functions have fkl < 0 over some input ranges.
●Assuming diminishing RTS means that we are assuming that MPl and MPk diminish rapidly enough to compensate for any possible negative cross-productivity efects.
15 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Example 9.2 A Diminishing RTS ●Production function in Example 9.1 is q = f (k, l) = 600k2l2 −k3l3.
●Marginal productivity functions are MPl = fl = ∂q ∂l = 1200k2l −3k3l2 MPk = fk = ∂q ∂k = 1200kl2 −3k2l3 Both will be positive for values of k and l for which kl < 400.
●Because fll = 1200k2 −6k3l fkk = 1200l2 −6kl3, Tus function exhibits diminishing marginal productivities for sufciently large values of k and l. fll, fkk < 0 if kl > 200.
16 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●However, even within the range 200 < kl < 400 where the marginal productivity behave “normally," this production function may not necessarily have a diminishing RTS.
●Cross diferentiation of either of the marginal productivity functions yields fkl = flk = 2400kl −9k2l2 which is positive only for kl < 2400/9 ≐266 17 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Returns to Scale ●How does output respond to increases in all inputs together?
Suppose that all inputs are doubled, would output double?
●Tis is a question of the returns to scale exhibited by the production function that has been of interest to economists ever since Adam Smith intensively studies the production of pins.
●A doubling of scale permits a greater division of labor and specialization of function.
●Doubling of the inputs also entails some loss in efciency because managerial overseeing may become more difcult.
18 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Returns to scale. If the production function is given by q = f (k, l) and if all inputs are multiplied by the same positive constant t (where t > 1), then we classify the returns to scale of the production function by Efect on Output Returns to Scale f (tk, tl) = t f (k, l) = tq Constant f (tk, tl) < t f (k, l) = tq Decreasing f (tk, tl) = t f (k, l) = tq Increasing ●It is possible for a function to exhibit constant returns to scale for some levels of input usage and increasing or decreasing returns for other levels.
●Te degree of returns to scale is generally defned within a fairly narrow range of variation in input usage.
19 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Constant returns to scale ●Constant returns-to-scale production functions are homogeneous of degree one in inputs because f (tk, tl) = t1 f (k, l) = tq ●If a function is homogeneous of degree k, its derivatives are homogeneous of degree k −1.
●Te marginal productivity functions derived from a constant returns to scale production are homogeneous of degree zero.
20 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Tat is, MPk = ∂f (k, l) ∂k = ∂f (tk, tl) ∂k , MPl = ∂f (k, l) ∂l = ∂f (tk, tl) ∂l , for any t > 1. Let t = 1/l, then MPk = ∂f (k/l, 1) ∂k , MPl = ∂f (k/l, 1) ∂l , ●Te marginal productivity of any input depends on the ratio of capital and labor, not on the absolute levels of these inputs.
21 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Homothetic production functions ●Te RTS(= MPl/MPk) for any constant-returns-to scale production will depend only on the ratio of k and l since RTS = MPl MPk = ∂f (k/l,1) ∂l ∂f (k/l,1) ∂k , not on the absolute level of k and l.
●Tis is a homothetic function, and its isoquants will be radial expansion of one another.
●However, a production function can have a homothetic isoquant map even if it does not exhibit constant returns to scale.
●As shown in Chapter 2, this property of homotheticity is retained by any monotonic transformation of a homogeneous function.
22 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Figure 9.2 Isoquant Map for a Constant Returns-to-Scale Production Function 23 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●For example, if f (k, l) is a constant returns-to-scale production function, let F(k, l) = f (k, l)γ, where γ is any positive exponent. If γ > 1 then F(tk, tl) = f (tk, tl)γ = tγ f (k, l)γ = tγF(k, l)>tF(k, l) for any t > 1. F exhibits increasing returns to scale and γ captures the degree of the increasing returns to scale.
●An identical proof can show that the function F exhibits decreasing returns to scale for γ < 1.
●In these cases, changes in the returns to scale will just change the labels on the isoquants rather than their shapes.
24 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Te n-input case ●Te defnition of returns to scale can be generalized to a production function with n inputs, q = f (x1, x2, ⋅, xn) If all inputs are multiplied by t > 1, we have f (tx1, tx2, ⋅, txn) = tk f (x1, x2, ⋅, xn) = tkq ●If k = 1, the production function exhibits constant returns to scale.
●Decreasing and increasing returns to scale correspond to the cases k < 1 and k > 1, respectively.
25 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Te Elasticity of Substitution ●An important characteristic of the production function is how “easy" it is to substitute one input for another.
●Tis is a question about the shape of a single isoquant rather than about the whole isoquant map.
●Te elasticity of substitution measures the proportionate change in k/l relative to the proportionate change in the RTS along an isoquant.
●For the production function q = f (k, l), σ = %∆(k/l) %∆RTS = d(k/l) dRTS × RTS k/l = d ln(k/l) d ln RTS = d ln(k/l) d ln(fl/fk) ●Te value of σ will always be positive because k/l and RTS move in the same direction.
26 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Figure 9.3 Graphic Description of the Elasticity of Substitution 27 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●If σ is high, the RTS will not change much relative to k/l, the isoquant will be close to linear.
●If σ is low, the RTS will change by a substantial amount as k/l changes, the isoquant will be sharply curved.
●In general, it is possible that σ will vary as one moves along an isoquant and as the scale of production changes.
●Ofen, it is convenient to assume that σ is constant along an isoquant.
28 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Four Simple Production Functions Case 1: Linear (σ = ∞) q = f (k, l) = αk + βl ●All isoquants are straight lines with slope −β/α.
●Constant returns to scale.
●RTS is constant.
29 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Case 2: Fixed proportions (σ = 0) q = min(αk, βl), α, β > 0 ●Capital and labor must always be used in a fxed ratio.
●Te frm will always operate along a ray where k/l is constant.
●Because k/l is constant, σ = 0.
30 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Case 3: Cobb-Douglas (σ = 1) q = f (k, l) = Akαlβ f (tk, tl) = A(tk)α(tl)β = tα+β f (k, l) 31 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ● RTS = fl fk = = βAkαl β−1 αAkα−1l β = β α ⋅k l ln(RTS) = ln(β α) + ln(k l ) σ ≡ ∂ln(k/l) ∂ln(RTS) = 1 ●Tis production function can exhibit any returns to scale, depending on whether α + β ⋛1.
●Te Cobb-Douglas production function is useful in many applications because it is linear in logarithms: ln q = ln A + α ln k + β ln l α is the elasticity of output with respect to k while β is the elasticity of output with respect to l.
32 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Case 4: CES production function ●Te constant elasticity of substitution (CES) production was frst introduced by Arrow et al. in 1961. It is given by q = f (k, l) = (kρ + l ρ)γ/ρ for ρ ≤1, ρ ≠0 and γ > 0. Since f (tk, tl) = [(tk)ρ + (tl)ρ]γ/ρ = tγ f (k, l) For γ > 1 the function exhibits increasing returns to scale, whereas for γ < 1 it exhibits decreasing returns.
33 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●From the production function and marginal product of k and l, q = f (k, l) = (kρ + l ρ)γ/ρ MPl = ∂f ∂l = (kρ + l ρ)γ/ρ−1 ⋅ρl ρ−1 MPk = ∂f ∂l = (kρ + l ρ)γ/ρ−1 ⋅ρkρ−1, we have RTS = MPl MPk = (k l ) 1−ρ ln RTS = (1 −ρ)ln(k l ) Terefore, σ = ∂ln(k/l) ∂ln RTS = 1 1 −ρ ●Te linear, fxed-proportions, and Cobb-Douglas cases correspond to ρ = 1, ρ = −∞and ρ = −0, respectively.
34 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Ofen the CES function is used with a distributional weight, α(0 ≤α ≤1), to indicate the relative signifcance of the inputs: q = f (k, l) = [αkρ + (1 −α)l ρ]γ/ρ ●With constant returns to scale and ρ = 0, this function converges to the Cobb-Douglas form q = f (k, l) = kαl1−α.
35 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Example 9.3 A Generalized Leontief Production Function ●Suppose that the production function is given by q = f (k, l) = k + l + 2 √ kl.
Tis is a special case of a class of functions named for the Russian-American economist Wassily Leontief.
●Tis function clearly exhibits constant returns to scale.
●Marginal productivities are fk = 1 + (k/l)−0.5, fl = 1 + (k/l)0.5.
●RTS = fl/fk diminishes as k/l falls, so the isoquants have the usual convex shape.
36 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Two ways to fnd the elasticity of substitution for this production function. First, the function can be factored as q = f (k, l) = k + l + 2 √ kl = ( √ k + √ l)2 = (k0.5 + l0.5)2 which makes clear that this function has a CES form with ρ = 0.5 and γ = 1. Hence σ = 1/(1 −ρ) = 2.
●Another more exhaustive approach is to apply the defnition in footnote 6 directly.
σ = fk fl f ⋅fkl = [1 + (k/l)−0.5][1 + (k/l)0.5] q ⋅0.5(kl)−0.5 = 2 + (k/l)−0.5 + (k/l)0.5 0.5(k/l)0.5 + 0.5(k/l)−0.5 + 1 = 2 37 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●For the production function q = f (k, l), it can be shown that the elasticity of substitution σ = d ln(k/l) d ln(fl/fk) can be derived to be σ = (k fk + l fl)fk fl kl(−fkk f 2 l + 2fk fl fkl −fll f 2 k ).
●If the production function exhibits constant returns to scale, then the elasticity of substitution can be reduced to σ = fk ⋅fl f ⋅fkl 38 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Proof: σ ≡d ln(k/l) d ln(fl/fk) = fl/fk k/l ⋅d(k/l) d(fl/fk) Total diferentiating d(k/l) and d(fl/fk), along with the fact that −dk dl = fl fk and thus dl = −fk fl dk gives d(k/l) = 1 l2(ldk −kdl) = 1 l2 (l+k fk fl ) dk = 1 l2(k fk + l fl)dk fl d(k/l) k/l = 1 kl (k fk + l fl)dk fl 39 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions d ( fl fk ) = ∂(fl/fk) ∂k ⋅dk + ∂(fl/fk) ∂l ⋅dl = (∂(fl/fk) ∂k + ∂(fl/fk) ∂l ⋅(−fk fl )) dk = (fl ∂(fl/fk) ∂k −fk ∂(fl/fk) ∂l ) dk fl d ( fl fk ) fl/fk = fk fl (fl ∂(fl/fk) ∂k −fk ∂(fl/fk) ∂l ) dk fl 40 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Terefore, σ = d(k/l) k/l d(fl/fk) fl/fk = fl(k fk + l fl) fkkl (fl ∂(fl/fk) ∂k −fk ∂(fl/fk) ∂l ) Since ∂(fl/fk) ∂k = 1 f 2 k (fk flk −fl fkk) ∂(fl/fk) ∂l = 1 f 2 k (fk fll −fl fkl) fl ∂(fl/fk) ∂k −fk ∂(fl/fk) ∂l = 1 f 2 k (fl fk flk −f 2 l fkk −f 2 k fll + fk fl fkl) = 1 f 2 k (−f 2 l fkk −f 2 k fll + 2fk fl fkl) Finally, σ = fk fl(k fk + l fl) kl (−f 2 l fkk −f 2 k fll + 2fk fl fkl) 41 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●For a general production function q = f (k, l), the elasticity of substitution is σ = fk fl(k fk + l fl) kl (−f 2 l fkk −f 2 k fll + 2fk fl fkl) ●If f (k, l) exhibits constant returns to scale, or, f (k, l) is homogeneous of degree one, the the marginal products of the inputs, fk, and fl, are homogeneous of degree zero. Tus, according to the Euler’s theorem, fkk + fll = f fkkk + fkll = 0 ⇒fkk = −l k fkl flkk + flll = 0 ⇒fll = −k l flk 42 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Tus, σ = fk fl f kl (−f 2 l (−l k fkl) −f 2 k ( k l flk) + +2fk fl fkl) = fk fl f (l2 f 2 l + k2 f 2 k + 2kl fk fl) fkl = fk fl f (k fk + l fl)2 fkl = fk fl f ⋅fkl 43 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Technical Progress ●Following the development of superior production techniques, the same level of output can be produced with fewer inputs. Te isoquant shifs inward.
Figure 9.5 Technical Progress 44 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Measuring technical progress ●Te frst observation to be made about technical progress is that historically the rate of growth of output overtime has exceeded the growth rate that can be attributed to the growth in conventionally defned inputs.
●Suppose that we let q = A(t)f (k, l) be the production function for some good, where A(t) represents all infuences that go into determining q other than k and l.
●Changes in A over time represent technical progress. A is shown as a function of time (t), dA/dt > 0.
45 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Growth accounting ●Diferentiating the production function with respect to time, dq dt = dA dt ⋅f (k, l) + A ⋅d f (k, l) dt = dA dt ⋅q A + q f (k, l) (∂f ∂k ⋅dk dt + ∂f ∂l ⋅dl dt) ●Dividing by q gives dq/dt q = dA/dt A + ∂f /∂k f (k, l) ⋅dk dt + ∂f /∂l f (k, l) ⋅dl dt = dA/dt A + ∂f ∂k ⋅ k f (k, l) ⋅dk/dt k + ∂f ∂l ⋅ l f (k, l) ⋅dl/dt k 46 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Let Gx denote the proportional rate of growth of variable x per unit of time, (dx/dt)/x, then the previous equation can be rewritten as Gq = GA + eq,kGk + eq,lGl, where ●eq,k = ∂f ∂k ⋅ k f (k,l) is the elasticity of output with respect to capital.
●eq,l = ∂f ∂l ⋅ l f (k,l) is the elasticity of output with respect to labor.
47 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●In a pioneer study of U.S. economy between the years 1909 and 1949, R. M. Solow recorded the following values: Gq = 2.75% per year Gl = 1.00% Gk = 1.75% eq,l = 0.65 eq,k = 035 Consequently, GA = Gq −eq,lGl −eq,kGk = 275 −0.65 ⋅1.00 −0.35 ⋅1.75 = 1.50 ●More than half (1.50/2.75 ≐55%) of the growth in real output could be attributed to technical change (A).
48 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Example 9.4 Technical Progress in the Cobb-Douglas Production Function ●Te Cobb-Douglas production function with technical progress provides an especially easy avenue for illustrating technical progress.
●Assuming constant returns to scale, the production function is q = A(t)f (k, l) = A(t)kαl1−α.
●Also assume that technical progress occurs at a constant exponential (θ), A(t) = Aeθt, then the production function becomes q = Aeθtkαl1−α.
49 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Taking logarithm and diferentiate with respect to t gives ln q = ln A + θt + α ln k + (1 −α)ln l ∂ln q ∂t = ∂q/∂t q = Gq = θ + α ⋅∂ln k ∂t + (1 −α) ⋅∂ln l ∂t = θ + αGk + (1 −α)Gl.
●Suppose A = 10, θ = 0.03, α = 0.5 and that a frm uses an input mix of k = l = 4.
●At t = 0, output is 40 (=10 ⋅40.5 ⋅40.5).
●Afer 20 years (t = 20), the production function becomes q = 10e0.03⋅20k0.5l0.5 = 10 ⋅1.82 ⋅k0.5l0.5 = 18.2 ⋅k0.5l0.5 With k = l = 4, q = 72.8.
50 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Input-augmenting technical progress. A plausible approach to modeling improvements in labor and capital separately is to assume that the production function is q = A(eψtk)α(eεtl)1−α, where ψ represents the annual rate of improvement in capital input and ε represents the annual rate of improvement in labor input.
●However, because of the exponential nature of the Cobb-Douglas function, this would be indistinguishable from our original example: q = Ae[αψ+(1−α)ε]tkαl1−α = Aeθtkαl1−α where θ = αψ + (1 −α)ε.
51 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Many-Input Production Functions E9.1 Cobb-Douglas Te many-input Cobb-Douglas production function is given by q = n ∏ i=1 xαi i a. Tis function exhibits constant returns to scale if ∑n i=1 αi = 1.
b. αi is the elasticity eq,xi. Since 0 ≤αi ≤1, each input exhibits diminishing marginal productivity.
c. Any degree of increasing returns to scale can be cooperated into this function, depending on ε = n ∑ i=1 αi.
52 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions d. Te elasticity of substitution between any two inputs in this production function is 1.
σij = ∂ln(xi/xj) ∂ln(fj/fi) fj fi = αjx α j−1 j Πi≠jxαi i αixαi−1 j Πj≠ix α j j = αj αi ⋅xi xj Hence, ln( fj fi ) = ln(αj αi ) + ln(xi xj ) and σij = 1.
●Because the parameter is so constrained, the function is generally not used in econometric analyses of microeconomic data on frms. However, the function has a variety of general uses in macroeconomics.
53 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions Te Solow growth model ●Te Solow model of equilibrium growth can be derived using a two-input constant returns-to-scale Cobb-Douglas function of the form q = Akαl1−α, where A is a technical change factor that can can be represented by exponential growth of the form A = eat Dividing both sides by l yields ˆ q = eatˆ kα where ˆ q = q/l, ˆ k = k/l.
54 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions ●Solow shows that economies will evolve toward an equilibrium value of ˆ k. Hence cross-country diferences in growth rates can be accounted for only by diferences in the technical change factor a.
●However, the equation is incapable of explaining the large diferences in per capita output (ˆ q) observed around the world.
●A second shortcoming is that it ofers no explanation of the technical change parameter a. By adding additional factors, it becomes easier to understand how the parameter a may respond to is easier to economic incentives. Tis is the key insight of literature on “endogenous" growth theory. (Romer 1996) 55 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions E9.2 CES Te many-input constant elasticity of substitution (CES) production function is given by q = (∑αixρ i ) γ/ρ , ρ ≤1.
a. Tis function exhibits constant returns to scale for γ = 1. For γ > 1, the function exhibits increasing returns to scale.
b. Tis function exhibits diminishing marginal productivities for each inout when γ ≤1.
c. Te elasticity of substitution is given by σ = 1 1 −ρ , and this applies to substitution between any two of the inputs.
56 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions E9.3 Nested production functions ●In some applications, Cobb-Douglas and CES production functions are combined into a “nested" single function.
●For example, there are three primary inputs, x1, x2, x3.
Suppose that x1 and x2 are relatively closely related in their use (e.g. capital and energy), whereas the third input (labor) is relatively distinct.
●One can use a CES aggregator function to construct a composite input for capital services of the form x4 = [γxρ 1 + (1 −γ)xρ 2]1/ρ.
●Ten the fnal production might take a Cobb-Douglas form q = xα 3 xβ 4 ●Nested production functions have been used in studies that seek to measure the precise nature of the substitutability between capital and energy inputs.
57 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions E9.4 Generalized Leontief q = n ∑ i=1 n ∑ j=1 αij √xixj, where αij = αji, ●Te function exhibits constant returns to scale.
●Because each input appears both linearly and under the radical, the function exhibits diminishing marginal productivities to all inputs.
●Te restriction αi j = αji is used to ensure symmetry of the second-order partial derivatives.
58 / 59 Outline Marginal Productivity RTS Returns to Scale Elas. of Substi.
4 Production Functions Technical Progress Extensions E9.5 Translog ln q = α0 + n ∑ i=1 αi ln xi + 0.5 n ∑ i=1 n ∑ j=1 αij ln xi ln xj, αi j = αji ●Cobb-Douglas function is a special case of this function where α0 = αij = 0 for all i, j.
●Tis function may assume any degree of returns to scale. If n ∑ i=1 αi = 1, n ∑ j=1 αij = 0 for all i, then this function exhibits constant returns to scale.
●Te condition αij = αji is required to ensure equality of the cross-partial derivatives.
●Translog production function has been used to study the ways in which newly arrived workers may substitute for existing workers.
59 / 59 |
3596 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8152179/ | Delirium Tremens in the Older Adult - PMC
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J Neurosci Nurs
. Author manuscript; available in PMC: 2021 Dec 1.
Published in final edited form as: J Neurosci Nurs. 2020 Dec;52(6):316–321. doi: 10.1097/JNN.0000000000000543
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Delirium Tremens in the Older Adult
Malissa A Mulkey
Malissa A Mulkey
1 University of North Carolina-Rex Hospital, Raleigh, NC.
Find articles by Malissa A Mulkey
1, DaiWai M Olson
DaiWai M Olson
2 University of Texas Southwest, Dallas, TX.
Find articles by DaiWai M Olson
2
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1 University of North Carolina-Rex Hospital, Raleigh, NC.
2 University of Texas Southwest, Dallas, TX.
✉
Email: malissa.mulkey@duke.edu
PMC Copyright notice
PMCID: PMC8152179 NIHMSID: NIHMS1697785 PMID: 33156592
The publisher's version of this article is available at J Neurosci Nurs
Abstract
INTRODUCTION:
Caring for patients experiencing alcohol withdrawal syndrome can be challenging. Patients 65 and older are at increased risk for alcohol withdrawal syndrome related complications. The higher prevalence of co-morbidities, including cognitive impairment, longer drinking history and greater sensitivity to alcohol withdrawal syndrome treatment are the result of decreased ability of the brain to adapt to stressors such as illness, trauma, or surgery.
DELIRIUM TREMENS:
Symptoms may appear earlier from the last drink and present with a wide range of symptoms. The most effective interventions require high-quality nursing care delivery to prevent, decrease the severity and shorten the duration of delirium.
NURSING IMPLICATIONS:
Strategies that help minimize these challenges starts with obtaining the patient’s selfreport of their alcohol use history. Nurses should be diligent in their monitoring for signs of active alcohol withdrawal. Screening and assessment tools such as the Clinical Institute Withdrawal Assessment for Alcohol–Revised should guide pharmacological management. To support nurses in identifying delirium tremens, this manuscript seek to describe the underlying pathophysiology, key assessment components and nursing management of delirium tremens in the older adult.
Keywords: alcohol withdrawal, delirium, delirium tremens, geriatric, older adult
There is a robust and growing body of literature discussing assessment and identification of delirium in the older adult.1-3 Delirium is characterized by an acute onset of confusion that fluctuates throughout the day. Delirium tremens (DTs) is a unique form of delirium characterized by acute confusion in the setting of agitation, tachycardia, tremors, and hypertension occurring secondary to withdrawal from alcohol.4,5 It is the most severe level of alcohol with-drawal, common in the acute and critical care environment, and may be particularly difficult to discern in the setting of traumatic brain injury and new-onset seizures.4-6 Literature discussing alcohol-related DTs is well established.7 Despite age being a risk factor for both delirium and DTs, there is limited discussion of differentiating DTs, in the geriatric patient.
In hospitalized patients, DTs can be difficult to manage. Approximately 5% of patients with alcohol use disorder develop DTs.8 Although the underlying etiology is not well understood, there seems to be a relationship between duration of alcohol use and severity of withdrawal symptoms. Delirium tremens prolongs hospital length of stay, increases healthcare cost, and triples mortality risk.9 Effective management requires early recognition and aggressive treatment.6,7 Because of their continuous presence and relationships with patients and families, nurses are in an ideal position to detect the signs and symptoms of alcohol withdrawal.4 Therefore, it is important for nurses to elicit information regarding alcohol use and critically assess for signs of withdrawal.5 This article seeks to describe the underlying pathophysiology, key assessment components, and nursing management of DTs in the older adult.
Alcohol depresses the central nervous system by increasing the activity of γ-aminobutyric acid (GABA), an inhibitory neurotransmitter responsible for downregulation of excitatory neurotransmitters (ie, glutamate). Downregulation is due to inhibition of N-methyl-D-aspartate (NMDA) receptors similar to a quickly progressing NMDA encephalitis.4,10 Prolonged use further increases GABA production, function, and alteration in the associated receptors.7,11 Because of the multifactorial complex nature, primary treatment is to alleviate the cause.
Stages of Alcohol Withdraw
Although all patients will not progress sequentially through the 3 stages of withdrawal, the stages describe how alcohol withdrawal symptoms evolve and de-fine the clinical features (Table 1). Most patients only progress to level 1. Stage 1 can occur within hours, if an individual stops or greatly reduces alcohol consumption, presenting with signs of a hangover including headache, nausea and vomiting, tremors, and anxiety. With resumed consumption, symptoms typically subside. If the individual does not consume enough alcohol, within 24 hours, he/she may progress on to stage 2. During the second stage, increased sympathetic system activity leads to tachycardia, hypertension, tremors, and diaphoresis. The individual’s demeanor may change as he/she becomes agitated but temporarily settles down when directed. Administering sedatives at this point may delay progression. Keeping these individuals in bed and preventing therapy interruption such as removing tubes, lines, and monitoring equipment may become increasingly challenging due to hyperactivity, restlessness, and frigidity. Advanced stage 2 becomes apparent when hallucinations, delusions, and nightmares develop. A distinguishing feature of these hallucinations is the ability to maintain lucidity and recognize altered perceptions that are related to alcohol withdrawal. Aggressive management with sedation and close monitoring for respiratory decline are needed at this point. Approximately one-third of patients require intensive care, especially older adults. Although more commonly seen in patients with a history of epilepsy, development of tonic-clonic seizures marks progression to stage 3. Although patients experiencing seizures may include DTs in the differential, DTs are not seizures.12 Progression to stage 3 occurs in less than 10% of patients and usually takes more than 2 days.12
TABLE 1.
Stages of Alcohol Withdrawal
| Stage | Onset | Duration | Physical Symptoms | Cognitive Symptoms |
:---: :---:
| 1 | 0-8 h | Up to 1 d | Headache, nausea and vomiting, and tremors | Anxiety |
| 2 | 2-48 h | Up to 1 wk | Tachycardia, hypertension, tremors, and diaphoresis | Agitation Hyperactivity, restlessness, and frigidity Hallucinations, delusions, and nightmares Anhedonia, dysphoric behavior, and cravings |
| 3 | 3-7 d | Up to 2 wk | Tonic-clonic seizures | Delirium tremens |
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Similar to onset and progression, recovery is also proportional to severity. Patients receiving interventions will typically recover within 24 hours of outpatient treatment. When patients progress beyond stage 1, the recovery is more complex and often requires hospitalization for a few days to several weeks. On-going symptoms including anhedonia, dysphoric behavior, and cravings are thought to be the result of alterations in dopamine within the central nervous system and can be more frequent in older adults.13
Patients 65 years and older are at an increased risk for AWS-relatedcomplications because of the higher prevalence of comorbidities, including cognitive impairment, longer drinking history, and greater sensitivity to AWS treatment. Older patients who have less physiologic reserve may become symptomatic sooner as a result of the brain’s limited ability to adapt to stressors such as illness, trauma, or surgery. However, these notions are not universally true, and a higher index of suspicion for AWS should be raised. Symptoms may appear earlier (6-12 hours) from the last drink and can present with a wide range of symptoms.14 When symptoms appear, severity of AWS scores may be higher and duration is likely to be longer than that of younger individuals.15
When an acutely ill older patient develops DTs, there is a greater hyperdynamic state and electrophysiological changes such as a prolonged QT interval, tendency for greater increases in cardiac enzymes, and the odds of dying increase by 1.56.16 Hypokalemia and hypomagnesemia secondary to renal and extrarenal losses increase the risk for arrhythmias including torsade de pointes, sustained ventricular tachycardia, atrial fibrillation, and supraventricular tachycardia, causing a supply-and-demand imbalance.16
Identifying and Determining the Severity
Like other forms of delirium, AWS-related delirium identification can be challenging. In addition, because of the increased risk for seizure with advanced age, AWS may be masked as a possible seizure disorder.12 Monitoring older patients for complaints or signs of insomnia, depression, anxiety, memory loss, aches and pains, poor diet, and decreased libido may indicate the presence of an AUD but are often viewed as part of the aging process. Furthermore, the hallmark sign of becoming “nonfunctional” such as missing work or reporting intoxicated is not likely to occur in this age group because many are retired or live alone.13
Several bedside assessment tools are available to assist with early detection. When screening tools are part of the routine process, patient injury and risk of readmission are reduced.17,18 In an effort to prevent DTs, screening tools are used to evaluate the frequency and quantity of alcohol consumption. Screening tool use is important and should be initiated at the point of hospital entry including emergency departments, preoperative surgical areas, and inpatient units. These screening tools are effective when used short-term or with acute drinkers. However, exercise caution with patients considered to be severely alcoholic because tool effectiveness in this population remains debatable.
Anytime a patient presents with an alcohol history, one should try to quantify the use.19 Additional assessments should include risk factors such as history of DTs, previous withdrawal or detoxification, fluid and electrolyte disturbances, nutritional status, advanced age, and coexisting illnesses. Patients experiencing DTs often present with severe dehydration (as much as 10-L deficits). Hypoglycemia, potassium, magnesium, and phosphorous levels may also be critically low. The untoward effects of alcohol on the various organ systems may also be more pronounced. For example, a reduction in gastric enzymes may allow for higher concentrations of alcohol to enter the blood stream, and tolerance may be reduced. Adding alcohol to regular diuretic use produces a synergistic effect in the presence cardiac disease and may, therefore, result in more significant hypovolemia and orthostatic hypotension.13 Falls associated with alcohol-induced gait impairment are more likely to result in fractures due to preexisting osteoporosis.
Nursing Management
During the admission process, nurses are in a perfect position to obtain a patient’s medical history, including alcohol consumption. Nurses are often the first clinicians to observe alterations and elicit information from the patient and family regarding alcohol use and abuse. When patients have a known or suspected drinking history, nurses should attempt to quantify alcohol use with a validated screening tool.20 Exercise caution with family reports because only 1 study has found family reports to be equivalent to the patient’s self-report.20
Nonpharmacological Interventions
In Memorandum of Nursing, Nightingale21 described the importance of direct sunlight. Through her belief in the environmental theory, she advocated for hospitalized patients to be moved within their room to follow the sun rather than being stationary in dim rooms. Using light, a patient’s circadian rhythm could be restored. Unfortunately, in current practice, rooms and beds are designed with the placement of medical equipment being the higher priority. The more serious the patient’s condition, the greater the reliance on artificial light.21
Interventions that target the patient’s specific risk factors through staff education, nonpharmacologic intervention protocolization, and improvement in the patient’s environment have been shown to be the most effective intervention. These interventions require high-quality nursing care delivery to prevent, decrease the severity, and shorten the duration of delirium.22 In addition, studies have shown that these interventions are cost-effective in acute care. Interventions should focus on cognitive and sensory impairment, hydration and nutrition, oxygenation, infection prevention, early mobility, sleep, and relief of pain. Routine nursing care should promote daily activity; a quiet, well-lit environment; and minimizing room and bed changes. Additional nursing interventions should include use of hearing and visual aids, care-giver continuity, and encouraging personal items. Because patients with delirium-associated withdrawal typically experience hyperactive delirium, interventions should focus on decreasing sensory overload. These interventions might include limiting visits, removing items associated with painful stimulus such as urinary catheters, limiting monitoring with medical devices, and testing such as routine daily laboratory tests and frequent vital signs.
An individualized plan of care must address signs and symptoms for patients experiencing AWS. This plan should include diligent monitoring, astute assessment, and use of clinical judgment to detect subtle changes in patient condition. Subtle changes include early identification of worsening withdrawal and differentiating these changes from a worsening medical condition and aggressive medication management. Use of a severity assessment scale such as the CIWA-Ar in conjunction with clinical assessment and critical thinking is recommended to guide the therapy. Although a specific delirium scale cannot be recommended, the Society of Critical Care Medicine recommends using the CAM-ICU or the ICDSC in critical care units. Delirium screening may help assess patients experiencing AWS requiring mechanical ventilation.23 Early identification and preventative therapy are thought to reduce the risk of DTs and mortality.24 Hypokalemia on admission may identify an increased risk for DTs, even after the potassium level is corrected. Kartasheva et al24 found that CIWA-Ar scores of less than 10 indicate minimal alcohol with-drawal, whereas scores of 10 to 20 indicate moderate alcohol withdrawal, and scores greater than 20 indicate severe alcohol withdrawal; of those with a severe score, 24% developed DTs.
Pharmacological Management
Benzodiazepine administration remains the corner-stone for pharmacologic treatment of DTs. There are several common strategies for administering ben-zodiazepines: fixed, loading, and symptom-triggered. Fixed dosing means a predetermined dose is administered on a set and regular schedule. When using the loading strategy, a long-acting benzodiazepine is administered until the symptoms significantly improve, at which point the patient “self-tapers” as the benzodiazepine and associated metabolites slowly clear from his/her system.24 This method is frequently used in emergency departments. Symptom-triggered therapy uses a scale like the CIWA-Ar to determine the dose of benzodiazepines. Most available US and European research has found symptom-triggered therapy superior to routine fixed dose when patients are not medically complicated, psychiatric, or multiple substance users and should be used cautiously in these populations.8,17
Adjunct Therapies
Antipsychotics
Using antipsychotics such as haloperidol (phenothiazines and butyrophenones) in patients with AWS can be considered as an adjunct therapy.25,26 When added to the treatment regimen, it is usually because symptoms such as agitation, hallucinations, or delirium are not adequately managed with benzodiazepines alone. Antipsychotics should be used with caution because they can prolong the cardiac QT interval, increasing the risk for cardiac arrhythmias and mortality.27 In addition, they lower the seizure threshold, which may be problematic, especially in the first 48 hours when seizure risk is highest.
Antiepileptic Agents
Levetiracetam has a high affinity for GABAergic and glutamatergic receptors. It is thought that levetiracetam has a protective effect by reducing excessive neuronal activity.28 However, its mechanism of action in the treatment of AWS remains unclear. Despite this lack of understanding, along with benzodiazipines, antiepileptics are widely used in the treatment of AWS. Because there are a limited number of quality studies to adequately evaluate the effects of antiepileptic agents, the Cochrane Review did not recommend their use in the treatment of AWS.29 Although additional research is needed, available studies have shown levetiracetam shows the most promise by providing rapid stable clinical improvement. Significant study limitations and concerns with research design prevent recommendations related to carbamazepine, valproate, and gabapentin. Antiepileptic superiority to benzodiazepines remains unknown and, therefore, cannot be recommended for routine clinical practice and should be used with caution.28
Sedatives and Barbiturates
Dexmedetomidine decreases sympathetic overdrive and norepinephrine release and therefore causes a “cooperative sedation.”28,30 Because of its lack of GABAergic activity, it can only be used as an adjunct therapy to decrease autonomic hyperactivity not managed by benzodiapines alone. Dexmedetomidine use has been shown to reduce the amount of benzodiazepines needed to manage AWS. Propofol enhances the effects of GABA A receptor and decreases NMDA activity. Although only a few case studies have evaluated propofol, these cases have shown a rebound withdrawal when the drug is discontinued. Although barbiturates have been shown to be as effective as benzodiazepines, because of their narrow therapeutic index and lack of toxicity antidote, they are rarely used to treat AWS. Conversely, baclofen interacts with GABA B receptors. With limited research, its use cannot be recommended; however, the few available studies have indicated that baclofen may rapidly decrease severe AWS symptoms and cravings.28,31
Additional Therapies to Consider
Magnesium plays an important role in inhibiting the release of neurotransmitters and may decrease the hyperexcitability associated with NMDA.28 Chronic alcohol use has been associated with abnormal metabolism. Therefore, patients experiencing AWS are frequently given prophylactic doses of magnesium. However magnesium levels should be monitored to maintain normal levels. Like many of the other treatments, research is limited, leading to an inability of the Cochrane Review to recommend its routine use.31
Wernicke’s encephalopathy occurs as a result of lesions in the central nervous system. Thiamine depletion leads to lesions in the central nervous system.31 Thiamine deficiency causes the classic triad of symptoms including confusion, ataxia, and ophthalmoplegia seen with WE. The European Federation of Neurological Societies guidelines call for at least two of the following symptoms for a WE diagnosis: dietary deficiencies, eye signs, cerebellar dysfunction, and either an altered mental state or a mild memory impairment. Although WE rarely occurs, it has a high morbidity and mortality rate. According to the European Federation of Neurological Societies guidelines, patients experiencing AWS should receive parenteral thiamine supplementation regardless of the initial presentation. Thiamine has been found to prevent the development of WE and should be administered as early as possible to hasten recovery. In addition, thiamine should be administered parenterly before administering carbohydrate-containing fluids. This can only occur when clinicians obtain a thorough alcohol use history during the admission process.24
Impact
Older adults with alcohol abuse are at an increased risk for developing withdrawal symptoms earlier. Because of age-related changes and the potential for a longer abuse history, nurses need to obtain a reliable drinking history. Nursing care should include diligent monitoring using a standardized assessment tool and tailored interventions. Multidisciplinary team collaboration and careful consideration regarding vitamin, mineral, and electrolyte deficit replacement as well as adequate nutrition are critical. The frequency of hypertension and hypokalemia has been seen more frequently in the older adult experiencing AWS and may require more aggressive management.15 Symptom duration and severity may likely be increased with younger adults typically reaching symptom resolution around day 7, whereas older adults may still be symptomatic at day 9.15 Appropriate medication management and good basic nursing care are the most effective interventions.22 Greater restraint use, often seen in older adults, may contribute to prolonged symptom duration.15 Unfortunately, this is contrary to research-based recommendations that suggest restraint use in older adults is inappropriate and should be rarely used.15
Conclusion
Caring for patients experiencing AWS can be challenging. Strategies that help minimize these challenges start with obtaining the patient’s self-report of their alcohol use history. Nurses should be diligent in their monitoring for signs of active alcohol with-drawal. For patients experiencing AWS, screening and assessment tools such as the CIWA-Ar should guide pharmacological management.4 Although nurses do not prescribe medical management, their knowledge, expertise, and collaboration with providers are important when treatment decisions are being made. Nurses’ constant presence at the bedside provides them with the unique ability to detect and report changes in the patient’s condition essential to determining the appropriate diagnosis and treatment.
TABLE 2.
DSM-5 Diagnostic Criteria (American Psychiatric Association, 2013)
Cessation of or reduction in alcohol use in patients with a history of heavy and prolonged use
Two (or more) of the following occur within a few days of reduced alcohol use and are clinically significant enough to cause distress or impairment in social, occupational, or other important areas of functioning Autonomic hyperactivity such as diaphoresis and tachycardia
Increased hand tremor
Insomnia
Nausea or vomiting
Hallucinations or illusions
Agitation and/or anxiety
Tonic-clonic seizures
The symptoms are not due to a general medical condition and cannot be explained by another mental disorder.
Open in a new tab
Abbreviation: DSM-5, Diagnostic and Statistical Manual of Mental Disorders (Fifth Edition).
Acknowledgments
The authors thank Beverly Murphy, MLS, AHIP, FMLA, Medical Librarian, for assistance with the literature review.
Footnotes
The authors declare no conflicts of interest.
Contributor Information
Malissa A. Mulkey, University of North Carolina-Rex Hospital, Raleigh, NC.
DaiWai M. Olson, University of Texas Southwest, Dallas, TX.
References
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3597 | https://www.khanacademy.org/science/up-class-11th-biology/x6cdb38ba1d131d88:breathing-and-exchange-of-gases/x6cdb38ba1d131d88:mechanism-of-breathing/v/mechanism-of-breathing | Mechanism of Breathing (video) | Khan Academy
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UP Biology Grade 11
Course: UP Biology Grade 11>Unit 14
Lesson 1: Mechanism of breathing
Meet the lungs!
Mechanism of Breathing
The respiratory system
Pulmonary volume
Mechanism of breathing
Respiratory volumes and capacities
Science>
UP Biology Grade 11>
Breathing and exchange of gases>
Mechanism of breathing
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Mechanism of Breathing
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How do we breathe? What causes air to enter the lungs when we breathe in and air to leave the lungs when we breathe out? Watch this video to understand the mechanism of breathing.Created by Nivedhitha Suresh.
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3598 | https://www.youtube.com/watch?v=rBz5k2T7p4A | What is 0^0? Is 0 a natural number? MATHEMATICIANS DISAGREE!
Dr. Trefor Bazett
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Read the full survey on math conventions from Thomas Lam here:
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Transcript:
Math conventions mathematicians don't always agree even about some of the most basic definitions used in mathematics so Thomas lamb created a survey asking 2500 mathematicians a 100 math convention questions and the results are kind of fascinating and get into some really interesting mathematics I want to begin with the ever controversial zero to the 0^0 is indeterminate? power of zero you might think mathematicians just have one answer to what Z to the power of zero is but it turns out they disagree a little under half say the number is one but then there's also big groups about a quarter say indeterminate and about a quarter say undefined so what is going on here so what's the controver Verity to begin with if I look at something like 0 to the power of a this is clearly zero like 0 squar would be 0 0 that's clearly zero but in contrast if I look at something like B to the power of zero this is always equal to one here I've graphed B to the X this is either exponential growth if B is greater than one or exponential decay if B is less than one but either way when I zoom in on x equal to Z I get a height of one so I've got one argument for why it should be zero another argument for why it should be one which is it now ultimately the way you answer this probably depends on the discipline of mathematics to which you're being exposed let me take the calculus students perspective first in calculus we might say well let's look at a function like x to the X this is a lovely function and as you can see as we go toward Zero from the right hand side this is going to approach the value of one and in calculus the way we write this down formally we'd say that the limit as X Goes To Zero from the right is equal to one but that's just one function I could consider all sorts of functions of the type f ofx to the power of G ofx when f and g were themselves both going to zero x to the x is one special case but like here's a completely different special case how about the base could be e to the 1x^2 the the top could be X by the laws of exponents I can multiply the X's through this gives me e to the 1/x and here's that plot and as you can see as X goes to zero you get the value of zero and so for this example we have our limit being equal to zero so the point is if I'm going to consider the general context of f to the^ power of G where my f is going to a zero in a limit and my G is going to a zero in limit well the answer could be zero it could be one it could be Infinity it could be Pi you can make an example that will be anything that you want and that's why we say it's indeterminant if I'm interpreting Z to Z as a sort of limiting process in this calculus sense then we're going to call it indeterminate and indeed that's what I teach my calculus students if you think well no I don't want to interpret this problem in this limit sense I just notice there's these tensions I might just call it undef find instead of 0^0=1? indeterminate but then why are some people saying the value is one well let's switch context a bit do you remember Pascal's triangle this is the triangle you put ones all the way along then if I say focus on the pink thing what I do here is I I look at the two numbers right above it one and one I add those to get two I can go down 1 plus 2 is 3 2+ 1 is 3 1 plus 3 is 4 and I can fill out this whole triangle this is Pascal's triangle it's a lovely object with all sorts of lovely patterns one in specific is it's really helpful for expanding binomials like if I take 1 + x^ of 5 well the polinomial that this represents has these coefficients 1 5 10 10 5 1 that shows up directly as this row on Pascal's triangle and there's a formula for this the binomial theorem says that 1 plus x ^ of N is a sum of the way I say this is n choose K this is the number of ways I can choose chose K objects from within n items it can be expressed in terms of factorials and then multipli by X the K so there's this lovely formula for it and if I look on the left hand side I imagine plugging in x equal to Z there's nothing wrong with plugging x equal to Z on the left hand side right that's just 1 plus 0 to the power of n 1 to the N but on the right hand side if I was to plug in x equal to0 then in the k equal to0 case I'd have 0 to the 0 appearing there's no reason the left hand side expression should stop making sense at x equal to Z and so the right hand side should should not stop making sense at zero as well so in that special case when I plug in x equal to 0 I should have 1 is equal to 0 to the 0 so the point is this defining 0 to the 0 to be equal to 1 makes this formula true in all of the cases and I don't have to have this extra restriction written down except in the case when X is equal to Zer only then does it not make sense because 0 to Z is undefined or indeterminant based on our previous arguments and there's all sorts of places in mathematics where making this choice of Z to the Z equal one is just really helpful ultimately I don't care what you say whether you say it's indeterminate whether you say it's undefined whether you say it's one I'm not actually not aware of a good argument for why it should be zero but maybe a couple people think that what's more important is to say that in whatever context you might be in there's a convention within that specific context and if people aren't sure you can specify what you mean and then continue it's not like there's some deep disagreement here about the ideas it's just what's the definition going to be or not be in our specific context I'll give you another example it blows my 0 is a natural number? mind how even this is the natural numbers are like 1 2 3 4 five and so on but do they include zero well a little over half of people think yes they do include zero a little under half think no they don't include zero is it 0 One Two Three or is it just 1 2 3 honestly if you've got a new textbook you kind of just have to go in and look and see how they Define the natural numbers if it becomes relevant at some point zero has actually a really long and kind of interesting history and all sorts of different cultures whether you want to call it natural or not well it's led to different historical Traditions if you want to avoid this controversy you can use things for example non negative integers that's going to include zero or you could say positive integers that's going to start it at one there's ways to avoid around it but it's still common still natural if you don't mind me saying to just talk about the naturals and sometimes people mean that they've got zero in it and sometimes they don't I'll give you one argument for why I like to include zero that's my own personal preference here here if you imagine I'm counting objects like this is a box with one ball in it a box with two balls a box with three balls if I'm doing this counting talking about cardinal numbers here then I also want to refer to like how many items are there what's the size of a box with nothing in it so seems very natural to abuse the language here to think of zero as being a natural number it's referring to the size of an empty box okay this 1/x is a continuous function? one might really mess with your mind is the function one over X continuous and it turns out that well about a third of people think the answer to that is yes so the case for not continuous is may be obvious just says well look at the graph it it's got a vertical ASM toote right that's not continuous duh but let's be a bit more careful if I go to say one of the big calculus textbooks uh Thomas's calculus a standard book I use it in my University it defines a continuous function as one that is continuous at every point and here's the catch in its domain zero isn't in the domain of 1 /x 1 / Z is not defined there's no controversy about that and so if I look at the graph again in all the spots where it's defined that is where X is not equal to zero the function is perfectly continuous and thus it is a continuous function and again it's it's not that this really matters so much these these conventional differences are not some really important sort of aspect of mathematics usually if there's any ambiguity people are just going to tell you what they mean so so these conventional differences don't mean that people start getting mad and angry at each other it just means that if you're working through exercises in your Calculus checkbooks well you have to be a little bit careful about what precisely it was defined in your specific example here's another one also f(x)=1 is increasing? for the for the calculus students uh is the function f ofx equal to one which is completely a horizontal line is that increasing well again about a third of people say yes and and 2third say no the argument for not increasing well let's flat it's just not going up right but but let's be precise again but what does increasing mean so here's the definition going back to Thomas's calculus it says that the function is increasing if the function value one point is strictly less than the function value at another Point whenever X1 is less than X2 so if you if you have two inputs and one's less than the other then the outputs have one being less than the other a sensible notion of increasing but notice the strict inequality here F of X1 is less than F of x2 in other sources like for example go to wol from right now they say less than or equal to here and so according to Thomas this is not an increasing function according to Wolfram it is an increasing function people using the the Wolfram definition would say the Thomas definition is of strictly increasing so there's increasing and then strictly increasing the the special case when it's strictly less than Note by the way that well I haven't written on the screen both of these definitions are about an interval of points where X1 is less than X2 the standard calculus student error is to confuse the definition of increasing with the derivative at a specific point being 3x+1 is linear? positive okay this one's going to annoy people uh is the function 3x + 1 linear oh okay apparently again it keeps on being about 1/3 23 for so many of these one3 say no this is not a linear function you might be like what do you mean it's not a linear function graph it it's a line that's linear what's there to talk about but there's a really important algebraic notion of linearity and it goes like this it says if I have a function and I take a linear combination so F of ax plus b y a linear combination then the output is the linear combination of the outputs it's a f ofx plus b f of Y so this is a lovely algebraic property and note that this demands that F of 0 is zero so 3x + 1 is not linear according to this definition it doesn't go through the origin and the terminology that you might use here is that something like 3x which does go through the origin and does satisfy this property is linear and then 3x + 1 which is sort of like linear but then shifted is an apine transformation in the subject of linear algebra we really study this algebraic property in a lot of detail we can generalize this for example to a higher number of dimensions and talk about linear transformations of the plane or three-dimensional space there's a lot of really lovely work and primarily you're focused on that concept where it's linear and not apine this Axiom of Choice? next one is not just about basic conventions it actually gets to some deeper issues within mathematics and this is the axium of choice and in the axiomatic foundations of mathematics and there's multiple different ways to do this the the Axiom of choice is an axiom that most mathematicians tend to accept as you can see here it's about 85% and then the rest have some combination of either rejecting it entirely or maybe they reject the axium of choice but they accept a weaker notion for example the axium of countable Choice the axum of Dependable choice is another option that's not here in the particular poll but the point is most people accept it okay so so so what is the axum of choice imagine I've got like a bag it's got a bunch of marbles inside of it and then I have another bag with another bunch of marbles another bag another bag and then I want to imagine that this goes on to Infinity if I only have a finite number of bags it's very easy to say well I can pick one marble from the first one marble from the second and one marble from the third but if I have infinitely many can I keep on doing that association with infinitely many bags is there a way to come up with the choice function that picks one marble out of every single one of these bags that you can do this is called the Axiom of choice this is often paired together with other axioms in something called zero Franklin set theory which is one of the sort of foundations of mathematics and this maybe intuitive Maybe not maybe you tempted to agree that this is obvious maybe you're tempted to say no it's not obvious people do genuinely disagree about it and the types of mathematics that you get unlike these previous just definitional disagreements the actual mathematics that you get really changes depending on whether you do or do not reject the axium of choice 6/2(1+2)? now the final one I'm going to leave you with here I have shared my thoughts to like a half million of you before in a video is the ever viral 6 divided 2 uh 3 + 1 and well well I'll just uh show the results here and and I'll let you make your own conclusions about what the average mathematician thinks of this Brilliant.org/TreforBazett problem now if you're interested not just in worrying about silly math conventions but actually learning mathematics then I'd strongly encourage you to check out the sponsor of today's video which is brilliant.org brilliant has thousands of lessons across mathematics computer science Ai and more and Brilliant helps you learn by doing all of their lessons are delightfully interactive so that you are the one in the driver's seat actually doing the math it builds up complexity and layers so you can be confident with your understanding on one step before you're jumping on to the next one and it is constantly providing feedback and opportunities to self assess one of the big challenges that I had to deal with with both as a math YouTuber and as a professor is students who try to learn a lot of math just by watching a lot of videos which can be great but if you're not actually doing the mathematics it's really hard to learn it deeply and so I really appreciate the approach that brilliant takes to learning and that's why I'm so proud to be sponsored by brilliant to try everything that brilliant has for free for a full 30 days go to brilliant.org Trevor bazet or click the link down in the description and clicking that link will give you an additional 20% off and annual premium subscription with that said and done I hope you enjoyed the video if you have your own little math controversies leave them down in the comments below and we'll do some more math in the next video |
3599 | https://www.youtube.com/watch?v=KmuWR_LriQU | TRANSLATING WORDS INTO ALGEBRAIC EXPRESSIONS!
Mashup Math
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Posted: 28 Jul 2015
On this lesson, you will learn how to translate words into algebraic expressions and how to translate algebraic expressions into words!
Not sure of when to use parenthesis? Wondering what a switch word is? Perplexed by seeing the alphabet in math? We got you on this one!
Join us as we explore the ins-and-outs of translating algebraic expressions and equations into words :) For more MashUp Math content, visit and join our free mailing list! :)
This lesson answers the questions: How do I write an algebraic expression in words? How do I translate a verbal expression? How do I translate a verbal equation? How do I model a function in words?
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Transcript:
[Music] hello everyone and welcome to this lesson on algebraic translation now our aim for this lesson is how can we translate written expressions and equations into numerical form so basically we see it in written form and we want to write it in terms of numbers and variables now before we can do that we have to lay the groundwork here and there's a few things that we need to know so that we can use them when we get to doing the actual translations now if we had something like 7 + 5 verbally we could express this for example as the sum of seven and five and we would be done but what if instead of a seven we had a variable we had the letter X there instead of saying the sum of seven and five we would say the sum of a number and five because X is a variable and it could represent any number so we're just going to call it a number and even though X is the most commonly used letter this applies to any letter in the alphabet that can be used to represent a variable which is just some unknown value so let's translate 10 plus a number so again it's an expression so we have the number 10 plus we know is just addition and then a number is just some unknown we'll call it X so that translates to 10 + x now what if we add the statement 10 more than a number now we should know that more than is associated with addition so our sign is not going to change so let's think about this one differently we have a number some unknown number some variable and this phrase represents 10 more than whatever that number is so in this case we're going to start with the variable and then add 10 to it now since this was addition and because addition is commutative the order didn't actually matter but let's take a look at an example Now using subtraction so now let's look at six less than a number so again we see that word then so we have three parts here we have the number six we have less than which we know is subtraction and then a number we'll call it n now this phrase represents a value that is six units smaller than whatever our number n is so to find that number we would have to take n and subtract six from it so in cases like this the second term comes first and the first term comes second we have to switch the order now this is the case when we see the word then and we're going to say that then is a switch word which means that the operator stays in the middle but the order of the first term and the last term is Switched so the way that it's written the actual expression will be in reverse order cool moving on now if we had the phrase the difference of three and the number we could easily translate this difference is subtraction and we have three minus some number we'll call it P so 3 minus P would represent this phrase but what if instead of the difference of three and a number we had the difference of end twice a number + one so now instead of P we have to represent that whole expression which we can call 2 p + 1 and we can enclose this individual expression in parentheses so when performing algebraic translations you can use parentheses to separate independent [Music] groupings and now let's look at a number divided 3 is 8 so we have a number let's call it t and we're dividing it by three so let's use it as a fraction T over3 and when we say is 8 that just means that it equals 8 so be aware that the word is or is equivalent to just means equals to so a quick recap before we get into the examples remember that a number is just a variable the word then we call a switch word use parentheses for independent groupings and is means equal to so now we're ready for a few examples so first let's translate the phrase the product of a number and nine so we know that product just means to multiply and what we're multiplying together is a number we'll call it n and the number 9 now n 9 is fine but we'll more commonly see it written as 9 N without the multiplication sign but that means 9 n next we have the phrase half a number decreased by 12 decrease by means to subtract and now the first term is half a number we'll call that number X and half of it means dividing it by two so we have X over 2 - 12 so that expression is the translation of that verbal phrase next we want to translate the phrase three times the difference of a number and one so basically we're multiplying three by another expression whatever the difference of a number and one is so you have two separate parts here so we have three multiplied by an independent group in this case the difference of a number in one which we'll call P minus one so this is an example of how you can use parentheses to separate independent groupings and if you don't use the parentheses here it will not be [Music] correct our next example is 20 times a number less than two so we have three parts here we have 20 times a number less than and two notice the phrase less than we know that less than means subtraction we need to remember that then is a switch word so we're going to switch the order of the terms so our translation is going to be the value of 2 minus 20 times a number which we can call 20 y so our translation will be 2us 20 y okay and for our last example we have the sum of five and the Square < TK of 8 a number is 12 we know that sum means addition so we're going to need a plus sign and we are basically adding two terms together and that is going to be equal to some other value so we just have to plug in here now so 5 plus the square root of 8 times a number we'll call it 8 R is equal to 12 remember that is just means equal to and now we have translated this phrase into an algebraic equation not an expression remember equations have an equal sign so that's it for this lesson I know that we explored a lot in this one so you probably want to go back and redo those examples again remember practice makes perfect the more you go through them the more you think about these Concepts the better you'll understand them and the easier it will become so I'll leave you guys off with a joke what does a nosy pepper do get jalapeno business sorry guys I'll see you next time thank you again everyone for joining us and please reach out to us on Twitter at mashup maath we are dying to hear from you so please share some love all right we're done here |
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