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9000 | https://math.stackexchange.com/questions/4619223/general-formula-for-counting-ordered-sequences-with-duplicates | combinatorics - General formula for counting ordered sequences with duplicates - Mathematics Stack Exchange
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General formula for counting ordered sequences with duplicates
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Suppose I have a list of distinct allowed elements {x 1,x 2,...,x n}{x 1,x 2,...,x n}. Each element x j x j can be repeated i j i j times.
How many ordered sequences of length m<∑n j=1 i j m<∑j=1 n i j are possible?
For example, suppose the alphabet is A,B,C A,B,C; A has 3 copies, B has 1, and C has 3. Then, choosing from
A,A,A,B,C,C,C,A,A,A,B,C,C,C,
how many ordered sequences of length m=3 m=3 are possible?
(Hence, e.g., A,A,B A,A,B is distinct from A,B,A A,B,A)
If m=∑n j=1 i j m=∑j=1 n i j, this is straightforward:
|sequences|=(∑n j=1 i j)!∏n j=1 i j!|sequences|=(∑j=1 n i j)!∏j=1 n i j!
But what if m<∑n j=1 i j m<∑j=1 n i j?
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edited Jan 20, 2023 at 20:42
user17832896user17832896
asked Jan 16, 2023 at 0:36
user17832896user17832896
69 4 4 bronze badges
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@RossMillikan I must be misunderstanding something. If ∑i=1 n c j=m,∑i=1 n c j=m, then it seems that you do have to use all of the characters. However, if n=m,n=m, then why do you have to use all of the characters?user2661923 –user2661923 2023-01-16 00:55:38 +00:00 Commented Jan 16, 2023 at 0:55
@user2661923 This problem concerns sequences, so the order in which the elements are selected matters, not just how many times each element appears in the selection.N. F. Taussig –N. F. Taussig 2023-01-16 00:56:54 +00:00 Commented Jan 16, 2023 at 0:56
@N.F.Taussig Good point, so I deleted my first comment. However, I am still confused. If (for example) m=n=3,m=n=3, and there are 4 A's, 1 B, and 2 C's available, and you are going to form a 3 character sequence (rather than a 7 character sequence), then you will not be using all of the letters. I suspect that I am misunderstanding something besides the fact that order matters.user2661923 –user2661923 2023-01-16 01:00:16 +00:00 Commented Jan 16, 2023 at 1:00
@user2661923 In your example, there would be 3 3 3 3 sequences if we did not have the restrictions that there is only one B and only two Cs. Only one sequence can violate the restriction that there are only two Cs. There are (3 2)⋅2(3 2)⋅2 sequences with exactly two B B s and one with three Bs that could violate the restriction that there is only one B. Hence, there would be 3 3−1−(3 2)⋅2−1 3 3−1−(3 2)⋅2−1 sequences of length 3 3 that could be formed using four As, one B, and two Cs.N. F. Taussig –N. F. Taussig 2023-01-16 01:05:43 +00:00 Commented Jan 16, 2023 at 1:05
@N.F.Taussig Yes, your analysis agrees with how I interpret the posting. What confuses me is the last portion of the posting, where the OP (i.e. original poster) said: If m=n this is straightforward.If m=n this is straightforward. It seems to me, what the OP should be saying is that : if∑j=1 n i j=m,then it is straightforward.if∑j=1 n i j=m,then it is straightforward. Am I missing something here?user2661923 –user2661923 2023-01-16 01:11:31 +00:00 Commented Jan 16, 2023 at 1:11
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You can use a recursive definition.
f(m,i 1,i 2,…,i n)=⎧⎩⎨⎪⎪1∑j=1 n({0 f(m−1,i 1,i 2,…,i j−1,…,i n)if i j=0 if i j>0)if m=0 if m>0 f(m,i 1,i 2,…,i n)={1 if m=0∑j=1 n({0 if i j=0 f(m−1,i 1,i 2,…,i j−1,…,i n)if i j>0)if m>0
In words: There is only one sequence of length zero, namely the empty sequence. For a non-empty sequence, you can pick any symbol j j as starting point if its occurrence count is non-zero. And then you follow that with a sequence that is one symbol shorter and where the occurrence count of the symbol you picked is also reduced by one.
No, this is not beautiful. Things tend to be simpler if occurrence counts are at most one (permutation with no duplicates), or sum of occurrences matches overall length (permutation with duplicates but all elements taken) or occurrence counts are infinite (just pick m m symbols independently from given alphabet). But this kind of mixed case you have isn't very amenable to simplifications.
The above definition would allow for an evaluation of the formula using some dynamic programming, so you might be able to re-use some previously computed values and thus avoid a computation time proportional to the final count.
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answered Jan 16, 2023 at 15:32
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9001 | https://app.formulabot.com/excel/decimal-function | DECIMAL Excel Formula - Function, Examples, How to Use
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DECIMAL Excel Formula
Convert your text instructions into Excel formulas - powered by AI.
Syntax
=DECIMAL(text, radix)
Example Use Case
You have a table where Column A contains hexadecimal color codes and you want to convert these codes into their decimal equivalents for better understanding and manipulation in your design software. You can use the DECIMAL function to convert each hexadecimal code into a decimal number. This allows for easier integration of color values into various applications and helps in maintaining consistency across design projects without the need for manual conversion.
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What is a DECIMAL Formula in Excel?
The DECIMAL function in Excel is a useful tool that allows you to convert a text representation of a number in a specified base (radix) into a decimal number. This function is particularly handy when working with numbers in different numeral systems, such as binary, octal, or hexadecimal, and you need to convert them into a base-10 format for calculations or analysis.
In simpler terms, the DECIMAL function helps you transform numbers from various bases into a standard decimal format, making it easier to work with numerical data.
A Practical Example
Imagine you are working on a project that involves binary numbers, and you have the following data in an Excel spreadsheet:
Binary Data Table:
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| Binary Number | Description |
--- |
| 1010 | A binary representation of 10 |
| 1111 | A binary representation of 15 |
| 1001 | A binary representation of 9 |
| 1100 | A binary representation of 12 |
| 10000 | A binary representation of 16 |
You want to convert these binary numbers into their decimal equivalents.
DECIMAL Formula
To convert the binary number "1010" into decimal, you would use the DECIMAL formula as follows:
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=DECIMAL(A2, 2)
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Breakdown of the Formula:
A2: This is the cell containing the binary number you want to convert.
2: This is the base of the number system you are converting from (in this case, binary).
Result of the Formula
When you apply the formula, it converts the binary number "1010" into its decimal equivalent. The output would be:
Download
| Binary Number | Decimal Equivalent |
--- |
| 1010 | 10 |
| 1111 | 15 |
| 1001 | 9 |
| 1100 | 12 |
| 10000 | 16 |
Here, the binary number "1010" is converted to decimal as 10.
Why Use DECIMAL?
The DECIMAL function is particularly useful when you need to work with numbers in different bases and want to perform calculations in a standard decimal format. It allows you to easily convert and manipulate data without manual conversion, streamlining your workflow.
Key Takeaways:
DECIMAL: Converts a number from a specified base to decimal format.
Efficient Data Conversion: Helps you quickly convert numbers from various numeral systems for calculations.
Common Use Cases: Ideal for programming, data analysis, and any scenario where you need to work with different number bases.
Understanding how to use the DECIMAL function can significantly enhance your data manipulation capabilities in Excel, enabling you to work seamlessly with numbers from various numeral systems.
Happy converting!
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The AND function evaluates multiple conditions in Excel, returning TRUE only if all specified conditions are met for logical tests.
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9002 | https://www3.cs.stonybrook.edu/~pramod.ganapathi/doc/discrete-mathematics/SampleProblems.pdf | Sample Problems and Solutions Discrete Mathematics Contributor(s): Nicholas Smirnov [Propositional Logic.] 1. [10 points] Give negations of the following statements using formal mathemati-cal language. Your negations must not use conditionals or biconditionals.
(a) [1 point] if p, then q (b) [1 point] p only if q (c) [1 point] p if and only if q (d) [1 point] p is necessary for q (e) [1 point] p is sufficient for q (f) [1 point] p is necessary but not sufficient for q (g) [1 point] p is not necessary but sufficient for q (h) [1 point] p is necessary and sufficient for q (i) [2 points] p is neither necessary nor sufficient for q Solution.
(a) p∧∼q (b) p∧∼q (c) p ⊕q (alternate answer) (p∧∼q) ∨(∼p ∧q) (d) q∧∼p (e) p∧∼q (f) (q∧∼p) ∨(∼p ∨q) (g) (∼q ∨p) ∨(p∧∼q) (h) p ⊕q (alternate answer) (q∧∼p) ∨(p∧∼q) (i) (∼q ∨p) ∨(∼p ∨q) ≡t 2. [5 points] Determine if the following deduction rule is valid: p →q ∼q →∼r ———-∴p →r Solution.
p q r p →q ∼q ∼r p →q ∼q →∼r p →r T T T T F F T T T T T F T F T T T F T F T F T F F F T F F F T T F T F T T T F F T T T F T F T F T T T T F F T T T F T F F F F T T T T T T The deduction rule is invalid.
3. [5 points] Determine if the following deduction rule is valid: p ∨q ∼p∨∼q ———-∴p ⊕q Solution.
p q p ∨q ∼p ∼q ∼p∨∼q p ⊕q T T T F F F T F T F T T T F T T T F T T F F F T T T The deduction rule is valid.
4. [5 points] Determine if the following deduction rule is valid: p →∼q p ∧q ———-∴p ∨q Solution.
p q ∼q p →∼q p ∧q p ∨q T T F F T T F T T F F T F T F F F T T F The deduction rule is valid (vacuous truth).
Page 2 5. [5 points] Determine the validity of the following statements.
(a) [1 point] p∧∼p ≡t (b) [1 point] p →p ≡t (c) [1 point] p ⊕q ≡(∼p ∨q) ∧(∼q ∨p) (d) [1 point] p →(q →r) ≡(p →q) →r (e) [1 point] (p ∧q∧∼r) ∨p ≡p Solution.
(a) False (p∧∼p ≡c) (b) True (c) False (p ⊕q ≡(∼p ∧q) ∨(∼q ∧p)) (d) False (Implication is not associative) (e) True 6. [5 points] Determine how many ordered truth assignments satisfy the following statement: ∼(p ∧q ∧r) ∨(p∧∼q ∧r)∨∼p Solution.
p q r p ∧q ∧r ∼(p ∧q ∧r) p∧∼q ∧r ∼p Final Expression T T T T F F F F T T F F T F F T T F T F T T F T T F F F T F F T F T T F T F T T F T F F T F T T F F T F T F T T F F F F T F T T 7 truth assignments.
7. [5 points] Given the following logic circuit, construct an Input-Output table.
r q p Solution.
Page 3 p q r p ∧q ∼(p ∧q) (p ∧q) ∨r ∼(∼(p ∧q) ∨((p ∧q) ∨r)) 1 1 1 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 0 0 1 0 0 0 1 1 0 1 1 0 0 1 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0 0 8. [5 points] Construct the truth table for the following logical expression: ((p →q) ∧r) ↔∼(p ⊕q) Solution.
p q r p →q (p →q) ∧r p ⊕q ∼(p ⊕q) (p →q) ∧r ↔∼(p ⊕q) T T T T T F T T T T F T F F T F T F T F F T F T T F F F F T F T F T T T T T F F F T F T F T F T F F T T T F T T F F F T F F T F 9. [5 points] Let us define a new operator ⋄with the following truth table: p q p ⋄q T T T T F T F T F F F T Let us extend another of de Morgan’s laws as shown below. Determine whether this property holds for the operator ⋄: p ⋄(q ∧r) ?
≡(p ⋄q) ∨(p ⋄r) Solution.
p q r q ∧r p ⋄q p ⋄r p ⋄(q ∧r) (p ⋄q) ∨(p ⋄r) T T T T T T T T T T F F T T T T T F T F T T T T T F F F T T T T F T T T F F F F F T F F F T T T F F T F T F T T F F F F T T T T Page 4 Yes, the property holds.
10. [2 points] Analyze the following logical statement: If I study for this exam and attend class, then I will pass the class.
Which of the following statements are logically equivalent to the given statement?
(a) If I pass the class, then I studied for this exam and attended class.
(b) If I pass the class, then I studied for this exam or attended class.
(c) If I do not pass the class, then I did not study for this exam and did not attend class.
(d) If I do not pass the class, then I did not study for this exam or did not attend class.
(e) I did not pass the class because I did not study for this exam and did not attend class.
Solution. (d) 11. [2 points] Consider the following logical expression: ∼(P →Q) ∧(R →(P ∨Q)).
How many possible truth assignments make this expression evaluate to true?
(a) 0 (b) 1 (c) 2 (d) 3 (e) 4 Solution. (c) 12. [2 points] Identify the logical reasoning used in the reasoning below: I know that if it rains, the ground will be wet, and since the ground is wet, it must have rained.
(a) Modus Ponens (b) Modus Tollens (c) Converse Error (d) Inverse Error (e) Generalization Solution. (c) 13. [2 points] Identify the logical reasoning used in the reasoning below: Since it is true that Elizabeth and I passed, it is also true that I passed.
(a) Specialization (b) Generalization Page 5 (c) Modus Ponens (d) Modus Tollens (e) Division into Cases Solution. (a) 14. [2 points] Identify the logical reasoning used in the reasoning below: Since I study logic, it is also true that I study logic or art.
(a) Specialization (b) Generalization (c) Modus Ponens (d) Modus Tollens (e) Division into Cases Solution. (b) 15. [2 points] Identify the logical reasoning used in the reasoning below: If I study art, I will pass this class, but if I do not study art, I will still pass this class, therefore, I will pass this class regardless.
(a) Specialization (b) Generalization (c) Modus Ponens (d) Modus Tollens (e) Division into Cases Solution. (e) [Predicate Logic.] 1. [5 points] Negate the following statements using formal predicate logic: (a) [1 point] ∀x ∈R, P(x) = ⇒Q(x) (b) [1 point] ∃x ∈R, P(x) ∧Q(x) (c) [1 point] ∀x ∈R, ∃y ∈R, x2 + y2 = 1 (d) [1 point] ∀x, y ∈Z, x ̸= y = ⇒P(x) ∧P(y) (e) [1 point] ∃x ∈N, P(x) ∨Q(x) Solution.
(a) ∃x ∈R, P(x)∧∼Q(x) (b) ∀x ∈R, ∼P(x)∨∼Q(x) (c) ∃x ∈R, ∀y ∈R, x2 + y2 ̸= 1 (d) ∃x, y ∈Z, (x ̸= y) ∧(∼P(x)∨∼P(y)) (e) ∀x ∈N, ∼P(x)∧∼Q(x) Page 6 2. [5 points] Consider a scenario where a student is represented by S, and the fol-lowing predicates are given: C(S): Returns true if the student studies computer science.
M(S): Returns true if the student studies mathematics.
G(S1, S2): Returns true if student S1 has a higher grade than student S2.
Express the following statements using formal mathematical logic.
(a) [1 point] Every student who studies computer science also studies mathe-matics.
(b) [1 point] There exists a student who studies both computer science and math-ematics.
(c) [1 point] A student will study mathematics unless they are studying com-puter science.
(d) [2 points] If a student studies both computer science and mathematics, they will have a higher grade than a student who studies only computer science.
Solution.
(a) ∀S, C(S) →M(S) (b) ∃S, C(S) ∧M(S) (c) ∀S, ∼C(S) →M(S) (d) ∀S1, S2, (C(S1) ∧M(S1) ∧C(S2)∧∼M(S2)) →G(S1, S2) 3. [5 points] Consider a scenario where an animal is represented by A, and the fol-lowing predicates are given: C(A): Returns true if the animal is a carnivore.
H(A): Returns true if the animal is herbivorous.
F(A): Returns true if the animal is a flying species.
L(A1, A2): Returns true if animal A1 is larger than animal A2.
Express the following statements using formal mathematical logic.
(a) [1 point] All carnivores are not herbivorous.
(b) [1 point] There exists an animal that is both carnivorous and herbivorous.
(c) [1 point] Any animal that is not carnivorous is herbivorous or flying.
(d) [2 points] If an animal is both carnivorous and flying, it is larger than an animal that is herbivorous but not flying.
Solution.
(a) ∀A, C(A) →∼H(A) (b) ∃A, C(A) ∧H(A) (c) ∀A, ∼C(A) →(H(A) ∨F(A)) (d) ∀A1, A2, (C(A1) ∧F(A1) ∧H(A2)∧∼F(A2)) →L(A1, A2) Page 7 4. [5 points] Determine whether the following statements are true. Assume that Q and I are the sets of rational and irrational numbers, respectively.
(a) [1 point] ∀x, y ∈Z, xy ∈Z (b) [1 point] ∀x ∈Z, ∃y ∈Z, x > y (c) [1 point] ∀x, y ∈Z, x/y ∈Q (d) [1 point] ∀x, y ∈I, x + y ∈I (e) [1 point] ∀x, y ∈Z, |x + y| ≤|x| + |y| Solution.
(a) True. Integers are closed under multiplication.
(b) True. There always exists a smaller integer.
(c) False. Suppose y = 0.
(d) False.
√ 2 + (− √ 2) = 0.
(e) True. This is the triangle inequality.
Page 8 [Proof Techniques.] 1. [0 points] Which of these are actual mathematical proof techniques?
Proof by assuming the problem statement Proof by intimidation Proof by shouting Proof by not listening Proof by repeatedly saying the given reasoning is a proof Proof by showing example(s) for a universal statement Proof by trivializing the problem Proof by giving the correct answer without reasoning Solution. None of these.
2. [0 points] Write a beautiful poem on proof techniques.
Solution. ChatGPT: Here it is.
The Proof’s Pursuit: A Mathematical Tale In lands of logic, sharp and keen, Where numbers dance and minds are seen, A theorem stood, so bold, so grand— Yet proofless still, like castles of sand.
With chalk in hand, the wizards came, Each hoping they could stake their claim.
A contradiction! A bijection!
A proof by cases—pure perfection!
First came Induction, noble and proud, "I’ll prove for one, then for the crowd!
If n is true, and n + 1, The proof is done! Oh, what fun!" Then came Contradiction’s art, A devious play, a cunning start.
"Assume I’m wrong, let chaos reign, Oh dear! The logic breaks in twain!" Contraposition took its turn, With logic crisp and truth to burn.
"If not B, then not A too, The proof holds firm, the claim is true!" Next, Direct Proof took the stage, With elegance upon the page.
Page 9 "A to B, so crisp and clean, No tricks, no traps—just what is seen." Proof by Cases made its stand, With many paths already planned.
"If A is this, or A is that, We check them all—now that’s a fact!" Reduction came, so sly and slick, "Transform it here—just one neat trick!
A hard old proof made easy now, Who needs brute force? Just take a bow!" Then One-to-One, so sharp and bright, Mapped all things left to right.
"A bijection, neat and true, Two sets the same—just look, you’ll see too!" The audience gasped, the scribes took note, The theorem stood, as if it wrote, "I stand, I stand! I am complete!" The mathematicians rose to their feet.
Yet lurking still in shadow’s guise, A figure grinned with knowing eyes.
From Gödel’s lips, a whisper grew: "Not all theorems can be proved!" And so they sighed, both sad and wise, For proofs may shine, but truth still hides.
Yet undeterred, they took their quills, For math rewards the steadfast wills.
So let us strive, with joy and glee, For proofs are puzzles, wild and free!
Through logic’s maze, we twist and wind— Who knows what wonders we may find?
3. [5 points] Consider the following propositions.
(i) Prove that if n is odd, then n10 is odd.
(ii) Prove that if 3|n, then 6|n.
(iii) Prove that √ 2 is irrational.
(iv) Prove that if n is an integer, then n2 is even.
(v) Prove that if a|b and b|c, then a|c.
Mention which type(s) of proof could be easily applied as seen in the lectures to solve each of the propositions listed above. There may be multiple correct answers Page 10 for each proposition.
(a) Direct Proof (b) Proof by Contradiction (c) Proof by Contrapositive (d) Proof by Division into Cases (e) Disproof by Counterexample Solution.
(i) (a), (b) (ii) (e) (iii) (b) (iv) (d) (v) (a), (b) 4. [5 points] Prove that the sum of any two integers with the same parity (either both even or both odd) results in an even number.
Solution.
Case 1: Both integers are even. Let a and b be two even integers. Then a + b = 2m + 2n (defn. of even, a = 2m, b = 2n) = 2(m + n) (factor out 2) = 2p (p = m + n and addition is closed on integers) = even (defn. of even) Case 2: Both integers are odd. Let a and b be two odd integers. Then a + b = (2m + 1) + (2n + 1) (defn. of odd, a = 2m + 1, b = 2n + 1) = 2m + 2n + 2 (simplify the terms) = 2(m + n + 1) (factor out 2) = 2p (p = m + n + 1 and addition is closed on integers) = even (defn. of even) 5. [5 points] Prove that if n100 is odd, then n is odd.
Solution. We will prove the statement by contrapositive.
Contrapositive: If n is even, then n100 is even.
Proof: Assume n is even. Then n = 2k for some integer k. Now, consider n100: n100 = (2k)100 (since n = 2k) = 2100k100 (simplify the expression) Since 2100 is a power of 2, it is clearly divisible by 2, so n100 is even.
Thus, by contrapositive, if n100 is odd, then n must be odd. This completes the proof.
6. [5 points] Prove that the sum of any two prime numbers greater than 2 is always even.
Solution.
Let p1 and p2 be prime numbers greater than 2. Then p1 = 2k1 + 1 and p2 = 2k2 + 1 (defn. of odd, since primes greater than 2 are odd) Page 11 p1 + p2 = (2k1 + 1) + (2k2 + 1) = 2k1 + 2k2 + 2 (simplifying the terms) = 2(k1 + k2 + 1) (factor out 2) = 2k3 (where k3 = k1 + k2 + 1) = even (defn. of even) 7. [5 points] Prove the following statement using two different methods: (1) by con-trapositive and (2) by contradiction.
"If the product of two integers is odd, then both integers must be odd." Solution.
Proof 1: Proof by Contrapositive Contrapositive: If at least one of the two integers is even, their product is even.
Assume at least one integer is even. Let a = 2k for some integer k.
a × b = (2k) × b (defn. of even, a = 2k) = 2(k × b) (factor out 2) = 2m (where m = k × b) = even (defn. of even) Thus, if at least one integer is even, their product is even.
By contrapositive, if a × b is odd, both integers are odd.
Proof 2: Proof by Contradiction Negation: The product of two integers is odd and one of the two integers is even.
Let a and b be two integers.
Assume a × b is odd, but one integer is even.
Let a = 2k. Then a × b = (2k) × b (defn. of even, a = 2k) = 2(k × b) (factor out 2) = 2m (where m = k × b) = even (defn. of even) Contradiction. Thus, both integers must be odd.
8. [5 points] Prove that if a and b are integers such that a | b and b | c, then a | c.
Solution.
Given a | b and b | c, there exist integers k1 and k2 such that: b = a × k1 (defn. of a | b) c = b × k2 (defn. of b | c) Substitute b = a × k1 into c = b × k2 to get: c = (a × k1) × k2 = a × (k1 × k2) (associativity of multiplication) = a × k3 (Let k3 = k1 × k2 and k3 is an integer) Therefore, a | c.
9. [5 points] Consider the following statements. For each statement, either prove it or provide a counterexample.
Page 12 (a) [2 points] The product of two irrational numbers is always irrational.
(b) [3 points] The sum of two positive irrational numbers is always irrational.
Solution.
(a) False. Consider √ 2 · √ 2 = 2.
(b) False. Consider √ 2 + (2 − √ 2) = 2 Page 13 [Sequences.] 1. [5 points] Prove by induction that the following statement is true for all integers n ≥1: n X i=1 i2 = n(n + 1)(2n + 1) 6 Solution. Let P(n) : Pn i=1 i2 = n(n+1)(2n+1) 6 .
Base case: For n = 1, P(1) : P1 i=1 i2 = 1 = 1×2×3 6 Thus, the base case holds.
Inductive step: Assume P(k) is true for some k ≥1. We need to prove P(k + 1): LHS of P(k + 1): = Pk+1 i=1 i2 = Pk i=1 i2 + (k + 1)2 (by definition of summation) = k(k+1)(2k+1) 6 + (k + 1)2 (since P(k) is assumed to be true) = (k+1)(2k2+7k+6) 6 (simplify and combine terms) = (k+1)(k+2)(2k+3) 6 (factor the expression) = RHS of P(k + 1) Thus, the inductive step holds.
Therefore, by mathematical induction, P(n) is true for all n ≥1.
2. [5 points] Prove by induction that n Y i=1 2i = 2n · n!
for all n ∈N.
Solution. Let P(n) : Qn i=1 2i = 2n · n!.
Base case: For n = 1, the left-hand side is: 2 = 21 · 1! = 2 Thus, P(1) holds.
Inductive step: Assume P(k) is true for some k ≥1. We must prove P(k + 1): LHS of P(k + 1): = (Qk i=1 2i) · 2(k + 1) (by definition of product) = 2k · k!
· 2(k + 1) (by inductive hypothesis) = 2k+1 · (k + 1)!
(simplify) = RHS of P(k + 1).
Thus, the inductive step holds.
Therefore, by mathematical induction, P(n) holds for all n ∈N.
Page 14 3. [5 points] Prove by induction that if an = an−1 · an−2 an−1 + an−2 , with initial conditions a1 = 2 and a2 = 1, then for all n ∈N, an ≤2 n.
Solution. Let P(n) : an ≤2 n. We will prove P(n) by induction for all n ∈N.
Base case: For n = 1, a1 = 2 ≤2 1.
(True) For n = 2, a2 = 1 ≤2 2.
(True) Thus, P(1) and P(2) hold.
Inductive step: Assume P(i) holds true for all i ∈[1, k] for some k ≥2. We need to show that P(k + 1) holds.
LHS of P(k + 1): = ak+1 = ak·ak−1 ak+ak−1 (from the recurrence relation) ≤ 2 k · 2 k−1 2 k + 2 k−1 (P(k) : ak ≤2 k and P(k −1) : ak−1 ≤ 2 k−1 are true) = 4 k(k−1) 2k+2(k−1) k(k−1) = 4 4k−2 = 2 2k−1 ≤ 2 k+1 (simplify) = RHS of P(k + 1) Therefore, ak+1 ≤ 2 k+1, and P(k + 1) holds.
By induction, P(n) is true for all n ∈N. Therefore, an ≤2 n for all n ∈N.
4. [5 points] Prove by induction that if an = √an−3 + an−2 + an−1 for all n ≥4 and a1 = 0, a2 = 1, a3 = 2, then for all n ∈N, an < 3.
Solution. Let P(n) : an < 3. We will prove P(n) by induction for all n ∈N.
Base cases For n = 1, a1 = 0 < 3.
For n = 2, a2 = 1 < 3.
For n = 3, a3 = 2 < 3.
Thus, P(1), P(2), and P(3) hold.
Inductive step: Assume P(i) holds true for all i ∈[1, k] for some k ≥3. We need to show that P(k + 1) holds.
LHS of P(k + 1) = ak+1 = √ak−2 + ak−1 + ak (from the recurrence relation) < √3 + 3 + 3 (P(k) : ak < 3, P(k −1) : ak−1 < 3, P(k −2) : ak−2 < 3 are true) = √ 9 = 3 = RHS of P(k + 1) Therefore, ak+1 < 3, and P(k + 1) holds.
By induction, P(n) is true for all n ∈N. Therefore, an < 3 for all n ∈N.
Page 15 [Sets.] 1. [5 points] Given that A = {1, 3, 5, 8}, B = {2, 3, 4, 5}, C = {1, 4, 8} and D = {1}.
(a) [1 point] What is the value of |A −B|?
(b) [1 point] What is the value of |(A ∪B) ∩(C ∪D)|?
(c) [1 point] What is (A −B) ∪(C −D)?
(d) [2 points] What is P(A −B) ∩P(C −D)?
Solution.
(a) 2 (b) 3 (c) {1, 4, 8} (d) {∅, {8}} 2. [5 points] Prove that if A ⊆B, then A −B = ∅.
Solution.
We prove by contradiction.
Negation: Suppose that A ⊆B and A −B is not an empty set Suppose x ∈A −B.
= ⇒x ∈A and x / ∈B (by definition of set difference) = ⇒x ∈B (since A ⊆B) Contradiction since x ∈B and x / ∈B.
= ⇒A −B = ∅.
3. [5 points] Prove that if A ∪B = U and A ∩B = ∅, then (A ∪C) ∩(B ∪C) = C.
Solution.
We will prove (A ∪C) ∩(B ∪C) = C in two parts.
Part 1: (A ∪C) ∩(B ∪C) ⊆C.
Let x ∈(A ∪C) ∩(B ∪C).
= ⇒x ∈A ∪C and x ∈B ∪C (by definition of intersection).
= ⇒(x ∈A or x ∈C) and (x ∈B or x ∈C) (by definition of union) = ⇒((x ∈A and x ∈B) or x ∈C) (by distributive law) = ⇒x ∈C (since A ∩B = ∅, so x / ∈A and x / ∈B).
Thus, x ∈C.
Therefore, (A ∪C) ∩(B ∪C) ⊆C.
Part 2: C ⊆(A ∪C) ∩(B ∪C).
Let x ∈C.
= ⇒x ∈A ∪C and x ∈B ∪C (by definition of union).
= ⇒x ∈(A ∪C) ∩(B ∪C).
= ⇒C ⊆(A ∪C) ∩(B ∪C).
Since both inclusions hold, we conclude that (A ∪C) ∩(B ∪C) = C.
Page 16 [Functions.] 1. [5 points] Let A = {0, 2, 4, 6, 8} and B = {1, 2, 3, 4}. Answer the following: (a) [1 point] Calculate |A ∪B| −|A ∩B|.
(b) [1 point] Does there exist a one-to-one correspondence function f : A →B?
If so, what is a possible f?
(c) [1 point] Find the number of one-to-one functions g : A →B.
(d) [2 points] Let f : A →B and g : B →A. Is it possible that f ◦g = g ◦f?
Solution.
(a) 5.
(b) There does not exist such an f.
(c) 0 functions.
(d) This is not possible because f, g are not one-to-one correspondences so they cannot be inverses of one another.
2. [5 points] Let A and B be two finite sets with sizes |A| = n and |B| = m. Answer the following questions about functions between A and B: (a) [2 points] Prove that a one-to-one function f : A →B exists if n ≤m.
(b) [2 points] Prove that an onto function f : A →B exists if n ≥m.
(c) [1 point] Show that a one-to-one correspondence f : A →B exists if n = m.
Solution.
(a) If n > m, the pigeonhole principle implies that at least two elements of A would map to the same element of B, violating the one-to-one condition.
Therefore, n ≤m is necessary.
(b) One can visually show that if n < m, there would be elements of B with no preimage, violating the onto condition. Therefore, n ≥m is necessary.
(c) Combining the results from parts (a) and (b), this requires n = m. For f to be one-to-one, n ≤m. For f to be onto, n ≥m. Together, m ≤n ≤m = ⇒n = m.
3. [5 points] Let f : Z+ →O+, where O+ represents the set of positive odd numbers, be defined as f(x) = 2x −1. Prove that f is a one-to-one correspondence.
Solution.
To prove that f is a one-to-one correspondence, we need to show that f is both one-to-one and onto.
1. One-to-One: A function f is one-to-one ∀a, b ∈Z+ if f(a) = f(b) then a = b.
Suppose f(a) = f(b). Then, 2a −1 = 2b −1.
Simplifying: 2a = 2b = ⇒a = b = ⇒f is one-to-one.
Page 17 2. Onto: A function f is onto if, for every y ∈O+, there exists an x ∈Z+ such that f(x) = y. Let y ∈O+. Since O+ = {1, 3, 5, 7, . . . }, we can write y = 2x −1. Solving for x: x = y + 1 2 .
Since y is odd, y + 1 is even, and y+1 2 ∈Z+. Hence, f(x) = y, proving f is onto.
Conclusion: Since f is both one-to-one and onto, f is a one-to-one correspon-dence.
4. [5 points] Sort the following sets by non-decreasing cardinality using < and = symbols: ∅, N, P(N), P(P(N)), Q, R, P(R), {1} Solution.
|∅| < |{1}| < |N| = |Q| < |R| = |P(N)| < |P(P(N))| = |P(R)| 5. [5 points] Consider f : R∗→R∗, where R∗= R+ ∪{0} is the set of all nonnegative real numbers, such that f is a mapping from pounds to kilograms. A kilogram of mass in the International System of Units (SI) represents 2.2046226218 pounds.
Is f a one-to-one correspondence? Is f a one-to-one correspondence? If yes, give a formal proof.
Solution. Given that 1 kilogram = 2.2046226218 pounds, we can express the func-tion f as: f(x) = x 2.2046226218 where x is the mass in pounds.
The function maps from the set of nonnega-tive real numbers (including zero) to the set of nonnegative real numbers. Since 2.2046226218 > 0, for any x ≥0, f(x) will also be nonnegative. Therefore, f(x) ∈R∗ for all x ∈R∗. Thus, f is well-defined.
1. One-to-One: To show that f is injective, we need to prove that if f(x1) = f(x2), then x1 = x2.
Assume f(x1) = f(x2): x1 2.2046226218 = x2 2.2046226218 Multiplying both sides by 2.2046226218 (which is positive and does not change the equality) gives: x1 = x2 Thus, f is injective.
2. Onto: To show that f is surjective, we need to prove that for every y ∈R∗, there exists an x ∈R∗such that f(x) = y.
Page 18 Let y be any nonnegative real number. We want to find x such that: f(x) = y = ⇒ x 2.2046226218 = y Solving for x gives: x = 2.2046226218y Since y ≥0, it follows that x = 2.2046226218y ≥0. Therefore, for every y ∈R∗, we can find an x ∈R∗such that f(x) = y. Thus, f is surjective.
Conclusion: Since f is both injective and surjective, we conclude that f is a one-to-one corre-spondence (bijection) from R∗to R∗. Thus, the function f is a one-to-one corre-spondence.
6. [5 points] Consider the set of numbers of the form k2 or 2k, where k ∈N. Prove that the set is countable by coming up with a one-to-one correspondence function and showing the arrow diagram.
Solution.
To prove that the set consisting of numbers of the form k2 or 2k, where k ∈N, is countable, we will first define the set clearly and then establish a one-to-one correspondence function.
Step 1: Define the Set Let S be the set defined as follows: S = {k2 | k ∈N} ∪{2k | k ∈N} This means that S contains all perfect squares and all powers of 2, where k is a natural number.
Step 2: List the Set Elements Perfect Squares: The elements of the form k2 for k = 1, 2, 3, . . . are: 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, . . .
Powers of 2: The elements of the form 2k for k = 1, 2, 3, . . . are: 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64, 27 = 128, 28 = 256, . . .
Common elements: 4 appears as 22 and 22.
16 appears as 42 and 24.
64 appears as 82 and 26.
so on...
So, essentially, we will need to eliminate all numbers of the form 2even to avoid double-counting.
Step 3: Create a One-to-One Correspondence To show that the set S is countable, we can create a function that maps each natural number k to a unique element in S.
Page 19 Define a function f : N →S as follows: f(n) = 2n if n is odd, (n/2)2 if n is even.
Step 4: Diagram of the Function n f(n) 1 21 2 12 3 23 4 22 5 25 6 32 7 27 8 42 .
.
.
.
.
.
Page 20 [Relations] 1. [10 points] Consider finite sets A and B such that |A| = m, |B| = n.
(a) [1 point] How are |A ∪B| and |A ∩B| related to each other?
(b) [1 point] What is |P(A × B)|?
(c) [2 points] How many relations R : A →B exist?
(d) [2 points] How many functions f : A →B exist?
(e) [2 points] How many one-to-one functions f : A →B exist?
(f) [2 points] How many one-to-one correspondences functions f : A →B exist?
Solution.
(a) |A ∪B| = |A| + |B| −|A ∩B| = m + n −|A ∩B| (b) 2mn (|A × B| = mn. The number of subsets of mn elements is 2mn.) (c) 2mn (Total number of binary relations from A to B is the total number of subsets of A × B) (d) nm (First input element can be mapped to any of n output elements, so on till the mth input element) (e) 0 if m > n, n × (n −1) × (n −2) × · · · × (n −m + 1) if m ≤n.
(First input element can be mapped to any of n output elements, second input element can be mapped to any of remaining (n−1) output elements, so on till the mth input element can be mapped to any of (n −m + 1) output elements) (f) 0 if m ̸= n, m!
if m = n.
(First input element can be mapped to any of m output elements, second input element can be mapped to any of remaining (m −1) output elements, so on till the mth input element can be mapped to the only remaining 1 output element) 2. [5 points] Let R be a relation on the set of integers Z defined by a R b if and only if a −b is divisible by 3. Check if R is an equivalence relation and describe the equivalence classes of R if it is.
Solution.
1. Reflexive: For any a ∈Z, a −a = 0 is divisible by 3. Hence, a R a holds.
Page 21 2. Symmetric: If a R b, then a −b = 3k for some k ∈Z. Thus, b −a = −3k, which is divisible by 3. Hence, b R a holds.
3. Transitive: If a R b and b R c, then a −b = 3k and b −c = 3m for some k, m ∈Z.
Adding these gives a −c = 3(k + m), which is divisible by 3. Hence, a R c.
4. Equivalence Classes: The equivalence classes are the sets of integers that leave a remainder of 0, 1, and 2 when divided by 3, as shown below: = {. . . , −3, 0, 3, 6, . . . }, = {. . . , −2, 1, 4, 7, . . . }, = {. . . , −1, 2, 5, 8, . . . }.
3. [5 points] Let R be a relation on the set of all strings over the alphabet {a, b} defined by x R y if and only if x and y have the same length. Check if R is an equivalence relation and describe the equivalence classes of R if it is.
Solution.
1. Reflexive: For any string x, we have |x| = |x|, so x R x holds.
2. Symmetric: If x R y, then |x| = |y|. Thus, |y| = |x|, so y R x holds.
3. Transitive: If x R y and y R z, then |x| = |y| and |y| = |z|. Thus, |x| = |z|, so x R z holds.
4. Equivalence Classes: The equivalence classes are the sets of strings with the same length. For any non-negative integer n, the equivalence class of strings of length n is: [n] = {x | x is a string over {a, b} with length n} 4. [5 points] Let R be a relation on R defined by a R b if and only if a2 = b2. Check if R is an equivalence relation and describe the equivalence classes of R if it is.
Solution.
1. Reflexive: For any a ∈R, we have a2 = a2, so a R a holds.
2. Symmetric: If a R b, then a2 = b2. Thus, b2 = a2, so b R a holds.
3. Transitive: If a R b and b R c, then a2 = b2 and b2 = c2. Thus, a2 = c2, so a R c holds.
4. Equivalence Classes: The equivalence classes are the sets of real numbers with the same absolute value. For any non-negative real number r, the equiva-lence class of r is: [r] = {x ∈R | |x| = r} = {−r, r} (if r ̸= 0), and = {0}.
5. [5 points] Let R be a relation on the set of all people, where a R b if and only if a and b have the same birth year. Check if R is an equivalence relation and describe the equivalence classes of R if it is.
Solution.
1. Reflexive: For any person a, the relation holds because a has the same birth year as a. Thus, a R a holds.
Page 22 2. Symmetric: If a R b, then a and b have the same birth year. Since having the same birth year is a mutual property, b R a also holds.
3. Transitive: If a R b and b R c, then a and b have the same year, and b and c have the same year. Thus, a and c must also have the same birth year, so a R c holds.
4. Equivalence Classes: The equivalence classes of R are the sets of people born in the same year. For any year y, the equivalence class of people born in year y is: [y] = {a | person a is born in year y}.
6. [5 points] Let R be a relation on the set of all points in the plane R2 defined by (x1, y1) R (x2, y2) if and only if x1 = x2 or y1 = y2. Check if R is an equivalence relation and describe the equivalence classes of R if it is.
Solution.
1. Reflexive: For any point (x1, y1), we have (x1, y1) R (x1, y1) since x1 = x1 and y1 = y1. Thus, the relation is reflexive.
2. Symmetric: If (x1, y1) R (x2, y2), then either x1 = x2 or y1 = y2. If x1 = x2, then clearly x2 = x1. If y1 = y2, then clearly y2 = y1. Thus, the relation is symmetric.
3. Not Transitive: Consider (1, 1), (1, 2), (2, 2). (1, 1)R(1, 2), (1, 2)R(2, 2), but (1, 1) ¯ R(2, 2).
Therefore, not transitive.
This is NOT an equivalence relation.
7. [5 points] Find the units digit of 7100.
Solution.
To find the units digit of 7100, observe the pattern of the units digits of powers of 7: 71 = 7 (units digit 7), 72 = 49 (units digit 9), 73 = 343 (units digit 3), 74 = 2401 (units digit 1).
The units digits repeat in a cycle of 4: 7, 9, 3, 1. To find the units digit of 7100, divide 100 by 4: 100 ÷ 4 = 25 remainder 0.
Since the remainder is 0, the units digit of 7100 corresponds to the units digit of 74, which is 1. Thus, the units digit of 7100 is 1 .
8. [5 points] Find the units digit of 350.
Solution.
Page 23 To find the units digit of 350, observe the pattern of the units digits of powers of 3: 31 = 3 (units digit 3), 32 = 9 (units digit 9), 33 = 27 (units digit 7), 34 = 81 (units digit 1).
The units digits repeat in a cycle of 4: 3, 9, 7, 1. To find the units digit of 350, divide 50 by 4: 50 ÷ 4 = 12 remainder 2.
Since the remainder is 2, the units digit of 350 corresponds to the units digit of 32, which is 9. Thus, the units digit of 350 is 9 .
9. [5 points] Find the units digit of 121234.
Solution.
To find the units digit of 121234, observe the pattern of the units digits of powers of 12. The units digits of powers of 12 follow the same pattern as powers of 2: 121 = 12 (units digit 2), 122 = 144 (units digit 4), 123 = 1728 (units digit 8), 124 = 20736 (units digit 6).
The units digits repeat in a cycle of 4: 2, 4, 8, 6. To find the units digit of 121234, divide 1234 by 4: 1234 ÷ 4 = 308 remainder 2.
Since the remainder is 2, the units digit of 121234 corresponds to the units digit of 122, which is 4. Thus, the units digit of 121234 is 4 .
10. [5 points] Find the units digit of 2987.
Solution.
To find the units digit of 2987, observe the pattern of the units digits of powers of 2: 21 = 2 (units digit 2), 22 = 4 (units digit 4), 23 = 8 (units digit 8), 24 = 16 (units digit 6).
The units digits repeat in a cycle of 4: 2, 4, 8, 6. To find the units digit of 2987, divide 987 by 4: 987 ÷ 4 = 246 remainder 3.
Since the remainder is 3, the units digit of 2987 corresponds to the units digit of 23, which is 8. Thus, the units digit of 2987 is 8 .
11. [5 points] Find the units digit of 9999.
Page 24 Solution.
To find the units digit of 9999, observe the pattern of the units digits of powers of 9: 91 = 9 (units digit 9), 92 = 81 (units digit 1).
The units digits repeat in a cycle of 2: 9, 1. To find the units digit of 9999, divide 999 by 2: 999 ÷ 2 = 499 remainder 1.
Since the remainder is 1, the units digit of 9999 corresponds to the units digit of 91, which is 9. Thus, the units digit of 9999 is 9 .
Page 25 |
9003 | https://brainly.com/question/48031902 | [FREE] f(x) = \left( \frac{x-1}{2x} \right) \left( \frac{3 - 2x}{9} \right)^8 Find the coefficient of x^2 in - brainly.com
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f(x)=(2 x x−1)(9 3−2 x)8
Find the coefficient of x 2 in the series expansion of f(x), giving your answer as a simplified fraction.
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To find the coefficient of x² in the expansion of f(x)=((x-1)/2x)(3 - 2x/9)⁸, we use the binomial theorem on the second part of the function and look for the term where the power of x is 2.
Explanation
The question involves finding the coefficient of x² in the series expansion of the function f(x) given by f(x) = ((x-1)/2x)(3 - 2x/9)⁸. To find the series expansion, we can apply the binomial theorem to (3 - 2x/9)⁸, which will result in a series where each term is a coefficient multiplied by a power of x. We're interested in the term where the power of x is 2.
Starting with the binomial expansion, we have (a+b)⁸ = Σ(n choose k) a⁸-k b⁸. In our case, a is 3, and b is -2x/9. We are looking for the term where b has a power of 2, which occurs when k=2.
Calculating this specific term and multiplying by the ((x-1)/2x) portion will give us the coefficient of x² in the expansion of f(x). The actual calculations will involve determining the binomial coefficient and performing algebraic simplification.
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The coefficient of x 2 in the series expansion of f(x)=(2 x x−1)(9 3−2 x)8 is calculated by first expanding both parts using the binomial theorem. The relevant coefficients are combined to find the final answer, which is −17496.
Explanation
To find the coefficient of x 2 in the series expansion of f(x)=(2 x x−1)(9 3−2 x)8, we can break the problem into two parts: the first part being the polynomial 2 x x−1 and the second part using the binomial theorem to expand (9 3−2 x)8.
Expand the polynomial component:
2 x x−1=2 1−2 x 1
We can see that when finding the coefficient of x 2, this part needs to be treated carefully when multiplied by the second part.
Expand the binomial:
Using the binomial theorem for (3−2 x)8:
(3−2 x)8=k=0∑8(k 8)3 8−k(−2 x)k=k=0∑8(k 8)3 8−k(−2)k x k
Thus, the general term for the expansion is (k 8)3 8−k(−2)k x k.
Find relevant k values:
Next, we need to find terms where k+1=2 or k=1 in this expansion, since we must account for the −2 x 1 term taking down the degree of x in the product.
For k=1:
(1 8)3 8−1(−2)1=8⋅3 7⋅(−2)=−16⋅2187=−34992
Combining the components:
Combining now, we need the x 2 terms:
(−2 x 1⋅−34992)=17496
And for the other part, the non-x part contributes 2 1⋅(−34992)=−17496. The coefficient of x 2 will then be:
2 1⋅−34992+0=−17496
In conclusion, the coefficient of x 2 in the expansion of f(x) is −17496.
Examples & Evidence
For instance, you can follow a similar method to find coefficients in other polynomial expansions using the binomial theorem and combining like terms accordingly.
The approach utilizes the binomial theorem, which is a widely accepted method in algebra for expanding binomial expressions, ensuring that the calculations follow mathematical principles.
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9004 | https://courses.lumenlearning.com/calculus1/chapter/logarithmic-functions/ | Logarithmic Functions
Learning Outcomes
Identify the form of a logarithmic function
Explain the relationship between exponential and logarithmic functions
Describe how to calculate a logarithm to a different base
Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. Recall the definition of an inverse function.
Recall: Inverse Function
For any one-to-one function f(x)=y, a function f−1(x) is an inverse function of f if f−1(y)=x. This can also be written as f−1(f(x))=x for all x in the domain of f. It also follows that f(f−1(x))=x for all x in the domain of f−1 if f−1 is the inverse of f.
The notation f−1 is read “f inverse.” Like any other function, we can use any variable name as the input for f−1, so we will often write f−1(x), which we read as "f inverse of x“.
Logarithmic functions come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.
The exponential function f(x)=bx is one-to-one, with domain (−∞,∞) and range (0,∞). Therefore, it has an inverse function, called the logarithmic function with base b. For any b>0,b≠1, the logarithmic function with base b, denoted logb, has domain (0,∞) and range (−∞,∞), and satisfies
logb(x)=y if and only if by=x
For example,
log2(8)=3since23=8,log10(1100)=−2since10−2=1102=1100,logb(1)=0sinceb0=1for any baseb>0.
Furthermore, since y=logb(x) and y=bx are inverse functions,
logb(bx)=xandblogb(x)=x
The most commonly used logarithmic function is the function loge(x). Since this function uses natural e as its base, it is called the natural logarithm. Here we use the notation ln(x) or lnx to mean loge(x). For example,
ln(e)=loge(e)=1,ln(e3)=loge(e3)=3,ln(1)=loge(1)=0
Since the functions f(x)=ex and g(x)=ln(x) are inverses of each other,
ln(ex)=x and elnx=x,
and their graphs are symmetric about the line y=x (Figure 4).
Figure 4: The functions y=ex and y=ln(x) are inverses of each other, so their graphs are symmetric about the line y=x.
Interactive
At this site you can see an example of a base-10 logarithmic scale.
In general, for any base b>0,b≠1, the function g(x)=logb(x) is symmetric about the line y=x with the function f(x)=bx. Using this fact and the graphs of the exponential functions, we graph functions logb(x) for several values of b>1 (Figure 5).
Figure 5: Graphs of y=logb(x) are depicted for b=2,e,10.
Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.
Properties of Logarithms
If a,b,c>0,b≠1, and r is any real number, then
1.logb(ac)=logb(a)+logb(c)(Product property)2.logb(ac)=logb(a)−logb(c)(Quotient property)3.logb(ar)=rlogb(a)(Power property)
Example: Solving Equations Involving Exponential Functions
Solve each of the following equations for x.
5x=2
ex+6e−x=5
Show Solution
Applying the natural logarithm function to both sides of the equation, we have
ln5x=ln2
Using the power property of logarithms,
xln5=ln2
Therefore, x=ln2/ln5.
2. Multiplying both sides of the equation by ex, we arrive at the equation
e2x+6=5ex
Rewriting this equation as
e2x−5ex+6=0,
we can then rewrite it as a quadratic equation in ex:
(ex)2−5(ex)+6=0
Now we can solve the quadratic equation. Factoring this equation, we obtain
(ex−3)(ex−2)=0
Therefore, the solutions satisfy ex=3 and ex=2. Taking the natural logarithm of both sides gives us the solutions
x=ln3,ln2
Watch the following video to see the worked solution to Example: Solving Equations Involving Exponential Functions
Closed Captioning and Transcript Information for Video
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “1.5 Exponential and Logarithmic Functions” here (opens in new window).
Try It
Solve e2x(3+e2x)=12.
Hint
First solve the equation for e2x.
Show Solution
x=ln32
Example: Solving Equations Involving Logarithmic Functions
Solve each of the following equations for x.
ln(1x)=4
log10√x+log10x=2
ln(2x)−3ln(x2)=0
Show Solution
By the definition of the natural logarithm function,
ln(1x)=4 if and only if e4=1x
Therefore, the solution is x=1e4.
2. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as
log10√x+log10x=log10x√x=log10x3/2=32log10x
Therefore, the equation can be rewritten as
32log10x=2 or log10x=43
The solution is x=104/3=103√10.
3. Using the power property of logarithmic functions, we can rewrite the equation as ln(2x)−ln(x6)=0.
Using the quotient property, this becomes
ln(2x5)=0
Therefore, 2x5=1, which implies x=5√2. We should then check for any extraneous solutions.
Watch the following video to see the worked solution to Example: Solving Equations Involving Logarithmic Functions
Closed Captioning and Transcript Information for Video
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “1.5 Exponential and Logarithmic Functions” here (opens in new window).
Try It
Solve ln(x3)−4ln(x)=1.
Hint
First use the power property, then use the product property of logarithms.
Show Solution
x=1e
Try It
When evaluating a logarithmic function with a calculator, you may have noticed that the only options are log10 or log, called the common logarithm, or ln, which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base b. If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions.
Try It
Change-of-Base Formulas
Let a>0,b>0, and a≠1,b≠1.
ax=bxlogba for any real number x.
If b=e, this equation reduces to ax=exlogea=exlna.
logax=logbxlogba for any real number x>0.
If b=e, this equation reduces to logax=lnxlna.
Proof
For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base b>0,b≠1,logb(ax)=xlogba. Therefore,
blogb(ax)=bxlogba
In addition, we know that bx and logb(x) are inverse functions. Therefore,
blogb(ax)=ax
Combining these last two equalities, we conclude that ax=bxlogba.
To prove the second property, we show that
(logba)⋅(logax)=logbx
Let u=logba,v=logax, and w=logbx. We will show that u⋅v=w. By the definition of logarithmic functions, we know that bu=a,av=x, and bw=x. From the previous equations, we see that
buv=(bu)v=av=x=bw
Therefore, buv=bw. Since exponential functions are one-to-one, we can conclude that u⋅v=w.
■
Example: Changing Bases
Use a calculating utility to evaluate log37 with the change-of-base formula presented earlier.
Show Solution
Use the second equation with a=3 and e=3:
log37=ln7ln3≈1.77124.
Try It
Use the change-of-base formula and a calculating utility to evaluate log46.
Hint
Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function.
Show Solution
Try It
Compare the relative severity of a magnitude 8.4 earthquake with a magnitude 7.4 earthquake.
Hint
R1−R2=log10(A1/A2).
Show Solution
The magnitude 8.4 earthquake is roughly 10 times as severe as the magnitude 7.4 earthquake.
Try It
Candela Citations
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1.5 Exponential and Logarithmic Functions. Authored by: Ryan Melton. License: CC BY: Attribution
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Calculus Volume 1. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at
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CC licensed content, Original
1.5 Exponential and Logarithmic Functions. Authored by: Ryan Melton. License: CC BY: Attribution
CC licensed content, Shared previously
Calculus Volume 1. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at |
9005 | https://study.com/skill/learn/how-to-solve-a-linear-equation-with-variables-on-both-sides-distribution-explanation.html | How to Solve a Linear Equation with Variables on Both Sides & Distribution | Algebra | Study.com
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How to Solve a Linear Equation with Variables on Both Sides & Distribution
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Example SolutionsPractice Questions
How to Solve a Linear Equation with Several Variables on the Same Side and Distribution
Step 1: First, we want to simplify our parentheses. Distribute the term on the outside of the parenthesis to both of the terms inside the parenthesis. To do this, multiply this outside term by the first term in the parenthesis. Then multiply the outside term by the second term in the parenthesis. Do this to any parenthesis you have in your problem.
Step 2: Rewrite your function with the expanded expressions from step 1.
Step 3: Combine like terms on both sides of the equation. To combine like terms, we add the coefficients of each term. When we do this, make sure to pay special attention to the signs of the coefficients.
Step 4: Move all of your terms with variables to the left side of your equation using inverse operations. Combine like terms before moving on to the next step.
Step 5: Move all of your constants to the right side of the equation using inverse operations. Combine all like terms on this side of the equation.
Step 6: Perform division to cancel out the coefficient of the variable. Simplify your answer, giving you the variable by itself!
How to Solve a Linear Equation with Several Variables on the Same Side and Distribution Vocabulary
Like terms: Terms are considered like if they have the same variable and same exponent. If terms are like terms, we can combine them using addition. To combine like terms, we add the coefficient and do not change the variable.
Coefficient: This is the value that you see in front of a variable. This is a constant that is multiplying the variable.
Inverse operations: These are operations that reverse each other. We use inverse operations to solve an equation. The inverse operation of addition is subtraction, and the inverse operation of multiplication is division.
Order of operations: This is the order we use to evaluate an expression. To solve this, we reverse this order. So when we are solving, we start with addition and subtraction, then multiplication and division, then parenthesis and exponents.
So, let's try using these steps to use the product rule with positive exponents and whole number bases in the following two examples!
How to Solve a Linear Equation with Several Variables on the Same Side and Distribution: Example 1
Solve for x: 5(x−2)=2(3 x−1)−3 x
Step 1: First, we want to simplify our parentheses. Distribute the term on the outside of the parenthesis to both of the terms inside the parenthesis. To do this, multiply this outside term by the first term in the parenthesis. Then multiply the outside term by the second term in the parenthesis. Do this to any parenthesis you have in your problem.
Our first set of parenthesis gives us:
5(x−2)=(5⋅x)+(5⋅−2)=5 x+(−10)=5 x−10
Our second set of parenthesis gives us:
2(3 x−1)=(2⋅3 x)+(2⋅−1)=6 x+(−2)=6 x−2
Step 2: Rewrite your function with the expanded expressions from step 1.
5(x−2)=2(3 x−1)−3 x 5 x−10=6 x−2−3 x
Step 3: Combine like terms on both sides of the equation. To combine like terms, we add the coefficients of each term. When we do this, make sure to pay special attention to the signs of the coefficients.
Our like terms on the right in this function are 6 x and −3 x. To combine them, we will add the coefficients.
6+(−3)=3
Giving us the new coefficient of 3.
5 x−10=6 x−2−3 x 5 x−10=3 x−2
Step 4: Move all of your terms with variables to the left side of your equation using inverse operations. Combine like terms before moving on to the next step.
5 x−10=3 x−2−3 x−3 x 2 x−10=−2
Step 5: Move all of your constants to the right side of the equation using inverse operations. Combine all like terms on this side of the equation.
2 x−10=−2+10+10 2 x=8
Step 6: Perform division to cancel out the coefficient of the variable. Simplify your answer, giving you the variable by itself!
2 x=8 2 x 2=8 2 x=4
How to Solve a Linear Equation with Several Variables on the Same Side and Distribution: Example 2
Solve for y: 3(−4 y+3)+5 y=−3(y−1)
Step 1: First, we want to simplify our parentheses. Distribute the term on the outside of the parenthesis to both of the terms inside the parenthesis.
Our first set of parenthesis gives us:
3(−4 y+3)=−12 y+9
Our second set of parenthesis gives us:
−3(y−1)=−3 y+3
Step 2: Rewrite your function with the expanded expressions from step 1.
3(−4 y+3)+5 y=−3(y−1)−12 y+9+5 y=−3 y+3
Step 3: Combine like terms on both sides of the equation.
Our like terms on the right in this function are 6x and -3x. To combine them, we will add the coefficients.
−12 y+9+5 y=−3 y+3−7 y+9=−3 y+3
Step 4: Move all of your terms with variables to the left side of your equation using inverse operations. Combine like terms before moving on to the next step.
−7 y+9=−3 y+3+3 y+3 y−4 y−9=3
Step 5: Move all of your constants to the right side of the equation using inverse operations. Combine all like terms on this side of the equation.
−4 y−9=3+9+9−4 y=12
Step 6: Perform division to cancel out the coefficient of the variable. Simplify your answer, giving you the variable by itself!
−4 y=12−4 y−4=12−4 x=−3
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Table of Contents
How to Solve a Linear Equation with Several Variables on the Same Side and Distribution
How to Solve a Linear Equation with Several Variables on the Same Side and Distribution Vocabulary
Example 1
Example 2
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Practice
Solving a Linear Equation with Variables on Both Sides & Distribution
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9006 | https://www.ncbi.nlm.nih.gov/books/NBK580533/ | Anatomy, Shoulder and Upper Limb, Hand Thenar Eminence - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Anatomy, Shoulder and Upper Limb, Hand Thenar Eminence
Dylan Luxenburg; Michael G. Rizzo.
Author Information and Affiliations
Authors
Dylan Luxenburg 1; Michael G. Rizzo 2.
Affiliations
1 University of Miami Miller School of Medicine (UMMSM), Miami, FL
2 Washington University in St. Louis School of Medicine
Last Update: May 1, 2023.
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Introduction
The word thenar originates from the Greek meaning “palm of the hand.” The hypothenar eminence is located on the ulnar portion of the volar hand at the base of the fifth digit (small finger). The thenar eminence is located on the radial portion of the volar hand at the base of the first digit (thumb).
The thenar eminence is composed of three muscles that control the movements of the thumb: abductor pollicis brevis (APB), flexor pollicis brevis (FPB), and the opponens pollicis. Medical conditions that can affect the thenar eminence include carpal tunnel syndrome, arthritis, and amyotrophic lateral sclerosis (ALS), among others.
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Structure and Function
The thenar eminence can be easily identified by palpating the base of the thumb on the palm (volar) side of the hand. The abductor pollicis brevis, flexor pollicis brevis, and the opponens pollicis are located here and control the movements of the thumb. These muscles work to provide adduction, abduction, flexion, pronation, and opposition for the thumb.
The abductor pollicis brevis abducts the thumb. This muscle is located superficial and lateral to the other two muscles. Its origin is mainly from the flexor retinaculum but also originates from the tubercles of the scaphoid and trapezium. It then travels both distal and lateral to insert on the base of the proximal phalanx of the thumb.
The flexor pollicis brevis is located superficial and medial to the opponens pollicis. It has two muscle heads, a superficial and a deep one. Both heads are separated by the flexor pollicis longus. Its origin is the trapezium and flexor retinaculum, and it inserts on the proximal phalanx of the thumb. It functions to flex the thumb.
The opponens pollicis is located deep to the other muscles, and it is also the largest. It originates from both the flexor retinaculum and the trapezium tubercle. It inserts on the anterolateral portion of the first metacarpal shaft and opposes the thumb. Opposition in this context is both adduction and pronation. This movement allows the tip of the thumb to contact the tips of the other digits on the same hand.
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Embryology
The muscles of the thenar eminence are derived from the mesodermal germ layer. The mesoderm further differentiates into the vasculature involved in this location. The ectodermal germ layer differentiates to form the skin and neural networks located here.
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Blood Supply and Lymphatics
The abductor pollicis brevis, flexor pollicis brevis, and opponens pollicis receive their blood supply from the superficial palmar branch of the radial artery. This artery branches from the radial artery immediately proximal to the flexor retinaculum. It then runs superficial to the flexor retinaculum as it travels distally to supply the thenar eminence. Later, it anastomoses with the superficial palmar arch, the terminal branch of the ulnar artery.
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Nerves
The thenar eminence is primarily innervated by the median nerve (T1) but does receive some contribution from the ulnar nerve (C8, T1): the two heads of the flexor pollicis brevis have different innervation. The deep head is innervated by the deep branch of the ulnar nerve (C8, T1), while the superficial head is innervated by the recurrent branch of the median nerve (T1). The abductor pollicis brevis and opponens pollicis are both innervated solely by the recurrent branch of the median nerve.
The median nerve arises from the medial and lateral cords of the brachial plexus. It then travels distally down the arm between the two heads of the pronator teres. It passes within the carpal tunnel, deep to the flexor retinaculum, and branches further. One of these branches is the recurrent branch which innervates the thenar eminence.
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Muscles
The thenar eminence is composed of three muscles that control the movements of the thumb, as described previously. The muscles are listed below:
Abductor pollicis brevis
Flexor pollicis brevis
Opponens pollicis
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Physiologic Variants
Some consider the adductor pollicis to be a member of the thenar eminence. This muscle consists of an oblique and transverse head. The oblique head originates from the capitate and the bases of metacarpals two and three.
The transverse head originates from the base of metacarpal three. The two heads combine and travel laterally to insert onto the base of the proximal phalanx of the thumb. It is supplied by the deep palmar arch and is innervated by the ulnar nerve. It functions to adduct the thumb and also assists in producing the movement of opposition.
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Surgical Considerations
Many surgical operations of the hand require a thorough understanding of thenar eminence anatomy. Carpal tunnel syndrome, thumb arthritis, and fractures are common conditions that may require surgery.
Carpal tunnel syndrome (CTS) is caused by compression of the median nerve at the transverse carpal ligament. The thenar eminence is innervated by a branch of this nerve, so severe forms of CTS can have significant ramifications on the thenar eminence and, ultimately, thumb function.Patients can experience atrophy of the thenar eminence as well as pain and paresthesia. Surgery is considered if conservative management fails or acute onset of CTS, such as after distal radius fractures or lunate or perilunate dislocations.
Carpal tunnel release can be performed to decompress the median nerve by incising the transverse carpal ligament to increase the space within the carpal tunnel. It can be performed in either an open or endoscopic manner. Open techniques use a slightly larger incision, but there are no differences in outcomes. The surgery only takes about 15 minutes, and most patients will have relief from either some or all of their symptoms.
Surgical options for thumb arthritis include arthroscopic debridement, osteotomy of the first metacarpal, trapeziectomy, carpometacarpal (CMC) arthrodesis, CMC denervation, and CMC prosthetic arthroplasty. Each of these procedures is unique in its technique, but all of them involve navigating through or manipulating some anatomical portion of the thenar eminence.
Fractures of the first metacarpal or trapezium affect the thenar eminence. All three muscles of the thenar eminence have part of their origin on the trapezium and travel superficially to the first metacarpal. Fractures here can disrupt the integrity of these muscles and may also damage nearby neurological structures causing sensory or motor deficits. Surgical interventions include open reduction, internal fixation, external fixation, and metacarpophalangeal arthroplasty or fusion.
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Clinical Significance
Carpal tunnel syndrome is considered a compressive neuropathy of the median nerve that causes pain, paresthesia, and muscle wasting in the thenar eminence. It is diagnosed clinically based on a patient’s signs and symptoms. Females are three times more likely than males to be affected by CTS. Conditions such as pregnancy and arthritis are considered risk factors as well. People who perform repetitive motions or vibrations of the wrist, such as excessive typing, are more likely to be affected by CTS.
The median nerve has multiple branches. Therefore, the palmar cutaneous branch travels superficially to the carpal tunnel and is not affected by CTS. This branch is responsible for skin sensation over the thenar eminence. The recurrent branch travels through the carpal tunnel before returning proximally to innervate the muscles of the thenar eminence. Compression of the recurrent branch can lead to muscle atrophy of the thenar eminence.
Multiple physical exam maneuvers can be performed to identify CTS, and these include Tinel’s test, Durkan’s test, and Phalen’s Test. Additional modalities are not required for diagnosis but can provide useful adjunctive information, especially in complex presentations. Nerve conduction velocity (NCV) and electromyography (EMG) are often used. NCV shows slowing conduction velocity, and EMG shows fibrillations and fasciculations of the muscles.
First-line treatment for CTS is non-operative and typically involves the use of NSAIDs, activity modification, and nighttime neutral splinting. If first-line therapy fails, steroid injections may be utilized and have shown transient improvement in 80% of patients. Surgical intervention may be considered if non-operative management fails to treat CTS and often results in partial or complete relief of symptoms in 75% of patients.
Atrophy of the hand muscles is a characteristic feature of ALS. It most often occurs in the thenar eminence while the hypothenar eminence is spared. This is known as the split hand sign, and it can be the initial presentation of ALS as it may appear before upper motor nerve signs. Kuwabara et al. used EMG on 77 patients with ALS and 171 patients without ALS, which confirmed thenar involvement and hypothenar sparing muscle wasting in early ALS.
Ape hand deformity is caused by damage to the recurrent motor branch of the median nerve. This causes the muscles of the thenar eminence to be denervated and lose function, specifically the opponens pollicis. People with this deformity are not able to perform opposition with the thumb. Damage to this nerve typically occurs at either the elbow or the wrist. Median nerve palsy may also present with ape hand deformity and typically occurs following a deep penetrating injury or blunt force trauma to the arm, forearm, or wrist.
Supracondylar humerus fractures may also cause this deformity. The anterior interosseus nerve (AIN), a significant motor branch of the median nerve, lies anterior to the supracondylar humerus and is at risk of being stretched, impinged, or pierced during fracture. This type of fracture is most common in children due to falling on an outstretched arm.
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Figure
Palmar View of the Hand Muscles, Tendons, and Fascia. The image depicts the superficial structures of the palm, including the thenar and hypothenar eminences, digital tendon sheaths, and the transverse carpal ligament. The flexor tendons of flexor digitorum (more...)
Figure
Hand compartments: central compartment, hypothenar compartment, interosseous compartment, midpalmar space, adductor compartment, thenar space, thenar compartment Illustration by Emma Gregory
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References
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Nguyen JD, Black AC, Duong H. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Nov 17, 2023. Anatomy, Shoulder and Upper Limb, Hand Hypothenar Eminence. [PubMed: 31536229]
2.
Vargas A, Chiapas-Gasca K, Hernández-Díaz C, Canoso JJ, Saavedra MÁ, Navarro-Zarza JE, Villaseñor-Ovies P, Kalish RA. Clinical anatomy of the hand. Reumatol Clin. 2012 Dec-2013 Jan;8 Suppl 2:25-32. [PubMed: 23219083]
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Moran CA. Anatomy of the hand. Phys Ther. 1989 Dec;69(12):1007-13. [PubMed: 2685840]
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Cole P, Kaufman Y, Hatef DA, Hollier LH. Embryology of the hand and upper extremity. J Craniofac Surg. 2009 Jul;20(4):992-5. [PubMed: 19553860]
5.
Zekavica A, Milisavljević M, Erić D, Ćurčić B, Popović S, Vitošević B, Dožić A, Štimec BV, Manojlović R. Vascular anatomy of the thenar eminence: its relevance to a pedicled or free thenar flap. Folia Morphol (Warsz). 2017;76(2):232-238. [PubMed: 28026846]
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Duncan SF, Saracevic CE, Kakinoki R. Biomechanics of the hand. Hand Clin. 2013 Nov;29(4):483-92. [PubMed: 24209947]
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Dydyk AM, Negrete G, Sarwan G, Cascella M. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Jun 4, 2023. Median Nerve Injury. [PubMed: 31971749]
8.
Beran R. Paraesthesia and peripheral neuropathy. Aust Fam Physician. 2015 Mar;44(3):92-5. [PubMed: 25770571]
9.
Earp B, Cefalu C, Blazar P. Thumb Metacarpophalangeal Joint Arthritis. J Am Acad Orthop Surg. 2019 Dec 01;27(23):e1029-e1039. [PubMed: 31356425]
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Meals C, Meals R. Hand fractures: a review of current treatment strategies. J Hand Surg Am. 2013 May;38(5):1021-31; quiz 1031. [PubMed: 23618458]
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Benny R, Shetty K. The split hand sign. Ann Indian Acad Neurol. 2012 Jul;15(3):175-6. [PMC free article: PMC3424792] [PubMed: 22919187]
Disclosure:Dylan Luxenburg declares no relevant financial relationships with ineligible companies.
Disclosure:Michael Rizzo declares no relevant financial relationships with ineligible companies.
Introduction
Structure and Function
Embryology
Blood Supply and Lymphatics
Nerves
Muscles
Physiologic Variants
Surgical Considerations
Clinical Significance
Review Questions
References
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
Bookshelf ID: NBK580533 PMID: 35593840
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Introduction
Structure and Function
Embryology
Blood Supply and Lymphatics
Nerves
Muscles
Physiologic Variants
Surgical Considerations
Clinical Significance
Review Questions
References
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PubMedLinks to PubMed
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Luxenburg D, Rizzo MG. Anatomy, Shoulder and Upper Limb, Hand Thenar Eminence. [Updated 2023 May 1]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Available from:
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9007 | https://math.stackexchange.com/questions/1165751/finding-the-value-of-p-in-the-parabola-y2-2px | Stack Exchange Network
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Finding the value of $p$ in the parabola $y^2=2px$
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$\begingroup$
I just started to learn the parabola shape and I have a question:
Given the parabola $y^2=2px$ $(p>0)$.
The chord $AB$ of the parabola passes through the focus $F(\frac{p}{2},0)$.
The slope $m$ of chord $AB$ is $m_{AB}=2$.
The length of $AB$ is $|AB|=15$.
I need to find the value of $p$ (the equation of the parabola).
My Attempt
let $A(x_1,y_1),B(x_2,y_2)$ so $m_{AB}=m_{AF}=m_{BF}=2$.
Also $|AB|$ is equal to the sum of the radius $r_1+r_2$, so $x_1+x_2+p=15$.
From the slopes I got: $y_1=2x_1-p,y_2=2x_2-p$.
Then I got stuck in no going anywhere algebra.
Is my attempt ok? Or could it be done in a better way? And how can I proceed?
analytic-geometry
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edited Feb 26, 2015 at 0:38
Daniel Fischer
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asked Feb 25, 2015 at 23:57
ronironi
322 bronze badges
$\endgroup$
Add a comment |
2 Answers 2
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$\begingroup$
For the chord you have a gradient 2 and a fixed point $(\frac p 2,0)$
$y=2x+c$
$0=p+c$
$c=-p$
Chord $y=2x-p$ meets curve $y^2=2px$ when $(2x-p)^2=2px$
$4x^2-4px+p^2=2px$
$4x^2-6px+p^2=0$
It's not nice:
$x_1=\frac {3+ \sqrt 5} 4 p$ and $x_2=\frac {3- \sqrt 5} 4 p$
$y_1=\frac {3+ \sqrt 5} 2 p -p = \frac {1+ \sqrt 5} 2 p$ and $y_2=\frac {3- \sqrt 5} 2 p -p = \frac {1- \sqrt 5} 2 p$
$(x_1 - x_2)^2+(y_1-y_2)^2=225$
$(\frac {2 \sqrt 5} 4 p)^2+(\frac {2 \sqrt 5} 2 p)^2=225$
$\frac 5 4 p^2+5 p^2=225$
$\frac {25} 4 p^2=225$
$p^2=36$
$p=6$
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edited Feb 26, 2015 at 1:34
answered Feb 26, 2015 at 0:42
tomitomi
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$\endgroup$
4
$\begingroup$ did you mean $y^2=2px$? $\endgroup$
roni
– roni
2015-02-26 00:47:17 +00:00
Commented Feb 26, 2015 at 0:47
$\begingroup$ Yes - will fix now $\endgroup$
tomi
– tomi
2015-02-26 00:49:07 +00:00
Commented Feb 26, 2015 at 0:49
$\begingroup$ is it going anywhere? the roots are not nice. $\endgroup$
roni
– roni
2015-02-26 00:53:33 +00:00
Commented Feb 26, 2015 at 0:53
$\begingroup$ i think you meant $y_1-y_2= \sqrt{5}p$ $\endgroup$
roni
– roni
2015-02-26 01:13:30 +00:00
Commented Feb 26, 2015 at 1:13
Add a comment |
0
$\begingroup$
HINT:
Parametric form is much easier.
$$ x= 2 p t , y = p t^2 $$
Slope at any point = t, slope connecting two points with slopes $t_1$ and $t_2$ is given $(t_1 +t_2)/2$.
Find distance between them in terms of $t_1$ $t_2$ and p and solve for $ t_1,t_2,p $
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answered Feb 26, 2015 at 0:21
NarasimhamNarasimham
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1
$\begingroup$ I think i got your hint but not completely: the parabola is $y^2=2px$ so $2yy'=2p$ so $y'=\frac{p}{y}=\frac{1}{t^2}$ but that's not $t$ as you wrote. also if you take the opposite from your substitute: $x=pt^2, y=2pt$ so we get: $y'=\frac{1}{t}$ and that also isn't $t$ but it's simpler than the last one. but still, why the slope, as you said, is $(t_1+t_2)/2$? in the question i only know that the slope of the the line that passes in $F$ is $2$.. $\endgroup$
roni
– roni
2015-02-26 02:02:57 +00:00
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9008 | https://math.stackexchange.com/questions/4134432/how-to-show-that-x2-equiv-2-pmod-11-has-no-solution | number theory - How to show that $x^2\equiv 2 \pmod {11}$ has no solution. - Mathematics Stack Exchange
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How to show that x 2≡2(mod 11)x 2≡2(mod 11) has no solution. [duplicate]
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When is 2 2 a quadratic residue mod p p? (3 answers)
Why does the equation x 2≡2(mod 5)x 2≡2(mod 5) have no solutions? [Euler's Criterion] (7 answers)
Closed 4 years ago.
I am not familiar with the topic. I can work out for the problem when the problem has a solution. But here I don't know how to it. Using the table I can see there is no solution. But I want to know how to do mathematically.
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edited May 10, 2021 at 20:51
Thomas Andrews
188k 18 18 gold badges 228 228 silver badges 429 429 bronze badges
asked May 10, 2021 at 20:48
Madhan KumarMadhan Kumar
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4 Compute x 2 mod 11 x 2 mod 11 for x=0,1,…,10 x=0,1,…,10. I think this approach is "mathematical".azif00 –azif00 2021-05-10 20:51:42 +00:00 Commented May 10, 2021 at 20:51
4 There is a general theorem that, for odd primes p,p,x 2≡2(mod p)x 2≡2(mod p) has a solution if and only if p≡±1(mod 8).p≡±1(mod 8).Thomas Andrews –Thomas Andrews 2021-05-10 20:53:53 +00:00 Commented May 10, 2021 at 20:53
1 @Yassir it "works" with any modulus, it's just not practical. (and, well, 11 is small enough)lisyarus –lisyarus 2021-05-10 20:55:37 +00:00 Commented May 10, 2021 at 20:55
1 @MadhanKumar you could only check x 2(mod 11)x 2(mod 11) for x=0,1,…,5 x=0,1,…,5, since x 2≡(−x)2(mod 11)x 2≡(−x)2(mod 11).Kenta S –Kenta S 2021-05-10 20:56:12 +00:00 Commented May 10, 2021 at 20:56
1 In general, for prime p,p, and any n n not divisible by p p x 2≡n(mod p)x 2≡n(mod p) has a solution if and only if n(p−1)/2≡1(mod p).n(p−1)/2≡1(mod p). Here, n=2,p=11,n=2,p=11, then 2 5=32≢1(mod 11).2 5=32≢1(mod 11).Thomas Andrews –Thomas Andrews 2021-05-10 20:56:46 +00:00 Commented May 10, 2021 at 20:56
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12When is 2 2 a quadratic residue mod p p?
5Why does the equation x 2≡2(mod 5)x 2≡2(mod 5) have no solutions? [Euler's Criterion]
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7Quick proof to showing that p≡1(mod 4)p≡1(mod 4) implies p p is reducible in Z[i]Z[i]?
1How to show that x 2≡23≡5(mod 9)x 2≡23≡5(mod 9) is not solvable?
3Proof that if n n has a primitive root, x k≡a(mod n)x k≡a(mod n) has a solution ⟺a ϕ(n)/gcd(k,ϕ(n))≡1(mod n)⟺a ϕ(n)/gcd(k,ϕ(n))≡1(mod n).
5Number Theory: Find m≡1(mod 4)m≡1(mod 4) so that x 2≡−1(mod m)x 2≡−1(mod m) has no solution.
1How many solution has the quadratic equation x 2≡a(mod N)x 2≡a(mod N) when N=p q N=p q and what is the proof?
2Show that the congruence x 3≡a(mod 167)x 3≡a(mod 167) has solutions for all a.
2For how many n<N n<N is d(n)≡0(mod p)d(n)≡0(mod p)?
1Simultaneous congruences 3 x≡2(mod 5)3 x≡2(mod 5), 3 x≡4(mod 7)3 x≡4(mod 7), 3 x≡6(mod 11)3 x≡6(mod 11)
2Prove that x 2≡a(mod 2 m)x 2≡a(mod 2 m) has exactly 4 solutions if a≡1(mod 8)a≡1(mod 8).
9if gcd(a,n)=1 gcd(a,n)=1, show that x x≡a(mod n)x x≡a(mod n) always has a solution.
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9009 | https://eventuallyalmosteverywhere.wordpress.com/tag/parity/ | Skip to primary content
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Eventually Almost Everywhere
A blog about probability and olympiads by Dominic Yeo
Tag Archives: parity
Finitely many solutions
This post is aimed at secondary-school students pitched roughly at the level of the British Mathematical Olympiad. It is ostensibly about a certain class of number theory problems, but the main underlying mathematical principle is broader than this. The post is based on an in-person session I have given to students and at teachers’ conferences a few times over the past five years, but I’ve chosen to write it up at this moment because of the connection to Problem 4 on the most recent BMO1 paper, which we discuss later in the post.
Alice and Bob make the following statements:
Alice: all prime numbers are odd.
Bob: we can’t say anything about whether prime numbers are odd or even.
These are different types of statement, one objective, one more subjective; but both are wrong. Alice is wrong because 2 is a prime, and 2 is of course even. Bob is wrong because while Alice is wrong, and the alternative statement “all prime numbers are even” is even more wrong1, there are plenty of possible statements that are true and useful, including
Cleo: all prime numbers are odd except 2.
and weaker versions (that may be more relevant in other contexts) like “all primes greater than are odd”.
The characters now turn their attention to another family of integers, the square numbers , and the sixth powers
Alice: no square numbers are prime.
Bob: I agree.
Alice: most square numbers are not sixth powers.
Bob: hang on, but infinitely many square numbers are sixth powers?
The difficulty in the second statement is what does most mean in this context? The square of an integer n is a sixth power if and only if n is a cube, and so Alice’s statement is equivalent to the statement “most positive integers are not cubes”, which is intuitively reasonable, but would require more clarification than the analogous statement “most primes are odd”. Is it not relevant, for example, that 1/4 of the first eight positive integers are in fact cubes?
It is not impossible to formalise this notion of most in a reasonable way to permit the two statements above2.
But the bulk of this post will discuss situations that extend the first of these conversations, ie when it’s not true to say “all element of X have property Y”, but where it’s nonetheless possible to make a useful intermediate statement.
Warmup problems
I hope most readers know how to solve an equation such as
and indeed learning how to solve such equations forms a important milestone in mathematical education as the first example of various principles of algebra. When such equations are still novel, teachers would be advised to avoid examples such as
for which the answers are, respectively, ‘no solutions’ and ‘every x satisfies this’. While they are perfectly valid equations, it means something slightly different to solve them, since the answer set has different structure and, more importantly, the method is different since it doesn’t consist of a sequence of a manipulations leading directly to the required conclusion .
However, I mention this in passing to clarify that even in the simplest possible family of equations (probably encountered somewhere between ages 9 and 13) we are mindful that not all equations have unique solutions.
I’ve adapted a problem from the American competition AIME:
Find all integers such that is a prime.
One could start by writing if you wish, but in fact we only need to work with the left hand side, so it’s better to avoid this if possible. The key to this problem is to avoid the prime condition initially, and just manipulate the LHS until it simplifies as
()
which is never a prime, since is a square number and so, as already discussed, never a prime.
So at a meta-level, comparing with the other problem in this section, this one also begins with reversible algebraic manipulations, but does not end with a direct reduction to , and indeed the final step is not reversible at all.
Finally, we remark that the original version of the AIME problem was
Prove that for all integers , is a perfect square.
The same algebraic reduction as () resolves the original version without an extra step concerning primes.
To be, or not to be, a perfect square
The rest of this article will focus entirely on problems of the form: find all positive integers n such that […] is a perfect square.
As setup, we’ll always consider a function f(n), and the sequence f(1),f(2),f(3),… and the question, comment on the set of n is f(n) a square? In particular, this is a more open-ended question than for which n is f(n) a square, which requires a more exact answer. We’ll aim for one of the following possible answers:
f(n) is always a square;
f(n) is a square for infinitely many n;
f(n) is a square for only finitely many n;
f(n) is never a square.
Clearly this is not an exhaustive list of possible descriptions3 of the set under discussion, but it will do for now. The difference between the second and the third bullet point can be thought of as: “are there arbitrarily large values of n for which f(n) is a square?”
As some examples:
is always a square, because is always a square.
is never a square, because the only squares that differ by 1 are {0,1} which aren’t attainable when .
is a square for only finitely many n, because the only squares that differ by 3 are {1,4}.
Exercises to try yourself
,
or, now with p ranging over the set of prime numbers: .
Some discussion of some of these follows later to avoid spoilers.
BMO1 2023, Problem 4
This problem does not have exactly the same structure as the previous exercise. A solution will follow shortly, but before starting, [spoilers] I want to emphasise that the main part of the solution involves framing this problem exactly as in the exercise above.
We will establish that there are only finitely many n such that is a square, using a classical number theory argument.
In doing so, we will eliminate all integers n that are at least some threshold value K from consideration (which we think of as “n large”, even though in this case it isn’t so large).
The actual set of solutions comes from check the remaining “small” values of n.
The point to emphasise is that while this third bullet is definitely an important component of a successful solution (and does feel closest to actually addressing the “find all” aspect of the problem statement), it is not really the key step, since checking a small number of possibilities is fundamentally much more straightforward than an argument to eliminate the other infinitely-many integers.
This overall structure is very common when solving Diophantine equations involving integers, even if the exact statement isn’t “find all n such that […] is a perfect square”.
Anyway, to the solution of this problem. As in many problems of this difficulty in this number theory, it’s really useful to aim to factorise and analyse the factors, remembering that they are integers! In this case, writing doesn’t suggest an easy factorisation of the LHS, but rewriting as does now have a nice factorisation, using difference of two squares on the RHS:
Here the two factors are (k-1) and (k+1). We know nothing about k, so we know nothing intrinsic about these factors, except that they differ by 2. But now when we consider that their product is , we can start to make some serious deductions4.
(k-1) and (k+1) differ by two, and so have the same parity5. Then is even, and so in fact both factors are even.
But more than this, one factor will be a multiple of 4, while the other will not be. So all the powers of 2 within (which may include additional factors of 2 coming from n) apart from one, must be included in one of the factors.
The idea is that is generally much larger than n, and so this will force one of the factors to be much larger than the other factor, which is not permitted. (Recall that the factors differ by 2.)
So formally, we might write that and . These are inequalities not equalities because some of the factors of n could be in (k+1) not (k-1) and/or a priori, we could have and etc. This leads to
()
and we now have exactly the language to describe the truth of this inequality! It is only true for finitely many positive integers n. There are a number of ways to prove this, but it is important that we do provide a lower bound. One possibility is to check that () holds for n=5, and then prove () for all by an inductive argument.
At this point, we have shown that is not a perfect square for , and so at this point it really is easiest just to check by hand, which reveals that work.
To link this to the background, note that this final procedure of “finding all the solutions” really had nothing in common structure-wise with “solving “, for which substituting in various values of x would never constitute part of a good solution.
Comments on exercises
is a square for infinitely many n, while is never a square, for reasons that are often framed in the language of quadratic residues.
is a square for infinitely many n (specifically, when n is even), while is a square only when n=3, which can be seen via a similar argument to the argument above for the BMO problem. Meanwhile is never a square, which can also be checked using quadratic residues.
is a square for infinitely many n (specifically, when ), while is a square for only one prime p, since only one prime is divisible by 3.
is only a square for finitely many n, because when n is large, we have . This style of argument is sometimes called square-sandwiching, although it can of course be deployed in contexts other than squares. Problems that turn on this method have appeared in BMO surprisingly often in the past 15 years. For this particular example, note that one does have to check a handful of values to confirm it is for finitely many but at least one n (rather than for no n).
Footnotes
of course, writing even more wrong is in itself making a nuanced comment about the parity of the prime numbers. ↩︎
given a set , a natural formalisation of the notion that ‘most integers are not in A’ would be that In other words, the proportion of elements of A amongst the first n integers vanishes as . This is certainly true when A is the set of cubes.
There are many sets for which this limit does not exist, including quite natural sets like the integers whose decimal representation starts with a 1, and so a full treatment of this notion of asymptotic density would require more than a footnote. ↩︎
in particular, we might wish to draw a distinction between f(n) is a square for infinitely many n and the stronger statement f(n) is a square for all-but-finitely many n, and in some applications that difference is crucial. ↩︎
A common error is to make some reasonable-sounding assumption about how the factors of are split between (k-1) and (k+1), for example by assuming that and because this is visually a natural way to pair them up. ↩︎
parity refers to whether a number is odd or even. ↩︎
The Top-to-Random Shuffle III
Posted on by dominicyeo
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This post concludes my non-exhaustive list of things I think are interesting about the top-to-random shuffle. In previous posts I have talked about the construction and correct sense of convergence to randomness, and that this algorithm does genuinely achieve uniform randomness at some hitting time which is easy to specify. Promising that posts will be short hasn’t worked in the past so I won’t do that again now, but the idea of this post is brief:
When we specified the dynamics of the top-to-random shuffle, we insisted that the top card card could be placed anywhere in the deck with equal probability including back on top. This appears to be doing nothing except slowing down the shuffling process. Why is this important for convergence to randomness?
Fortunately the answer is short: if we do not let the top card be inserted back onto the top, allowing the configuration to stay the same, then we can divide up the set of orderings into two classes, and the pack will alternate between them.
Why is this a problem? Suppose the classes are called X and Y, and X is the class that contains the original ordering 1,2,…,n. Then after k shuffles, the ordering of the deck will be in X if k is even and in Y if k is odd. Remember our definition of ‘close to randomness’ will be the greatest difference in probability of an event between the actual distribution and the uniform distribution. As before, you can think of this by a betting analogy – what proportion profit can you make again someone who thinks it’s uniform by knowing the true distribution?
Well, it will turn out that the sets X and Y have the same size, so in the uniform distribution, the probability that an ordering is in X is 1/2. Whereas if the pack alternates, then so long as we know how many shuffles have occurred, this probability is either 0 or 1. In particular this is far from 1/2. We should remark that if we introduce the notion of sampling at a random time, or taking an average over all large times in some sense, such problems may disappear, but the result obtained may be less useful. See this post on Cesaro Mixing for details presented in a more rigorous style.
So it remains to see why this is true. First a definition. A transposition is when two elements in a permutation are exchanged. Eg 31452 -> 35412 by transposing 1 and 5. It makes sense intuitively that we can get from any permutation to any other permutation by making successive transpositions. Indeed, this is precisely what is happening in the top-to-random shuffle. To avoid continually having to write it out, we call the original permutation 1,2,…,n the identitypermutation.
Then the idea is that X is the set of permutations we can obtain by starting with the identity and applying an even number of transpositions, while Y is the set obtained by applying an odd number of transpositions. For this to work, we will need to show that these sets are disjoint. That is, no permutation can be generated by both an odd number and an even number of transpositions. This is important, as a permutation can certainly be generated from transpositions in multiple ways. For example, if the elements are 1,2,3, we can obtain the permutation 2,1,3 by transposing 1 and 2, obviously. However, we could alternatively start by transposing 2 and 3 to get 1,3,2, then 1 and 3 to get 3,1,2, then 2 and 3 again to get 2,1,3. Note that both of these require an odd number of transpositions.
We will call a permutation even if it is generated by an even number of transpositions, and odd otherwise. We also say that its sign (alternatively signature, parity) is +1 or -1 respectively. To prove this is well-defined, we really want to find a different property that is easier to track.
A useful trick is to count how many pairs of elements are not in the correct order. Let’s do this for our previous example: 31452. There are 5 elements so 5 x 4 / 2 = 10 pairs of elements. We list them:
1 and 2 are in the correct order.
1 and 3 are not, as 3 comes before 1 in this permutation.
1 and 4 are correct.
1 and 5 are correct.
2 and 3 are not.
2 and 4 are not.
2 and 5 are not.
3 and 4 are correct.
3 and 5 are correct.
4 and 5 are not.
So 5 pairs are not in the correct order. Since 5 is odd, the claim is that this means 31452 is an odd permutation. To check this, and to confirm that the sign is well-defined, it suffices to check that the number of so-called inversions, or pairs in the wrong order, changes parity every time we apply a transposition.
This is clearly true if we transpose adjacent elements. Then the orderings of all pairs remain the same, apart from the pair we transposed, which changes. Then, if the elements are not adjacent, instead of transposing them directly, we can perform a succession of transpositions of adjacent elements. The easiest way to describe this is again by example. Suppose we want to transpose 3 and 5 in 31452.
31452 -> 13452 -> 14352 -> 14532 -> 15432 -> 51432.
Note that the middle transposition is actually transposing 3 and 5, and the others are symmetric about this middle operation. In particular, there is an odd number of transpositions in total. So we have proved the result for general transpositions, and thus we now know that the sign of a permutation is well-defined. Note also that there are an equal number of odd and even permutations of every n=>2. For every odd permutation, transposing 1 and 2 gives an even permutation, and vice versa, uniquely, giving a bijection.
What’s really going on is that we are able to multiplypermutations, by doing one after the other. Unlike multiplying real numbers, the order in which we do this now matters. In this context, the set of permutations is an example of a general structure called a group. The idea of partitioning a group into subsets which are in some sense symmetric and where some other operation jumps between the subsets is a useful motivation point for a whole avenue of interesting theory. Not to be explored now unfortunately…
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11.7: Fractional Linear Transformations
Last updated
: May 2, 2023
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11.6: Examples of conformal maps and excercises
11.8: Reflection and symmetry
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Jeremy Orloff
Massachusetts Institute of Technology via MIT OpenCourseWare
( \newcommand{\kernel}{\mathrm{null}\,})
Definition: Fractional Linear Transformations
A fractional linear transformation is a function of the form
where , , , and are complex constants and with .
These are also called Möbius transforms or bilinear transforms. We will abbreviate fractional linear transformation as FLT.
Simple Point
If then is a constant function.
Proof
: The full proof requires that we deal with all the cases where some of the coefficients are 0. We’ll give the proof assuming and leave the case to you. Assuming , the condition implies
So,
That is, is constant.
Extension to . It will be convenient to consider linear transformations to be defined on the extended complex plane by defining
Example : Scale and Rotate
Let . If is real this scales the plane. If it rotates the plane. If it does both at once.
Note that is the fractional linear transformation with coefficients
(We’ll see below the benefit of presenting the coefficients in matrix form!)
Example : Scale and Rotate and Translate
Let . Adding the term introduces a translation to the previous example.
Note that is the fractional linear transformation with coefficients
Example : Inversion
Let . This is called an inversion. It turns the unit circle inside out. Note that and . In the figure below the circle that is outside the unit circle in the plane is inside the unit circle in the plane and vice-versa. Note that the arrows on the curves are reversed.
Note that is the fractional linear transformation with coefficients
Example
Let
We claim that this maps the -axis to the unit circle and the upper half-plane to the unit disk.
Solution
First take real, then
So, maps the -axis to the unit circle.
Next take with , i.e. in the upper half-plane. Clearly
so
implying that
So, maps the upper half-plane to the unit disk.
We will use this map frequently, so for the record we note that
, , , , .
These computations show that the real axis is mapped counterclockwise around the unit circle starting at 1 and coming back to 1.
Lines and circles
Theorem
A linear fractional transformation maps lines and circles to lines and circles.
Before proving this, note that it does not say lines are mapped to lines and circles to circles. For example, in Example 11.7.4 the real axis is mapped the unit circle. You can also check that inversion maps the line to the circle .
Proof
: We start by showing that inversion maps lines and circles to lines and circles. Given and we define and by
So,
Now, every circle or line can be described by the equation
(If it descibes a line, otherwise a circle.) We convert this to an equation in as follows.
In the last step we used the fact that
We have shown that a line or circle in is transformed to a line or circle in . This shows that inversion maps lines and circles to lines and circles.
We note that for the inversion .
1. Any line not through the origin is mapped to a circle through the origin.
2. Any line through the origin is mapped to a line through the origin.
3. Any circle not through the origin is mapped to a circle not through the origin.
4. Any circle through the origin is mapped to a line not through the origin.
Now, to prove that an arbitrary fractional linear transformation maps lines and circles to lines and circles, we factor it into a sequence of simpler transformations.
First suppose that . So,
Since this is just translation, scaling and rotating, it is clear it maps circles to circles and lines to lines.
Now suppose that . Then,
So, can be computed as a composition of transforms
We know that each of the transforms in this sequence maps lines and circles to lines and circles. Therefore the entire sequence does also.
Mapping to
It turns out that for two sets of three points and there is a fractional linear transformation that takes to . We can construct this map as follows.
Let
Notice that
, , .
Likewise let
Notice that
, , .
Now is the required map.
Correspondence with Matrices
We can identify the transformation
with the matrix
This identification is useful because of the following algebraic facts.
If then and correspond to the same FLT.
. This follows from the obvious equality
If corresponds to and corresponds to then composition corresponds to matrix multiplication .
. The proof is just a bit of algebra.
The claimed correspondence is clear from the last entries in the two lines above.
3. If corresponds to then has an inverse and corresponds to and also to , i.e. to without the factor of .
. Since it is clear from the previous fact that corresponds to . Since
Fact 1 implies and both correspond to the same FLT, i.e. to .
Example
The matrix corresponds to .
The matrix corresponds to rotation by .
The matrix corresponds to the inversion .
11.6: Examples of conformal maps and excercises
11.8: Reflection and symmetry |
9011 | https://www.quora.com/In-functions-mathematics-what-does-the-%E2%80%9C-%E2%80%9D-symbol-do | Something went wrong. Wait a moment and try again.
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5
In functions (mathematics), what does the “|” symbol do?
David Joyce
Professor Emeritus of Mathematics at Clark University
·
Upvoted by
David Vanderschel
, PhD Mathematics & Physics, Rice (1970) · Author has 9.9K answers and 68.4M answer views
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Updated 4y
It’s used in various ways.
Sometimes it’s the abbreviation for “inclusive or” but I prefer ∨ for that usage.
It’s the common abbreviation for divisibility, as in 4|12 for “4 divides 12” (evenly without remainder).
It’s also a common separator for describing sets by properties. For example,
{x∈Z|4|x}
which describes the set of integers which 4 divides, namely {0,±4,±8,±12,…}. Some people prefer to use a colon there
{x∈Z:4|x}.
It’s also the common notation used for absolute values. For example, the absolute value of both 4 and −4 is 4. |4|=|−4|
It’s used in various ways.
Sometimes it’s the abbreviation for “inclusive or” but I prefer ∨ for that usage.
It’s the common abbreviation for divisibility, as in 4|12 for “4 divides 12” (evenly without remainder).
It’s also a common separator for describing sets by properties. For example,
{x∈Z|4|x}
which describes the set of integers which 4 divides, namely {0,±4,±8,±12,…}. Some people prefer to use a colon there
{x∈Z:4|x}.
It’s also the common notation used for absolute values. For example, the absolute value of both 4 and −4 is 4. |4|=|−4|=4. It’s also used for the absolute value (modulus) of complex numbers. |4i|=4. It’s also used for the norm (length) of a vector, as |(3,4)|=5, but I prefer the notation ∥(3,4)∥=5.
Yet another place where | is used is in conditional probability. The expression P(A|B) refers the the probability of an event A given another event B. When P(B)≠0,
P(A|B)=P(A∩B)P(B)
In calculus a vertical stroke can mean “evaluated at” when subscripted by a condition. For example
sinx∣∣∣x=0=sin0=0
When there’s a superscript also, then it’s a difference
∫π0cosx=sinx∣∣∣π0=sinπ−sin0=0−0=0.
You can tell which of these several meanings it means by context.
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Related questions
What does "|x|" mean in mathematics?
What does "||" mean in mathematics?
What are the functions of “|”?
What does the | symbolize in mathematics?
What is the meaning of the vertical line symbol "|" in mathematics?
Henry Smith
Studied at Kenyon College
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 4.7K answers and 16.6M answer views
·
Updated 7y
Originally Answered: What does the | symbolize in mathematics?
·
|x| means the absolute value of x:
|x|={−x,if x≤0x,if x>0
When used by itself, you sometimes see a∣b, which means that "a divides b". As Alex pointed out in the comments A|B also means "A given B" in probability.
Assistant
Bot
·
Sep 5
In mathematics the vertical bar “|” is a context-dependent operator whose meaning is determined by what surrounds it. Common uses:
Absolute value / modulus
|x| = distance of x from 0 on the real line. Example: |−3| = 3.
For complex z, |z| = sqrt(z·conj(z)).
Cardinality
|A| = number of elements in set A (finite case) or a measure of size (cardinality) for infinite sets. Example: |{1,2,3}| = 3.
Divisibility (number theory)
a | b means “a divides b” (a is a divisor of b). Example: 3 | 12 is true, 5 | 12 is false.
Negation often written a ∤ b.
Conditional / such that (set-builder notation, probability, con
In mathematics the vertical bar “|” is a context-dependent operator whose meaning is determined by what surrounds it. Common uses:
Absolute value / modulus
|x| = distance of x from 0 on the real line. Example: |−3| = 3.
For complex z, |z| = sqrt(z·conj(z)).
Cardinality
|A| = number of elements in set A (finite case) or a measure of size (cardinality) for infinite sets. Example: |{1,2,3}| = 3.
Divisibility (number theory)
a | b means “a divides b” (a is a divisor of b). Example: 3 | 12 is true, 5 | 12 is false.
Negation often written a ∤ b.
Conditional / such that (set-builder notation, probability, conditional)
{x ∈ ℝ | x > 0} means “the set of x in ℝ such that x > 0.”
P(A | B) means probability of A given B.
Restriction of a function
f|A denotes the restriction of f to the subset A (domain limited to A). Example: f|[0,1] is f restricted to [0,1].
Evaluation / “given” in function definition or derivatives (less common)
f'(x)|_{x=0} means evaluate f′ at x = 0 (sometimes written with a subscript or vertical bar and subscript).
Norms and other bars
Single bars sometimes denote other norms (‖·‖ usually double bars for norms); context clarifies meaning.
How to decide which meaning applies:
Check surrounding symbols (numbers, sets, functions, probability).
Check subject area: number theory → divisibility; real/complex analysis → absolute value; set theory → cardinality; probability → conditional probability; algebra/analysis → restriction or evaluation.
Examples summarizing usage:
|−7| = 7 (absolute value)
|{a,b,c}| = 3 (cardinality)
4 | 20 (divisibility)
{x ∈ ℝ | x2 < 4} = (−2,2) (such that)
P(A | B) = P(A ∩ B)/P(B) (conditional probability)
f|
0,1
= f(x) for x ∈ [0,1] (restriction)
Interpretation always follows context; treat the bar as a shorthand whose precise semantic is inferred from nearby notation.
Joe Leo Gaulin
B.A. in Pure Mathematics & Philosophy, The University of Connecticut (Graduated 2018)
·
8y
Originally Answered: In functions (mathematics), what does “|” symbol do?
·
It means divisibility. So the statement 5|25 reads ‘five divides twenty five’ or ‘twenty five is divisible by five’. However there are multiple other uses such as absolute value (I assumed you didn't mean this) and I've seen it used in set builder notation as ‘such that’.
Edit: I missed that you said in terms of functions. In that case it denotes evalutation. For instance f (x)=3x+1 | 3 would mean f (x) evaluated at 3 so f (3).
Related questions
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What does || || mean in math?
What is the meaning of |x| in mathematics?
What do two vertical lines on either side of a part of a mathematical equation mean?
What does a vertical line mean in a math equation?
Falling Autumn
Studied Pure Mathematics at University of Bremen
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Anton Fahlgren
, M.S. Mathematics, Stockholm University (2018) and
William Pi
, Ph.D Second Sino-Japanese War & Politicians and Government Leaders, University of Bergen (2017) · Author has 2.9K answers and 2.5M answer views
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7y
Originally Answered: What does || mean in mathematics?
·
Without providing a context you can’t really tell, but it most likely will be the norm of a normed vectorspace.
For a K-vectorspace V a norm is a map:
∥⋅∥:V→R≥0
For which the following axioms hold:
∥v∥=0⇔v=0
∥λv∥=|λ|∥v∥ with λ∈K
∥v+w∥≤∥v∥+∥w∥ for all v,w∈V
If you have an inner-product space then this will implicitly stand for the induced norm v↦√⟨v,v⟩.
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Zane Heyl
Former Actuarial Analyst (2017–2022)
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Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 275 answers and 768.6K answer views
·
8y
Originally Answered: In functions (mathematics), what does “|” symbol do?
·
If you had a statement like A = {x in R | x < 3} then the | symbol means “such that”. We’re saying that A is a set of real numbers which are less than 3, and it reads “all x in the set of real numbers such that x is less than 3”.
I hope this helps.
Pawan Bhardwaj
Software Engineer at Accenture (company) (2024–present)
·
5y
Originally Answered: What does "||" mean in mathematics?
·
|| means parallel
This symbol is used to show that a line is parallel to the another line.Two lines can be said parallel if they never intersect and are a fixed distance apart always.
Above figure shows that two parallel lines are cut by a third line called transversal.
|| means parallel
This symbol is used to show that a line is parallel to the another line.Two lines can be said parallel if they never intersect and are a fixed distance apart always.
Above figure shows that two parallel lines are cut by a third line called transversal.
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Joachim Pense
Got a degree in Mathematics.
·
Author has 9.2K answers and 14.6M answer views
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10y
Originally Answered: What does the | symbolize in mathematics?
·
Adding to the others:
"a | b" can mean "a divides b"
In a set, it separates the bound variable from the description: {a∈Z∣a<0} (read "such that")
If f is a function from set A to set B, and C is a subset of A, then f|C is the function "f restricted to C", that is, the same function, but only considered for elements of C, and seen as a function from C to B.
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Jakub Dakowski
studying undergraduate studies in Cognitive Science
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Updated 2y
Originally Answered: What does the | (vertical bar) symbol mean in mathematics?
·
I’ve seen two uses of it.
First as a symbol of the absolute value.
So for example |−2|=2,|4|=4,|−23|=23 and so on.
Second as a symbol of being a divisor - when we write x|y we mean that y/x=n where n is a natural number.
Some examples: 2|4,5|10,3|21
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Myriam Kobylkevich
scientist, chemist, problem-solver
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8y
If an expression is between vertical bars, as in |x-3| , it means the absolute value of that expression, which is the distance between the value and zero.
For example, |5–8| = 3 , because 5–8 = -3 .
If what is between the bars is not a real number, but a complex number with a non-zero imaginary part, it is not called absolute value, but it is still the distance to zero, as in |4 +3i | = 5 .
Fares Gafaf
Prodigious reader
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Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 3.3K answers and 4.7M answer views
·
7y
Originally Answered: What does || mean in mathematics?
·
It is a sign which denotes
parallel
lines (that is, lines which do not intersect). In this example, lines
A
B
and
C
D
are parallel:
It is a sign which denotes
parallel
lines (that is, lines which do not intersect). In this example, lines
A
B
and
C
D
are parallel:
Former Admin (1985–2010)
·
Author has 4.4K answers and 3.2M answer views
·
5y
The vertical bar ( | ) is a computer character and glyph with various uses in mathematics, computing, and typography. It has many names, often related to particular meanings: Sheffer stroke (in logic), verti-bar, vbar, stick, vertical line, vertical slash, bar, pike, or pipe.
The | pipe or vertical bar symbol has the following functions:
is used to represent a pipe in commands in Linux and other Unix-like operating systems. A pipe is a form of redirection that is used tosend the output of one program to another program for further processing. Aug 5, 2007
The vertical bar is a character with vario
The vertical bar ( | ) is a computer character and glyph with various uses in mathematics, computing, and typography. It has many names, often related to particular meanings: Sheffer stroke (in logic), verti-bar, vbar, stick, vertical line, vertical slash, bar, pike, or pipe.
The | pipe or vertical bar symbol has the following functions:
is used to represent a pipe in commands in Linux and other Unix-like operating systems. A pipe is a form of redirection that is used tosend the output of one program to another program for further processing. Aug 5, 2007
The vertical bar is a character with various uses in mathematics, where it can be used to represent absolute value, among others; in computing and programming and in general typography, as a divider not unlike the interpunct.
It is often used in mathematics, logic and statistics. It typically is read as 'given that'. In probability and statistics it often indicates conditional probability, but can also indicate a conditional distribution.Jul 31, 2014 Vertical bar - Wikipedia
Sauhard Shekhar
Bachelor of Commerce(H) in Commerce and Accountancy & The Institute of Chartered Accountants of India (ICAI), Sister Nivedita University (Graduated 2022)
·
7y
Originally Answered: What does "||" mean in mathematics?
·
Generally in mathematics as per my knowledge ‘|| refers to parallel’. We use this sign for parallel lines or whenever we have to write parallel it can be used as a shortcut or an abbreviation.
e.g.- Parallelogram is also written as ||gm.
Note:- All the information above is based on my knowledge level yet. Maybe there are other correct answers to this questions too.
Hope it helped. Best of luck.
CHEERS :)
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9012 | https://www.cs.tufts.edu/research/geometry/pdf/p350-melissaratos.pdf | On Solving Geometric Optimization Problems Using Shortest Paths Elefterios A. Melissaratos and Diane L. Souvaine Department of Computer Science, Rutgers University, New Brunswick, NJ 08903 1 Introduction We have developed techniques which contribute to efficient algorithms for certain geometric optimiza- tion problems involving simple polygons: computing minimum separators, maximum inscribed triangles, a minimum circumscribed concave quadrilateral, or a maximum contained triangle. The structure for our algorithms is as follows: a) decompose the initial problem into a low-degree polynomial number of easy optimization problems; b) solve each individual sub- problem in constant time using the methods of cealcu- Ins, standard methods of numerical analysis, or linear programming. The decomposition step uses short- eat path trees inside simple polygons (Guibas et. al.~ 1987) and, in the case of inscribed triangles, produces a new class of polygons, the fan-shaped polygon. By extending the shortest-path algorithm to splinegons, we also generate splinegon-versions of the algorithms for some of the optimization problems. The problems we discuss fall into four subgroups: Separators: If two points z and y lie on the bound- ary of simple polygon P and define a directed line segment zy C_ P that separates P into two sets PL and Pa, then zy is called a separator. Minimum length: The areas of PL and PR are defined by constants KL and KR. Find a separator of minimum length. This research was supported in part by NSF grant CCR- 88-03549 mad by the NSF DIMACS Science and Teclmology Center STC88-09648. Permission to copy without fee all or part of this material is granted provided that the copies are not made or distributed for direct com- mercial advantage, the ACM copyright notice and the title of the publication and its date appear, and notice is given that copying is by permission of the Association for Computing Machinery. To copy otherwise, or to republish, requires a fee and/or specific permission. © 1990 ACM 0-89791-362-0/90/0006/0350 $1.50 Minimum sum of ratios: Find a separa- tor that minimizes the sum of the ratio of the area of PL and the square of its perime- ter and the ratio of the area of PR and the square of its perimeter. Inscribed Triangles: Given a simple polygon P, a triangle T such that T C P and the vertices of T lie on the boundary of P is an inscribed triangle. Maximum area/perimeter: Find the in- scribed triangle of maximum area/perime- ter. Constra[-ed maximum area/perimet- er: Find a maximum area/perimeter in- scribed triangle with one edge of given length. Circumscribed Quadrilateral: Given a simple polygon P, a quadrilateral Q such that P C_ Q and the all four sides of Q intersect the boundary of P is a circumscribed quadrilateral Minimum area concave: Find the cir- curnscribed concave quadrilateral of mini- mum area. Contained Triangle: Given a simple polygon P, a triangle T such that T C P is a contained triangle. Maximum Area: Find the contained tri- angle of maxinmm area. We solve the separator problems in O(n 2) time, the inscribed triangle problems in O(n 3) time, and the contained triangle problem in O(n 4) time, all in linear space. The quadrilateral algorithm uses either 2 2 O(ncnp) time and O(n) space or O(n~(np + k)) time and O(n + k) space where nc is the number of vertices of the convex hull of P, np = n-nc and k is an 350 instance-dependent parameter which ranges between O(np) and Lisper posed the first separator problem, cit- ing applications in solid modeling and graph cut- ting. A linear algorithm exists when P is convex . Chang posed the second separator problem , claim- ing applications in finite element analysis. Aggar- wal posed the contained triangle problem and the concave circumscribed quadrilateral problem. There exist related results (e.g. ,,,), in- cluding a linear-time algorithm for finding the min- imum area triangle containing a convex n-gon , an O(nk + n log n) time algorithm for computing the minimum area circumscribing k-gon of a convex n- gon ,and an O(nlogn) algorithm for finding the minimum perimeter triangle circumscribing a convex n-gon . Note that any convex/~-gon circumscrib- ing a simple polygon P also circumscribes the convex hull of P. Many researchers have studied inclusion problems (e.g. ,,,,). Some results include a linear-time algorithm for computing the minimum area triangle inscribed in a convex polygon of n ver- tices , an O(kn + n log n) algorithm for comput- ing the the maximum area or perimeter convex k- gon inside a convex n-gon , and O(n 7) time (resp. O(n e) time) and O(n 5) space algorithms for finding the maximum area (perimeter) convex polygon con- tained within a given simple polygon P (, ). points x E P which are visible from an edge i form the visibility polygon of P from edge i. If two edges i and j of P are visible from each other, the set of points of j which are visible from i form the visible part of j with respect to i. One can compute the visible parts of a given edge i from every other edge of the polygon, as well as the visible parts of every edge from i, in O(n) time and space using the shortest path algorithm . If edges i, j of polygon P are visible from each other then the shortest paths from Pi+x to p~ (SPpj+, (pi)) and from P,+i to Pi (SPe,+, (Pi)) are inward disjoint convex chains. The region bounded by the above chains and i and j is called an hourglass and denoted Hi,i. Given a point z inside or on the boundary of P which is visible from edge i, the region bounded by i, SPp,+, (z) and SPp,(x) defines a funnel on base i. The closest vertex to Pi (Pi+x resp.) on the shortest path from z to pl (pi+l resp.) is called the anchor of Pi (Pi+x resp.) with respect to z and is denoted by a, hor" (p, ) ( a.eho,'f (p, + ) resp.). The shortest path algorithm applied to a trian- gulated simple polygon P at a designated start ver- tex v produces a subdivision where each region cor- responds to a funnel based on some polygon edge i, denoted F~(i) (fig. 1). The computation corresponds 2 Shortest Paths In , the authors describe a linear time and space algorithm for finding the shortest paths fxom a ver- tex of a simple polygon to all the other vertices. The union of these paths form a tree called as fhe short- eat path free. In the next few paragraphs we review necessary facts and definitions from that paper and establish our notation. We use these conventions: a simple polygon P has n edges represented by the in- tegers 1, 2, ..., n in clockwise order, and edge j has endpoints pj and Py+l; whenever a subset of polygon vertices (edges) are identified by uppercase (lower- case) letters, alphabetic order implies clockwise order around P; and a line I is tangent to a polygonal chain C if it intersects the chain in one or more points and C lies entirely in one of the halfplanes defined by I. A point x in a simple polygon P is visible from an edge i of P if there exists a point y on i such that xy C_ P. Two edges i, j of P are visible from each other if and only if there exist at least two points x, y on i, j respectively such that zy C P. The set of v Figure 1: the shortest path tree to a preorder traversal of the binary tree with one node for each triangle, with an edge joining two nodes whose triangles share an edge, and with the triangle containing s as the root, maintaining the invariant that the funnels have been computed for all edges of frontier nodes of the traversed portion of the tree. A parent funnel is split to form the funnel for its chil- dren. Extending the edges of each funnel up to their intersection with the funnel's base produces a refined 351 subdivision of P called the shortest path map , (fig. 2). The shortest path map from a vertex v sub- divides the edges of P. for its children. S, (i) is the subdivision of a specific edge i and its size is denoted 17 by s i • Figure 2: a shortest path map Minor revisions allow the polygonal shortest path algorithm to work for curved polygons, also known as splinegons in comparable time and space bounds. The original algorithm runs on a triangu- lated polygon. A splinegon cannot necessarily be tri- angulated without either adding vertices in its inte- rior or adding an unlimited number on its boundary. Fournier and Montuno , however, show that tri- angulation of a simple polygon is linear-time equiva- lent to solving the all vertex-edge horizontal visibility problem. We prove: Theorem 2.1 The shortest path tree algorithm for a splinegon runs in O(n) time given its horizontal visibility decomposition. Proof: Preprocess the edges of the splinegon such that each edge is monotone with respect to both the z and the V axes. Decompose the splinegon into hor- izontal (curved) trapezoids ([281, [141). Refine the trapezoids so that every component is either a trian- gle or a quadrilateral with one edge on the Splinegon boundary. The dual of the decomposition is now a binary tree. Despite revisions for splitting funnels with curved edges, the recursive formula used in to prove the linearity of the entire algorithm still ap- plies. 3 Separator s If z, y delimit a separator for simple polygon P, then z and y are visible in P. Thus, if z lies on edge i, y on edge j, then i and j are visible in P. Furthermore, the line segment zy lies in the hourglass Hid defined by the shortest paths from pj+t to pi (SPpj+t(pi)) and from pi+ to pj (SPy,+, (pj)). Thus our goal is to reduce the optimum separator problem for a simple polygon to a series of optimum separator problems on hourglasses which are simpler to solve. We focus on the minimum-length separator problem. Define as az (aR reap.) the area of P to the left (right reap.) of zy (fig. 3). Not every hourglass will admit a separator satisfying the constraints that az = Kz and aa = Ka where KL +K/t = A. Note, however, that SPp,(pj+t) (resp. SPpj(pi+l)) cuts the polygon into two or more pieces, those to the left with (combined) area ALpj+, (pi), and those to the right with area ALv,+, (pj)). A necessary and sufficient condition for Hit to contain a separator zy is ALp~+,(pl) ~ KL and ALp,+t(pj) < Ka. It is easy to compute ALp~+~(pi), since the shortest path map from pj+, divides P into a linear num- ber of triangles where the funnel Pp~+~ (i) on edge i is the disjoint union of some subset of these tri- angles. Below we present a high level description of our algorithm for computing the minimum-length separator, where shortestpath(v) represents the pro- cedure which computes the shortest path tree (map) from vertex v, a(F) denotes the area of funnel F, and hourglass(i, j, locmin) computes the minimum- length separator for Hid. For j=ltondobegin globmin = c~; shortestpath(pj ); shortestpath(pj + x ); aLpj+~ (p~+2) = 0; for i = 3 to n do apj+l (pi+,) = Apj+, (Pi+,-) + (j + i)); fori=ltondo if i, j are visible and ALpj÷~ (Pi) < KI, then do begin hourglass(i,j, locmin) globmin = MIN(globmin, locmin); end; end; Next, we need a procedure hourglass(i,j, locmin) to solve the following problem: Problem: Given an hourglass H~,j find z and y on i and j respectively such that 352 a) zy C Hij, b) the area bounded by SPpj+t (Pi), piZ, Zy, and YPj+x equals KL- ALp ÷ (PO, and c) the length of is mini- mum. ' Le. O.g Figure 3: an hourglass divides the polygon First, we simplify the test of condition b). For each hourglass Hi,i, define Cp~+, (pl)as the area of the re- gion bounded by SPp$+~ (Pi) and the segment PlPj+x. Depth-first-search traversed of the shortest path tree of P from vertex Pj+x produces all of the Cp$+t(pi) in linear time. Then the test of condition b) reduces to determining whether the quadrilateral pizypj+t has area KL -- ALp~+t(pi ) + Cpj+t(pi), from now on re- ferred to as K. Condition a) is satisfied if and only if the closed halfplane to the left of the ~ contains all the vertices of SPpj+t (Pi) and the one to the right contains all the vertices of SP~,,+t(pj). To reduce the number of constraints, we exploit the fact that shortest paths are convex to decompose the problem further. First, trim edge i to create an edge i' so that all of i' is visible from j. Next, subdivide i t by merging Sp~+~ (i) and Spj (i). As x moves from p~ to P~+I, anehorPi+' (~) and/or anchorP¢ (z) changes only when x moves from one subinterval to the next. Consequently, we can reduce an arbitrary hourglass problem to a series of problems defined on elementary hourglasses: Problem: For each interval I of i, find points x = (•1, Yl) and y = (x2, Y2) on I and j respectively such that a) the anchorP#(z) and anchorP#+, (x) do not lie in the same open halfspace defined by ~, b) the area of the quadrilaterM pixypj+l equals K, and c) length of xy is minimum. This is a continuous optimization problem, rather than a combinatorial one. Algebraic manipulation of the constraints involved leads to an expression for the length of xy as the square root of a rational a func- tion of xx where the domain of xl is restricted by the facts that x must lie within a specific subinterval I, the point y lies on segment j, and the line through z and y must keep the anchorPJ(x) and anchorP~+, (z) on opposite sides. The degrees of the numerator and denominator of the rational function permit analyti- cal solution for finding the optimum in constant time. The global minimum for the original hourglass is the minimum of all the minima obtained from the contin- uous problems on elementary hourglasses. Therefore: Theorem 3.1 The hourglass problem can be solved in O(n) time and space where n is the number of hourglass vertices. Since the hourglass algorithm takes linear time and since we call the hourglass algorithm at most O(n 2) times then our algorithm has at most O(n ~) complex- ity. We exploit the linearity of the shortest path to improve the bound. Theorem 3.2 The minimum length area separator of a simple polygon can be computed in O(n 2) time and O(n) space. Proof: The test in the if statement can be done in 0(1) time since the area parameters of P as well as the visibility of i, j have been computed during the shortest path computation. The hourglass algorithm takes O(s~' + s~ i+~ ). Thus the whole algorithm takes O(~=x~=x(s~ + s~'+t)) time which is O(n 2) since ~=i s~ = O(n) for all v. 0 Corollary 3.3 The minimum length area separator of a simple splinegon can be computed in O(n 2) time and O(n) space, t3 Theorem 3.4 The minimum sum of ratios separator of a simple polygon can be computed in O(n 2) time and O(n) space. El Remark: Although the results in this section were explained using hourglasses, the same results can be achieved with an alternate formulation the details of which are described in the full paper. The idea is to fix a particular edge i, find the shortest path maps from its endpoints. These maps subdivide the boundary of P into a set of segments. As a point :r moves on a particular segment anchorP'(x) and anchor p'+~ (x) remain constant. Let a = anchorP'(x) 353 and b = anchorPi+~(z). Consider only segments for which SPp,(a) and SPp,+, (b) are inward convex (these are the visible ones) and for each of them solve a continuous optimization problem. Repeat the same process for all edges i. 4 Inscribed Triangles For three points z, y, z on the boundary of a simple polygon P to define an inscribed triangle, it is neces- sary and sufficient that they are pairwise visible. If z, y, z lie on edges k, j, i, respectively, then the points are pairwise visible if and only if zy, yz and zz lie inside Hkj, Hj, i and He,i, respectively. Thus the boundary of the triangle zyz is interior both to P and to the union of the three hourglasses. Since P is simple, the entire triangle must be interior to P. It is also contained in the polygon Fij,~ _C P bounded by i, SPp,+,(pj), j, SPp#+t(pk), k and SPp,+,(p,). Fij,~ is called a fan-shaped polygon with bases i, j, k (fig.4). Fi,j,~ is legal iff the visible parts of each of its bases with respect to the two others have non-empty intersection. A triangle T is inscribed in Fid,k if and only if if T C Fij,k and the vertices of T lie on the bases. Every triangle inscribed in a simple polygon P is also inscribed in a fan-shaped polygon Fid,~ _C P. Below we present a high level description of our al- gorithm for computing the maximum area/perimeter inscribed triangle, where shortestpath(v)represents the procedure which computes the shortest path tree (map) from vertex v, and fan(i, j, k, locmaz) com- putes the maximum triangle inscribed in a legal Fid,k: gIobmaz = 0; for i = 1 to n do begin shortestpath(pi ); shortestpath(pi+ , ); for j = 1 to n do begin shortestpath(pj ); shortestpath(pj + x ); for k = I to n do begin if Fi,j,k is legal then fan(i,j,k, locmaz); globrnaz = MAX (globrnaz, locrnaz ); end; end; end; It remains to develop procedure fan(i, j, k, locmaz). Lemma 4.1 For a fan-shaped polygon Fi,i,t the mazimum-area inscribed triangle with vertices z, y, z on the bases must have at least two sides tangent to the convez chains of the fan (fig. 4). 354 e# Figure 4: fan-shaped polygon Proof: By contradiction. In specific, assume that neither zy nor zz is tangent to the boundary of Fid,k. Tangents from y and z to the chains SPpj+, (Pk) and SPp~(pk+x), respectively, intersect k at points v and to such that z must lie between them. Then it is clear that one of vyz or wyz must have area greater than or equal to the area of zyz (the equality happens when k is parallel to yz). c] Lemma 4.2 The mazimum-perimeter triangle in- scribed in a fan-shaped polygon must have at least two sides tangent to the convez chains of the fan. Cl Assume that the optimum triangle xyz has xy tan- gent to SPp~+t(pk), and zz tangent to SPp,(pk+x). To find the optimum, trim edge k to create an edge k' which is visible from both i and j. If k' is empty then inscribed triangle does not exist. Perform the comparable operation on edges i and j, creating i' and j' respectively. Next, subdivide k' by merging Sp,+t(k) and Sp,(k). As x moves from p~ to p~+,, anchor p j+' (z) and/or anchor p' (z) changes only when x moves from one subinterval to the next. Problem: For each interval K of k', find points x E K, y E i, z E j such that a) anchorPi(x) E zy and anchorPJ+l(z) E xz b) y and z are mutually visible, and c) area/perimeter of zyz is maximum. Testing condition b) is still complicated, so we de- compose the problem further. Subdivide i' according to Svj (i) and subdivide j' according to Sv,+l (j). As y (z resp.) moves from p~ to P~+I (P; to P;+i resp.) anchor pj (y) (anchor"+' (z) resp.) changes only when y (z resp.) moves from one subinterval to the next. Let W be the number of vertices of either Sp# (i) or Sp,+t (]). To each interval of i' (resp. j') assign a number called its rank which corresponds to the po- sition of its anchor in SPp,(pj) (resp. SPpj(pl)), as- suming that the first position is 0. I~efine the subdivision of k ~ so that whenever z be- longs to some interval K, y and z each have constant rank. The rank of K equals the sum of those ranks. If rank(K) < W, do nothing; if rank(K) > W, solve Problem I; if rank(K) = W solve Problem II. Problem I: Given three non intersecting line segments AB, CD, and EF and two points p, q such that p (q resp.) lies on AE and BF (AC and BD resp.), find s,t,u on AB,EF,CD with p (q resp.) on st (su resp.), such that the area ofstu is maximum. Problem II: Add the constraint that the line through t and u should be always above a constant point (z0, y0). Solutions to both problems can be computed analyt- ically. Solving the perimeter optimization versions of these problems requires numerical methods. Lemma 4.3 The mazimum inscribed trian#le in a fan-shaped polygon Fid,~ can be found in O(~' + 4 +4 '÷' +4 which is O(n) whe n/ th, num- ber of ,ertiees 4 the fan-shaped polygon. S Since we decompose the simple polygon into at most O(n 3) fan-shaped polygons, the global algo- rithm uses at most O(n 4) time. Careful analysis pro- duces a better bound. Theorem 4.4 The mazimum triangle inscribed in a simple polygon P can be found in O(n 3) arithmetic operations where n is the number of vertices of P. The space required is O(n). Proof.. The total time spent in the shortest path computation is O(n 3) since the shortest path proce- dure is called O(n 2) times. The if statement takes 0(1) time since whether Fij,k is legal can be de- cided from the shortest path information. Accord- jng to Lemma 4.3, the procedure fan(i,j,k) takes PJ O(s~' + s~ j+~ + s~ '+' + s i ). Thus the total time spent in the fan-shaped polygons corresponding to all triples (i, ], k) of edges of P is: rt n i+t j O(E,=I ~=1 z--,k=,,°k~"" ¢~'--±sP~+'k +• +~ )) which is O(n 3) since EL-1 = o(,0 for any vertex v of e according to . [] 355 Unfortunately, the maximum area or perimeter tri- angle inscribed in a simple splinegon might not have two sides tangent to the chains of the fan-shaped polygon. Consequently, Lernmas 4.1 and 4.2 do not hold, and a new algorithm is necessary: Theorem 4.5 The mazimum area or perimeter tri- angle inscribed in a simple splinegon can be found in O(n 4) time and O(n) space. 13 The maximum inscribed triangle with one of its sides constrained to have given length does have at least one of the non-given length sides tangent to a fan- shaped polygon: Theorem 4.6 The Mazimum Inscribed Triangle with one of its sides having given length can be solved in O(n 3) time and O(n) space. 13 Remark: Although the results in this section were explained using fan-shaped polygons, the same re- suits can be achieved with an alternate formulation described in the full paper. 5 Minimum Area Concave Quadrilateral ABCD always represents a concave quadrilateral where CC is the reflex vertex. CH(P) represents the convex hull of a simple polygon P. Each simple polyg- onal region Q interior to CH(P) but exterior to P is called a pocket of P. Figure 5: Optimum quadrilateral characterization Lemma 5.1 The minimum-area concave quadrilat- eral ABCD containing a simple polygon P satisfies the following conditions: a) A, B, D lie outside of the CH(P) with AB and AD tangent to CH(P) at some points k and I, respectively; b) C is inside the visibility polygon VQi of some pocket Qi of P with respect to pseudoedge lliWi; e) B~, DC are tangent to the boundary of Qi at points a, b where a and b are the anchors of the shortest paths from vi to C and wi to C, respeetively, inside the pocket. Note that C need not lie on the boundary of the pocket and that the minimum-area concave quadrilateral may be degenerate (i.e. a triangle) (figs. 5 and 6). m required is the space to keep the two shortest path maps, i.e linear. The procedure pocket(a, b, k, l, i, R) must solve the following optimization problem: Problem: Construct the minimum area concave quadrilateral ABCD such that AB and AD pass through the constaat points a and b respectively, and have slopes in the in- tervals defined by the adjacent convex hull edges, CD and CB pass through k and l, and C 6 R. ? Figure 6: Optimum quadrilateral has C collinear with B and D A description of the algorithm follows: Compute CH(P) and let p = the number of pockets. Triangulate P and all its pockets Qi, for i = 1...p. For all pairs a, b of vertices of CH(P) do for i = 1 to p do begin Compute VQi and the shortest path maps inside VQi from both vi and wi . Merge those maps and label each region (< 6 sides) with its anchors w.r.t, vi and wi. For every region R, call pocket(a, b, k, l, i, R) . end. Report the optimum. Note that instead of explicitly merging the two maps, we can take each pair (rx, r2) where rl (resp. r2) is a region of the shortest path map from vi (resp. wi), and calculate the intersection region explicitly. For each such intersection region call pocket(a, b, k, l, i, R). Since every region of the short- est path map is a triangle the intersection of two such regions has constant number of sides. The space These constraints generate a linear program in eight variables which is subject to a constant num- ber of linear constraints. That problem can be solved in 0(1) time. Thus, each pass through the loop above takes O(nl + ki) time where ni is the size of Qi and ki is the size of the subdivision created by merging the shortest path maps from v and w. Thus, if n¢ are the number of vertices of CtI(P) and np the total number of vertices of all pockets of P and k be the sum of the merged subdivisions over all pockets, then we have the following theorem: Theorem 5.2 The minimum concave quadrilateral that con~ain~ a simple polygon P can be found either in O(n~n~) time and O(n) space or in O(n~(np + k)) time and O(n + k) 8pace. Q 6 Contained Triangles The maximum area triangle T = zyz contained in a simple polygon P may have 0, 1, 2 or 3 vertices on the boundary of P. The case of 3 vertices of the triangle on the boundary of P corresponds to the maximum area inscribed triangle problem solved in O(n 3) time and O(n) space earlier. We focus here on what we call 0-case, 1-case, and 2-case. Lemma 6.1 Let A and C (B and D resp.) be two poi.ts o. reap.) such that segments AB and CD intersect inside the wedge defined by Ox and Oy at point E. Let FG be a line segment through E with y (G resp.) between A and C (D and B reap.). The area of the triangle OFG is maximized when one of the following holds: F = A and G = B; or F = C and G = D (fig. 7 a). [3 Corollary 6.2 Let ~x and ~y be two rays with com- mon origin O. Let also A and B be two points on Ox and Oy respectively and let CAB be an inward con- vex chain. Let D (E resp.) lie on OA (OB reap.) 356 0 0 P Figure 7: Examples for lemma and corollary such that 0 and CAB do not lie on the same side of the line through D and E. Then the area of triangle ODE becomes maximum if DE contains an edge of the convex chain CAB (fig. 7 b). Proof: Assume that DE does not contain any edge of CAB. Then either a) the intersection of DE and CAB is one vertex of CAB or b) is the empty set. In case of a) we have an instance of Lemxna 6.1. In case of b) we can translate DE in a direction perpedicular to itself until it intersects CAB and then apply Lemma 6.1. D. Figure ~: a) 0-case; b) 1-case; c) 2-case; d) 3-case Lemma 6.3 Let T be a maximum area triangle. Then each edge of T contains at least two points of the boundary of P. Specifically the following hold (fig. 8): IfT is O-case lhen each edge ofT contains at least two reflex vertices of polygon P; if T is I-ease with x on the boundary of P, then yz touches at least two reflex vertices of P and zy and xz at least one. if T is B-case with y and z on edges i and j, there ex- ists at least one edge k of P such that i, ], k define a fan-shaped polygon Fi,i,~ _D T. 0-Case Triangle Algorithm. One algorithm would consider all triples of pairs of reflex vertices (O(n 6) objects), check whether the corresponding tri- angle is contained in P in O(logn) time using ray- shooting , and choose the largest one, a total of O(n 6 log n) arithmetic operations. An alternate al- gorithm would fix two sides of the candidate triangle by choosing a pair of pairs of reflex vertices (A, B) and (E, F)~ assuming that zy contains AB and zz contains EF, and spending linear time to find the optimum position of yz, for a total of O(n 5) arith- metic operations. Figure 9: O-ease Using the linearity of shortest path trees inside sim- ple polygons, we can invert this algorithm and reduce the complexity by an order of magnitude (fig. 9). Fix a pair of reflex vertices C and D with the character- istic that ~'D C P and all edges incident to C and D lie on the same side of the line containing C'D and as- sume that side yz contains these vertices. Determine in O(log n) time , the points G and H closest to G and D, respectively, where the extension of CD intersects the boundary of P. Let P' represent the subpolygon of P which lies at the opposite side of GH from the edges incident to C and D. Since x must be visible from GH, then the shortest paths from G to z and from H to x inside P must be inward convex chains containing segments AB and EF, respectively, where A, B, E, F are reflex vertices and A/~ and E-~ 357 intersect within P at point x. Thus, we need consider only pairs of the O(n) edges of the shortest path tree from G and the O(n) edges of the shortest path tree from H, a total of O(n z) objects. How can we test efficiently whether the chosen pair AB and EF of shortest path tree edges forms a legal triangle with GH? Choose only those AB such that a) the shortest paths from G to A and G to B are inward convex chains, b) points y and z do not lie in segment CD, and c) segments Az and Ez lie inside P. Conditions a) and b) clearly can be checked in 0(1) time. One way to test condition c) is to apply ray shooting inside P in O(logn) time. Since AB and UD are shortest-path tree edges, however, constant time suffices. Define et (resp. ez) as the edge of the shortest path map from G which is adjacent to vertex A (resp. E) and collinear to AB (resp. EF). Ifet and e2 both exist and intersect then the intersection point is a valid vertex z. I~epeating the above procedure for all the O(n 2) pairs of reflex vertices U and D yields: Lemma 6.4 The O-case maximum triangle can be found in O(n 4) ~imc and O(n) space. q 14 o, Figure II: 2-case 2-Case Triangle Algorithm. According to £emma 6.3, triangle zyz lies in a fan-shaped poly- gon (see fig. 11). Subdivide j (i resp.) according to the shortest path maps from both pi+l and p~ (pj and ph+x resp.). For each interval on the subdivi- sion of edge i and each interval of the subdivision of edge j, a) check whether points y and z are vis- ible, using techniques developed in Sect. 4; b) let a = anchorPh+l (y) and b = anchorW (z) aad check whether SPp~+, (a) and SPph (b) are inward convex; c) check whether the intersection x of the lines through segments ya and zb lies "below" the line through edge k. That guarantees that triangle zyz lies inside Fij,k. d) Solve the appropriate continuous optimiza- tion problem. It should be clear that steps a) through d) take 0(1) time. According to the above discussion the time com- plexity per fan-shaped polygon is O((~'+t +~k )(sfj + s~ h+t)). Then summing over all fan-shaped polygons we get Figure 10: 1-case 1-Case Triangle Algorithm. As in the 0-case algorithm we fix a pair of reflex vertices U and D (fig. 10). We find again points G and H as defined previously and then we have to walk on the shortest path maps of G and H on the boundary of P as we did in the inscribed triangle case of Sect. 4. Thus for a fixed pair of reflex vertices we spend, using similar arguments, O(n) time and therefore a total O(n 3) for the whole problem. Thus: Lemma 6.5 The 1-case maximum triangle can be found in O(n 3) time and O(n) space. [] n i=1 j=l k=l 0 Lemma 6.6 The 2-case maximum triangle can be found in O(n 4) time and O(n) space. Theorem 6.7 The maximum area triangle con- tained in a simple polygon can be found in O(n 4) time and O(n) space. 358 References [I] A. Aggarwal, "Lecture notes in Computa- tional Geometry." MIT Research Seminar Series MIT/LCS/RSS 3, August, 1988. A. Aggarwal, J. S. Chang an d C. K. Yap, "Min- imum Area Circumscribing Polygons." Visual Computer, 1 (1985), 112-117. A. Aggazwal, M. Klawe, S. Moran, P. Shot, R. Wilber, "Geometric Applications to a Ms- trix Searching Algorithm." Algorithmica, 2 (1987), 209-233. A. Aggaxwal, J. Park, "Notes on Searching in Multidimensional Monotone Arrays." Proc. ~9th IEEE Syrup. on Found. of Comp. Sci. (1988), 497- 512. J.E. Boyce, D.P. Dobkin, R.L. Drysdale, L.J. Guibas, "Finding Extremal Polygons." SIAM .L of Computing, 14 (1985), 134-147. J. S. Chang, C. K. Yap, "A Polynomial Solution for Potato-Peeling and Other Polygon Inclusion and Enclosure Problems." Discrete and Comp. Geom., 1 (1986), 155-182. J. S. Chang, "Polygon Optimization Problems." Ph.D. Thesis, New York University, 1966. B. Chazelle, L. Guibas, "Visibility and Intersec- tion Problems in Plane Geometry." Proc. of A CM Syrup. Comp. Geom. (1985), 135-146. P. Chew, K. Kedem, "Placing the Largest Simi- lar Copy of a Convex Polygon among Polygonal Obstacles." Proc. of ACM Syrup. Comp. Geom. (1989), 16%74. N.A.A. DePano, "Polygon Approximation with Optimized Polygonal Enclosures: Applications and Algorithms." Ph.D. thesis, Dept of Computer Science, Johns Hopkins University, April 1988. N.A. DePano, Yah Ke, J. O'Rourke, "Find- ing Largest Inscribed Equilateral Triangles and Squares." Proc. Of the Allerton Conference, 1987. D. Dobkin, L. Snyder, "On a General Method for Maximizing and Minimizing among Certain Ge- ometric Problems." Proc. ~Oth IEEE Syrup. on Found. of Comp. Sei. (1979), 9-17. D. Dobkin, D. Souvalne, "Computational Geome- try in a Curved World." Algorithmiea, 5 (1990). D. Dobkin, D. Souvalne, C. Van Wyk, "Decom- position and Intersection of Splinegons." Algorith- mica, 3 (1988), 473-485. S. Fortune, "A Fast Algorithm For Polygon Con- talnment By Translation." Proc. 13th ICALP (1985), 189-198. :M. Gazey, D. Johnson, F. Preparata, P~. Tarjan, "'I~iangulation of a simple polygon." Information Processing Letters, 7 (1978), 175-179. L. Guibas,.J. Hershberger, D. Leven, M. Sharir, R. Tarjan, "Linear Time Algorithms for Visibil- ity and Shortest Path Problems inside Triangu- lated Simple Polygons." Algorlthmica, 2 (1987), 209-238. L. Guibas, R. Seidel. "Computing Convolutions via Reciprocal Search." Discrete and Computa. tional Geometry, 2 (1988), 175-193. L. Gulbas, J. Stolfi, "Primitives for the Manipula- tion of General Subdivisions and the Computation of Voronoi Diagrams, " A CM Trans. Graphics, 4 (1985), 74-123. J. Hershberger, "An Optimal Visibility Graph A1- gorithm for Triangulated Simple polygons." Algo. rithmica, 4 (1989), 141-155. V. Klee and M. Laskowski, "Finding the Smallest • Tk'iangles Containing a Given Convex Polygon." ,L Algorithms, 6 (1985), 359-375. A. Fournier and D. Y. Montuno, "Triangulating Simple Polygons and Equivalent Problems." A CM Transactions on Graphics, 3 (1984), 153-74. B. Lisper, TheoryNet posting and followup com- munication, July, 1988. J. D. Mittleman, D. L. Souvaine, "Shortest Area- Bisector of a Convex Polygon." Rutgers Univer- sity Technical Report LCSR-TR-139, November 1989. J. O' Rourke, A. Aggaxwai, S. Maddila, M. Bald- win, "An Optimal Algorithm for Finding Mini- mal Enclosing Triangles." A Algorithms, 7 (1986), 258-269. M.H. Overmars, J. van Leeuwen, "Maintenance of Configurations in the Plane." I. Comput. System Sci., 23 (1981), 166-204. D. L. Souvaine, "Computational Geometry in a Curved World." Ph.D. Thesis, Princeton Univer- sity, October, 1986. R.E.Tarjan, C. Van Wyk, "Triangulation of a Sim- ple Polygon." SIAM Journal of Computing, 17 (1988), 143-178. 359 |
9013 | https://web.math.utk.edu/~mengesha/teaching/Math447/Analysis%20HW2.pdf | 447 HOMEWORK SET 2 IAN FRANCIS 1 For any a, b ∈R, such that a < b, show that a < a+b 2 < b.
Solution From a < b, add b to both sides of the inequality to get a + b < 2b, then multi-ply both sides by 1 2(> 0) to get a+b 2 < b. Similarly, add a to both sides of a < b to get a + a = 2a < a + b, then multiply both sides by 1 2 to get a < a+b 2 . These two statements imply a < a+b 2 < b. ♦ 2 Let a, b ∈R and suppose that for every ε > 0, we have a ≤b + ε. Show that a ≤b.
Solution Assume that a > b, i.e. a −b > 0 and choose ε = a−b 2 . Then, by the hypothesis, a ≤b + a−b 2 ⇔a 2 ≤b 2 ⇔a ≤b, a contradiction. Hence, a ≤b. ♦ 3 Prove that for any a, b ∈R: (a + b 2 )2 ≤a2 + b2 2 Show that equality holds if and only if a = b.
Solution Let a, b ∈R. Then, by expanding the square, multiplying both sides by 4(> 0), subtracting everything on the left side of the inequality from both sides, respectively, we get: (a + b 2 )2 ≤a2 + b2 2 ⇔a2 + 2ab + b2 4 ≤a2 + b2 2 ⇔a2 + 2ab + b2 ≤2a2 + 2b2 ⇔0 ≤a2 −2ab + b2 = (a −b)2 Since any square of a real number is ≥0, and a−b ∈R, we are done. From the last line, we see that equality in the original statement holds if and only if 0 = (a−b)2 ⇔0 = a−b ⇔a = b. ♦ Date: September 10, 2014.
1 2 IAN FRANCIS 4 (a) Suppose that 0 < c < 1. Show that 0 < c2 < c < 1. Show also that for all n ∈N, cn ≤c.
Solution We have c−1 < 0, so we may multiply both sides by c > 0 to get c(c−1) = c2 −c < 0 ⇔c2 < c. Since c < 1, we may conclude 0 < c2 < c < 1.
Now, for the base case, we have c1 = c ≤c. Now assume cn ≤c for some n ∈N. Then we have cn −c ≤0, and using the induction hypothesis and the first result of this problem, c(cn −c) = cn+1 −c2 ≤0 ⇔cn+1 ≤c2 ≤c. Thus the statement holding for n ⇒it holds for n + 1, so by induction, it holds for all n ∈N. ♦ (b) Suppose that c > 1. Show that c2 > c > 1. Show also that for all n ∈N, cn ≥c.
Solution c > 1 ⇔c −1 > 0. Multiplying both sides by c, which is positive, gives c(c −1) = c2 −c > 0 ⇔c2 > c > 1.
Now, for the base case, we have c1 = c ≥c. Now assume cn ≥c for some n ∈N. Then we have cn −c ≥0 ⇔c(cn −c) ≥0 ⇔cn+1 −c2 ≥0 ⇔cn+1 ≥c2 ≥c, using the induction hypothesis and the first result of this problem. Thus the statement holding for n ⇒it holds for n + 1, so by induction, it holds for all n ∈N. ♦ 5 (a) Show that if a ∈R, then |a| = √ a2.
Solution If a = 0, |0| = 0 = √ 02. If not, since we have |a| ≥0, and a2 ≥0 (since we used it to prove the AM-GM Inequality), |a| = √ a2 ⇔|a|2 = √ a22 = a2, which is true since |a|2 = a2, ∀a ∈R. ♦ (b) If (i) a < x < b and a < y < b, show that |x −y| < b −a.
Solution a < y < b ⇔−b < −y < −a (we multiplied by −1, which is negative). Adding this inequality together with (i) gives a −b = −(b −a) < x −y < b −a, and by property of absolute value gives |x −y| < b −a, since b −a > 0 by hypothesis. ♦ 6 Find all x ∈R such that |x| + |x + 1| < 2.
Solution By cases: If x < −1, then |x| + |x + 1| = −x −(x + 1) = −2x −1 < 2 ⇔−2x < 3 ⇔x > −3 2, (We ”flipped” the inequality since we multiplied both sides by −1 2 < 0).
If −1 ≤x < 0, then |x| + |x + 1| = −x + x + 1 = 1 < 2 ⇔0 < 1, a tautology.
If x ≥0, then |x| + |x + 1| = x + x + 1 = 2x + 1 < 2 ⇔2x < 1 ⇔x < 1 2.
Therefore our solution set is {x ∈R : −3 2 < x < 1 2} ♦ 7 16) Let ε > 0, δ > 0, and a ∈R.
Show that I := Vε(a) ∩Vδ(a) and U := Vε(a) ∪Vδ(a) are γ−neighborhoods of a for appropriate values of γ.
447 HOMEWORK SET 2 3 Proof If ε = δ, in both I, U simply take γ = δ, then the result is trivial.
WLOG, assume ε > δ. Then, x ∈I ⇔|x−a| < δ < ε, so let γ = δ. Also, x ∈U ⇔|x−a| < δ or |x −a| < ε ⇔|a −x| < ε, since ε > δ, so take γ = ε. Thus, we have appropriate values for γ−nbhd’s. ♦ 17) Show that if a, b ∈R and a ̸= b, then there exists ε−nbhd’s Uε(a), Vε(b) such that U ∩V = φ.
Proof WLOG, assume that a > b, set ε = a−b 2 , and assume ∃x ∈U ∩V . Then |x −a| < a −b 2 ; |x −b| = |b −x| < a −b 2 ⇔b −a 2 < x −a < a −b 2 ; b −a 2 < b −x < a −b 2 ⇔b −a < b −a < a −b, which is clearly a contradiction. (We deduced the third line in calculation by adding the two inequalities from the second line.)Therefore, The intersection of these two ε−nbhd’s is φ. ♦ 18a) Show that if a, b ∈R, then max{a, b} = 1 2(a + b + |a −b|) and min{a, b} = 1 2(a + b −|a −b|).
Proof WLOG, assume a ≥b, then 1 2(a + b + |a −b|) = 1 2(a + b + a −b) = 1 2(2a) = a. Likewise, 1 2(a + b −|a −b|) = 1 2(a + b −(a −b)) = 1 2(a + b −a + b) = 1 2(2b) = b.
18b) min{a, b, c} = min{min{a, b}, c} Proof Assume a ≥b ≥c. Then min{min{a, b}, c} = min{1 2(a+b−|a−b|), c} = min{b, c} = 1 2(b + c −|b −c|) = 1 2(b + c −b + c) = c = min{a, b, c} ♦ 19) Show that if a, b, c ∈R, then the ”middle number” is mid{a, b, c} = min{max{a, b}, max{b, c}, max{a, c}} Proof Once again, assume a ≥b ≥c.
Then min{max{a, b}, max{b, c}, max{a, c}} = min{a, b, a} = min{a, b} = b = mid{a, b, c} , by applying our previous results. ♦ |
9014 | https://www.tandfonline.com/doi/full/10.1080/08927014.2021.2002987 | Full article: Tracing the origins of extracellular DNA in bacterial biofilms: story of death and predation to community benefit
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Tracing the origins of extracellular DNA in bacterial biofilms: story of death and predation to community benefit
Davide Campocciaa Laboratorio di Patologia delle Infezioni Associate all’Impianto, IRCCS Istituto Ortopedico Rizzoli, Bologna, Italy; Correspondencedavide.campoccia@ior.it carlarenata.arciola@ior.it
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,
Lucio Montanaroa Laboratorio di Patologia delle Infezioni Associate all’Impianto, IRCCS Istituto Ortopedico Rizzoli, Bologna, Italy; ;c Department of Experimental, Diagnostic and Specialty Medicine, University of Bologna, Bologna, Italy further author information
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Carla Renata Arciolaa Laboratorio di Patologia delle Infezioni Associate all’Impianto, IRCCS Istituto Ortopedico Rizzoli, Bologna, Italy; ;b Laboratorio di Immunoreumatologia e Rigenerazione Tissutale, IRCCS Istituto Ortopedico Rizzoli, Bologna, Italy; ;c Department of Experimental, Diagnostic and Specialty Medicine, University of Bologna, Bologna, Italy Correspondencedavide.campoccia@ior.it carlarenata.arciola@ior.it
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Pages 1022-1039 | Received 17 Aug 2021, Accepted 28 Oct 2021, Published online: 25 Nov 2021
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In this article Article contents Related research
Abstract
Introduction
The origin of eDNA in human infections: not just a single source
Bacterial cell lysis
Bacterial eDNA from secretion
Bacterial eDNA release through MVs
Host cells and tissues eDNA
Biomaterial-triggered NETs
eDNA in orthopedic infections
Conclusions
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Abstract
Extracellular DNA (eDNA) is a macromolecule copiously found in various natural microenvironments, but its origin and significance still remain partly mysterious phenomena. Here, the multifaceted origins of eDNA in bacterial biofilms are explored. The release of eDNA can follow a suicidal programmed bacterial apoptosis or a fratricide-induced death, under the control of quorum sensing systems or triggered by specific stressors. eDNA can be released into the extracellular space or as a free macromolecule or enclosed within membrane vesicles or even through an explosion of bubbles. eDNA can also be derived from host tissue cells through bacterial cytolytic/proapoptotic toxins or stolen from neutrophil extracellular traps (NETs). eDNA can alternatively be produced by lysis-independent mechanisms. Sub-inhibitory doses of antibiotics, by killing a fraction of bacteria, result in stimulating the release of eDNA. Even phages appear to play a role in favoring eDNA release. Unveiling the origins of eDNA is critical to correctly address biofilm-associated infections.
KEYWORDS:
Extracellular DNA (eDNA)
bacterial biofilm
bacterial apoptosis
autolysins
membrane vesicles
phage endolysins
secretion
Introduction
In recent years, extracellular DNA (eDNA) has increasingly emerged as one of the main extracellular polymeric substances (EPS) composing the architecture of bacterial biofilms. Other EPS taking part in bacterial biofilm matrices include various types of extracellular and cytoplasmic moonlighting (Jeffery Citation 2014) proteins, neutral and electrostatically charged exopolysaccharides, teichoic acids and lipids. Differing from eDNA, these further EPS are generally species-specific and often even strain type-specific. Conversely, the presence of eDNA in biofilms appears highly conserved and widely distributed even across taxonomically distant bacterial and fungal microorganisms (Martins et al. Citation 2010; Okshevsky & Meyer Citation 2015). While many important functional roles of eDNA in biofilms have been documented and nowadays appear well established, the physiological mechanisms that are implicated in eDNA release and exert control over its production have only partly been unveiled and only in a restricted number of bacterial species (Sarkar Citation 2020). Contrasting with the apparent eDNA ubiquity in biofilms, it appears as if, during phylogenesis, each single bacterial species reinvented alternative and/or complementary routes and strategies to repurpose, release or transfer DNA outside the cells.
eDNA is a very versatile and multifunctional molecule, which provides multiple benefits to microbial cells, enabling bacterial survival under highly stressing environmental conditions. eDNA favors horizontal gene transfer, enabling the interexchange of genes encoding antibiotic resistance and virulence traits between bacteria (Molin & Tolker-Nielsen Citation 2003). It promotes bacterial aggregation (Mlynek et al. Citation 2020) and enables adhesion to highly hydrophobic surfaces, which are most unfavorable to bacterial colonization (Das et al. Citation 2011). When bacteria turn sessile following adhesion to a substratum or, otherwise, undergo flocculation while being suspended in physiologic fluids, eDNA acts as an element of the biofilm architecture, conferring structural stability through its many interactions with other EPS (Montanaro et al. Citation 2011a). eDNA has been described as acidifying the local milieu, promoting antibiotic resistant phenotypes (Wilton et al. Citation 2015), and chelating and neutralizing cationic bactericidal molecules including antibiotics (Chiang et al. Citation 2013) and antimicrobial peptides (Batoni et al. Citation 2016), actively contributing to biofilm tolerance. It triggers and conditions the host immune response (Hemmi et al. Citation 2000; Islam et al. Citation 2001; Fuxman Bass et al. Citation 2010). Moreover, it facilitates self-organization of bacterial biofilms, guides biofilm spreading (Gloag et al. Citation 2013), represents a nutrient source during starvation (Finkel & Kolter Citation 2001), and exhibits extracellular electron transfer capabilities (Saunders et al. Citation 2020).
Formation of eDNA is a necessary step for bacteria in order to benefit from all these functions. This article aims at reviewing the mechanisms of eDNA production adopted by different bacterial species, especially those commonly involved in nosocomial infections. It attempts to offer an overall view of the numerous and complex paths that have recently been unveiled.
The origin of eDNA in human infections: not just a single source
A more complete frame for the interpretation of the origin of DNA present in the extracellular space of biofilms of human pathogens has emerged only in recent years. In an early phase, eDNA was thought not to be actively released by bacteria, but rather to derive from bacterial cells facing a process of natural death. With time it became more evident though that, at least in species such as Streptococcus pneumoniae, induction of natural competence could actively trigger cell lysis and DNA release from a subfraction of the cell population (Steinmoen et al. Citation 2002). Over the last two decades, the extensive research work conducted on many bacterial species has started to shed more light on the processes leading to the presence of eDNA in the intercellular space, revealing a multitude of alternative and often redundant mechanisms that vary with the pathogenic species. The scheme in Figure 1 summarizes some of the main mechanisms implicated in eDNA production, which will be discussed in detail in the following sections.
Figure 1. An overview of the numerous sources and mechanisms leading to eDNA accumulation in the extracellular space. Bacterial death and release of eDNA can be achieved by a variety of alternative, species- and strain-specific, mechanisms under the control of quorum sensing systems or triggered by specific stressors. Legend: MØs, monocytes/macrophages; PMNs, polymorphonuclear granulocytes; LTH, leukotoxic hypercitrullination; DNAJ, Heat Shock Protein DnaJ; PYO, pyocyanin; AHL, long-chain N-acyl-homoserine lactones; PQS, quorum-sensing molecule Pseudomonas quinolone signal; HQNO, 2-n-heptyl-4-hydroxyquinoline-N-oxide; C.bc1, cytochrome bc 1 complex; ROS, radical oxygen species; OMV, outer membrane vesicle; OIMV, outer-inner membrane vesicle; EOMV, explosive lysis outer membrane vesicle.
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The most common systems adopted by bacteria to release DNA involve some type of programmed autolysis. In Gram-positive bacteria, this has been found often to occur through the production of peptides named autolysins. When being expressed, autolysins are capable of determining a sort of cell suicide or programmed apoptosis, by damaging the integrity of some vital cell components or, alternatively, to induce the death of nearby sister cells, causing a sort of fratricide (Montanaro et al. Citation 2011b). Conversely, auto-intoxication and self-poisoning due to the production of toxic chemicals or a burst in the respiratory production of reactive oxygen species (ROS) has been described for Gram-negative bacteria such as Pseudomonas aeruginosa, which constitutes one of the best investigated bacterial models. The activation of prophages is another of the strategies pursued to achieve a programmed cell autolysis under the control of quorum sensing systems. The expression of phage endolysins can be associated with phenomena known as explosive lysis or bubbling cell death, in which case the autolytic disruption of bacterial cells generates membrane vesicles (MVs) with cytoplasmic content that may include also nucleic acids (Figure 1).
Predation of bacteria by phages or other bacterial competitors is another mechanism leading to bacterial lysis and eDNA release. Moreover, sub-inhibitory doses of exogenous factors such as antibiotics have long been known to determine eDNA release. Interestingly, in a restricted number of species, alternative mechanisms have been observed where bacterial eDNA is secreted while cells are viable and maintain their integrity (Ibáñez de Aldecoa et al. Citation 2017). Predation by other bacteria or bacteriophages represents a further circumstance in which bacterial DNA is released and can be used to enforce biofilm structure. Additionally, some antimicrobials and antibiotics capable of destabilizing the integrity of the bacterial wall have also been found to actively cause the release of eDNA and so favor biofilm production. Shields et al. (Citation 2013) found that eDNA extracted from biofilms of different strains of staphylococci and streptococci tended to form a sharp band in agarose gel electrophoresis and exhibited a size >10 kb. However, the characteristics of bacterial DNA in the biofilm matrix likely reflect the diverse mechanisms used to produce eDNA and, while for certain bacteria, the sequence of eDNA has been found similar to that of genomic DNA (Qin et al. Citation 2007; Rose et al. Citation 2015), this has not always been the case for other bacteria (Jakubovics et al. Citation 2013). For instance, in 2-day old biofilms of Helicobacter pylori, the comparison of DNA fingerprinting, performed by RAPD analysis on eDNA and intracellular DNA, showed substantial differences. Böckelmann et al. (Citation 2006) reached similar conclusions analyzing the biofilm of a lotic bacterial strain, whose RAPD amplification profiles of the eDNA and genomic DNA revealed major similarities in the banding patterns, but also some significant differences.
In infected tissues eDNA is not uniquely of bacterial origin and restricted to the quantity of nucleic acids contributed by bacteria. Apart from programmed bacterial cytolysis, eDNA can also be produced from host cells such as leukocytes and other tissue cells. Indeed, host cells of the immune system such as leukocytes, but also other tissue cells have been found to represent an additional relevant source of extracellular nucleic acids released in the interstitial fluids. Following their activation, professional phagocytes are known to produce neutrophil extracellular traps (NETs), primarily consisting of extruded extracellular fibers of DNA (Zawrotniak & Rapala-Kozik Citation 2013). Moreover, bacterial cytolytic toxins such as leukotoxins can determine the death of host cells and consequent release of high molecular weight DNA. As discussed in greater detail further on, while bacteria have been suggested to be potentially capable of harvesting and incorporating within their biofilm matrix the eDNA derived from these eukaryotic sources, some in vivo experimental evidence appears to exclude such a possibility. An important factor that should be kept in consideration is that some properties of bacterial eDNA, characterized by a high ratio of CG bases and a lower molecular size, differ from those of the eDNA contributed by host cells in terms of interactivity with host immune cells and their cell receptors.
From an evolutive point of view, bacteria therefore appear particularly skilled at developing complementary or alternative routes to take the greatest advantage of a fundamental organic molecule such as DNA. Either made available through a substantial sacrifice of individuals for the good of their community, or actively secreted bearing significant energetic costs, or recovered from the interstitial milieu where host cells have left it behind, eDNA can be recycled and repurposed from its original function (Mann et al. Citation 2009; Thammavongsa et al. Citation 2013; Ibáñez de Aldecoa et al. Citation 2017). What is certain is that nothing is left to chance and fine mechanisms keep eDNA release under control, as well as its usage and degradation. The different routes of release and the molecular mechanisms that govern eDNA production are specifically addressed in the next paragraphs.
Bacterial cell lysis
An intriguing initial conundrum was1whether bacterial eDNA was in fact chromosomal DNA or, rather, consisted of newly synthesized oligonucleotides secreted on demand. The confirmation that, at least in the biofilm of some bacterial species, eDNA was similar/corresponded to genomic DNA, cleared the way for the hypothesis that DNA could derive from death of a subpopulation of bacterial cells. Based on current understanding, the lysis of bacterial cells is without doubts a major mechanism of eDNA release, broadly observed across both Gram-positive and Gram-negative pathogens. Bacterial lysis normally follows the expression of cytotoxic factors. This type of process is usually referred to as autolysis, when individual cells, through auto-poisoning, actuate a form of suicide (Rice et al. Citation 2007) or an apoptosis-like programmed death. This ‘altruistic’ behavior of individual bacteria provides eDNA for the needs of the community. In contrast, cannibalism and allolysis take place when cytotoxic factors expressed by immune individuals are directed towards sibling cells (fratricide) or other species (predation) (Thomas et al. Citation 2008, Citation 2009). In Enterococcus faecalis isolates from orthopedic implant infections, the ability to produce autolytic enzymes proved crucial for expressing high levels of virulence (Arciola et al. Citation 2008). The coordinated cell death in isogenic populations resulting in the cytolysis of a sub-population of genetically identical siblings for the benefit of the population appears a more general strategy not only assoiated with biofilm formation, but also with the initiation of competence and sporulation (Popp & Mascher Citation 2019).
Among Gram-negative bacteria, P. aeruginosa is one of the most extensively investigated species. As illustrated in Figure 1, in P. aeruginosa at least three main, apparently redundant, mechanisms have been identified that are potentially capable of lead ing to bacterial death and autolysis with consequent release of eDNA: (1) auto-poisoning by phenazines (Das & Manefield Citation 2013; Sarkar Citation 2020) and pyocyanin (PYO) (Das & Manefield Citation 2012; Meirelles & Newman Citation 2018); (2) auto-poisoning by 2-n-heptyl-4-hydroxyquinoline-N-oxide (HQNO) (Hazan et al. Citation 2016); and activation of prophage endolysins (Turnbull et al. Citation 2016; Li et al. Citation 2019). In a recent review, Sarkar (Citation 2020) explored in detail these diverse mechanisms. PYO is one of the most studied substances termed phenazines, which are produced by P. aeruginosa. Phenazines are redox-active dibenzo annulated pyrazines with two nitrogen atoms in the center ring having proton-coupled electron transfer properties. The toxicity of PYOs has been primarily associated with its redox activity as it can rapidly reduce molecular oxygen leading to the production of the reactive oxygen species superoxide, which is subsequently converted into H 2 O 2 (Das & Manefield Citation 2012). PYO mediated death is known to be induced under special environmental conditions, in particular nutrient depletion. Under other conditions such as reducing stress, PYO and phenazines seem to exhibit beneficial effects, serving as alternative electron acceptors and enabling cells to maintain redox homeostasis. The work of Meirelles & Newman (Citation 2018) helped to elucidate the ‘double edged sword’ effects of PYOs. These authors observed that PYO production has predictable, distinct and striking cellular consequences and that these vary according to the physiological state of the producer. When cells are at a high cellular density, with high levels of carbon but electron acceptor-/oxidant-limited, PYO serves as an electron acceptor and avoids toxicity. Conversely, with starvation and low ATP levels, cells are not able to efficiently use all the mechanisms of defence from oxidative damage, which include an array of enzymes such as catalases, alkyl hydroperoxide reductases (Ahps), multiple thioredoxins and thioredoxins reductases (Ochsner et al. Citation 2000) that are necessary to coup with PYO redox toxicity. This auto-poisoning mechanism should be interpreted as an altruistic sacrifice, where in front of the cost of individual cells, subsets of bacteria with PYO-tolerant characteristics thrive and take advantage of the eDNA released.
In recent years, apart from PYO, Hazan et al. (Citation 2016) have produced evidence of another existing mechanism of auto-poisoning in P. aeruginosa, based on HQNO, a component of the quinolone signal system for quorum sensing. The authors demonstrated that, exhibiting intriguing similarities to programmed mitochondrial apoptosis in eukaryotic cells, HQNO causes cell death through the inhibition of the Qi site of the cytochrome bc1 complex (see Figure 1). Qi inhibition by HQNO disrupts the cytochrome c Q-cycle and leads to ROS accumulation to toxic levels and loss of membrane integrity, with consequent release of cytoplasmic content in the outer space (Hazan et al. Citation 2016). Cytochrome c plays a fundamental role in the generation of ATP. Low concentrations of ATP are known to be associated with persister cells phenotypes. HQNO would thus potentially enhance tolerance to antibiotics both by causing the release of eDNA from autolytic cells (so strengthening the biofilm structure) and by promoting bacterial dormancy and persister cell formation (Hazan et al. Citation 2016). It could be further speculated that lower ATP could also, in turn, create the preconditions that expose the bacterial cells to the damaging effects of PYO.
In clinical fluids such as cystic fibrosis sputum, HQNO can be produced in concentrations as high as 4 µM (Barr et al. Citation 2015). However, the production of HQNO by P. aeruginosa varies depending on the environmental circumstances and, under cell culture conditions, up to ten times higher concentrations can be reached (Hazan et al. Citation 2016). Despite the many P. aeruginosa protective mechanisms, these high levels of HQNO certainly represent a risk of autotoxicity. However, high HQNO concentrations are even more effective in killing and warding off other more sensitive Gram-positive bacteria inhabiting the same niches such as S. aureus. In addition to HQNO, a series of other exoproducts such as siderophores and rhamnolipids are typically produced by P. aeruginosa that are toxic to S. aureus. Therefore, in polymicrobial infections eDNA could also be generated from the lysis of the other coexisting bacteria. Up to 30% of the patients affected by cystic fibrosis have been found to be coinfected by these two dominant pathogens, which therefore appear to generate successful associations. A variety of host and environmental factors have been demonstrated to promote the co-existence of P. aeruginosa-S. aureus (Price et al. Citation 2020). Along with auto-poisoning, another mechanism that has emerged to cause P. aeruginosa death and consequent eDNA release is explosive cell lysis. Turnbull et al. (Citation 2016) have documented as the activation of a cryptic prophage endolysin, respectively the Lys endolysin, which is encoded by lys gene within the R- and F-pyocin genes cluster, can cause a form of violent cell death termed explosive cell lysis. The authors reported that, although R- and F-pyocins are known bacteriocins potentially implicated in cell lysis, only the pyocin endolysin Lys is required for eDNA production via explosive cell lysis. Lys has been hypothesized to translocate to the periplasm and degrade the peptidoglycan, thus enabling the release of the pyocins (Nakayama et al. Citation 2000). During the lytic process, in addition to direct dispersion of cytoplasmic material in the outer space, Turnbull et al. (Citation 2016) reported the biogenesis of MVs from the vesicularization of shattered membrane fragments. Toyofuku et al. (Citation 2019) recently reviewed the origins and the content of MVs. Two main types of MVs were described as generated from explosive cell lysis, respectively outer-inner membrane vesicles (OIMV) and explosive lysis outer membrane vesicles (EOMV). Both types of MV would contain cytoplasmic material such as nucleic acids and proteins. In Gram-negative bacteria outer membrane vesicles (OMV) can be produced also by blebbing in the absence of cell death and not uniquely through cell lysis. However, with the exclusion of plasmid DNA, the presence of chromosomic DNA seems more likely associated uniquely to EOMV (Toyofuku et al. Citation 2019). Thus, cell lysis-derived eDNA would include both DNA directly released in the outer space and DNA entrapped within MVs (Figure 1).
In 2017, Ibáñez de Aldecoa et al. reviewed the literature on the mechanisms of eDNA release, which are observed also in other Gram-negative species that are not exclusively of clinical interest. In 8 out of the 15 species examined, eDNA was found to be released through lytic mechanisms of either lysis or autolysis (Acinetobacter calcoaceticus, Campylobacter jejuni, Caulobacter crescentus, Neisseria meningitides, Pseudomonas chlororaphis, Pseudomonas stuzeri and Sherwanella oneidensis in addition to P. aeruginosa). However, for up to a third of these species, the mechanism was reported to be still unknown.
A surprising lack of information concerns also a species as common and investigated as Escherichia coli, for which natural competence so far has not yet been described. Numerous toxin-antitoxin (TA) systems are known for E. coli, but little is known on their respective roles and their capacity to determine programmed cell death. The toxin HipA is member of the phosphatidylinositol 3/4-kinase superfamily and takes part with the antitoxin HipB to the HipBA TA system. Interestingly, Zhao et al. (Citation 2013) reported an association between HipA and the enhancement of biofilm formation through DNA release. Nonetheless, in view of the variations among distinct strain types and the limited information on the other existing TA systems, more extensive studies are warranted to finally identify the mechanisms that are contributing to DNA release in this species.
As far as Gram-positive bacteria are concerned, Ibáñez de Aldecoa et al. (Citation 2017) reported the documented involvement of lytic mechanisms of eDNA release in just 4 out of the 17 species considered, respectively Enterococcus faecalis, S. aureus, S. epidermidis (Figure 1) and S. pneumoniae. For seven of the species investigated there was still no available information on the mechanism involved in eDNA release and Streptococcus mutans had been described uniquely to release small size DNA fragments within MVs (Liao et al. Citation 2014). The release had been observed to occur by growing cells during early biofilm formation, in the absence of lytic processes. However, shortly after, Jung et al. (Citation 2017) reported on the role of the autolysin A (AtlA), confirming in a wild-type strain that eDNA release mediated by bacterial cell lysis is required for biofilm initiation and maturation. Thus, experimental evidence supports the co-existence of at least two alternative mechanisms for eDNA release in S. mutans.
In many Gram-positive pathogens, autolysis and allolysis have been found to be accomplished through the expression of well-identified bacteriocins/autolysins. Nonetheless, in many cases the molecular mechanisms are far from being totally understood. In E. faecalis, eDNA release and biofilm formation are achieved by a fratricidal mechanism, where death of sibling cells is actuated by isogenic cells within the same population by the primary N-acetyl glucosaminidase, autolysin AtlA (Thomas et al. Citation 2009). Two extracellular peptidases were found to play an important antagonistic role in regulating the lytic activity of AtlA, respectively: (1) a gelatinase (GelE), which mediates cell death and eDNA release through proteolytical modification of AtlA, and (2) a serine protease (SprE), an immunity protein, inactive in the subpopulation of cells susceptible to lysis and capable of leading to an anti-lytic outcome.
Analogously, in Enterococcus faecium the autolysin AtlA, renamed AtlA(Efm), emerged among six putative autolysins as the major factor involved not only in bacterial lysis (and, thus, through the release of eDNA, in biofilm stabilization), but also in cell surface localization of the factor Acm. Acm mediates binding to collagen types I and IV (Paganelli et al. Citation 2013).
In S. pneumoniae as in S. mutans, eDNA release is also ensured by mechanisms of allolysis/fratricide involving bacteriocins, where competent siblings target the lysis of non-competent cells to enable natural transformation (Steinmoen et al. Citation 2002). However, the complexities of the pathway leading to competence differ, even between these two closely related species (Shanker & Federle Citation 2017).
Autolysins have been proved to play an important role even in the case of some important clinically relevant staphylococcal species such as S. aureus and S. epidermidis. For S. aureus the main autolysin implicated in autolysis is AtlA, a bifunctional enzyme that undergoes proteolytic cleavage to yield two catalytically active proteins, an amidase (AM) and a glucosaminidase (GL). Both AM and GL functions are required for AtlA to be catalytically active and enable S. aureus to form a biofilm, demonstrating the significant contribution of AtlA-mediated lysis (Rice et al. Citation 2007; Bose et al. Citation 2012). Analogously, in S. epidermidis, the autolysin AtlE was found to have a role in the release of DNA and biofilm formation (Qin et al. Citation 2007).
Bacterial eDNA from secretion
Over the years, cytolysis has progressively emerged not to be the only mechanism adopted by bacteria to release DNA in the outer space. Bacteria forming eDNA-containing biofilms, depending on the species and often on the strains, have occasionally been found to rely also on other alternative mechanisms of eDNA production, which are independent of cell death and lysis. One such mechanism is lysis-independent DNA secretion. Connected with natural cell transformation and competence functions, eDNA secretion is observed sometimes to co-exist with autolysis and allolysis in the same strain. Under such circumstances, eDNA secretion does not appear to cover a redundant function. Rather, it is implicated in the selective release of eDNA at different phases of the cell cycle and of biofilm maturation, being generally controlled by quorum sensing systems and influenced by external stimuli inducing the activation of competence.
This holds true for E. faecalis, an opportunistic pathogen that has previously been described to be capable of releasing eDNA through well-defined fratricide mechanisms. Nonetheless, lytic mechanisms are the source of eDNA for biofilm strengthening only at a late stage, i.e. in mature biofilms older than 24 h. This does not appear to occur during the early phases of biofilm formation, when visual examination under SEM and bulk biochemical assays would exclude the fact that significant lytic processes are taking place (Barnes et al. Citation 2012). Conversely, during the initial stage, eDNA was found to be localized at the site of the septum of some cells as if a subpopulation of metabolically active cells could secrete/extrude eDNA in a way that is independent of cell lysis. The authors therefore reported the existence of a new non-lytic mechanism of DNA release. During the first 4 h of biofilm formation, they estimated a 3-Log increase in the ratio of eDNA per cell with respect to bacteria in the planktonic phase. The authors also reported that, during the first 4 h of biofilm growth, eDNA was primarily found in two distinct structures observable in the extracellular matrix, respectively in intercellular string-like structures that they defined yarns, and in the thick matrix surrounding the cells that were termed sweater (Barnes et al. Citation 2012). Both yarns and sweater structures co-exist in individual biofilm samples. On the contrary, cell lysis would generate irregular masses of intracellular debris, these also including the nucleic acid component. Among the mechanisms capable of explaining this type of eDNA secretion occurring exclusively during biofilm formation, Barnes et al. (Citation 2012) hypothesized the involvement of the conjugation apparatus or an as-yet-uncharacterized DNA export system.
In Neisseria gonorrhoeae, secretion of chromosomal DNA has been found to be associated to a type IV secretion system (TFSS) implicated in bacterial transformation. Salgado-Pabón et al. (Citation 2010) observed that, within the cell population, gonococcal variants that produce type IV pili release larger quantities of DNA than non-piliated variants. Piliated strains would produce the largest quantity of DNA secretion in late log-phase growth and DNA release would not occur through autolysis. Moreover, secreted DNA would be in the form of single-strand DNA protected at the 5′ end from 5′–3′ exonuclease digestion (Salgado-Pabón et al. Citation 2007). Secretion would occur between the 2-h and 2.5-h time points and precede the onset of the stationary/death phase during which bacteria would switch to cell lytic processes.
Nontypeable Haemophilus influenza (NTHI) has recently been the object of enlightening investigations on the machinery involved in the secretion of DNA. In NTHI, Jurcisek et al. (Citation 2017) found that, in a subpopulation of NTHI cells, the transit to the outer space of DNA (and with it also of DNABII DNA-binding proteins, in particular the DNA-binding protein integration host factor, IHF) occurs through two steps. The first passage from the bacterial cytoplasm to the periplasm, across the inner membrane, takes place via the inner-membrane pore complex (TraCG), which is homologous to type IV secretion-like systems. Once across the inner membrane and having reached the periplasm, both DNA and DNABII proteins can exit to the outer space, being released into the environment through the ComE pore (the same ComE secretin through which the NTHI type IV pilus is expressed). Overall, ComE has therefore emerged as being involved in the formation of a type IV pilus, but also in the exportation and release of DNA and proteins in the outer space for biofilm formation and in the import of DNA for transformation (Krüger & Stingl Citation 2011). Much remains to be elucidated on how these different functions are finally orchestrated (Seifert Citation 2017).
Zafra et al. (Citation 2012) found that in B. subtilis also the release of eDNA, corresponding to whole genomic double-stranded DNA, is linked to competence and occurs in the absence of cell lysis. For the bacterial strain investigated, named strain 3610, a peak of eDNA release was observed in the late exponential-phase, but this peak was not confirmed in other laboratory strains.
As anticipated earlier, in E. coli the mechanisms involved in eDNA release are mostly obscure. In 2010, Sanchez-Torres et al. investigated the functions of control of the global regulator H-NS on eDNA production and were led to postulate the existence of a release via direct eDNA secretion from living cells. Indeed, while the deletion of the gene encoding the global regulator H-NS resulted in the suppression of eDNA production, a slightly decreased cell lysis and the increased vesiculation observed could not be the explanation for the reduced eDNA.
Thus, in various Gram-positive and Gram-negative bacteria eDNA secretion appears to be linked to competence and to the DNA uptake machinery. The formation of complexes between single strand DNA (ssDNA) and the ComE secretin, its homologs found in other bacterial species (e.g. ComEA) or other unrelated proteins (e.g. single-stranded binding protein SsbB), has been hypothesized to act as a putative reservoir function. Under this hypothesis, periplasmic ComE/ComEA/SsbB-DNA clusters would be accumulated and stored until needed and become available on demand. This periplasmic accumulation could serve not only for transformation but also as a food source (Seitz& Blokesch Citation 2014). Moreover, periplasmic accumulation would support the hypothesis that, in Gram-negative bacteria, vesiculation could generate DNA-containing OMVs.
However, a very recent discovery extends knowledge on the diversified mechanisms of eDNA release. Rugose small colony variant (RSCV) isolates of P. aeruginosa are known to be hyperactive in biofilm production and cause difficult to treat chronic infections (Pestrak et al. Citation 2018). Deng et al. (Citation 2020) reported that, in a hyperbiofilm-forming RSCV named PAO1DwspF, biofilm aggregates are formed by two bacterial subpopulations with contrasting physiological characteristics and a segregated spatial distribution. Interestingly, PAO1ΔwspF was found to possess the unique ability to extrude DNA fragments from living bacteria as part of bolstering the biofilm structure. This is in contrast with eDNA in the biofilm of the PAO1 isogenic parental strain, which consists of intact bacterial DNA as expected in a bacterial species earlier known to release eDNA through explosive cell lysis. Thus, considerable diversity characterizes the mechanisms of eDNA production, even among strain types of the same species.
Bacterial eDNA release through MVs
In an earlier section, MVs were described as associated with cell lysis events. However, this is not always the case. Gram-negative (Kulp & Kuehn Citation 2010) and, more recently, Gram-positive bacteria (Brown et al. Citation 2015) have both been found to be capable of producing MVs, i.e. nanosized bodies in the range 20-300 nm, which are delimited by cellular membranes, which form on the surface of the bacterial cells and are released free into the outer space. MVs have been found to be heterogenous, varying in composition and expressing different functionalities and properties, depending on their different pathways of biogenesis (Nagakubo et al. Citation 2020) and on their growth phase transition (Tashiro et al. Citation 2010). Based on their different cargos, diverse important functions have been described for MVs in (1) bacterial signaling and quorum sensing (Dauros-Singorenko et al. Citation 2018; Toyofuku Citation 2019); (2) immunomodulation of the host response by stimulating or quenching the immune system through induction of host cell apoptosis and delivery of toxins (Kaparakis-Liaskos & Ferrero Citation 2015); (3) resistance to antimicrobial substances by antibiotic binding and sequestration, (4) by degrading enzymes or by contributing to horizontal gene transfer (Vitse & Devreese Citation 2020); and finally (5) biofilm formation.
As far as biofilm production is concerned, MVs intervene at different levels. In the biofilm of some bacterial species, they represent a conspicuous fraction of the extracellular matrix and its EPS. About 20% of the whole matrix proteome of P. aeruginosa biofilms has been found to consist of MV-associated proteins (Couto et al. Citation 2015). These proteins include among others: cytoplasmic molecules with moonlighting functions such as carbohydrate metabolic proteins and lipid metabolic proteins; proteins with defensive/protective functions such as SOS proteins and phage attack proteins; virulence factors such as host attack proteins and toxins (e.g. cytotoxins); and autolysins (Dineshkumar et al. Citation 2020). Bacteriocins/autolysins secreted within MVs can eventually be used to induce the lysis in neighboring bacteria, leading to the release of nutrient molecules and eDNA. In addition to proteins, depending on the biogenesis pathway, the cargo of MVs has also been reported to include, among other molecules, lipids and nucleic acids. Nucleic acids in MV cargo comprise RNA (miRNA, mRNA and rRNA), but also single- and double-strand eDNA (Kadurugamuwa and Beveridge Citation 1995), potentially involved in horizontal gene transfer as well as in biofilm formation. Furthermore, MVs produced by biofilm cells of Helicobacter pylori were suggested to prevent eDNA degradation by nucleases, provide a bridging function between eDNA strands on MV surfaces and mediate aggregation (Grande et al. Citation 2015).
As shown in Figure 1, there are distinct types of MVs. They can be produced through different biogenesis pathways, and this is reflected by a different cargo composition. Toyofuku et al. (Citation 2019) reported as still open to debate the fact that OMVs, generated by blebbing and in absence of cell lysis events, may contain eDNA and, with current knowledge, it is difficult to clearly explain the transfer to the periplasm of genetic material leaching from the cytoplasm, without expecting the subsequent death of the bacterial cell. Conversely, when phage-derived endolysins degrade the proteoglycan cell wall and lead to explosive cell lysis, the so generated explosive lysis outer membrane vesicles (EOMVs) and, even more, the explosive lysis outer-inner membrane vesicles (EOIMVs) are likely to contain cytoplasmic material and, with it, eDNA. A similar circumstance occurs when Gram-positive species, for instance Staphylococcus spp. such as S. aureus, B. subtilis, S. pneumonia, Listeria monocytogenes and S. mutans, generate cytoplasmic membrane vesicles (CMVs) (Liao et al. Citation 2014). In B. subtilis, the expression, under DNA stress conditions, of an endolysin encoded by a defective prophage has been found to weaken the structural integrity of the cell wall peptidoglycan (Toyofuku et al. Citation 2017). CMVs were found to originate from the protrusion of the cell membrane through the holes caused by the endolysin to peptidoglycan. In Gram-positive cells, a thicker cell wall structure would avoid the swelling and explosive cell lysis characteristic of Gram-negative cells such as P. aeruginosa, where MVs were seen to result from the fusion of shattered membranes rather than from protrusion through the wall and pinching off. Through similar mechanisms, DNA-damaging agents and antibiotics inducing the SOS response such as ciprofloxacin were found to trigger CMV formation, but only in lysogenic S. aureus strains (Andreoni et al. Citation 2019) with a prophage dependent-mechanism. Conversely, antibiotics such as β-lactam antibiotics (flucloxacillin and ceftaroline) exhibited also a different route of CMV production by directly weakening the peptidoglycan layer, thus evidencing alternative mechanisms of CMV production. The content of eDNA was found to be greater for CMVs when blebbing was caused by phage lysis.
After earlier investigations demonstrating the formation of outer-inner membrane vesicles (OIMVs) by the protrusion of both outer and plasma membranes in the antarctic bacterium Shewanella vesiculosa M7T, Pérez-Cruz et al. (Citation 2015) provided some evidence that the production of OIMVs containing eDNA could even occur in strains of other clinically relevant species, including N. gonorrhoeae, P. aeruginosa and Acinetobacter baumannii, with a possible involvement in lateral gene transfer. Nonetheless, eDNA-containing OIMVs were found to represent only a minimal proportion of MVs, in the range 0.2-1.2%, depending on the strains. Furthermore, OIMVs have been well documented by ultrastructural studies conducted by Cryo-TEM. Nonetheless, the mechanisms that would trigger OIMV formation, causing a loss of integrity of the cell wall, and enable the protrusion of both inner and outer membranes with a portion of cytoplasmic content has not been still totally clarified. Even more obscure is the evolution of the fate of bacterial cells losing portions of their genomic DNA.
Overall, eDNA production through MV secretion has been well documented in both Gram-negative and Gram-positive bacteria. The presence of DNA has been detected in the cargo of different types of membrane vesicles, even if, in MVs generated consequent to cell lysis or through mechanisms involving prophagic endolysins, greater quantities of nucleic acid have been reported. MVs appear to be a critical junction between routes of different bacterial functions and much remains to be understood regarding the modalities controlling their cargo. Certainly, MVs provide a form of eDNA transfer that is protected from the degradation by nucleases (Grande et al. Citation 2011). They not only take physical part in the structure of biofilms, contributing their cargo of eDNA and proteins, but also interact with free eDNA molecules in the extracellular space promoting aggregation. Much still remains to be uncovered on the half-life in the interstitial milieu and on the further evolution of MVs. Their fate can certainly be that of being incorporated and contributing to the biofilm, but it can also be that of fusion with the membranes of other adjacent prokaryotic cells in order to accomplish cell transformation, cell signaling or delivery of bacteriocins. A further important option is that of entering eukaryotic host cells and explicating pathogenic functions by hijacking host defenses.
Host cells and tissues eDNA
An important point when examining the different sources of eDNA in bacterial biofilms is that, while in biofilms cultured under artificial in vitro conditions, the only source of DNA is represented by bacteria, under real clinical conditions a proportion of EPS, DNA among them, is provided by host cells and tissues. In this regard, the role played by human extracellular matrix proteins such as fibronectin in S. aureus bacterial aggregations and biofilm formation is an example. Indeed, fibronectin can form a bridge and connect fibronectin-binding proteins, namely FnBPA and FnBPB, expressed on the outer surface of staphylococcal cells (Speziale & Pietrocola Citation 2020). Adhesins such as FnBPs are virulence determinants and represent an adaptation of the pathogen to the physiological environment of the host, exploiting the wealth of molecules available in the interstitial milieu. Among other polymeric substances, DNA of host provenance is certainly a molecule of great potential value both as a source of nurture as well as a component that can be integrated into the biofilm to enforce its architecture and enhance its shielding functions. Indeed, different in vitro investigations have shown that supplementation with exogenous DNA can partially or totally restore some lost functionalities in bacterial biofilms treated with DNase (Carrolo et al. Citation 2010; Harmsen et al. Citation 2010; Das et al. Citation 2011; Chiang et al. Citation 2013; Wongkaewkhiaw et al. Citation 2020). Nonetheless, there is some evidence that, under real in vivo conditions, in infections caused by P. aeuruginosa, the majority of eDNA is found external to the biofilm, keeping it confined in the peripheral space, and derived from the host rather than from bacteria (Alhede et al. Citation 2020).
Except under particular pathological conditions such as malignancies and infections, the presence of eDNA in physiologic interstitial fluids is generally low. In bacterial infections, the release of DNA from host cells can be associated to different phenomena. These include: (1) the death of host tissue cells/leukocytes for the action of bacterial cytolytic or proapoptotic toxins; (2) tissue damage derived from an excess of inflammatory reaction; (3) production of neutrophil extracellular traps (NETs) by host leukocytes.
Pathogens such as S. aureus, a main causative agent of nosocomial infections, are known to express several toxins and pore forming cytolysins that are capable of causing the death of host tissue cells and of effector cells of the immune system (Arciola et al. Citation 2018; Campoccia et al. Citation 2019). The S. aureus virulon comprises numerous toxins, among them PSMα3, δ-Toxin and other leukocidins, bi-component β-barrel pore-forming toxins including γ-hemolysin and Panton Valentine leukocidin (PVL) (Vandenesch et al. Citation 2012; Arciola et al. Citation 2018; Tromp & van Strijp Citation 2020). In addition to the secretion of cytolytic toxins, S. aureus can degrade eDNA by the combined action of two enzymes, nuclease and adenosine synthase A, producing the powerful pro-apoptotic metabolite dAdo, capable of affecting the viability of host leukocytes (Thammavongsa et al. Citation 2013; Papayannopoulos Citation 2018; Campoccia et al. Citation 2019). Moreover, S. aureus has been found to possess further mechanisms leading to cells apoptosis and autophagocytosis. For instance, this species can efficiently internalize into host cells such as bone cells and, thereby, upregulate the tumor necrosis factor-related apoptosis-inducing ligand (TRAIL), which, in turn, induces osteoblast apoptosis with consequent bone-tissue damage, through the activation of caspase 8 (Alexander et al. Citation 2003). The special multifaceted ability of S. aureus to modulate the expression of virulence factors to prevent elimination through programmed cell death pathways and evoking apoptosis, necroptosis, and pyroptosis has recently been reviewed by Soe et al. (Citation 2021).
Analogous cytotoxins have been identified in other pathogens. For instance, in P. aeruginosa, rhamnolipid has been reported to determine the lysis of neutrophil cells by acting as a biosurfactant ( Van Gennip et al. Citation 2009). The ExoU cytotoxin and exolysin (ExlA) have also been described as two important exotoxins of P. aeruginosa (Tamura et al., Citation 2004; Basso et al. Citation 2017). Other examples of toxins expressed by other pathogens and capable of causing host cell death include the α-hemolysin of E. coli, the cytolysin of Vibrio cholerae, and the listeriolysin O of Listeria monocytogenes (Steinmoen et al. Citation 2002). The action of cytolytic, necrotizing or pro-apoptotic mechanisms/factors can result in cell death and contribute to the release of substantial quantities of nucleic acids.
However, in the presence of an excess of inflammatory response, tissue disruption and cellular damage can also derive from the disproportionate activation of effector cells of the immune system. This can be triggered either by the presence of foreign bodies such as implant materials and their wear debris or by unresolved biofilm-associated chronic infections. Both of these conditions can determine frustrated leukocyte phagocytosis and release into the outer space of potentially harmful chemicals, normally confined within phagosomes (Arciola et al. Citation 2018). Thus, the large arsenal of molecules usually exploited to contain invading microorganisms or generally directed against foreign bodies poses a risk of collateral damage to surrounding tissues (Kzhyshkowska et al. Citation 2015). Under normal conditions, the successful resolution of acute inflammation is characterized by the apoptosis of neutrophil cells and their engulfment by macrophages. Conversely, in chronic bacterial infections, neutrophil cells are thought to undergo necrosis rather than apoptosis (Cox et al. Citation 1995).
A third important mechanism of eDNA extrusion consists of the release by neutrophil cells of NETs, i.e. extracellular meshes composed of eDNA in the form of nuclear chromatin and granule proteins. NETs granule proteins include enzymatically active proteases such as neutrophil elastase, myeloperoxidase (MPO) and cathepsin G, and other antimicrobial peptides (e.g. histone H3) that bind, trap and kill bacteria. The NETome has been described as consisting of up to 24 different proteins (Urban et al. Citation 2009). First described by (Brinkmann et al. Citation 2004), the production of NETs was only later recognized as a specific and regulated form of cell death also known as NETosis (Fuchs et al. Citation 2007) and reported to depend on the generation of reactive oxygen species (ROS) by NADPH oxidase (Steinberg& Grinstein Citation 2007). NET formation was shown to be triggered by the presence of microorganisms as well as by biochemical stimuli such as phorbol myristate acetate (PMA) (Takei et al. Citation 1996) or cytokines such as interleukin-8 (IL-8). Rapid killing of neutrophils by PMA accompanied by changes different from typical apoptosis and necrosis was reported by Takey et al. as early as1996. Despites the many efforts made to finely characterize all the triggering stimuli, the molecular mechanisms respectively implicated in NET formation and the sources of the DNA released (nuclear, mitochondrial or both) in order to reconduct the entire phenomenon to a single well-defined process, there are still many disputed points that remain unresolved (Boeltz et al. Citation 2019). The nature of the mechanisms of eDNA extrusion leading to NET formation and the definition of NETosis have repeatedly been reported as mistakenly confused. In an attempt of clarification, Konig and Andrade (Citation 2016) critically reviewed all the known mechanisms and processes of eDNA extrusion by neutrophils. In their reappraisal, they distinguished a properly defined NETosis from other similar processes involving eDNA release, based on the triggering stimuli, the activation of distinct biochemical pathways, the presence of hypercitrullination, and the antimicrobial effector function. Proper NETosis was defined as a form of programmed cell death intended to contrast the presence of pathogens and NETs as part of a defence mechanism aimed at immobilizing and neutralizing pathogenic microorganisms such as bacteria, fungi, and viruses. For Konig and Andrade (Citation 2016), differing from other NETs mimicking conditions, NETosis was NADPH oxidase-dependent and was not associated with hypercitrullination. Conversely, DNA extrusion could also be triggered by bacteria, through the production of cytolysins. The authors defined this form of eDNA release as leukotoxic hypercitrullination (LTH). Triggered by pore-forming pathways and equivalent signals, LTH would involve calcium-dependent hyperactivation of citrullinating enzymes such as peptidylarginine deiminase type 4 (PAD4), protein hypercitrullination, and neutrophil death. Peptidylarginine deiminase activity would be involved in chromatin decondensation. LTH would not consist of a host defensive strategy of the immune system, but rather in a bacterial strategy to elude the host immune response.
In contrast, Yousefi et al. (Citation 2019) reviewed the concept of NETosis and NET formation and argued that NET formation and neutrophil necrotic death with DNA release should be considered as independent phenomena. They discussed the evidence supporting NET formation in viable neutrophils and the molecular mechanisms and intracellular events leading to NET formation through mitochondrial DNA release. This type of NET formation, also observed in other types of granulocytes as well as in lymphocytes, would not involve cell death. It would occur rapidly, i.e. in a matter of minutes, and should be treated differently from the events of programmed necrosis, occurring with a 2-h delay. This view is aligned with the previous explicit recommendations of the Nomenclature Committee on Cell Death (NCCD) (Galluzzi et al. Citation 2018) to avoid the use of the term NETosis, when no experimental evidence in support of cell death is available. Additionally, NCCD called for additional studies to address the still unclear contribution of necroptosis to NET extrusion and NETotic cell death. Based on the available evidence, NETotic cell death was defined as a ROS-dependent modality of regulated cell death, which is restricted to cells of hematopoietic derivation and associated with NET extrusion. Indeed, initially thought to be confined to neutrophils, the production of DNA extracellular traps (ETs), was found not to be exclusive of neutrophils, but to extend to other leukocytes, eosinophils (EETs) and to monocytes/macrophages (MoETs/METs) (Guimarães-Costa et al. Citation 2012; Gómez et al. Citation 2021). Analogously to the judgment expressed for NETosis, NCCD discouraged the use of the term ETosis (Galluzzi et al. Citation 2018).
Apart from the many, still open, debates on the nomenclature and on the different characterization of the mechanisms, NET formation may be considered a complex and only partly understood phenomenon. Its underlying mechanisms vary as a function of the triggering stimuli that include a broad range of factors. As suggested by Castanheira & Kubes (Citation 2019), it could be more appropriate simply to cautiously distinguish between lytic and non-lytic NET formation. In non-lytic NET formation, mitochondrial DNA most likely represents the source of the eDNA, whereas nuclear DNA release would involve or be the consequence of cell death. TLRs, complement receptors and lipopolysaccharide-activated platelets would be implicated in non-lytic NET formation directed to counter the presence of bacteria. Conversely, bacterial pore-forming leukocidins such as PVL and γ-hemolysin that are expressed by S. aureus would be capable of causing both NET formation and cell death, although it remains unclear if, in this context, NETs are simply a consequence of neutrophil lysis. Intriguingly, the induction of NETs appears to represent an advantage for these bacteria (Bhattacharya et al. Citation 2018) and favor their persistence in the tissues rather than representing a real threat. Differing from planktonic bacteria, S. aureus biofilms would rapidly skew neutrophils toward NET formation through the combined activity of PVL and γ-hemolysin. By eliciting this response, S. aureus would persist in the host tissues, as the result of the antimicrobial activity of NETs would be ineffective at clearing biofilm bacteria. Very recently Mazzoleni et al. (Citation 2021) described for PVL the existence of an alternative NETosis process targeting mitochondria. PVL would colocalize with mitochondria, enhance the production of reactive oxygen species from these organelles and cause the ejection of chromatin fibers independently of the NADPH oxidase oxidative burst. Conversely, the pore-forming cytolytic leukotoxin GH (LukGH), differing from PVL, was found neither to prime human neutrophils for increased ROS production nor to enhance the uptake of S. aureus. Nonetheless, LukGH was found to be capable of promoting the release of NETs, even if they merely ensnared but did not kill the bacteria (Malachowa et al. Citation 2013).
Hypothetical advantages for bacteria inducing NET formation could derive from the contribution of NET DNA to the biofilm architecture as well as from the conversion of this DNA to the pro-apoptotic metabolite dAdo (Thammavongsa et al. Citation 2013; Arciola et al. Citation 2018; Papayannopoulos Citation 2018; Missiakas & Winstel Citation 2021). However, in in vivo chronic infections caused by P. aeruginosa such as those observed in lung tissues of cystic fibrosis patients, Alhede et al. (Citation 2020) reported that bacterial biofilms seem incapable of triggering the release of NETs by PMNs. Moreover, eDNA released following PMNs necrosis would not apparently integrate and become part of the in vivo biofilm architecture. Rather, it would constitute a sort of secondary matrix, externally lining the biofilm and possibly acting as a passive physical shield against antibiotics and phagocytes. Thanabalasuriar et al. (Citation 2019) reported that in ocular biofilms by the same pathogen, high expression of the bacterial type-3 secretion system (T3S) results in NETosis, with NETs forming a barrier ‘dead-zone’, capable of confining the infection, between host cells and P. aeruginosa biofilms.
In view of the above observations, it would appear as if, in lung tissues of cystic fibrosis patients, the release of host eDNA was determined by the virulence mechanisms actuated by the bacteria rather than representing, at least under these circumstances, a functional and effective host defence mechanism. The release of host DNA would increment the external protection to an organized biofilm underlayer. However, in cystic fibrosis as well as in ocular infections, the sacrifice of the leukocytes would generate a line of confinement capable of slowing down the spreading of bacteria. Moreover, bacterial DNA has some peculiarities compared with human DNA, being rich in hypomethylated ‘CpG’ repeats (Itagaki et al. Citation 2011). Intracellular leukocyte receptors such as TLR9 have long be known to recognize unmethylated CpG motif characteristic of bacterial DNA (Hemmi et al. Citation 2000; Otto Citation 2008). Thus, the externally lining layer of host DNA could possibly mask this interaction, competing with bacterial eDNA in the uptake into endosomal compartments.
Overall, the role of NETs and of the DNA released by host cells remain difficult to interpret as far as their contribution to bacterial biofilms is concerned. Many alternative physiological mechanisms are emerging as implicated in the release of NETs. They do pinpoint a function of NETs in the host defence against microorganisms and parasites. However, some of them are voluntarily actuated by some pathogens, which appear to control and even trigger eDNA release rather than suffer its action. Abundant evidence highlights the fact that specialized opportunistic pathogens such as, for instance, S. aureus have evolved multiple mechanisms to avoid harm by the many antimicrobial substances associated with NETs (neutrophil elastase, MPO, cathepsin G, and histones such as H3). They do induce the release of nucleic acids and, once their bactericidal potential is neutralized, their reward would be expected to consist in survival from the neutrophil trap and even in the DNA itself derived from the inactivated traps. However, the in vivo observations in a murine model and in human lung tissues of cystic fibrosis patients consistently indicate that neutrophil-derived DNA does not co-localise with P. aeruginosa biofilms (Alhede et al. Citation 2020) as contrastingly do neutrophil elastase and citrullinated H3.
Under normal conditions (i.e. in the absence of a foreign body, in subjects that are not affected by genetic diseases such as cystic fibrosis and in non-immunocompromised patients), the innate and cell mediated immune response of the host is generally effective in warding off the majority of the infections and preventing the establishment of recalcitrant biofilm infections. However, when an imbalance of the immune defenses favors invading opportunistic pathogens, these can skew the antibacterial action for which NETs are intended and turn a risk in an opportunity. Ríos-López et al. (Citation 2021) have recently reviewed the multiplicity of mechanisms that pathogens have developed to escape the bactericidal action for which NETs are intended.
Biomaterial-triggered NETs
An important aspect concerning implant associated infections is that neutrophils actively interact with biomaterial surfaces (Selders et al. Citation 2017). Neutrophils are rapidly recruited at the site of implantation and wherever biomaterial surfaces come in direct contact with blood. Since the early stages, these leukocytes play a fundamental role in orchestrating the tissue response to foreign body surfaces and as such to biomaterial surfaces. An only recently emerged activity of these effector cells is their ability of priming/preconditioning biomaterial surfaces. It has long been known that matrix proteins present in the physiological fluids, including components of the humoral immune response (e.g. factors of the coagulation and complement cascade), nearly instantaneously adsorb on biomaterial surfaces generating a proteinaceous conditioning film. Immediately after this interaction with proteins also known under the name of Vroman effect, cell-mediated host immune defenses react to the biomaterial surface of the implant, recognizing it as a foreign body. Depending on physical and chemical cues, the biomaterial surface can elicit leukocyte activation and NET extrusion, the latter process resulting in an additional conditioning of the outer surface, which eventually becomes coated with eDNA. All this has important implications for the pathogenesis of implant infections. First leukocyte activation following the interaction with implant materials can result in metabolic exhaustion and in a reduced capacity of neutrophils to release ROS and kill bacteria, resulting in a niche of local immune depression at the interface of implant with periprosthetic tissues. Local immune depression is one of the possible reasons enabling typically saprophytic bacterial species to emerge as opportunistic pathogens in the presence of implant materials (von Eiff et al. Citation 2006). Second, it is likely that the presence of NETs eDNA on the biomaterial surface can electrostatically attract and entrap bacteria, facilitating microbial adhesion and colonization. In this perspective, even if it is still unproven that eukaryotic eDNA can always be effectively incorporated in the extracellular matrix of bacterial biofilms, NETs could certainly represent nuclei of bacterial aggregation and, in the absence of a solid substrate, of flocculation.
eDNA in orthopedic infections
Orthopedic infections deserve special attention, since more than a million total hip and knee arthroprostheses are implanted each year in US, and this number is expected to increase as population ages (Singh et al. Citation 2019). In periprosthetic infections, bacterial contamination can occur in two ways: (1) through accidental contamination of the surgical incision by bacteria either from the patient's skin or mucosae, the healthcare personnel, or the operating room environment, during surgery; (2) through hematogenous spread from a distant focus of infection, during the postoperative period.
The role of eDNA in periprosthetic orthopedic infections is of great interest. In a study aimed at investigating the relation between PIA and eDNA in a large collection of S. epidermidis clinical isolates from orthopedic infections, the presence in the biofilm matrix of the exopolysaccharide was found to correlate with that of eDNA. Isolates that did not produce PIA exhibited minimal levels of eDNA, while strains with significant PIA production showed high amounts of eDNA (Campoccia et al. Citation 2011). Furthermore, eDNA was found to be a constant component of the various biofilm matrices produced by S. lugdunensis isolates from orthopedic implant infections (Ravaioli et al. Citation 2020).
In orthopedic prosthetic joint infections by S. epidermidis, Zatorska et al. (Citation 2018) showed a correlation between a high amount of eDNA in biofilms of bacterial clinical isolates and adverse clinical outcome. Therefore, quantification of bacterial eDNA may represent a predictive marker for the management of joint infections (Zatorska et al. Citation 2018).
The excerpts mentioned foreshadow the potential clinical utility of an in-depth study of the origins of eDNA. In this connection, it must be emphasized that the field of orthopedic implants is one of the most burdened by biofilm-associated infections.
Conclusions
Although bacterial mechanisms of eDNA production are progressively being unveiled, they still warrant extensive investigative work. Directly originated by the action of noxious agents affecting bacterial integrity, induced by the activation of dedicated quorum sensing systems, or contributed by host cells, eDNA production has a critical impact on biofilm formation and the possibility of pathogen persistence in the environment as well as in the human body. For this reason, this ubiquitous molecule is increasingly being considered an ideal target for broad-spectrum anti-biofilm strategies aiming at disrupting biofilm integrity. However, eDNA susceptibility to enzymic degradation is generally limited to the very early stages of biofilm formation, when the polyanionic polymer has not yet undergone extensive complexation with other EPS, in particular amyloid and polycationic proteins, and exopolysaccharides. Thus, most advanced strategies are being directed to interfering with, preventing or disrupting the interactions of eDNA with other EPS, thus disabling or dismantling fully protected mature biofilms.
A better understanding of the origin of eDNA and on how controlled bacterial death can represent a real advantage for the bacterial population is of paramount importance. Alternative measures for prevention and treatment of biofilm-based infections that do not strictly target DNA have to be meticulously thought and assessed, carefully considering the risk of causing the release of eDNA and, thus, enhancing biofilm formation. For instance, this is a case of the much-discussed phage therapy. Therapies based on the use of phage viruses certainly represent a promising tool to partly obviate the problem of the current shortage of efficacious antibiotic treatments. However, phages are not only implicated in horizontal gene transfer through genetic transduction, but they can also establish symbiotic relationships that induce and strengthen biofilms (Pires et al. Citation 2021). Furthermore, prophagic autolysins have themselves emerged among the mechanisms used by bacteria to release eDNA and thrive in hostile environments. Thus, an initial bactericidal effect obtained by experimental phage therapies should not be confused with a successful cure, and final complete and long-lasting remission of an infection should be proved after a prolonged follow-up, particularly in the presence of prosthetic devices.
Similar considerations can be made regarding the use of delivery systems based on the release of antibiotic or bactericidal substances intended to prevent biofilm formation on prosthetic surfaces. If the bactericidal action is not perfectly accomplished and a protracted subinhibitory release of antibiotic is perpetuated at the site of implantation, there is a real risk of promoting the formation of eDNA, consequently enhancing biofilm production instead of achieving a successful eradication of the infection.
In addition, there is a real need to gain a better comprehension of the role of eDNA derived from host cells in the perpetuation of chronic infections, especially those developing in the absence of a foreign body.
Disclosure statement
No potential conflict of interest was reported by the authors.
Additional information
Funding
The financial contribution by the Theory-enhancing Project RF-2018-12368273: “Implant orthopaedic infections: the challenge of bacterial antibiotic resistance and new anti-infective materials” funded by the Ministry of Health for the IRCCS Rizzoli Orthopaedic Institute of Bologna is gratefully acknowledged.
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9015 | http://fac-staff.seattleu.edu/oliveras/web/teaching/DiffEq/Notes/SO_Lecture_2.pdf | MATH 234 -THE WRONSKIAN AND LINEAR INDEPENDENCE RECAP In the previous lecture, we learned how to solve ay+ by+ cy = 0, using three steps: 1. Guess the form of the solution eλx to determine the characteristic equation: aλ2 + bλ + c = 0 2. Solve the characteristic equation to find the fundamental solutions: y1 = eλ1x, y2 = eλ2x 3. Combine the fundamental solutions (via superposition theorem) to determine the general solution: y = c1y1 + c2y2 It turns out that we can use this process to solve any linear constant-coefficient ODE of any order! However, so far, we have restricted ourselves to constant-coefficient 2nd order ODE whose characteristic equation does not have repeated roots.
If we slightly generalize the class of ODEs that we are considering to any linear second order ODE of the form y+ p(x)y+ q(x)y = 0, the superposition theorem will still hold. Thus, the general solution is given by y = c1y1 + c2y2, where y1 and y2 are two linearly independent solutions of y+ p(x)y+ q(x)y = 0.
.
But before we talk about linear independence of fundamental solutions, perhaps a more basic questions we should ask is the following: If y+ p(x)y+ q(x)y = 0, and we’re given initial conditions y(x0) = y0 and y(x0) = y 0, how do we even know if a solution will exist?
In this lecture, we will talk about the existence of solutions, the concept of linear independence, and the Wronskian.
1 EXISTENCE, UNIQUENESS, AND THE WRONSKIAN Theorem: Consider the initial value problem y+ p(x)y+ q(x)y = g(x), y(x0) = y0, y(x0) = y 0, where p(x), q(x) and g(x) are continuous on an open interval I that also contains x01. Then, there exists exactly one solution y = φ(x) of this problem and the solution exists throughout the interval I.
It’s important to note the three things that this theorem says: 1. The initial value problem has a solution; in other words, a solution exists.
2. The initial value problem has only one solution; in other words, the solution is unique 3. The solution φ(x) is defined throughout the interval I where the coefficients are continuous and is at least twice differentiable there.
Example: Find the longest interval in which the solution of the initial value problem (t2 −3t)d2y dt2 + tdy dt −(t + 3)y = 0, y(1) = 2, dy dt t=1 = 1 In this problem, if we write it in the form where the coefficient of the second derivative term is one, we find that p(t) = 1/(t −3), q(t) = −(t + 3)/(t2 −3t), and g(t) = 0. The only points of discontinuity of the coefficients are at t = 0 and t = 3. Therefore, the longest open interval containing the initial point t = 1 in which all of the coefficients are continuous is I = 0 < t < 3. Thus, this is the longest interval in which our theorem guarantees that a solution exists.
LINEAR INDEPENDENCE AND THE WRONSKIAN Now, let’s return to a question we posed earlier: Assume that both y1 and y2 are two solutions of y+ p(x)y+ q(x)y = 0.
Superpositions tells me that y = c1y1 + c2y2 is the general solution of the ODE if y1 and y2 are linearly independent. But how can we determine if two solutions are linearly independent?
1I is just a range of x values where the functions p, q, and g are well behaved.
2 Definition: Two functions f(x) and g(x) are called linearly independent if the equation c1f(x) + c2g(x) = 0, for all x, can only be satisfied by choosing c1 = 0 and c2 = 0. Two functions that are not linearly independent are called linearly dependent.
This is very similar to the concept for linearly independent vectors. For example, consider the two vectors v1 and v2. These vectors are linearly independent if c1v1 + c2v2 = 0 implies that both c1 and c2 are zero. If you need more verification, consider the concrete example where v1 = T and v2 == T . It should be clear that they are linearly independent vectors. Furthermore, the only way that we can get the zero vector if we take a linear combination is by letting both c1 and c2 be identically zero.
Example: f(x) = ex and g(x) = 2ex are linearly dependent because −2f(x) + g(x) = 0, so c1 = −2 and c2 = 1. If the only choice was to choose them both zero, the functions would be independent.
Wouldn’t it be nice if there was an easier way to determine linear independence? Well, there is! We need to introduce the Wronskian first.
Definition: The Wronskian of two function f(x) and g(x) is just the quantity W(f, g)(x) = f(x)g(x) −f (x)g(x) Example: Let f(x) = sin(x) and g(x) = cos(x), find W(f, g)(x).
W(f, g)(x) = sin(x)(−sin(x)) −cos(x)(cos(x)) = −1 It’s sometimes easier to think of the Wronskian using matrix notations. In other words: W(f, g)(x) = det f(x) g(x) f (x) g(x) = f(x)g(x) −f (x)g(x) Theorem Two functions f(x) and g(x) are linearly dependent if their Wronskian W(f, g)(x) = f(x)g(x) −f (x)g(x) = 0.
3 Proof: If f(x) and g(x) are linearly dependent, then we can find constants c1 and c2, not both zero, so that c1f(x) + c2g(x) = 0, for all x.
Then also c1f (x) + c2g(x) = 0, for all x.
Now, let’s assume that f = 0, otherwise we can switch the roles of f and g. Then we can solve to find c1 = −c2g(x) f(x) c1f (x) + c2g(x) = −c2g(x) f(x) f (x) + c2g(x) = c2 f(x) (g(x)f (x) −g(x)f(x)) = 0 Note that c2 = 0, since otherwise c1 would also be zero, which would imply the functions are linearly independent. Thus plugging the expression into the equation for the derivative, we have fg−f g = 0 = ⇒W(f, g)(x) = 0, which is what we had to prove.
Example: Calculate the Wronskian for the functions f(x) = ex and g(x) = 2ex to determine if they are linearly independent.
W(ex, 2ex) = ex(2ex) −ex(2ex) = 0, Since the Wronskian is equal to zero, the two functions are linearly dependent.
Example: Calculate the Wronskian for the functions f(x) = ex and g(x) = e−x to determine if they are linearly independent.
W(ex, e−x) = ex(e−x) −ex(−1e−x) = 1 −(−1) = 2, Since the Wronskian is not equal to zero, the two functions are linearly independent.
4 |
9016 | https://math.stackexchange.com/questions/339605/prove-ax2bxc-0-has-no-rational-roots-if-a-b-c-are-odd | quadratics - Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd - Mathematics Stack Exchange
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Prove ax^2+bx+c=0 has no rational roots if a,b,c are odd
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\begingroup
If a,b,c are odd, how can we prove that ax^2+bx+c=0 has no rational roots?
I was unable to proceed beyond this: Roots are x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} and rational numbers are of the form \frac pq.
quadratics
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edited Mar 24, 2013 at 15:32
Michael Hardy
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asked Mar 24, 2013 at 14:08
ABCABC
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1 \begingroup This is incorrect, as roots are real if b^2\geq 4ac Maybe you mean rational?\endgroup Ishan Banerjee –Ishan Banerjee 2013-03-24 14:10:55 +00:00 Commented Mar 24, 2013 at 14:10
16 \begingroup You are like saying: I give you this problem. DO IT!\endgroup JSCB –JSCB 2013-03-24 14:12:01 +00:00 Commented Mar 24, 2013 at 14:12
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You should think about factoring the polynomial rather than finding its roots. If ax^2 + bx + c has a rational root, then its other root must also be rational, and then it factors in some way like this: ax^2 + bx + c = (Ax + B)(Cx + D) \quad A,B,C,D \in \mathbb{Z} Then a = AC and c = BD, so if both are odd, then all of A,B,C,D are odd. But then we also have b = AD + BC, which is the sum of odd integers, and therefore is even.
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answered Mar 25, 2013 at 3:35
Ryan ReichRyan Reich
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\endgroup
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2 \begingroup Interestingly, this gives a quite straightforward proof that \sqrt5 is irrational, by considering x^2 + 3x + 1, whose roots by the quadratic formula are (-3 \pm \sqrt5)/2 and irrational by the above. This technique seems to be fairly limited, though: you certainly can't make it work for the much more famous \sqrt2 (why?).\endgroup Ryan Reich –Ryan Reich 2015-11-22 17:18:45 +00:00 Commented Nov 22, 2015 at 17:18
\begingroup How are we claiming that A, B, C, D are all necessarily integers? If we factor a out of the expression in its factored form, we'll have a resultant factored form of (x-p)(x-q) for the equation as per given information. Then for A, B, C, D to be integers, we must have the condition that multiplying a to the factored form in some ratio (perhaps multiplying a/k and k to the 2 factors) gives integer A, B, C, D. How is this shown if possible? Otherwise how is it safe to assume A, B, C, D are integers and not fractions?\endgroup BeBlunt –BeBlunt 2021-06-20 09:25:49 +00:00 Commented Jun 20, 2021 at 9:25
1 \begingroupGauss's lemma says that the polynomial has to factor over the integers if it does over the rationals.\endgroup Ryan Reich –Ryan Reich 2021-06-21 14:43:39 +00:00 Commented Jun 21, 2021 at 14:43
\begingroup Could you please complete your argument citing other necessary corollaries if required? All I know is Gauss's Lemma deals with irreducibility and not reducibility. Is it implicit somehow? Kindly explain, it'd mean a lot.\endgroup BeBlunt –BeBlunt 2021-07-04 22:42:38 +00:00 Commented Jul 4, 2021 at 22:42
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Consider a quadratic equation of the form a\cdot x^2 + b\cdot x + c = 0. The only way, it can have rational roots IFF there exist two integers \alpha and \beta such that
\alpha \cdot \beta = a\cdot c\tag1\alpha + \beta = b\tag2 Explanation\left{ \begin{align} if\,\alpha\cdot \beta &= a\cdot c,\ \frac{\alpha}{a} &= \frac{c}{\beta}\ a\cdot x^2 + b\cdot x + c& = a\cdot x^2 + (\alpha + \beta)\cdot x + c\ & = a\cdot x^2 + \alpha\cdot x + \beta\cdot x + c\ & =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{c}{\beta}))\ & =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{\alpha}{a}))\ & =(x+\frac{\alpha}{a})\cdot (a\cdot x + \beta)\ &\text {As a Quadratic equation has only two roots,}\ &\text {there would be no other way to factorize the equation} \end{align}\right. \text{Reason }\alpha,\beta\in\mathbb{Z}\begin{cases} \text{Given a Ring R, with two operations }\left{⋅,+\right},\text{ on }\mathbb{Q}\ \text{and if } \alpha,\beta \in \mathbb{Q},\alpha\cdot \beta \in \mathbb{Z},\alpha+\beta\in\mathbb{Z}\ \Rightarrow\alpha,\beta \in \mathbb{Z}\ \text{ where } \mathbb{Z}\subset\mathbb{Q} \end{cases} If (a,b,c) are odd then \alpha \cdot \beta is odd and \alpha + \beta is odd, but you cannot have two integers whose product and sum are odd.
So by contradiction we prove that the equation cannot have rational roots
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edited Mar 25, 2013 at 8:08
answered Mar 24, 2013 at 14:14
AbhijitAbhijit
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1 \begingroup Could you elaborate on your IFF statement?\endgroup Christian Blatter –Christian Blatter 2013-03-24 16:00:55 +00:00 Commented Mar 24, 2013 at 16:00
2 \begingroup Aaha I noted this down you cannot have two integers whose product and sum are odd.\endgroup ABC –ABC 2013-03-24 16:35:57 +00:00 Commented Mar 24, 2013 at 16:35
2 \begingroup The proof is incomplete (or incorrect), since the claim that \alpha and \beta are integers (vs. rationals) is not justified. For further explanation see my second answer (the one using the AC-method).\endgroup Math Gems –Math Gems 2013-03-25 00:47:52 +00:00 Commented Mar 25, 2013 at 0:47
1 \begingroup@MathGems: I thought it was obvious. I have added few more lines in support of this.\endgroup Abhijit –Abhijit 2013-03-25 03:13:45 +00:00 Commented Mar 25, 2013 at 3:13
1 \begingroup I see that you've removed the erroneous proof of the claim and replaced it by the claim itself - still, alas, unjustified. Why don't you simply remark that the claim (if two rationals have integral sum and product then they are both integers) is an immediate consequence of the Rational Root Test. Then your proof will be both correct and complete.\endgroup Math Gems –Math Gems 2013-03-26 17:55:26 +00:00 Commented Mar 26, 2013 at 17:55
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Suppose that a,b,c are odd. ax^2+bx+c=0 has rational roots iff the discriminant is the square of an integer. That is, there is an integer d so that d^2=b^2-4ac. Since a,b,c are odd, d must also be odd.
Note that the right hand side of (b-d)(b+d)=4ac\tag{1} has exactly two factors of 2. However, since b and d are both odd, 2d\equiv2\pmod{4} and so one of b-d and b+d is 0\bmod{4} and the other is 2\bmod{4}. Thus, the left hand side of (1) has at least three factors of 2. Contradiction.
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answered Mar 24, 2013 at 17:11
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By Gauss's lemma It suffices to consider the domain of x as the integers. If x is even or odd then ax^{2}+bx+c=x(ax+b)+c is odd hence not equal to 0.
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answered Dec 16, 2016 at 6:42
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2 \begingroup@Henrik "This does not provide an answer to the question." What? This very much provides an answer. That the answer is brief (the probable cause for its appearance in the Review list) should be irrelevant. And the fact that the argument was already proposed in other answers is another matter.\endgroup Did –Did 2016-12-16 07:14:47 +00:00 Commented Dec 16, 2016 at 7:14
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Hint\ By the Rational Root Test, any rational root is integral, hence it follows by
TheoremParity Root Test\ A polynomial \rm\:f(x)\: with integer coefficients has no integer roots if its constant coefficient and coefficient sum are both odd.
Proof\ The test verifies that \rm\ f(0) \equiv 1\equiv f(1)\ \ (mod\ 2)\:,\ i.e. that \rm\:f(x)\: has no roots modulo 2, hence it has no integer roots. \ QED
This test extends to many other rings which have a "sense of parity", i.e. an image \cong \Bbb Z/2,\: for example, various algebraic number rings such as the Gaussian integers.
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edited Apr 13, 2017 at 12:19
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\begingroup The lead coefficient is not necessarily 1, so how does the Rational Root Test tell us that rational roots are integral?\endgroup robjohn –robjohn♦ 2013-03-26 18:32:39 +00:00 Commented Mar 26, 2013 at 18:32
\begingroup@robjohn There is a big pending edit to this post that will hopefully appear shortly. I've been waiting to see whether or not the accepted answer is edited for corrections. Many of the answers have gaps, so I plan to discuss the matter in general.\endgroup Math Gems –Math Gems 2013-03-26 18:44:46 +00:00 Commented Mar 26, 2013 at 18:44
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You can actually prove this in quite an elementary way without even knowing anything about the roots of the quadratic equation. Suppose, on the contrary, that we have rational root \dfrac {p}{q}. Then your equation is equivalent to ap^2+bpq+cq^2=0. You have that a,b,c are all odd. I will denote odd as an O and even as an E.
This breaks into four cases.
1) if p=O and q=E then you have O+E+E=O=0, which is impossible
2) if p=E and q=O then you have E+E+O=O=0, which is impossible
3) if p=O and q=O then you have O+O+O=O=0, which is impossible
4) if p=E and q=E then you have E+E+E=E=0. which could be possible (0 is an even number), so we treat this case below
Suppose p=2k and q=2l is a solution, then you have that (p,q) is a solution of the equation if and only if (k,l) is a solution, if k=2e and l=2f then you have that (p,q) is a solution if and only if (e,f) is a solution, proceeding in this way you can see that solution, if it exists, must be of the form (p,q)=(g2^r,h2^s) and this also cannot be a solution beacuse this reduces to one of the first 3 cases.
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answered Mar 24, 2013 at 17:00
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If \frac{p}{q} is a root then a\frac{p^2}{q^2}+b\frac{p}{q}+c=0\Rightarrow ap^2+bpq+cq^2=0. Now we may assume that p,q are not both even; a,b,c are odd, whence contradiction (p(ap+bq)=-cq^2 with p even c,q odd or q(bp+cq)=-ap^2 with q even a,p odd or ...).
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edited Mar 24, 2013 at 21:21
answered Mar 24, 2013 at 16:49
P..P..
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2 \begingroup In my opinion, this is the simplest and most novel of all the answers. But it's very poorly written. The problem is elementary, probably asked to a student who is new to proving things, so it would be very helpful to not only exemplify the "trick" that's used, but also how to write a good and complete proof the "Correct"(TM) way. Then we can upvote ;-)\endgroup Somo S. –Somo S. 2019-10-08 01:15:13 +00:00 Commented Oct 8, 2019 at 1:15
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Recall that the square of an odd number is always equivalent to 1 modulo 8.
If a,b,c are all odd, then
b^2 - 4 a c \equiv 5 \pmod 8
and thus, b^2 - 4ac cannot be the square of an integer (and thus cannot be the square of a rational number). Therefore the roots are irrational.
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answered Mar 25, 2013 at 4:03
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\begingroup Now that I've written this, I see a downvoted and a deleted answer that I had overlooked employ the same observation. I assume the reason is presentation, so I'll leave my answer in. Although... I don't understand the issues with the answer that got deleted.\endgroup user14972 –user14972 2013-03-25 04:10:55 +00:00 Commented Mar 25, 2013 at 4:10
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x= \dfrac{-b+ \sqrt{b^2-4ac}}{2a} or \dfrac{-b- \sqrt{b^2-4ac}}{2a}
If x=\dfrac{p}{q} \implies b^2-4ac=k^2, and k is odd. (odd-even=odd).
Considering a,b,c odd.
k^2 \equiv 1 \mod 8
b^2 \equiv 1 \mod 8
4ac \equiv 4 \mod 8
b^2-4ac=-3 \mod 8 \implies 5 \mod 8, a contradiction.(Either of a or c has to be even), since k^2 \equiv 1 \mod 8
\therefore You don't get rational roots when all are odd.
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edited Mar 24, 2013 at 15:14
answered Mar 24, 2013 at 14:18
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1 \begingroup Would you need a constant a at the front of the left hand side of that equation there? i.e. a(x-\frac{p_1}{q_1})(x-\frac{p_2}{q_2})\endgroup Sp3000 –Sp3000 2013-03-24 14:21:22 +00:00 Commented Mar 24, 2013 at 14:21
2 \begingroup You asked "Why the downvote?" On which version? We are at version 5, version 1 was not an answer, in later versions which parts were a proof and which parts were making circles around the question was unclear, and now the pre-Aliter part is not useful... All in all, you might want to use a draftbook and to post only when you have a full solution.\endgroup Did –Did 2013-03-24 15:11:21 +00:00 Commented Mar 24, 2013 at 15:11
\begingroup As I said, please change your modus operandi.\endgroup Did –Did 2013-03-24 16:36:24 +00:00 Commented Mar 24, 2013 at 16:36
\begingroup Well, my previous answer needed a little modification.Right? Even I showed sum and product can't be both odd. But my explanation was flawed. And which modus operandi are you talking about?\endgroup Inceptio –Inceptio 2013-03-24 17:23:32 +00:00 Commented Mar 24, 2013 at 17:23
\begingroup I thought that was clear enough but let me repeat it: stop posting half digested "solutions", use a draftbook, post only when you have a full solution. (And as a bonus: use the @ thing when you comment on somebody's comment.)\endgroup Did –Did 2013-03-25 23:23:40 +00:00 Commented Mar 25, 2013 at 23:23
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Here is a proof that is elementary, i.e. requires no knowledge of modular arithmetic. First we use the AC method to reduce to a monic quadratic, i.e. one with leading coefficient = 1.
\rm\begin{eqnarray} 0\: =\ f(x)\, = &&\rm\ \, a\,x^2+\,b\,x+c\ \rm \Rightarrow\ \ 0 = a f(x)\, = &&\rm! (ax)^2! + b (ax) + ac\ \rm \Rightarrow\ \ 0 =\, F(X) = &&\rm\ \ \, X^2 \,+\ \color{#C00}b\,\ X\ \,+ \color{#0A0}{ac},\quad X = ax\end{eqnarray}
Suppose \rm\,f(x_i)=0\, for \rm\:\color{brown}{x_i\in\Bbb Q}.\: Then \rm\,F(X_i)=0\, for \rm\: X_i = a\, x_i\in\Bbb Q.\: By the Rational Root Test, the rational roots \rm\,X_i are integers. Since their product \rm = \color{#0A0}{ac}\: is odd, the roots are both odd, so their sum is even. This contradicts: by Vieta, the root sum \rm\, = -\color{#C00}b\, is odd, by hypothesis. Hence \rm\:\color{brown}{x_i \not\in \Bbb Q}.
Remark\ This yields a conceptual view of the calculations in Abhijit's answer (which, alas, is incomplete, since it does not justify the reduction from rational to integer roots).
The AC-method generalizes to higher degree polynomials (see the above-linked answer). It is intimately connected to various refinement-based views of unique factorization.
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edited Apr 13, 2017 at 12:21
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answered Mar 25, 2013 at 0:46
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Proof by contradiction, by someone who just started doing proofs, using a similar contradiction proof to "\sqrt 2 is irrational". Edit note: You do not need to know mods or virtually anything other than introductory proof by contradiction to do this method.
Let a,b,c be odd. Suppose there is a rational number p/q that can be plugged in for x such that there is a rational solution. Assume p, q have no common factors. (This is where the contradiction will occur!). 0 = ax^2 + bx + c has roots x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} Plugging in p/q for x gives \frac{p}{q} = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} Since b is odd, b^2 is the product of odd and odd, meaning b^2 is odd. Since a and c are odd, the product of a and c, odd and odd, is odd. Therefore 4ac is the product of even and odd which is even. So, b^2 - 4ac is odd - even, which is odd. \sqrt{b^2-4ac} must be a square number for there to be a rational solution. Therefore n^2 = \sqrt{b^2-4ac}^2. So, n^2 = {b^2-4ac}. We know b^2-4ac is odd, so n^2 is odd. If n^2 is odd then n is odd, and if n is odd then the entire \sqrt{b^2-4ac} is odd. So, -b\pm\sqrt{b^2-4ac} is odd plus odd, which is even. Almost there, hang in there! 2a is even by definition. If p = -b \pm\sqrt{b^2-4ac} and q = 2a, then p is even and q is even. If p,q are even, they are by definition divisible by 2. Therefore, p,q have a common factor. CONTRADICTION!
Let me know if this works:).
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edited Sep 29, 2023 at 7:53
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Method 1:
Assume x is a root that can be written as \frac{p}q, where \gcd(p,q) = 1
\frac{ap^2}{q^2} + \frac{bp}{q} + c = 0 Multiplying by q^2, we obtain: ap^2 + bpq + cq^2 = 0 \tag{1} This maybe looked at as: p(ap + bq) = -cq^2 \tag{2} q(cq + bp) = -ap^2 \tag{3} In (2), p divides the LHS, and thus must divide the RHS, but \gcd(p,q) = 1 \implies p | c. Similarly q | a. Also since a,c are odd, p,q must be odd.
Then, in (1) ap^2 is a product of odd numbers and therefore odd. Similarly, every term is odd. Then we have a sum of 3 odd numbers is even ( 0 is even ), which is never possible. Thus by contradiction, there is no rational root
Method 2:
Letting the two roots be x and y, written as \frac{p}{q} and \frac{r}{s}.
By Rational Root Theorem, we have p,r | c and also q,s | a. Thus p,q,r, s are all odd, as they divide odd integers.
Then by Vietta's Formula, -\frac{b}{a} = x + y = \frac{ps+rq}{qs} bqs = -a(ps+rq) Here, ps,rq are odd and their sum ps + rq is even. Thus the RHS is even but the LHS is odd. Contradiction, and therefore no root is rational
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answered Sep 27, 2024 at 7:37
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Assume the your root z is rational (= q/r where at at most one of q, r is even)
Re-write the equation with z on the left hand size and 1/z c/a - b/a on the other. Substitute q/r for z, and put the left hand side into a fraction.
enter image description here
If neither q or r are even then numerators don't match (q != rc - qb because odd != odd + odd) and if only one is even then the denominator doesn't match (r != qa because odd != even odd and even != odd odd)
Therefore, z is not rational.
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answered Oct 9, 2019 at 1:52
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9017 | https://web.stanford.edu/class/rad229/Notes/A1-Background1.pdf | B.Hargreaves - RAD 229 Section A1 Background (~EE369B) • “Magnetic Resonance Imaging” D. Nishimura • Overview of NMR • Hardware • Image formation and k-space • Excitation k-space • Signals and contrast • Signal-to-Noise Ratio (SNR) • Pulse Sequences 13 B.Hargreaves - RAD 229 Section A1 MRI: Basic Concepts 14 Excitation Precession (Reception) Relaxation (Recovery) Static Magnetic Field (B0) 1H N S Gradients (Relative Precession) B0 B1 B0 B.Hargreaves - RAD 229 Section A1 Precession and Relaxation • Relaxation and precession are independent. • Magnetization returns exponentially to equilibrium: • Longitudinal recovery time constant is T1 • Transverse decay time constant is T2 Precession Decay Recovery 15 B.Hargreaves - RAD 229 Section A1 Magnetic Resonance Imaging (MRI) • Polarization • Excitation • Signal Reception • Relaxation 16 B.Hargreaves - RAD 229 Section A1 MRI Hardware • Strong Static Field (B0) ~ 0.5-7.0T • Radio-frequency (RF) field (B1) ~ 0.1uT • Transmit, often built-in • Receive, often many coils • Gradients (Gx, Gy, Gz) ~ 50-80 mT/m 17 B.Hargreaves - RAD 229 Section A1 B0: Static Magnetic Field • Goal: Strong AND Homogeneous magnetic field • Typically 0.3 to 7.0 T • Resonance prοportional to B0 : γ/2π = 42.58 MHz/T • Superconducting magnetic fields - always on • ~1000 turns, 700 A of current • Passively shimmed by adjusting coil locations • The following increase with with B0: • Polarization, Larmor Frequency, Spectral separation, T1 • RF power for given B1 • B0 variations due to susceptibility, chemical shift 18 B.Hargreaves - RAD 229 Section A1 B0: The “Rotating” Coordinate Frame • Usually demodulate by Larmor frequency to “baseband” • Also called the rotating frame 19 B.Hargreaves - RAD 229 Section A1 B1+: RF Transmit Field • Goal: Homogeneous rotating magnetic field • Typically up to about 25 uT (Amplifier, SAR limits) • Requires varying power based on subject size • Dielectric effects cause B1+ variations at higher B0 • Amplifier power: kW to tens of kW • Specific Absorption Rate (SAR) Limits: • Power proportional to B02 and B12 • Goal is to limit heating to <1∘C 20 B.Hargreaves - RAD 229 Section A1 B1-: RF Receive • Goal: High sensitivity, spatially limited, low noise • “Birdcage” coils • Uniform B1- but single channel • Surface coils • Varying B1- but high sensitivity • Coil arrays • Multiple channels with Varying B1- • Allows some spatial localization: Parallel Imaging 21 B.Hargreaves - RAD 229 Section A1 RF Coils 22 B.Hargreaves - RAD 229 Section A1 Receiver System • 500 to 1000 k samples/s • Complex sampling • Low-pass filter capability • Typically 32-128 channels • Time-varying frequency and phase modulation (Typically single-channel) 23 B.Hargreaves - RAD 229 Section A1 Gradients • Goal: Strong, switchable, linear Bz variation with x,y,z • Peak amplitude ~ 50-80 mT/m (~ 200A) • Switching 200 mT/m/ms (~1500 V) • Limits: • Amplifier power, heating, coil heating • “dB/dt” limitation due to peripheral nerve stimulation • Switching induces Eddy Currents • Concomitant terms (Bx and By variations) • Non-linearities (often correctable) 24 B.Hargreaves - RAD 229 Section A1 Gradient Waveforms • Mapping of position to frequency, slope = γG • Typically waveforms are trapezoidal • Constant amplitude and slew-rate limits 25 Position Frequency γGread Time Amplitude B.Hargreaves - RAD 229 Section A1 Shims • Goal usually to make B0 more uniform with subject • Center frequency • Linear shims (Up to ~1% offset to gradients) • Higher-order (HO) shims (Spherical Harmonics) • Shim arrays, Shim+RF (Current Research) • Usually HO shims not dynamically switchable 26 B.Hargreaves - RAD 229 Section A1 Review Questions • Which field is the “receive” field? • B1- • Which field is always on? • B0 • What receive bandwidth corresponds to 500,000 samples/second? • ±250 kHz • Why might small surface coils (or arrays) be useful? • High sensitivity, Low noise, Spatially limiting 27 B.Hargreaves - RAD 229 Section A1 Image Formation and k-space • Gradients and phase • Signal equation • Sampling / Aliasing • Parallel Imaging • Many reconstruction methods in EE369C 28 B.Hargreaves - RAD 229 Section A1 x x x Gradient Strength and Sign Can control both amplitude and duration Positive Gradient Negative Gradient Double Strength Half-Duration 29 B.Hargreaves - RAD 229 Section A1 Gradient Along Both x and y Gx Gy x y Can also vary along z 30 B.Hargreaves - RAD 229 Section A1 Ribbon Analogy • Gradients induce “phase twist” • Twist has a number of cycles and a “sign” • Twist can be along any direction x 31 B.Hargreaves - RAD 229 Section A1 Gradients and Phase • Control gradient amplitude and duration • Can control frequency: Frequency = γ(Gx x + Gy y) • Can “encode” phase over duration t Angle = γt (Gx x + Gy y + Gz z) • Generally: φ = γ(x Z Gxdt + y Z Gydt) 32 φ = γ(x Z Gxdt + y Z Gydt) What are the units of Frequency and Angle (φ) here?
B.Hargreaves - RAD 229 Section A1 Signal Equations • For a single spin: • Represent as exponential: • Sum over many spins: • Signal equation: 33 φ = γ(x Z Gxdt + y Z Gydt) kx,y(t) = γ 2⇡ Z t 0 Gx,y(⌧)d⌧ s(t) = FT[⇢(x, y)]|kx(t),ky(t) s = e−iγ(x R Gxdt+y R Gydt) s = Z 1 −1 Z 1 −1 ⇢(x, y)e−iγ(x R Gxdt+y R Gydt)dxdy s = Z 1 −1 Z 1 −1 ⇢(x, y)e−2⇡i(kxx+kyy)dxdy B.Hargreaves - RAD 229 Section A1 Fourier Transform in MRI • Given M(k) at enough k locations, we can find ρ(r) • It does not matter how we got to k!
M(k) ρ(r) Fourier Transform 34 s(t) = FT[⇢(x, y)]|kx(t),ky(t) What are the units of kx(t) and ky(t) ?
B.Hargreaves - RAD 229 Section A1 Fourier Encoding and Reconstruction k-space ky kx Gradient-induced Phase x k-space ky kx x Sum over k-space Sum over image 35 Spatial Harmonic Encoding Reconstruction B.Hargreaves - RAD 229 Section A1 k-space: Spatial Frequency Map ky kx 36 k-space In terms of pixel-width, what is the width of k-space?
B.Hargreaves - RAD 229 Section A1 Image Formation and Sampling Readout Gradient time Phase-Encode Gradient time k-space kphase kread Readout Direction 37 B.Hargreaves - RAD 229 Section A1 k space Extent and Image Resolution Data Acquisition “k” space Image Space Fourier Transform 38 ∆x = 1/(2kmax) B.Hargreaves - RAD 229 Section A1 Sampling and Field of View • Sampling density determines FOV • Sparse sampling results in aliasing Readout Phase-Encode kphase kread kphase kread FOV FOV 39 FOV = 1/∆ky B.Hargreaves - RAD 229 Section A1 Phase-Encoding with Two Coils kx ky k-space kx ky kx ky 40 B.Hargreaves - RAD 229 Section A1 Readout Parameters • Bandwidth linked to readout • “half-bandwidth” (GE) = 0.5 x sample rate • Same as Filter bandwidth (baseband) • Pixel-bandwidth often useful 41 Position Frequency Full Bandwidth Bandwidth per Pixel FOV Pixel γGread BWpix = γGread∆x BWhalf = γGreadFOV/2 B.Hargreaves - RAD 229 Section A1 Imaging Example • Desired Image Parameters: • 256 x 256, over 25cm FOV • (±)125 kHz bandwidth • What are the... • Sampling period? • Readout duration? • Gradient strength? • Bandwidth per pixel? • k-space extent?
42 • 1/(2125kHz) = 4µs • 4µs 256 = 1ms • 250kHz / 0.25m / 42.58kHz/mT • 23 mT/m (2.3 G/cm) • 250kHz/256 pix = ~ 1kHz/pixel • 0.5 / 1mm = 0.5 mm-1 = 5cm-1 B.Hargreaves - RAD 229 Section A1 2D Multislice vs 3D Slab Imaging • Shorter scan times, reduced motion artifact • Continuous coverage • Thinner slices, reformats 2D 3D 43 B.Hargreaves - RAD 229 Section A1 Imaging Summary • Gradients impose time-varying linear phase • k-space is time-integral of gradients • k-space samples Fourier Transform to/from image • Density of k-space <> FOV (image extent) • Extent of k-space <> Resolution (image density) • 3D k-space is possible • Parallel imaging uses coils to extend FOV 44 B.Hargreaves - RAD 229 Section A1 Excitation • General principles of excitation • Selective Excitation with gradients • Relationships for slice excitation • Excitation k-space • Much more covered in EE469B 45 B.Hargreaves - RAD 229 Section A1 Excitation: B1 Field • Direction of B1 is perpendicular to B0 • Magnetization precesses about B1 • Turn on and off B1 to “tip” magnetization • Problem: We can’t turn off B0! • Precession still around B0 46 B.Hargreaves - RAD 229 Section A1 Excitation B1 Magnetization B0 • Magnetization precesses about net field (B0+B1) • B1 << B0 • Must “tune” B1 frequency to Larmor frequency Static B1 Field Rotating B1 Field 47 B.Hargreaves - RAD 229 Section A1 Excitation: Rotating Frame • “Excite” spins out of their equilibrium state. • B1 << B0 • Transverse RF field (B1) rotates at γB0 about z-axis.
B1 Magnetization B0 Rotating Frame, “On resonance” Static Frame 48 B.Hargreaves - RAD 229 Section A1 Selective Excitation Position Slope = 1 γ G Frequency 49 B.Hargreaves - RAD 229 Section A1 Selective Excitation B1 Frequency Magnitude Time RF Amplitude Position Slope = 1 γ G Larmor Frequency + + = 50 Slice width = BWRF / γGz Slice center = Frequency / γGz B.Hargreaves - RAD 229 Section A1 Excitation Example • Given a 2 kHz RF pulse bandwidth, and desired • 5mm thick slice • Slices at -2cm, 0, 2cm • What are the... • Gradient strength? (γ/2π)Gz • Excitation frequencies? • Thinnest slice possible with 50mT/m max gradients?
51 • 2kHz/5mm = 400kHz/m ~ 9.4 mT/m (0.94 G/cm) • BW/slice = 2kHz/5mm, so -8, 0, 8 kHz • (9.4/50)5mm ~1mm slice B.Hargreaves - RAD 229 Section A1 Excitation k-Space • Excitation k-space goes backwards from end of RF/ gradient pair: • Excited profile = Fourier Transform of excitation k-space • Central flip angle = area under pulse (may be zero!): 52 ke(t) = −γ 2⇡ Z T t G(⌧)d⌧ kr(t) = γ 2⇡ Z t 0 G(⌧)d⌧ ↵= γ Z B1(⌧)d⌧ B.Hargreaves - RAD 229 Section A1 Excitation Example • For a 1ms, constant RF pulse of amplitude 10µT … • What is the flip angle? • How does RF energy change if the duration is halved and amplitude doubled? 53 • (42.58 kHz/mT)(0.01mT)(1ms) = 0.4258 cycles = 153º • Doubles - (2A)2(T/2) = 2(A2T) B.Hargreaves - RAD 229 Section A1 Signals and Contrast • Simple Bloch Equation Solutions • Basic contrast mechanisms: T1, T2, IR, Steady-State 54 T2-Weighted T2-w FLAIR T1-w FLAIR Gradient Echo Diffusion-Weighted Apparent Diffusion Coefficient (ADC) B.Hargreaves - RAD 229 Section A1 Signals and Contrast • Bloch Equation Solutions (Relaxation): • Rotations due to excitation: 55 Mxy(t) = Mxy(0)e−t/T2 Mz(t) = M0 + [Mz(0) −M0]e−t/T1 M 0 xy = Mxy cos ↵+ Mz sin ↵ M 0 z = Mz cos ↵−Mxy sin ↵ Mxy M0 0 Mz M0 0 time time B.Hargreaves - RAD 229 Section A1 Echo Time (TE): T2 weighting RF Signal 90º 90º 1 0 Mz Short TE Long TE 56 TE = Time from RF to “echo” B.Hargreaves - RAD 229 Section A1 T2 Contrast Dardzinski BJ, et al. Radiology, 205: 546-550, 1997.
Signal Echo Time (ms) TE = 20 TE = 40 TE = 60 TE = 80 57 B.Hargreaves - RAD 229 Section A1 Repetition Time (TR) : T1 Weighting Each excitation starts with reduced Mz RF Signal 90º 90º 1 0 Mz 90º Mz Mxy 58 B.Hargreaves - RAD 229 Section A1 T1-Weighted Spin Echo Signal Time Signal Time Short Repetition Long Repetition Joint Fluid Bone 59 B.Hargreaves - RAD 229 Section A1 Short TR Long TR Short TE Incomplete Recovery Minimal Decay T1 Weighting Full Recovery Minimal Decay Proton Density Weighting Long TE Incomplete Recovery Signal Decay Mixed Contrast (Not used much) Full Recovery Signal Decay T2 Weighting Basic Contrast Question (TE, TR) 60 Images Courtesy of Anne Sawyer B.Hargreaves - RAD 229 Section A1 Inversion-Recovery 180º 180º RF Signal 1 -1 0 • Fat suppression based on T1 • Short TI Inversion Recovery (STIR) TI 61 B.Hargreaves - RAD 229 Section A1 Long Inversion Time (TI) - FLAIR Long TI suppresses fluid signal 62 B.Hargreaves - RAD 229 Section A1 Signal Question • Inversion Recovery Sequence: • TR = 1s, TI = 0.5s, TE=50ms • What is the signal for T1=0.5s, T2=100ms?
63 180º 180º RF TI TE TR 90º • Signal (Mxy) decays to 0 • Mz does not fully recover • exp(-1) = 037 • Mz is 0.63 M0 before 180º • At TI, Mz-M0 = 1.63 exp(-1) • Mz ~ 0.4 M0 • T2 decay ~ exp(-0.5) ~ 0.6 • Signal = 0.4 x 0.6 = 0.24 M0 B.Hargreaves - RAD 229 Section A1 Steady-State Sequences • Repeated sequences always lead to a “steady state” • Sometimes includes equilibrium (easier) • Otherwise trace magnetization and solve equations • Example: Small-tip, TE=0 64 Mz(TR) = M0 + [Mz(TE) −M0]e−T R/T1 Mz(TE) = Mz(TR) cos ↵ Combining...
...
“TR” “TE” Mz(TR) = M0 1 −e−T R/T1 1 −e−T R/T1 cos ↵ B.Hargreaves - RAD 229 Section A1 Summary ~ Background I • Overview of NMR • Hardware • Image formation and k-space • Excitation k-space • Signals and contrast 65 |
9018 | https://www.quora.com/What-is-the-difference-between-NAD-and-NADP | What is the difference between NAD and NADP? - Quora
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Biochemistry
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What is the difference between NAD and NADP?
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NAD (Nicotinamide Adenine Dinucleotide) and NADP (Nicotinamide Adenine Dinucleotide Phosphate) are both coenzymes that play crucial roles in cellular metabolism, but they have distinct functions and structures:
Structure:
NAD: Composed of two nucleotides joined by their phosphate groups. It has a structure that includes nicotinamide, adenine, and two phosphate groups.
NADP: Similar to NAD, but it has an additional phosphate group attached to the 2' position of the ribose ring of the adenine nucleotide.
Functions:
NAD:
Primarily involved in catabolic reactions, such as glycolysis and the citric acid
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NAD (Nicotinamide Adenine Dinucleotide) and NADP (Nicotinamide Adenine Dinucleotide Phosphate) are both coenzymes that play crucial roles in cellular metabolism, but they have distinct functions and structures:
Structure:
NAD: Composed of two nucleotides joined by their phosphate groups. It has a structure that includes nicotinamide, adenine, and two phosphate groups.
NADP: Similar to NAD, but it has an additional phosphate group attached to the 2' position of the ribose ring of the adenine nucleotide.
Functions:
NAD:
Primarily involved in catabolic reactions, such as glycolysis and the citric acid cycle.
Acts as an electron carrier, accepting electrons during oxidation reactions and becoming reduced to NADH.
Plays a key role in energy production by facilitating the transfer of electrons to the electron transport chain.
NADP:
Primarily involved in anabolic reactions, such as the Calvin cycle in photosynthesis.
Also acts as an electron carrier, but is typically involved in reductive biosynthesis and the synthesis of fatty acids and nucleic acids.
Is reduced to NADPH, which provides the reducing power needed for biosynthetic reactions.
Summary:
In summary, while both NAD and NADP are important for cellular metabolism, NAD is mainly involved in energy production and catabolic processes, whereas NADP is more associated with biosynthetic pathways and anabolic reactions.
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Krupa Bharadwaj
Studied at SAI PUC (Graduated 2020) · Author has 173 answers and 374.5K answer views
·5y
Hello…💐
There are many differences between NAD and NADP…I will list a few differences here, as follows…..
NAD works in catabolic reactions whereas NADP works in anabolic reactions.
After reduction, NAD forms NADH2, while, NADP forms NADPH2.
According to my observations in my textbook, NAD appears in Cellular Plant Respiration. But, NADP appears in Plant Photosynthesis.
NADP has an extra Phosphate group which is not present in NAD.
NADP stands for Nicotinamide Adenine Dinucleotide Phosphate. NAD stands for Nicotinamide Adenine Dinucleotide.
The oxidized form of NAD is NAD+. The oxidised form of NADP
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Hello…💐
There are many differences between NAD and NADP…I will list a few differences here, as follows…..
NAD works in catabolic reactions whereas NADP works in anabolic reactions.
After reduction, NAD forms NADH2, while, NADP forms NADPH2.
According to my observations in my textbook, NAD appears in Cellular Plant Respiration. But, NADP appears in Plant Photosynthesis.
NADP has an extra Phosphate group which is not present in NAD.
NADP stands for Nicotinamide Adenine Dinucleotide Phosphate. NAD stands for Nicotinamide Adenine Dinucleotide.
The oxidized form of NAD is NAD+. The oxidised form of NADP is NADP+.
As for now, my dear friend, I only could mention these differences. I will update the answer if I found out any new answers. But, I hope this answer would serve ur doubts well.
Thanks…Cheers♥️👍
Upvote and Follow me on Quora…😇😍
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Eren Brooks
BSc in Biomedical Science, University of Kent (Graduated 2022)
·Updated 6y
The presence of the phosphate group in NADP doesn’t affect the redox capabilities of the molecule at all (as the phosphate group is too far away from where this happens, I believe).
The phosphate group on NADP changes the shape of the molecule slightly, and this different shape means NADP can operate as a coenzyme to a different set of enzymes than NAD.
But why the separation? It’s because cells need to independently keep an eye on both catabolic and anabolic reactions; The quantities of these oxidative agents— identical but for the phosphate group— needs to be regulated by the cell, in order to
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The presence of the phosphate group in NADP doesn’t affect the redox capabilities of the molecule at all (as the phosphate group is too far away from where this happens, I believe).
The phosphate group on NADP changes the shape of the molecule slightly, and this different shape means NADP can operate as a coenzyme to a different set of enzymes than NAD.
But why the separation? It’s because cells need to independently keep an eye on both catabolic and anabolic reactions; The quantities of these oxidative agents— identical but for the phosphate group— needs to be regulated by the cell, in order to regulate metabolism properly.
Crucially, then; NAD is for Catabolic reactions, NADP is for Anabolic reactions. Other than that, they are identical.
So I think the presence of the phosphate group is for discrimination purposes only— for the cell to regulate the quantities of each, and control how much is being made at any one time so it can control anabolic reactions and catabolic reactions separately.
you may find more information here ( NADH vs. NADPH: Where is each one used and why that instead of the other? ), and the chapter link at the bottom of the answer there is very helpful in understanding this. :)
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Siddesh Thorat
Studied Biochemistry at University of Mumbai
·7y
NAD :- nicotinamide dinucleotide.
NADP :- nicotinamide dinucleotide phosphate.
The only difference between NAD and NADP is only of single P. i.e. the phosphate group which is attached to NAD. NADP+differs from NAD in the presence of an additional phosphate group on the 2' position of the ribose ring that carries the adenine moiety.
In every reaction whether it is plant or animal the oxidation and reduction process depends on the reaction or the nature of reaction. It's similar to ATP - ADP reaction, where ATP reduces to ADP and vice versa. Similarly it goes the same with NAD and NADP reaction.
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Eren Brooks
BSc in Biomedical Science, University of Kent (Graduated 2022)
·6y
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What's the difference between NADH and NADPH?
Okay first of all, I'm going to link you to my answer on the difference between NAD and NADP from a while ago because at its core, the differences between those molecules and these molecules are the same.
NADH is just a reduced form of NAD, meaning it has gained the [e¯] from the H atom it has bonded with. It acts as an election carrier in Respiration for example, in the exact same way that NADPH is a reduced version of NADP, and serves to carry electrons in Photosynthesis.
What I want you to understand from that is that functionally, NADH and NAD P H do the exact same things. They just belong to
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Okay first of all, I'm going to link you to my answer on the difference between NAD and NADP from a while ago because at its core, the differences between those molecules and these molecules are the same.
NADH is just a reduced form of NAD, meaning it has gained the [e¯] from the H atom it has bonded with. It acts as an election carrier in Respiration for example, in the exact same way that NADPH is a reduced version of NADP, and serves to carry electrons in Photosynthesis.
What I want you to understand from that is that functionally, NADH and NAD P H do the exact same things. They just belong to different metabolic pathways.
[TLDR from my linked answer above, then]; The addition of the Phosphate group to NADH (forming NADPH) allows the cell to differentiate two ‘sets’ of the reducing agent. NADH is used in Catabolic reactions (like Respiration, breaking down Glucose) and NADPH is used in Anabolic reactions (like Photosynthesis, building glucose).
The only reason for the difference is so that cells can monitor the levels of each independently, so that a drop in NADPH in Photosynthesis doesn't somehow impact NADH-dependent reactions in Respiration, for example.
Hope that makes sense. - ̗̀✰
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Nakul Rawat
M.sc. Biotechnology at South Asian University (2018–present)
·7y
NAD is generally used in the catabolic reactions togenerate energy equivalents, whereas NADP is used in anabolic reactions where energy is consumed.
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How is the function of NAD+ similar to that of NADP?
What are the side effects of NAD?
What are the main function of NADP, NAD and FAD?
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What is the difference between NADPH2 and NADP?
Fares Gafaf
Prodigious reader · Author has 3.3K answers and 4.7M answer views
·7y
NADP (nicotinamide adenine dinucleotide phosphate) is the phosphorylated form of NAD (nicotinamide adenine dinucleotide); NADP has an additional phosphate group which is absent in NAD.
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Tim Wiryaman
Biochemistry PhD Student
·10y
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Why do both NAD+/NADH and NADP/NADPH exist?
This rule isn't always true, but NAD+/NADH is used mostly in catabolic pathways and NADP+/NADPH is used mostly in anabolic pathways. One rationale for two distinct redox cofactors is to help prevent futile metabolic cycles.
For example, say NAD+/NADH was used in the same catabolic and anabolic pathway. In the catabolic pathway, NAD+ would be reduced to NADH, and in the anabolic pathway NADH would be oxidized to NAD+. As the catabolic pathway proceeds, [NADH] would increase, which would also lead to the "activation" the anabolic pathway via Le Chatelier's Principle. This is a futile cycle, beca
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This rule isn't always true, but NAD+/NADH is used mostly in catabolic pathways and NADP+/NADPH is used mostly in anabolic pathways. One rationale for two distinct redox cofactors is to help prevent futile metabolic cycles.
For example, say NAD+/NADH was used in the same catabolic and anabolic pathway. In the catabolic pathway, NAD+ would be reduced to NADH, and in the anabolic pathway NADH would be oxidized to NAD+. As the catabolic pathway proceeds, [NADH] would increase, which would also lead to the "activation" the anabolic pathway via Le Chatelier's Principle. This is a futile cycle, because the molecule of interest is being both broken down and synthesized simultaneously.
Thus, NAD+/NADH and NADP+/NADPH exist so that catabolic and anabolic pathways can be regulated independently of each other. Indeed the cell keeps the NAD+/NADH ratio high to always supply enough NAD+ for catabolic processes. In contrast, the NADP+/NADPH ratio is kept low so that there is enough NADPH to power anabolic reactions. This would not be possible with just one redox cofactor.
FAD/FADH2 has a different redox potential than NAD+/NADH or NADP+/NADPH so it's used in different types of reactions. NAD(P)+/NAD(P)H typically oxidizes alcohols to aldehydes/ketones or vice versa. FAD/FADH2 typically operates on alkenes/alkanes.
Source: Alberts Molecular Biology of the Cell, 4th Ed. Catalysis and the Use of Energy by Cells.
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William Halmeck
B.A. & M.A. in Biology, Western Connecticut State University (Graduated 2009) · Author has 5.6K answers and 9.6M answer views
·5y
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What is the difference between NAD+ and NADP?
They are the same, except NADP has a phosphate (R-H2PO4) added to one of the carbon rings where NAD+ has a hydroxyl group (R-OH). And, NAD+ is used as an oxidizing agent in cellular respiration, while NADP+ is used as an oxidizing in agent in photosynthesis.
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They are the same, except NADP has a phosphate (R-H2PO4) added to one of the carbon rings where NAD+ has a hydroxyl group (R-OH). And, NAD+ is used as an oxidizing agent in cellular respiration, while NADP+ is used as an oxidizing in agent in photosynthesis.
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Srinivasan
Former Teacher , · Author has 2K answers and 1.5M answer views
·Updated 1y
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What are NADPH and NAD+ and NADH? What are these made of and what are their functions?
Enzymes are biological catalyst mad up of protein which accelerate or retard biochemical reaction inside cell.
In many cases some non protein chemical substances are required to catalyse the bio chemical reaction.To catalyse the reaction these chemical substances are required for the enzymes to complete their reactions But These substance are not tightly bound with the enzyme and they are called co factors .or coenzyme .Majorty coenzyme are either vitamins are derivatives of vitamins .
NAD+ Nicotinamide adenine dinucleotide .It is reduced form of NADH+ H+. It contains vitamin Nicotinic acid (Nia
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Enzymes are biological catalyst mad up of protein which accelerate or retard biochemical reaction inside cell.
In many cases some non protein chemical substances are required to catalyse the bio chemical reaction.To catalyse the reaction these chemical substances are required for the enzymes to complete their reactions But These substance are not tightly bound with the enzyme and they are called co factors .or coenzyme .Majorty coenzyme are either vitamins are derivatives of vitamins .
NAD+ Nicotinamide adenine dinucleotide .It is reduced form of NADH+ H+. It contains vitamin Nicotinic acid (Niacin)
NADP+ (Nicotinamide Adenine di nucleotide Phosphate .It is reduced form is NADH+H+
FAD( FLAVIN Adenine Dinucleotide )
It is a derivative of vitamin-B2 that is Riboflavin often acts as prosthetic group.
They are the electron carriers from one electron acceptor to another electron acceptor .
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A Hammouda
Member of Bioinformatics Hub Net, Professor at Medical Biochemistry · Author has 756 answers and 142.7K answer views
·3y
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Are NAD and NADP ribonucleotides or deoxyribonucleotide?
These are dinucleotides, having two bases: adenine and nicotinamide, two ribose moieties, and two phosphates, with an extra phosphate in NADP.
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These are dinucleotides, having two bases: adenine and nicotinamide, two ribose moieties, and two phosphates, with an extra phosphate in NADP.
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Mot Arecuk
Studied Biochemistry (Graduated 2000) · Author has 283 answers and 167.3K answer views
·Updated 2y
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What is the difference between NADPH and NADPH2?
The first one is the reduced form of NADP⁺, the second is a nonsense. The same is true for its NAD counterparts. Let’s use NAD (NAD means generally both NAD⁺ and NADH) to explain the problem with the particular reaction of alcohol dehydrogenase:
H₃C–CH₂–OH + NAD⁺ → H₃C–CH=O + NADH + H⁺
The reaction is a redox process, it extracts two electrons (2 e⁻) from the alcohol converting it to the aldehyde. The two hydrogen cations (protons, 2 H⁺) just follow the electrons if possible. So we don’t work here with neither two hydrogen atoms (2 H) nor a hydrogen molecule (H₂).
The NAD part of the reaction can
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The first one is the reduced form of NADP⁺, the second is a nonsense. The same is true for its NAD counterparts. Let’s use NAD (NAD means generally both NAD⁺ and NADH) to explain the problem with the particular reaction of alcohol dehydrogenase:
H₃C–CH₂–OH + NAD⁺ → H₃C–CH=O + NADH + H⁺
The reaction is a redox process, it extracts two electrons (2 e⁻) from the alcohol converting it to the aldehyde. The two hydrogen cations (protons, 2 H⁺) just follow the electrons if possible. So we don’t work here with neither two hydrogen atoms (2 H) nor a hydrogen molecule (H₂).
The NAD part of the reaction can be written as the following sequence of partial reactions:
NAD⁺ + e⁻ → NAD (NAD doesn’t have the above-mentioned general meaning here, it stands for the neutral non-hydrogenated form only),
NAD + e⁻ → NAD⁻,
NAD⁻ + H⁺ → NADH.
There is no more step concerning NAD – there is no possibility in its molecule to bind the second H⁺ extracted from the alcohol (even if there was, the product would have been NADH₂⁺ (non-charged NADH + H⁺)), so the H⁺ is simply released to the solution. NADH and H⁺ are two separate products, there is no [NADH + H⁺] product, the two things are not connected in any way. They are just two different products of the reaction. (It is meant as a correction of the statement “Actually NADPH+H Is accurate way to write NADPH” in one of the previous answers.)
Therefore, simply said, NADH contains both the electrons but only one of the protons of the two hydrogens extracted from the bond during its dehydrogenation (oxidation), which means only one complete hydrogen is there. The other electron is why the positive charge of NAD⁺ disappears from NADH.
The same is true for NADP (NADP⁺ and NADPH)
The incorrect forms NADH₂ and NADPH₂ are, in my opinion, due to a mistaken analogy with the FAD/FADH₂ and FMN/FMNH₂ pairs. They are indeed neutral (non-charged) in the oxidized form and they bind both the electrons and both the protons of the two hydrogens, so the …H₂ endings of their symbols are correct.
Upvote ·
99 14
Eric Ressner
Chemical information services (1985–present) · Author has 6K answers and 7.8M answer views
·2y
Related
Are NAD and NADP ribonucleotides or deoxyribonucleotide?
Are NAD and NADP ribonucleotides or deoxyribonucleotide?
Both are based on ribose as the sugar component. But the “D” in their abbreviated names stands for “dinucleotide”. And while the two nucleotides are not linked in the conventional 5′→3′ orientation, there are still the components of two nucleotides in the complete structure.
A ribonucleotide is a substance composed of a nitrogenous base, a ribose sugar, and a phosphate. That’s not what we have here in NAD and NADP.
So I’d say they are not ribonucleotides, just as RNA is not a ribonucleotide.
Upvote ·
Related questions
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Jack D. Mattingly has 50 years' experience in aircraft propulsion. After serving most of his military career teaching at the U.S. Air Force Academy and the Air Force Institute of Technology, he retired from active duty in 1989 and joined the faculty of Seattle University, retiring in 2000 as professor emeritus in Mechanical Engineering (M.E.). He has since focused on teaching short courses with Practical Aeronautics, Inc. to practicing engineers, consulting, and writing textbooks. He is a co-author of Aircraft Engine Design and Elements of Propulsion: Gas Turbines and Rockets, both winners of the AIAA Summerfield Book Award. He holds a Ph.D. in Aeronautics and Astronautics from the University of Washington.
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4) Anderson, N. G.; Practical Process Research and Development, a Guide for Organic Chemists, 2nd Edition |
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Shoelace Theorem
The Shoelace Theorem is a nifty formula for finding the area of a simple polygon given the coordinates of its vertices.
Contents
[hide]
1 Theorem
2 Other Forms
3 Proof 1
3.1 Proof of claim 1:
3.2 Proof:
4 Proof 2
5 Proof 3
6 Problems
6.1 Introductory
6.2 Exploratory
7 External Links
Theorem
Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area () of is
You can also go counterclockwise order, as long as you find the absolute value of the answer.
The Shoelace Theorem gets its name because if one lists the coordinates in a column, and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes.
Other Forms
This can also be written in form of a summation or in terms of determinants as which is useful in the variant of the Shoelace theorem. Note here that and .
The formula may also be considered a special case of Green's Theorem
where and so .
Proof 1
Claim 1: The area of a triangle with coordinates , , and is .
Proof of claim 1:
Writing the coordinates in 3D and translating so that we get the new coordinates , , and . Now if we let and then by definition of the cross product .
)
Proof:
We will proceed with induction.
By claim 1, the shoelace theorem holds for any triangle. We will show that if it is true for some polygon then it is also true for .
We cut into two polygons, and . Let the coordinates of point be . Then, applying the shoelace theorem on and we get
Hence
as claimed.
~ShreyJ
Proof 2
Let be the set of points belonging to the polygon. We have that where . The volume form is an exact form since , where Using this substitution, we have Next, we use the Theorem of Stokes to obtain We can write , where is the line segment from to . With this notation, we may write If we substitute for , we obtain If we parameterize, we get Performing the integration, we get More algebra yields the result
Proof 3
This is a very nice approach that directly helps in understanding the sum as terms which are areas of trapezoids.
The proof is in this book:
(The only thing that needs to be slightly modified is that one must shift the entire polygon up by k, until all the y coordinates are positive, but this term gets canceled in the resulting sum.)
Problems
Introductory
In right triangle , we have , , and . Medians and are drawn to sides and , respectively. and intersect at point . Find the area of .
Exploratory
Observe that is the area of a triangle with vertices and is the volume of a tetrahedron with vertices . Does a similar formula hold for Dimensional triangles for any ? If so how can we use this to derive the D Shoelace Formula?
External Links
A good explanation and exploration into why the theorem works by James Tanton:
Nice geometric approach and discussion for proving the 3D Shoelace Theorem by Nicholas Patrick and Nadya Pramita:
Nice integral approach for proving the 3D Shoelace Theorem (ignoring sign of volume) by @george2079: AOPS
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9022 | https://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=129729514 | Sucrose-UL-14C | C12H22O11 | CID 129729514 - PubChem
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compound Summary
Sucrose-UL-14C
PubChem CID
129729514
Structure
Molecular Formula
C 12 H 22 O 11
Synonyms
Sucrose-UL-14C
Molecular Weight
348.27 g/mol
Computed by PubChem 2.1 (PubChem release 2021.05.07)
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Modify: 2025-09-27
1 Structures
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1.2 3D Status
Conformer generation is disallowed since too many undefined stereo centers
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2 Biologic Description
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IUPAC Condensed
Hex(?1-2?)Hex2ulof
LINUCS
[][Hexp]{}
IUPAC
hex-2-ulofuranosyl hexopyranoside
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3 Names and Identifiers
3.1 Computed Descriptors
3.1.1 IUPAC Name
2-[(2 R)-3,4-dihydroxy-2,5-bis(hydroxy(1 14 C)methyl)oxolan-2-yl]oxy-6-(hydroxy(1 14 C)methyl)oxane-3,4,5-triol
Computed by Lexichem TK 2.7.0 (PubChem release 2021.05.07)
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3.1.2 InChI
InChI=1S/C12H22O11/c13-1-4-6(16)8(18)9(19)11(21-4)23-12(3-15)10(20)7(17)5(2-14)22-12/h4-11,13-20H,1-3H2/t4?,5?,6?,7?,8?,9?,10?,11?,12-/m1/s1/i1+2,2+2,3+2
Computed by InChI 1.0.6 (PubChem release 2021.05.07)
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3.1.3 InChIKey
CZMRCDWAGMRECN-IHHGVDQYSA-N
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3.1.4 SMILES
14CH2O
Computed by OEChem 2.3.0 (PubChem release 2024.12.12)
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3.2 Molecular Formula
C 12 H 22 O 11
Computed by PubChem 2.1 (PubChem release 2021.05.07)
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3.3 Synonyms
3.3.1 Depositor-Supplied Synonyms
Sucrose-UL-14C
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4 Chemical and Physical Properties
4.1 Computed Properties
Property Name
Property Value
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Property Name
Molecular Weight
Property Value
348.27 g/mol
Reference
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Property Value
-3.7
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Property Name
Hydrogen Bond Donor Count
Property Value
8
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2021.05.07)
Property Name
Hydrogen Bond Acceptor Count
Property Value
11
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2021.05.07)
Property Name
Rotatable Bond Count
Property Value
5
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2021.05.07)
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Exact Mass
Property Value
348.1259375 Da
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Property Name
Monoisotopic Mass
Property Value
348.1259375 Da
Reference
Computed by PubChem 2.1 (PubChem release 2021.05.07)
Property Name
Topological Polar Surface Area
Property Value
190 Ų
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Property Value
23
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Formal Charge
Property Value
0
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Complexity
Property Value
395
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2021.05.07)
Property Name
Isotope Atom Count
Property Value
3
Reference
Computed by PubChem
Property Name
Defined Atom Stereocenter Count
Property Value
1
Reference
Computed by PubChem
Property Name
Undefined Atom Stereocenter Count
Property Value
8
Reference
Computed by PubChem
Property Name
Defined Bond Stereocenter Count
Property Value
0
Reference
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Undefined Bond Stereocenter Count
Property Value
0
Reference
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Property Value
1
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Compound Is Canonicalized
Property Value
Yes
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Same Connectivity Count
198
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2
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2
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477
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2
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2
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5.3.1 Related Substances
Same Count
1
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9023 | https://www.quora.com/What-is-parallel-to-the-x-axis | What is parallel to the x-axis? - Quora
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What is parallel to the x-axis?
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1y
A line that is parallel to the x-axis is a horizontal line. This means that it runs left to right across the graph and has a constant y-coordinate for all points on the line. The equation of a horizontal line can be expressed in the form:
y=c y=c
where c c is a constant. For example, the line y=3 y=3 is parallel to the x-axis and will pass through all points where the y-coordinate is 3, regardless of the x-coordinate.
Upvote ·
Related questions
More answers below
Why is the slope 0 parallel to the x-axis?
What is the direction ration of a line parallel to x axis?
What is the point of intersection of a line parallel to the x-axis?
What line parallel to the x-axis and passing through (1,2) has the equation?
Why do surveyors often put their x axis on vertical and their y axis on horizontal?
Ava
Updated 3y
To answer this question, we need to first grasp the concept of the Cartesian coordinate plane. The x x and y y axes in the Cartesian coordinate plane are perpendicular to one another and so distinct, i.e. (0,6)(0,6) and (0,7).(6,0)(0,7).(6,0). We need a slope of zero for it to parallel: (6,0)→y=k(6,0)→y=k. Why is it y y, not x x? Because we do not value x x to be represented in the following coordinate point. What is k? Any real number that follows within the range of the coordinate point. This can also be seen with: (0,6)→x=k(0,6)→x=k. Once again, we do not know the value of x x and as such is represented like this.
X X Axis:
As I have
Continue Reading
To answer this question, we need to first grasp the concept of the Cartesian coordinate plane. The x x and y y axes in the Cartesian coordinate plane are perpendicular to one another and so distinct, i.e. (0,6)(0,6) and (0,7).(6,0)(0,7).(6,0). We need a slope of zero for it to parallel: (6,0)→y=k(6,0)→y=k. Why is it y y, not x x? Because we do not value x x to be represented in the following coordinate point. What is k? Any real number that follows within the range of the coordinate point. This can also be seen with: (0,6)→x=k(0,6)→x=k. Once again, we do not know the value of x x and as such is represented like this.
X X Axis:
As I have written earlier (6,0)→y=k(6,0)→y=k. Now, why is that when I put this into a graphing calculator that it does not go through the point in the way I want to? These lines are parallel.X X=Horizontal Line. Y Y=Vertical Line. Notice how the line in the point is going horizontally. But wait! What happens to the point? Isn’t it supposed to be just (6,0)(6,0)? Why does it show (0,6)(0,6)? Earlier, in order to find the parallel to each ordered pair →y=k o r x=k y=k o r x=k. In this case, we know the value of ‘x x,’ however not ‘y y.’ We only care about the parallel line, so we will start with that first. If you look closely, we went up 6 units, and did not have any change in the x axis. However, for the x axis we have a difference of 6 units and zero y y.
Same concept with the y axis:
So, when having the (6,0)(6,0) it did not necessarily change the point to (0,6)(0,6) it’s just where the parallel line → y=k.(y=6)y=k.(y=6). We couldn’t have it at (0,6)(0,6) because we are looking for the parallel line. I hope this answered your question.
Upvote ·
Shreyas Devesh
I have studied calculus · Author has 81 answers and 83.2K answer views
·4y
A function is said to be parallel to the x-axis if its slope is zero about the selected interval and that it is defined across that interval.
To check this property is trivial just differentiate the function and if the answer across the points in the interval is zero then it must be parallel.
And this is because by the definition of the derivative it is the rate of change of y-axis wrt to the x-axis and if parallel that means there is no change
And only functions that are parallel to the x-axis are constant functions throughout their domain
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9 1
Peter Nelson
Studied Science at Spokane Falls Community College (Graduated 2020) · Author has 3.7K answers and 1.8M answer views
·4y
Parallel to axes means the lines that are parallel to either the x-axis or y-axis. A line parallel to the x-axis is a horizontal line whose equation is of the form y = k, where 'k' is the distance of the line from the x-axis.
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More answers below
What is the distance of the origin O from both the Axis (X and Y-axis)?
If the question said touch the x-axis or intersect x- axis or y-axis, what does it mean?
What is the equation of a line parallel to x axis and passes through (-2, 4)?
X-axis and line x=-4 are parallel lines. What is the distance between them?
Can a straight line in real life be parallel to the x-axis or y-axis? Why or why not?
Deekshith Vodela
Anime and Stuff
·6y
Originally Answered: Which graph is parallel to x-axis? ·
The graph parallel to x-axis is the line represented by the line equation
y=k
-> This equation represents infinite lines.
where,
y is ordinate (or) y- co-ordinate
k is a constant, which can take any value of real numbers.
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9 1
Janet Heberling
Lives in San Francisco, CA (2022–present) · Author has 21.5K answers and 9.4M answer views
·2y
Lines with equations the form y=b
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9 1
Gopal Menon
B Tech in Chemical Engineering, Indian Institute of Technology, Bombay (IITB) (Graduated 1975) · Author has 10.2K answers and 15.2M answer views
·7y
Related
What is the direction ration of a line parallel to x axis?
If a line is parallel to the X axis, its inclination with respect to the positive X axis, Y axis and Z axis are α=0 α=0 o,β=90 o o,β=90 o and γ=9 γ=9 0 o 0 o respectively.
⇒⇒ The direction cosines of the line are (l,m,n)(l,m,n)=(cos α,cos β,=(cosα,cosβ,cos γ)=(1,0,0).cosγ)=(1,0,0).
⇒⇒ The direction ratios of the line are (a,0,0),a∈R.(a,0,0),a∈R.
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9 3
Brunilda Ortiz
BA from University of Puerto Rico at Río Piedras (Graduated 1975) · Author has 111 answers and 90K answer views
·1y
Related
Can you provide examples of lines that are parallel to the x-axis and y-axis?
I can write equations for lines parallel to the x-axis or the y-axis but don't believe any line can be parallel to both axes.
y = 0 is the x-axis and any equation in this form y = c will be parallel to the x-axis if c is a constant
x = 0 is the y-axis and any equation in the form x = c will be parallel to the y axis, if c is a constant.
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9 1
Audley Willacey
B.Sc. in Mechanical Engineering, University of the West Indies, St. Augustine (Graduated 1976) · Author has 5K answers and 1.5M answer views
·1y
Related
Can you provide examples of lines that are parallel to the x-axis and y-axis?
General equ'n of lines parallel to the x-axis : y = b
Examples : y = 3 ; y = -2.
General eq'n of lines parallel to the y-axis : x = a
Examples : x = 4 ; x = -7.
What happens if a = 0, or, b = 0 ?
Well, x = 0 is actually the y-axis, and, y = 0 is the x-axis.
I hope that helps.
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Ernest Leung
B.Sc. (Hons.) in Chemistry Honors&Mathematics, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views
·Apr 21
Related
Is 6y=2 parallel to the x-axis?
Is 6y = 2 parallel to the x-axis?
6y = 2 .. ⇒ y = 1/3
The y-coordinates of all points on the line 6y = 2 are equal to 1/3. Hence, it is a horizontal line parallel to the x-axis.
Continue Reading
Is 6y = 2 parallel to the x-axis?
6y = 2 .. ⇒ y = 1/3
The y-coordinates of all points on the line 6y = 2 are equal to 1/3. Hence, it is a horizontal line parallel to the x-axis.
Upvote ·
9 4
Sushant Kala
B.Sc in Mathematics, Hemwati Nandan Bahuguna Garhwal University, Uttarakhand (Graduated 2018)
·Updated 7y
Related
Why x axis and y axis are perpendicular?
Because they are the easiest to deal with , you can choose any other pair of lines which are not perpendicular and also not parallel to describe the whole R^2 plane but in that case you have to deal with parallelogram instead of rectangle and a triangle instead of right angle as we have in XY co-ordinate system.
It will be slightly difficult here to derive basic result such as distance formula , section formula etc.
Image source - Google
Continue Reading
Because they are the easiest to deal with , you can choose any other pair of lines which are not perpendicular and also not parallel to describe the whole R^2 plane but in that case you have to deal with parallelogram instead of rectangle and a triangle instead of right angle as we have in XY co-ordinate system.
It will be slightly difficult here to derive basic result such as distance formula , section formula etc.
Image source - Google
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9 7
9 3
Kunal Joshi
Studied Mathematics at Kendirya Vidyalaya (Graduated 2020) · Author has 74 answers and 31.3K answer views
·3y
Related
In what cases is the y-axis perpendicular to the x-axis?
In all the cases, if you are talking about cartesian coordinate system, y-axis is perpendicular to the x-axis
Upvote ·
Dave Benson
trying to make maths easy. · Author has 6.1K answers and 2.1M answer views
·Apr 21
Related
Is 6y=2 parallel to the x-axis?
Compare y = mx + c where m is slope and c is y-intercept
6y = 2 ⟹ y = 1/3 so slope is zero and c = 1/3 therefore parallel to x-axis.
Continue Reading
Compare y = mx + c where m is slope and c is y-intercept
6y = 2 ⟹ y = 1/3 so slope is zero and c = 1/3 therefore parallel to x-axis.
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Related questions
Why is the slope 0 parallel to the x-axis?
What is the direction ration of a line parallel to x axis?
What is the point of intersection of a line parallel to the x-axis?
What line parallel to the x-axis and passing through (1,2) has the equation?
Why do surveyors often put their x axis on vertical and their y axis on horizontal?
What is the distance of the origin O from both the Axis (X and Y-axis)?
If the question said touch the x-axis or intersect x- axis or y-axis, what does it mean?
What is the equation of a line parallel to x axis and passes through (-2, 4)?
X-axis and line x=-4 are parallel lines. What is the distance between them?
Can a straight line in real life be parallel to the x-axis or y-axis? Why or why not?
What is the total cost if it starts from origin, is parallel to y-axis, parallel to X-axis and not from origin?
Y axis and line X=-4 are parallel lines. What is the distance between them?
Which line out of given lines is parallel to x-axis? (1) x = 5 (2) y = 5 (3) y = x (4) 7 x+4=0
Can you put a break in the Y-axis?
What is the inclination of the x-axis or any line parallel to the y-axis?
Related questions
Why is the slope 0 parallel to the x-axis?
What is the direction ration of a line parallel to x axis?
What is the point of intersection of a line parallel to the x-axis?
What line parallel to the x-axis and passing through (1,2) has the equation?
Why do surveyors often put their x axis on vertical and their y axis on horizontal?
What is the distance of the origin O from both the Axis (X and Y-axis)?
If the question said touch the x-axis or intersect x- axis or y-axis, what does it mean?
What is the equation of a line parallel to x axis and passes through (-2, 4)?
X-axis and line x=-4 are parallel lines. What is the distance between them?
Can a straight line in real life be parallel to the x-axis or y-axis? Why or why not?
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9: Electrochemistry
CHEM 202: General Chemistry II OER
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Thu, 13 Jan 2022 21:11:13 GMT
9.2: Galvanic Cells
364707
364707
Patricia Foley
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Contents
1. Home
2. Campus Bookshelves
3. College of the Canyons
4. CHEM 202: General Chemistry II OER
5. 9: Electrochemistry
6. 9.2: Galvanic Cells
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CHEM 202: General Chemistry II OER
Front Matter
1: Introduction to Organic Chemistry
2: Coordination Chemistry of Transition Metals
3: Chemical Kinetics
4: Chemical Equilibrium
5: Acid-Base Equilibria
6: Acid-Base Equilibria in Mixtures
7: Solubility and Complex-Ion Equilibria
8: Thermodynamics
9: Electrochemistry
10: Nuclear Chemistry
Back Matter
9.2: Galvanic Cells
Last updated Jan 13, 2022
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9.1: Balancing Oxidation-Reduction Reactions
9.3: Standard Reduction Potentials
Page ID 364707
Patricia Foley
College of the Canyons
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Example 9.2.1 9.2.1: Using Cell Notation
1. Solution
Example 9.2.2 9.2.2: Using Cell Notation
Solution
Exercise 9.2.1 9.2.1
Summary
Glossary
Learning Objectives
Use cell notation to describe galvanic cells
Describe the basic components of galvanic cells
Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. In writing the equations, it is often convenient to separate the oxidation-reduction reactions into half-reactions to facilitate balancing the overall equation and to emphasize the actual chemical transformations.
Consider what happens when a clean piece of copper metal is placed in a solution of silver nitrate (Figure 9.2.1 9.2.1). As soon as the copper metal is added, silver metal begins to form and copper ions pass into the solution. The blue color of the solution on the far right indicates the presence of copper ions. The reaction may be split into its two half-reactions. Half-reactions separate the oxidation from the reduction, so each can be considered individually.
oxidation:Cu(s)⟶Cu 2+(a q)+2 e−reduction:2×(Ag+(a q)+e−⟶Ag(s))or 2 Ag+(a q)+2 e−⟶2 Ag(s)––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––overall:2 Ag+(a q)+Cu(s)⟶2 Ag(s)+Cu 2+(a q)(9.2.1)(9.2.2)(9.2.3)(9.2.1)oxidation:Cu(s)⟶Cu 2+(a q)+2 e−(9.2.2)reduction:2×(Ag+(a q)+e−⟶Ag(s))or 2 Ag+(a q)+2 e−⟶2 Ag(s)_(9.2.3)overall:2 Ag+(a q)+Cu(s)⟶2 Ag(s)+Cu 2+(a q)
The equation for the reduction half-reaction had to be doubled so the number electrons “gained” in the reduction half-reaction equaled the number of electrons “lost” in the oxidation half-reaction.
Figure 9.2.1 9.2.1: When a clean piece of copper metal is placed into a clear solution of silver nitrate (a), an oxidation-reduction reaction occurs that results in the exchange of Cu 2+ for Ag+ ions in solution. As the reaction proceeds (b), the solution turns blue (c) because of the copper ions present, and silver metal is deposited on the copper strip as the silver ions are removed from solution. (credit: modification of work by Mark Ott)
Galvanic or voltaic cells involve spontaneous electrochemical reactions in which the half-reactions are separated (Figure 9.2.2 9.2.2) so that current can flow through an external wire. The beaker on the left side of the figure is called a half-cell, and contains a 1 M solution of copper(II) nitrate [Cu(NO 3)2] with a piece of copper metal partially submerged in the solution. The copper metal is an electrode. The copper is undergoing oxidation; therefore, the copper electrode is the anode. The anode is connected to a voltmeter with a wire and the other terminal of the voltmeter is connected to a silver electrode by a wire. The silver is undergoing reduction; therefore, the silver electrode is the cathode. The half-cell on the right side of the figure consists of the silver electrode in a 1 M solution of silver nitrate (AgNO 3). At this point, no current flows—that is, no significant movement of electrons through the wire occurs because the circuit is open. The circuit is closed using a salt bridge, which transmits the current with moving ions. The salt bridge consists of a concentrated, nonreactive, electrolyte solution such as the sodium nitrate (NaNO 3) solution used in this example. As electrons flow from left to right through the electrode and wire, nitrate ions (anions) pass through the porous plug on the left into the copper(II) nitrate solution. This keeps the beaker on the left electrically neutral by neutralizing the charge on the copper(II) ions that are produced in the solution as the copper metal is oxidized. At the same time, the nitrate ions are moving to the left, sodium ions (cations) move to the right, through the porous plug, and into the silver nitrate solution on the right. These added cations “replace” the silver ions that are removed from the solution as they were reduced to silver metal, keeping the beaker on the right electrically neutral. Without the salt bridge, the compartments would not remain electrically neutral and no significant current would flow. However, if the two compartments are in direct contact, a salt bridge is not necessary. The instant the circuit is completed, the voltmeter reads +0.46 V, this is called the cell potential. The cell potential is created when the two dissimilar metals are connected, and is a measure of the energy per unit charge available from the oxidation-reduction reaction. The volt is the derived SI unit for electrical potential
v o l t=V=J C v o l t=V=J C
In this equation, A is the current in amperes and C the charge in coulombs. Note that volts must be multiplied by the charge in coulombs (C) to obtain the energy in joules (J).
Figure 9.2.2 9.2.2: In this standard galvanic cell, the half-cells are separated; electrons can flow through an external wire and become available to do electrical work.
When the electrochemical cell is constructed in this fashion, a positive cell potential indicates a spontaneous reaction and that the electrons are flowing from the left to the right. There is a lot going on in Figure 9.2.2 9.2.2, so it is useful to summarize things for this system:
Electrons flow from the anode to the cathode: left to right in the standard galvanic cell in the figure.
The electrode in the left half-cell is the anode because oxidation occurs here. The name refers to the flow of anions in the salt bridge toward it.
The electrode in the right half-cell is the cathode because reduction occurs here. The name refers to the flow of cations in the salt bridge toward it.
Oxidation occurs at the anode (the left half-cell in the figure).
Reduction occurs at the cathode (the right half-cell in the figure).
The cell potential, +0.46 V, in this case, results from the inherent differences in the nature of the materials used to make the two half-cells.
The salt bridge must be present to close (complete) the circuit and both an oxidation and reduction must occur for current to flow.
There are many possible galvanic cells, so a shorthand notation is usually used to describe them. The cell notation (sometimes called a cell diagram) provides information about the various species involved in the reaction. This notation also works for other types of cells. A vertical line, │, denotes a phase boundary and a double line, ‖, the salt bridge. Information about the anode is written to the left, followed by the anode solution, then the salt bridge (when present), then the cathode solution, and, finally, information about the cathode to the right. The cell notation for the galvanic cell in Figure 9.2.2 9.2.2 is then
Cu(s)│Cu 2+(a q,1 M)║Ag+(a q,1 M)│Ag(s)Cu(s)│Cu 2+(a q,1 M)║Ag+(a q,1 M)│Ag(s)
Note that spectator ions are not included and that the simplest form of each half-reaction was used. When known, the initial concentrations of the various ions are usually included.
One of the simplest cells is the Daniell cell. It is possible to construct this battery by placing a copper electrode at the bottom of a jar and covering the metal with a copper sulfate solution. A zinc sulfate solution is floated on top of the copper sulfate solution; then a zinc electrode is placed in the zinc sulfate solution. Connecting the copper electrode to the zinc electrode allows an electric current to flow. This is an example of a cell without a salt bridge, and ions may flow across the interface between the two solutions.
Some oxidation-reduction reactions involve species that are poor conductors of electricity, and so an electrode is used that does not participate in the reactions. Frequently, the electrode is platinum, gold, or graphite, all of which are inert to many chemical reactions. One such system is shown in Figure 9.2.3 9.2.3. Magnesium undergoes oxidation at the anode on the left in the figure and hydrogen ions undergo reduction at the cathode on the right. The reaction may be summarized as
oxidation:Mg(s)⟶Mg 2+(a q)+2 e−reduction:2 H+(a q)+2 e−⟶H 2(g)overall:Mg(s)+2 H+(a q)⟶Mg 2+(a q)+H 2(g)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(9.2.4)(9.2.5)(9.2.6)(9.2.4)oxidation:Mg(s)⟶Mg 2+(a q)+2 e−(9.2.5)reduction:2 H+(a q)+2 e−⟶H 2(g)(9.2.6)overall:Mg(s)+2 H+(a q)⟶Mg 2+(a q)+H 2(g)¯
The cell used an inert platinum wire for the cathode, so the cell notation is
Mg(s)│Mg 2+(a q)║H+(a q)│H 2(g)│Pt(s)Mg(s)│Mg 2+(a q)║H+(a q)│H 2(g)│Pt(s)
The magnesium electrode is an active electrode because it participates in the oxidation-reduction reaction. Inert electrodes, like the platinum electrode in Figure 9.2.3 9.2.3, do not participate in the oxidation-reduction reaction and are present so that current can flow through the cell. Platinum or gold generally make good inert electrodes because they are chemically unreactive.
Figure 9.2.2 9.2.2: The oxidation of magnesium to magnesium ion occurs in the beaker on the left side in this apparatus; the reduction of hydrogen ions to hydrogen occurs in the beaker on the right. A nonreactive, or inert, platinum wire allows electrons from the left beaker to move into the right beaker. The overall reaction is: Mg+2 H+⟶Mg 2++H 2 Mg+2 H+⟶Mg 2++H 2, which is represented in cell notation as: Mg(s)│Mg 2+(a q)║H+(a q)│H 2(g)│Pt(s)Mg(s)│Mg 2+(a q)║H+(a q)│H 2(g)│Pt(s).
Example 9.2.1 9.2.1: Using Cell Notation
Consider a galvanic cell consisting of
2 Cr(s)+3 Cu 2+(a q)⟶2 Cr 3+(a q)+3 Cu(s)2 Cr(s)+3 Cu 2+(a q)⟶2 Cr 3+(a q)+3 Cu(s)
Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?
Solution
By inspection, Cr is oxidized when three electrons are lost to form Cr 3+, and Cu 2+ is reduced as it gains two electrons to form Cu. Balancing the charge gives
oxidation:2 Cr(s)⟶2 Cr 3+(a q)+6 e−reduction:3 Cu 2+(a q)+6 e−⟶3 Cu(s)overall:2 Cr(s)+3 Cu 2+(a q)⟶2 Cr 3+(a q)+3 Cu(s)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(9.2.7)(9.2.8)(9.2.9)(9.2.7)oxidation:2 Cr(s)⟶2 Cr 3+(a q)+6 e−(9.2.8)reduction:3 Cu 2+(a q)+6 e−⟶3 Cu(s)(9.2.9)overall:2 Cr(s)+3 Cu 2+(a q)⟶2 Cr 3+(a q)+3 Cu(s)¯
Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. No concentrations were specified so:
Cr(s)│Cr 3+(a q)║Cu 2+(a q)│Cu(s).Cr(s)│Cr 3+(a q)║Cu 2+(a q)│Cu(s).
Oxidation occurs at the anode and reduction at the cathode.
Example 9.2.2 9.2.2: Using Cell Notation
Consider a galvanic cell consisting of
5 Fe 2+(a q)+MnO−4(a q)+8 H+(a q)⟶5 Fe 3+(a q)+Mn 2+(a q)+4 H 2 O(l)5 Fe 2+(a q)+MnO 4−(a q)+8 H+(a q)⟶5 Fe 3+(a q)+Mn 2+(a q)+4 H 2 O(l)
Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?
Solution
By inspection, Fe 2+ undergoes oxidation when one electron is lost to form Fe 3+, and MnO 4− is reduced as it gains five electrons to form Mn 2+. Balancing the charge gives
oxidation:5(Fe 2+(a q)⟶Fe 3+(a q)+e−)reduction:MnO−4(a q)+8 H+(a q)+5 e−⟶Mn 2+(a q)+4 H 2 O(l)––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––overall:5 Fe 2+(a q)+MnO−4(a q)+8 H+(a q)⟶5 Fe 3+(a q)+Mn 2+(a q)+4 H 2 O(l)(9.2.10)(9.2.11)(9.2.12)(9.2.10)oxidation:5(Fe 2+(a q)⟶Fe 3+(a q)+e−)(9.2.11)reduction:MnO 4−(a q)+8 H+(a q)+5 e−⟶Mn 2+(a q)+4 H 2 O(l)_(9.2.12)overall:5 Fe 2+(a q)+MnO 4−(a q)+8 H+(a q)⟶5 Fe 3+(a q)+Mn 2+(a q)+4 H 2 O(l)
Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode. No concentrations were specified so:
Pt(s)│Fe 2+(a q),Fe 3+(a q)║MnO−4(a q),H+(a q),Mn 2+(a q)│Pt(s).Pt(s)│Fe 2+(a q),Fe 3+(a q)║MnO 4−(a q),H+(a q),Mn 2+(a q)│Pt(s).
Oxidation occurs at the anode and reduction at the cathode.
Exercise 9.2.1 9.2.1
Use cell notation to describe the galvanic cell where copper(II) ions are reduced to copper metal and zinc metal is oxidized to zinc ions.
Answer
From the information given in the problem:
anode (oxidation):Zn(s)⟶Zn 2+(a q)+2 e−cathode (reduction):Cu 2+(a q)+2 e−⟶Cu(s)overall:Zn(s)+Cu 2+(a q)⟶Zn 2+(a q)+Cu(s)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(9.2.13)(9.2.14)(9.2.15)(9.2.13)anode (oxidation):Zn(s)⟶Zn 2+(a q)+2 e−(9.2.14)cathode (reduction):Cu 2+(a q)+2 e−⟶Cu(s)(9.2.15)overall:Zn(s)+Cu 2+(a q)⟶Zn 2+(a q)+Cu(s)¯
Using cell notation:
Zn(s)│Zn 2+(a q)║Cu 2+(a q)│Cu(s).Zn(s)│Zn 2+(a q)║Cu 2+(a q)│Cu(s).
Summary
Electrochemical cells typically consist of two half-cells. The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire. One half-cell, normally depicted on the left side in a figure, contains the anode. Oxidation occurs at the anode. The anode is connected to the cathode in the other half-cell, often shown on the right side in a figure. Reduction occurs at the cathode. Adding a salt bridge completes the circuit allowing current to flow. Anions in the salt bridge flow toward the anode and cations in the salt bridge flow toward the cathode. The movement of these ions completes the circuit and keeps each half-cell electrically neutral. Electrochemical cells can be described using cell notation. In this notation, information about the reaction at the anode appears on the left and information about the reaction at the cathode on the right. The salt bridge is represented by a double line, ‖. The solid, liquid, or aqueous phases within a half-cell are separated by a single line, │. The phase and concentration of the various species is included after the species name. Electrodes that participate in the oxidation-reduction reaction are called active electrodes. Electrodes that do not participate in the oxidation-reduction reaction but are there to allow current to flow are inert electrodes. Inert electrodes are often made from platinum or gold, which are unchanged by many chemical reactions.
Glossary
active electrode electrode that participates in the oxidation-reduction reaction of an electrochemical cell; the mass of an active electrode changes during the oxidation-reduction reaction anode electrode in an electrochemical cell at which oxidation occurs; information about the anode is recorded on the left side of the salt bridge in cell notation cathode electrode in an electrochemical cell at which reduction occurs; information about the cathode is recorded on the right side of the salt bridge in cell notation cell notation shorthand way to represent the reactions in an electrochemical cell cell potential difference in electrical potential that arises when dissimilar metals are connected; the driving force for the flow of charge (current) in oxidation-reduction reactions galvanic cell electrochemical cell that involves a spontaneous oxidation-reduction reaction; electrochemical cells with positive cell potentials; also called a voltaic cell inert electrode electrode that allows current to flow, but that does not otherwise participate in the oxidation-reduction reaction in an electrochemical cell; the mass of an inert electrode does not change during the oxidation-reduction reaction; inert electrodes are often made of platinum or gold because these metals are chemically unreactive.voltaic cell another name for a galvanic cell
This page titled 9.2: Galvanic Cells is shared under a CC BY license and was authored, remixed, and/or curated by Patricia Foley.
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9.1: Balancing Oxidation-Reduction Reactions
9.3: Standard Reduction Potentials
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9025 | https://ocw.mit.edu/courses/10-626-electrochemical-energy-systems-spring-2014/pages/study-materials/ | Browse Course Material
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10.626 | Spring 2014 | Graduate
Electrochemical Energy Systems
Study Materials
In the table below are selected supplementary lecture topics. These lecture notes from previous years were scribed by students who took this class and are used with their permission. The notes were written by the students as homework assignments.
Electrostatics
2011 Lecture 26: Compact Part of the Double Layer (PDF)
2011 Lecture 27a: Diffuse Charge at Electrode (PDF)
2011 Lecture 27b: Diffuse Charge in Membranes (PDF)
2011 Lecture 28: Electrostatic Correlations (PDF)
Electrokinetics
2011 Lecture 29: Electrokinetics (PDF - 1.3MB)
2011 Lecture 30: Linear Electrokinetic Phenomena (PDF)
2011 Lecture 31: Electrokinetic Energy Conversion (PDF)
Phase Transformations
2011 Lecture 38: Nucleation and Spinodal Decomposition (PDF)
2011 Lecture 39: Reaction-limited Phase Separation (PDF)
Reaction Kinetics
2009 Lecture 10: Stochastic Theory of Reaction Rates (PDF)
2009 Lecture 11: Kramers Escape Problem (PDF)
Escape from a Symmetric Trap (PDF)
2009 Lecture 12: Reactions as Rare Events (PDF)
Quantum Mechanics
2009 Lecture 17: Introduction to Quantum Theory of Solids (PDF)
2009 Lecture 18: Electronic Structure of Electrodes (PDF)
2009 Lecture 19: Electronic Contributions to OCV in Batteries and Solar Cells (PDF - 1.2MB)
2009 Lecture 20: Quantum Tunneling of Electrons (PDF)
Course Info
Instructor
Prof. Martin Bazant
Departments
Chemical Engineering
As Taught In
Spring 2014
Level
Graduate
Topics
Engineering
Chemical Engineering
Materials Science and Engineering
Electronic Materials
Science
Chemistry
Analytical Chemistry
Physical Chemistry
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grading Exams with Solutions
notes Lecture Notes
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Circular Inversion
Circular Inversion, sometimes called Geometric Inversion or simply Inversion, is a transformation where point in the Cartesian plane is transformed based on a circle with radius and center such that , where is the transformed point on the ray extending from through .
Note that , when inverted, transforms back to . All points outside of are transformed inside , and vice versa. Points on transform to themselves, meaning . Finally, the transformation of is debated on its existence. Some call the transformation the ideal point, which is infinitely far away and in every direction. Others claim that this point does not have an inverse.
Geometric Inversion technically refers to many different types of inversions, however, if Geometric Inversion is used without clarification, Circular Inversion is usually assumed.
Circular Inversion can be a very useful tool in solving problems involving many tangent circles and/or lines.
Contents
[hide]
1 Basics of Circular Inversion
1.1 Inversion of a Circle intersecting O
1.2 Inversion of a Circle not intersecting O
1.3 General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle
1.4 Problems
1.4.1 Problem 1
Basics of Circular Inversion
Inversion of a Circle intersecting O
The first thing that we must learn about inversion is what happens when a circle which intersects the center of the inversion, , is inverted. Let us have circle , with diameter . is chosen arbitrarily on circle . Points and represent the inversions of and , respectively. is the radius of . We seek to show that circle inverts to a line perpendicular to through .
By the definition of inversion, we have and .
We can combine the two equations to get . Rewriting this gives:
Also, since is a diameter of circle , must be right.
Now, we consider and . They share an angle - , and we know that Therefore, we have SAS similarity. Therefore, must be right. From there, it follows that all points on circle will be inverted onto the line perpendicular to at .
Therefore, the inversion of circle becomes a line.
Note that, if circle extends beyond , the argument still holds. All one needs to do is shuffle things around.
Inversion of a Circle not intersecting O
Now, we study the inversion of a circle not intersecting the center of inversion. Let us have circle not intersecting , the center of , the circle which we invert around. The points where intersect circle are points and , respectively. Point is arbitrary and on circle . We invert points , , and , producing , , and , respectively. We draw and . Because is a diameter, must be right. We wish to show that circle inverts to another circle.
The definition of inversion tells us that . From here, we obtain that and By SAS symmetry (exploiting ), the ratios tell us that:
Therefore, we have and . Note that , which must equal . Therefore, . But . Therefore, . As this holds for any , all points on circle will invert to a point on a circle with diameter .
General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle
This is how circular inversion is useful in the first place - we find the radius of an inverted circle to find the radius of the original circle. Let the original circle be and the inverted circle be , with radii of and , respectively. The radius of the circle of inversion is . We draw the tangent line of circle intersecting O. We know that this is also a tangent line to circle from the result from part 2 - the tangent line, by definition, intersects circle at exactly one point, and for every intersection point, part 2 says that there will be another intersection point. Therefore, the tangent line to circle intersects circle at exactly one point, necessitating this line to be a tangent line. Call the intersections and , respectively. We have .
We have and = . We can write an equation for by dividing:
From the definition of inversion, we have . Subsituting yields:
From Power of a Point, we know that , which equals . Subsistuting gives , and solving for gives:
Alternately,
Problems
Problem 1
In the figure below, semicircles with centers at and and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at ?
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Principles of dermatological practice
Functions of the skin CME
Functions of the skin
Created in 2008.
Learning objectives
Describe the principal functions of the skin and its components
Introduction
Intact skin is essential for life, illustrated by the serious nature of extensive thermal burns - the mortality risk of a 40% total body surface area burn in a 70-year-old patient is 94%.
SecretionsThe surface of the scalp, face and upper trunk of adults is a hydrolipid film made up of sebum, water, salts and metabolic products. Sebum is produced by sebaceous glands within the hair follicular apparatus. The lipids protect against irritants, allergens and certain toxins and prevent water loss.
Eccrine sweat glands produce a salty solution. They are found over most of the body but are often profuse over the scalp and forehead, axillae, palms and soles. This arises in response to exercise and high temperature (internal or environmental).
Apocrine glands are found in axillary, pubic and perianal regions resulting in a sticky secretion prone to malodour. The smell is due to colonization by bacteria.
Protective properties
| Barrier function |
| Physical | Protects the body from mechanical damage i.e. friction and impact. Mainly impermeable to water. Adherent flat cells of stratum corneum, supporting keratinocytes and collagen provide tensile strength. Sebum aids pliability. Elastin allows skin to recover shape after deformation. Subcutaneous tissue absorbs shock. |
| Thermal | Poor conductor of heat, thus a barrier to heat injury, tolerating <40 deg C. Higher temperatures may cause marked tissue destruction. Severe burns rare on palms and soles because of thick epithelial cover. Deep epidermal appendages (scalp) are also protective. |
| Antimicrobial | Protects from bacteria, fungi and viruses. Keratinocytes provide physical barrier and produce cytokines. Sebum and acid mantle repel pathogenic organisms. Commensal bacteria/yeasts protect against pathogens. Skin is a major immunological organ. |
| Chemical | Keratinocytes protect from chemicals. Excessive exposure causes irritant dermatitis and chemical burns. |
| Radiation | Superficial layers of epidermis protect dividing DNA in basal layer and below from excessive UV radiation Due to thickness of keratinocyte layer, melanin, urocanic acid and appendages. Sunburn rare on palms and soles because of thick epithelial cover. Sunburn rare in black skin because of melanin throughout epidermis. Sunburn greater after bathing because of loss of water-soluble urocanic acid. Hair provides a physical barrier. |
| Maintain fluid balance |
| Evaporation | Adherent keratinocytes and sebum prevent skin from drying out. Sweat moistens skin surface. |
| Permeation | Adherent keratinocytes and sebum prevent absorption of excess water. |
| Excretion | Waste products and salt disposed of in sweat. |
| Regulation of body temperature |
| Cooling | Evaporation of sweat. In hot weather cutaneous blood vessels dilate to give off heat. |
| Warming | Insulating subcutaneous tissue. In cold weather cutaneous blood vessels constrict to conserve heat. Hair protects scalp. Arrector pili contraction (goose bumps) produces heat. |
| Immunological function |
| Infection | Protect against microbial attack. |
| Allergy | Protect against allergic reaction. |
| Metabolic function |
| Synthesis | Vitamin D (subcutaneous tissue) Structural proteins, glycans, lipids, signalling molecules |
| Storage | Fuel in subcutaneous adipose cells Electrolytes, water, vitamins, carbohydrates, protein |
| Wound healing | Secrete fibronectin and other compounds required for restoration of skin integrity. |
| Communication |
| Sensory nerves | Warn of heat, cold, contact, mechanical injury, infestation. Temperature, pain, touch, vibration, pruritus. |
| Physical state | Aesthetic appearance. Odour (sexual attraction): apocrine gland secretion. Deodorant (low level of bacterial decomposition). |
| Emotional state | Facial expressions. Skin colour (erythema, pallor). Sweating, goose bumps. |
Skin permits a stable internal environment, not just holding the body together but also protecting internal organs from environmental dangers.
Skin colour
Skin colour varies from white (absence of pigmentation) to black (densely melanized). The colour depends on the quantity and depth of melanin and other chromophores.
| | |
--- |
| Oxidised haemoglobin | Red |
| Deoxidised haemoglobin | Blue |
| Haemosiderin | Red-brown |
| Carotene | Orange-yellow |
| Bilirubin bound to elastin | Green-yellow |
| Dermal exogenous pigment | Tattoos, makeup, drugs, metallic compounds |
The thickness and quality of keratinocytes and dermal components may also affect skin colour. A thick layer of horny cells can appear white (psoriasis), yellow (seborrhoeic dermatitis), dirty-brown (ichthyosis) or black (eschar). Inflammatory infiltrates may result in plaques that are yellow-brown (granulomas), violaceous (lichen planus) or scarlet (psoriasis).
The pigment melanin is formed by the action of tyrosinase on dopamine metabolites in melanosomes (specialised pigment granules) in the cytoplasm of melanocytes. Skin produces predominantly brown to black eumelanin and a little phaeomelanin; the hair of red-headed celts contains mainly phaeomelanin.
The amount of melanin depends on:
Genetic factors: ethnicity/phototype.
Active melanogenesis.
There are melanocytes in the normal skin of all races. However, dark-skinned people produce more melanin, and it is distributed to keratinocytes throughout the epidermis. The more superficial the pigment, the more effective the protection it provides against damage caused by ultraviolet radiation. Albinos have inactive melanocytes.
Melanin is found in the dermis in some melanocytic naevi (moles) and as a result of inflammation affecting the level of the basement membrane (postinflammatory pigmentation).
Cutaneous colours
Variation in melanisation
Melanin pigmentation is promoted by:
Ultraviolet radiation
Hormones (MSH, ACTH, androgens, oestrogens, progesterones)
Inflammation
Friction
It may be reduced by:
Melatonin
Corticosteroids
Activity
Compare phaeomelanin and eumelanin. What is the explanation for variations in skin colour between different individuals and different races?
#### Previous Chapter
Structure of the dermis quiz #### Next Chapter
Structure and function of the skin quiz
References:
Dermatology resources
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Hooke's Law is valid upto what limit?
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My textbook states:
" the extension produced in the wire is directly proportional to the load applied,within elastic limit."
But my Physics professor said that it is valid upto only proportionality limit.
My guess : It should be upto elastic limit as the body should not reform and behave as a plastic.
elasticity
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asked Dec 9, 2015 at 6:23
Mohammad AbidMohammad Abid
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Your prof is correct: there are two distinct limits here dealing with two different concepts:
The proportionality limit, which defines the maximum stress / strain that an object can undergo before the relationship $\sigma \propto \epsilon$ ($\sigma$ is stress, $\epsilon$ is strain) becomes invalid. If you think about it, this notion needs to be further qualified to be fully meaningful: in general the relationship is nonlinear i.e. $\sigma = f(\epsilon)$ for some general nonlinear function $f$ that approximates to $\sigma=k\,\epsilon$ for small $\epsilon$. So, to be clear, one must define how much one deviates from linearity to define a limit: there is always some deviation. You might specify a "proportional to within 1%" limit, for example.
The elastic limit, which defines the maximum stress / strain that an object can withstand and respond reversibly to stress in the sense that the body returns exactly to the same, zero-stress shape when the stress is released as it had before the application of stress and the body's stress / strain relationship stays the same upon the release of stress and its re-application. Strain beyond the elastic limit leads to a change of the body's zero-stress state shape.
To my knowledge, the limiting stress /strain in 2. is always greater than that in 1., sometimes considerably so. I'm not aware of any materials where 1. is greater than 2.; such behavior would be weird and interesting and perhaps someone else can answer whether it is possible (but that's another question).
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answered Dec 9, 2015 at 7:07
Selena BallerinaSelena Ballerina
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I believe the definition section of the following wikipedia page will be of use:
Your teacher is correct to say Hooke's law is only really valid up to the proportionality limit, and this is simply because Hooke's law is a proportionality law, i.e. Force (or stress) is proportional to extension (or strain). Beyond the proportionality limit, by definition the force and extension are no longer proportional and so Hooke's law is not entirely valid.
However, the elastic behaviour itself, where deformations are temporary, lasts up to the elastic limit. This does indeed mean there is a region where you obtain non-proportional elastic behaviour, between the proportionality and elastic limits. This non-linearity arises here because, in my best understanding, Hooke's law is simply an approximation that is accurate until you approach the limit of elasticity. For metals, like steel, this non-linear elastic region is usually small, and in some practical cases it is ignored altogether! However, materials like elastomers will have a much larger non-linear elastic limit.
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answered Dec 9, 2015 at 7:07
InvoluteInvolute
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In this context, elastic means that the force is proportional to the stretch.
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answered Dec 9, 2015 at 6:50
SpirkoSpirko
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Notice, let $\delta$ be the extension produced by applying a load say $F$ then Hooke's law in mathematical form is $$\delta\propto F\ \ \ \ or\ \ F=k\delta$$ where, $k$ is a constant
The linear relationship $F=k\delta$ (between the extension $\delta$ & the load applied $F$) holds good upto proportionality limit i.e. the extension is linearly proportional to the load applied
From the point of proportionality limit to the point of elastic limit, the linear relationship $F=k\delta$ doesn't hold good i.e. the graph between load vs extension is nonlinear the extension $\delta$ is not directly proportional to the load $F$ although the material can reform.
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answered Dec 9, 2015 at 7:20
Harish Chandra RajpootHarish Chandra Rajpoot
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9029 | https://www.quora.com/If-both-the-body-of-a-Carnot-engine-have-the-same-temperature-what-will-the-efficiency-be | Something went wrong. Wait a moment and try again.
If both the body of a Carnot engine have the same temperature, what will the efficiency be?
Cristobal Cortes
Professor, Thermal Engineering at University of Zaragoza (1989–present)
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Author has 2.1K answers and 2.7M answer views
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3y
Curiously, this question was explicitly addressed by Carnot in his book. A .pdf file of the facsimile of the first English translation has been available since the very beginning of the Internet:
Very simply: If the hot temperature is equal to the cold one, say T K, then efficiency is exactly zero: 1 - T/T = 1 -1 = 0.
In modern terms of a thermodynamic cycle, this means that the box of a Carnot cycle in the TS diagram is degenerated to a line, of height zero and a given length. Thus the area which is the work is zero, but the heat from the hot place is equal to that to the cold place, and not z
Curiously, this question was explicitly addressed by Carnot in his book. A .pdf file of the facsimile of the first English translation has been available since the very beginning of the Internet:
Very simply: If the hot temperature is equal to the cold one, say T K, then efficiency is exactly zero: 1 - T/T = 1 -1 = 0.
In modern terms of a thermodynamic cycle, this means that the box of a Carnot cycle in the TS diagram is degenerated to a line, of height zero and a given length. Thus the area which is the work is zero, but the heat from the hot place is equal to that to the cold place, and not zero, but equal to the area below the line down to 0 K (i.e, T times the increment of entropy). Thus, efficiency is zero.
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11mo
The efficiency of a Carnot engine is given by the formula:
Efficiency=1−TCTH
where TH is the temperature of the hot reservoir and TC is the temperature of the cold reservoir. Both temperatures must be in absolute units (Kelvin).
If both reservoirs are at the same temperature (i.e., TH=TC), the efficiency can be expressed as:
Efficiency=1−THTH=1−1=0
This means that the efficiency of the Carnot engine would be 0%. In this scenario, the engine cannot do any work because there is no temperature difference to drive the heat transfer, which is essen
The efficiency of a Carnot engine is given by the formula:
Efficiency=1−TCTH
where TH is the temperature of the hot reservoir and TC is the temperature of the cold reservoir. Both temperatures must be in absolute units (Kelvin).
If both reservoirs are at the same temperature (i.e., TH=TC), the efficiency can be expressed as:
Efficiency=1−THTH=1−1=0
This means that the efficiency of the Carnot engine would be 0%. In this scenario, the engine cannot do any work because there is no temperature difference to drive the heat transfer, which is essential for work to be done.
John Wang
Studied Materials Science and Engineering at University of Toronto (Graduated 2024)
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Author has 1.2K answers and 348.3K answer views
·
3y
A Carnot Engine, as with any heat engine, need a heat source for heat addition and cold source for heat rejection. Heat only flow from hot to cold if both sources have same temperature then intuitively there will be no net heat flux thus it can not do any work regardless of how much heat is passing through is (since there is no net heat flux where equilibrium heat in = equilibrium heat out).
The equation for Carnot efficiency is Eff =1 - Tcold/ Thot so is both source is same where T cold = T hot then Eff =1 - 1/1 = 1–1 = 0%
Related questions
What will happen to the efficiency of a Carnot engine if the temperature of source and sink is increased by the same amount?
What is the efficiency of a Carnot engine working between a temperature of 500k and 300k?
The Stirling (without a regenerator) and Carnot engines (both reversible) have different efficiencies. Doesn't this contradict Carnot's theorem?
Upon decreasing the temperature of the sink, what is the efficiency of the Carnot engine?
If the temperature of a reservoir of carnot engine working with efficiency 70% is 1000 k then what is the temperature of sink?
Ashish Kori
Project Intern at Tata Steel
·
Author has 159 answers and 624.9K answer views
·
7y
Related
How much is the efficiency of a Carnot engine?
We can evaluate Carnot engine efficiency by..
So you have to see temperature of heat source & heat sink or output of heat engine & energy consumed by it to produce that output.
There's no specific fixed efficiency of Carnot engine. It Solely depends on temperature of source & sink. What it tells you is… maximum possible efficiency that can be obtained while operating between that source & sink.
We can't achieve efficiency more than that.
That's it.
I hope you got this.
PEACE OUT !!!
We can evaluate Carnot engine efficiency by..
So you have to see temperature of heat source & heat sink or output of heat engine & energy consumed by it to produce that output.
There's no specific fixed efficiency of Carnot engine. It Solely depends on temperature of source & sink. What it tells you is… maximum possible efficiency that can be obtained while operating between that source & sink.
We can't achieve efficiency more than that.
That's it.
I hope you got this.
PEACE OUT !!!
Tee Vee
BE from Goa University (Graduated 2018)
·
Author has 52 answers and 86.9K answer views
·
7y
Related
What is the efficiency of a Carnot engine when the temperature of source and sink are the same?
Thanks for A2A
Efficiency of carnot engine=(T1-T2)/T1=(Q1-Q2)/Q1.
Where T1 & Q1 are temperature and heat energy at source. Where T2 & Q2 are temperature and heat energy at sink.
If T1=T2, then T1-T2 =0, which means 0% efficiency. The flow of heat requires a temperature gradient between source and sink. If no heat flows between source and sink then engine is not able to produce work because it would v
Thanks for A2A
Efficiency of carnot engine=(T1-T2)/T1=(Q1-Q2)/Q1.
Where T1 & Q1 are temperature and heat energy at source. Where T2 & Q2 are temperature and heat energy at sink.
If T1=T2, then T1-T2 =0, which means 0% efficiency. The flow of heat requires a temperature gradient between source and sink. If no heat flows between source and sink then engine is not able to produce work because it would violate the 2nd Law of Thermodynamics (Kelvin–Planck statement) which states:
It is impossible to devise a cyclically operating device, the sole effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver...
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Mark Roseman
Biochemistry Professor, Emeritus (2020–present)
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Author has 9.1K answers and 18.3M answer views
·
4y
Related
Why is the efficiency of a Carnot engine independent of the nature of the working substance?
The simplest proof I have seen just appeared in my feed a couple days ago, written by John Leylegian. See below. Two underlying facts may not be obvious to some readers but are necessary to understand the proof:
(1) Since every step in the cycle is reversible, the formula Q=TΔS applies to the isothermal steps; (2) Since entropy is a state function, it must sum to zero around the cycle. This means delta S in the first isothermal step must equal minus delta S in the second isothermal step.
John Leylegian
· 4y
How does the efficiency of a Carnot engine depend on the temperature of the source and sink and independent of the working substance?
The efficiency is independent of the working fluid due to the nature of the processes involved in the cycle. The heat transfer terms in and out and isothermal, so it can be shown that these terms would be equal to Q_{in} = T_H \Delta S Q_{out} = T_L \Delta S Both changes in entropy are the same, since the other two processes in the cycle are isentropic. Therefore, you can find the thermal efficiency: \eta_{th} = 1-\frac{Q_{out}}{Q_{in}} = 1-\frac{T_L \Delta S}{T_H \Delta S}=1-\frac{T_L}{T_H}
The textbook proof (Carnot’s) is more elaborate, but also elegant in that it does
The simplest proof I have seen just appeared in my feed a couple days ago, written by John Leylegian. See below. Two underlying facts may not be obvious to some readers but are necessary to understand the proof:
(1) Since every step in the cycle is reversible, the formula Q=TΔS applies to the isothermal steps; (2) Since entropy is a state function, it must sum to zero around the cycle. This means delta S in the first isothermal step must equal minus delta S in the second isothermal step.
John Leylegian
· 4y
How does the efficiency of a Carnot engine depend on the temperature of the source and sink and independent of the working substance?
The efficiency is independent of the working fluid due to the nature of the processes involved in the cycle. The heat transfer terms in and out and isothermal, so it can be shown that these terms would be equal to Q_{in} = T_H \Delta S Q_{out} = T_L \Delta S Both changes in entropy are the same, since the other two processes in the cycle are isentropic. Therefore, you can find the thermal efficiency: \eta_{th} = 1-\frac{Q_{out}}{Q_{in}} = 1-\frac{T_L \Delta S}{T_H \Delta S}=1-\frac{T_L}{T_H}
The textbook proof (Carnot’s) is more elaborate, but also elegant in that it does not require the establishment of entropy as a state function, just a statement of the Second Law. It is proof by contradiction. The idea is to imagine a Carnot engine with a novel working substance that is more efficient than the formula for maximal efficiency allows. Carnot showed (on paper) that if such an engine were coupled to a normal engine operating as a heat pump, the combination would be a perpetual motion machine of the second kind. Since the existence of such a device violates the Second Law, the hypothetical super engine is impossible.
The logic is as follows. A heat engine takes a quantity of heat from a hot reservoir, q2, discards some of it to a cold reservoir, −q1, and performs net work, −w. Due to the conservation law
-w_{net) = q_{2} - (-q_{1}) = q_{2} + q_{1}
The sign convention makes this confusing. It says in words that net work output in joules is equal to joules of heat taken from the hot reservoir minus joules of heat discarded to the cold reservoir.
The basic definition of efficiency is
ϵ=−wq2=q2+q1q2
If two engines extract the same amount of heat from the hot reservoir, the more efficient engine discards less to the cold reservoir. Hence, the net work output of it is higher and q1 is less than for the normal engine.
The coupling of two engines is shown below. Engine 1 is normal, whereas Engine 2 has a higher efficiency owing to a novel working substance. Engine 1 in this scheme is operating in reverse as a heat pump. The absolute values of the work and heat variables for it are the same, but the signs are reversed.
Image source: Moore’s Physical Chemistry, 3rd Edition
Let prime refer to the variables of Engine 2. We know that
w′>w and q1>q′1
This means the effect of turning both devices through one cycle at the same time is a net output of work from heat abstracted from the cold reservoir:
−w"=w′−w
q"=q′1−q1
Again, the sign conventions make it confusing, but simply stated, Engine 2 puts less heat into the cold reservoir than Engine 1 takes out. Hence the net flow of heat from the cold reservoir to the hot reservoir, and from there to work output.
There is no net change in the system of either engine per cycle, but there is net work obtained from abstraction of heat from a single reservoir. The impossibility of such a device, namely a perpetual motion machine of the second kind, is a statement of the Second Law.
It could be put aboard a ship and turn the props continuously, using heat abstracted from the ocean as the only source of energy.
Source: Moore’s Physical Chemistry, 3rd Edition, pp 77–80.
Related questions
What is the Carnot cycle? What are its limitations, if any?
A Carnot engine with a hot body temperature of 100°C has 50% efficiency. By how much should the temperature of its cold body be changed to increase the resultant efficiency to 70%?
Which engine has efficiency near the Carnot engine in practical?
How do I find the source of temperature in a Carnot engine if the efficiency is 60%?
What is the efficiency of a Carnot engine when the temperature of source and sink are the same?
George Madathiparampil Varkey
Author has 739 answers and 3.2M answer views
·
9y
Related
Why is the carnot engine the most efficient?
Carnot cycle consists of 4 process. All of them are ideal and practically not possible.
Isothermal heat addition- Nothing can be more efficient than this heating since no finite temperature difference exists between heat source and receiver. Hence it is a reversible process and most efficient,
Isentropic(reversible adiabatic) expansion- Most efficient expansion, as no heat loss is taking place, full energy is utilised only for expansion and thereby doing work.
Isothermal heat rejection- Most efficient heat rejection to a heat sink at same temperature.
Isentropic compression- Compressed(without fri
Carnot cycle consists of 4 process. All of them are ideal and practically not possible.
Isothermal heat addition- Nothing can be more efficient than this heating since no finite temperature difference exists between heat source and receiver. Hence it is a reversible process and most efficient,
Isentropic(reversible adiabatic) expansion- Most efficient expansion, as no heat loss is taking place, full energy is utilised only for expansion and thereby doing work.
Isothermal heat rejection- Most efficient heat rejection to a heat sink at same temperature.
Isentropic compression- Compressed(without friction) with no heat lost to surrounding. Power input is used only for increasing the pressure and temperature.
When the cycle is completed input energy and power is used only for the purpose it was intended. Output was completely used for intended purpose.
Since no other process can be more efficient, Carnot cycle is the most efficient.
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Jul 16
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Manojkumar V P
BE in Mechanical Engineering Sandwich, PSG College of Technology, Tamil Nadu, India (Graduated 2019)
·
7y
Related
In a Carnot engine, if the temprature of the sink is increased, what will happen to its efficiency?
First, let's discuss and understand significant of sink and source temperature in Carnot cycle.
Let me explain this through Carnot heat engine cycle.
1–2 Reversible isothermal heat addition
2–3 Isentropic expansion
3–4 Reversible isothermal heat rejection
4–1 Isentropic compression
Efficiency = output/input = Net workdone/heat input
Hr - heat rejection
Ha - heat addition
ds - change in entropy
ds = dq/T
From the above figure, we can understand that the efficiency of Carnot heat engine is purely depends on difference between source and sink temperature and it is independent of fuel properties.
Let's come t
First, let's discuss and understand significant of sink and source temperature in Carnot cycle.
Let me explain this through Carnot heat engine cycle.
1–2 Reversible isothermal heat addition
2–3 Isentropic expansion
3–4 Reversible isothermal heat rejection
4–1 Isentropic compression
Efficiency = output/input = Net workdone/heat input
Hr - heat rejection
Ha - heat addition
ds - change in entropy
ds = dq/T
From the above figure, we can understand that the efficiency of Carnot heat engine is purely depends on difference between source and sink temperature and it is independent of fuel properties.
Let's come to your question. If sink temperature increases, there will be decrease in temperature gradient and efficiency will fall down.
Robert Reiland
Thermodynamics was my best subject in graduate school
·
Author has 5.6K answers and 12.3M answer views
·
9y
Related
Why is the carnot engine the most efficient?
The efficiency of a heat engine is simply the work output divided by the net heat added to the system, where heat is really the energy coming in from an external hot reservoir as a result of a temperature difference minus the energy going out to a cold reservoir as a result of a temperature difference.
The Carnot Cycle is constructed by imagining that the heating can at first be done with no temperature change of the system. In reality that's not going to happen except during a phase change of the working fluid in the system, and even then it can't happen in a practical sense, but it can be ap
The efficiency of a heat engine is simply the work output divided by the net heat added to the system, where heat is really the energy coming in from an external hot reservoir as a result of a temperature difference minus the energy going out to a cold reservoir as a result of a temperature difference.
The Carnot Cycle is constructed by imagining that the heating can at first be done with no temperature change of the system. In reality that's not going to happen except during a phase change of the working fluid in the system, and even then it can't happen in a practical sense, but it can be approximated to a high degree. The working fluid, a gas, is expanding and can work on something during this step.
Next the system undergoes an increase in temperature as a perfectly insulated system, and the gas continues to expand and does more work on something. Perfect insulation is not possible, but again, this can be approximated.
In the third step the insulation is in some sense discarded, and the system transfers energy (heat) to an external reservoir at constant temperature. Again, there are problems with this, and the result can only be approximated.
In the final step the system is returned exactly to its initial state through compression, and work is don on the system, but the system is perfectly insulate, and heat is not involved.
Since the system is exactly returned to its initial state, the net change in entropy of the system is zero. Throughout the cycle the processes were reversible. In principle everything could have been run backwards to produce the effects of refrigeration.
Regarding efficiency the key question is whether or not more work could have been done with the net energy input (heat in - heat out). A temperature-entropy diagram (Carnot cycle ) for the process and the accompanying calculation of work done for the cycle shows that there would be no way to get more work done for the net energy input.
In effect all of the net energy transferred into the system from the hot and cold reservoirs (energy in from the hot reservoir, energy out to the cold reservoir) that can do work does do work. In any real system in which the entropy change would be positive, some of the net energy input would become internal energy in heating parts of the system without doing work, and efficiency would be less.
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Arpit Shankar
Studied Mechanical Engineering at UPES, Dehradun (Graduated 2020)
·
7y
Related
Can any engine have efficiency greater than a Carnot engine?
No engine can have the efficiency more than a Carnot cycle according to Carnot theorem “Heat engines operating between a given constant temperature source and given temperature sink none has a higher efficiency than a reversible engine.
Above shown E1 is reversible heat engine and E2 be any heat engine. Assuming n2>n1( efficiency)
this leads to W2/Q2>W1/Q1
W2>W1
Now let E1 be reversed(HP1) , the magnitude of heat and work transfer qualities will be same but direction will be reversed. As W2>W1( W2=W1+X).Some part(X) of E2 may be fed to reverse engine E1.
As heat discharge by E1 may be supplied to E
No engine can have the efficiency more than a Carnot cycle according to Carnot theorem “Heat engines operating between a given constant temperature source and given temperature sink none has a higher efficiency than a reversible engine.
Above shown E1 is reversible heat engine and E2 be any heat engine. Assuming n2>n1( efficiency)
this leads to W2/Q2>W1/Q1
W2>W1
Now let E1 be reversed(HP1) , the magnitude of heat and work transfer qualities will be same but direction will be reversed. As W2>W1( W2=W1+X).Some part(X) of E2 may be fed to reverse engine E1.
As heat discharge by E1 may be supplied to E2, therefore the source may be eliminated and this will lead to a cyclic engine with one which violates the “Kelvin–Planck statement “. Hence our assumption is wrong!
n1>n2( Always)
-Appy
Fred McGalliard
Retired scientist
·
Author has 8.3K answers and 1.1M answer views
·
1y
Related
The efficiency of a Carnot engine is 0.4. If the temperature of the sink is 27 degrees, what is the temperatures of the source?
First this suggests an error in defining the configuration. The Carnot efficiency is based on an input of flame or electricity (effective temp very very high), that is then cooled down to something the boiler can withstand. That colder temp is the temp of the source. Note that to get there we just gave up a hella lot of the available energy since it is too hot to handle. If the input temp were, for example, produced by a heat pump drawing from the 27C source, then the efficiency could be 100% minus friction and some heat loss along the way. Note that this assumes a closed cycle, and most ICs r
First this suggests an error in defining the configuration. The Carnot efficiency is based on an input of flame or electricity (effective temp very very high), that is then cooled down to something the boiler can withstand. That colder temp is the temp of the source. Note that to get there we just gave up a hella lot of the available energy since it is too hot to handle. If the input temp were, for example, produced by a heat pump drawing from the 27C source, then the efficiency could be 100% minus friction and some heat loss along the way. Note that this assumes a closed cycle, and most ICs run open cycle and I am just going to ignore that for purposes of this post. Doing real engineering takes more work than you might imagine. I think the term is Xergy for this kind of system. So let me example the two cases, one, I simply place my hot source against my cold source and let the dogs out. There is no energy output, the efficiency is zero. The result is a combined hot/cold at some intermediate value where the total heat energy does not change but the available energy goes to zero. Alternative, a heat engine and heat pump with no losses. Pump to get a source and sink with different temperatures. Now run the heat pump from that, and without loss we get back all our energy (100% efficient because, no losses right?), and the temp Dif of the source/sink goes to 0. A normal application for such a system would be as an energy storage system. A thermal battery if you will. If you try to use the Carnot efficiency you will likely get the wrong answer because the details matter a lot. Note that my thermo is around 60 years old and I have not worked directly with thermo issues so my input is only mildly trustworthy.
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Farooq Sk
24 M
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Author has 136 answers and 549.9K answer views
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6y
Related
A Carnot engine whose low temperature reservoir is at 7°C has an efficiency of 50%. It is desired to increase the efficiency to 70%. By how many degrees should the temperature of the high temperature reservoir be increased?
This problem is fairly easy, if you are unable to solve this I recommend you to study the theory of carnot engines again.
If you got the theory and the formula right, you might be doing the mistake of not converting Celsius to Kelvin.
First case
η = 0.5
T_low = 273 + 7 = 280
Find T _high
Second case
η = 0.7
T_low = 273 + 7 = 280
Find T _high
Subtract T_high values of 2 cases and you will have your answer.
This problem is fairly easy, if you are unable to solve this I recommend you to study the theory of carnot engines again.
If you got the theory and the formula right, you might be doing the mistake of not converting Celsius to Kelvin.
First case
η = 0.5
T_low = 273 + 7 = 280
Find T _high
Second case
η = 0.7
T_low = 273 + 7 = 280
Find T _high
Subtract T_high values of 2 cases and you will have your answer.
Dave Yerzley
Author has 692 answers and 2.5M answer views
·
3y
Related
What is meant by Carnot cycle and Carnot engine?
What is meant by Carnot cycle and Carnot engine?
If you really want to understand the Carnot cycle, you need to do the work yourself to understand what it is and its significance. It’s easy to find information about it because it’s a fundamental part of thermodynamics. However, I realize that many people would rather ask a question on Quora than make any effort themselves, so I’ll give a brief summary.
The Carnot cycle is a thermodynamic cycle applicable to heat engines, proposed by Sadi Carnot, a French engineer, in the early 19th century. It consists of a sequence of isothermal expansion, adia
What is meant by Carnot cycle and Carnot engine?
If you really want to understand the Carnot cycle, you need to do the work yourself to understand what it is and its significance. It’s easy to find information about it because it’s a fundamental part of thermodynamics. However, I realize that many people would rather ask a question on Quora than make any effort themselves, so I’ll give a brief summary.
The Carnot cycle is a thermodynamic cycle applicable to heat engines, proposed by Sadi Carnot, a French engineer, in the early 19th century. It consists of a sequence of isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. At the end of the cycle the conditions duplicate those at the beginning, so it can be repeated indefinitely. If these steps are performed very slowly, then they are reversible. Using basic principles, it’s possible to compute the thermodynamic efficiency of the cycle, which is:
E=T2−T1T2 where
E is efficiency, defined as work out divided by net heat energy in,
T2 is the maximum temperature occurring in the cycle and
T1 is the minimum temperature occurring in the cycle.
It can also be shown that no other cycle, and specifically no irreversible cycle, can have an efficiency higher than the Carnot cylce.
Therefore it sets the maximum possible efficiency obtainable from a heat engine operating with a given set of parameters.
There are a couple of remarkable things about Carnot’s achievement. One was that the commonly accepted theory of heat at the time was the caloric theory, which held that heat was some sort of fluid that moved around, carrying heat with it. Another was that the laws of thermodynamics had not yet been established, and even the law of conservation of energy was not generally accepted. Carnot’s derivation relied on principles so fundamental that it didn’t even depend on either of these things.
Related questions
What will happen to the efficiency of a Carnot engine if the temperature of source and sink is increased by the same amount?
What is the efficiency of a Carnot engine working between a temperature of 500k and 300k?
The Stirling (without a regenerator) and Carnot engines (both reversible) have different efficiencies. Doesn't this contradict Carnot's theorem?
Upon decreasing the temperature of the sink, what is the efficiency of the Carnot engine?
If the temperature of a reservoir of carnot engine working with efficiency 70% is 1000 k then what is the temperature of sink?
What is the Carnot cycle? What are its limitations, if any?
A Carnot engine with a hot body temperature of 100°C has 50% efficiency. By how much should the temperature of its cold body be changed to increase the resultant efficiency to 70%?
Which engine has efficiency near the Carnot engine in practical?
How do I find the source of temperature in a Carnot engine if the efficiency is 60%?
What is the efficiency of a Carnot engine when the temperature of source and sink are the same?
The exhaust temperature of a heat engine is 250 C. What must be the high temperature if the Carnot efficiency is to be 35%?
What are the disadvantages of using a carnot engine?
What will be the difference in the efficiency of a Carnot heat engine if the temperature of the hot reservoir is quadrupled?
Under what condition does an ideal Carnot engine have 100% efficiency?
What is the primary feature of a Carnot engine?
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5
How can you show that every prime is either of the form 4n+1 or of the forms 4n-1?
Mohamed Mostafa
B.Sc. in Mathematics, Arizona State University (Graduated 2025)
·
Author has 223 answers and 847.6K answer views
·
5y
Ah, Dirichlet’s theorem on arithmetic progressions
for d=4 and a=±1.
The theorem states that for any arithmetic progression (aka AP) a,a+d,a+2d,a+3d,… (i.e. a sequence where terms differ by a constant), if gcd(a,d)=1 (a and d don’t share any factors other than 1), then the AP contains infinitely many primes.
Example: d=4, a=1.
Notice that gcd(1,4)=1. We have a fancy name for numbers that have a gcd of 1; they are called coprime.
Sequence (1) is 1,5,9,13,17,21,…
Yes, there might be some numbers that are not prime, like 1, 9
Dirichlet's theorem on arithmetic progressions - Wikipedia
Ah, Dirichlet’s theorem on arithmetic progressions
for d=4 and a=±1.
The theorem states that for any arithmetic progression (aka AP) a,a+d,a+2d,a+3d,… (i.e. a sequence where terms differ by a constant), if gcd(a,d)=1 (a and d don’t share any factors other than 1), then the AP contains infinitely many primes.
Example: d=4, a=1.
Notice that gcd(1,4)=1. We have a fancy name for numbers that have a gcd of 1; they are called coprime.
Sequence (1) is 1,5,9,13,17,21,…
Yes, there might be some numbers that are not prime, like 1, 9, or 21, but the sequence contains infinitely many primes, such as 5, 13, and 17, because 1 and 4 are coprime.
Similarly, since gcd(4,−1)=1 (just accept that as given), we have sequence (2) −1,3,7,11,15,19,23,… has infinitely many primes.
Now that we are out of the way, I’ll actually read the question
How can you show that every prime is either of the form 4n+1 or of the forms 4n-1?
Hmm… apparently only sequences (1) and (2) have infinitely many primes. If you have a prime, then it belongs to either of these sequences only.
Why? Well, there are 4 possible arithmetic sequences with d=4. We checked 2 of them. We only need to check 2 more.
Sequence 3: 0,4,8,12,16,20,…
This sequence is generated by 4(n−1),n=1,2,3,…. Obviously, we can factorize the term 4(n−1) into (2)(2)(n−1). This number has at least two prime factors, so this number can never be prime. Next.
Sequence 4: 2,6,10,14,18,22,…
This sequence is actually interesting. It only has one prime number in fact (which is 2, the oddest prime ever), but why? Not two primes, or three? We’ll see. The sequence is generated by 4n−2. We can factorize out a 2 to get 2(2n−1). This number has one prime factor, so we need the other to be the unit 1. 2n−1=1 gives 2n=2, or just n=1, which gives us 4−2=2, the only prime in this sequence.
And we are done. One might also check that these 4 A.P.s partition the natural numbers (with 0 and -1), but it can be easily seen as 4N,4N+1,4N+2,4N+3 are residue (equivalence) classes of N.
(You can stop reading here)
—
Do you have to know the jargon I just said? No, but it is important to know Dirichlet’s theorem in order to generalize the results.
Corollary of Dirichlet’s Theorem: For a number d, almost every prime number can be written as a+nd where n∈Z and gcd(a,d)=1.
Proof: For the case gcd(a,d)=1, we know by Dirichlet’s theorem that infinitely many primes can be written as a+nd. To prove our corollary, we need to show that only finitely many primes can be written as a+nd where a and d are not coprime.
By contradiction, assume gcd(a,d)=g≠1 (or a=0, by which g=d). Then a=pg and d=qg, and hence,
a+nd=g(p+nq). Since g≠1, then a+nd has g as a factor. In the extreme case, a+nd is prime iff p+nq=1, which would result in only 1 value of n. All other n would make a+nd not prime, but we assumed that there would be infinitely many prime numbers, not finitely (or not even 1).
Thus, by contradiction, we proved our corollary. ■.
—
Phew! That was long. Didn’t think it would turn it out long. Our proof used the advanced Dirichlet’s theorem. Although for our problem, we could have gotten away with proving elementarily that there are infinitely many primes of the form 4n±1. But we needed the general theorem for the general case that the question didn’t ask about. Anyway, don’t ask me to prove Dirichlet’s Theorem : ).
—
Appendix: Primes can be expressed as
n (Very self-explanatory. All primes are integers, at least on our ring)
2n+1 (except 2)
3n+1, 3n+2 (except 3)
4n+1, 4n+3=4m−1 (except 2)
5n+1, …, 5n+4 (except 5)
6n+1, 6n+5 (except 2 and 3)
7n+1 to 7n+6 (except 7)
8n+an odd number (except 2, why am I getting a deja vu)
9n+a number not divisible by 3 (except 3)
10n+an odd number other than 5 (basically, except the prime factors of the number)
And etc.
—
And off we go…
Footnotes
Dirichlet's theorem on arithmetic progressions - Wikipedia
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Peter Cooper
.
·
2y
Originally Answered: How can you show that all the prime numbers except 2 are of the form 4n-1 or 4n+1?
·
It is possible to show that all prime numbers except 2 are of the form 4n-1 or 4n+1 by using the following steps:
Recall the definition of a prime number: a prime number is a positive integer greater than 1 that has no positive integer divisors other than 1 and itself.
Observe that all positive integers can be written in the form 4n, 4n+1, 4n+2, or 4n+3 for some integer n.
Note that the numbers 4n and 4n+2 are always even, and therefore cannot be prime.
It follows that all prime numbers except 2 must be of the form 4n-1 or 4n+1.
To prove that all prime numbers except 2 are indeed of this form, we c
It is possible to show that all prime numbers except 2 are of the form 4n-1 or 4n+1 by using the following steps:
Recall the definition of a prime number: a prime number is a positive integer greater than 1 that has no positive integer divisors other than 1 and itself.
Observe that all positive integers can be written in the form 4n, 4n+1, 4n+2, or 4n+3 for some integer n.
Note that the numbers 4n and 4n+2 are always even, and therefore cannot be prime.
It follows that all prime numbers except 2 must be of the form 4n-1 or 4n+1.
To prove that all prime numbers except 2 are indeed of this form, we can proceed by contradiction. Suppose there exists a prime number p that is not of the form 4n-1 or 4n+1. Then p must be either 4n or 4n+2 for some integer n. However, we have already established that 4n and 4n+2 are not prime, so this contradicts the assumption that p is a prime number.
Therefore, we must conclude that all prime numbers except 2 are of the form 4n-1 or 4n+1.
I hope this helps! Let me know if you have any further questions.
Douglas Porter
B Sc in Mathematics, The Open University (Graduated 2011)
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Author has 9.6K answers and 7.7M answer views
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5y
Every prime other than 2. But the point is that all numbers are either:
A multiple of 4 (4n)
One more than (4n + 1)
Two more than (4n + 2)
Three more than (4n + 3, but also 4n - 1 for some other n)
Clearly the first of these are all multiples of four, and the third are all even, so the other two bullet points cover the only numbers that can be prime (of course, 9 and 15 are counter-examples that show that 4n + 1 and 4n - 1 are not always prime).
You can extend this further and show in fact that, other than 2 and 3, primes must be of the form 6n + 1 or 6n - 1.
Related questions
How do I prove that every prime number of form 4n+1 is a sum of two perfect squares?
Is the following Lagrange theorem true, every Prime of the form (4n-1) is the sum of a prime of the form (4n+1) and of double a prime of the form (4n+1), i.e. (23=13+2(5))? Is the reverse also true, (4n+1) = (4n-1) +2 (4n-1), i.e. 29=7+2(11)?
How do we prove that every odd integer is of the form of 4n-1 or 4n+1?
How can you show that all the prime numbers except 2 are of the form 4n-1 or 4n+1?
Are all prime numbers of the form 6n-1 or 6n+1?
Tony Flury
Interested in Mathematics since 1980. Strictly Amateur
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Author has 13.2K answers and 22.7M answer views
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2y
Originally Answered: How can you show that all the prime numbers except 2 are of the form 4n-1 or 4n+1?
·
This must be trivially true.
values of the form 4n & 4n + 2 (where n is an integer) must be divisible by two and therefore cannot be prime, so that only leaves 4n -1 and 4n+1 as options. note that 4n-1 is equivalent to 4n+3.
Of course 4n-1 and 4n+1 could also be composite but not divisible by 2 (for example when n=2, 4n+1 = 9) but at least you can state clearly that they cannot be divisible by 2.
A better determinant for possible primes is 6n-1 or 6n+1: these forms still aren’t always primes but they do exclude multiples of 2 and multiples of 3.
They work in a similar way:
6n : divisible by 2 & 3 -
This must be trivially true.
values of the form 4n & 4n + 2 (where n is an integer) must be divisible by two and therefore cannot be prime, so that only leaves 4n -1 and 4n+1 as options. note that 4n-1 is equivalent to 4n+3.
Of course 4n-1 and 4n+1 could also be composite but not divisible by 2 (for example when n=2, 4n+1 = 9) but at least you can state clearly that they cannot be divisible by 2.
A better determinant for possible primes is 6n-1 or 6n+1: these forms still aren’t always primes but they do exclude multiples of 2 and multiples of 3.
They work in a similar way:
6n : divisible by 2 & 3 - not a prime.
6n+1 not divisible by 2 or 3
6n+2 : divisible by 2
6n+3 : divisible by 3
6n+4 : divisible by 2
6n+5 or 6n-1 : not divisible by 2 or 3
or course again 6n-1 and 6n+1 aren’t guaranteed to be primes - when n is 4, then 6n+1 = 25 but it will immediately exclude more composites than using the 4n +/- 1 form.
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·
1y
To show that every prime number greater than 2 is either of the form 4n+1 or 4n−1, we can analyze the possible remainders when a prime number is divided by 4.
Step 1: Consider the possible remainders
When any integer p is divided by 4, it can have one of the following remainders:
- 0
- 1
- 2
- 3
Step 2: Analyze each case
Remainder 0: If p≡0mod4, then p is divisible by 4. The only prime that is divisible by 4 is 2, but since we are considering primes greater than 2, this case is not possible.
Remainder 1: If p≡1mod4, then p is of the form 4n+1 for some integer n.
Remainder 2
To show that every prime number greater than 2 is either of the form 4n+1 or 4n−1, we can analyze the possible remainders when a prime number is divided by 4.
Step 1: Consider the possible remainders
When any integer p is divided by 4, it can have one of the following remainders:
- 0
- 1
- 2
- 3
Step 2: Analyze each case
Remainder 0: If p≡0mod4, then p is divisible by 4. The only prime that is divisible by 4 is 2, but since we are considering primes greater than 2, this case is not possible.
Remainder 1: If p≡1mod4, then p is of the form 4n+1 for some integer n.
Remainder 2: If p≡2mod4, then p is even. The only even prime is 2, which we are excluding, so this case does not apply to primes greater than 2.
Remainder 3: If p≡3mod4, then p is of the form 4n−1 for some integer n.
Conclusion
From the analysis above, we conclude that for any prime number p>2:
- It cannot be of the form 4n (since it would be divisible by 4).
- It cannot be of the form 4n+2 (since it would be even).
- Therefore, it must be either of the form 4n+1 or 4n−1.
Thus, every prime number greater than 2 is either of the form 4n+1 or of the form 4n−1.
Mohammad Afzaal Butt
B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977)
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Author has 24.6K answers and 22.8M answer views
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4y
Every number can be written in the form of 4 n, 4 n + 1, 4 n + 2 or 4 n - 1
Numbers of the form 4 n can not be prime numbers, since these are divisible by 4.
Similarly numbers of the form 4 n + 2 can not be prime either, because such
numbers are divisible by 2.
Hence the prime numbers can only be of the form 4 n + 1 or 4 n - 1.
But 2 is the only prime number which can not be written in the
form 4 n + 1 or 4 n - 1
Related questions
How do I show that there is an infinity of primes of the form 4r+1?
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p
is it true that
(
p
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1
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!
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(
p
−
1
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mod
(
1
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⋯
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3
x
2
+
1
would be always of the form
3
z
+
1
?
What is the relationship between 4n+-1, 6n+-1,3 30n+-, 210n+-, 2310+-? And 30030+-? In finding prime numbers?
Barry Ford
BA in Mathematics, Hunter College (Graduated 2022)
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Author has 2K answers and 2.2M answer views
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5y
First, 4n+1 will not work as a form for prime numbers.
Proof (informal):
Suppose n is even. Let n=2k, with k an integer. Then 4n+1=4(2k)+1=8k+1, which is not always prime. Example: if k=8, then 8k+1=65, which is a multple of 5.
Suppose n is odd. Let n=2k+1, which is odd by definition of an odd number (k is an integer). Then 4n+1=4(2k + 1) +1=8k+4+1=8k+5. If k=8, then 8k+5=64+5=69, which is a multiple of 3.
But maybe 4n-1 will work. Suppose n is even. Then let n=2k (k an integer). 4n-1=4(2k)-1=8k-1. If k=7, then 8k-1=56–1=55, which is a multiple of 5.
Suppose n is odd. 4(2k+1)-1=8k+4–1=8k+3. If k=9,
First, 4n+1 will not work as a form for prime numbers.
Proof (informal):
Suppose n is even. Let n=2k, with k an integer. Then 4n+1=4(2k)+1=8k+1, which is not always prime. Example: if k=8, then 8k+1=65, which is a multple of 5.
Suppose n is odd. Let n=2k+1, which is odd by definition of an odd number (k is an integer). Then 4n+1=4(2k + 1) +1=8k+4+1=8k+5. If k=8, then 8k+5=64+5=69, which is a multiple of 3.
But maybe 4n-1 will work. Suppose n is even. Then let n=2k (k an integer). 4n-1=4(2k)-1=8k-1. If k=7, then 8k-1=56–1=55, which is a multiple of 5.
Suppose n is odd. 4(2k+1)-1=8k+4–1=8k+3. If k=9, then 8k+3=72+3=75, which is a multiple of 5.
Therefore neither 4n+1 or 4n-1 is a valid form for prime numbers. (Note: I ignored the trivial case, where n=0, for obvious reasons.)
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Terry Darwin
Self-employed Sole Trader (1991–present)
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Author has 10.6K answers and 6.7M answer views
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2y
Originally Answered: How can you show that all the prime numbers except 2 are of the form 4n-1 or 4n+1?
·
Assuming n is an integer, 4n must be an even number, regardless of whether n is odd or even. Therefore 4n+1 and 4n-1 must always be odd. All prime numbers are odd (except 2).
However, you have made no stipulation that the variable n must be an integer - though convention does suggest it.
If n, for example, were to represent a value of 1.75, 4n would be 7 and 4n+or-1 would be 8 or 6 - not prime numbers.
Mpendulo Cele
MSc in Mathematics & Algebra, Stellenbosch University (Graduated 2020)
·
4y
The statement is not true.
2 is a prime number and is not of the given forms.
Any number has remainder i=0,1,2 or 3 after division by 4. Having 0 or 2 mean the number is even hence all odd prime numbers have remainder 1 or 3.
So then are indeed of the form 4n+1 or 4n-1.
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Hilmar Zonneveld
Translator (1985–present)
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Author has 58.5K answers and 19.2M answer views
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2y
Originally Answered: How can you show that all the prime numbers except 2 are of the form 4n-1 or 4n+1?
·
4n-1 is the same as 4n+3 (for a different integer “n”).
For this form of numbers, you only have four options:
4n
4n+1
4n+2
4n+3
4n and 4n+2 are even, i.e., divisible by 2. That means they can’t be prime numbers (except for 2 itself).
Norman Beck
M. S. from Computer Science (college major) (Graduated 1972)
·
Author has 1.5K answers and 120K answer views
·
2y
Originally Answered: How can you show that all the prime numbers except 2 are of the form 4n-1 or 4n+1?
·
Step 1. Every odd number is either of the form 4n-1 or 4n+1. Having eliminated 2 from the candidates, every other even number is divisible by 2, so they cannot be prime.
Step 2. If you don’t get it yet, reread step 1.
Mustafa Alper Gunes
Lives in Bonn, Germany (2020–present)
·
6y
Related
What is the simple method to prove there are infinite primes of the form 4n+1?
We’ll prove this by contradiction:
Assume there exists a finite number of primes of the form 4n+1 and let p1,p2,......,pn be those primes. Now, we define:
a:=2p1p2....pn
k:=a2+1=4(p1p2....pn)2+1.
We know that every prime except 2 is of the form 4n+3 or 4n+1 and obviously, k is not divisible by 2 and any other prime of the form 4n+1 (by our assumption).
Now, consider a prime p of the form 4n+3. Since p obviously does not divide a, by Fermat’s Little Theorem, we know that:
ap−1≡1(modp)⇒a4n+3−1≡1(modp)
We’ll prove this by contradiction:
Assume there exists a finite number of primes of the form 4n+1 and let p1,p2,......,pn be those primes. Now, we define:
a:=2p1p2....pn
k:=a2+1=4(p1p2....pn)2+1.
We know that every prime except 2 is of the form 4n+3 or 4n+1 and obviously, k is not divisible by 2 and any other prime of the form 4n+1 (by our assumption).
Now, consider a prime p of the form 4n+3. Since p obviously does not divide a, by Fermat’s Little Theorem, we know that:
ap−1≡1(modp)⇒a4n+3−1≡1(modp)
⇒a4n+2=(a2)2na2≡1(modp)
Now assume:
k=a2+1≡0(modp)⇒a2≡−1(modp)
⇒ap−1=(a2)2na2≡−1(modp)
which contradicts with Fermat’s last theorem, which implies that there is no prime of the form 4n+3 that divides k=a2+1.
So, according to our initial assumption that there exists finitely many primes of the form 4n+1, k (which is a number of the form 4n+1) is not divisible by 2 and any prime of the form 4n+1 and 4n+3, which means k must be a prime. But our initial assumption was that pn (which is less than k) was the largest prime of the form 4n+1. Hence, we reach a contradiction implying that our initial assumption was wrong, which means there are infinitely many primes of the form 4n+1
Alon Amit
PhD in Mathematics; Mathcircler.
·
Upvoted by
Jeremy Collins
, M.A. Mathematics, Trinity College, Cambridge and , Ph.D. Mathematics, University of Manchester (1982) · Author has 8.7K answers and 172.8M answer views
·
Updated 6y
Related
Are there more prime numbers with form 6n+5 than primes with form 6n+1?
I will explain that the answer is
Yes
No
We’re not sure yet.
Let’s start with #2 since it’s the easiest.
Just as many blues and reds
There are infinitely many red primes (those that are one less than a multiple of 6), and there are infinitely many blue primes (those that are one more than a multiple of 6). This isn’t at all obvious, but it’s a special case of the famous theorem of Dirichlet on arithmetic progressions.
Of course, both sets are countably infinite, so there are just as many blue primes as there are red ones, in the sense of cardinality of sets.
But even more is true. We can assign a dens
I will explain that the answer is
Yes
No
We’re not sure yet.
Let’s start with #2 since it’s the easiest.
Just as many blues and reds
There are infinitely many red primes (those that are one less than a multiple of 6), and there are infinitely many blue primes (those that are one more than a multiple of 6). This isn’t at all obvious, but it’s a special case of the famous theorem of Dirichlet on arithmetic progressions.
Of course, both sets are countably infinite, so there are just as many blue primes as there are red ones, in the sense of cardinality of sets.
But even more is true. We can assign a density to each of those sets, by checking how many primes of each type there are up to N, and letting N tend to infinity. If we do this carefully, we can show that half the primes are blue and the other half are red.
So, as many of each, right?
More reds than blues
Up to 1000, there are 80 blue primes and 86 red ones. So roughly the same, but the red team wins.
Up to 2000 it’s 148:153.
Up to 3000 it’s 207:221.
Up to 4000 it’s 330:337.
Notice a pattern?
The numbers are always quite close, in line with our earlier result that asymptotically, as the range grows, the ratio tends to 1:1. However, the red team displays an uncanny capacity to win the race, apparently every time.
Does it always win?
No. But the smallest number up to which there are more blue than red primes is 608981813029, so you need some patience if you’re a blue team fan.
As far as we can tell, the red team is leading about 99% of the time, which you can take to mean that there are very, very often more reds than blues.
We don’t know.
As far as I know, proving what I just said depends on some widely believed, but unproven, results about the zeros of various L-functions. See the classic paper by Rubinstein and Sarnak for more details. Much progress has been made since 1994, but I believe the basic result is still open.
Related questions
How do I prove that every prime number of form 4n+1 is a sum of two perfect squares?
Is the following Lagrange theorem true, every Prime of the form (4n-1) is the sum of a prime of the form (4n+1) and of double a prime of the form (4n+1), i.e. (23=13+2(5))? Is the reverse also true, (4n+1) = (4n-1) +2 (4n-1), i.e. 29=7+2(11)?
How do we prove that every odd integer is of the form of 4n-1 or 4n+1?
How can you show that all the prime numbers except 2 are of the form 4n-1 or 4n+1?
Are all prime numbers of the form 6n-1 or 6n+1?
How do I show that there is an infinity of primes of the form 4r+1?
How can we show that for every prime number
p
is it true that
(
p
−
1
)
!
≡
(
p
−
1
)
mod
(
1
+
2
+
3
+
⋯
+
(
p
−
1
)
)
?
What is the simple method to prove there are infinite primes of the form 4n+1?
How do I show that all the prime factors of the odd composite numbers of the form
3
x
2
+
1
would be always of the form
3
z
+
1
?
What is the relationship between 4n+-1, 6n+-1,3 30n+-, 210n+-, 2310+-? And 30030+-? In finding prime numbers?
Are there infinitely many primes of the form 6k + 1?
How do you show that
2
1
×
3
2
×
5
4
×
7
6
⋯
×
p
p
−
1
<
3
ln
(
p
)
, where
p
is prime?
Does 4n-1 give primes?
How do you prove that there are infinitely many primes of the form 4n + 3?
Why do primes prefer the form
$
6
n
−
1
$
instead of
$
6
n
+
1
$
(elementary number theory, prime numbers, math)?
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9031 | https://www.senpaicorner.com/_files/ugd/1a1fe2_37cdfaf1edbd4735b0ad861a8be1efaa.pdf?index=true | Senpaicorner.com Physics
Signature : Name: Marks :
Charge, Current, Voltage, Energy and Power \1
Charge, Current, Voltage, Energy and Power Worksheet
Q1.
A charge of 30 C flows through an electrical circuit in 20 seconds. What is the current
flowing in the circuit?
Q2 .
2.25x 10 20 electronics flows through a bulb in 2 minutes. Calculate the electric current that
flows through the bulb. (Charge on one electron is 1.6x10 -19 C)
Q3.
Calculate the number of electrons that flow through a wire if a steady current of 1.5 A flows
through the wire for 2 minutes. (Charge on an electron is 1.6x10 -19 )Senpaicorner.com Physics
Signature : Name: Marks :
Charge, Current, Voltage, Energy and Power \2
Q4 .
How many electrons are there in 0.8 C of charge? (Charge on one electron is 1.6x10 -19 C)
Q5 .
When lightning strikes, 500 C of charge is transferred and 1.6x10 9 J of energy is
dissipated. What is the potential difference between the cloud and the earth?
Q6.
The potential difference across two metal plates is 400 V. How much energy is required to
transfer 50 C of charge across the two plates? Senpaicorner.com Physics
Signature : Name: Marks :
Charge, Current, Voltage, Energy and Power \3
Q7 .
What is the number of electrons transferred across two metal plates of potential
difference 420 V if 1.26x10 4 J of energy is dissipated during the process? (Charge of an
electron = 1.6x10 -19 C)
Q8 .
A current of 0.2 A flows in a conductor when a potential different of 3 V is applied across
it. Calculate the resistance of the conductor.
Q9 .
Calculate the current flowing across a conductor with resistance of 50 Ω when a potential
difference of 12 V is applied across the conductor. Senpaicorner.com Physics
Signature : Name: Marks :
Charge, Current, Voltage, Energy and Power \4
Q10.
A constantan wire of length 3x and diameter 2d has the resistance R. What is the
resistance, in terms of R, for another constantan wire of length x and diameter d?
Q11.
State four factors that affect the resistance of a conductor.
Q12.
When a resistor is connected to a 6 V supply, 6.25x10 19 electrons flow across the resistor
in every 2 seconds. Calculate the energy dissipated and electrical power. (Charge of
electron = 1.6x10 -19 C)Senpaicorner.com Physics
Signature : Name: Marks :
Charge, Current, Voltage, Energy and Power \5
Q13.
A 48 Ω resistor was connected to a 240 V power supply. Calculate the amount of energy
dissipated in the resistor after 2 minutes.
Q1 4.
When a bulb is connected to a power supply of 12 V for 3 minutes, 1080 J of energy is
dissipated. Calculate
a) the resistance of the bulb
b) the amount of energy dissipated when the same bulb is connected to a power supply
of 20 V for 5 minutes. Senpaicorner.com Physics
Signature : Name: Marks :
Charge, Current, Voltage, Energy and Power \6
Q1 5.
A water heater is rated 240 V, 1500 W. Calculate the resistance of the heating element
and the current when the water heater is connected to a 240 V power supply .
Q1 6.
When an electrical kettle is connected to a 240 V power supply, 216 kJ of energy is
dissipated in 5 minutes. Calculate
a) the power of the electrical kettle
b) the resistance of the heating element
c) the current that flows when the 240 V supply is connected. Senpaicorner.com Physics
Signature : Name: Marks :
Charge, Current, Voltage, Energy and Power \7
Q1 7.
A water heater is rated 240 V, 1000 W heats up 120 g of water from 25 0C to 38 oC in 8
seconds. Calculate the efficiency of the water heater. (Specific heat capacity of water =
4200 J kg -1 oC-1)
Q1 8.
The diagram shows a light bulb which glows brightly when electrical current flows through
it.
a) What is the meaning of current?
b) Name one instrument that can be used to measure the magnitude of current flow.
c) A constant current of 3.0 A flows through the light bulb for 0.5 minute. Calculate
the amount of charge that flows through the bulb. Senpaicorner.com Physics
Signature : Name: Marks :
Charge, Current, Voltage, Energy and Power \8
Q19 .
Diagram 2 shows an electric iron with power rating 240 V, 1800 W,
Diagram 2
a) State one suitable material that can be used as the heating element in the electric
iron.
b) The electric iron in Diagram 2 is connected to a 240 V power supply. Calculate
I. The current that flows through the heating element
II. The resistance of the heating element
c) A student carried out an investigation to compare the heating effect of the heating
elements P, Q and R. The electric iron is switched on until it reaches a certain fixed
temperature. The table below shows the results of the investigation.
Heating element Potential
Difference /V
Current /A Time required to
reach a fixed
temperature /
minutes
P 240 8.0 2.0
Q 240 5.0 5.0
R 240 3.0 3.0
I. State the energy change the occurs when the electric iron is switched on.
II. Calculate the energy supplied by each of the heating elements P, Q and R to reach
the temperature required.
III. Based on your answers in (c)II., suggest the most suitable heating element to be
used in the electric iron. Give one reason for your answer. |
9032 | https://dspace.mit.edu/bitstream/handle/1721.1/104426/6-042j-spring-2010/contents/exams/MIT6_042JS10_quiz06_sol.pdf | Massachusetts Institute of Technology 6.042J/18.062J, Spring ’10: Mathematics for Computer Science May 5 Prof. Albert R. Meyer revised May 15, 2010, 1221 minutes Solutions to Mini-Quiz May 5 Problem 1 (6 points). You would like to give a bouquet for Mother’s Day. You find an online service that will make bouquets of lilies, roses and tulips, subject to the following constraints: • there must be at most 3 lilies, • there can be any number of roses, • there must be a multiple of four tulips. Example: A bouquet of 4 tulips, 5 roses and no lilies satisfies the constraints. Let fn be the number of possible bouquets with n flowers that fit the service’s constraints. Express F (x), the generating function corresponding to hf0, f1, f2, . . . i, as a quotient of polynomials (or products of polynomials). You do not need to simplify this expression. Solution. Generating function for the number of ways to choose lilies: FL(x) = 1 + x + x 2 + x 3 = 1 − x4 1 − x Generating function for the number of ways to choose roses: 1 FR(x) = 1 + x + x 2 + x 3 + x 4 + = · · · 1 − x Generating function for the number of ways to choose tulips: 1 FW (x) = 1 + x 3 + x 7 + = · · · 1 − x4 By the Convolution Property, the generating function for fn is therefore the product of these func tions, namely, F (x) = FL(x)FR(x)FW (x) (1 + x + x2 + x3) = (1 − x)(1 − x2) (1 − x4) = (1 − x)2(1 − x4) 1 = .
(1 − x)2 Creative Commons 2010, Prof. Albert R. Meyer. 2 Solutions to Mini-Quiz May 5 Problem 2 (8 points). [A Baseball Series] The New York Yankees and the Boston Red Sox are playing a two-out-of-three series. (In other words, they play until one team has won two games. Then that team is declared the overall winner and the series ends.) Assume that the Red Sox win each game with probability 2/3, regardless of the outcomes of previous games. Answer the questions below using the four step method. You can use the same tree diagram for all three problems. (a) What is the probability that only 2 games are played? Solution. This means that either the Yankees win both games, which occurs with probability (1/3)2 or the Red Sox win both games, which occurs with probability (2/3)2 . Summing these yields (1/3)2 + (2/3)2 = 5/9 (b) What is the probability that the winner of the series loses the first game? Solution. We use the four step method with the graph below. (c) What is the probability that the Red Sox loses the series? Solution. A tree diagram is worked out below. Sox Win Winner Lost First Game x x x 3 Solutions to Mini-Quiz May 5 From the tree diagram, we get: 4 2 6 Pr {Winner Lost First Game} = + = 27 27 27 3 2 2 7 Pr {Sox Lost} = + + = 27 27 27 27 Problem 3 (6 points). The following combinational identity is known as Pascal’s Triangle: n n − 1 + n − 1 = k k k − 1 (a) Give a combinatorial proof that for Pascal’s Triangle. Solution. Both sides are equal to: n! k! (n − k)! (b) Verify this combinational proof by giving an algebraic proof of this same fact. Solution. This is identical to problem 16.12.1 in the notes, reproduced below: Jay figures that n people (including himself) are competing for spots on a team and only k will be selected. As part of maneuvering for a spot on the team, he needs to work out how many different teams are possible. There are two cases to consider: • Jay is selected for the team, and his k − 1 teammates are selected from among the other n − 1 competitors. The number of different teams that can be formed in this way is: n − 1 k − 1 • Jay is not selected for the team, and all k team members are selected from among the other n − 1 competitors. The number of teams that can be formed this way is: n − 1 k 4 Solutions to Mini-Quiz May 5 All teams of the first type contain Jay, and no team of the second type does; therefore, the two sets of teams are disjoint. Thus, by the Sum Rule, the total number of teams is: n − 1 + n − 1 k k − 1 Another way of thinking about it is that n people (including himself) are trying out for k spots. Thus, the number of ways to select the team is simply: n k Each method correctly counts the number of teams, so the answers must be equal. So we know: n n − 1 + n − 1 = k k k − 1 MIT OpenCourseWare 6.042J / 18.062J Mathematics for Computer Science Spring 2010 For information about citing these materials or our Terms of Use, visit: |
9033 | https://www.youtube.com/watch?v=Tx8AHDMzoGo | This Integral Went Viral in USA | Integral xfloor(1/x)dx
Problem Analysis
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Posted: 3 Oct 2024
math505 #calculus #floor #integral
Calculus Lecture on convergence criteria of series of real numbers
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Transcript:
okay let's solve this integral from 0 to 1 x the floor function of 1 /x DX I want to make a little note here depending on the literature you use the floor function can have a different notation notations most commonly used are this one in the square brackets 1 /x like we have in this case or the other notation is 1 /x but in this lower part of the squared brackets of course the ceiling function would be notated as the upper part of the square brackets of 1 /x and you can write the ceiling function as a floor function of 1 /x and then add one to it now when it comes to solving an actual integral from 0 to 1 x the floor function of 1 /x DX many students just cancel out X and 1 /x and then the by cancelling out they get integral from 0 to 1 DX which is x in the borders from 0 to 1 which is only one and of course this is wrong 1 /x is an argument of the floor function you cannot cancel out something which multiplies a function with the argument of that function so this is completely wrong a proper way to solve an integral from one from 0 to 1 x a floor function of 1 /x DX is as follows we are going to use a substitution that t is equal to 1 /x and then we get that differential of T is equal to 1 / x² the differential of X well this is equal to -1 /x and then entire part squared differential of X and that is equal to - t^ 2 DX from that we get that negative differential of T / t^2 is equal to a differential of X now for the borders of this integral we have from 0 to 1 we have a division in this part 1 /x and division by y0 is not possible so that means that we have an improper integral for the lower border of an integral we have that X approaches zero from the side of positive numbers as T is equal 1 /x when X approaches zero from the positive side T is going to approach Infinity from the positive side now for X which is is equal to 1 we have that t is equal to 1 / X and that is going to give us only 1 then our integral which we have started from becomes an integral lower border is going to be positive infinity and the upper border is going to be one and when we use the substitution we get 1 / T the floor function of T -1 over t^ 2 differential of T we can use this minus sign here to flip the borders so we have from one to positive Infinity integral and that is going to give us 1 / T cubed times the floor function of T differential of T we can now deal with this improper integral by using limits so this is a limit when M approaches Infinity from the side of positive numbers integral from 1 to m 1 / T cubed times the floor function of T differential of T now m is a positive real number as the limit exists for every positive real number then it exists for natural numbers now this is the way we can simplify the problem by looking at number M as a natural number now that is useful because we can split an integral as the sum of natural numbers so we get this to be equal limit when M approaches Infinity from the side of positive numbers some where K goes from 1 to n -1 integral from K to k + 1 1 / T cubed floor function of T differential of T so the problem which we have to solve now is the limit when M approaches Infinity from the side of positive numbers sum from K goes from 1 to M minus 1 integral of K to k + 1 1 / T cubed floor function of T differential of T now the graph of the floor function looks like like this we have the de cards coordinate system we have the origin and the numbers 1 2 and so on we have -1 -2 and -3 on this part Min - 1 - 2 - 3 on this part and 1 2 3 and so on on this part this is the Y AIS this is the X AIS now from zero to 1 the floor function is going to take values from 0 to one but not including number one then from this part right here from one to it is going to take the number one now for the input from one to two the floor function is going to take number one and it is going to approach the dot two dot two is going to take the value two and all numbers higher than number two are going to have images which are number two and of course this is where number three is so when we take a look from the K to k + 1 interval the K is going to have an image at number K which is this dot right here and everything to the k + 1 is going to have an image at K except the k + 1 k + 1 is going to have an image at the number k + 1 so integrals go respectful of the variable T from K to k + one and on the interval from K to k + 1 which is this interval the floor function is going to take the value of K so this is equal to K this is then equal to a limit when M approaches Infinity from the side of positive numbers sum where K goes from 1 to n - 1 K time an integral from K to k + 1 T the^ -3 differential of T and that is equal to a limit when M approaches Infinity from the positive side sum where K goes from 1 to M -1 K t^ -2 / by -2 and this is going to be in the borders where T goes from K to k + 1 so this is then equal to a limit when M approaches Infinity from the side of positive numbers 1/2 we are going to take outside of the sum sum is going to be from K is = 1 to M - 1 we are left with - K times and in Brackets we have 1 / k + 1 2 - 1 / K SAR we can use this minus to flip the values inside the brackets so we get limit when M approaches infin infity from the positive side/ 12 of the sum where K goes from 1 to M -1 K 1 / k^ 2 - 1 / k + 1 2ar and we can write that as a limit when M approaches Infinity from the positive side of the 12 the sum K goes from 1 to m - 1 k / k^ 2 - k / k + 1 squared this is now going to be equal to a limit when M approaches Infinity from the positive side 12 of the 1 / 1 2 - 1 / 2^ 2 + 2 / 2 2 - 2 / 3 2 + 3 / 3 2 - 4 uh - 3 / 4 2ar + 4 / 4 2 - 4 / 5 2ar + and so on plus M - 2 / M - 2 sared - M - 2 / M -1 2 + m -1 / M -1 2 - M -1 / m² now we can use the telescoping trick and we get that this is equal to a limit when M approaches Infinity from the positive side of 12 and in Brackets we have 1 / 1 sared and and this is where we are going to use the telescopic approach so we get + 1 / 2^ 2+ again using the telescopic approach 1 / 3 SAR and of course 1 / by 4^ 2ar and so on the last term is going to be 1 / M -1 squared we get it from this point right here and we are going to rewrite - M -1 / m 2 so this is then equal to a limit when M approaches Infinity from the positive side of 1/2 we have the sum where K goes from 1 to n - 1 1 / n 2 - M -1 / m² now we can distribute the limit because both of these are convergent I have a calculus lecture on various criteria of the convergence of series of real numbers here on the channel and that lecture is linked Below in the description of this video so I urge you to go and watch that so this is then equal to 1/ 12 the limit when M approaches Infinity from the positive side of sum K goes from 1 to n -1 1 / n 2 uh M -1 and of course minus 12 the limit from M -1 divided m sared in the numerator we have M of course M approaches Infinity from the positive side in the numerator we have M minus1 so the degree of M is 1 here and in the denominator we have m² the degree is 2 so this entire part is going to approach zero when M approaches Infinity so this is only equal to 12 and we can do this as the sum where K goes from one and limit we can put up here as the positive Infinity of 1 / n² so this is then equal to/ 12 if and this sum is equal to p² / 6 which is finally going to give us the result of p^ 2 / 12 now if you want to see one more amazing integral which students have a lot of problems with I suggest that you watch this video next please like share and subscribe to the channel I'm looking forward to reading your comments it's |
9034 | https://www.cnmflowers.org/FlowerShapes_Mobile.htm | Flower Shapes
Flower Shapes
Examples of Simple flower shapes
Generally, this means the flower has a rotational symmetry. This produces flowers with an over-all shape which is round, tubular, urn, or bell like.
Examples of Odd flower shapes
Generally, this means the flower has a mirror symmetry (or no obvious symmetry at all.) Again, this applies to the individual flower, not the inflorescence (that is, not the flower cluster). For example, the first two plants shown below, the Plantain and the Clover, have an inflorescence that is simply shaped, but the tiny individual flowers are odd.
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Adverbs Modifying Verbs
What is an Adverb that Modifies a Verb?
An adverbis a part of speech that modifies a verb, an adjective, or another adverb. When modifying verbs, adverbs answer questions about where, when, how, and to what extent an action took place.
Some examples of adverbs modifying verbs would be the following:
I rode a bus here. (Where did I ride a bus?)
Sasha barely passed her science quiz. (To what extent did Sasha pass her science quiz?)
Grant silentlywrote in his journal. (How did Grant write in his journal?)
I will exercise tomorrow. (When will I exercise?)
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Why You Should Use Adverbs That Modify Verbs in Your Writing
Why shouldn’t you use adverbs in your writing?! Adverbs provide so much depth and color to your writing, but for now, we’ll mention two primary reasons for using adverbs: description and precision.
Description: Adverbs modify (or alter) verbs, adjectives, and other adverbs. In a similar fashion, they are to verbs (and adjectives and adverbs) what adjectives are to nouns. In other words, they “describe” verbs, adjectives, and other adverbs. Notice the difference between the sentence with an adverb and without an adverb:
Without an Adverb: She sang the song.
With an Adverb: She sang the song beautifully.
Precision: Adverbs help you as a writer provide very specific detail so that you can convey your thoughts much more precisely.
Without an Adverb: I broke the glass window.
With an Adverb: I intentionally broke the glass window.
Notice the difference between the two sentences? The second sentence conveys much more detail than the first and reveals a good bit more context for your reader.
Use adverbs to keep your writing descriptive and precise!
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9036 | https://www.pasco.com/curriculum/essential-physics/BookInd-719?srsltid=AfmBOorptsuLa9Gw0WFeyMG3z_yaleixQMuD_qCDz7_Ad7L5nHyjSb3F | Setting Up Velocity, Time, and Position Problems - Essential Physics| PASCO
Essential Physics: Sample Chapter 3Learn more about PASCO's curriculum solutions »
Setting up velocity, time, and position problems
What does the sign of velocity mean? Velocity can be positive or negative depending on how you define direction. If moving to the right is defined to be positive, then a negative velocity, such as −2 m/s, describes moving to the left. You should realize that this is a choice and not a rule of physics. You can choose any direction to be positive, including to the left instead of right, or down instead of up. You must be consistent with your choice, however, and not change it in the middle of solving a problem! When do I use velocity instead of speed? Mathematically, speed is the absolute value of velocity. Velocity has the same units as speed and has the same meaning in terms of describing how fast an object is moving. Velocity has the additional capability of indicating the direction. For one-dimensional motion, the direction of motion is described as positive or negative. What does “constant speed” mean? Many problems include the term constant speed or constant velocity. This means the value of the velocity v does not change over time. For example, if a problem gives a “constant speed of 10 m/s,” then the speed stays 10 m/s for all times of interest in the problem. Of course, no real speed stays constant for very long; in many circumstances, however, it is a good approximation. Velocity vs. speedThe variable for speed is the letter v (for velocity). Velocity →v is the vector equivalent of speed. Speed is the absolute value or magnitude of velocity. As an example of the difference, the speedometer in a car measures the car’s speed, such as 55 km/hr, but it says nothing about which direction the car is going. To know your velocity you need to also specify the direction, such as 55 km/hr south. In one-dimensional motion the velocity is positive for one direction and negative for the opposite direction. For example, if the velocity for an object traveling north was +55 km/hr then −55 km/hr would mean the same speed but traveling south. With velocities, a sign convention is always implied that defines one direction as positive and the other as negative. In this book, velocity vectors are drawn as black arrows. Initial time In many physics problems, it is convenient to set the initial time t i to zero. In such cases, the time interval becomes Δ t=t. The initial position (at time zero) may be written as x=x 0 or, alternatively, as x i. Example with positive and negative velocitiesA cart travels to the right at a speed of 0.5 m/s for 15 s, then travels to the left at a speed of 0.3 m/s for 18 s. What is the final position of the cart if it started at x=0? Asked: change in position Δ x (or final position since x i=0) Given: v=0.5 m/s for 15 s, then v=−0.3 m/s for 18 s. The second speed is expressed as a negative velocity because it is in the opposite direction. Relationships: Δ x = v Δ t Solution: This problem is solved by calculating two displacements and then adding them.Δ x 1 = v 1 Δ t 1 = (0.5 m/s)(15 s) = 7.5 m Δ x 2 = v 2 Δ t 2 = (−0.3 m/s)(18 s) = −5.4 m Δ x = Δ x 1 + Δ x 2 = 7.5 m − 5.4 m = 2.1 m Since the cart started at position x 0=0, its final position is:x=0+Δ x=2.1 m Answer: The final position of the cart is +2.1 m. Test your knowledge Ryan moves to the right with a positive velocity of 5 m/s for 1 s, then to the left with a negative velocity of −5 m/s for 1 s. What is Ryan’s displacement after 2 s? 1. 5 m 2. 10 m 3. 0 m 4. −5 m Answer c is correct, because Ryan’s displacement after 2 s is zero (0 m).
79
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9037 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11098264/ | Global perspectives on the burden of sexually transmitted diseases: A narrative review - PMC
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. 2024 May 17;103(20):e38199. doi: 10.1097/MD.0000000000038199
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Global perspectives on the burden of sexually transmitted diseases: A narrative review
Chukwuka Elendu
Chukwuka Elendu, BSc, MD
a Federal University Teaching Hospital, Owerri, Nigeria
Find articles by Chukwuka Elendu
a,, Dependable C Amaechi
Dependable C Amaechi, MBBS
b Igbinedion University, Okada, Nigeria
Find articles by Dependable C Amaechi
b, Ijeoma D Elendu
Ijeoma D Elendu, BNSc, RN, RM, RPHN
c Imo State University, Owerri, Nigeria
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c, Tochi C Elendu
Tochi C Elendu, BNSc, RN, RM, RPHN
c Imo State University, Owerri, Nigeria
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c, Emmanuel C Amaechi
Emmanuel C Amaechi, MBBS
d Madonna University, Elele, Nigeria
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d, Emmanuel U Usoro
Emmanuel U Usoro, MBBS
e Babcock University, Ilishan-Remo, Nigeria
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e, Nkechi L Chima-Ogbuiyi
Nkechi L Chima-Ogbuiyi, BSc, RN, MSN
f Herzing University, Madison, WI
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f, Divine B Arrey Agbor
Divine B Arrey Agbor, MD
g Richmond University Medical Center, Staten Island, NY
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g, Chukwunnonso J Onwuegbule
Chukwunnonso J Onwuegbule, MBBS
e Babcock University, Ilishan-Remo, Nigeria
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e, Eniola F Afolayan
Eniola F Afolayan, MBBS
e Babcock University, Ilishan-Remo, Nigeria
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e, Benjamin B Balogun
Benjamin B Balogun, MBBS
h University of Ilorin, Ilorin, Nigeria.
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h
Author information
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Copyright and License information
a Federal University Teaching Hospital, Owerri, Nigeria
b Igbinedion University, Okada, Nigeria
c Imo State University, Owerri, Nigeria
d Madonna University, Elele, Nigeria
e Babcock University, Ilishan-Remo, Nigeria
f Herzing University, Madison, WI
g Richmond University Medical Center, Staten Island, NY
h University of Ilorin, Ilorin, Nigeria.
Correspondence: Chukwuka Elendu, Federal University Teaching Hospital, Owerri 460281, Nigeria (e-mail: elenduchukwuka@yahoo.com).
Received 2023 Nov 4; Received 2024 Feb 14; Accepted 2024 Apr 19; Collection date 2024 May 17.
Copyright © 2024 the Author(s). Published by Wolters Kluwer Health, Inc.
This is an open access article distributed under the Creative Commons Attribution License 4.0 (CCBY), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC11098264 PMID: 38758874
Abstract
Sexually transmitted diseases (STDs) pose a significant global health challenge with far-reaching social, economic, and public health implications. These infections have haunted humanity from ancient times to today, transcending geographical boundaries and cultural contexts. This article explores the multifaceted landscape of STDs, delving into their epidemiology, pathophysiology, clinical manifestations, and global response strategies. The global prevalence of STDs is staggering, with millions of new cases reported annually. Prominent among these infections is HIV/AIDS, which remains a major global health crisis, affecting over 38 million people worldwide.
Additionally, bacterial STDs like chlamydia, gonorrhea, and syphilis continue to pose significant health risks, with millions of new cases reported yearly. Beyond the physical manifestations, STDs have profound social and economic implications. They can result in severe reproductive health issues, stigma, discrimination, and psychological distress, burdening healthcare systems and affecting individuals’ quality of life. The global response to STDs has been multifaceted, with international organizations and governments implementing various prevention and control strategies, including sexual education programs and scaling up access to testing and treatment. However, challenges persist, including disparities in healthcare access, sociocultural factors influencing transmission, and evolving pathogens with increasing resistance to treatment. Through case studies and real-world examples, we illuminate the human stories behind the statistics, highlighting the lived experiences of individuals grappling with STDs and the complex interplay of factors shaping their journeys. Ultimately, this review calls for continued research, innovative strategies, and sustained global commitment to mitigating the burden of STDs and promoting sexual health and well-being for all.
Keywords: global burden, prevalence, prevention and control, public health, regional perspectives, sexually transmitted diseases, sociocultural factors
1. Introduction and background
Sexually transmitted diseases (STDs) represent a significant global health challenge with profound social, economic, and public health implications. These infections, caused by various bacterial, viral, and parasitic agents, are primarily transmitted through sexual contact but can also be transmitted vertically from mother to child during childbirth or through contaminated needles and blood products. Picture Maria, a 24-year-old college student from a rural community. Despite her cautious approach to sexual encounters, Maria finds herself grappling with the stigma and uncertainty of a positive chlamydia diagnosis. Her journey reflects the stark reality faced by millions worldwide, where the fear of judgment and lack of access to comprehensive sexual health services perpetuate the cycle of infection and silence. The global prevalence of STDs is alarmingly high, with millions of new cases reported yearly. According to the World Health Organization (WHO), it is estimated that over 1 million sexually transmitted infections (STIs) are acquired each day globally. Prominent among these infections is human immunodeficiency virus (HIV)/AIDS, which remains a major global health crisis, affecting over 38 million people worldwide. Consider the case of David, a 35-year-old man living in an urban center, who discovers he is HIV positive during a routine health checkup. His diagnosis shatters his world, confronting him with the reality of living with a chronic condition that requires lifelong management. David’s story highlights the devastating impact of HIV/AIDS on individuals, families, and communities, underscoring the urgent need for comprehensive prevention, testing, and treatment initiatives. Meet Ahmed, a migrant worker in a bustling urban center, illustrates the silent spread of hepatitis B among his community. His journey echoes the intersecting dynamics of migration, poverty, and healthcare access, illustrating the pervasive reach of STIs across diverse populations and settings.
Additionally, bacterial STDs like chlamydia, gonorrhea, and syphilis continue to pose significant health risks, with millions of new cases reported annually. The burden of STDs extends far beyond the physical health of individuals. It has profound social and economic implications. STDs can result in severe reproductive health issues, including infertility, ectopic pregnancies, and adverse pregnancy outcomes. Moreover, they can lead to stigma, discrimination, and psychological distress, often resulting in delayed diagnosis and treatment due to fear of social consequences. Economically, the cost of treating STDs and their complications is substantial. Direct medical expenses, including diagnostics, treatment, and follow-up care, burden healthcare systems considerably. The indirect costs related to lost productivity and reduced quality of life are substantial, particularly in resource-limited settings. The global response to STDs has been multifaceted. International organizations such as the WHO, UNAIDS, and UNICEF have launched comprehensive initiatives to combat these infections. National and regional governments have also implemented various prevention and control strategies, including sexual education programs, promotion of condom use, and scaling up access to testing and treatment. Despite these efforts, challenges persist, including disparities in access to healthcare resources, varying sociocultural factors influencing transmission, and evolving pathogens with increasing resistance to treatment. These challenges underscore the need for continued research, innovative strategies, and sustained global commitment to mitigating the burden of STDs.
2. Objectives of study
To Quantify the Global Burden of STDs using disability-adjusted life years (DALYs) and Standardized Metrics: This objective aligns closely with our enhanced methodology, which incorporates DALYs and other standardized metrics to quantify the health loss attributable to STDs. By applying rigorous epidemiological methods and data analysis techniques, we aim to provide a robust assessment of the overall burden of STDs, including morbidity, mortality, and disability, across different populations and geographic regions.
To Identify High-Risk Populations and Geographic Areas with Elevated STD Burden: Our review aims to go beyond merely quantifying the burden of STDs and seeks to identify vulnerable populations and geographic areas disproportionately affected by these diseases. By disaggregating data by age, sex, and geography, we can pinpoint areas with the highest disease burden and target interventions accordingly. This objective supports health planning efforts by guiding resource allocation and prioritizing interventions where they are most needed.
To Evaluate the Effectiveness of STD Prevention and Control Strategies: In addition to quantifying the burden of STDs, our review aims to assess the effectiveness of existing prevention and control strategies. We seek to identify best practices and gaps in current approaches by critically appraising the literature and analyzing intervention outcomes. This objective is crucial for informing health planning efforts and guiding the development of evidence-based interventions tailored to local contexts.
To Provide Evidence-Based Recommendations for Improving STD Prevention and Control Efforts: Our review aims to translate findings into actionable recommendations for improving STD prevention and control efforts globally. We can offer practical guidance for policymakers, healthcare practitioners, and public health officials by synthesizing and aligning the latest evidence with stakeholder priorities. These recommendations will support health planning efforts by promoting effective strategies, enhancing resource allocation, and advancing the global agenda for STD prevention and control.
3. Discussion
3.1. Methodology
To assess the global burden of STDs comprehensively, we employed a multifaceted approach that integrated various measures and indicators to quantify health loss and inform health planning efforts. Our methodology used DALYs as a primary metric, alongside other standardized indicators, to capture the full extent of morbidity and mortality attributable to STDs across different populations and geographic regions.
3.2. Measurement of health loss
DALYs served as a valuable tool for quantifying the overall burden of STDs by accounting for both years of life lost due to premature mortality and years lived with disability. By applying DALYs to STDs, we captured the impact of these diseases on individuals’ quality of life and mortality rates, providing a comprehensive understanding of their public health significance.
3.3. Data sources and age-sex breakdown
Data sources for our analysis included a combination of global health databases, epidemiological studies, and surveillance reports from reputable institutions such as the WHO, the Centers for Disease Control and Prevention (CDC), and academic research publications. These sources provided information on STD prevalence, incidence, and associated health outcomes.
To ensure a thorough analysis, we stratified the burden of STDs by age and sex, recognizing the differential impact of these diseases across different demographic groups. By disaggregating the data, we could identify vulnerable populations, such as adolescents, women of reproductive age, and men who have sex with men, who may bear a disproportionate burden of STD-related morbidity and mortality.
3.4. Geographic breakdown
Geographic variations in the burden of STDs were also explored, focusing on assessing disparities between regions and countries. We utilized global health maps and spatial analysis techniques to visualize the distribution of STD prevalence and incidence, highlighting areas with the highest disease burden and disparities in access to healthcare services.
3.5. Quality assurance and limitations
Quality assurance measures were implemented to ensure the reliability and validity of the data used in our analysis. Sensitivity analyses and validation checks were conducted to assess the robustness of our findings and address any potential biases or uncertainties.
While our methodology provided valuable insights into the global burden of STDs, it is essential to acknowledge certain limitations. These include variations in data availability and quality across different regions, underreporting of STD cases, and challenges in estimating the actual burden of asymptomatic infections. In addition to our comprehensive methodology for assessing the global burden of STDs, we have developed a practical algorithm aimed at enhancing clarity and applicability for health planning and intervention strategies. This algorithm, presented in Table 1, serves as a valuable tool for healthcare practitioners, policymakers, and public health officials to guide decision-making and resource allocation in preventing and controlling STDs.
Table 1.
Practical algorithm for STD prevention and control.
| Step | Action | Description |
:---:
| 1 | Surveillance and monitoring | Establish robust surveillance systems to monitor STD prevalence, incidence, and trends. |
| 2 | Risk assessment | Identify high-risk populations and geographic areas with elevated STD burden for targeted interventions. |
| 3 | Education and awareness | Implement comprehensive sex education programs to promote safe sexual practices and raise awareness about STD prevention. |
| 4 | Screening and testing | Offer routine STD screening for asymptomatic individuals, especially among high-risk groups. Ensure access to confidential and affordable testing services. |
| 5 | Treatment and management | Provide prompt diagnosis and appropriate treatment for individuals diagnosed with STDs. Ensure access to affordable medications and follow-up care. |
| 6 | Partner notification and contact tracing | Encourage individuals diagnosed with STDs to notify their sexual partners for testing and treatment. Conduct contact tracing to interrupt transmission chains. |
| 7 | Vaccination | Promote vaccination against vaccine-preventable STDs such as HPV and hepatitis B to reduce disease incidence and transmission. |
| 8 | Integration of services | Integrate STD prevention, testing, and treatment services with existing healthcare systems to improve access and utilization. |
| 9 | Community engagement | Engage communities and stakeholders in STD prevention efforts through outreach, education, and advocacy initiatives. |
| 10 | Evaluation and feedback | Regularly evaluate the effectiveness of STD prevention and control measures and adjust strategies based on feedback and surveillance data. |
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Practical implementation: This algorithm provides a step-by-step approach to addressing the burden of STDs and offers practical guidance for implementing evidence-based interventions. It emphasizes the importance of a comprehensive, multi-sectoral approach encompassing surveillance, education, screening, treatment, and community engagement.
Visual enhancement: Visual aids such as flowcharts or infographics can be incorporated to illustrate the sequential nature of the steps and facilitate comprehension to enhance the practicality of the algorithm. Additionally, explanatory notes can be included to provide further context and rationale for each action.
Conclusion: By integrating this practical algorithm into STD prevention and control efforts, stakeholders can streamline their approach, allocate resources more effectively, and ultimately reduce the burden of STDs on individuals and communities worldwide. This algorithm serves as a valuable resource for guiding evidence-based interventions and advancing the global agenda for STD prevention and control.
The practical algorithm presented in Table 1 offers a structured framework for addressing the challenges posed by STDs and underscores the importance of comprehensive prevention and control strategies. We encourage stakeholders to utilize this algorithm in their efforts to combat STDs and promote sexual health.
STDs = sexually transmitted diseases.
3.6. Epidemiology of STDs
STDs pose a significant global health burden, transcending geographical boundaries and socioeconomic strata. In a world increasingly interconnected through travel and technology, the prevalence and distribution of STDs have become intricate and multifaceted.
Across the globe, an estimated 1 million individuals acquire an STD each day, underscoring the pervasive nature of these infections. While the burden is shared, regional variations in prevalence and incidence rates exist, reflecting disparities in healthcare access, education, cultural norms, and sexual behavior.
In high-income countries, such as the United States and those in Western Europe, STD rates have witnessed a resurgence in recent years, driven by factors like decreased condom use, inadequate screening, and the emergence of drug-resistant strains. In contrast, low- and middle-income countries bear a disproportionate burden, grappling with challenges like limited healthcare infrastructure, stigma, and barriers to accessing prevention and treatment services.
Disparities in age, gender, and sexual orientation mark the epidemiology of STDs. Young people aged 15 to 24 years are particularly vulnerable, accounting for a substantial proportion of new infections globally. Moreover, women, especially those in resource-limited settings, face heightened risks due to biological factors, gender-based violence, and limited agency in sexual decision-making.
The advent of globalization has facilitated the spread of STDs across borders, with travel and migration serving as conduits for transmission. Urbanization further exacerbates this phenomenon, as densely populated areas create environments conducive to rapidly disseminating infections.
Emerging trends, such as the intersection of STDs with other public health crises like HIV/AIDS and viral hepatitis, underscore the interconnectedness of health outcomes. Additionally, the advent of dating apps and online platforms has reshaped sexual networks, presenting both opportunities for prevention and challenges in tracing contacts and delivering care.
3.7. Pathophysiology of STDs
At the heart of the pathophysiology of STDs lies a complex interplay between infectious agents and the human body’s defense mechanisms. STDs encompass diverse pathogens, including bacteria, viruses, parasites, and fungi, each with its unique transmission mode, pathogenesis, and clinical manifestations.
Upon exposure to an infectious agent through sexual contact, the pathogen enters the body through mucosal surfaces, such as the genital tract, oral cavity, or rectum. The transmission efficiency varies depending on factors like pathogen virulence, host susceptibility, and predisposing factors such as genital trauma or inflammation.
Once inside the body, the pathogen establishes infection by adhering to host cells, evading immune surveillance, and replicating within the local tissue. For bacterial STDs like gonorrhea and chlamydia, this process often involves colonization of the epithelial cells lining the genital tract, leading to inflammation, tissue damage, and the formation of characteristic clinical lesions.
Viruses, such as herpes simplex virus (HSV), human papillomavirus (HPV), and HIV, employ diverse strategies to perpetuate infection and evade host defenses. HSV, for instance, establishes latency within the sensory ganglia following primary infection, periodically reactivating to cause recurrent episodes of genital ulceration and shedding. HPV, on the other hand, integrates into the host genome, predisposing to the development of genital warts and cervical, anal, and oropharyngeal cancers over time.
Parasitic STDs, including trichomoniasis and pubic lice infestation, thrive in the warm, moist environment of the genital region, where they reproduce and elicit local inflammation and irritation. Fungal infections, such as candidiasis, exploit disruptions in the vaginal microenvironment, such as alterations in pH or hormonal imbalances, to increase and cause symptomatic disease.
The pathophysiology of STDs extends beyond local genital manifestations to encompass systemic sequelae and complications. Disseminated infections, ascending spread to the upper reproductive tract, and chronic inflammation can lead to severe consequences like pelvic inflammatory disease (PID), infertility, ectopic pregnancy, and an increased risk of HIV acquisition and transmission.
Furthermore, STDs can exert a profound impact on reproductive and maternal-child health outcomes, with implications for pregnancy outcomes, neonatal health, and vertical transmission of infections. For example, untreated syphilis during pregnancy can result in congenital syphilis, causing stillbirths, neonatal death, and a spectrum of congenital abnormalities.
3.8. Types of STIs
STIs encompass a diverse array of pathogens, each with its unique characteristics, modes of transmission, and clinical presentations. From bacterial culprits to viral assailants and parasitic invaders, the world of STIs is as varied as it is, presenting global challenges to public health efforts.
Bacterial STIs, such as gonorrhea and chlamydia, are notorious for their stealthy nature and propensity to cause silent infections, particularly in women. Despite their asymptomatic course in many cases, untreated bacterial STIs can lead to severe complications, including PID, infertility, and ectopic pregnancy. Neisseria gonorrhoeae, the bacterium responsible for gonorrhea, has exhibited alarming rates of antimicrobial resistance, posing a significant challenge to treatment efforts.
Syphilis, caused by the bacterium Treponema pallidum, is a classic example of a bacterial STI with protean manifestations, progressing through distinct stages if left untreated. Primary syphilis presents as painless genital ulcers or chancres, while secondary syphilis manifests as a systemic illness with rash, fever, and mucocutaneous lesions. Tertiary syphilis, although rare in the era of antibiotics, can result in devastating cardiovascular and neurologic complications.
Viral STIs, including HSV, HPV, and HIV, exert a significant global health burden, with implications for morbidity, mortality, and quality of life. HSV infection presents with recurrent episodes of painful genital ulcers, accompanied by stigma and psychosocial distress. HPV, the most common viral STI worldwide, encompasses a multitude of genotypes, some of which are oncogenic and predispose to the development of cervical, anal, and oropharyngeal cancers.
HIV/AIDS, perhaps the most formidable of all STIs, continues to take a heavy toll on individuals and communities worldwide. Despite advances in treatment and prevention, HIV/AIDS remains a significant public health challenge, with millions of new infections occurring annually. The virus not only undermines the immune system, leading to opportunistic infections and malignancies but also perpetuates social and economic inequalities, exacerbating the burden of disease among marginalized populations.
Parasitic STIs, though less prevalent than their bacterial and viral counterparts, nonetheless pose significant health risks and challenges to control efforts. Trichomoniasis, caused by the protozoan parasite Trichomonas vaginalis, can cause vaginal discharge, itching, and discomfort, increasing the risk of HIV acquisition and transmission. Additionally, pubic lice infestation, or “crabs,” although more of a nuisance than a serious health threat, can cause itching and inflammation of the genital area.
3.9. Clinical manifestations of STIs
STIs are notorious for their diverse and often elusive clinical presentations, ranging from asymptomatic carriage to debilitating symptoms and complications. The manifestations of STIs are influenced by factors such as the type of pathogen, site of infection, host immune response, and individual risk factors, making diagnosis and management challenging.
Bacterial STIs, such as gonorrhea and chlamydia, can manifest as a spectrum of symptoms or remain entirely asymptomatic, particularly in women. When symptomatic, gonorrhea typically presents with urethritis, characterized by dysuria, urethral discharge, and, in some cases, epididymitis in men. Chlamydia infection, on the other hand, may cause urethritis, cervicitis, or PID, leading to lower abdominal pain, vaginal discharge, and dyspareunia.
Syphilis, a systemic bacterial infection caused by Treponema pallidum, progresses through distinct stages if left untreated. Primary syphilis manifests as painless genital ulcers or chancres, often accompanied by regional lymphadenopathy. Secondary syphilis presents with a wide array of systemic symptoms, including rash, fever, sore throat, and mucocutaneous lesions. Tertiary syphilis, although rare in the era of antibiotics, can result in devastating cardiovascular and neurologic complications.
Viral STIs, such as HSV and HPV, are characterized by recurrent episodes of symptomatic disease interspersed with periods of asymptomatic shedding. HSV infection presents with painful genital ulcers or lesions, often accompanied by systemic symptoms such as fever, malaise, and lymphadenopathy. HPV infection, on the other hand, may cause genital warts, cervical dysplasia, or asymptomatic infection, depending on the viral genotype and host immune response.
Hepatitis refers to liver inflammation, which various factors, including viruses, alcohol, drugs, toxins, and autoimmune diseases can cause. However, when discussing hepatitis in the context of STIs, the focus is primarily on viral hepatitis, specifically hepatitis B virus (HBV) and hepatitis C virus (HCV), which can be transmitted through sexual contact.
HBV and HCV are bloodborne pathogens that can cause acute and chronic liver infections. While both viruses can be transmitted through exposure to infected blood, HBV can also be transmitted through sexual contact, sharing of contaminated needles, and from mother to child during childbirth. Sexual transmission of HBV is more likely to occur among individuals engaging in unprotected sex, particularly those with multiple sexual partners or individuals with a history of other STIs.
Hepatitis B infection can manifest as acute or chronic hepatitis, with symptoms ranging from mild flu-like illness to severe liver damage, cirrhosis, and hepatocellular carcinoma (liver cancer). Acute hepatitis B infection may present with symptoms such as fatigue, jaundice (yellowing of the skin and eyes), dark urine, abdominal pain, and loss of appetite. Most adults with acute HBV infection recover completely, but a small percentage develop chronic hepatitis B, which can lead to long-term complications and liver damage if left untreated.
Chronic hepatitis B infection is a significant cause of liver cirrhosis and liver cancer worldwide. It requires long-term management with antiviral medications to suppress viral replication and reduce the risk of disease progression. Additionally, vaccination against hepatitis B is highly effective in preventing new infections. It is recommended for individuals at increased risk of exposure, including sexually active adults, healthcare workers, and individuals with certain medical conditions.
Hepatitis C virus (HCV) is primarily transmitted through exposure to infected blood, such as the sharing of contaminated needles, blood transfusions, or organ transplants. However, sexual transmission of HCV can also occur, particularly among individuals engaging in high-risk sexual behaviors such as unprotected anal intercourse, sex with multiple partners, and sex work. While the risk of sexual transmission of HCV is lower than that of HBV, it can still occur, especially in the presence of genital ulcers or mucosal trauma.
Chronic hepatitis C infection is a leading cause of liver cirrhosis, liver failure, and hepatocellular carcinoma worldwide and requires early detection and treatment to prevent complications. Direct-acting antiviral medications have revolutionized the treatment of hepatitis C, offering high cure rates (>95%) with minimal side effects and shorter treatment durations. Screening for hepatitis C is recommended for individuals at increased risk of infection, including people who inject drugs, individuals with a history of blood transfusions or organ transplants, and those with high-risk sexual behaviors.
HIV/AIDS, perhaps the most formidable of all STIs, can present with a spectrum of clinical manifestations, ranging from acute retroviral syndrome to advanced immunodeficiency and opportunistic infections. Acute HIV infection may mimic infectious mononucleosis, with symptoms such as fever, sore throat, rash, and lymphadenopathy. As the disease progresses, HIV/AIDS can cause opportunistic infections, malignancies, and neurologic complications, leading to significant morbidity and mortality if untreated.
Parasitic STIs, although less common than bacterial and viral infections, can nonetheless cause significant morbidity and complications. Trichomoniasis, caused by the protozoan parasite Trichomonas vaginalis, presents with vaginal discharge, itching, and discomfort in women and urethritis in men. Pubic lice infestation, or “crabs,” manifests as itching and inflammation of the genital area, often accompanied by visible lice or nits.
3.10. Uncommon STIs
Uncommon STIs may not be as well-known as their more prevalent counterparts. However, they still pose significant challenges to sexual health and require attention from healthcare providers and public health authorities. While less common, these infections can have severe consequences if left untreated and may be associated with unique clinical presentations, diagnostic challenges, and management considerations.
One example of an uncommon STI is lymphogranuloma venereum (LGV), a bacterial infection caused by specific serotypes of Chlamydia trachomatis. LGV primarily affects the lymphatic system, causing inflammation and swelling of the lymph nodes in the genital or anorectal region. Unlike other types of chlamydia infections, which typically present with urethritis or cervicitis, LGV may manifest as painful genital ulcers, tender inguinal lymphadenopathy (swollen glands), or proctitis (inflammation of the rectum). LGV can also lead to complications such as rectal strictures, fistulas, and genital elephantiasis if left untreated.
Another uncommon STI is granuloma inguinale, also known as donovanosis, caused by the bacterium Klebsiella granulomatis. This infection is characterized by painless, beefy-red ulcers or nodules in the genital, perineal, or anorectal region, which can progress to extensive tissue destruction and disfigurement if untreated. Granuloma inguinale is rare in developed countries but remains endemic in certain regions, mainly tropical and subtropical areas with poor access to healthcare and sanitation.
Additionally, sexually transmitted parasites such as Trichomonas vaginalis and Phthirus pubis (pubic lice) can cause uncommon STIs that may present with distinctive clinical features. Trichomoniasis, caused by the protozoan parasite T. vaginalis, can cause symptoms such as vaginal discharge, genital itching, dysuria (painful urination) in women and urethritis or prostatitis in men. While often asymptomatic, trichomoniasis can increase the risk of adverse pregnancy outcomes and HIV transmission if left untreated. Pubic lice infestation, or “crabs,” is characterized by intense itching and irritation in the genital area due to infestation with tiny parasitic insects that feed on human blood. While pubic lice infestation is typically considered a nuisance rather than a serious health threat, it can be sexually transmitted and may coexist with other STIs.
Moreover, emerging STIs such as Mycoplasma genitalium and Zika virus have garnered attention in recent years due to their potential to cause reproductive complications and public health concerns. Mycoplasma genitalium is a bacterial infection associated with urethritis, cervicitis, and PID in women and urethritis and prostatitis in men. While less common than other STIs, Mycoplasma genitalium has been implicated in infertility, ectopic pregnancy, and adverse pregnancy outcomes. Zika virus, primarily transmitted through mosquito bites, can also be sexually transmitted, posing risks of congenital Zika syndrome (CZS) in pregnant women and neurological complications such as Guillain-Barré syndrome in adults.
Bowenoid papulosis is a relatively uncommon STIcharacterized by small, reddish-brown papules or plaques on the genital skin or mucous membranes. This condition is caused by HPV infection, specifically high-risk strains such as HPV types 16 and 18, which are also associated with cervical and anal cancer.
Bowenoid papulosis typically affects young, sexually active adults, particularly those who engage in unprotected sexual activity with multiple partners or have a history of genital warts. The lesions of Bowenoid papulosis are usually asymptomatic or may cause mild itching or irritation, leading some individuals to seek medical attention for cosmetic reasons or concerns about STIs.
Clinically, the lesions of Bowenoid papulosis appear as raised, reddish-brown papules or plaques on the genital skin, shaft of the penis, vulva, or perianal area. The lesions may be solitary or multiple and can vary from a few millimeters to several centimeters in diameter. Unlike genital warts, which often have a characteristic cauliflower-like appearance, the lesions of Bowenoid papulosis are typically flat or slightly elevated and may have a smooth or rough surface.
Diagnosis of Bowenoid papulosis is based on clinical examination and may be confirmed through biopsy and histopathological examination of the affected tissue. Histologically, Bowenoid papulosis is characterized by atypical keratinocytes (abnormal skin cells) with enlarged nuclei and increased mitotic activity, similar to the changes seen in precancerous lesions such as Bowen’s disease or squamous cell carcinoma in situ.
Management of Bowenoid papulosis typically involves conservative approaches such as topical treatments or observation, as most lesions spontaneously regress over time without causing complications. Topical treatments may include podophyllotoxin, imiquimod, or trichloroacetic acid (TCA), which can help reduce the size and appearance of lesions and promote resolution. Surgical interventions such as cryotherapy or laser therapy may be considered for persistent or symptomatic lesions, mainly if there is concern for malignant transformation or cosmetic reasons.
The prognosis for Bowenoid papulosis is generally favorable, with most lesions resolving spontaneously within months to years without causing long-term complications. However, individuals with Bowenoid papulosis should be counseled about the potential risk of HPV transmission to sexual partners and the importance of regular follow-up and screening for other HPV-related conditions, including cervical, anal, and oropharyngeal cancer. Additionally, vaccination against HPV can help prevent new infections and reduce the risk of HPV-related diseases, including Bowenoid papulosis.
Finally, while uncommon STIs may not receive as much attention as more prevalent STIs, they still represent significant challenges to sexual health and require vigilance from healthcare providers and public health authorities. By raising awareness, improving diagnostic capabilities, and implementing targeted prevention and control strategies, stakeholders can work towards reducing the burden of uncommon STIs and promoting optimal sexual health for all individuals and communities.
3.11. Factors contributing to the global burden of STDs
The global burden of STDs is a multifaceted challenge shaped by an intricate interplay of biological, social, economic, and structural factors. At the heart of this burden lies a complex web of determinants, each contributing to the perpetuation and exacerbation of STDs on a global scale.
Biological factors play a pivotal role in driving the transmission and persistence of STDs. Pathogens such as bacteria, viruses, parasites, and fungi exploit vulnerabilities in the human body’s defenses, adapting and evolving to evade immune responses and antimicrobial therapies. Moreover, the intricate biology of STDs, including modes of transmission, incubation periods, and asymptomatic carriage, complicates efforts to control and prevent their spread.
Social and behavioral factors significantly influence patterns of STD transmission and acquisition, reflecting the complex dynamics of human relationships, sexual practices, and cultural norms. Factors such as multiple sexual partners, inconsistent condom use, transactional sex, and substance abuse contribute to heightened risk behaviors and facilitate the spread of infections within and between communities. Moreover, stigma, discrimination, and lack of access to comprehensive sexual health education and services perpetuate misconceptions, hinder prevention efforts, and exacerbate the burden of STDs among marginalized populations.
Economic disparities and structural determinants of health further compound the global burden of STDs, exacerbating inequalities in access to prevention, screening, diagnosis, and treatment services. Limited healthcare infrastructure, inadequate funding for public health programs, and competing priorities divert resources away from STD control efforts, perpetuating cycles of transmission and disease burden. Moreover, poverty, unemployment, housing instability, and social marginalization create environments conducive to the spread of STDs, disproportionately impacting vulnerable populations and hindering efforts to achieve health equity.
Globalization and interconnectedness have facilitated the spread of STDs across borders, transcending geographical boundaries and exposing populations to novel pathogens and transmission dynamics. Travel, migration, and international trade contribute to the dissemination of infections, while urbanization and population mobility create environments conducive to the rapid transmission of STDs within densely populated areas. Furthermore, the emergence of digital technologies, including dating apps and online platforms, has reshaped sexual networks and behaviors, presenting both opportunities for prevention and challenges in reaching at-risk populations with targeted interventions.
3.12. Global efforts in STD prevention and control
Efforts to prevent and control STDs on a global scale are characterized by a myriad of initiatives, partnerships, and strategies aimed at addressing the complex and multifaceted nature of these infections. From grassroots campaigns to multinational collaborations, the fight against STDs encompasses various stakeholders, interventions, and innovations.
At the forefront of global efforts in STD prevention and control are comprehensive sexual health education programs, which aim to empower individuals with knowledge, skills, and resources to make informed decisions about their sexual health. These programs promote healthy behaviors, encourage condom use, and foster open dialogue about consent, communication, and reproductive rights. By addressing underlying social norms and attitudes surrounding sexuality, sexual health education plays a crucial role in reducing risk behaviors and promoting positive sexual health outcomes.
Screening and testing initiatives form another cornerstone of STD prevention efforts, enabling early detection, diagnosis, and treatment of infections. Routine screening for STDs, particularly among high-risk populations such as sexually active adolescents, men who have sex with men, and individuals living with HIV, helps identify asymptomatic infections and prevent onward transmission. Furthermore, innovative approaches to screening, such as self-testing kits, mobile clinics, and community-based outreach programs, expand access to testing services and reach underserved populations with limited healthcare access.
Treatment and management of STDs are integral components of comprehensive prevention and control efforts aimed at reducing morbidity, preventing complications, and interrupting transmission chains. Antimicrobial therapy, tailored to the specific pathogen and clinical presentation, is central to the management of bacterial STDs such as gonorrhea, chlamydia, and syphilis. Similarly, antiviral medications play a crucial role in the treatment of viral STDs such as HSV, HPV, and HIV, prolonging survival, improving quality of life, and reducing transmission risk.
Pre-exposure prophylaxis (PrEP) and post-exposure prophylaxis (PEP) offer additional strategies for preventing HIV transmission among high-risk individuals, including serodiscordant couples, injection drug users, and individuals with a history of condomless sex or STI diagnosis. PrEP, in particular, has emerged as a game-changer in HIV prevention, providing a highly effective biomedical intervention for individuals at risk of HIV acquisition. By combining antiretroviral medications with behavioral counseling and support services, PrEP programs offer a holistic approach to HIV prevention, addressing both biological and behavioral determinants of risk.
Community engagement and advocacy efforts are essential for mobilizing resources, raising awareness, and fostering political commitment to STD prevention and control. Grassroots organizations, civil society groups, and advocacy networks play a crucial role in amplifying the voices of affected communities, advocating for policy change, and holding governments and policymakers accountable for addressing the needs of marginalized populations. By centering the experiences and priorities of those most affected by STDs, community-led initiatives help ensure that prevention and control efforts are equitable, inclusive, and responsive to the diverse needs of diverse communities.
Innovation and research drive progress in STD prevention and control, fostering the development of new technologies, interventions, and approaches to tackling the challenges posed by these infections. From novel diagnostic tests and treatment modalities to behavioral interventions and digital health solutions, innovation can revolutionize STD prevention and control, making services more accessible, efficient, and responsive to the evolving needs of populations worldwide.
3.13. Regional perspectives on STDs
Across different regions of the world, the landscape of STDs varies significantly and is shaped by unique cultural, social, economic, and epidemiological factors. From high-income countries with well-established healthcare systems to low- and middle-income countries facing resource constraints, each region grapples with challenges and opportunities in preventing and controlling STDs.
In high-income countries, such as those in North America, Western Europe, and parts of Asia-Pacific, STD control efforts are characterized by robust healthcare infrastructure, comprehensive sexual health education programs, and widespread access to screening, diagnosis, and treatment services. Despite these advantages, high-income countries face challenges such as increasing rates of antibiotic resistance, resurgence of syphilis and gonorrhea, and persistent disparities in access to care among marginalized populations, including ethnic minorities, LGBTQ + individuals, and immigrants.
In contrast, low- and middle-income countries, particularly those in sub-Saharan Africa, South Asia, and Latin America, bear a disproportionate burden of STDs, driven by factors such as limited healthcare resources, poverty, gender inequality, and stigma. In these settings, challenges such as inadequate funding for STD control programs, shortages of trained healthcare personnel, and barriers to accessing preventive services hinder efforts to address the epidemic effectively. Additionally, cultural norms, taboos surrounding sexuality, and misconceptions about STDs contribute to delayed diagnosis, untreated infections, and ongoing transmission within communities.
In regions affected by humanitarian crises, such as conflict-affected areas, refugee camps, and areas of natural disaster, the burden of STDs is exacerbated by disruptions to healthcare systems, displacement, and breakdowns in social support networks. Humanitarian organizations and non-governmental organizations play a crucial role in delivering essential sexual and reproductive health services, including STD prevention, testing, and treatment, to affected populations, often in challenging and resource-constrained environments.
In some regions, such as parts of Eastern Europe and Central Asia, injection drug use drives the spread of bloodborne infections, such as HIV and hepatitis C, highlighting the intersection of drug use, harm reduction, and infectious disease control. Harm reduction interventions, including needle exchange programs, opioid substitution therapy, and access to sterile injecting equipment, are essential components of comprehensive STD prevention efforts in these settings, helping to reduce transmission risk and improve health outcomes among people who inject drugs.
Indigenous populations, particularly in regions such as Australia, Canada, and the Pacific Islands, face unique challenges in STD prevention and control, including cultural and linguistic barriers, historical trauma, and disparities in healthcare access. Culturally sensitive approaches that respect Indigenous knowledge, values, and traditions are essential for building trust, engaging communities, and addressing the underlying determinants of STDs effectively.
3.14. Challenges and future directions in STD prevention and control
The landscape of STD prevention and control is marked by a dynamic interplay of challenges and opportunities, reflecting the complex nature of these infections and the evolving needs of populations worldwide. From persistent barriers to access and stigma to emerging threats such as antimicrobial resistance and technological advancements, stakeholders face a multitude of challenges in their efforts to curb the spread of STDs and improve sexual health outcomes.
One of the foremost challenges in STD prevention and control is the persistent stigma and discrimination associated with these infections, which hinder individuals from seeking testing, treatment, and support. Addressing stigma requires multifaceted approaches that challenge misconceptions, promote empathy and understanding, and foster inclusive healthcare environments where individuals feel safe and supported to access sexual health services without fear of judgment or discrimination.
Limited access to comprehensive sexual health education is another significant barrier to STD prevention, particularly among young people and marginalized populations. Effective sexual health education programs should go beyond mere anatomy and physiology to address topics such as consent, communication, healthy relationships, and contraception. By empowering individuals with accurate information and skills to make informed decisions about their sexual health, education programs can help reduce risk behaviors and promote positive sexual health outcomes.
The rise of antimicrobial resistance poses a significant threat to the effectiveness of existing treatment regimens for bacterial STDs such as gonorrhea and syphilis. As pathogens evolve and develop antibiotic resistance, the arsenal of effective treatment options diminishes, exacerbating the challenge of managing and controlling these infections. To address this threat, stakeholders must prioritize antimicrobial stewardship, surveillance, and research into new treatment modalities, such as novel antibiotics and alternative therapies, to combat resistant strains effectively.
Emerging technologies, including telemedicine, mobile health applications, and point-of-care testing devices, offer promising avenues for expanding access to STD prevention and control services, particularly in remote and underserved areas. Telemedicine platforms enable individuals to access confidential consultations, testing, and treatment services remotely, reducing barriers to care and reaching populations who may face challenges accessing traditional healthcare settings. Similarly, mobile health applications can provide personalized sexual health information, reminders for testing and medication adherence, and support for behavior change, empowering individuals to take control of their sexual health in their own hands.
Integrating STD services into primary care settings and other healthcare delivery platforms represents a promising strategy for expanding access to prevention, testing, and treatment services. By embedding sexual health services within existing healthcare infrastructure, such as family planning clinics, community health centers, and antenatal care programs, stakeholders can leverage existing resources, maximize efficiency, and reach individuals who may not otherwise seek out specialized STD services. Furthermore, integration facilitates comprehensive and holistic approaches to sexual health, addressing the full spectrum of reproductive and sexual health needs of individuals and communities.
3.15. Research and Innovation in STD prevention and control
In the ongoing battle against STDs, research and innovation serve as vital weapons, driving progress in prevention, diagnosis, treatment, and public health strategies. From groundbreaking discoveries in vaccine development to innovative technologies for screening and surveillance, the field of STD research is dynamic and multifaceted, characterized by continuous efforts to improve outcomes and address evolving challenges.
One area of significant progress in STD research is vaccine development, with efforts underway to develop safe and effective vaccines against viral pathogens such as HPV and HSV. HPV vaccines, in particular, have revolutionized cervical cancer prevention by targeting the strains responsible for the majority of cervical cancers and genital warts. Ongoing research seeks to expand vaccine coverage to additional HPV genotypes and explore the potential for therapeutic vaccines to treat existing infections and reduce transmission.
Innovations in diagnostic technologies have also transformed STD prevention and control efforts, enabling rapid, accurate, and accessible testing for a wide range of pathogens. Point-of-care tests, molecular assays, and nucleic acid amplification techniques have revolutionized STD screening by providing results in real-time, facilitating timely treatment, and reducing the risk of onward transmission. Furthermore, advances in serological testing and biomarker discovery promise to improve early detection of infections and monitor treatment response.
Behavioral interventions and digital health technologies offer innovative approaches to promoting sexual health and preventing STD transmission among at-risk populations. Mobile health applications, social media platforms, and online interventions provide opportunities for delivering tailored sexual health information, promoting condom use, and facilitating partner notification and testing. Additionally, social network analysis and geospatial mapping tools help identify high-risk populations and target interventions where they are most needed.
Advances in treatment modalities and antimicrobial therapies have expanded the options for managing bacterial and viral STDs, improving outcomes for affected individuals, and reducing transmission risk. Combination therapies, novel antibiotics, and alternative treatment regimens offer hope for addressing antimicrobial resistance and ensuring effective management of infections such as gonorrhea and syphilis. Similarly, antiviral medications and immune-based therapies hold promise for controlling viral STDs such as HIV and HSV, prolonging survival, and reducing the burden of disease.
Epidemiological research and modeling play a crucial role in informing STD prevention and control strategies, providing insights into transmission dynamics, risk factors, and the impact of interventions. Mathematical modeling, phylogenetic analysis, and data visualization techniques help forecast disease trends, evaluate the effectiveness of prevention programs, and guide resource allocation decisions. Furthermore, collaborative research networks and global surveillance systems facilitate data sharing, standardization, and harmonization of methodologies, enabling cross-country comparisons and informing worldwide policy and programming.
3.16. Global partnerships and collaboration in STD prevention and control
In the fight against STDs, global partnerships and collaboration play a pivotal role in mobilizing resources, sharing knowledge, and coordinating efforts to address the complex challenges posed by these infections on a worldwide scale. From multinational organizations to grassroots initiatives, STD prevention and control is characterized by a web of interconnected stakeholders, each contributing their unique expertise, resources, and perspectives to the collective effort.
At the forefront of global STD prevention and control efforts are multinational organizations such as the WHO, UNAIDS, and the United Nations Population Fund (UNFPA). These organizations are vital platforms for coordinating international action, setting global health priorities, and mobilizing political commitment and financial resources to support STD control programs in low- and middle-income countries. Through initiatives such as the Global Fund to Fight AIDS, Tuberculosis, and Malaria and the President’s Emergency Plan for AIDS Relief (PEPFAR), these organizations provide funding, technical assistance, and capacity-building support to national governments and local partners to strengthen STD prevention and control efforts.
Regional partnerships and collaborations are crucial in tailoring STD prevention and control strategies to different regions’ unique epidemiological, cultural, and socioeconomic contexts. Regional organizations such as the European Centre for Disease Prevention and Control (ECDC), the Pan American Health Organization (PAHO), and the African Union (AU) facilitate information sharing, capacity building, and cross-border collaboration among member states to address common challenges and achieve shared health goals. Countries can leverage collective expertise and resources through regional networks to strengthen surveillance systems, harmonize standards and protocols, and implement evidence-based interventions responsive to local needs and priorities.
Non-governmental organizations (NGOs), civil society groups, and community-based organizations (CBOs) play a critical role in STD prevention and control efforts, particularly in reaching vulnerable and marginalized populations who may face barriers to accessing traditional healthcare services. These organizations provide frontline services such as HIV testing and counseling, condom distribution, and outreach to key populations such as sex workers, men who have sex with men, and people who inject drugs. Furthermore, NGOs and CBOs serve as advocates for policy change, human rights, and social justice, amplifying the voices of affected communities and holding governments and policymakers accountable for addressing the needs of those most at risk of STDs.
Academic institutions, research organizations, and public-private partnerships contribute valuable expertise and resources to STD prevention and control efforts, drive innovation, advance scientific knowledge, and translate research findings into practice. Collaborative research networks such as the HIV Prevention Trials Network (HPTN), the International AIDS Vaccine Initiative (IAVI), and the Global Antibiotic Research and Development Partnership (GARDP) facilitate interdisciplinary collaboration, data sharing, and capacity building to address priority research questions and develop new tools and strategies for STD prevention, diagnosis, and treatment.
3.17. Policy and advocacy in STD prevention and control
In the realm of STD prevention and control, policy and advocacy efforts serve as powerful tools for driving change, shaping priorities, and promoting the rights and well-being of individuals and communities affected by these infections. From advocating for increased funding and resources to pushing for legislative reforms and challenging stigma and discrimination, policy and advocacy play a critical role in shaping the landscape of STD prevention and control at local, national, and global levels.
One of the primary goals of policy and advocacy in STD prevention and control is to mobilize political commitment and financial resources to support comprehensive and evidence-based interventions. Advocacy organizations, civil society groups, and affected communities play a crucial role in raising awareness about the importance of sexual health, highlighting the impact of STDs on individuals, families, and communities, and calling for increased investment in prevention, testing, treatment, and support services. By engaging policymakers, legislators, and government officials, advocates can influence budget allocations, policy decisions, and program priorities to ensure that STD prevention and control efforts are adequately resourced and prioritized within broader health agendas.
Policy advocacy also plays a crucial role in challenging discriminatory laws, policies, and practices that impede access to sexual health services and perpetuate stigma and discrimination against affected populations. Laws criminalizing HIV transmission, punitive regulations targeting sex workers and people who inject drugs, and discriminatory policies related to sexual orientation and gender identity hinder efforts to reach key populations with essential prevention and care services and fuel the spread of STDs. Advocates work tirelessly to repeal discriminatory laws, promote human rights and dignity, and create enabling environments that support access to comprehensive sexual health services for all individuals, regardless of their sexual orientation, gender identity, or social status.
Moreover, policy advocacy efforts seek to promote evidence-based approaches to STD prevention and control that prioritize the needs and preferences of affected communities. Advocates push for policies and programs grounded in scientific evidence, respect human rights, and address the social determinants of health, including poverty, gender inequality, and structural violence. By championing approaches such as harm reduction, comprehensive sexuality education, and community-led interventions, advocates can help ensure that STD prevention and control efforts are effective, equitable, culturally sensitive, and responsive to the diverse needs and realities of different populations.
International advocacy and diplomacy play a critical role in shaping global norms and standards for STD prevention and control, fostering collaboration among countries, and promoting best practices in policy and programming. Through forums such as the United Nations General Assembly Special Session (UNGASS) on HIV/AIDS, the International Conference on Population and Development (ICPD), and the World Health Assembly (WHA), advocates can influence global health agendas, advocate for the rights of affected populations, and hold governments accountable for their commitments to STD prevention and control. Furthermore, international partnerships and collaborations facilitate knowledge sharing, capacity building, and resource mobilization to support STD control efforts in low- and middle-income countries, where the burden of disease is often most significant.
3.18. Intersectionality and STD prevention and control
In the realm of STD prevention and control, the principles of intersectionality and health equity are crucial for understanding and addressing the complex and intersecting factors that influence STD risk, vulnerability, and access to care. Intersectionality acknowledges that multiple intersecting factors, including race, ethnicity, gender, sexual orientation, socioeconomic status, immigration status, and disability, among others, shape individuals’ experiences of health and well-being. By recognizing diverse populations’ unique experiences and needs, intersectionality provides a framework for promoting health equity and addressing disparities in STD prevention and control efforts.
One key aspect of intersectionality in STD prevention and control is the recognition of how social determinants of health intersect with STD risk and vulnerability. Structural factors such as poverty, racism, sexism, homophobia, and discrimination contribute to disparities in access to sexual health information, resources, and services, exacerbating STD risk and limiting individuals’ ability to protect themselves from infection. For example, marginalized communities may face barriers such as lack of insurance coverage, transportation, or childcare, which hinder their ability to access STD testing, treatment, and prevention services. By addressing these structural inequities, stakeholders can create more equitable and inclusive healthcare systems that ensure all individuals have access to the resources they need to maintain sexual health and well-being.
Moreover, intersectionality highlights the importance of recognizing and addressing the unique needs and experiences of key populations who face disproportionate burdens of STDs. Groups such as racial and ethnic minorities, LGBTQ + individuals, sex workers, people who inject drugs, incarcerated individuals, and adolescents may experience heightened vulnerability to STDs due to a combination of social, economic, and structural factors, including stigma, discrimination, violence, and marginalization. By centering the experiences and priorities of these populations, stakeholders can develop culturally competent and responsive interventions that meet their specific needs and promote health equity.
Intersectionality also underscores the importance of taking a holistic and comprehensive approach to STD prevention and control that addresses the intersecting social, economic, and health-related needs of individuals and communities. This approach recognizes that STDs do not occur in isolation but are intertwined with broader issues such as poverty, housing instability, mental health, substance use, and access to education and employment. By integrating STD services with other essential health and social services, stakeholders can address the root causes of health inequities and promote holistic well-being among marginalized populations.
Furthermore, intersectionality emphasizes the importance of engaging affected communities in designing, implementing, and evaluating STD prevention and control efforts. Community-led approaches that prioritize participatory decision-making, cultural humility, and mutual respect empower individuals and communities to take ownership of their sexual health and advocate for their needs. By building partnerships with community-based organizations, grassroots activists, and affected individuals, stakeholders can ensure that STD prevention and control efforts are responsive to local contexts, culturally appropriate, and effective in addressing the needs of diverse populations.
3.19. Community engagement and empowerment in STD prevention and control
Community engagement and empowerment are fundamental principles in the realm of STD prevention and control, emphasizing the importance of involving affected communities as active partners in identifying priorities, shaping interventions, and driving change. By recognizing the expertise, knowledge, and resilience of communities and fostering collaboration and shared decision-making, community engagement, and empowerment efforts aim to build trust, strengthen resilience, and promote sustainable solutions to the complex challenges of STDs.
At the heart of community engagement and empowerment is recognizing that communities affected by STDs are not passive recipients of services but active agents of change with unique insights and experiences to contribute. By engaging affected communities in the design, implementation, and evaluation of STD prevention and control efforts, stakeholders can ensure that interventions are responsive to local needs and culturally appropriate and effectively address the root causes of STD transmission. Community-based organizations, grassroots activists, and peer educators are crucial in mobilizing community members, raising awareness, and advocating for policies and programs that promote sexual health and well-being.
Community engagement involves various activities to build trust, foster dialogue, and promote collaboration among diverse stakeholders. Community forums, town hall meetings, and focus groups allow community members to voice their concerns, share their experiences, and contribute ideas for addressing STD-related challenges. Similarly, participatory research approaches, such as community-based participatory research (CBPR), involve community members as active participants in the research process, empowering them to shape research questions, methods, and outcomes and ensuring that research findings are relevant and actionable.
Empowerment is central to community engagement efforts, emphasizing the importance of building the capacity of communities to advocate for their needs, mobilize resources, and drive change. Empowerment approaches focus on strengthening individual and collective skills, knowledge, and self-efficacy through education, training, and leadership development initiatives. By providing opportunities for community members to build confidence, assertiveness, and advocacy skills, stakeholders can empower communities to challenge stigma, discrimination, and social inequities that fuel the spread of STDs and advocate for policies and programs that promote sexual health and rights.
Moreover, community engagement and empowerment efforts prioritize the inclusion of marginalized and vulnerable populations who may face intersecting barriers to accessing sexual health services and information. By centering the voices and experiences of groups such as racial and ethnic minorities, LGBTQ + individuals, sex workers, people who inject drugs, and adolescents, stakeholders can ensure that interventions are equitable, inclusive, and responsive to the diverse needs of diverse communities. Culturally competent approaches that respect diversity, foster trust, and build partnerships with community-based organizations are essential for reaching marginalized populations and addressing disparities in STD outcomes.
3.20. Surveillance and monitoring
Surveillance and monitoring are vital components of STD prevention and control efforts, providing essential data and insights into the epidemiology, trends, and impact of STDs on populations worldwide. By systematically collecting, analyzing, and disseminating information on STD prevalence, incidence, risk factors, and outcomes, surveillance systems enable stakeholders to track disease trends, identify emerging threats, target interventions, and evaluate the effectiveness of prevention and control strategies.
One key aspect of STD surveillance is the monitoring of disease burden and trends over time. Surveillance systems collect data on the number of reported cases, prevalence rates, and demographic characteristics of affected individuals, providing insights into the distribution and dynamics of STDs within and between populations. Trends in STD incidence and prevalence can reveal patterns of transmission, identify high-risk populations, and inform targeted prevention and control efforts in communities where STD rates are highest. Furthermore, longitudinal surveillance data enable stakeholders to monitor the impact of interventions, track progress towards national and global targets, and identify areas where additional resources and support may be needed.
Surveillance systems also play a crucial role in detecting outbreaks and clusters of STDs, facilitating timely responses to emerging threats, and preventing further transmission within communities. By monitoring trends in STD diagnoses, test positivity rates, and geographic distribution, surveillance systems can identify clusters of cases, detect changes in transmission patterns, and alert public health authorities to potential outbreaks. Early detection of outbreaks enables stakeholders to implement targeted interventions such as enhanced screening, contact tracing, and community outreach to control transmission and prevent further spread of STDs.
In addition to monitoring disease burden and trends, STD surveillance systems track key risk factors and determinants of STD transmission, providing insights into the social, behavioral, and environmental factors that contribute to STD risk and vulnerability. Data on sexual behaviors, condom use, substance use, healthcare access, and socioeconomic status help identify populations at the highest risk of STD acquisition and transmission and guide the development of tailored interventions and targeted prevention strategies. Furthermore, surveillance data on comorbidities, coinfections, and syndemics such as HIV/AIDS, viral hepatitis, and substance use disorders enable stakeholders to address the complex interactions between different health conditions and develop integrated approaches to care and prevention.
Surveillance and monitoring efforts also support international collaboration and knowledge sharing, facilitating data exchange, standardization of methodologies, and harmonization of surveillance practices across countries and regions. Through initiatives such as the Global AIDS Monitoring (GAM) system, the Global Gonococcal Antimicrobial Surveillance Program (GASP), and the WHO’s Global Health Observatory (GHO), stakeholders can access standardized data on STD epidemiology, trends, and responses, enabling cross-country comparisons, benchmarking, and evaluation of progress towards global targets and commitments.
3.21. Resilience and adaptation
Resilience and adaptation are essential principles in the realm of STD prevention and control, emphasizing the ability of individuals, communities, and healthcare systems to respond effectively to changing circumstances, emerging threats, and unexpected challenges. In the face of evolving epidemiological trends, social determinants of health, and environmental factors, resilience and adaptation enable stakeholders to navigate uncertainty, overcome obstacles, and sustain progress toward reducing the burden of STDs and promoting sexual health and well-being.
One key aspect of resilience and adaptation in STD prevention and control is the ability of healthcare systems to respond effectively to shifts in disease burden, transmission dynamics, and healthcare needs. By building flexible and responsive healthcare systems that adapt to changing epidemiological trends, stakeholders can ensure that essential STD prevention, testing, treatment, and support services remain accessible and available to all individuals, regardless of their circumstances. This may involve strengthening primary care infrastructure, integrating STD services with other health services, and adopting innovative service delivery models like telemedicine and mobile clinics to reach underserved populations and remote areas.
Furthermore, resilience and adaptation require a proactive approach to addressing emerging threats and vulnerabilities, such as antimicrobial resistance, changing sexual behaviors, and the impact of environmental factors such as climate change and urbanization. By monitoring trends in STD epidemiology, conducting risk assessments, and engaging in scenario planning, stakeholders can anticipate future challenges and develop proactive strategies to mitigate risks, prevent outbreaks, and protect the health of populations. This may involve strengthening surveillance systems, enhancing laboratory capacity, and investing in research and development of new diagnostics, treatments, and prevention technologies.
Community resilience and empowerment are also essential for effective STD prevention and control, enabling communities to mobilize resources, support each other, and advocate for their needs in the face of adversity. By fostering social cohesion, building support networks, and promoting community-led approaches to health promotion and advocacy, stakeholders can empower communities to address the social determinants of health, challenge stigma and discrimination, and create environments that promote sexual health and well-being for all individuals. This may involve supporting community-based organizations, providing training and capacity-building opportunities, and facilitating collaboration and knowledge sharing among diverse stakeholders.
Moreover, resilience and adaptation require a commitment to learning from past experiences, evaluating the effectiveness of interventions, and continuously improving strategies and approaches based on feedback and evidence. By fostering a culture of learning, innovation, and continuous improvement, stakeholders can identify gaps in services, address barriers to access, and optimize the impact of STD prevention and control efforts over time. This may involve conducting program evaluations, engaging in quality improvement initiatives, and sharing best practices and lessons learned with other stakeholders to inform future action.
3.22. Social and economic implications of STDs
The social and economic implications of STDs are profound and multifaceted, encompassing a wide range of impacts on individuals, families, communities, and societies as a whole. From the direct costs of healthcare services to the broader societal consequences of stigma, discrimination, and lost productivity, STDs have far-reaching effects that extend beyond the realm of public health.
At the individual level, STDs can have significant physical, emotional, and financial consequences for affected individuals. The direct costs of seeking testing, treatment, and care for STDs can be substantial, particularly for those who lack health insurance or access to affordable healthcare services. Additionally, STD-related complications such as pelvic inflammatory disease, infertility, chronic pain, and certain types of cancer can have long-term health effects and impair individuals’ quality of life. Furthermore, the stigma and shame associated with STDs can lead to social isolation, relationship difficulties, and psychological distress, exacerbating the emotional toll of living with these infections.
The economic burden of STDs extends beyond the individual level to impact families, communities, and societies as a whole. The costs of treating STDs, including medications, laboratory tests, and healthcare provider visits, impose a significant financial burden on healthcare systems and taxpayers. Moreover, the indirect costs of STDs, such as lost productivity due to illness, disability, and absenteeism from work or school, can have a substantial impact on economic productivity and growth. Additionally, the social and economic consequences of untreated STDs, such as reduced educational attainment, lower workforce participation, and higher rates of poverty, can perpetuate cycles of disadvantage and inequality within communities.
Furthermore, the social and economic implications of STDs are often exacerbated by intersecting factors such as gender, race, ethnicity, sexual orientation, socioeconomic status, and geographic location. Marginalized populations, including racial and ethnic minorities, LGBTQ + individuals, sex workers, people who inject drugs, and adolescents, are disproportionately affected by STDs due to structural inequities such as poverty, lack of access to healthcare, discrimination, and stigma. These populations may face barriers to accessing sexual health services, experience higher rates of STD-related morbidity and mortality, and be more vulnerable to the social and economic consequences of living with STDs.
Addressing the social and economic implications of STDs requires a comprehensive and multi-sectoral approach that addresses the underlying determinants of health, promotes equity and social justice, and fosters collaboration among diverse stakeholders. This may involve expanding access to affordable healthcare services, implementing comprehensive sexual health education programs, promoting condom use and other prevention strategies, and reducing stigma and discrimination against affected individuals and communities. Additionally, investing in research, surveillance, and monitoring of STD trends and impacts can help inform evidence-based interventions and allocate resources effectively to address the social and economic burden of STDs.
3.23. Innovations in the treatment and care of STDs
In the ever-evolving landscape of STD prevention and control, innovations in treatment and care play a pivotal role in improving outcomes for individuals affected by these infections. From advancements in medication development to innovative approaches in patient-centered care, these innovations transform how STDs are diagnosed, treated, and managed, offering hope for better health outcomes and enhanced quality of life for affected individuals.
One significant area of innovation in STD treatment is the development of new medications and treatment modalities for bacterial and viral infections. For bacterial STDs such as gonorrhea and syphilis, the emergence of antimicrobial resistance has posed a significant challenge to traditional treatment regimens. Researchers are also exploring alternative antibiotics, combination therapies, and novel treatments, such as phage therapy, to combat resistant strains and effectively manage these infections. Similarly, for viral STDs such as HIV and HSV, advances in antiviral medications, immune-based therapies, and gene editing technologies hold promise for controlling viral replication, reducing transmission risk, and improving long-term outcomes for affected individuals.
In addition to medication development, diagnostic technology innovations are revolutionizing how STDs are detected and diagnosed. Traditional laboratory-based tests for STDs such as chlamydia and gonorrhea often require time-consuming sample collection, processing, and analysis, delaying diagnosis and treatment initiation. However, advances in point-of-care testing (POCT), molecular diagnostics, and rapid testing platforms enable real-time detection of STDs at the point of care, allowing for same-day diagnosis and treatment initiation. These rapid tests offer benefits such as increased convenience, reduced time to results, and improved patient satisfaction, facilitating timely intervention and reducing the risk of onward transmission.
Furthermore, innovations in patient-centered care are enhancing the delivery of STD services and improving the patient experience. Traditionally, STD care has been delivered in clinical settings such as hospitals, clinics, and private practices, which may be inaccessible or stigmatizing for specific populations. However, innovative approaches such as telemedicine, mobile health clinics, and community-based outreach programs are expanding access to STD services and reaching underserved populations where they are. Telemedicine platforms enable remote consultations, counseling, and prescription refills, eliminating barriers such as transportation and childcare and empowering individuals to take control of their sexual health from the comfort of their own homes.
Moreover, advancements in holistic and integrated care models are addressing the complex needs of individuals affected by STDs, including their physical, emotional, and social well-being. Integrated care models, which combine STD services with other essential health and social services such as mental health care, substance use treatment, and social support, offer comprehensive support to individuals living with STDs and address the underlying determinants of health that contribute to STD risk and vulnerability. By addressing the social, behavioral, and structural factors that influence STD transmission and outcomes, these innovative care models promote holistic well-being and improve health outcomes for affected individuals and communities.
3.24. Psychosocial impacts and mental health
The psychosocial impacts of STDs extend far beyond the physical symptoms of infection, profoundly affecting the mental health and well-being of individuals and communities. From the moment of diagnosis to the long-term management of chronic infections, STDs can have a significant impact on emotional well-being, relationships, and overall quality of life.
Upon receiving an STD diagnosis, individuals may experience a range of emotional responses, including shock, fear, shame, and guilt. The stigma associated with STDs, fueled by societal attitudes and misconceptions, can exacerbate these feelings and lead to social isolation, self-blame, and internalized stigma. Additionally, concerns about disclosure, rejection, and judgment from partners, family members, and peers can further compound the emotional burden of living with an STD.
The psychological impact of STDs can extend beyond the individual to affect interpersonal relationships and sexual behavior. For individuals in committed relationships, the disclosure of an STD diagnosis can pose challenges to trust, communication, and intimacy, leading to feelings of betrayal, resentment, and uncertainty about the future of the relationship. Similarly, for individuals who are dating or seeking new sexual partners, the fear of rejection or transmission can create barriers to forming intimate connections and lead to avoidance of sexual relationships altogether.
Furthermore, the management of chronic STDs such as HIV/AIDS, HSV, and HPV can pose ongoing challenges to mental health and well-being. The daily regimen of medication adherence, regular medical appointments, and monitoring of viral load or disease progression can be emotionally taxing and disruptive to daily life. Additionally, the uncertainty about long-term health outcomes, the potential for disease progression, and the risk of transmitting the infection to others can create feelings of anxiety, depression, and existential distress.
Moreover, the psychosocial impacts of STDs are often intertwined with broader social determinants of health, including socioeconomic status, race, ethnicity, gender identity, and sexual orientation. Marginalized populations, including racial and ethnic minorities, LGBTQ + individuals, and people living in poverty, may face additional barriers to accessing mental health support and coping resources, exacerbating disparities in mental health outcomes. Furthermore, intersecting forms of discrimination and oppression, such as racism, homophobia, and transphobia, can compound the emotional toll of living with an STD and contribute to feelings of social exclusion, marginalization, and identity-related stress.
Addressing the psychosocial impacts of STDs requires a comprehensive and multidisciplinary approach that integrates mental health support, counseling, and psychosocial interventions into STD care and prevention programs. Providing individuals with access to confidential, non-judgmental counseling and support services can help alleviate feelings of shame, guilt, and isolation and empower individuals to navigate the emotional challenges of living with an STD. Additionally, promoting destigmatization and education about STDs can help reduce stigma and discrimination, foster supportive social networks, and create environments that promote acceptance, empathy, and understanding.
3.25. Health disparities and health equity
Health disparities and health equity are critical issues in the realm of STDs, highlighting the unequal distribution of STD burden and access to care among different populations. A complex interplay of social, economic, and structural factors, including race, ethnicity, gender, sexual orientation, socioeconomic status, geographic location, and access to healthcare, influences these disparities.
Racial and ethnic minorities, particularly Black, Hispanic/Latinx, and Indigenous communities, experience disproportionate rates of STDs compared to white populations. Structural inequities such as poverty, lack of access to healthcare, discrimination, and racism contribute to higher rates of STD transmission and poorer health outcomes among these populations. Additionally, historical and ongoing injustices, such as the legacy of slavery, colonization, and systemic racism, have created environments where individuals from marginalized racial and ethnic groups face increased vulnerability to STDs due to limited access to education, economic opportunities, and healthcare services.
Gender disparities also play a significant role in shaping STD burden and access to care. Women, particularly young women and transgender women, are disproportionately affected by certain STDs, such as chlamydia and HPV, due to biological factors, gender-based violence, and unequal power dynamics in sexual relationships. Additionally, LGBTQ + individuals, including gay and bisexual men and transgender individuals, face higher rates of STDs compared to heterosexual individuals, often due to stigma, discrimination, and barriers to accessing culturally competent and affirming healthcare services.
Socioeconomic status is a significant determinant of STD risk and vulnerability, with individuals living in poverty experiencing higher rates of STDs compared to those with higher incomes. Economic disparities limit access to preventive services such as STD testing, contraception, and healthcare coverage, exacerbating STD transmission and perpetuating cycles of disadvantage and inequality. Additionally, individuals experiencing homelessness, incarceration, or substance use disorders face unique challenges in accessing sexual health services. They may be at heightened risk of STD acquisition and transmission due to unstable housing, incarceration, and injection drug use.
Geographic disparities also contribute to variations in STD burden and access to care, with certain regions and communities experiencing higher rates of STDs compared to others. Rural communities, in particular, face challenges such as limited access to healthcare services, shortage of healthcare providers, and stigma related to seeking sexual health services. Additionally, urban areas with high population density and socioeconomic deprivation may also experience elevated rates of STDs due to factors such as poverty, overcrowded housing, and limited access to education and healthcare.
Addressing health disparities and promoting health equity in STD prevention and control requires a multifaceted and intersectional approach that addresses the underlying determinants of health and prioritizes the needs of marginalized and underserved populations. This includes increasing access to culturally competent and linguistically appropriate sexual health services, implementing comprehensive sexuality education programs, expanding access to contraception and reproductive healthcare services, and addressing social determinants of health such as poverty, discrimination, and lack of access to education and economic opportunities.
Furthermore, promoting community engagement, empowerment, and leadership among affected populations is essential for developing interventions responsive to diverse communities’ unique needs and priorities. This may involve partnering with community-based organizations, grassroots activists, and affected individuals to co-create and implement interventions that address the social, economic, and structural factors that contribute to STD risk and vulnerability.
3.26. Policy and legal frameworks
Policy and legal frameworks play a crucial role in shaping the landscape of STD prevention and control, influencing everything from access to healthcare services to the implementation of prevention strategies and the protection of individuals’ rights. These frameworks encompass a wide range of laws, regulations, policies, and guidelines at the local, national, and international levels, aimed at addressing STDs from a public health and human rights perspective.
At the national level, governments develop policies and laws to guide STD prevention and control efforts, often in collaboration with public health agencies, healthcare providers, community organizations, and other stakeholders. These policies may include guidelines for STD testing and treatment, regulations for reporting STD cases to public health authorities, and requirements for healthcare providers to offer STD counseling, testing, and treatment to their patients. Additionally, national governments may allocate funding and resources to support STD prevention programs, research initiatives, and public awareness campaigns aimed at reducing STD transmission and promoting sexual health.
Furthermore, legal frameworks play a critical role in protecting the rights and well-being of individuals affected by STDs, ensuring access to confidential and nondiscriminatory healthcare services, and preventing discrimination based on STD status. Laws such as the Health Insurance Portability and Accountability Act (HIPAA) in the United States, for example, protect the privacy and confidentiality of individuals’ health information, including STD test results and medical records. Similarly, anti-discrimination laws prohibit discrimination based on STD status in employment, housing, education, and healthcare settings, ensuring that individuals living with STDs are not unfairly stigmatized or denied access to essential services.
Internationally, organizations such as the WHO and the United Nations (UN) develop guidelines and recommendations for STD prevention and control, which member states can adopt and implement within their own legal and policy frameworks. These international guidelines help ensure consistency and harmonization of STD prevention strategies across countries and regions, facilitate knowledge sharing and collaboration among stakeholders, and promote a coordinated global response to the STD epidemic.
Moreover, legal frameworks also play a role in regulating behaviors and practices that contribute to STD transmission, such as sex work, drug use, and sexual violence. Laws and regulations related to these activities can have a significant impact on STD risk and vulnerability, influencing access to prevention services, harm reduction interventions, and legal protections for marginalized populations. In some cases, laws criminalizing certain behaviors, such as sex work or drug possession, can exacerbate stigma, drive these activities underground, and hinder efforts to reach at-risk populations with essential health services.
Addressing the complex legal and policy issues surrounding STD prevention and control requires a multi-sectoral and evidence-based approach that balances public health objectives with respect for individual rights and autonomy. This may involve conducting policy analysis and advocacy to identify gaps in existing legal frameworks, promote reforms that support STD prevention and care, and challenge laws and regulations that perpetuate stigma, discrimination, and barriers to access. Additionally, engaging with affected communities, civil society organizations, and policymakers can help ensure that legal and policy responses to STDs are responsive to the diverse needs and priorities of all individuals and communities affected by these infections.
3.27. Ethical Considerations
Ethical considerations are paramount in the realm of STD prevention and control, guiding decision-making, policy development, and practice in ways that prioritize the rights, dignity, and well-being of individuals affected by STDs. From ensuring confidentiality and informed consent to promoting autonomy and justice, ethical principles inform every aspect of STD care, research, and public health interventions.
One fundamental ethical principle in STD prevention and control is respect for individual autonomy and informed consent. Individuals have the right to make informed decisions about their sexual health, including whether to undergo STD testing, treatment, and disclosure of their STD status to partners or healthcare providers. Healthcare providers must provide accurate information about STDs, treatment options, and potential risks and benefits, allowing individuals to make autonomous decisions that align with their values, preferences, and circumstances. Additionally, ensuring the confidentiality and privacy of an individual’s health information is essential for fostering trust and maintaining the integrity of the patient-provider relationship.
Another key ethical consideration is the principle of beneficence, which requires healthcare providers and public health practitioners to act in the best interests of their patients and the broader community. This may involve providing timely and appropriate STD testing, treatment, and counseling services to individuals affected by STDs, regardless of their ability to pay or access to healthcare. Additionally, promoting preventive interventions such as vaccination, condom use, and harm reduction strategies can help reduce the risk of STD transmission and protect the health and well-being of individuals and communities.
Furthermore, the principle of non-maleficence underscores the importance of avoiding harm and minimizing risks to individuals affected by STDs. This may involve ensuring that STD testing, treatment, and prevention services are provided in a culturally competent, non-judgmental, and respectful manner, free from stigma, discrimination, and coercion. Additionally, healthcare providers must minimize the risk of adverse outcomes associated with STDs, such as complications, transmission to partners, and psychosocial distress, through appropriate medical management, counseling, and support services.
Justice is another ethical principle central to STD prevention and control efforts, emphasizing fairness, equity, and inclusivity in distributing resources and access to care. Addressing health disparities and promoting health equity requires a commitment to addressing the underlying social determinants of health that contribute to STD risk and vulnerability, including poverty, discrimination, and lack of access to education and healthcare services. Additionally, ensuring that STD prevention and control efforts are accessible, affordable, and culturally relevant for all individuals and communities, particularly those who are marginalized or underserved, is essential for promoting justice and reducing health inequities.
Moreover, ethical considerations extend to research involving STDs, including issues related to informed consent, privacy, and protection of human subjects. Researchers are responsible for obtaining informed consent from study participants, protecting their privacy and confidentiality, and minimizing risks and burdens associated with participation in research studies. Additionally, researchers must ensure that study findings are disseminated transparently and responsibly, contributing to advancing knowledge and informing evidence-based interventions and policies.
3.28. Global health security and pandemic preparedness
Global health security and pandemic preparedness are critical considerations in the context of STDs, highlighting the interconnectedness of STD prevention and control efforts with broader efforts to prevent and respond to emerging infectious diseases and public health threats. While STDs may not always be at the forefront of discussions on global health security, they pose significant challenges to health systems, economies, and social well-being, particularly in the context of pandemics such as HIV/AIDS and COVID-19.
One of the key lessons learned from past pandemics, such as HIV/AIDS, is the importance of early detection and rapid response to emerging infectious diseases. STD surveillance systems play a vital role in monitoring trends in STD prevalence, incidence, and antimicrobial resistance, providing early warning signals for potential outbreaks and informing targeted interventions to control transmission by integrating STD surveillance into broader infectious disease surveillance systems, public health authorities can detect and respond to emerging threats more effectively, preventing the spread of STDs and reducing their impact on population health.
Furthermore, the HIV/AIDS pandemic has underscored the importance of community engagement, empowerment, and leadership in pandemic response efforts. Community-based organizations, grassroots activists, and affected individuals played a crucial role in advocating for access to HIV testing, treatment, and prevention services, challenging stigma and discrimination, and mobilizing resources to support those affected by the epidemic. Similarly, in the context of STD prevention and control, empowering communities to take ownership of their sexual health, advocate for their needs, and participate in decision-making processes is essential for building resilient and responsive health systems.
Additionally, pandemic preparedness efforts must address the intersecting vulnerabilities and social determinants of health that contribute to STD risk and vulnerability, particularly among marginalized and underserved populations. Addressing issues such as poverty, lack of access to healthcare, discrimination, and stigma is essential for reducing health disparities and promoting health equity in STD prevention and control efforts. Moreover, integrating STD prevention and control into broader health systems strengthening initiatives can help build resilience and capacity to respond to emerging infectious diseases and other public health threats, ensuring a more holistic and sustainable approach to global health security.
In the context of the COVID-19 pandemic, STD prevention and control efforts have faced unique challenges and opportunities. The pandemic has disrupted healthcare services, including STD testing, treatment, and prevention programs, leading to disruptions in care and increased risk of STD transmission. However, the COVID-19 pandemic has also highlighted the importance of innovative approaches such as telemedicine, mobile health clinics, and self-testing kits for expanding access to STD services and reaching underserved populations where they are. Additionally, the pandemic has underscored the need for resilient and adaptable health systems that can respond effectively to emerging threats, prioritize the needs of vulnerable populations, and promote equity and justice in healthcare delivery.
3.29. Diagnosis and management of STDs
Diagnosis and management are crucial aspects of effectively addressing STDs, ensuring timely detection, appropriate treatment, and prevention of further transmission. Healthcare providers employ a variety of approaches and techniques to diagnose STDs accurately and manage them effectively, tailored to the specific characteristics of each infection and the individual needs of patients.
One of the primary methods for diagnosing STDs is through laboratory testing of biological specimens, such as blood, urine, genital swabs, or tissue samples. These tests can detect the presence of pathogens, antibodies, or genetic material associated with STDs, allowing for accurate diagnosis and targeted treatment. Common laboratory tests for STDs include nucleic acid amplification tests (NAATs), enzyme immunoassays (EIAs), polymerase chain reaction (PCR) assays, and culture-based methods, each with its advantages and limitations depending on the specific pathogen being targeted.
In addition to laboratory testing, healthcare providers may use clinical examination and history-taking to diagnose STDs, particularly infections with characteristic symptoms or signs. Physical examination may involve visual inspection of the genital area, palpation of lymph nodes, and evaluation of mucosal surfaces for lesions, discharge, or other abnormalities. Obtaining a thorough sexual history, including information about sexual practices, partners, and risk factors, is also essential for guiding diagnostic and management decisions and identifying individuals at increased risk of STD acquisition.
Once a diagnosis is made, the management of STDs typically involves a combination of pharmacological treatment, behavioral counseling, partner management, and preventive interventions. Pharmacological treatment may include antibiotics, antiviral medications, or other antimicrobial agents, depending on the type of STD and the susceptibility of the causative pathogen. Behavioral counseling aims to promote safer sexual practices, enhance adherence to treatment regimens, and reduce the risk of STD transmission to partners or reinfection. Partner management involves notifying and treating sexual partners of individuals diagnosed with STDs to prevent the further spread of infection and reduce the risk of complications.
Moreover, preventive interventions play a critical role in managing STDs and reducing their impact on individuals and communities. These interventions may include vaccination against STDs such as HPV, HBV, and certain strains of meningococcus and pneumococcus, which can prevent acquiring and transmitting these infections. Additionally, promoting condom use, regular STD testing, pre-exposure prophylaxis (PrEP) for HIV prevention, and harm reduction strategies for individuals at increased risk of STD acquisition, such as people who inject drugs or engage in transactional sex, can help reduce transmission and improve health outcomes.
In essence, treating STIs often involves the use of antimicrobial drugs to eradicate the causative pathogens and alleviate symptoms. These drugs target the specific mechanisms of action of the pathogens, inhibit their growth or replication, and ultimately help restore the individual’s health and prevent further transmission of the infection.
One commonly used class of antimicrobial drugs for STIs is antibiotics, which are effective against bacterial infections such as chlamydia, gonorrhea, and syphilis. For example, azithromycin and doxycycline are frequently prescribed antibiotics for the treatment of chlamydia infections. Azithromycin inhibits bacterial protein synthesis, while doxycycline disrupts bacterial protein production by binding to the ribosome. These antibiotics have relatively long half-lives, with azithromycin having a half-life of approximately 68 hours and doxycycline ranging from 12 to 25 hours. A typical course of treatment for chlamydia infection involves a single dose of azithromycin or a week-long course of doxycycline.
For gonorrhea, the recommended treatment often involves a combination of antibiotics to combat increasing antimicrobial resistance. Ceftriaxone, a third-generation cephalosporin, is commonly used with azithromycin or doxycycline to treat gonorrhea infections. Ceftriaxone inhibits bacterial cell wall synthesis, while azithromycin and doxycycline target bacterial protein synthesis. Ceftriaxone has a half-life of approximately 5 to 9 hours and is typically administered as a single intramuscular injection, while azithromycin and doxycycline are administered orally.
For syphilis, penicillin remains the drug of choice for treatment, particularly for primary and secondary syphilis infections. Penicillin disrupts bacterial cell wall synthesis, ultimately leading to cell death. The duration of treatment for syphilis varies depending on the stage of the infection, with early-stage infections typically requiring a single intramuscular injection of benzathine penicillin G and more advanced infections necessitating multiple doses over a longer duration.
In addition to antibiotics, antiviral drugs are used to treat viral STIs, such as HSV and HIV. Antiviral drugs such as acyclovir, valacyclovir, and famciclovir are commonly prescribed for HSV infections to reduce the frequency and severity of recurrent outbreaks and alleviate symptoms. These antiviral drugs work by inhibiting viral DNA synthesis and replication, ultimately suppressing viral activity and reducing the duration and severity of symptoms. They have relatively short half-lives, with acyclovir having a half-life of approximately 2.5 to 3.3 hours and valacyclovir and famciclovir being converted to acyclovir in the body.
For HIV/AIDS, combination antiretroviral therapy (ART) is the standard of care for treatment, consisting of a combination of drugs from different classes to suppress viral replication and restore immune function. These drugs include nucleoside reverse transcriptase inhibitors, non-nucleoside reverse transcriptase inhibitors, protease inhibitors, integrase strand transfer inhibitors, and entry inhibitors. ART targets different stages of the HIV life cycle, preventing the virus from replicating and reducing viral load in the body. The duration of treatment with ART is lifelong, as discontinuation of therapy can lead to viral rebound and disease progression. Table 2 presents clinical guidelines for the management of STDs.
Table 2.
A broad table summarizing essential clinical guidelines for the management of STDs from various national and international organizations.
| Organization | Guideline name | Focus areas | Highlights |
:---: :---: |
| Centers for Disease Control and Prevention | STD Treatment Guidelines | Diagnosis, treatment, and management of STDs | - Evidence-based recommendations for STD testing, treatment regimens, partner management, and prevention strategies - Regularly updated to reflect changes in STD epidemiology and clinical practice |
| World Health Organization | Guidelines for the Management of Sexually Transmitted Infections | Diagnosis, treatment, and prevention of sexually transmitted infections worldwide | - Global recommendations for syndromic management of STIs, laboratory diagnosis, antimicrobial resistance, and prevention strategies such as condom use and vaccination against HPV and hepatitis B |
| European Centre for Disease Prevention and Control | Guidelines for the Management of Chlamydia trachomatis Infections | Diagnosis, treatment, and partner management of chlamydia infections | - Recommendations for diagnosis, treatment, and partner management of chlamydia infections, as well as strategies for prevention and control at the population level |
| British Association for Sexual Health and HIV | British Association for Sexual Health and HIV Guidelines | Diagnosis, treatment, and management of sexually transmitted infections in the United Kingdom | - Clinical guidelines developed by expert panels of healthcare professionals in the UK. - Cover a wide range of STDs and related conditions |
| International Antiviral Society-USA | Guidelines for the Management of HIV and Other Viral Infections | Management of HIV/AIDS and viral infections commonly associated with HIV | - Recommendations for managing viral infections such as HSV and HPV - Widely regarded as authoritative in the field of HIV/AIDS |
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These guidelines serve as valuable resources for healthcare providers, policymakers, and public health officials, providing evidence-based recommendations for diagnosing, treating, and preventing STDs. By standardizing clinical practice and promoting consistency in care, these guidelines contribute to improved health outcomes for individuals affected by STDs and support efforts to reduce the burden of STDs at the population level.
HIV = human immunodeficiency virus, HSV = herpes simplex virus, HPV = human papillomavirus, STDs = sexually transmitted diseases.
4. Conclusion
The global burden of STDs represents a significant contribution to the field of public health and international health planning. By synthesizing the latest evidence and employing rigorous methodology, we have gained valuable insights into the magnitude and implications of STDs worldwide.
Our findings underscore the substantial impact of STDs on individuals, communities, and healthcare systems, highlighting the urgent need for effective prevention and control strategies. Through the quantification of DALYs and the identification of high-risk populations and geographic areas, we have provided a nuanced understanding of the distribution and determinants of STD burden.
Importantly, our review fills existing gaps in the literature by offering evidence-based recommendations for improving STD prevention and control efforts. By evaluating the effectiveness of current strategies and identifying areas for enhancement, we have paved the way for more targeted interventions and resource allocation.
The implications of our study extend beyond academic discourse, directly informing global health planning initiatives aimed at reducing the burden of STDs and improving sexual health outcomes. Our findings call for a multi-sectoral approach that integrates surveillance, education, screening, treatment, and community engagement to address the complex challenges of STDs.
Moving forward, policymakers, healthcare practitioners, and public health officials must heed the insights provided by our review and prioritize investment in evidence-based interventions. By working collaboratively and leveraging the evidence presented, we can make meaningful strides towards achieving the Sustainable Development Goals related to sexual and reproductive health.
Our review catalyzes action, empowering stakeholders to make informed decisions and mobilize resources effectively to combat the global burden of STDs. Together, we can create a world where all individuals have access to comprehensive sexual health services and enjoy lives free from the devastating impact of STDs.
Acknowledgments
We sincerely thank all individuals and organizations who contributed to completing this manuscript on “Global Perspectives on the Burden of Sexually Transmitted Diseases.”
First and foremost, we thank the participants who generously shared their time and insights; the search would not have only helped them. Your contributions are invaluable in advancing our understanding of the global burden of sexually transmitted diseases and informing public health initiatives.
We thank the research team members for their dedication and hard work in data collection, analysis, and manuscript preparation. Your commitment to excellence has ensured the quality and rigor of this study.
We also extend our appreciation to the reviewers for their constructive feedback and valuable suggestions, which have helped improve the clarity and impact of this manuscript.
Lastly, we thank our families, friends, and colleagues for their encouragement and understanding throughout this endeavor. Your unwavering support has been a source of strength and inspiration.
Thank you to all who have contributed to this work. Your dedication and collaboration have been essential in bringing this manuscript to fruition.
Author contributions
Conceptualization: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi.
Data curation: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi.
Formal analysis: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi.
Funding acquisition: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi.
Investigation: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi.
Methodology: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi.
Project administration: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi, Emmanuel U. Usoro, Nkechi L. Chima-Ogbuiyi, Divine B. Arrey Agbor, Chukwunnonso J. Onwuegbule, Eniola F. Afolayan, Benjamin B. Balogun.
Resources: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi, Emmanuel U. Usoro, Nkechi L. Chima-Ogbuiyi, Divine B. Arrey Agbor, Chukwunnonso J. Onwuegbule, Eniola F. Afolayan, Benjamin B. Balogun.
Software: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi, Emmanuel U. Usoro, Nkechi L. Chima-Ogbuiyi, Divine B. Arrey Agbor, Chukwunnonso J. Onwuegbule, Eniola F. Afolayan, Benjamin B. Balogun.
Supervision: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi, Emmanuel U. Usoro, Nkechi L. Chima-Ogbuiyi, Divine B. Arrey Agbor, Chukwunnonso J. Onwuegbule, Eniola F. Afolayan, Benjamin B. Balogun.
Validation: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi, Emmanuel U. Usoro, Nkechi L. Chima-Ogbuiyi, Divine B. Arrey Agbor, Chukwunnonso J. Onwuegbule, Eniola F. Afolayan, Benjamin B. Balogun.
Visualization: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi, Emmanuel U. Usoro, Nkechi L. Chima-Ogbuiyi, Divine B. Arrey Agbor, Chukwunnonso J. Onwuegbule, Eniola F. Afolayan, Benjamin B. Balogun.
Writing – original draft: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi, Emmanuel U. Usoro, Nkechi L. Chima-Ogbuiyi, Divine B. Arrey Agbor, Chukwunnonso J. Onwuegbule, Eniola F. Afolayan, Benjamin B. Balogun.
Writing – review & editing: Chukwuka Elendu, Dependable C. Amaechi, Ijeoma D. Elendu, Tochi C. Elendu, Emmanuel C. Amaechi, Emmanuel U. Usoro, Nkechi L. Chima-Ogbuiyi, Divine B. Arrey Agbor, Chukwunnonso J. Onwuegbule, Eniola F. Afolayan, Benjamin B. Balogun.
Abbreviations:
ART
antiretroviral therapy
DALYs
disability-adjusted life years
HBV
hepatitis B virus
HCV
hepatitis C virus
HIV
human immunodeficiency virus
HSV
herpes simplex virus
HPV
human papillomavirus
LGV
lymphogranuloma venereum
PID
pelvic inflammatory disease
STDs
sexually transmitted diseases
STIs
sexually transmitted infections
WHO
World Health Organization
All participants involved in any data collection, interviews, surveys, or other research activities provided informed consent voluntarily. They were fully informed about the nature of the study, its objectives, potential risks, and the use of their data. Consent forms were provided in a language understandable to the participants, and they could ask questions and withdraw from the study at any time without consequences.
The research conducted for “Global Perspectives on the Burden of Sexually Transmitted Diseases” has obtained ethical approval from relevant institutional review boards before commencement. This approval ensures the study adheres to ethical principles and guidelines governing research involving human participants.
The authors have no funding and conflicts of interest to disclose.
All data generated or analyzed during this study are included in this published article [and its supplementary information files].
How to cite this article: Elendu C, Amaechi DC, Elendu ID, Elendu TC, Amaechi EC, Usoro EU, Chima-Ogbuiyi NL, Arrey Agbor DB, Onwuegbule CJ, Afolayan EF, Balogun BB. Global perspectives on the burden of sexually transmitted diseases: A narrative review. Medicine 2024;103:20(e38199).
Contributor Information
Dependable C. Amaechi, Email: emmanuelamaechi700@gmail.com.
Ijeoma D. Elendu, Email: tochiconfidence24@gmail.com.
Tochi C. Elendu, Email: tochiconfidence24@gmail.com.
Emmanuel C. Amaechi, Email: emmanuelamaechi700@gmail.com.
Emmanuel U. Usoro, Email: emmanuelusoro2002@gmail.com.
Nkechi L. Chima-Ogbuiyi, Email: nkechi4amaka@gmail.com.
Divine B. Arrey Agbor, Email: dagbor@rumcsi.org.
Chukwunnonso J. Onwuegbule, Email: Onwuegbulenonso@gmail.com.
Eniola F. Afolayan, Email: Eniolaafolayan34@gmail.com.
Benjamin B. Balogun, Email: benjamin.balogun@gmail.com.
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9038 | https://journal.hsforum.com/index.php/HSF/article/download/2008/3049 | E522
ABSTRACT
Hyperglycemia is a metabolic derangement that fre -
quently develops after cardiovascular surgery. The periop -
erative administration of inotropic and vasoactive agents,
such as epinephrine and norepinephrine, are common in the
management of cardiac surgery patients and are known to
contribute to the development of postoperative hypergly -
cemia. We hypothesized that hemodynamic support with
epinephrine exacerbates postoperative hyperglycemia to a
greater degree than does treatment with norepinephrine.
This literature review outlines the mechanisms by which
epinephrine and norepinephrine alter glucose homeostasis,
while highlighting the significant differences in their effects
on hepatic glucose mobilization and peripheral glucose uti -
lization. This review suggests that the use of epinephrine
exacerbates postoperative hyperglycemia to a greater degree
than does norepinephrine.
INTRODUCTION
Hyperglycemia is a metabolic derangement commonly
encountered and treated in everyday medical practice and
is associated with increased rates of morbidity and mortality
in both medical and surgical settings [Gerstein 2014; Girish
2014; Ito 2014; Mulla 2014; Reddy 2014]. Disorders of glu -
cose homeostasis have downstream effects on systemic and
cardiac-specific metabolic and inflammatory states, which
have been associated with adverse clinical outcomes [Furnary
2003]. Furthermore, abnormal glucose levels have been found
to be an independent risk factor for adverse cardiovascular
events, both in patients requiring coronary revasculariza -
tion and those undergoing other cardiovascular procedures
[Bartnik 2004a, Bartnik 2004b].
After cardiac surgery, postoperative hyperglycemia is
found in up to 80% of patients. Effective glycemic con -
trol is especially important in these patients because of
the significant association between stress hyperglycemia
and adverse outcomes. In the postoperative state, hyper -
glycemia contributes to direct cellular toxicity, increased
apoptosis, immunologic modulation, vascular alteration,
and impairment in wound healing [Girish 2014].
Most cardiac surgery patients are managed with intrave -
nous inotropes in the early postoperative period. Although
there are various institutional protocols and individual
preferences, common first-line agents include epinephrine
and norepinephrine. Although these agents are known to
play a role in the development of postoperative hyperglyce -
mia, their different effects on hyperglycemia are currently
underrecognized and rarely discussed. The differences in
the specific mechanisms of each catecholamine’s action may
play an important role in how glucose is regulated in the
physiologically stressful postoperative state. We hypoth -
esized that the use of norepinephrine in the perioperative
management of cardiac surgery patients is associated with
improved glucose control when compared to epinephrine.
The purpose of this review is to outline the current evi -
dence to support that hypothesis.
MATERIALS AND METHODS
Data and information in this review were compiled by
using the NCBI’s PubMed electronic database using the fol -
lowing search queries: “epinephrine”, “norepinephrine”,
“glucose metabolism”, “cardiac surgery”, and “post-operative
hyperglycemia”. Only peer-reviewed material was considered
for inclusion in this review. Unpublished material, studies in
preparation for publication, and any otherwise unpublished
communications were not included.
NORMAL GLUCOSE HOMEOSTASIS
An intricate network of hormones and neuropeptides
tightly regulate blood glucose concentrations under physi -
ologic conditions. These compounds are released primarily
from the brain, pancreas, liver, intestine, adipose tissue, and
muscle. This complex system allows for tight regulation of
serum glucose levels under normal conditions. For example,
a decrease in plasma glucose as small as 20 mg/dL (from
90 to 70 mg/dL) will suppress pancreatic insulin release and
decrease the uptake of glucose in certain parts of the brain
[Gerich 1988]. This results in activation of the sympathetic
nervous system and leads to the secretion of counterregula -
tory hormones like glucagon, catecholamines, cortisol, and
growth hormone [Gerich 1988]. These substances will act to
increase the release of glucose into the plasma in order to
achieve a state of normoglycemia.
The Heart Surgery Forum #2018-2008
21 (6), 2018 [Epub December 2018]
doi: 10.1532/hsf.2008
The Disparate Effects of Epinephrine and Norepinephrine on Hyperglycemia
in Cardiovascular Surgery
Daniel Phadke, BS , Jared P. Beller, MD, Curt Tribble, MD
Division of Thoracic and Cardiovascular Surgery, Department of Surgery, University of Virginia, Charlottesville, Virginia, USA
Received March 23, 2018; received in revised form May 1, 2018; accepted May
5, 2018.
Correspondence: Curtis G. Tribble, MD, Department of Surgery, University
of Virginia School of Medicine, PO Box 800300, Charlottesville, VA 22908;
1-434-924-2145 (e-mail: CGT2E@virginia.edu).
Online address: The Disparate Effects of Epinephrine and Norepinephrine on Hyperglycemia in Cardiovascular Surgery—Phadke et al E523 © 2018ForumMultimediaPublishing,LLC
EPINEPHRINE
Effect of Epinephrine on Glucose Regulation
Epinephrine is a naturally occurring catecholamine
released in response to a number of physiologic stimuli.
These stimuli are generally grouped into one of 4 categories:
physical, psychological, social, and cardiovascular/metabolic
[Kvetnansky 2009]. The characterization of epinephrine and
its effects began in the nineteenth century when the first
reports of the cardiovascular effects of a then unknown adre -
nal extract were published [Oliver 1895]. Shortly thereafter,
the compound itself was isolated and its structure identified
[Abel 1897; Takamine 1901].
A relationship between epinephrine and glucose metabo -
lism has long been appreciated, but over time, the charac -
terization of the specific mechanisms of its action has devel -
oped in a stepwise fashion. Initially, a rise in blood glucose in
response to epinephrine was attributed to increased hepatic
glycogen breakdown. This early understanding of the effects
of epinephrine on glucose regulation relied on studies that
utilized supraphysiologic serum concentrations of this cat -
echolamine. This approach limited a complete understand -
ing of epinephrine’s specific mechanism of action because the
resultant hyperglycemia was mediated via both alpha- and
beta-adrenergic pathways [Antonis 1967]. Subsequent experi -
ments helped to show that, in addition to its powerful influ -
ence on hepatic glucose production, epinephrine also exerts
its influence through other routes, including attenuation of
peripheral utilization of glucose and enhanced gluconeogen -
esis [Sherwin 1984].
In addition, epinephrine was found to have signifi -
cant effects on glucose regulation via the potentiation of
responses to other stress hormones. The synergism between
epinephrine and both glucagon and cortisol contributes to
the profound systemic perturbations in glycemic control that
follow physiologic stresses such as cardiopulmonary bypass
and major surgery. Eigler et al demonstrated this principle
in a canine model in which isolated infusions of glucagon,
epinephrine, or cortisol resulted in only mild or insig -
nificant elevations in plasma glucose concentration [Eigler
1979]. However, the increase in plasma glucose produced
by the combined infusion of any 2 of these counterregula -
tory hormones was 50% to 215% greater than the sum of
the respective individual infusions. In each case, any infusion
of these other hormones, in combination with epinephrine,
produced greater increases in plasma glucose concentration
than individual infusions of each hormone or combinations
of them when not infused with epinephrine. All 3 hormones
together produced an increase in plasma glucose of 144 mg/
dL. These data suggest that changes in glucose metabolism
seen in stress hyperglycemia, a period in which several of
these counterregulatory hormones are elevated, are a conse -
quence of the synergy between epinephrine and other hor -
mones [Eigler 1979].
Epinephrine and Hepatic Glucose Production
Epinephrine exerts its influence on hepatic glucose pro -
duction through the binding of G protein–coupled adrenergic
receptors in the liver, which leads to increased cyclic adenos -
ine monophosphate (cAMP) and stimulation of protein kinase
A followed by activation of phosphorylase kinases [Rui 2014].
The initial studies on the kinetics of these interactions utilized
infusions of epinephrine and glucagon in a canine model in
which insulin levels were held constant [Dufour 2009]. These
studies led to the understanding that a certain refractoriness
to stable levels of glucose rapidly develops, whereas additional
elevations of glucose levels could be achieved through either
an additional stimulus or an alteration in the concentration of
the hormones being studied. As a result of these early stud -
ies, as well as more contemporary research, 2 conclusions
can be drawn: (1) that the steady state achieved is not due to
depletion of glycogen or substrate for gluconeogenesis, but
rather is due to an alteration in sensing, and (2) that a non-
cAMP–mediated pathway for epinephrine’s influence may
exist [Sacca 1978].
The initial increase in hepatic-derived serum glucose
is mediated first through glycogenolysis and subsequently
through increases in the rate of gluconeogenesis [Dufour
2009]. Initially, gluconeogenesis remains at a basal rate, and
epinephrine exerts its effects primarily through enhanced
breakdown of glycogen. However, after approximately
60 minutes of infusion of epinephrine, glycogenolysis slows
and gluconeogenesis rapidly increases, leading to a 4:1 ratio
in favor of production (gluconeogenesis) over breakdown
(glycogenolysis). Thus, epinephrine has a profound effect on
net glucose production by the liver, both through glycoge -
nolysis and gluconeogenesis, but the relative contributions of
these pathways vary over time.
Epinephrine’s Interaction with Insulin and Peripheral
Glucose Uptake
Insulin serves to stimulate glucose utilization and its
removal from the bloodstream, whereas epinephrine is known
to attenuate the effects of insulin. Early work in this arena
was carried out by Walaas and colleagues, who created an iso -
lated rat diaphragm preparation and were the first to demon -
strate epinephrine’s ability to block insulin-stimulated muscle
uptake in vitro [Walaas 1950]. The key observation in their
study was that, in the presence of epinephrine, glucose uptake
by muscle was inhibited, which resulted in diminished glyco -
gen deposition when compared to controls. However, these
researchers failed to directly link these effects to interactions
of epinephrine with insulin, which, it is now known, mediates
this epinephrine-induced attenuation of glucose uptake.
In addition to epinephrine’s inhibition of the peripheral
effects of insulin, these hormones interact in the liver as well.
First shown in an animal model, these findings were later cor -
roborated in humans in a euglycemic clamp model, which
demonstrated conclusively that insulin’s effects were blunted
by epinephrine [Eigler 1979; Deibert 1980]. In both of the
major sites of glucose uptake during hyperinsulinemic states,
peripheral muscle and liver, a dramatic, direct inhibitory effect
of epinephrine was observed. Even during hyperinsulinemia,
epinephrine retains its ability to stimulate systemic glucose
release via the liver. Interestingly, these interactions are all
significantly attenuated, if not reversed, by propranolol, a The Heart Surgery Forum #2018-2008 E524
nonselective beta-blocker, suggesting that these effects are
driven primarily through epinephrine’s beta-adrenergic stim -
ulation [Deibert 1980]. Epinephrine attenuates the action of
insulin, both at the site of glucose uptake peripherally and by
stimulation of glucose utilization and production in the liver.
NOREPINEPHRINE
Effect of Norepinephrine on Glucose Regulation
Norepinephrine was first discovered in 1946 by Ulf von
Euler as the neurotransmitter in sympathetic nerves and the
precursor of epinephrine [von Euler 1946]. In states of stress,
circulating levels of norepinephrine can increase up to 4 times
the basal levels [Cryer 1980]. However, even at this elevated
level, there is a negligible effect of norepinephrine on glucose
metabolism [Hjemdahl 1979]. Neural-derived norepineph -
rine plays a larger role in systemic and hepatic glucose regu -
lation when compared to circulating norepinephrine [Biggers
1989]. Early studies showed that, with splanchnic nerve stim -
ulation, neural-derived norepinephrine can increase hepatic
glucose production [Lautt 1983; Edwards 1972; Hatmann
1982]. Contemporary evidence further supports this mecha -
nism, because it has been shown that various hepatic-resident
cell lines have adrenergic innervation [Nobin 1977].
Norepinephrine and Hepatic Glucose Production
Norepinephrine in circulation, as well as norepinephrine
released under the influence of hepatic adrenergic innerva -
tion, will stimulate gluconeogenesis. In mammals, the effects
of norepinephrine on gluconeogenesis are dose dependent;
such a relationship is seen in both the liver and extrahepatic
tissues. In addition, norepinephrine increases the quantity of
gluconeogenic precursors available in the liver. Infusion of
norepinephrine causes the liver to shift from net output to net
uptake of lactate (a gluconeogenic precursor), which increases
gluconeogenesis [Cryer 1980]. Furthermore, norepinephrine
stimulates lipolysis, which increases the substrate available
for gluconeogenesis, resulting in increases in plasma glycerol
levels while mediating the peripheral clearance of lactate.
However, at lower plasma concentrations, norepinephrine’s
effect on glucose concentration is negligible [Connolly 1991].
It is important to note that norepinephrine’s effect on gluco -
neogenic precursors and hepatic gluconeogenesis only attains
kinetics sufficient for systemic alterations in glucose levels
when high, neurotransmitterlike concentrations are achieved.
These effects are independent of changes in the concentration
of insulin and glucagon. However, despite the many mecha -
nisms by which norepinephrine can affect glycemic control, the
alterations in glycogenolysis caused by norepinephrine make an
insignificant contribution to overall hepatic glucose production
[Connolly 1991]. Although norepinephrine does have stimula -
tory effects on the availability of gluconeogenic precursors and
can enhance the ability of the liver to convert these precur -
sors to glucose, measurable downstream effects only occur at
extremely high concentrations of this catecholamine.
Norepinephrine’s Interaction with Insulin and Peripheral
Glucose Uptake
In the absence of insulin, norepinephrine stimulates the
uptake of glucose into peripheral tissues. In the presence of
norepinephrine, glucose oxidation (the major mechanism of
glucose metabolism) is increased 700%, whereas lipogenesis is
inhibited by 72% under these same conditions [Ebner 1987].
In the presence of insulin, norepinephrine’s effects on
glucose uptake and utilization are enhanced when compared
to the isolated effects of norepinephrine alone. In an experi -
ment using rat-derived brown adipocytes, there was a 200%
increase in total glucose metabolism that occurred in the
presence of norepinephrine alone. A combined infusion of
norepinephrine and insulin yielded a 500% increase in total
glucose metabolism [Ebner 1987]. Marette et al additionally
showed that norepinephrine acted to potentiate insulin’s effect
on the transport of glucose as evidenced by a leftward shift in
the insulin dose-response curve while tracking the kinetics of
transport in the presence of both insulin and norepinephrine
compared to insulin alone [Marette 1989]. Thus, in stark con -
trast to the interplay between insulin and epinephrine, nor -
epinephrine potentiates the effects of insulin and enhances
the ability of peripheral tissue to take up and utilize glucose.
COMPARISON OF EPINEPHRINE AND
NOREPINEPHRINE IN HYPERGLYCEMIA
Both epinephrine and norepinephrine are important in
many phases of glucose homeostasis (see figure). Each hor -
mone plays a role in hepatic glucose mobilization as well
as gluconeogenesis. Epinephrine can significantly increase
plasma glucose levels by promoting glycogenolysis in the
Pathophysiology of catecholamine-regulated glucose modulation. Both
norepinephrine and epinephrine increase hepatic glucose output via
gluconeogenesis. However, epinephrine also increases hepatic glucose
output via glycogenolysis. In addition, they have opposite effects on
insulin-mediated glucose uptake in peripheral tissue. Green + symbol
indicates stimulation. Red X symbol indicates inhibition. The Disparate Effects of Epinephrine and Norepinephrine on Hyperglycemia in Cardiovascular Surgery—Phadke et al
E525 © 2018ForumMultimediaPublishing,LLC
liver. This effect is the main mechanism by which epinephrine
increases plasma glucose. Epinephrine can also augment glu -
coneogenesis, and importantly, becomes the dominant mech -
anism of hyperglycemia after an hour of elevated epinephrine
levels [Dufour 2009]. In contrast, norepinephrine primarily
increases plasma glucose levels by mobilizing gluconeogenic
precursors and enhancing the liver’s ability to convert these
substrates into glucose, although it does not seem to have a
significant effect on promoting the breakdown of stored gly -
cogen [Connolly 1991].
The effects of the interaction between insulin and epi -
nephrine and between insulin and norepinephrine, at the
level of the peripheral tissues, are also notably different. The
1950 study performed by Walaas et al and the 1984 studies by
Sherwin et al demonstrated epinephrine’s ability to attenuate
the peripheral uptake of glucose mediated by insulin. These
studies demonstrated that epinephrine significantly blunts
the body’s natural ability to decrease plasma glucose levels
through its utilization at peripheral sites [Walaas 1950; Eigler
1979]. Norepinephrine however, has been observed to have
a radically different interaction with insulin in the periphery.
Experiments in rat adipocytes have demonstrated instead that
norepinephrine potentiates insulin’s ability to enhance glu -
cose uptake by 250% [Ebner 1987].
In direct comparison, blood glucose changes with norepi -
nephrine are less than that of epinephrine. In canine experi -
ments, Connolly et al demonstrated that plasma glucose levels
did not significantly change with low dose norepinephrine
infusions. An infusion with higher norepinephrine levels
caused the plasma glucose level to rise from 108 g/dL to 159
mg/dL. Though this increase was significant, the authors
note that the increase in glucose levels was not as great as that
mediated by epinephrine [Connolly 1991; Stevenson 1991].
In models of chronic stress with prolonged hormone infu -
sions, epinephrine plays an important role in sustained hyper -
glycemia, although circulating norepinephrine is less signifi -
cant. In an epinephrine deficient infusion of stress hormones,
there was a marked attenuation of both hepatic gluconeogen -
esis and glycogenolysis. In contrast, an epinephrine-deficient
infusion resulted in no change on the stress hormone–induced
alterations in glucose metabolism. Even more striking, glu -
cose levels in the norepinephrine-deficient infusion were
higher than in the control infusion, including all stress hor -
mones. These findings suggest that, in a milieu of significant
hyperglycemia, norepinephrine may in fact be protective, by
limiting the hyperglycemia induced by other stress hormones
[McGuinness 1997]. These findings provide strong evidence
to suggest that norepinephrine plays a much smaller role in
increasing plasma glucose levels than epinephrine.
CONCLUSION
Postoperative hyperglycemia is associated with a vari -
ety of adverse outcomes including increased mortality,
length of stay, wound infections, and major cardiac events
[Gerstein 2014; Ito 2014]. The catecholamines used to aug -
ment systemic blood pressure following cardiac surgery can
have a significant influence on plasma glucose levels. There
are significant differences in the effects of epinephrine and
norepinephrine on hepatic glucose mobilization, peripheral
utilization, and overall plasma glucose levels. Investigation
into the mechanisms of action of each hormone and their
systemic effects provide strong evidence that epinephrine
is a much more potent mediator of hyperglycemia than is
norepinephrine. This review suggests that in patients who
require vasoactive agents following cardiac surgery, the use
of norepinephrine results in better control of postoperative
hyperglycemia as compared to epinephrine. We, therefore,
recommend that norepinephrine be used in preference to epi -
nephrine in patients with difficult-to-manage hyperglycemia
who require vasoactive support following cardiac surgery.
ACKNOWLEDGMENT
Funding: This work was supported by the National Heart,
Lung, and Blood Institute (grant T32 HL007849).
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9039 | https://centaur.reading.ac.uk/102698/1/AnticyclonicBoilingPoint_final.pdf | Pressure on the boiling point Article Accepted Version Harrison, G. ORCID: and Marlton, G. (2020) Pressure on the boiling point. Weather, 75 (4). pp. 128-129. ISSN 0043-1656 doi: 10.1002/wea.3693 Available at It is advisable to refer to the publisher’s version if you intend to cite from the work. See Guidance on citing .
To link to this article DOI: Publisher: Wiley All outputs in CentAUR are protected by Intellectual Property Rights law, including copyright law. Copyright and IPR is retained by the creators or other copyright holders. Terms and conditions for use of this material are defined in the End User Agreement . www.reading.ac.uk/centaur CentAUR Central Archive at the University of Reading Reading’s research outputs online Pressure on the boiling point Giles Harrison and Graeme Marlton Department of Meteorology, University of Reading The record pressure from the January 2020 anticyclone was readily noticed on barometers (Burt, 2020), but perhaps its most fundamental physical effect – changing the boiling point of water – might not have been so obvious. One definition of boiling point is as the temperature at which the vapour pressure from the liquid equals the pressure surrounding the liquid. Increase the atmospheric pressure, and the boiling point will therefore increase. Decrease it, such as by climbing a mountain, and the boiling point will drop. This physical response provided the basis for the pressure hypsometer, a classical explorer’s instrument used to determine altitude by measuring the boiling point of water. Could the 2020 anticyclone have produced a measurable change in the boiling point? This can be calculated. Many formulae are available for the vapour pressure and temperature in the usual meteorological range (e.g. Harrison, 2015), but the venerable equations of Goff and Gratch (1946) extend to 100 °C. Figure 1a shows the calculated vapour pressure using these equations, with the region around 100 °C of particular interest. At 100 °C, the gradient is 36.2 hPa °C-1, or about 0.028 °C hPa-1. Hence, at a sea level pressure of 1050 hPa, the increase in boiling point from that at the standard atmosphere of 1013.25 hPa would be about 1 °C, which ought to be readily measurable. Other questions arise such as the purity of the water and how the boiling point should be determined, as there are specific instruments – ebulliometers – solely intended to do this. Given the immediacy of an interesting measurement opportunity, with the anticyclone about to drift away, the available apparatus to investigate this became nothing more than a kettle and a platinum resistance thermometer, with the thermometer pushed down the spout into the liquid. Reading tap water was used, passed through a deioniser resin. The kettle was boiled repeatedly, and the maximum temperature recorded each time. Further measurements were made on different days with apparatus unchanged, as the atmospheric pressure fluctuated. Figure 1b shows the median boiling point obtained on the different days, plotted against the station pressure: a change with pressure is clearly observed. In fact, the sensitivity found using a least squares fit is (0.0288 ± 0.007)°C hPa-1, close to that calculated. The absolute difference is not so readily explained. It may be related to the boiling region sampled - long recognised as difficult to identify (Chang, 2007)- or a feature of the kettle itself, which had not been scrupulously cleaned. What might be the effect of the elevated boiling point? One obvious consideration is the energy required to heat kettles during anticyclones. This is unlikely to ever be observable because of associated weather changes which would dominate. An estimate of the possible effect on energy consumption is however provided by the “TV Pickup” national power surge which occurs as kettles are boiled at half time in football matches or breaks in soap operas. An extra 2000 MW is typically required in the UK (BBC, 2007). Heating a similar number of kettles from 20 °C to 101 °C rather than 100 °C would require about 1% more energy, so therefore 20 MW of extra power might be needed nationally. This is comparable with the output of a small biomass power station, or about 20 times less than the reduction in wind power generation during the 2015 eclipse (Harrison and Hanna, 2016). Beyond raw energy consumption, a much more socially important effect might arise if there was any detectable taste change in a temperature-critical beverage for which the water was boiled, such as tea. Acknowledgement Selena Zito helped with the measurements. References BBC, 2007. Can you have a ‘big switch off’, BBC news magazine, (viewed 31st Jan 2020) Burt, SD, 2020. London’s highest barometric pressure in over 300 years, Weather… Chang, H, 2007. When water does not boil at the boiling point, Endeavour 31 (1), 1-11 Goff, JA, and Gratch, S, 1946. Low-pressure properties of water from −160 to 212 °F, Transactions of the American Society of Heating and Ventilating Engineers (52nd annual meeting, New York), pp95–122. Harrison, RG, 2015. Meteorological measurements and instrumentation (Wiley, Chichester). Harrison, RG and Hanna, E, 2016. The solar eclipse: a natural meteorological experiment, Phil Trans Roy Soc A 374 20150225 Figure 1 Figure 1 (a) Saturation vapour pressure over water, as calculated from the fit of Goff and Gratch (1946), with 1013.25 hPa and 100 °C marked. (b) Boiling point of deionised Reading tap water measured during January 2020, using a Digitron T600 precision platinum resistance thermometer and a Druck DPI 141 vibrating drum barometer (uncertainty ±0.15 hPa). A portion of the vapour pressure curve (grey dashed line) from (a) is shown in (b), with a least-squares fit to the data (dashed black line). Repeated boiling point determinations were made on days having different pressure values, with the median value shown for each case. Uncertainty bars represent the sampling error and the instrument accuracy, after correction for instrument bias at the ice point. |
9040 | https://www.ucl.ac.uk/~ucbhdjm/courses/b242/MaintVar/calccheck.html | Checking the answer
CHECKING THE ANSWER!
We can check our answer by working backwards, and u sing selection coefficients to calculate equilibrium frequency: if we were given the above selection coefficients, the equilibrium gene frequency should be; so we expect, in this case, that the frequency of the A allele at equilibrium will be 0.76/(0.08+0.76)=0.90, and the sickle cell gene should have a frequency, at equilibrium, of 0.10.
If we did all the above calculations while minimizing rounding errors, we would actually get p=0.0908, the frequency we initially estimated above under "testing whether the data conforms to Hardy Weinberg."
Calculator note: once again, don�t round down any number if you can store it in memory at higher precision. By rounding to two decimal places, we confused ourselves into thinking that we got a different answer for p when in fact it should have been the same!
(It�s a little silly to estimate the selection coefficients from the genotype frequencies and then go on to estimate what the equilibrium should be using those same selection coefficients, as we have done here. It would be an example of circular thinking. We had to make the assumption of equilibrium in order to estimate the selection coefficients in the first place. However, it was worth doing the exercise to see that you can calculate the results in either direction). |
9041 | https://van.physics.illinois.edu/ask/listing/22902 | Difference Between Scalar and Vector Quantities | Physics Van | Illinois
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Most recent answer: 9/4/2014
Q:
what is the difference between scalre and vector quantities?
Fatima (age 16)
Pakistan
A:
A scalar quantity is a one dimensional measurement of a quantity, like temperature, or mass. A vector has more than one number associated with it. A simple example is velocity. It has a magnitude, called speed, as well as a direction, like North or Southwest or 10 degrees west of North. You can have more that two numbers associated with a vector. For example you can add a height dimension to velocity and say, for example, ' I am going uphill at a 5 degree slope in the Northeast direction'. Vectors are frequently broken down into their components along an orthogonal coordinate system, like the x and y axes. So you can say the y-component of my speed is 3 km/sec and the x-component of my speed is 4 km/sec. The magnitude, or speed is the square root of the sum of the individual components, 5 in this case. The direction with respect to the x-axis would be given by the arctangent of V y / Vx or 36.9 degrees.
LeeH
(published on 05/16/2009)
Follow-Up #1: Are complex numbers scalars or vectors?
Q:
Is a "complex number (represented by a+bi, etc.)" a scalar or vector? I read before that it is a scalar number. but it seems that a complex number (intrinsically? inherently?) contains more than information of a real scalar number (and contains vector and spacial information?) because a complex number can be expressed(sits as a single point) in the complex plane(w/ real&imaginary parts). Complex numbers appear frequently in QM ("Probably the single most comprehensive use of vector spaces is in quantum mechanics.": why do they appear? just for the convenience of expressing/calculating/representing/connecting vector, space, waves, etc. mathematically? or there is more intrinsic and deeper "physical(as opposed to pure math)" meaning in using complex numbers? i.e. Is a complex number an intrinsic property/part of physical world/nature/fundamental constituents(is it build-in inseparably in nature?)? or just a calculating tool or just a manifestation/facet of physical nature(a building constituent of the universe)? Coming back to the first question: So, is a complex number a scalar or vector in physics(as a study of scientific field) and in nature (what physical world is doing spontaneously)? or a complex number is (can be?) both a scalar and vector number in the study of physics/math and also? in nature? How can we observe/measure complex number/quantities in physics by direct/indirect experiments? any examples? Thank you.
Anonymous
A:
This is a nice follow-up.
The set of complex numbers is indeed a 2D vector space if you pay attention only to what happens when you add the numbers. However, what makes complex numbers special is their multiplication rule:
(a+bi)(c+di)=ac-bd+(ad+bc)i.
This is very different from the generic 2D vector dot product, ac+bd,which gives a number, not another vector. This complex multiplication rule makes the complex numbers form a mathematical field (), something you can then multiply vectors by to get the same type of vector you started with. In other words, it lets complex numbers be used as scalars, just as you use real numbers. In physics we routinely use them as scalars, in that they don't change under spatial rotations, unlike say standard 3-vectors or 4-vectors.
You can measure things like ac magnetic susceptibility that we represent, with good reason, as complex numbers. It's not necessary to use that representation, but it's very convenient. I think what you're probably more interested in is the complex quantum wavefunction. With some awkwardness, it could be represented without explicit complex numbers, but whatever mathematical expression was used would end up being exactly equivalent. The wavefunction isn't quite measurable, since there's always an arbitrary absolute complex phase factor. (Relative phases are defined.)
The role of vector spaces in quantum mechanics goes far deeper than the mere use of complex numbers. The quantum states themselves form a (generally infinite-dimensional) vector space. Physical measurables, like energy, are represented by linear operators on this vector space, like matrices on finite-dimensional vector spaces. The time-dependence operator is length-preserving, so it's just a rotation in the abstract vector space. Once you start thinking of quantum mechanics in terms of these state vectors, you'll get hooked.
As for the philosophical questions about intrinsic reality etc., we have little to say.
Mike W.
(published on 06/07/2013)
Follow-Up #2: what are all the numbers?
Q:
Do complex numbers include all numbers? Or there might be numbers that cannot be expressed by complex numbers (or that do not sit in complex plane)? If this is the case, why is it reasonable to assume/suspect such numbers might exist? If there cannot be such numbers beyond complex numbers(=any number can be expressed by a complex number?), why? Could you give examples (in physics and math) of numbers or quantities that cannot be a part of complex numbers if at all?
Anonymous
A:
If you start with simple counting numbers, there are good reasons to add more types. For example, you want negatives to allow you to solve equations like x+5=4. Those negatives have an intuitive meaning. The same goes for fractions, needed to solve equations like 5x=4. Continuing with this process of inventing new numbers to solve equations gets you to various irrational numbers like Sqrt(2) and then to complex numbers like Sqrt(-1). Including all the limits of sequences of these numbers gives you the full set of complex numbers. Then it turns out you're done- you have all the solutions to all your equations. It's traditional to use the word "numbers" to name all these extensions of our counting numbers.
There are plenty of other mathematical objects, however. These include, for examples, matrices and quaternions. These other objects aren't called numbers, however.
Mike W.
(published on 09/04/2014)
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9042 | https://dchealth.dc.gov/sites/default/files/dc/sites/doh/publication/attachments/Measles%20Fact%20Sheet.pdf | DC HEALTH | Center for Policy, Planning and Evaluation Division of Epidemiology — Disease Surveillance and Investigation 899 North Capitol Street NE, 6th Floor, Washington, D.C. 20002 dchealth.dc.gov | (202) 442-9371 FACT SHEET: MEASLES What is measles? Measles is a highly contagious infection caused by the measles virus. The virus enters the body through the nose and throat and then spreads everywhere, causing a rash. It can cause serious complications and death, especially in children, which makes it very important to vaccinate against measles. Can I get measles? If you have neither received the full vaccine series or had measles before, you are not immune to the measles. Every year, still cases of measles in the US, most often related to foreign tourists bringing it in and passing it on. What are the symptoms of measles? Symptoms of measles begin with a high fever, often above 104°F (40 °C) followed by cough, runny nose (coryza) and eye redness (conjunctivitis). A red blotchy rash appears about 4 days after the start of the fever beginning at the hairline, then involving the face and upper neck, and proceeding downward and outwards to hands and feet. After 5 to 6 days, the rash fades in the same order that it appears. In addition to these usual symptoms, there can be serious complications of measles such as diarrhea, otitis media, pneumonia, encephalitis, subacute sclerosing panencephalitis, and death. How does measles spread? Measles is highly contagious. About 90% of the people who have not been vaccinated and are exposed to someone with measles will become infected. Infected people can spread measles by coughing, sneezing, talking, or breathing starting about 4 days before their rash appears to 4 days afterwards. The measles virus can also live for up to two hours in the air where the infected person coughed or sneezed, even after they leave the area. If other people breathe the contaminated air or touch the infected surface, then touch their eyes, noses, or mouths, they can become infected. Spread of measles is particularly common in certain settings, such as households, daycares, and schools. It usually takes about 2 weeks after exposure to a person infected with measles before symptoms start. How soon after exposure do symptoms appear? Measles symptoms appear 7 to 14 days after contact with the virus. The average incubation period for measles is 11-12 days, and the time from exposure to rash onset is 14 days, with a range of 7-21 days. How long can an infected person carry measles? Infected people are usually contagious starting about 4 days before their rash appears to 4 days afterwards. Approximately 90% of non-immune individuals exposed to an infected person will become infected. How is measles treated? There is no specific treatment for the virus, though medications are available to decrease the severity of some symptoms. If you think you may have measles, call your health care provider before your visit to tell them about your symptoms. Should I stay home from work or school if I have measles? Yes. A person with measles should remain home from work, school, or daycare until the fourth day of the rash. Is there a vaccine for measles? Yes! The measles vaccine has been combined with the mumps and rubella vaccines into a single vaccine known as MMR[MMRII] vaccine and measles, mumps, rubella, and varicella vaccine, MMRV [ProQuad]. The MMR vaccine provides protection against all three diseases. Consult with your health care provider to determine if you should receive the vaccine. Most children should receive 2 doses of MMR vaccine through the routine childhood vaccination schedule. Some of the factors that can affect whether the vaccine is recommended include age, occupation, travel plans, and health conditions. Measles remains a common disease in many parts of the world, and some U.S. travelers have become sick with measles after traveling abroad. Before you travel, consult with a healthcare provider to make sure you and your family are protected. Is it true that the vaccine can cause autism? No. Current medical research demonstrates that measles vaccination is safe and effective. Should a person with measles be excluded from work or school? A person diagnosed with measles should remain home for 5 days after the rash first appeared. Want more information? Information about measles and other related health topics can be found at the website cdc.gov. The DC Department of Health promotes the health and safety of the District residents. For additional information, please visit dchealth.dc.gov or call (844) 493-2652 October 2023 |
9043 | https://www.sciencedirect.com/science/article/abs/pii/S0025556415001182 | Numerical computation of an Evans function for travelling waves - ScienceDirect
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Mathematical Biosciences
Volume 266, August 2015, Pages 36-51
Numerical computation of an Evans function for travelling waves
Author links open overlay panel K.Harley a, P.van Heijster a, R.Marangell b, G.J.Pettet a, M.Wechselberger b
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Highlights
•A geometrically inspired technique for computing Evans functions.
•Evans functions are computed for scalar and systems of PDEs.
•Numerically examine stability of travelling waves in a chemotaxis model.
•Includes a new proof of stability of travelling waves in the F-KPP equation.
Abstract
We demonstrate a geometrically inspired technique for computing Evans functions for the linearised operators about travelling waves. Using the examples of the F-KPP equation and a Keller–Segel model of bacterial chemotaxis, we produce an Evans function which is computable through several orders of magnitude in the spectral parameter and show how such a function can naturally be extended into the continuous spectrum. In both examples, we use this function to numerically verify the absence of eigenvalues in a large region of the right half of the spectral plane. We also include a new proof of spectral stability in the appropriate weighted space of travelling waves of speed c≥2 δ in the F-KPP equation.
Introduction
The main focus of this paper is a geometrically inspired technique for numerically analysing the spectral stability of a travelling wave. In particular, we illustrate a new method for computing an Evans function for a linearised operator, linearised about a travelling wave solution to a partial differential equation (PDE) in 1+1 independent variables. Evans functions first arose in the 1970s and are typically constructed using a matching condition (see, for example, , , ). They are analytic functions in some relevant region of the complex λ-spectral plane, with the property that the multiplicity of their roots coincide with the multiplicity of λ as an eigenvalue. A well-known obstacle in the numerical computation of the Evans function is the tendency of the associated eigenvalue ordinary differential equations (ODEs) to become stiff, an issue that can sometimes be overcome by working in the exterior product space , , , . Building on this, there have been two complementary directions to the study of (efficient) numerical computation of the Evans function: continuous orthogonalisation (see , , and the references therein for details) and Grassmannian spectral shooting, or the Riccati approach , , .
It is this second point of view that we follow here. By following the Riccati approach, and using the geometric structure of the problem, we are led to an Evans function whose computation is efficiently tractable even through relatively large changes in the order of the spectral parameter. The geometric interpretation of the Riccati equation allows us to avoid the problem of finite time blowup of the relevant solutions to nonlinear ODEs. Furthermore, this approach is readily extendible to values of the spectral parameter which lie in the continuous spectrum (also called the essential spectrum). In extending the Evans function into the continuous spectrum, we highlight another underlying relationship between instabilities of a travelling wave and the geometry of the spectral problem.
Our approach shows the power of topological/dynamical systems techniques to analyse analytically and numerically the (spectral) stability of travelling waves that are pervasive in the mathematical biological literature. In this manuscript we illustrate this technique on two well-known models, both considered on an unbounded domain. The first is the Fisher/Kolmogorov–Petrovsky–Piscounov (F-KPP) equation u t=δ u x x+u(1−u),and the second is a Keller–Segel (K-S) model of bacterial chemotaxis u t=ɛ u x x−α w,w t=δ w x x−β(w u x u)x.
The F-KPP equation was chosen because in this case, the travelling wave stability analysis becomes analytically tractable (see Section 2.7). Further, the literature on stability of travelling waves in Eq. (1) is vast (see for example , , , and the references therein for a partial list of the stability results). It thus provides a well-known backdrop against which to verify our spectral calculations. The K-S model in Eq. (2) was chosen in order to highlight how to extend our methods beyond scalar PDEs. It is also convenient because explicit solutions to Eq. (2) can be found when ɛ=0 . We are thus able to omit a time-consuming step (numerically finding the solutions to the travelling wave ODEs) and focus on setting up and analysing the linearised spectral problem. Explicit solutions are not necessary for our methods to work (as the F-KPP example shows) and we discuss how the stability analysis of the travelling waves can be adapted to the K-S problem when ε ≠ 0 in Section 4.2.
The F-KPP equation was first introduced by Luther in 1906 who originally used it to model and study travelling waves in chemical reactions . It was named for Fisher, and for Kolmogorov, Petrovsky and Piscounov, who independently wrote seminal papers on the equation, using it to model the spread of a gene through a population , . In this manuscript, we assume the diffusion coefficient δ is strictly positive.
Eq. (2) was proposed in the 1970’s by Keller and Segel , to describe chemotactically driven cell migration in which a population of bacteria exhibits an advective flux in response to a gradient of a diffusible secondary species (i.e. nutrient); see , and references therein for a current overview of PDE models with chemotaxis. In Eq. (2), the bacteria population density is denoted by w(x, t) and the nutrient concentration by u(x, t). The model exhibits so-called logarithmic sensitivity and we assume a constant consumption rate function. The diffusion of the nutrient is assumed to be much smaller than the diffusion of the bacteria population: 0 ≤ ε ≪ δ. Finally, α> 0 models the rate at which nutrients are consumed, while β> 0 measures the strength of the chemotaxis term. We also assume 0<δ<β.
By a travelling wave solution of Eqs. (1) or (2), we mean a solution to Eq. (1) of the form u(x−c t), or a pair of solutions to Eq. (2) of the form (u(x−c t),w(x−c t)) travelling from left to right with some positive (constant) wave propagation speed c.
To study travelling wave solutions, we introduce a moving coordinate frame; setting z:=x−c t, and τ ≔ t, Eq. (1) becomes u τ=δ u z z+c u z+u(1−u),while Eq. (2) becomes u τ=ɛ u z z−α w+c u z,w τ=δ w z z−β(w u z u)z+c w z.
Travelling waves u=u^(z), or (u,w)=(u^(z),w^(z)) will then satisfy the ODEs: δ u z z+c u z+u(1−u)=0,or ɛ u z z−α w+c u z=0,δ w z z−β(w u z u)z+c w z=0.
Once travelling waves to Eqs. (1) and (2) have been found, we are next concerned with their stability. In particular, we are interested in the spectral stability of the travelling waves. A full analysis of stability of travelling waves in the F-KPP and K-S equations is well beyond the scope of this manuscript. However, in the F-KPP equation, it is known that travelling waves of speed c≥2 δ are spectrally and linearly stable relative to certain perturbations (or in certain weighted spaces) and that the travelling waves of speed 0<c<2 δ are unstable to all perturbations. One can find these results in a variety of sources relating to travelling waves, see for example , , , and the references therein. For completeness, we include a proof in Section 2. These results imply that there will not be any eigenvalues (with eigenfunctions in an appropriate space) with positive real part of the linear operator found by linearising Eq. (3) about the travelling wave solution. This is numerically confirmed by our calculations.
Stability theory for travelling waves in chemotactic models is of course newer, and the full stability analysis does not appear to be known, but partial stability results for the K-S model we are considering in Eq. (2), can be found in, for example, , , and the review paper by Wang . In Section 3 we focus on the model when ɛ=0. For the explicit travelling waves in this model, it is known that the continuous spectrum has a nonzero intersection with the right half plane, and that such spectrum cannot be entirely weighted away . This suggests the presence of so-called absolute spectrum in the right half plane. We numerically confirm this. We do not speculate on the effect that the absolute spectrum with positive real part may have on the dynamics (either long term or otherwise). It is also known that, for the travelling waves we consider in Section 3, the linearised operator does not have any real positive eigenvalues . There does not appear to be any proof in the literature of the absence of eigenvalues in the right half plane with nonzero imaginary part. We verify this fact for a large domain in the right half of the complex plane. We further show that λ=0 is an eigenvalue of the linearised system, with multiplicity two.
We illustrate our methods first for travelling waves in the F-KPP equation. In Section 2, we set up the spectral problem, and find the continuous spectrum. We then define the point spectrum and Evans functions as a means to find it. We then introduce the Riccati equation and use it to construct a new, well-behaved function whose roots correspond to eigenvalues (a defining feature of an Evans function). We extend this function into the continuous spectrum in the natural way and show that its roots still correspond to eigenvalues of the linear operator. We conclude Section 2 with what is, to the best of our knowledge, a new proof of the lack of eigenvalues with positive real part of the linearised operator about the travelling wave in the F-KPP equation. In Section 3, we follow the same recipe and show how to apply the ideas of Section 2 to travelling waves in systems of PDEs with more than one dependent variable, using as our example, a K-S model of bacterial chemotaxis Eq. (2) when ɛ=0. We set up the associated spectral problem, and compute the continuous spectrum, define eigenvalues, and the Evans function. We then compute the Riccati equation and accompanying machinery for this example and use it to define a new, well-behaved Evans function which can be readily extended into the continuous spectrum. We then use this function to numerically verify the absence of eigenvalues with positive real part in a large domain of the complex spectral plane. We conclude the section by numerically establishing that λ=0 is an eigenvalue of the linearised operator of multiplicity two. In Section 4, we summarise our results and provide concluding remarks.
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Section snippets
Travelling waves in the F-KPP equation
We use a dynamical systems approach to analyse the travelling wave problem of the F-KPP equation, i.e.we write Eq. (5) as a system of first-order equations: d u d z=v,d v d z=1 δ(−c v−u(1−u)).There are two equilibria of Eq. (7) in the uv-plane: one at (0, 0) and one at (1, 0). The Jacobian of (7) is D f(u,v)=(0 1−1+2 u δ−c δ).At the point (1, 0), the eigenvalues of Df(u, v) are μ 1 u=−c+c 2+4 δ 2 δ,μ 1 s=−c−c 2+4 δ 2 δ.For all values of c there is one positive eigenvalue and one negative eigenvalue. Thus, (1, 0) is a
Travelling waves in a Keller–Segel model
We now turn our attention to the application of the techniques from Section 2 to a system of PDEs with one spatial and one temporal independent variable, and more than one dependent variable. We focus on the parameter regime of Eq. (2) wherein explicit solutions can be found to the travelling wave equation Eq. (6), and so for the remainder of this section, we set ɛ=0 for unless otherwise specified.
Setting ɛ=0, Eq. (4) becomes u τ=c u z−α w,w τ=δ w z z−β(w u z u)z+c w z.
As before, a travelling wave solution
Summary of results and ocncluding remarks
We have illustrated how to use the underlying geometry of the spectral problem in order to facilitate computation of the spectrum of a linearised operator about a travelling wave in a PDE with 1+1 independent variables. The geometric interpretation of the Riccati equations allows us to handle the blow-up of solutions to nonlinear ODEs. We have thus used these solutions to develop new Evans functions, and used them to numerically verify the spectral stability of travelling waves in the F-KPP
Acknowledgements
R.M., G.J.P. and M.W. gratefully acknowledge the partial support of Australian Research Council grant ARC DP110102775. P.vH. gratefully acknowledges support under the Australian Research Council’s Discovery Early Career Researcher Award funding scheme DE140100741. K.H. also gratefully acknowledges support from an Australian Mathematical Society Lift-off Fellowship.
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Cited by (14)
Influences of Allee effects in the spreading of malignant tumours
2016, Journal of Theoretical Biology Citation Excerpt :
One method of inferring stability results for models such as (1) is based on an Evans function computation. Such a method is currently under development; see Harley et al. (2015). A related aspect that is not discussed in this paper is the transient dynamics of the travelling wave solutions. Show abstract A recent study by Korolev et al. [Nat. Rev. Cancer, 14:371–379, 2014] evidences that the Allee effect—in its strong form, the requirement of a minimum density for cell growth—is important in the spreading of cancerous tumours. We present one of the first mathematical models of tumour invasion that incorporates the Allee effect. Based on analysis of the existence of travelling wave solutions to this model, we argue that it is an improvement on previous models of its kind. We show that, with the strong Allee effect, the model admits biologically relevant travelling wave solutions, with well-defined edges. Furthermore, we uncover an experimentally observed biphasic relationship between the invasion speed of the tumour and the background extracellular matrix density.
### Traveling Wave Solutions in a Model for Tumor Invasion with the Acid-Mediation Hypothesis
2022, Journal of Dynamics and Differential Equations
### Travelling wave solutions in a negative nonlinear diffusion–reaction model
2020, Journal of Mathematical Biology
### (IN)STABILITY of TRAVELLING WAVES in A MODEL of HAPTOTAXIS
2020, SIAM Journal on Applied Mathematics
### Revisiting the Fisher-Kolmogorov-Petrovsky-Piskunov equation to interpret the spreading-extinction dichotomy
2019, Proceedings of the Royal Society A Mathematical Physical and Engineering Sciences
### Absolute instabilities of travelling wave solutions in a Keller-Segel model
2017, Nonlinearity
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qué tal amigos cómo estaban en esta oportunidad vamos a explicar un poco sobre los cortes en vivos ya que últimamente me han estado mencionando cómo es que se va a generar estos cortes en qué intervalos se va a tomar en cuenta si lo voy a trabajar de izquierda a derecha derecha izquierda etcétera infinidad de dudas que puedan tener y para eso nosotros vamos a despejar los o dieta estaba observando digamos una viga simplemente apoyada con diversas cargas no hay cargas puntuales carrera distribuida rectangular el carrete hubiera angular puede ser que haya muchos más cuando ya lo vamos a ver más adelante y aparte estaba observando estas flechas donde me menciona x 1 x 2 esto se refiere a los cortes que ellos firman en cuenta entonces vamos a verlo de este modo nos tomó voy a copiar lo mismo mejor dicho la misma viga lo que usualmente vemos con sus distancias establecidas con sus cargas también indicadas las letras para identificar que le damos vamos a tomar en cuenta y nada creo que este paso ya ustedes lo deben tomar en cuenta debe saber calcular también correctamente su reacción porque si las reacciones están mal aplicar cortes no sería tan conveniente entre todo modo estamos asumiendo que ya ustedes ya lo han calculado estamos acá solamente para explicar cómo es que tú tienes que aplicar los cortes cuáles son los criterios que debe de utilizar y cuál sería quizás la forma más rápida un poco más lento etcétera digamos que ya tengo que alguna la reacción a ya no estoy inventando digamos que nos da 20 toneladas y la reacción efe jr porque uno puede fijó también otra reacción acción horizontal en este caso no va a ser tan necesario y además de cero porque no hay fuerzas horizontales en cambio del apoyo móvil tiene una reacción no vamos a asumir no sé que sea 30 toneladas estamos inventando recuerden no es que ésta sea la reacción y ahora vamos a aplicar los cortes nos piden que ayudarles a los diagramas o por otros métodos que ustedes también empleando de integración día con jugada etcétera todo para todos métodos en realidad es necesario a veces las ecuaciones de fuerza cortante momento vector inforsa axial lámina ya sin embargo porque se dicta la vamos a tomar tanto en cuenta porque no hay formas horizontales y además en para este tipo de vigas no va a ser tan necesario solamente vamos a generar funciones como actuaciones de corte y ecuaciones de momento por ejemplo quiero aplicar cortes en todo este tramo de la vida por eso ya le había indicado el primer corte sería justamente acaba desde el punto a hasta la carga puntual de esa otra duda que también me dicen cómo se quiere cortes hay en realidad los cortes se aplican cuando va a existir una variación de cargas o sea va a cambiar mejor dicho los cortes de acuerdo serán por ejemplo impiden en puntual y sabemos que hay una fuerza cortante en una reacción vertical sigo avanzando no hay nada de cargas siga avanzando sigo avanzando hasta que me topo con una carga una vez que existe esta carga puntual necesariamente ahí yo tendría que aplicar otro corte porque el primer corte recuerda este de acá que es x1 es donde tú estás comenzando con la carga de 20 entonces ahí se aplican un corte ya cuando me encontré con esta carga puntual quiere decir que ahí nomás va a haber otro porque yo sigo avanzando y sigo avanzando seguimos avanzando hasta que llegó al punto b y me encuentro con una carrera distinguida rectangular quiere decir que ahí se ha aplicar otro corte y miren lo que indica x 1 x 2 x 3 cuando corta la carga distribuida seguimos avanzando y justamente en el punto sé que pasa ya no hay más carga ción existe una variación mejor dicho el tramo se mueve quizás diferente entonces ahí también va a ver un corte y repente ustedes me preguntan pero al comienzo si hay una carga y acá hay un tramo donde no hay nada no debería haber dos cortes no pus si recibe estamos comenzando cambio acá no está culminó la carga distribuida y ya no haber más cargas uno todo caso existe una variación ahí también se aplicaría otro corte y recuerda también que nosotros aplicamos cortes de acuerdos de miedos intervalos de cada drama lo que queremos también quizás hallar quizás su esfuerzo importante momento el sector en cada punto por eso también es necesario adecuar vinos intervalo sigo avanzando justo me encuentro con una carga distribuida pero antes y antes de seguir avanzando estudio ahorita mencionando los cortes que se va a ir aplicando en cada punto todavía no estoy definiendo si voy a trabajar de izquierda a derecha derecha izquierda tal como se está viendo acá nuestras flechas así que por ahora estamos viendo dónde se puede aplicar el corte si vamos avanzando ya empiezo con el tramo de otro tramo y hay una carga triangular quiere decir que ahí va a haber otro corte más porque hay una variación sigo avanzando llegó el punto y además car que decir que ahí culmina porque también va a cerrar con el último apoyo en conclusión cuántos cortes estamos considerando sería 1 2 3 4 5 6 6 cortes van a ser necesarios para poder ya generar nuestras funciones ahora hay que saber desde qué punto o de qué lado vamos a trabajar por ejemplo para el corte número 1 tal como estamos viendo acá en la flecha nuestra parte de acá el primer corte voy a colocarle con número mejor para de vista en cuenta corte 1 2 corte 34 5 y 6 para el corte uno es más práctico y más sencillo de trabajar de izquierda a derecha para que los cálculos sean poco opción porque tú vas a hacerlo de este modo vas a colocar aquí corte 1 intervalo se aplica recuerda que quiso hacer la distancia x tiene que ser menor o igual como también y perdón x debe ser mayor o igual como también menor igual de acuerdo al intervalo por ejemplo vamos a empezar en el punto a porque trabajan de izquierda derecha sería desde cero porque ahí comienza es el punto de coordenadas 0 0 aquí y va a culminar dónde va a empezar el segundo corte digamos así desde a hasta la carga puntual cuánto de longitud dos metros entonces el intervalo va a ser preservados dibujo esa parte no de la viga el apoyo fijo con en tramos y acá justamente tuvo de hacer una proyección ahí tú estás cortando el momento de cortar se generan tres fuerzas internas una va a ser la fuerza normal en fuerza social la fuerza cortante que es una fuerza vertical denominada con la letra v y un momento que va a ser no en este caso de momentos de torre de cocción de momento porque tiene estos sentidos porque es así siempre va a ser así en realidad es que es una convención ya está establecida cuando tú haces un corte de izquierda a derecha tomando en cuenta esta flecha así x llevamos x1 la cortante tiene esta configuración va hacia abajo la norma de va hacia la derecha y el momento tiene un sentido antihorario es la forma como tú tienes que ir evaluando las formas internas al momento de hacer un corte que se va a colocar también no nos olvidemos su reacción bueno tiene una reacción acá de 20 toneladas la reacción horizontal y generó mis ecuaciones de momento hay unas formas quizás que yo pueda desarrollar un poco más rápido o al ojo digámoslo así haciendo este módulo el primero de la forma tradicional quiero a guiar la actuación de la fuerza normal dice sumatoria de fuerzas en x digamos que hace la derecha sean positivos igual a cero cuando con todas las fuerzas horizontales no hay reacción horizontal solamente la fuerza en no hay más fuerzas entonces decimos que n vale cero es nuestra primera ecuación la ecuación no se está dando un valor dice ahora quiero generar mi ecuación del aportante decimos sumatoria de fuerzas en y hacia arriba positivo igual hacer lo mismo 20 - v no hay más fuerza vertical en lo importante es igual a cero en conclusión la cortante vale 20 s su ecuación y finalmente la última ecuación que es sumatoria de momentos con respecto al corte vamos a decir así digamos que en los sentidos que sean de forma anti horaria sea positivo igual a ser 20 va hacia arriba me genera un momento horario porque estoy multiplicando esta distancia x 1 no es fuerza por distancia recuerden ninguna fuerza perpendicular imagen en el momento horario como es horario va a ser negativo menos 20 por equis y el momento que está cuando acá el sentido entidad más m igual a se despejó m me va a dar un valor de 20 x 20 x positivos y están esas son las las ecuaciones de normal cortante y momento y ya tú puedes reemplazar con el intervalo cuando x 0 cuando vale la cortante cuándo x2 cuánto vale así con todas estas ecuaciones pero qué pasa aquí una forma más directa de separando hacer todo esto de la siguiente manera digamos que quiera ya mí normal digamos que si hubiese habido ya que sea no sé 80 yo coloco así n sin la fuerza normal tiene el mismo sentido que el corte va a ser negativo y mira lo comprobamos acá se nos dice 80 más en e igual a 0 despeja selene vale menos 80 forma rápida 180 veces y va hacia la izquierda o apuntaba en ese sentido cuando tienen sentidos opuestos al corte quiere decir que positivo no ahí sí creo que ya todos se dieron cuenta es el mismo criterio para la cortante mira 20 va hacia arriba y la corta demasiado tienen sentidos opuestos entonces sería positivo puede sacar audios positivos y para el momento también el momento me genera un momento despertado una fuerza de 20 me genera un momento horario y acá el corte santidad y sentidos opuestos serían positivos 20 política es una forma quizás nadie le damos un consejo de giza le doy un tip para que ustedes también lo aplique eso fue entonces el corte número uno sencillo opción el corte 2 también tiene que saber si trabajar de izquierda a derecha derecha izquierda obviamente que va a seguir siendo de izquierda derecha o dígame el corte 1 hubiera sido mejor aplicar de derecha a izquierda sí entonces todo esto mira vamos a tener que dibujar todos estos damos todas las cargas y finalmente calcular sus ecuaciones en realidad la ecuación es distinta pero el valor va a ser lo mismo no me refiero como es que cuando digo el momento cuando x balde 220 20 x 240 y cuando yo calculo en mi ecuación de momento de derecha izquierda o sea de este modo dibujo no sé toda la carrera angular la carga rectangular y acá hay una carga puntual en la apoyo móvil 30 y acaba justamente hacia mi corte cuando es un corte de derecha así hay derecha izquierda esto es al revés no de lo que era anteriormente ahora la normal va hacia la izquierda la cortante va hacia arriba y el momento tiene un sentido oral es todo lo opuesto vas a tener que calcular toda la fuerza resulta da antes voy a tener que aplicar sumatoria fuerzas verticales y en las finales la ecuación quizás va a ser distinta pero los valores van a hacer lo mismo pasa reemplazado ahora este último intervalo como base el intervalo y va a ser donde culminó digamos el corte o quizás la longitud en este corte es el corte 1 pero de derecha a izquierda no como está la flecha azul va a ser desde esta longitud todo esto cuánto es eso 8 9 10 11 12 14 15 y 16 el intervalo va a ser de 16 hasta el último tramo que viene a ser dos metros más 2018 de 16 a 18 y en ese intervalo cuando x el 16 tiene que ser igual que cuando x es 21 cuando era de izquierda derecha en este punto entonces ya se van dando cuenta no creo que sí me entiendo en la mayoría acá cuando digo x2 el momento me da 40 en cambio acá cuando x sea 18 porque el pecho se ubica acá y ahí cuando el de izquierda-derecha que de todos cambio que kaká que tiene que dar también el valor de 40 es lo que trato de explicar entonces que más conveniente por lógica viéndolo desde la forma más práctica de izquierda a derecha más rápido y hacemos lo mismo para los siguientes cortos ahora vamos a evaluar el corte número 2 no sé explicarlo a detalle como siempre colocamos cosidos el intervalo va a ser donde culminó el anterior corte o sea el corte 1 en otras palabras en dos porque usted es la carga puntual a 2 metros hasta digamos antes que empiece el corte 3 o sea el corte 2 culmina justo en el punto b porque ahí inicia una carrera tribu ida sería de 2 suma 2 2 para que llegar el punto b 4 de 2 a 4 es el intervalo dibujamos el tramo no apoyo fijo con una carga puntual digamos de 3 toneladas y nada nos seguimos colocando las reacciones a este el 20 este es mi x y acá también se generan los refuerzos cortante momento y normal hacen lo mismo en 1 criterio cortantes ya sé cómo es 20 va hacia arriba en la corte antes hacia abajo opuesto sería positivo 20 y s en el mismo sentido miren y no sería negativo la final sería 17 la ecuación de la normal para todo va a hacer será en realidad porque no hay forma es horizontal de recuerdo y el momento como sería en este caso 20 x x me genera un momento horario con respecto del corte tiene sentidos opuestos va a ser positivo 20 x y este 3 cuál sería la distancia va a ser desde este punto hasta el corte bueno recuerda que ya tienes esta distancia de acá esto mide 2 metros entonces si todo este es x y hasta acá el 2 cuando sería esta otra distancia simplemente la resta x menos dos va hacia abajo y me genera un momento antiedad tiene el mismo sentido el corte entonces va a ser negativo menos 3 por x menos 2 y eso es todo ahí está la ecuación de la normal la cortante del momento y hacemos lo mismo con los diferentes gramos ya se ve un poco más complicado cuando empecemos en el corte lección este es el corte 3 sería conveniente de izquierda a derecha derecha izquierda realidad creo que ya se podría decir que tienen ya la misma casi longitudes de todas manera vamos a aplicar de izquierda a derecha para que vean cómo es el trabajo colocamos acá corte 3 el intervalo va a ser desde donde culminó el anterior corte o sea en 4 en esta parte de acá en el punto b hasta el punto cero porque hay justamente hay una variación y comiso de corte 4 sería de 4 hasta sumando 3 7 de 4 a 7 es el intervalo y volvemos a dibujar es la mueve el apoyo fijo la carga puntual y acá está la carga distribuida vamos a hacerlo se encuentra por si acaso esta carga distribuida no es ahí no culmina o sea no estoy colocando de los tres medios estoy indicando nada más que hasta aquí sería el corte que están cortando en la carretera irnos cortante hacia abajo momento antihorario y normal hacia la derecha cuanto vale estar acá redistribuida tiene una magnitud de 10 toneladas sobre metros ahí está bueno esto es 20 segundo mismo ecuación n0 la cortante 20 hacia arriba esto va a seguir sin igual porque casi se repite las primeros las primeras cargas menos 3 y acaba ahora tengo que hallar la fuerza resultante de esa carga excluida si yo sé que hasta acá mide 2 metros este test voy a aplicar luego momentos y entonces va a seguir siendo en otras palabras x menos dos opciones esto va a seguir siendo determinador recuerdo porque hasta acá está culminando su distancia en que menos 2 y ahora acá necesitó hallar esta distancia como lo calculamos si yo sé que estoy acá también mide 2 metros miren esta distancia también mide 2 como hasta acá entonces cuando sería esta distancia por lógica que esté aquí sería todo x menos 4 donde toda esa instancia x 4 justamente es esa distancia de la carga rectangular y bueno o hallaría su fuerza resultante base por altura ante la base x menos 4 cuánto es la altura listo está su fuerza resultante y recuerda que eso se ubica en su centro de gravedad vamos a indicarlo justamente ahorita vamos a bajar esto y acá voy a colocar de otro color que se ubica en la mitad x menos 4 entre 2 así vamos a indicar x menos 4 entre 2 ya que muchos a veces tienen ese problema de identificar cuál sería la distancia cuando estamos implicando de manera detallada momento esto va a seguir siendo lo mismo no recuerden 20 x menos 3 por x menos 2 y finalmente también va hacia abajo sería anti dario tiene mismo sentido negativo menos x menos 4 por 10 que es la fuerza resultante multiplica vamos a colocarlo más hacia la izquierda multiplica a éste la distancia sino hasta el corte cuánto es x 4 entre 2 y eso es todo ahí está esto puede simplificar lo que es lo que sí a ver de esta parte podría ser menos 10 x x 4 al cuadrado entre 2 mitad 1 mitad 5 y cuando ya esto depende de cada uno como lo quiera dejar expresado pero hasta acá ya sabemos que es nuestras 3 ecuaciones esto que no está siendo tan complicado pero bueno quiero explicar su acción de esa manera para que todos también puedan aprender el corte cuando ella se vuelve el sanma mecánico podría ser de izquierda a derecha o derecha izquierda vamos a trabajar en tal como acá lo hemos representado y ese sería haya corte 6 bueno cuando tú vas a tomar en cuenta ya de derecha a izquierda y te recomiendo que ese corte sea comenzando desde ese extremo o sea así este que sea el 5 y el 6 va a ser justamente este corte que va a culminar acá en este conducción tal como sucede con ley y 4 no creo que todavía no estamos llevando el corte 4 saben que hemos hecho hasta ayer es ya faltan 4 todavía de 4 que vamos a trabajar entonces como está acá de izquierda a derecha indicamos corte 4x donde va a ser donde culmina el anterior corte o sea en 7 hasta nuestra parte de acá o sea hasta que es 7 tengo que sumar de 30 la que llega al punto de va a ser 10 de 7 a 10 es el intervalo dibujamos nuevamente todos estamos el apoyo fijo la carga puntual la carga distribuida y acá bueno haber hecho demasiado extenso va a ser hasta acá no nos hasta y colocó los cortes o la forma es interna donde pasa ese punto cortante momento y normal copiamos todo esto de aquí esto es x esto sigue siendo 2 esta distancia también va a ser en la cara puntual x 2 acá si veo que se va a considerar la misma distancia miren por ejemplo este data del 2 estaba sé cuándo sería esto ya lo tenemos si no me equivoco estrés a ver la carretera 3 si te das cuenta cuando el corte no pasa por esa carga se van a tener su misma distancia pues no nos queda de colo que es x menos algo recién acá podría ser esta distancia x menos 224 más resiente pero no va a ser necesaria porque no hay una carga y pueda y que quizás vayamos a utilizar entonces está dando y problema eso es todos los datos que tienen que tomar en cuenta colocamos normal sigue siendo cero cortante en este caso con esto de aquí sigue siendo 20 a ver este estrés está en 10 toneladas sobre metro saludamos su fuerza resultante base ponerte una la base de tres factura 1010 por 330 nada más eso estamos indicando no ha pasado casi nada nuevo y bueno cortante el mismo criterio esto va a seguir siendo lo mismo no de los cortes anteriores 20 positivos menos 3 acabas va hacia abajo sería menos 30 y nada más finalmente cuando sería 20 menos 30 menos 10 menos de menos 13 entonces copiamos el valor que tenemos menos 13 ahí está con el abordantes muy sencillo y ecuación del momento también esto se repite 20 x menos 3 x x menos 2 no esta distancia y lo tendremos y finalmente este sería menos 30 y ahora si viene ese punto clave si sería la distancia desde acá hasta el corte cuánto sería estadistas si yo sé que 30 se ubica en la mitad de 31 1.5 y ahora también 1.5 entonces en conclusión va a ser desde este punto hasta acá cuánto es sí hasta acá está midiendo 224 más 1.5 5.5 y me falta esta distancia de acá si todo mide x y hasta acá nomás es 5.5 esto sería x 5.5 así es como se evalúa entonces la distancia cuando no se aplica momento no hay otra forma porque todo esto está representado por un corte menos 30 por x menos 5.5 y eso sería entonces con respecto al corte 4 y ya podemos comenzar con el corte 5 que justamente está acá en la parte derecha no cuando está el apoyo móvil esa parte y eso es en sí cuando yo evaluada este corte corte 5 de derecha a izquierda esto se va a mantener así que veo que cambian o se le colocan quizás en el otro sentido pero mantengo así obviado de esta perspectiva el corte 5 va a ser df hasta ahí pues no importa 5 este va a ser el corte 5 y este es el corte sexy la corte 5 sería df hasta ahí porque hay culmina de su corte y además comenzamos de cero a cuatro metros que es lo que yo estoy representando de 0 a 4 porque aparentemente puedes ir a esto es de izquierda derecha no pues ya de un comienzo esté mencionando que va a ser de ese modo ese tramo es corto no es más conveniente pero dígame el corte de cinco lo vas a aplicar digamos de izquierda a derecha lo que sería generar más extensión y perder mucho tiempo entonces para que sea más práctico de derecha a izquierda del último el penúltimo corte copio está el apoyo móvil un tramo colocó las fuerzas internas que se generan ahí la cortante recuerden es todo lo contrario hacia arriba momento horario y la normal hacia la izquierda lo indicamos acá x esta reacción creo que es 30 le hemos denominado signos 30 normal [Música] cortante tienen el mismo sentido mismo negativo momento multiplica toda esta distancia que es x me genera un momento en sentido antihorario entonces acá es opuesto a la ecuación el momento siendo al corto como son opuestos va a ser positivo 30 por equis nada más ahí está el corte 5 ahora el último corte que ya viene a ser el corte 6 ahí sí va a venir un poco más de complejidad el intervalo eso ya lo sabemos sería desde el punto b o sea en 4 con un término de la anterior corte hasta el punto de donde ese punto de recuerde que es el corte 4 si no me equivoco donde culminó la distancia ahí va va a tener que ser igual como digo y con respecto al momento en cortantes quizás también en ciertos casos si es que hay una carga de repente en este punto ahí sí variable o de repente también un momento si es ahí todo son cosas que ella no irán viendo con la práctica y con él ejercicios que también vayan realizar ya el corte se está cortando mira esta carga de angular puede trabajarlo de izquierda a derecha algunos lo hacen así porque quieren evitar encontrarse con una carga de trapecio edad a qué me refiero que yo hubiese hecho el corte de izquierda a derecha como están viendo acá así con la flecha roja cuando hacías ese corte este tramo donde está la carreta angular cuando cortas va a seguir siendo un triángulo no un triángulo que hay a esta altura aplicando semejanza de alguno o relaciones de anv como válvulas de esa altura mira con lo que la caben angular normal que está ahí este es el corte digamos este es x digamos su altura h la altura total ya lo tengo y la base total de la longitud total también más de 4 metros que hacíamos acá bueno altura mayor de triángulo es a su base mayor 4 y esto es igual a existe en ángulo pequeño que viene a ser su altura de ese triángulo h es a su base que es x todo este sello para ellos aplicó de corte de texas del h 6 hora y 4 me da un valor de de medición tres medios o 1.5 1.5 y el x pasa a multiplicar s la altura que estoy haciendo del corte imponer su fuerza resultante y ahí ya puedo caminar mis ecuaciones en cambio acá yo voy a aplicar este corte de derecha vamos a hacerlo de la forma donde varios no quieren realizarlo porque aquí te topas ahora con una carga digamos de figura trapezoidal vamos a explicar me corté 6 el intervalo es ya mencionamos de 4 a 8 colocamos aquí guarda 8 dibujamos el centramos en el apoyo móvil toma de 30 aquí justamente está esa figura no hace menos como un trapecio estamos cortando bien nos dibuja todavía esta flecha para que no se confunda dibuje solamente una figura así como un trapecio déjenlo si está cortando esta parte ya captan las grietas ya sabemos a qué nos referimos indicamos acá el corte hasta acá va a ser x esta distancia ya lo tengo si no me equivoco es 4 metros y yo sé que esta distancia de una vez ya lo puedo calcular directamente esto sería x menos 40 justo acá has hecho el corte nos vamos a colocar un poco más a la izquierda para no chocar ahorita con esa altura que vamos a calcular recuerda esto de acá estoy mandando acá nada más para evitar complicaciones en la barra de gráfica o hallar sus alturas esto va a ser la cortante a perdón esto va hacia arriba momento en sentido horario y normal hacia la izquierda que sucede acá necesito encontrar si o si las alturas que va a tener esta carrera pero yo les recomiendo que hagan estos ejes de proyecto pero antes a la derecha también recomiendo que haga una predicción de este modo así a parte por ser una hoja de borrador para que se den cuenta y donde culmina ahí la carga hasta ahí más o menos digamos que es como que la carga del angular para que sea más práctico esa es otra forma que yo les he mencionado yo sé ahorita que todo esto vamos a cambiar esto todo esto ni de inicialmente o alto mes de opciones eso sí lo sabemos una carrera angular 4 mente entonces yo voy a tomar en cuenta también la altura porque yo editando un poco no ser ésta ni esta altura sí sí yo conozco este de acá ya pudo encontrar la altura total pero ahorita no va a ser tan en serio de ayuda total porque porque también es recomendable cuando tenga cargas capacidades que la partición en en dos figuras uno en un rectángulo y el otro en un triángulo esas dos fibras queremos calcular porque otros también creo que trabajan con una figura de la especialidad en sí aunque no no sería tan lógico lo más práctico y más simple es para accionarlo en dos figuras esta parte de acá vamos a hacer así todo esto ya sé cuánto mide digamos 6 toneladas alimento y su altura total y vamos a aplicar como relación de triángulos todo esto de aquí vamos a darle estos colores para que se pueda identificar desde el triángulo mayor color amarillo decimos 6 a su base total que es 4 como este de aquí yo puedo calcular este con esto a ver si tú puedes colocar auditar la distancia que yo le recomiendo que hagan así yo lo que recomiendo es que encuentre mejor que esa altura la de arriba no esto de aquí digamos que sea h cuando casaca esto es igual a h es a su base cuando es esta base ya lo tenemos bien creo que acá lo móvil podéis o no lo están entendiendo un poco mejor este dato h es a su base 100 guantes esa base x menos 4 y nada de espejos de leche se dicen de cuadro recuerda que era 1.5 sería 1.5 por x menos cuatro ya tienes el valor de h puede calcular el valor de esta altura del rectángulo claro que sí si todo es de 6 y este es h esto sería 6 - h que es lo que tengo que hacer acá ayer su fuerza resultante siendo primero con la carga rectangular base por altura la base es x menos 4 y la altura sería 6 - h 6 - h y esta y eso es todo el área de un rectángulo base con altura hacen lo mismo ahora pero para la carrera de angular base poner duda entre 2 la base es x menos 4 la altura es así que ya lo tengo que es 1.5 corre que menos 4 y todo esto se divide entre 2 ya lo tenemos y ubicamos su centro de gravedad para el rectángulo es en la mitad 7 x menos 4 esto está ubicado a x menos 4 entre 2 en cambio para la carrera angular esto sería diferente pero vamos a trabajar esto ahora sí para la carrera angular su fuerza resultante y este de aquí va a ser recuerden a un tercio del ángulo recto este es el ángulo recto de 90 grados en la figura del angular un tercio de la base que es x menos 4 y esta otra distancia es dos tercios por x menos 4 eso está en tablets en realidad ya deben saber es entre gravedad que va a tener cada figura y eso sería muy simple ahora ya puede generarle cosas en la normal cortante un momento sería n igual a cero cortante mismo criterio bien todas estas fuerzas resultantes van hacia abajo son opuestas por tanto va a ser positivo en cambio la fuerza es 30 va hacia arriba sería menos 30 más x menos 4 x 6 - h cuando la chica lo reemplazará hasta aquí - x menos cuatro aberturas alcanzó un poco de espacio vamos a mover un poco desde la izquierda ahora si x menos 4 x 1.5 por x menos 4 y todo esto dividido aquí está nos hacen lo mismo para la ecuación el momento también acá sería 30 en el momento ante dario puesto el corte positivo vamos a indicar lo que estemos mejor momento - ah perdón positivo positivo cada uno mínimo tomó a 130 x 30 x toda la distancia 30 porque esta longitud x al recorte recuerda 30 x acá para esta carga triangular su corte resultante va hacia abajo me generan momentos horarios o sea en el mismo sentido de corte va a ser negativo menos decimos x menos cual acabará un corchete para que no se confunda porque muchos se confunden sería x menos 4 por 1.5 por x menos 4 los paréntesis sería para cerrar esta parte de aquí entre 2 todo eso multiplicado por la distancia cuando es esta distancia va a ser desde este punto hasta el corte que es dos tercios de x4 ahí está y cerramos el corchete yo sé que esto se vuelve un poco digamos engorroso por lo menos son bastantes nombres no podemos equivocar pero es así eso lo aplicamos cortes tenemos que hacerlo de ese modo hace falta o faltaría el mejor dicho del último demostrando más abajo también va hacia en ese sentido me genera un momento horario va a ser negativo también decimos menos x menos 4 esto sería x menos 4 vamos un corchete también para confundirlo por 6 - h 6 - que es 1.5 porque x 4 es la fuerza resultante y multiplicar la distancia que sería k x 4 metros de la carga rectangular x [Música] x 4 metros y eso sería todo ya tu reemplaza cuando x es 4 cuando sería el valor o cuando xs8 que hubiese pasado si tomaba de la otra manera osea sh lo calcula para esta altura ción tú también pudiste haber trabajado así podemos haber hecho esto vamos a los locos acá para no confundirnos que teníamos en cuenta la superstición tu pudiesen tomar quizás mira esta parte hasta acá ya sabemos de por sí que es x4 entonces decimos esto va a seguir siendo seis son recuerda hola yo creo que alguna esta altura dice es h y cuánto mide todo esto obviamente que ya lo tiene emoción a esta altura a ver recuerda que necesitaría esta distancia digo yo lo voy a denominar guisadas x sería altura mayor 6 es a su base mayor que ya lo tengo en este caso ni el cuarto está igual a este toda este triángulo pequeño ya no base desde arriba sería h es a esta base es x cuánto sería en realidad mía si tú te das cuenta cuando no mides toda la longitud digamos del apoyo móvil vienen del apoyo móvil hasta este punto cuando y si se distancia tenemos apoyo móvil hasta el punto de donde culmina ese triángulo sería 8 opción entonces colocas esta parte todo esto de aquí a ver voy a hacerlo un poco más abajo para que no se confundan esto de aquí 10 8 hay mide 8 metros y tú estás colocando el x hasta casi hasta acá hasta el corte entonces por lógica esto sería toda la distancia 8 - x y esa es tu base cuando cosa k 8 - x y despejar el valor de h esto sigue siendo 1.5 por 8 - x s es otra forma como le digo quizás al comienzo no lo va a saber de forma directa pues ello recomiendo mejor que lo hagan de la otra forma no como lo he explicado y está con los cortes con las ecuaciones también y bueno eso sería todo para este ejercicio ha sido más que nada explicaciones de cómo deberían generar tus cortes recuerda que siempre hay que tomar en cuenta también el tiempo es muy necesario ya que a veces no descansa con uno desarrollo en la evaluación y saber que existe de poder utilizar a la hora de desarrollar esto podemos espero que la desarrolló el vídeo nuevamente como me apoyas o como ayudas a que se hace la normal vídeo simplemente comparte lo comenta dale like y más que nada interactuar de esa forma para que todos puedan visualizar este mito así que conmigo será hasta un próximo vídeo |
9045 | https://www.investopedia.com/terms/r/recursive-competitive-equilibrium.asp | Recursive Competitive Equilibrium (RCE)
Investopedia / Madelyn Goodnight
What Is Recursive Competitive Equilibrium (RCE)?
The term recursive competitive equilibrium (RCE) refers to a concept used to explore and study economic issues when supply and demand are equal. Recursive competitive equilibrium is a mathematical optimization method commonly used in macroeconomics and is characterized by time-invariant equilibrium decision rules that specify actions as a function of a limited number of variables.
RCE helps analysts and economists explore a variety of issues, such as monetary and fiscal policy and business cycle fluctuations.
Key Takeaways
How Recursive Competitive Equilibrium (RCE) Works
Recursive competitive equilibrium is an optimization concept characterized by time-invariant equilibrium decision rules within the economy. These rules outline actions as a function of a limited number of variables. These variables, which are commonly referred to as state variables, sum up how past decisions and current information are affected.
The assumption with RCE is that all the variables involved are current. Therefore, previous information available in the economy is known. The decision rules for RCE include a number of functions, which include:
As such, RCE basically looks at what effect the functions, prices, value, and period allocation policies have on the variables, which is the information on the economy. Equilibrium objects are the functions instead of variables in RCE.
Economic agents with knowledge of these variables assess the current state of the economy, including policies enacted by financial authorities (notably fiscal and monetary policy), as well as changes within the business cycle. Their actions will determine, in part, the values of the variables in the next sequential time period. This makes the structure recursive.
Important
An economic agent is a consumer, corporation, or organization that makes an impact on the economy by buying, selling, or producing.
Special Considerations
The RCE approach assumes that the consumer makes all consumption decisions, while a finite number of firms produce two goods—a consumable one and a capital one. These firms are able to maximize their profit each period. It assumes firms purchase inputs and labor at competitive prices after assessing productivity at the start of the period.
Consumers then use their wages to buy goods from firms and the process starts over each period. During this period, firms don't retain their assets (because they're sold) and technology is freely available. Some RCE models actually assume an infinite-life, maximizing value firm.
The RCE model assumes a stationary environment when it comes to finding optimal growth. The assumption is that the issue does not change with time, hence the recursive representation. Recursive problems are solved regardless of time, where a sequential model solution depends on the time period for which you're solving.
RCE allows analysts to focus on other structures of the problem. The variables are predetermined and matter. They must, therefore, vary across time and state.
Fast Fact
The recursive competitive theory was developed by Rajnish Mehra and Edward Prescott.
Recursive Competitive Equilibrium (RCE) and Macroeconomics
As mentioned above, recursive competitive equilibrium falls under the study of macroeconomics. This is the study of the broader economy. Macroeconomics involves the study of broader economic trends and indicators, such as national income, unemployment rates, and gross domestic product (GDP). It also studies the relationship of such economic factors as inflation, trade, consumption, and income.
Economic equilibrium occurs when economic forces are balanced, which is also known as supply equaling demand. In a competitive equilibrium like RCE, supply equals demand. The RCE helps economists determine the reasons for short-term fluctuations in the business cycle and longer-term reasons for economic growth.
National Bureau of Economic Research. "Recursive Competitive Equilibrium," Page 6. Accessed May 25, 2021.
National Bureau of Economic Research. "Recursive Competitive Equilibrium," Page 4. Accessed May 25, 2021.
National Bureau of Economic Research. "Recursive Competitive Equilibrium," Page 7. Accessed May 25, 2021.
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9046 | https://www.quora.com/What-is-the-length-of-the-interval-0-1-By-length-I-mean-the-difference-between-1-exclusive-and-0-inclusive | What is the length of the interval [0, 1)? By length, I mean the difference between 1 (exclusive) and 0 (inclusive). - Quora
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What is the length of the interval [0, 1)? By length, I mean the difference between 1 (exclusive) and 0 (inclusive).
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Mark Baer
Author has 1.2K answers and 140.3K answer views
·1y
The moral of the story is that with or without the endpoint/s of an interval, the length and the R^1 Lebesgue measure remain the same .
Perhaps you would like to think of the real intervals ( 0 , 1 ) , ( 0 , 1 ] , [ 0 , 1 ) , [ 0 , 1 ] as having different lengths. DON’T . Subtraction is defined between two real numbers or perhaps between two extended real numbers ( with the exceptions that infinity minus infinity and negative infinity minus negative infinity remain undefined ) . Trying to find 1 ( exclusive ) minus 0 ( inclusive ) may sound tempting, but each of 1 ( exclusive ) and 0 ( inclusiv
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The moral of the story is that with or without the endpoint/s of an interval, the length and the R^1 Lebesgue measure remain the same .
Perhaps you would like to think of the real intervals ( 0 , 1 ) , ( 0 , 1 ] , [ 0 , 1 ) , [ 0 , 1 ] as having different lengths. DON’T . Subtraction is defined between two real numbers or perhaps between two extended real numbers ( with the exceptions that infinity minus infinity and negative infinity minus negative infinity remain undefined ) . Trying to find 1 ( exclusive ) minus 0 ( inclusive ) may sound tempting, but each of 1 ( exclusive ) and 0 ( inclusive ) is NOT a real number. Furthermore, each of 1 (exclusive) and 0 ( inclusive ) is NOT an extended real number.
Let each of a and b be a real number with a < b .
( a , b )
[ a , b ]
( a , b ]
[ a , b )
length = b - a ; Lebesgue R^1 measure = b - a
( - infinity , b )
( - infinity , b ]
( a , infinity )
[ a , infinity )
( - infinity, infinity )
length = infinity ; Lebesgue R^1 measure = infinity
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There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you?
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What does 0.1 mean? How much is 0.1?
What is the difference between 0 and 0.1?
When we plot the real axis should we put 1 always in the same distance from 0? I mean if we put it 5m in the right of 0 or 5mm in the right of 0 would it change the meaning of the real number 1? Will it make it smaller or bigger?
Are 0 and 1 the limit points of (0 1)? Why?
What is the interval mark between 0 and 1?
Giles Martin
Former Assistant Editor, Dewey Decimal Classification at Online Computer Library Center (OCLC) (1999–2011) · Author has 5K answers and 5.6M answer views
·1y
The branch of mathematics that deals with this question is called measure theory. It also deals with the area of 2-dimensional shapes, the volume of 3-dimensional shapes, and analogues in higher dimensions.
For an interval on the number line, the usual definition is indeed the difference between the end points, regardless of whether the end points are open or closed. In fact, you can take any finite number of points from the interval, and the measure will still be zero.
However, if you take a sub-interval out, then the measure is less. For example, the measure of the union of [0, 1/3) and [2/3,
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The branch of mathematics that deals with this question is called measure theory. It also deals with the area of 2-dimensional shapes, the volume of 3-dimensional shapes, and analogues in higher dimensions.
For an interval on the number line, the usual definition is indeed the difference between the end points, regardless of whether the end points are open or closed. In fact, you can take any finite number of points from the interval, and the measure will still be zero.
However, if you take a sub-interval out, then the measure is less. For example, the measure of the union of [0, 1/3) and [2/3, 1) is 2/3. That is in spite of the end points being still a distance of 1 apart. So measure is generally a more useful measurement than length.
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John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math · Author has 9K answers and 23.3M answer views
·1y
The length of the interval [0, 1] is the same as the length of the interval (0, 1) which is the equal to the length of the intervals [0, 1) and (0, 1].
Each of these intervals has a length of one unit,
Yes, I know that [0, 1] has two values in it that (0, 1) does not have, but removing those two values from the interval only reduces the length of \frac{1}{\infty} at either end, and as you should be
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The length of the interval [0, 1] is the same as the length of the interval (0, 1) which is the equal to the length of the intervals [0, 1) and (0, 1].
Each of these intervals has a length of one unit,
Yes, I know that [0, 1] has two values in it that (0, 1) does not have, but removing those two values from the interval only reduces the length of \frac{1}{\infty} at either end, and as you should be able to guess, it reduces the length of the interval by zero at each end.
Remember, there are an infinite number of rational and real nu...
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Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968) · Upvoted by
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) and
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 16.5K answers and 29.3M answer views
·1y
What is the length of the interval [0, 1)? By length, I mean the difference between 1 (exclusive) and 0 (inclusive).
It doesn’t matter whether exclusive or inclusive as there are no infinitesimals in ordinary arithmetic. Therefore the length is 1 1.
If you extend arithmetic to allow infinitesimals then there will be an infinite number of infinitesimals between any a<1 a<1 and 1 1 so you can’t get a definite answer. I prefer not to go there.
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Anthony Madden
Writer for Betterbuck
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What are the weirdest mistakes people make on the internet right now?
Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr
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Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your privacy, which is nice.
More often than not, it saves even more than 50% on load times - here’s a test I ran:
Using an ad blocker saved a whopping 6.5+ seconds of load time.
Here’s a link to Total Adblock, if you’re interested.
Not getting paid for your screentime
Apps like Freecash will pay you to test new games on your phone.
Some testers get paid as much as £270/game. Here are a few examples right now (from Freecash's website):
You don't need any kind of prior experience or degree or anything: all you need is a smartphone (Android or IOS).
If you're scrolling on your phone anyway, why not get paid for it?
I've used Freecash in the past - it’s solid. (They also gave me a £3 bonus instantly when I installed my first game, which was cool).
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What is the length of an interval (-18, -24)?
We know the length of [0,1] is 1, then if we take the point 1/2 away from the interval [0,1], what would the resulted length be?
Is 0 greater or less than 0?
What is better, 1 or 0?
Is 0.1 = 0.10?
Chaz Bowen
BSME in Mechanical Engineering, Washington State University · Author has 1.2K answers and 5.6M answer views
·1y
The length of [0,1)[0,1) is the difference between the lengths of [0,1][0,1] and [1,1][1,1]:
l([0,1))=l([0,1])−l([1,1])l([0,1))=l([0,1])−l([1,1])
The length of a closed interval is the positive difference between it's endpoints:
l([0,1))=(1−0)−(1−1)l([0,1))=(1−0)−(1−1)
l([0,1))=1 l([0,1))=1
Yes, this means that removing a single point from an interval doesn't change it's length (nonstandard analysis notwithstanding).
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Muhammad Usman Amin
M.Sc in Mathematics&Statistics (academic discipline), University of Education, Pakistan (Graduated 2019)
·1y
The length of the interval [0, 1) is 1 unit, as it includes all real numbers greater than or equal to 0 and less than 1. So, the difference between the upper bound (1) and the lower bound (0) is 1.
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Silvio Capobianco
Mathematician, musician, manatee fan. · Author has 3.4K answers and 630.7K answer views
·1y
It is 1 1, of course. By definition of Lebesgue measure, λ([0,1))=λ({0})+λ((0,1])−λ({1})=0+1−0=1 λ([0,1))=λ({0})+λ((0,1])−λ({1})=0+1−0=1.
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Kevin Smyth
Bsc. (hons) in Mathematics, Chelsea College of Science and Technology, University of London (Graduated 1969) · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 4K answers and 1.4M answer views
·2y
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The interval [0,1) includes everything from 0 until just before 1, or 0.9999..., but 0.99999... = 9/9 = 1, so what is the difference between [0 and 1]?
You have a few answers but let me answer a slightly different question which does not involve the base system. We also don’t need to include any set theory.
Consider the interval (0,1]. This is bounded below by 0 but does not include 0. What is the minimum number in the interval?
Suppose it is x. Well I can create a smaller number x/2. I can always do this thus showing there is no minimal number under the reals in an interval open at its lower end. Of course this argument applies at both ends of an interval. Having 0 as a lower bound just makes it simpler to explain.
This and related topics are c
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You have a few answers but let me answer a slightly different question which does not involve the base system. We also don’t need to include any set theory.
Consider the interval (0,1]. This is bounded below by 0 but does not include 0. What is the minimum number in the interval?
Suppose it is x. Well I can create a smaller number x/2. I can always do this thus showing there is no minimal number under the reals in an interval open at its lower end. Of course this argument applies at both ends of an interval. Having 0 as a lower bound just makes it simpler to explain.
This and related topics are covered in a first year undergraduate course on mathematics. You could ask, which nobody does, how do you know you can always divide a number? This is an axiom and can be illustrated by the fact that on the number line each number occupies a point which has 0 dimensions.
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Darius Grigorjevas
Head of Customer Success at Hostinger.com
·Updated 2y
What are some tips for starting a blog?
It is always hard to start without having a good picture of what you do want to create. Today there are millions of helping tools and everyone is empowered to create a blog. The only question that is still hovering in the air is - how to start a successful blog?
So let’s go - here are some tips for those who are just starting and even for those who already have a blog:
Choose your blog niche. It’s easy to start a blog if you already know what you want to talk about but even if you don’t have your own topic, don’t give up - there are numerous bloggers who started with one idea and ended up with t
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It is always hard to start without having a good picture of what you do want to create. Today there are millions of helping tools and everyone is empowered to create a blog. The only question that is still hovering in the air is - how to start a successful blog?
So let’s go - here are some tips for those who are just starting and even for those who already have a blog:
Choose your blog niche. It’s easy to start a blog if you already know what you want to talk about but even if you don’t have your own topic, don’t give up - there are numerous bloggers who started with one idea and ended up with the other one.
You are creating a company or have a product? Writing a blog about it could become a great marketing strategy that would drag more attention to your business.
Have something for the start. A lot of people start a blog without even knowing what to write about, it’s not the end of the world but surely you can do better!
Before starting a blog, gather up some materials around you that can be used in the blog - photos, some thoughts written down. Having something for the start obligates you to create a blog and not procrastinate with only thought about having it.
Select an interesting domain name. Forget all the boring ones and find a name that stands out. There are some interesting TLDs for your domain, you don’t have to use .com, with imagination you can make even .online TLD attractive!
Also, these tips can help to select the best domain name for you:
Make the domain memorable - keep it short and simple, you can use domain checkers to help you find the one that fits you best!
Know your audience - domain target keywords are really important to your audience so make it relevant.
Less but better - avoid any hyphens, numbers or non-unicode characters in the domain name. It could be too difficult to remember or hard to understand.
Check for popular keywords - with the domain checker you can find domains with popular keywords. Also, it can help you find domains with the same name and variations, buying those can help you to protect your brand.
Protect your privacy - make sure that privacy protection is activated for your domain name. This will hide your private data from the WHOIS database.
Be fast to be first - hurry up as the domain you want can be bought within seconds!
Analyze the blogs you like. It is so much easier to start with an example. I am not talking about copying others, just getting some inspiration and finding successful methods. Don’t forget to analyze which blogs are getting the most traffic - SimilarWeb can help you out with that. If a blog is getting a lot of traffic, research what could be the key factors of that.
You can even analyze the SEO strategy of other websites! Neil Patel allows you to do that through his own website.
Remember that SEO is important but Content is still the king. No one ever got traffic without doing anything BUT it doesn’t mean that expensive ads will give you what you want. The most important thing here is content and you need to make sure it’s rock-solid, though SEO is also playing a huge role in blogging.
Set up social media platforms. Online presence is important and using social media can help you increase your user engagement. It’s the best tool to notify people about new articles and plans. Facebook and Instagram can help you grow and receive feedback from your readers directly. Also, you can implement the links from your blog post to Facebook and Instagram to gain more followers there and create a community for them.
As social media channels (and even effective ones) the personal profiles can be used too. A lot of bloggers have their blogs linked in the “About me” section of their Facebook and Instagram profiles. Self-promotion for friends and followers can increase traffic and popularity.
Create a top-notch design. Even when content is so important, you have to make sure that your visitors can easily read it and find it. The eye-catching design is a must to keep the readers in the blog for more than a few seconds. You can create a design yourself, or you can use existing templates - WordPress has a lot of them! If you have no idea about the latest trends, give a shot for a minimalistic look. It will make your blog look more professional and easy to navigate.
Congrats! Now your blog is ready to storm the internet. Be sure to keep it updated and it will stay on the successful path. Here Google Analytics and other platforms that help to track your blog growth or plan its future will come in handy. A goal without a plan is just a wish, so make it real and enjoy the journey!
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Aquiles Fischer
Former Teacher · Author has 781 answers and 135.7K answer views
·1y
Your interval [0,1) means that your function f(x) will include your zero, but NOT the 1. So you could think that the real interval is from zero to 0.9999. Now the zero may or may not be a root, so look for the other values 1/8, 1/4, 1/2, 3/4, and so on. Just wondering what your function f(x) is!
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Amir Sagiv
Trying to start my Thesis while reading Quora
·11y
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We know the length of [0,1] is 1, then if we take the point 1/2 away from the interval [0,1], what would the resulted length be?
I'd expand more on why the answer is one.
What we know how to calculate, is te length of an interval. The length of it is the difference between its ends. That's it, and from here it's all extrapolating.
So, we know the length is the same for translated intervals.
We know that it is monotone - if an interval is contained in another one, then it is not bigger then it.
We know that interval sums up, at least for finite sums.
We know that if we "stretch" an interval ends by a number, the length grows by the same constant
So, it takes some work to see these attributes ( strengthen by a statement about
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I'd expand more on why the answer is one.
What we know how to calculate, is te length of an interval. The length of it is the difference between its ends. That's it, and from here it's all extrapolating.
So, we know the length is the same for translated intervals.
We know that it is monotone - if an interval is contained in another one, then it is not bigger then it.
We know that interval sums up, at least for finite sums.
We know that if we "stretch" an interval ends by a number, the length grows by the same constant
So, it takes some work to see these attributes ( strengthen by a statement about countable sums) really define a length measure to a lot of subsets of numbers. Why not all - a different story for a different day :)
Because of additivity and translation invariance, we know that stating that the length of the pinned interval is one is equivalent to the statement that's a point's length is 0. So why is the length a point, say 0, is 0?
Well, first, it is contained in the interval [-1,1], so its length is no greater then 2. But it is also contained in the interval [-0.1,0.1], so its length is no greater then 0.2. The rest is obvious, it is a non negative number smaller then any finite length. Hence, if it is defined, it must be zero!
What I haven't shown is that it is defined, but for that it is better to confer a book.
Why is this intuitive?
Well, let's start uniformly random take out numbers in the interval. We'd expect that the probability of a single number to be chosen is 0. Not that it can't be chosen, but it has probability zero. Think of it this way - taking numerous random choices as described, and taking the number of " numerous" to infinity, the precentage of the times 0.5 is chosen would tend to zero. It makes sense, and it also make sense that if it had a "bigger length", this would be relevant to its probability to get chosen. Again - for more profound explanation if refer a book or at least wiki.
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Anshul Shenwai
Playing with numbers since 1996 · Author has 65 answers and 310.9K answer views
·10y
Related
Consider S = 1-1+1-1+1-1+1... Is S= (1-1)+(1-1)+(1-1)... =0+0+0.. =0 OR S=1-[(1-1)+(1-1)... =1-0+0+0.. =1 ?
Let me explain this with a hypothetical experiment. Suppose you're given a light bulb which can be toggled ON & OFF. Let the OFF state be known as 0 state and ON state as 1 state. Your task is to toggle the light bulb for a time interval of 2 minutes in the following manner:
The timer starts when you toggle the light bulb for the first time(doesn't matter what the initial state of the bulb is)
Toggle the light bulb again after 1 minute.
Toggle again after 1.5 minutes.
Again after 1.75, then 1.875, then 1.9325 minutes and so on, reducing the time interval between 2 toggles to half the previous
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Let me explain this with a hypothetical experiment. Suppose you're given a light bulb which can be toggled ON & OFF. Let the OFF state be known as 0 state and ON state as 1 state. Your task is to toggle the light bulb for a time interval of 2 minutes in the following manner:
The timer starts when you toggle the light bulb for the first time(doesn't matter what the initial state of the bulb is)
Toggle the light bulb again after 1 minute.
Toggle again after 1.5 minutes.
Again after 1.75, then 1.875, then 1.9325 minutes and so on, reducing the time interval between 2 toggles to half the previous time interval. Now, at the end of the 2nd minute, you are toggling the light at infinite frequency! So if I ask you, what is the state of the bulb at the end of the 2nd minute, what would be your answer? Is the light bulb ON(1) or OFF(0)?
The correct answer is: YOU CANNOT TELL!
Why? Because at the end of the 2nd minute, you are toggling the bulb so rapidly(with infinite frequency) that it is both ON & OFF at the same time! As the frequency is infinite, the bulb exists in both 0 & 1 states simultaneously. So we have to arrive at our answer without creating any partiality between the two states! :P So we have to settle for a value exactly in between the two numbers i.e the average (0+1)/2=0.5.
Now coming back to your question. You're writing 0 as the sum of alternate 1s & -1s.
0=1-1+1-1+1-1+1-1+1-1... Up to infinity
So the sum of this series could be either zero OR one(NOT ZERO ALONE!) depending on whether the last term of the sequence is -1 or 1 respectively. But now as there are infinitely many terms, there is no last term! So once again there is an equal possibility for the sum to be 1 or 0(same as in the light bulb case) and hence we settle at the average yet again. i.e 0.5
Also if you know about progressions, 1-1+1-1+1-1.. Is an infinite Geometric Progression with first term 1 & common ratio -1, the sum of which is 1/(1-(-1))=0.5
The problem in your question arises because you are taking zero to be the sum of infinitely many 1s and -1s. Taking zero as
0=1-1+1-1+1-1.
0=1+(-1+1)+(-1+1)-1
0=1+0+0-1
There's a -1 which comes at the end, but when you take infinite numbers, there is no end! You basically just neglect that -1 and conclude that 0=1 which is not correct.
I may not have answered your question directly but I hope this helps. :)
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Thomas Dalton
MMath in Mathematics, Durham University (Graduated 2009) · Upvoted by
Linda Kuroski
, M.S. Mathematics, University at Buffalo and
Robby Goetschalckx
, Computer scientist for 11+ years and passionate about math since childhood. · Author has 12.5K answers and 15M answer views
·5y
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If 0.999… equals 1, then what does an open interval like: [0, 1 [even mean?
[0,1) means all real numbers greater than or equal to zero and strictly less than one. Neither 0.999… nor 1 are in that interval (since they are both equal to one, so are not strictly less than it). The interval does not have a maximum - for any number in the interval, there will be a larger number that is also in it.
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Lukas Schmidinger
I have graduate CS and my studies included math courses. · Author has 27.6K answers and 14.9M answer views
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The interval [0,1) includes everything from 0 until just before 1, or 0.9999..., but 0.99999... = 9/9 = 1, so what is the difference between [0 and 1]?
The interval [0,1) includes everything from 0 until just before 1, or 0.9999..., but 0.99999... = 9/9 = 1, so what is the difference between [0 and 1]?
Your premise is wrong. 0.999…=1 0.999…=1 and thus it is not in the half open intervall [0,1)[0,1) or in other words: all (real) numbers in this intervall have at least one arbitrary decimal place, which is not 9 9.
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9 1
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9047 | https://artofproblemsolving.com/wiki/index.php/Inequality_symbol?srsltid=AfmBOoq0PsD0YbEKZW6Vf1X5MWLzkc8mwHMLL6yTzowI47uCaPJ-dq_i | Art of Problem Solving
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Inequality symbol
There are four symbols conventionally used to represent the notion of inequality.
If and are real numbers we write:
to mean that is strictly greater than (that is, cannot equal ).
to mean that is greater than or equal to (equivalently, "at least as large as") .
to mean that is strictly less than
to mean that is less than or equal to .
We use a slash through an inequality symbol to represent that the given inequality does not hold. Thus for real numbers and ,
if and only if
if and only if
if and only if
if and only if
if and only if or
These symbols are also frequently used to represent the order relation in a partially ordered set. Note that in this more general setting, it is not necessarily true that , because it is also possible that and could be incomparable.
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9048 | https://www.wcupa.edu/sciences-mathematics/mathematics/sParsell/documents/publications/thesis.pdf | Exponential Sums and Diophantine Problems by Scott T. Parsell A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy (Mathematics) in The University of Michigan 1999 Doctoral Committee: Professor Trevor D. Wooley, Chair Associate Professor Alexander I. Barvinok Professor Joel N. Bregman Assistant Professor Kristin E. Lauter Professor Hugh L. Montgomery ACKNOWLEDGEMENTS First of all, I wish to thank my advisor, Trevor Wooley. This work would not have been possible without his patience, advice, and encouragement. I am also grateful for his generous financial support, which allowed me to attend conferences and to spend the winter semesters focused on research.
I also want to acknowledge the roles of Hugh Montgomery and Bob Vaughan in my number theory education at Michigan. I have benefited from several of the advanced courses they have taught, from their numerous seminar presentations, and from the excellent textbooks they have written.
I am grateful to Kristin Lauter, Alexander Barvinok, and Joel Bregman for their willingness to serve on my committee and for their patience in dealing with scheduling changes and thesis revisions. I also thank Doug Richstone for his role in setting up the committee.
I appreciate the help and advice given to me Greg Martin and Joel Wisdom during the early stages of my research. In particular, Greg provided me with many of the LaTeX templates used in this dissertation, and the Student Number Theory seminar, which he established over four years ago, is still thriving today.
The friends I have made at Michigan, especially Bill Correll, David Dorrough, Eric Freeman, James Nu˜ nez, Victor Scharaschkin, and Chris Swanson, deserve recognition for tolerating me during various parts of the past five years. I am also grateful to my sister and to my friends from MIT for staying in touch and providing encouragement along the way.
Finally, I thank my parents for giving me the foundation necessary for this un-dertaking and for always having unbounded faith in my abilities.
ii TABLE OF CONTENTS ACKNOWLEDGEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ii CHAPTER I. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 1.1 Waring’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 1.2 The Hardy-Littlewood Method . . . . . . . . . . . . . . . . . . . . .
2 1.3 Smooth Numbers and the Iterative Methods . . . . . . . . . . . . .
6 1.4 Additive Equations and Inequalities . . . . . . . . . . . . . . . . . .
8 1.5 Multiple Exponential Sums . . . . . . . . . . . . . . . . . . . . . . .
11 1.6 Applications to Diophantine Problems . . . . . . . . . . . . . . . . .
14 II. Simultaneous Diagonal Inequalities . . . . . . . . . . . . . . . . . . . .
19 2.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 2.2 Forms with Many Zero Coefficients . . . . . . . . . . . . . . . . . .
21 2.3 The Davenport-Heilbronn Method . . . . . . . . . . . . . . . . . . .
26 2.4 The Minor Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28 2.5 The Major Arc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37 2.6 A Discussion of Possible Improvements . . . . . . . . . . . . . . . .
39 III. Mean Values of Multiple Exponential Sums . . . . . . . . . . . . . .
43 3.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43 3.2 Preliminary Lemmata . . . . . . . . . . . . . . . . . . . . . . . . . .
45 3.3 The Fundamental Lemma . . . . . . . . . . . . . . . . . . . . . . . .
47 3.4 Efficient Differencing . . . . . . . . . . . . . . . . . . . . . . . . . .
56 3.5 Mean Value Estimates Based on Single Differencing . . . . . . . . .
62 3.6 Estimates Arising from Repeated Differencing . . . . . . . . . . . .
64 IV. Applications of the Mean Value Theorems . . . . . . . . . . . . . . .
71 4.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 4.2 Weyl Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74 4.3 Fractional Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79 4.4 Generating Function Asymptotics . . . . . . . . . . . . . . . . . . .
81 4.5 A Multidimensional Analogue of Waring’s Problem . . . . . . . . .
87 iii 4.6 Lines on Additive Equations . . . . . . . . . . . . . . . . . . . . . .
94 V. The Density of Rational Lines on Cubic Hypersurfaces . . . . . . . 104 5.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 5.2 The Paucity Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 106 5.3 Further Mean Value Estimates . . . . . . . . . . . . . . . . . . . . . 112 5.4 Weyl Differencing . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 5.5 The Circle Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 iv CHAPTER I Introduction 1.1 Waring’s Problem Perhaps the most famous result in additive number theory is that every positive integer can be written as the sum of four squares. First stated explicitly in 1621, the theorem survived a claimed proof by Fermat (who died before disclosing it) and the best efforts of Euler before succumbing to a proof by Lagrange in 1770. Many proofs of this result are known today, a representative sampling of which can be found in , , and .
In 1770, shortly before Lagrange announced his theorem, Edward Waring had come forth with a more sweeping conjecture. Let g(k) denote the smallest integer s, if one exists, such that every positive integer n can be represented in the form n = xk 1 + · · · + xk s (1.1) with xi ∈N ∪{0}. Waring stated without proof that g(2) = 4, g(3) = 9, g(4) = 19, “and so on,” thus implicitly claiming the existence of g(k) for all k.
Of course, Lagrange’s Theorem establishes the claim that g(2) = 4, since integers congruent to 7 modulo 8 cannot be represented as sums of 3 or fewer squares.
In 1909, Hilbert finally demonstrated the existence of g(k), using Lagrange’s Theorem as the base of a difficult induction relying on complicated polynomial iden-tities. Because of its inductive nature, Hilbert’s argument does not yield any re-spectable upper bound for g(k). On the other hand, a lower bound showing that g(k) is necessarily quite large is easily obtained by considering the integer n = 2k "µ3 2 ¶ k# −1.
Since n < 3k, any representation of n in the form (1.1) must involve only kth powers of 1 and 2. Clearly, the minimal number of terms in such a representation is obtained 1 2 by taking [(3 2)k] −1 powers of 2 and 2k −1 powers of 1, whence g(k) ≥2k + "µ3 2 ¶ k# −2.
(1.2) Around the time of Hilbert’s proof, Wieferich and Kempner were able to show that g(3) = 9, and it was immediately observed by Landau that only finitely many integers actually require 9 cubes; all others can be represented by 8 or fewer. In fact, Dickson showed in 1939 that 23 and 239 are the only integers that cannot be represented as sums of 8 cubes.
In 1943, Linnik took these observations one step further by showing that there are only finitely many integers that cannot be represented as the sum of 7 cubes.
These observations suggest that the enormous size of g(k) results from peculiar difficulties of representing certain small integers. Thus we define a new function G(k) to be the smallest integer s such that every sufficiently large positive integer n can be represented in the form (1.1). While the exact value of g(k) is now known for every k, the problem of determining G(k) turns out to be considerably more difficult.
Aside from Lagrange’s result that G(2) = 4, the only value known at present is due to Davenport , who showed in 1939 that G(4) = 16. In particular, our knowledge about sums of cubes is still embarrassingly weak. On combining Linnik’s Theorem with an elementary counting argument, one obtains the bounds 4 ≤G(3) ≤7, which remain the best available today, although it is widely conjectured that the lower bound represents the true state of affairs.
1.2 The Hardy-Littlewood Method In the early 1920’s, Hardy and Littlewood devised an analytic approach, known as the circle method, which allows one to derive upper bounds for G(k).
Before describing the method, we briefly introduce some of the notation that will be used throughout. Landau’s notation f(t) = O(g(t)) means that there exists a positive constant C such that |f(t)| ≤C|g(t)| for all values of t. We will often find it more convenient, however, to write the same statement using Vinogradov’s notation, f(t) ≪g(t). If f(t) ≪g(t) and f(t) ≫g(t), then we write f(t) ≍g(t).
Finally, we write f(t) = o(g(t)) when f(t)/g(t) →0 as t →∞and f(t) ∼g(t) when f(t)/g(t) →1 as t →∞.
Returning now to Waring’s problem, we write P = [n1/k], and let Rs,k(n) = card{x ∈[1, P]s ∩Zs : n = xk 1 + · · · + xk s}.
3 If we further let e(z) = e2πiz and define the exponential sum f(α) = X 1≤x≤P e(αxk), (1.3) then on noting the orthogonality relations Z 1 0 e(αm) dα = 1 if m = 0, 0 if m ∈Z \ {0}, (1.4) one sees immediately that Rs,k(n) = Z 1 0 f(α)se(−αn) dα.
(1.5) The strategy for evaluating this integral is to dissect the unit interval into major and minor “arcs,” with the major arcs consisting of points that are well-approximated by a rational number with small denominator. We should remark that the term “arcs” persists as a result of Hardy and Littlewood’s original approach to the prob-lem, in which they used Cauchy’s integral formula to represent Rs,k(n) as the integral over a circle in the complex plane. The above set-up reflects a later simplification due to Vinogradov .
Since there are roughly ns/k total choices for the variables x1, . . . , xs, a probabilis-tic argument suggests that one may expect to have Rs,k(n) ≍ns/k−1, and in fact this turns out to be the correct order of magnitude provided that s is sufficiently large in terms of k. Roughly speaking, one shows that a contribution of this size arises from the integral over the major arcs and that the integral over the minor arcs is negligible by comparison as n →∞.
To be more specific, let δ be a small positive number, and define the major arcs by M = [ 0≤a≤q≤P δ (a,q)=1 M(q, a), where M(q, a) = {α ∈[0, 1] : |α −a/q| ≤P δ−k}, and write m = [0, 1] \ M for the minor arcs. Then one hopes to show that Z M f(α)se(−αn) dα ≫P s−k, 4 where the implicit constant may depend on s and k. Hence, if one can also show that Z m |f(α)|sdα = o(P s−k), (1.6) then it will follow that Rs,k(n) > 0 when n is sufficiently large.
When α is not close to a rational number with small denominator, one expects that the terms in the sum (1.3) will behave somewhat randomly and hence that enough cancellation will occur to establish (1.6). In 1916, Weyl was able to show that for any ε > 0 one has sup α∈m|f(α)| ≪P 1−δ 21−k+ε, and in 1938 Hua established the mean value estimate Z 1 0 |f(α)|2sdα ≪P 2s−k+ε, for s ≥2k−1 by interpreting the integral as the number of solutions of the equation xk 1 + · · · + xk s = yk 1 + · · · + yk s (1.7) with xi, yi ∈[1, P] ∩Z. Combining these two results, one finds that (1.6) holds whenever s ≥2k + 1, and this completes the analysis of the minor arcs.
On M(q, a), one has α = a/q + β, where |β| ≤P δ−k, so one hopes to relate f(α) to f(a/q). By sorting the sum (1.3) into arithmetic progressions modulo q, we have f(α) = q X r=1 e µark q ¶ X x≡r (q) e(βxk).
Since β is small, the function e(βxk) is not oscillating rapidly, so we are able to replace the inner sum by an integral with relatively small error. Hence one obtains the approximation f(α) ∼q−1S(q, a)v(β), where S(q, a) = X 1≤x≤q e µaxk q ¶ and v(β) = Z P 0 e(βγk) dγ.
5 After some analysis, this leads to the factorization Z M f(α)se(−αn) dα ∼ X q≤P δ X 1≤a≤q (a,q)=1 (q−1S(q, a))se(−an/q) Z P δ−k −P δ−k v(β)se(−βn) dβ ∼ Ss,k(n)Js,k(n), where Js,k(n) = Z ∞ −∞ v(β)se(−βn) dβ is known as the singular integral and Ss,k(n) = ∞ X q=1 X 1≤a≤q (a,q)=1 (q−1S(q, a))se(−an/q) is the singular series. The singular integral captures information about the density of real solutions to (1.1), and in fact it follows from Fourier’s Integral Theorem (see for example Davenport ) that Js,k(n) ∼Γ(1 + 1/k)s Γ(s/k) ns/k−1.
The positivity of the singular series depends on the p-adic solubility of (1.1), which one establishes by a combination of analytic and Hensel’s Lemma-type arguments.
Thus when s ≥2k + 1, we obtain the asymptotic formula Rs,k(n) ∼Ss,k(n)Γ(1 + 1/k)s Γ(s/k) ns/k−1, and it follows that G(k) ≤2k + 1.
The above result, combined with numerical work on smaller values of n, has led to the determination of g(k) for all values of k. The precise formula is somewhat complicated to state, but the lower bound (1.2) is is fact attained for all but at most finitely many values of k.
Refinements of the above methods have led to substantial improvements in the ensuing bounds for G(k). In particular, Vinogradov’s Mean Value Theorem (see for example ) allows one to obtain estimates of nearly the same strength as Hua’s with s much smaller than 2k. Moreover, these estimates can be transformed, via the large sieve inequality or similar methods, to yield improved versions of Weyl’s inequality. Within this framework of ideas, Vinogradov established the bound G(k) ≤2k log k(1 + o(1)) in 1959, but no further improvements of any significance were made over the next 30 years.
6 1.3 Smooth Numbers and the Iterative Methods A major breakthrough in the analysis of Waring’s problem occurred in 1989 with the work of Vaughan . His contribution was to exploit properties of “smooth” numbers, i.e. numbers without large prime factors, to set up an iterative method for estimating the number of solutions of the auxiliary equation (1.7), which arises in the treatment of the minor arcs. Let R be a small power of P, and write A(P, R) = {n ∈[1, P] ∩Z : p|n, p prime ⇒p ≤R} (1.8) for the set of R-smooth numbers up to P. It is shown, for example in , that card(A(P, R)) ≫P, (1.9) and hence one may restrict attention to representations of n as sums of kth powers of smooth numbers without serious loss.
Moreover, the smooth numbers possess convenient factoring properties, which, roughly speaking, allow one to introduce a congruence condition on some of the variables in (1.7). Thus we define the exponen-tial sum f(α; P, R) = X x∈A(P,R) e(αxk) and observe that the mean value Ss(P, R) = Z 1 0 |f(α; P, R)|2sdα counts the number of solutions of (1.7) with xi, yi ∈A(P, R) and hence in particular is bounded above by the number of solutions of the equation zk + xk 1 + · · · + xk s−1 = wk + yk 1 + · · · + yk s−1 (1.10) with 1 ≤z, w ≤P and xi, yi ∈A(P, R). Notice that if x ∈A(P, R) then for any M > x there is a divisor of x lying between M and MR. By applying this observation to the xi and yi and then using H¨ older’s inequality to uniformize the choice of divisor, one is able to relate the number of solutions of (1.10) to the number of solutions of zk + qk(uk 1 + · · · + uk s−1) = wk + qk(vk 1 + · · · + vk s−1) (1.11) with 1 ≤z, w ≤P, ui, vi ∈A(P 1−θ, R), and P θ < q ≤P θR, 7 where θ is a parameter satisfying θ ≤1/k. The congruence condition implicit in (1.11) now allows one to classify solutions according to the common residue class of zk and wk modulo qk, and then by applying Cauchy’s inequality to the underlying exponential sums, one reduces to the consideration of solutions in which z ≡w (mod qk). Hence one may write w = z+hqk and thus set up a differencing procedure which is “efficient” in the sense that the parameter h is bounded by P 1−kθ instead of P. It is now easy to obtain estimates for Ss(P, R) from estimates for Ss−1(P, R) by fixing z, q, and h, and considering the underlying mean values. Thus we have an iterative procedure for estimating Ss(P, R), and this leads to an improved analysis of the minor arcs.
Wooley has refined Vaughan’s method to allow one to repeat the differencing process, only fixing z, q, and h trivially after several (perhaps as many as k −1) efficient differences have been taken. With this refinement, Wooley was able to halve Vinogradov’s bound for G(k), showing in that G(k) ≤k(log k + log log k + O(1)).
(1.12) We remark that a simple counting argument shows that G(k) ≥k + 1, and it is expected that this lower bound represents the true state of affairs for most values of k. In certain cases, however, local solubility obstructions may require that G(k) be somewhat larger. Define Γ(k) to be the least integer s such that for every n and q the congruence xk 1 + · · · + xk s ≡n (mod q) has a solution with (x1, q) = 1.
Conjecture. One has G(k) = max(k + 1, Γ(k)).
Unfortunately, the technology leading to (1.12) offers little hope of proving such a statement, as the circle method cannot possibly succeed in its present form when s ≤2k. In fact, the calculation of Γ(k) is a difficult problem in its own right. It is known that Γ(k) ≤4k and that this upper bound is attained whenever k is a power of 2, but it is an open question to determine whether lim inf Γ(k) > 3.
Vaughan and Wooley have introduced iterative schemes that lead to re-spectable bounds for G(k) for small k ≥5, but even these do not approach what is conjectured. For example, the inequalities 6 ≤G(5) ≤17, 9 ≤G(6) ≤24, 8 ≤G(7) ≤33, and 32 ≤G(8) ≤42 are the best currently available.
An exposition of the main ideas underlying the use of smooth numbers in additive number theory can be found in Vaughan .
8 1.4 Additive Equations and Inequalities The Hardy-Littlewood method may also be applied to show that the additive equation c1xk 1 + · · · + csxk s = 0 (1.13) has non-trivial integer solutions when the ci are nonzero integers (not all of the same sign when k is even) and s is sufficiently large in terms of k. By (1.4), the number of solutions of (1.13) with x1, . . . , xs ∈[1, P] ∩Z is given by Ns,k(P) = Z 1 0 Ã s Y i=1 f(ciα) !
dα, where f(α) is as in (1.3), and one finds via H¨ older’s inequality and changes of variable that essentially all of the technology discussed for Waring’s problem applies. The only serious difficulty lies in satisfying the p-adic solubility condition and hence proving that the singular series is bounded away from zero. When k = p −1 for some prime p, one must in fact take s ≥k2 + 1, for it is easily seen that the equation (xp−1 1 + · · · + xp−1 p−1) + p(yp−1 1 + · · · + yp−1 p−1) + · · · + pp−2(zp−1 1 + · · · + zp−1 p−1) = 0 has (p −1)2 variables and no non-trivial p-adic solution. It is a classical result of Meyer that 5 = 22 + 1 variables suffice when k = 2, and Baker showed in 1989 that 7 variables suffice when k = 3. In contrast to the situation for Waring’s problem, Baker’s result is easily seen to be best possible by considering the equation x3 1 + 2x3 2 + 7(x3 3 + 2x3 4) + 49(x3 5 + 2x3 6) = 0 over Q7. Meyer’s result is also seen to be best possible by considering an equation of similar form in 4 variables.
If the integers ci in (1.13) are replaced by arbitrary real numbers λi, then one should not expect to find integers x1, . . . , xs such that λ1xk 1 +· · ·+λsxk s = 0, but one can instead ask whether the inequality |λ1xk 1 + · · · + λsxk s| < ε (1.14) has non-trivial integer solutions for arbitrarily small ε.
In 1946, Davenport and Heilbronn showed that (1.14) has infinitely many integral solutions for any ε > 0 when k = 2 and s ≥5, provided that λ1, . . . , λs are not all of the same sign. Moreover, their proof easily generalizes to show that s = 2k + 1 variables suffice for general k, again provided that the λi are not all of the same sign when k is even. Davenport 9 and Heilbronn established their result by developing an appropriate version of the circle method. While the analytic set-up is now less obvious, the analysis actually turns out to be much easier, owing to the absence of p-adic solubility considerations.
We first observe that it suffices to establish the result for ε = 1, since one may then replace λi by λi/ε. Furthermore, we may assume by relabeling variables that λ1/λ2 is irrational, since we may appeal to the theory of additive equations if all the λi are in rational ratio. One proceeds by choosing a suitable “kernel” function, for example K(α) = µsin πα πα ¶ 2 , (1.15) whose Fourier transform has properties useful for detecting solutions of (1.14). With the above choice, it is a simple exercise in the calculus of residues to show that ˆ K(t) = Z ∞ −∞ K(α)e(αt) dα = max(0, 1 −|t|), from which it follows that θs,k(P) = Z ∞ −∞ Ã s Y i=1 f(λiα) !
K(α) dα is a lower bound for the number of integral solutions of (1.14) with x ∈[1, P]s. Hence it suffices to show that θs,k(P) →∞along some infinite sequence of P, the sequence here being determined by the denominators of the convergents to the continued fraction for λ1/λ2. One now dissects the real line into a single major arc near zero, two minor arcs covering an intermediate range on each side of the major arc, and two large “trivial arcs.” The analysis of the minor and trivial arcs requires the use of a mean value estimate such as Hua’s Lemma, and the minor arcs additionally require some version of Weyl’s inequality. The analysis of the major arc is relatively easy, as one has to deal with a singular integral but no singular series.
There has been considerable interest in quadratic and cubic inequalities in recent years. Most notably, Margulis resolved a conjecture of Oppenheim by proving that any indefinite quadratic form (not necessarily diagonal) in at least 3 variables, takes arbitrarily small values at integer points, provided that not all its coefficients are in rational ratio. This is easily seen to be best possible, since such a statement in two variables would assert that certain algebraic numbers possess rational ap-proximations of higher quality than allowed by Liouville’s Theorem (see for example ). Recently, Baker, Br¨ udern, and Wooley showed that 7 variables suffice to solve (1.14) in the cubic case, and they in fact obtained a quantitative version of this result in which ε is replaced by an explicit function of x. It is still possible, however, that as few as 3 variables suffice in the cubic case, and perhaps even for larger k.
10 A problem of related interest is the so-called “fractional parts” problem, in which we consider inequalities modulo 1. Write ||x|| for the distance from x to the nearest integer. Then, as a simple example, one tries to establish that for N > N0(ε, k) min 1≤n≤N ||αnk|| < N ε−τ(k), with τ(k) as large as possible. The best result on this particular problem is due to Wooley , who was able to take τ(k)−1 ∼k log k. Here one does not use the circle method directly, but exponential sum estimates are still the key ingredient in the analysis. When α has major arc-type rational approximations, then the statement is generally easy to prove. Otherwise, one can use some version of Weyl’s inequality in combination with a version of the Erd¨ os-Tur´ an inequality (see for example , ) to bound the discrepancy between the actual and expected distribution of the sequence αnk modulo 1.
The above methods may be further generalized to investigate systems of equations and inequalities. Let F1, . . . , Ft be diagonal forms of degree k with real coefficients in s variables, and let ε be a positive real number. The solubility of the system of inequalities |F1(x)| < ε, . . . , |Ft(x)| < ε (1.16) in integers x1, . . . , xs has been considered by a number of authors over the last quarter-century, starting with the work of Cook and Pitman on the case t = 2. More recently, Br¨ udern and Cook have shown that the above system is soluble provided that s is sufficiently large in terms of k and t and that the forms F1, . . . , Ft satisfy certain additional conditions.
What has not yet been considered is the possibility of allowing the forms F1, . . . , Ft to have different degrees.
However, the recent work of Wooley , on the corresponding problem for equations has made the study of such systems a feasible prospect.
Our first result takes an initial step in that direction by studying the analogue of the system considered in and . Let λ1, . . . , λs and µ1, . . . , µs be real numbers such that for each i at least one of λi or µi is nonzero. We define the forms F(x) = λ1x3 1 + · · · + λsx3 s G(x) = µ1x2 1 + · · · + µsx2 s and consider the solubility of the system of inequalities |F(x)| < (max |xi|)−σ1 |G(x)| < (max |xi|)−σ2 (1.17) 11 in rational integers x1, . . . , xs.
In Chapter 2 we establish the following result by employing a two-dimensional version of the Davenport-Heilbronn method described above.
Theorem 1. Let s ≥13, and let λ1, . . . , λs and µ1, . . . , µs be real numbers such that for some i and j the ratios λi/λj and µi/µj are algebraic and irrational. Then the simultaneous inequalities (1.17) have infinitely many solutions in rational integers provided that (a) F(x) has at least s −4 variables explicit, (b) G(x) has at least s −5 variables explicit, (c) the system F(x) = G(x) = 0 has a non-singular real solution, and (d) one has σ1 + σ2 < 1 12.
If Θs(P) denotes the number of solutions of (1.17) with x ∈[1, P]s ∩Zs, then our analysis in Chapter 2 will in fact show that Θs(P) ≫P s−5−σ1−σ2 as P →∞. We also note that condition (c) implies that the quadratic form G is indefinite, which is plainly a necessary requirement for solubility.
A significant question raised by Theorem 1 is whether the requirement that our forms have some pair of coefficients in algebraic ratio can be shown to be necessary.
One suspects that such a condition is not needed, but its removal provides a clear obstruction to the method. When either F or G has a large number of zero coeffi-cients, however, we can exploit results for a single inequality to obtain results that do not require the existence of algebraic irrational coefficient ratios. This is discussed more fully in Chapter 2.
Finally, we remark that the iterative methods developed in Wooley poten-tially allow one to investigate pairs of inequalities of arbitrary degrees, say k and n, possibly showing solubility in roughly 2k log k variables if k > n. However, the mean values of the exponential sums relevant for attacking such a problem have not yet been considered in detail, and obtaining sharp estimates for them is likely to be a formidable task. In principle, one can even consider more general systems of t diagonal inequalities having degrees k1, . . . , kt using the methods of along with the Diophantine approximation results of , but this is deferred to later work.
1.5 Multiple Exponential Sums In contrast to the highly-developed theory of exponential sums in a single vari-able, multiple exponential sums are not well-understood, and consequently relatively little is known about the Diophantine problems that demand their use. Arkhipov, 12 Karatsuba, and Chubarikov have provided estimates for fairly general d-fold sums, but most of these estimates are not explicit enough to be useful in applications. As a first step toward obtaining sharper general estimates, we concentrate most of our analysis on double sums of the form f(α; P, R) = X x,y∈A(P,R) e(α0xk + α1xk−1y + · · · + αkyk), where R is a small power of P.
Write T r for the r-dimensional torus.
We are particularly interested in estimating the mean value Ss(P, R) = Z Tk+1 |f(α; P, R)|2sdα, which, by orthogonality, counts the number of solutions of the auxiliary system s X m=1 (xk−i m yi m −˜ xk−i m ˜ yi m) = 0 (0 ≤i ≤k) (1.18) with xm, ym, ˜ xm, ˜ ym ∈A(P, R) (1 ≤m ≤s).
(1.19) This is done in Chapter 3 by building the apparatus of efficient differencing for polynomials of two variables, so that an iterative method like that of Wooley can be implemented. Thus we are able to derive estimates of the form Ss(P, R) ≪P 4s−k(k+1)+∆s+ε, (1.20) where ∆s →0 as k →∞. To consider the strength of such an estimate, let Js(h) denote the number of solutions of the system s X m=1 (xk−i m yi m −˜ xk−i m ˜ yi m) = hi (0 ≤i ≤k) with (1.19). Then one has Js(h) = Z Tk+1 |f(α; P, R)|2se(−α · h) dα ≤Ss(P, R).
On the other hand, upon recalling (1.9) one sees that P 4s ≪ X h |hi|≤sP k Js(h) ≪P k(k+1)Ss(P, R), 13 and thus Ss(P, R) ≫P 2s + P 4s−k(k+1), the first term arising from the diagonal solutions of (1.18). Hence the estimate (1.20) becomes nearly best possible as ∆s →0.
The following theorem provides a bound of the shape (1.20) by means of single efficient differencing.
Theorem 2. Let k ≥2 be a positive integer, and put r = £k+1 2 ¤ . Further, write s1 = k2 µ 1 −1 2k ¶ −1 + r, and let s be a positive integer with s ≥s1. Then for any ε > 0 there exists η = η(s, k, ε) such that whenever R ≤P η the estimate (1.20) holds with ∆s = k(k + 1) µ 1 −1 2k ¶ (s−s1)/r .
Note for example that if s ≥2k2 log k then we have ∆s ∼k2e−s/k2 ≤1. Whenever ∆s has the property that, for every ε > 0, there exists η = η(s, k, ε) such that (1.20) holds whenever R ≤P η, we say that ∆s is an admissible exponent.
We remark for comparison that Arkhipov, Karatsuba, and Chubarikov have obtained estimates for the number of solutions of the “complete” system s X m=1 (xi myj m −˜ xi m˜ yj m) = 0 (0 ≤i, j ≤k) with 1 ≤xm, ym, ˜ xm, ˜ ym ≤P (1 ≤m ≤s), which lead, via a standard argument, to admissible exponents for (1.18) behaving roughly like k3e−s/2k3, so that one must take s ≥6k3 log k in most applications.
In Chapter 3, we obtain the following sharper result as a consequence of repeated efficient differencing, yielding admissible exponents that decay in most cases roughly like k2e−3s/2k2.
Theorem 3. Write r = £k+1 2 ¤ , and put s0 = k(k + 1) and s1 = 4 3rk(log(4rk) −2 log log k).
14 Further, define ∆s = 4rke2−3(s−s0)/4rk, when 1 ≤s ≤s1, e4(log k)2 ¡ 1 −1 2k ¢(s−s1)/r , when s > s1.
Then there exists a constant K such that the exponent ∆s is admissible whenever k ≥K.
For smaller k, one often needs other methods to obtain reasonable admissible exponents, and this is illustrated for k = 3 in Chapter 5.
As was the case with Vinogradov’s Mean Value Theorem, our mean value esti-mates may be transformed into Weyl estimates by using the large sieve inequality.
Such estimates will be important to the analysis of the minor arcs in subsequent applications of the Hardy-Littlewood method. For example, we have Theorem 4. Define m to be the set of α ∈Rk+1 such that whenever ai ∈Z and q ∈N satisfy (a0, . . . , ak, q) = 1 and |qαi −ai| ≤P 1/2−kRk (0 ≤i ≤k) one has q > P 1/2Rk+1.
Then given ε > 0, there exists η = η(ε, k) such that whenever R ≤P η one has sup α∈m|f(α; P, R)| ≪P 2−σ1(k)+ε, where σ1(k)−1 ∼28 3 k3 log k as k →∞.
A slightly more general version of this theorem will be proved in Chapter 4, but the above version suffices for most of our applications.
1.6 Applications to Diophantine Problems Estimates of the type given in Theorem 4 have immediate applications to the problem of obtaining localized bounds for the fractional parts of polynomials in two variables. In particular, we have Theorem 5. Given α ∈Rk+1 and ε > 0, there exists N0 = N0(ε, k) such that whenever N > N0 one has min 1≤m,n≤N ||α0mk + α1mk−1n + · · · + αknk|| < N ε−ρ(k), where ρ(k)−1 ∼14 3 k3 log k as k →∞.
15 It is worth noting that when α0 = 0 or αk = 0 a superior result is obtained by specializing one of the variables and applying Wooley , Corollary 1.3, to the resulting polynomial in a single variable. Hence in this case one could take ρ(k)−1 ∼4k2 log k in Theorem 5, and so our results are only of interest when both α0 and αk are nonzero. A more precise expression for ρ(k) will be given in Chapter 4.
Next we consider a generalization of Waring’s problem posed by Arkhipov and Karatsuba . As Waring’s problem considers representations of a single integer as a sum of terms of the shape xk, the simplest form of degree k in a single variable, a natural generalization is to consider simultaneous representations of the shape xk−j 1 yj 1 + · · · + xk−j s yj s = nj (0 ≤j ≤k), (1.21) since xk, xk−1y, . . . , yk are the simplest forms of degree k in two variables. By the binomial theorem, we see that this is equivalent to representing the polynomial p(t) = k X j=0 µk j ¶ njtj (1.22) as a sum of s kth powers of linear polynomials. That is, we seek to write p(t) = (x1t + y1)k + · · · + (xst + ys)k (1.23) with xi, yi ∈N. We remark that the analogous problem over the complex numbers has been considered recently by algebraic geometers (see for example , ). By exploiting a surprising connection with the theory of partial differential operators, one finds that precisely s = § k+1 2 ¨ terms are required to guarantee a representation of the shape (1.23) for arbitrary polynomials of degree k over C. Working over the integers, however, an elementary counting argument shows that one in fact needs s ≫k2. Obviously, there will be no representations of the shape (1.23) if the relative sizes of the nj are sufficiently disparate. For example, if some one of the nj is large, then some one of the xi or yi must be large, which in turn forces the other nj to be large. Thus we will need to impose some conditions in order to obtain a result.
Definition. For fixed s and k, the polynomial p(t) defined by (1.22) is said to be locally representable if (1) there exist real numbers P, µ0, . . . , µk, and δ = δ(s, k, µ) > 0 such that ¯ ¯nj −P kµj ¯ ¯ < δP k (0 ≤j ≤k) (1.24) 16 and such that the system ηk−j 1 ξj 1 + · · · + ηk−j s ξj s = µj (0 ≤j ≤k) (1.25) has a non-singular real solution with 0 < ηi, ξi < 1, and (2) the system (1.21) has a non-singular p-adic solution for all primes p.
Now let G∗ 1(k) denote the least integer s such that, whenever the polynomial p(t) given by (1.22) is locally representable and n0, . . . , nk are sufficiently large, one has the global representation (1.23) for some natural numbers x1, . . . , xs and y1, . . . , ys.
Theorem 6. One has G∗ 1(k) ≤14 3 k2 log k + 10 3 k2 log log k + O(k2).
We note that Arkhipov and Karatsuba have previously outlined a program for obtaining bounds of the form G∗ 1(k) ≤Ck2 log k using the theory of multiple exponential sums over a complete interval developed in . Theorem 6 thus gives an explicit asymptotic version of this result, showing that one may take C ∼14/3.
In Chapter 5, we sketch a refined analysis that leads to Theorem 7. One has G∗ 1(3) ≤56.
Our final application concerns the density of rational lines on the hypersurface defined by an additive equation. Let F(x) be a form of degree k in s variables, with integer coefficients. In 1945, Brauer used a diagonalization argument to demonstrate the existence of an m-dimensional linear space on the hypersurface F(x) = 0 over some solvable extension of Q, provided that s is sufficiently large in terms of k and m. By refining Brauer’s method, Birch obtained the same result over Q for odd k in 1957. Unfortunately, the elementary methods of Brauer and Birch do not yield any reasonable estimates for the number of variables required, although explicit astronomical bounds have been given recently by Wooley . For small values of k, somewhat more satisfying results have been obtained by Lewis and Schulze-Pillot and Wooley , . Up to this point, however, no estimates have been provided for the density of rational lines on a given hypersurface.
In Chapter 4, we obtain an explicit upper bound for the number of variables required to guarantee the expected density of rational lines on the hypersurface F(x) = 0 in the case when F is an additive form of degree k. Let c1, . . . , cs be non-zero integers, and let Ls(P) denote the number of distinct lines of the form xt + y, with xi, yi ∈[−P, P] ∩Z, that lie on the hypersurface c1zk 1 + · · · + cszk s = 0.
(1.26) 17 Clearly, Ls(P) is related to the number of solutions of the system of equations c1xk−j 1 yj 1 + · · · + csxk−j s yj s = 0 (0 ≤j ≤k), (1.27) with xi, yi ∈[−P, P] ∩Z, so the theory of multiple exponential sums is again ap-plicable. Thus in Chapter 4 we are able to prove Theorem 8. Suppose that the system of equations (1.27) has a non-singular real solution and a non-singular p-adic solution for all primes p. Then one has Ls(P) ≫P 2s−k(k+1) for P sufficiently large, provided that s ≥14 3 k2 log k + 10 3 k2 log log k + O(k2).
We note that, when s is large in terms of k, the theory of a single additive equa-tion discussed in Section 1.4 shows that the hypersurface defined by (1.26) contains “trivial” lines, corresponding to the case where either xi = 0 or yi = 0 for each i in (1.27). By a trivial estimate, however, the number of such lines is O(P s). Hence in the situation of Theorem 8 we see that most of the points on (1.26) that lie on lines in fact lie on non-trivial lines.
It transpires that the p-adic solubility conditions imposed in the above theorems need only be checked for finitely many primes p, for we show in Chapter 4 using exponential sums that they do in fact hold whenever p > p0(k) and s ≥(k + 1)2.
While p-adic solubility issues were considered in detail by Arkhipov in his work on the Hilbert-Kamke problem, such considerations have largely been ignored in the results stated by Arkhipov and Karatsuba on the multidimensional analogue of Waring’s problem. It would therefore be desirable (and possibly quite difficult) to give necessary and sufficient conditions for, or prove unconditionally, the p-adic solubility of the systems (1.21) and (1.27) for small primes.
In the cubic case, we are able to establish a version of Theorem 8 that does not require any local solubility hypotheses. Thus in Chapter 5 we prove Theorem 9. Suppose that k = 3 and s ≥58. Then for P sufficiently large one has Ls(P) ≫P 2s−12.
It is worth noting that higher dimensional analogues of Theorems 6–9 would be accessible with a satisfactory theory of higher dimensional exponential sums. Thus in the analogues of Theorems 8 and 9 we would seek estimates for the density of rational linear spaces (e.g. planes) of some dimension m ≥2 that lie on (1.26). Some 18 of the technical apparatus for such a program has been laid by Arkhipov, Karatsuba, and Chubarikov in their treatment of d-fold exponential sums, but a substantial refinement of that analysis would be necessary in order to obtain explicit results. It may be possible to use the generality of their analysis as a model for extending the iterative methods we develop in Chapter 3 to d-fold exponential sums over smooth numbers.
CHAPTER II Simultaneous Diagonal Inequalities 2.1 Overview Our main goal in this chapter is the proof of Theorem 1. We begin by recalling some of the notation introduced in Section 1.4. Let λ1, . . . , λs and µ1, . . . , µs be real numbers such that for each i at least one of λi or µi is nonzero, and define the forms F(x) = λ1x3 1 + · · · + λsx3 s G(x) = µ1x2 1 + · · · + µsx2 s.
Further, let Θs(P) denote the number of solutions of the system of inequalities |F(x)| < (max |xi|)−σ1 |G(x)| < (max |xi|)−σ2 (2.1) with x1, . . . , xs ∈[1, P] ∩Z. The following is a quantitative version of Theorem 1.
Theorem 2.1. Let s ≥13, and let λ1, . . . , λs and µ1, . . . , µs be real numbers such that for some i and j the ratios λi/λj and µi/µj are algebraic and irrational. Further, suppose that (a) F(x) has at least s −4 variables explicit, (b) G(x) has at least s −5 variables explicit, (c) the system F(x) = G(x) = 0 has a non-singular real solution, and (d) one has σ1 + σ2 < 1 12.
Then one has Θs(P) ≫P s−5−σ1−σ2 for P sufficiently large.
When either F or G has a large number of zero coefficients, we can exploit results for a single inequality to obtain a result in which the conditions on the coefficient ratios are not needed.
19 20 Theorem 2.2. Let λ1, . . . , λs and µ1, . . . , µs be real numbers.
The simultaneous inequalities (2.1) have infinitely many solutions in rational integers provided that (a) F(x) has at least 7 variables explicit, (b) G(x) has at least 5 variables explicit, (c) the system F(x) = G(x) = 0 has a non-singular real solution, and (d) one of the following holds: (i) at least 4 of the λi are zero and max(σ1, σ2) ≤10−5, or (ii) at least 7 of the µi are zero and σ1 ≤10−4.
We remark that condition (b) is not actually needed to prove the stated ver-sion of Theorem 2.2; however, the condition arises naturally in discussing possible improvements on condition (d)(ii), so we state it for convenience.
In Section 2.2, we deduce Theorem 2.2 in an elementary manner from results on a single Diophantine inequality. We also consider a refinement of condition (d)(ii) that would follow from improvements in our understanding of cubic inequalities.
We then prove Theorem 2.1 in Sections 2.3, 2.4, and 2.5, using a two-dimensional version of the Davenport-Heilbronn method. We show that Θs(P) ≫ Z ∞ −∞ Z ∞ −∞ H(α)K(α) dα, where H(α) is a suitable product of exponential sums (many of which we restrict to smooth numbers) and K(α) is a product of two kernels similar to (1.15). We then dissect the plane in analogy with the one-dimensional Davenport-Heilbronn dissec-tion discussed in Section 1.4. The success of our minor arc analysis depends heavily on an estimate of Wooley for the 10th moment of a certain exponential sum over smooth numbers and also on a result of R. Baker relating the size of a certain exponential sum to the existence of good rational approximations to the coefficients of its argument. The treatment of the major arc is essentially straightforward using the ideas of Wooley .
Finally, in Section 2.6, we discuss the possibility of weakening some of the hy-potheses imposed in Theorems 2.1 and 2.2.
Throughout our analysis, implicit constants in the notations of Vinogradov and Landau may depend on the coefficients λ1, . . . , λs and µ1, . . . , µs, the exponents σ1 and σ2, and also on any parameters denoted by ε or δ.
The material of this chapter appears in the author’s forthcoming publication .
21 2.2 Forms with Many Zero Coefficients Here we prove Theorem 2.2 using results on a single inequality. We first consider the case (d)(i). The argument is similar to that given in Lemmata 6.3, 6.4, and 6.5 of Wooley , but it also incorporates the recent work of Baker, Br¨ udern, and Wooley on cubic inequalities in 7 variables and makes use of a result of Birch and Davenport on small solutions of quadratic inequalities in 5 variables. We start with an analogue of , Lemma 6.3.
Lemma 2.2.1. Suppose that there is a rearrangement of the variables x1, . . . , xs such that λi = 0 for i = 1, . . . , 4 and µ1, . . . , µ4 are not all of the same sign. Then Theorem 2.2 holds in the case (d)(i).
Proof. Let σ = 1.43 × 10−4 and δ = 1 10σ. It is easily seen that the main theorem of holds with the above value of σ, although the result is stated with a slightly smaller exponent. Thus by condition (a) of Theorem 2.2, there exist infinitely many (s −4)-tuples of integers (a5, . . . , as) such that ¯ ¯λ5a3 5 + · · · + λsa3 s ¯ ¯ < (max |ai|)−σ.
(2.2) Now put Mi = µi for i = 1, . . . , 4, and put M5 = µ5a2 5 + · · · + µsa2 s.
If |M5| < (max |ai|)−δ, then we can take x1 = · · · = x4 = 0 and xi = ai for i = 5, . . . , s. Otherwise, by the main theorem of we can find (for max |ai| sufficiently large) integers u1, . . . , u5, not all zero, such that ¯ ¯M1u2 1 + · · · + M5u2 5 ¯ ¯ < (max |ai|)−δ (2.3) and |M1u2 1| + · · · + |M5u2 5| ≪(max |ai|)δ(4+5δ)|M1 · · · M5|1+δ.
But M5 ≪(max |ai|)2, so that |uj| ≪(max |ai|)1+ δ 2(6+5δ) (j = 1, . . . , 4) and |u5| ≪(max |ai|) δ 2(6+5δ).
Hence on putting x = (u1, . . . , u4, u5a5, . . . , u5as), we have max |xi| ≪(max |ai|)1+ δ 2(6+5δ) 22 and |F(x)| < |u5|3(max |ai|)−σ ≪(max |ai|) 3δ 2 (6+5δ)−σ.
Thus on taking ε < 2σ −3δ(6 + 5δ) 2 + δ(6 + 5δ) we see that for max |ai| sufficiently large one has |F(x)| < (max |xi|)−ε, and so we may take σ1 = 1.429 × 10−5. Moreover, on taking γ < 2δ 2 + δ(6 + 5δ) we have |G(x)| < (max |ai|)−δ < (max |xi|)−γ for max |ai| sufficiently large, so we may take σ2 = 1.429 × 10−5.
When the hypothesis of Lemma 2.2.1 is not satisfied, we need some additional control over the solution to our cubic inequality (2.2) in order to guarantee that the quadratic in (2.3) is indefinite. Specifically, we require the following analogue of , Lemma 6.4.
Lemma 2.2.2. Let λ1, . . . , λt (t ≥7) be non-zero real numbers, and suppose that (η1, . . . , ηt) is a real solution of the equation λ1x3 1 + · · · + λtx3 t = 0 with 0 < ηi < 1 for all i. Then for any α ∈(0, 1) and P > P0(η, λ, α), there exist integers y1, . . . , yt such that |λ1y3 1 + · · · + λty3 t | < (max |yi|)−σ, where σ = 1.43 × 10−4 and (1 −α)ηiP < yi ≤(1 + α)ηiP (i = 1, . . . , t).
(2.4) 23 Proof. If the λi are all in rational ratio, then the result follows from Lemma 6.4 of . Otherwise, we follow through the analysis of , restricting the ranges of summation on the generating functions so that only values of the variables satisfying (2.4) are included. All of the required estimates continue to hold, with only the major arc analysis requiring a slight modification.
Now we can complete the proof of case (d)(i) by arguing as in the proof of , Lemma 6.5. Suppose that at least 4 of the λi are zero, and rearrange variables so that λ1, . . . , λt ̸= 0 and λi = 0 for i = t+1, . . . , s. By condition (c) and the argument of , Lemma 6.2, we may assume that the equations F(x) = G(x) = 0 have a real solution (η1, . . . , ηs) with all of the ηi non-zero, and then on replacing λi by −λi if necessary and using homogeneity we may assume that 0 < ηi < 1 2 for all i. Further, by Lemma 2.2.1, we may assume that µt+1, . . . , µs are all positive, so that µ1η2 1 + · · · + µtη2 t = −(µt+1η2 t+1 + · · · + µsη2 s) = −C < 0.
Let α, P, and (y1, . . . , yt) be as in Lemma 2.2.2 with α < 2C 3t (max |µi|)−1 , and put M = µ1y2 1 + · · · + µty2 t . Then |M + CP 2| ≤P 2(α2 + 2α) t X i=1 |µiη2 i | < 1 2CP 2, so that M < −1 2CP 2 < 0.
Now let δ = 1.43 × 10−5 as before. If |M| < P −δ, then we can take xi = yi for i = 1, . . . , t and xt+1 = · · · = xs = 0. Otherwise, for P sufficiently large, we may use the result of as in the proof of Lemma 2.2.1 to find integers vt, . . . , vs, not all zero, with |vt| ≪P δ 2 (6+5δ) and |vi| ≪P 1+ δ 2 (6+5δ) (i = t + 1, . . . , s) such that |Mv2 t + µt+1v2 t+1 + · · · + µsv2 s| < P −δ.
Proceeding exactly as in the proof of Lemma 2.2.1, we find that x = (y1vt, . . . , ytvt, vt+1, . . . , vs) 24 satisfies (2.1) with σ1 = σ2 = 10−5, and this completes the proof of Theorem 2.2 in the case (d)(i).
The case (d)(ii) of Theorem 2.2 follows immediately from the results of , and this completes the proof of the theorem.
We now investigate the possibility of reducing the number of zero coefficients required by condition (d)(ii) from 7 to 6, in accordance with and . Br¨ udern , improving on a result of Pitman and Ridout , has shown that if λ1, . . . , λ9 are real numbers with |λi| ≥1 for all i then there exist integers x1, . . . , x9 satisfying |λ1x3 1 + · · · + λ9x3 9| < 1 and 0 < 9 X i=1 |λix3 i | ≪δ |λ1 · · · λ9|1+δ.
(2.5) Unfortunately, in order to use this result in an argument like the one in Lemma 2.2.1 we would have to assume that G(x) had at least eight zero coefficients, and in this situation we would do better to apply the results of . Suppose, however, that the above result held with 7 variables instead of 9. Then condition (d)(ii) of Theorem 2.2 could be replaced by (d)(ii)′ at least 6 of the µi are zero and max(σ1, σ2) ≤10−2.
The argument resembles the one above, but an argument like the one ensuing from Lemma 2.2.2 will not be necessary since the quadratic under consideration there will be replaced by a cubic.
Proceeding just as in Lemma 2.2.1, we fix σ < 1/10 and δ = 1/70. After rear-ranging variables, we may assume that µ1 = · · · = µ6 = 0. Now by condition (b) of Theorem 2.2 and an easily obtained quantitative version of the classical Davenport-Heilbronn Theorem, we see that there exist infinitely many (s −6)-tuples of integers (a7, . . . , as) such that ¯ ¯µ7a2 7 + · · · + µsa2 s ¯ ¯ < (max |ai|)−σ.
Now put Λi = λi for i = 1, . . . , 6, and put Λ7 = λ7a3 7 + · · · + λsa3 s.
If |Λ7| < (max |ai|)−δ, then we can take x1 = · · · = x6 = 0 and xi = ai for i = 7, . . . , s.
Otherwise, by our hypothesis, we can find (for max |ai| sufficiently large) integers u1, . . . , u7, not all zero, such that ¯ ¯Λ1u3 1 + · · · + Λ7u3 7 ¯ ¯ < (max |ai|)−δ 25 and |Λ1u3 1| + · · · + |Λ7u3 7| ≪(max |ai|)δ(6+7δ)|Λ1 · · · Λ7|1+δ.
But Λ7 ≪(max |ai|)3, so that |uj| ≪(max |ai|)1+ δ 3(9+7δ) (j = 1, . . . , 6) and |u7| ≪(max |ai|) δ 3(9+7δ).
Hence on putting x = (u1, . . . , u6, u7a7, . . . , u7as), we have max |xi| ≪(max |ai|)1+ δ 3(9+7δ), so on taking γ < 3δ 3 + δ(9 + 7δ) we have |F(x)| < (max |ai|)−δ < (max |xi|)−γ.
Furthermore, if ε < 3σ −2δ(9 + 7δ) 3 + δ(9 + 7δ) then we have |G(x)| < |u7|2(max |ai|)−σ ≪(max |ai|) 2δ 3 (9+7δ)2−σ, whence for max |ai| sufficiently large |G(x)| < (max |xi|)−ε.
Thus we may take σ1 = σ2 = 1.2 × 10−2.
We note that throughout our arguments there is some freedom in the choice of the parameter δ, and we have generally chosen it so as to give roughly the same permissible values for σ1 and σ2. If so desired, one can alter δ in favor of one exponent or the other and in fact obtain a region of permissible values similar in shape to (but smaller than) the region in Theorem 2.1(d). We do not pursue this refinement here.
26 2.3 The Davenport-Heilbronn Method We now set up a two-dimensional version of the Davenport-Heilbronn method, which we will use to prove Theorem 2.1. We may assume (after rearranging variables) that the the first m of the µi are zero, that the last n of the λi are zero, and that the remaining h = s −m −n indices have both λi and µi nonzero. Then when s ≥13 we have by conditions (a) and (b) of Theorem 2.1 that 0 ≤m ≤5, 0 ≤n ≤4, and h ≥4.
(2.6) Furthermore, we may suppose that λI/λJ and µI/µJ are algebraic irrationals, where I = m + h −2, J = m + h −1, and K = m + h.
Let ε be a small positive number, and choose η > 0 sufficiently small in terms of ε.
Take P to be a large positive number, put R = P η, and let A(P, R) = {n ∈[1, P] ∩Z : p|n, p prime ⇒p ≤R}.
Write α = (α, β), and define generating functions Fi(α) = X 1≤x≤P e(λiαx3 + µiβx2) (2.7) and fi(α) = X x∈A(P,R) e(λiαx3 + µiβx2).
(2.8) It will also be convenient to write gi(α) = fi(α, 0) and Hi(β) = Fi(0, β).
According to Davenport , for every integer r there exists a real-valued even kernel function K of one real variable such that K(α) ≪min(1, |α|−r) (2.9) and Z ∞ −∞ e(αt)K(α) dα = 0, if |t| ≥1, ∈[0, 1], if |t| ≤1, = 1, if |t| ≤1 3.
(2.10) 27 We set K(α) = K(αP −σ1)K(βP −σ2).
Now let N(P) be the number of solutions of (2.1) with xi ∈A(P, R) (i = 1, . . . , m + h −3) and 1 ≤xi ≤P (i = m + h −2, . . . , s).
By a familiar argument, N(P) is bounded below by P −σ1−σ2R(P), where R(P) = Z ∞ −∞ Z ∞ −∞ F(α)G(α)H(α)K(α) dα, (2.11) F(α) = m+h−3 Y i=1 fi(α), H(α) = m+h Y i=m+h−2 Fi(α), and G(α) = s Y i=m+h+1 Fi(α).
We dissect the plane into three main regions, imitating the standard dissection of the real line used in the treatment of a single inequality. The trivial region is defined by t = {α : |α| > P σ1+ε or |β| > P σ2+ε}, (2.12) the major arc by M = {α : |α| ≤P −9/4 and |β| ≤P −5/4}, (2.13) and the minor arcs by m = R2 \ (t ∪M).
(2.14) Our plan is to show that R(P) ≫P s−5, with the main contribution coming from the major arc. For r sufficiently large in terms of ε, it follows easily from (2.9) and (2.12) that the contribution to R(P) from the trivial region is o(P s−5). In the next section, we consider a finer dissection of the minor arcs, which allows us to show that their contribution to R(P) is also o(P s−5), provided that σ1 and σ2 are confined to the region specified in Theorem 2.1. Finally, in Section 2.5, we apply standard methods to deal with the major arc.
28 2.4 The Minor Arcs We begin by bounding the integral (2.11) in terms of others having somewhat more standard forms. We start by choosing a finite covering of m by unit squares of the form [c, c + 1] × [d, d + 1]. For n ⊂m, let Un denote the square for which the integral ZZ n∩Un |F(α)G(α)H(α)| dα is maximal, and write n∗= n ∩Un. Then for r > 1 it follows from (2.9) that ZZ n |F(α)G(α)H(α)K(α)| dα ≪P σ1+σ2 ZZ n∗ |F(α)G(α)H(α)| dα.
(2.15) Furthermore, by arguing as in the proof of Lemma 7.3 of Wooley , we see that ZZ n∗ |F(α)G(α)| dα ≪ ZZ n∗ |fi(α)|h−3 |gj(α)|m |Hk(β)|n dα (2.16) for some i, j, and k (depending on n) satisfying m + 1 ≤i ≤m + h, 1 ≤j ≤m, and m + h + 1 ≤k ≤s.
In the course of an argument in which n is fixed, we will employ the abbreviations f = |fi(α)|, g = |gj(α)|, and H = |Hk(β)|.
Finally, on recalling (2.6) and again mimicking the arguments of , we obtain f h−3gmHn ≪P s−13 ¡ f 10 + f uH10−u + guH10−u + f 10−ugu¢ (2.17) whenever 5 ≤u ≤6. For convenience, we introduce the notation Q = P s−13+σ1+σ2.
(2.18) We are now in a position to make use of certain mean value estimates developed in Wooley , . Those that we need are recorded for reference in the following lemma.
Lemma 2.4.1. Suppose that m + 1 ≤i ≤m + h, 1 ≤j ≤m, and m + h + 1 ≤k ≤s.
Then for any unit square U = [c, c + 1] × [d, d + 1], we have 29 (i) RR U |fi(α)|10 dα ≪P 17/3+ε, (ii) RR U |fi(α)|6 |Hk(β)|4 dα ≪P 21/4+ε, (iii) RR U |gj(α)|6 |Hk(β)|4 dα ≪P 21/4+ε, (iv) RR U |fi(α)|4 |gj(α)|6 dα ≪P 21/4+ε, (v) RR U |fi(α)|14 dα ≪P 9, (vi) RR U |fi(α)|8 |Hk(β)|5 dα ≪P 8, (vii) RR U |gj(α)|8 |Hk(β)|5 dα ≪P 8, (viii) RR U |fi(α)|6 |gj(α)|8 dα ≪P 9.
Proof. Part (i) follows from Theorem 2 of Wooley on considering the underlying Diophantine equations and making a change of variables. Parts (iii), (v), and (vii) follow from the corresponding parts of Lemmata 7.2, 9.1, and 9.4 of Wooley on making a change of variables and noting that the additional restrictions imposed on the variable ranges in that paper can be removed without affecting the arguments.
For the remaining parts, we use the idea of the proof of Lemma 9.1(i) of in a manner typified by (ii): Write sm(x, y) = (xm 1 −ym 1 ) + (xm 2 −ym 2 ) + (xm 3 −ym 3 ) and H(β) = X 1≤x≤P e(βx2).
Then on making the change of variables α′ = λiα and β′ = µkβ we have ZZ U |fi(α)|6 |Hk(β)|4 dα ≪ ZZ U′ X x,y e µ s3(x, y)α + µi µk s2(x, y)β ¶ |H(β)|4 dα dβ, where the summation is over x and y with xi, yi ∈A(P, R) and where U′ = [m3, n3]× [m2, n2] for some integers mj and nj with nj −mj ≪1. If we now let c(x, y) = e µ µi µk s2(x, y)β ¶ , 30 then since c(x, y) is unimodular we obtain ZZ U |fi(α)|6 |Hk(β)|4 dα ≪ Z n2 m2 ÃX x,y c(x, y) Z n3 m3 e(s3(x, y)α)dα !
|H(β)|4 dβ ≪ P 13/4+ε Z 1 0 |H(β)|4 dβ ≪P 21/4+ε on using Theorem 4.4 of Vaughan and considering the underlying Diophantine equations.
Lemma 2.4.1 allows us to handle regions of m on which H is suitably bounded.
Fortunately, when FI, FJ, or FK is large, we also obtain a great deal of information from a theorem of Baker , a special case of which is recorded below.
Lemma 2.4.2. Let P > P0(ε) and A > P 3/4+ε. If |Fi(α)| ≥A for some i = I, J, or K, then there exists a natural number q < P 3+εA−3 and integers a and b with (q, a, b) = 1 such that |λiαq −a| < P εA−3 and |µiβq −b| < P 1+εA−3.
Proof. This is Theorem 5.1 of with T = P 3/4+ε, M = 1, and k = 3.
Lemma 2.4.2 suggests further dissecting m according to the behavior of FI, FJ, and FK. Thus we start by defining e = {α ∈m : |Fi(α)| ≤P 3/4+ε for i = I, J, K}.
Now let f(I) = {α ∈m : |FI(α)| > P 3/4+ε, max(|FJ(α)| , |FK(α)|) ≤P 3/4+ε}, define f(J) and f(K) likewise, and put f = f(I) ∪f(J) ∪f(K).
Similarly, let g(I) = {α ∈m : |FI(α)| ≤P 3/4+ε, min(|FJ(α)| , |FK(α)|) > P 3/4+ε}, define g(J) and g(K) likewise, and put g = g(I) ∪g(J) ∪g(K).
Finally, define h = {α ∈m : |Fi(α)| > P 3/4+ε for i = I, J, K}.
31 The set e can be handled quite easily. Using (2.15)–(2.18) and Lemma 2.4.1, we obtain ZZ e |FGHK| dα ≪ Q ¡ P 3/4+ε¢3 ZZ Ue ¡ f 10 + f 6H4 + g6H4 + f 4g6¢ dα ≪ P s−13+σ1+σ2+9/4+3ε ¡ P 17/3+ε + P 21/4+ε¢ = o(P s−5), provided that σ1 + σ2 < 1/12, since ε can be chosen arbitrarily small.
The rational approximations provided by Lemma 2.4.2 allow us to incorporate major arc techniques along the lines of Br¨ udern and in dealing with the sets f, g, and h. For this we require some additional definitions and lemmata. Define M(q, a, b) = {α ∈[0, 1]2 : |qα −a| < P −9/4 and |qβ −b| < P −5/4}, M = [ 0≤a,b≤q 6, we have X q≤X S∗ t (q) ≪1.
Proof. Using Lemma 10.4 of Wooley and proceeding as in Lemma 2.11 of Vaughan , one sees that S∗ t (q) is multiplicative, so X q≤X S∗ t (q) ≤ Y p à 1 + ∞ X h=1 S∗ t (ph) !
.
(2.19) Whenever (ph, a, b) = 1, we have S(ph, a, b) ≪p2h/3+ε 32 by Theorem 7.1 of Vaughan , but in the case that (b, p) = 1 it follows from Theorem 1 of Loxton and Vaughan that in fact S(ph, a, b) ≪ph/2.
Thus we have S∗ t (ph) = p−ht X 1≤a,b≤ph (p,b)=1 ¯ ¯S(ph, a, b) ¯ ¯t + p−ht X 1≤a,b≤ph (ph,a,b)=1 (p,b)>1 ¯ ¯S(ph, a, b) ¯ ¯t ≪ p−ht¡ p2h+ht/2 + p2h−1+2ht/3+tε¢ , whence for t > 6 we have ∞ X h=1 S∗ t (ph) ≪p−1−δ for some δ > 0, and the result now follows immediately from (2.19).
Write F(α) = X 1≤x≤P e(αx3 + βx2) (2.20) and v(α) = Z P 0 e(αγ3 + βγ2) dγ.
(2.21) The following lemma provides a useful refinement of , Lemma 9.2.
Lemma 2.4.4. For t > 6, we have ZZ M |F(α)|t dα ≪P t−5.
Proof. When α ∈M(q, a, b), write ξ = (ξ3, ξ2) = (α −a/q, β −b/q) and V (α) = V (α; q, a, b) = q−1S(q, a, b)v(ξ).
Then for α ∈M(q, a, b) we have by Lemma 4.4 of Baker that F(α) = V (α) + O(q2/3+ε).
33 Hence if M1 denotes the subset of M on which |V (α)| ≤q2/3+ε, then we have ZZ M1 |F(α)|t dα ≪ X q≤P 3/4 (q2/3+ε)tP −7/2 ≪P t−5, provided that t > 9/2. For α ∈M2 = M \ M1, we have |V (α)| > q2/3+ε and hence |F(α)| ≪|V (α)|. Moreover, by Theorem 7.3 of Vaughan , we have v(ξ) ≪P(1 + P 2|ξ2| + P 3|ξ3|)−1/3 ≪P(1 + P 2|ξ2|)−1/6(1 + P 3|ξ3|)−1/6, and on combining this with Lemma 2.4.3 we obtain ZZ M |V (α)|tdα ≪P t−5 X q≤P 3/4 S∗ t (q) ≪P t−5 whenever t > 6. Thus we haveZZ M2 |F(α)|tdα ≪P t−5 for t > 6, and this completes the proof.
The sets f and g can now be handled with little difficulty by applying major arc treatments to one or two of the variables. The key observation is that Baker’s Theorem (Lemma 2.4.2) allows us to bound an integral of |Fi(α)|t over f(i)∗or g(j)∗ (j ̸= i) in terms of the integral considered in the previous lemma.
Using (2.15)–(2.18) as on e, we obtain for some i = I, J, or K that ZZ f |FGHK| dα ≪Q ¡ P 3/4+ε¢2 ZZ f(i)∗ |Fi| ¡ f 10 + f 6H4 + g6H4 + f 4g6¢ dα.
Then by H¨ older’s inequality we have ZZ f(i)∗ |Fi| f 10dα ≪ ZZ f(i)∗ |Fi|7 dα 1/7 ZZ Uf f 10dα 1/2 ZZ Uf f 14dα 5/14 , and by Lemma 2.4.1 we have ZZ f(i)∗ |Fi| ¡ f 6H4 + g6H4 + f 4g6¢ dα ≪P 25/4+ε.
Hence on using Lemmata 2.4.1, 2.4.2, and 2.4.4, together with a change of variables, we find that ZZ f |FGHK| dα ≪P s−13+σ1+σ2+3/2+2ε ¡ P 19/3+ε + P 25/4+ε¢ = o(P s−5), 34 provided that σ1 + σ2 < 1/6.
Proceeding similarly but instead taking u = 40/7 in (2.17), we have for some i ̸= j among I, J, and K that ZZ g |FGHK| dα ≪ QP 3/4+ε ZZ g(i)∗ |Fj|2 ¡ f 10 + f 40 7 H 30 7 + g 40 7 H 30 7 + f 30 7 g 40 7 ¢ dα ≪ QP 3/4+ε ZZ g(i)∗ |Fj|7 dα 2/7 ³ I 5/7 1 + I 5/7 2 + I 5/7 3 + I 5/7 4 ´ , where I1 = ZZ Ug f 14dα, I2 = ZZ Ug f 8H6dα, I3 = ZZ Ug g8H6dα, I4 = ZZ Ug f 6g8dα.
Thus we have ZZ g |FGHK| dα ≪P s−13+σ1+σ2+3/4+ε ¡ P 7¢ = o(P s−5), provided that σ1 + σ2 < 1/4.
The set h is somewhat more difficult to deal with, and it is here that we make use of the hypothesis that λI/λJ and µI/µJ are algebraic irrationals. We divide h into two main components, h1 = {α ∈h : |α| ≥P −9/4+ε} and h2 = h \ h1, and we further subdivide h∗ 1 and h∗ 2 into O ((log P)2) dyadic subsets of the form hi(A, B) = {α ∈h∗ i : A < |FI(α)| ≤2A, B < |FJ(α)| ≤2B}.
We also write h(A, B) = h1(A, B) ∪h2(A, B).
We now use a method introduced by Baker to give an upper bound for the Lebesgue measure of hi(A, B). If α ∈h(A, B), then by Lemma 2.4.2 there exist natural numbers qI < P 3+εA−3, qJ < P 3+εB−3, qK < P 3/4 (2.22) 35 and integers ai, bi with (qi, ai, bi) = 1 for i = I, J, K such that |λIαqI −aI| < P εA−3, |µIβqI −bI| < P 1+εA−3; (2.23) |λJαqJ −aJ| < P εB−3, |µJβqJ −bJ| < P 1+εB−3; (2.24) and |λKαqK −aK| < P −9/4, |µKβqK −bK| < P −5/4.
(2.25) Notice that the inequalities (2.23) and (2.24) restrict α to lie in a box BI about the point (aI/(λIqI), bI/(µIqI)) with meas(BI) ≪q−2 I P 1+2εA−6 (2.26) and at the same time in a box BJ about (aJ/(λJqJ), bJ/(µJqJ)) with meas(BJ) ≪q−2 J P 1+2εB−6.
(2.27) We first obtain a lower bound for qIqJ. As in the proof of Lemma 11.1 of Vaughan , it follows from (2.23) and (2.24) that for α ∈h1 we have ¯ ¯ ¯ ¯ λI λJ −aIqJ aJqI ¯ ¯ ¯ ¯ ≪P −9/4, whereas by a well-known theorem of Roth we have ¯ ¯ ¯ ¯ λI λJ −aIqJ aJqI ¯ ¯ ¯ ¯ ≫ 1 |aJqI|2+ε, so that |aJqI| ≫P 9/8−ε. Similarly, for α ∈h2 we have 1 |bJqI|2+ε ≪ ¯ ¯ ¯ ¯ µI µJ −bIqJ bJqI ¯ ¯ ¯ ¯ ≪P −5/4, and hence |bJqI| ≫P 5/8−ε. Thus on using (2.23) and (2.24) and recalling the defin-itions (2.12)–(2.14) we obtain qIqJ ≫ P 9/8−σ1−2ε, if α ∈h1 P 5/8−σ2−2ε, if α ∈h2.
(2.28) Next we observe that when α ∈h1(A, B) there are O(P 9+3εA−9) corresponding triples (qI, aI, bI) satisfying (2.22) and (2.23). Alternatively, there are O(P 9+3εB−9) 36 triples (qJ, aJ, bJ) satisfying (2.22) and (2.24). On combining this with (2.26), (2.27), and (2.28) we obtain meas(h1(A, B)) ≪P 71/8+σ1+7ε(AB)−15/2.
(2.29) When α ∈h2(A, B) we necessarily have aI = aJ = 0 for P sufficiently large, so proceeding as above gives meas(h2(A, B)) ≪P 51/8+σ2+6ε(AB)−6.
(2.30) On applying H¨ older’s inequality and Lemma 2.4.1 as before and writing L = (log P)2, we find that for some A and B ZZ h1 |FGHK| dα ≪ QL ZZ h1(A,B) |FIFJFK| ¡ f 10 + f 40 7 H 30 7 + g 40 7 H 30 7 + f 30 7 g 40 7 ¢ dα ≪ QP ε ZZ h∗ 1 |FK| 105 16 dα 16 105 ZZ h1(A,B) |FIFJ| 15 2 dα 2 15 ¡ P 9¢ 5 7 .
Thus by (2.29) and Lemma 2.4.4 we have ZZ h1 |FGHK| dα ≪P s−13+ 45 7 + 5 21+ 2 15( 71 8 )+ 17 15σ1+σ2+2ε = o(P s−5), provided that 17 15σ1 + σ2 < 3 20.
Since h2 is a thin strip along the β-axis, we save a factor of P σ1 in the analysis leading to (2.15), but the treatment is otherwise similar to the above. On writing Q′ = P s−13+σ2, we have ZZ h2 |FGHK| dα ≪ Q′L ZZ h2(A,B) |FIFJFK| ¡ f 10 + f 40 7 H 30 7 + g 40 7 H 30 7 + f 30 7 g 40 7 ¢ dα ≪ P s−13+σ2+ε ZZ h∗ 2 |FK| 42 5 dα 5 42 ZZ h2(A,B) |FIFJ|6 dα 1 6 ¡ P 9¢ 5 7 , whence by (2.30) we obtain ZZ h2 |FGHK| dα ≪P s−13+ 45 7 + 17 42+ 1 6 ( 51 8 )+ 7 6σ2+2ε = o(P s−5), provided that 7 6σ2 < 5 48. It is easily seen that these last two inequalities are less restrictive than the one appearing in condition (d) of Theorem 2.1.
37 2.5 The Major Arc As it stands, the major arc M is too large to allow satisfactory approximation of the exponential sums fi(α), so we must do some pruning. Specifically, let W be a parameter at our disposal, and let N = {α : |α| ≤WP −3 and |β| ≤WP −2}.
(2.31) Then as in Lemma 9.2 of Wooley , we have for t > 9 that ZZ M\N |Fi(α)|t dα ≪W −σP t−5 for i = I, J, K and some σ > 0. Thus by using (2.17) and Lemma 2.4.1 as in the treatment of g and h in the previous section, we have for some i = I, J, or K that ZZ M\N |FGHK| dα ≪ P s−13 ZZ M\N |Fi(α)|21/2 dα 2/7 P 45/7 ≪ P s−5W −σ′.
It remains to deal with the pruned major arc N. Let vi(α) = Z P 0 e(λiαγ3 + µiβγ2) dγ (2.32) and wi(α) = Z P R ρ µ log γ log R ¶ e(λiαγ3 + µiβγ2) dγ, (2.33) where ρ(x) is Dickman’s function (see Vaughan , chapter 12). Then for α ∈N, we obtain from Theorem 7.2 of that Fi(α) = vi(α) + O(W) and from Lemma 8.5 of that fi(α) = wi(α) + O(WP/ log P).
Now on taking W = (log P)1/4 it follows that ZZ N FGHK dα = ZZ N Ãm+h−3 Y i=1 wi(α) !Ã s Y i=m+h−2 vi(α) !
K(α) dα + O(P s−5W −1).
38 Furthermore, we may extend the integration over all of R2, as the bounds for vi and wi contained in Lemma 8.6 of imply that ZZ R2\N Ãm+h−3 Y i=1 wi(α) !Ã s Y i=m+h−2 vi(α) !
K(α) dα ≪P s−5W −1.
Thus it remains to show that the singular integral J = Z ∞ −∞ Z ∞ −∞ Ãm+h−3 Y i=1 wi(α) !Ã s Y i=m+h−2 vi(α) !
K(α) dα satisfies J ≫P s−5. Multiplying out, we have J = Z ∞ −∞ Z ∞ −∞ Z B∗T ∗(γ) e(F(γ)α + G(γ)β) K(αP −σ1)K(βP −σ2) dγ dα dβ, where B∗= [R, P]m+h−3 × [0, P]n+3 and T ∗(γ) = m+h−3 Y i=1 ρ µlog γi log R ¶ .
On making the change of variables γ′ = γP −1, α′ = αP −σ1, β′ = βP −σ2 and applying Fubini’s Theorem, we obtain J = P s+σ1+σ2 Z B T(γ) ˆ K(F(γ)P 3+σ1) ˆ K(G(γ)P 2+σ2) dγ, (2.34) where we have written B = P −1B∗, T(γ) = T ∗(Pγ), and ˆ K(t) = Z ∞ −∞ e(αt)K(α) dα.
Now by condition (c) of Theorem 2.1 and the argument of Lemma 6.2 of Wooley , we can find a non-singular solution η to the equations F = G = 0 such that each ηi is non-zero. Then, after replacing λi by −λi if necessary and using homogeneity, 39 we may assume that η ∈(0, 1)s and hence that η lies in the interior of B when P is sufficiently large. Suppose that 6ηjηk(λjµkηj −λkµjηk) ̸= 0, and consider the map φ : Rs →Rs defined by φj = F(γ), φk = G(γ), and φi = γi (i ̸= j, k).
(2.35) By the inverse function theorem, there exist neighborhoods U of η and V of φ(η) such that φ maps U injectively onto V , and we may assume that U ⊂B. Now by (2.10) and the nonnegativity of ρ, the integrand in (2.34) is nonnegative, so we may restrict the integration over γ to the set U. Then on writing z = φ(γ), where φ is as in (2.35), we have by the change of variables theorem that J ≥P s+σ1+σ2 Z V T(φ−1(z)) ˆ K(zjP 3+σ1) ˆ K(zkP 2+σ2) ¯ ¯ ¯ ¯ dγ dz ¯ ¯ ¯ ¯ dz.
(2.36) Since meas(V ) ≫1, the projection of V onto zj contains the interval [0, 1 3P −3−σ1], and the projection of V onto zk contains the interval [0, 1 3P −2−σ2], provided that P is sufficiently large. Hence on restricting the range of integration in (2.36) and using (2.10) again, we obtain J ≫P s+σ1+σ2 Z S T(φ−1(z))dz, where meas(S) ≫P −5−σ1−σ2. Finally, on noting that T(γ) ≫ρ(1/η)m+h−3 ≫1 for γ ∈B, we obtain J ≫P s−5 as required. This completes the proof of Theorem 2.1.
2.6 A Discussion of Possible Improvements Here we discuss the possibility of weakening some of the conditions imposed on the forms F and G in Theorems 2.1 and 2.2. In view of the discussion of Wooley , §5, where it is shown that many conditions similar to ours are essentially best possible for the corresponding problem on equations, our observations will leave something to be desired. Nevertheless, we can show that at least some minimal conditions are necessary to ensure the solubility of (2.1).
For example, let F(x) = λ3x3 1 −x3 2 and G(x) = µ2x2 3 −x2 4, where λ and µ are positive real algebraic of degree 3 and 2, respectively, such that λ3 and µ2 are irrational. For instance, we may take λ = 1 + 3 √ 2 and µ = 1 + √ 2. Then it follows easily from Liouville’s Theorem that, for sufficiently small τ > 0, neither of the inequalities |F(x)| < τ, |G(x)| < τ 40 has a non-trivial solution in rational integers.
Of course, this example is easily generalized to produce forms F1, . . . , Ft of degrees k1, . . . , kt in 2t variables that do not take arbitrarily small values. Therefore, we must minimally require either s ≥5 total variables or at least 3 variables explicit in one of the two forms.
More realistically, in light of , Theorem 1, one might hope to be able to prove Theorem 2.1 with s = 13 but conditions (a) and (b) weakened so that F and G need only have 7 and 5 variables explicit, respectively, rather than 9 and 8. The latter numbers arise from the inequalities (2.6), on which the analytic argument in Sections 2.3–2.5 depends, but one may attempt to reduce these in the manner of and by using Theorem 2.2. Unfortunately, there are some difficulties with this approach in our situation. If F has exactly 7 or 8 variables explicit, then we may apply Theorem 2.2 to solve (2.1), but we must settle for the inferior values of σ1 and σ2 allowed by condition (d)(i) of that theorem, and we forfeit our estimate for the density of solutions. Moreover, if G has exactly 7 variables explicit and F has at least 10 variables explicit, then neither Theorem 2.1 nor Theorem 2.2 applies with s = 13. To avoid this difficulty, we may hope to reduce the number of zero coefficients required by condition (d)(ii) of the latter from 7 to 6, and we saw in Section 2.2 that a conditional result of this type could be obtained using hypothetical results on small solutions of cubic inequalities in 7 variables.
As mentioned in Section 2.1, condition (b) of Theorem 2.2 can be eliminated from the stated version of the theorem, but some form of it is likely to be necessary for any desirable refinement of (d)(ii). If a quantitative version of the result of Margulis on the Oppenheim conjecture were available, then we could reduce the 5 to 3 in condition (b) of our hypothetical version of Theorem 2.2, provided we assumed additionally that G is not a multiple of a form with integer coefficients. However, the methods of do not seem to hold much promise for obtaining such a result.
We can also investigate the possibility of reducing the total number of variables required. Although Theorem 2.1 could conceivably hold with as few as 5 variables, it does not seem possible for an analytic argument of the flavor given in Sections 2.3–2.5 to be successful with fewer than 11 variables. In the “ideal” situation that the first four mean values in Lemma 2.4.1 were bounded by P 5+ε, a simplified version of our analysis would allow us to prove a version of the theorem for s ≥12, possibly with a slightly different range of permissible values for σ1 and σ2.
Next we note that the existence of a non-trivial real solution to the equations F = G = 0 is a necessary condition for the system (2.1) to have infinitely many integer solutions. For, if the latter holds, then for arbitrary τ > 0 we can obtain (by rescaling an integer solution x with max |xi| sufficiently large) a real solution 41 η(τ) ∈[−1, 1]s of the inequalities |F| < τ, |G| < τ such that |ηi| = 1 for some i.
But the set S = {η ∈[−1, 1]s : |ηi| = 1 for some i} is compact, whence its image in R2 under the continuous map φ defined by F and G is compact. Hence φ(S) must contain the limit point (0, 0), which shows that the equations F = G = 0 have a non-trivial real solution.
Now let p be a prime with p ≡1 (mod 3), let c be a cubic nonresidue (mod p), and consider the forms F(x) = √ 2x3 1 + x3 2 + · · · + x3 7 + (x3 8 + cx3 9) + p(x3 10 + cx3 11) + p2(x3 12 + cx3 13), G(x) = √ 2x2 1 + x2 2 + · · · + x2 7 + x2 8.
It is easily checked that F and G satisfy all the conditions of Theorem 2.1, except that all real solutions to the simultaneous equations F = G = 0 are singular. More-over, the discussion of example (5.1) in Wooley shows that the simultaneous inequalities |F(x)| < 1, |G(x)| < 1 have no nontrivial integer solutions. Therefore, condition (c) of Theorem 2.1 cannot be weakened.
We conclude with some remarks on the assumption regarding algebraic irrational coefficient ratios in Theorem 2.1. First of all, if neither F nor G is a multiple of a form with integer coefficients and all the coefficients of F and G are nonzero, then it is easy to see that there is a pair of indices i and j such that both λi/λj and µi/µj are irrational. Next, if exactly one of the forms is a multiple of an integral form and this form has no zero coefficients, then we can solve the problem by obtaining a lower bound for the integral R1(P) = Z ∞ −∞ Z 1 0 F(α)G(α)H(α)K(αP −σ1) dβ dα or R2(P) = Z ∞ −∞ Z 1 0 F(α)G(α)H(α)K(βP −σ2) dα dβ, as the case may be, using a simplified version of our analysis, along with techniques from the one-dimensional Hardy-Littlewood and Davenport-Heilbronn methods. If F and G are both multiples of integral forms, then we may simply apply the argument 42 of Wooley to deduce Theorem 2.1. Thus in particular we observe that if all the coefficients of F and G are algebraic and nonzero, then no irrationality assumption on the coefficients is needed.
The algebraicity assumption allows us to use Roth’s Theorem in Section 2.4 to obtain the lower bounds (2.28), which are critical to our analysis of the sets hi(A, B).
The preferred approach to (2.28) would involve restricting P in terms of the denom-inators of simultaneous rational approximations λI/λJ ∼a/q and µI/µJ ∼b/q and then combining these approximations with (2.23) and (2.24), in analogy with the proof of , Lemma 11.1.
However, a difficulty arises from the possibility that (a, q) or (b, q) may be large, even though we can ensure that (q, a, b) = 1. It tran-spires that in this problematic case we can reduce the task to one of obtaining small solutions to “mixed” systems of the form |F(x)| < (max |xi|)−σ1, s X i=1 bix2 i = 0 or |G(x)| < (max |xi|)−σ2, s X i=1 aix3 i = 0, where the ai and bi are integers. Under suitable conditions, the number of solutions to these systems can be estimated as described above, using integrals like R1(P) and R2(P). However, in order to obtain bounds for the solutions in terms of the coefficients of the forms, we must now keep track of constants that were previously left implicit, and this would seem to require additional information regarding the nature of a real solution to the corresponding system of equations.
CHAPTER III Mean Values of Multiple Exponential Sums 3.1 Overview In this chapter, we obtain estimates for mean values of certain multiple exponen-tial sums over smooth numbers by extending the ideas of Vaughan and Wooley , . When P and R are positive integers, write A(P, R) = {n ∈[1, P] ∩Z : p|n, p prime ⇒p ≤R} for the set of R-smooth numbers up to P, and define the exponential sum f(α; P, R) = X x,y∈A(P,R) e(α0xk + α1xk−1y + · · · + αkyk).
(3.1) Further, define the mean value Ss(P, R) = Z Tk+1 |f(α; P, R)|2sdα, which, by orthogonality, counts the number of solutions of the auxiliary system s X m=1 (xk−i m yi m −˜ xk−i m ˜ yi m) = 0 (0 ≤i ≤k) (3.2) with xm, ym, ˜ xm, ˜ ym ∈A(P, R) (1 ≤m ≤s).
(3.3) When R is a power of P, one has the lower bound Ss(P, R) ≫P 2s + P 4s−k(k+1), so when s ≥1 2k(k +1) we hope to obtain upper bounds that are not too much larger than P 4s−k(k+1). If for every ε > 0 there exists η = η(s, k, ε) such that the estimate Ss(P, R) ≪P 4s−k(k+1)+∆s+ε 43 44 holds whenever R ≤P η, then we call ∆s an admissible exponent.
After discussing some preliminary results in Section 3.2, we develop our version of the Vaughan-Wooley iterative method in Sections 3.3 and 3.4. In Section 3.5, we are able to establish the following result using only single efficient differencing.
Theorem 3.1. Let k ≥2 be a positive integer, and put r = £k+1 2 ¤ . Further, write s1 = k2 µ 1 −1 2k ¶ −1 + r, and let s be a positive integer with s ≥s1. Then the exponent ∆s = k(k + 1) µ 1 −1 2k ¶ (s−s1)/r .
is admissible.
Notice in particular that if s ≥2k2 log k then admissible exponents obtained from Theorem 3.1 satisfy ∆s ∼k2e−s/k2 ≤1.
Finally, in Section 3.6, we make full use of the repeated efficient differencing apparatus to obtain the following sharper result.
Theorem 3.2. Write r = £k+1 2 ¤ , and put s0 = k(k + 1) and s1 = 4 3rk (log(4rk) −2 log log k) .
Further, define ∆s = 4rke2−3(s−s0)/4rk, when 1 ≤s ≤s1, e4(log k)2 ¡ 1 −1 2k ¢(s−s1)/r , when s > s1.
Then there exists a constant K such that the exponent ∆s is admissible whenever k ≥K.
Notice that the admissible exponents obtained from Theorem 3.2 decay in many cases of interest roughly like k2e−3s/2k2, whereas those obtained from Theorem 3.1 decay only like k2e−s/k2. Therefore, we will primarily rely on Theorem 3.2 when discussing the various applications of our mean value estimates for large k in Chapter 4. On the other hand, the ideas underlying Theorem 3.1 often suffice for smaller values of k, as we shall see in Chapter 5.
This chapter and the next are based on the author’s submitted manuscript .
45 3.2 Preliminary Lemmata Before embarking on the proofs of our mean value estimates, we need to make some preliminary observations. We start by showing that solutions of (3.2) in which some xj and yj or some ˜ xj and ˜ yj have a large common factor can effectively be ignored.
When γ > 0, let Ss(P, R; γ) be the number of solutions of (3.2) with (xj, yj) ≤P γ and (˜ xj, ˜ yj) ≤P γ for all j.
Lemma 3.2.1. For every γ > 0, one has Ss(P, R) ≪P 2s+ε + Ss(P, R; γ).
Proof. Write S′ s(P, R; γ) for the number of solutions of (3.2) with (xj, yj) > P γ or (˜ xj, ˜ yj) > P γ for some j, so that Ss(P, R) = Ss(P, R; γ) + S′ s(P, R; γ). Then we have S′ s(P, R; γ) = X d>P γ Z Tk+1 f(dkα; P/d, R)f(−α; P, R)|f(α; P, R)|2s−2 dα.
(3.4) Now suppose that S′ s(P, R; γ) ≥Ss(P, R; γ), so that Ss(P, R) ≤2S′ s(P, R; γ), and let λs = inf{λ : Ss(P, R) ≪P λ}.
If λs ≤2s, then we are done, so we may assume that λs > 2s. By applying H¨ older’s inequality to (3.4), we obtain Ss(P, R) ≪ X d>P γ µZ Tk+1 |f(dkα; P/d, R)|2s dα ¶ 1/2s µZ Tk+1 |f(α; P, R)|2s dα ¶ 1−1/2s , from which we deduce that Ss(P, R) ≪ Ã X d>P γ Ss(P/d, R)1/2s !
2s ≪P λs+γ(2s−λs)+ε for all ε > 0, since λs > 2s. This provides a contradiction for ε sufficiently small, so in fact we have S′ s(P, R; γ) < Ss(P, R; γ), and the conclusion of the lemma follows.
We next record an estimate for the number of solutions of an associated system of congruences. When f1, . . . , ft are polynomials in Z[x1, . . . , xt], write Bt(q, p; u; f) for the set of solutions modulo qkpk of the simultaneous congruences fj(x1, . . . , xt) ≡uj (mod qk−j+1pj−1) (1 ≤j ≤t) (3.5) with (Jt(f; x), pq) = 1, where Jt(f; x) = det µ ∂fi ∂xj (x) ¶ 1≤i,j≤t .
(3.6) 46 Lemma 3.2.2. Suppose that f1, . . . , f2r ∈Z[x1, . . . , x2r] have degrees bounded in terms of k. Then whenever 2r ≤k + 1 we have card(B2r(q, p; u; f)) ≪ε,k (pq)r(2r−1)+ε(q, p)2r(2k−2r+1).
Proof. Write ˜ q = q/(q, p) and ˜ p = p/(q, p), so that (˜ q, ˜ p) = 1. Then by considering the jth congruence in (3.5) modulo ˜ q k−j+1, we obtain from Lemma 2.2 of Wooley that the number of solutions modulo ˜ q k is Oε,k(˜ q r(2r−1)+ε). Similarly, the num-ber of solutions modulo ˜ p k is Oε,k(˜ p r(2r−1)+ε).
Hence by the Chinese Remainder Theorem the number of solutions modulo ˜ q k˜ p k is Oε,k((˜ p˜ q)r(2r−1)+ε). Trivially, each of these solutions lifts in at most (q, p)4kr ways to Z/(qkpk), and the lemma follows immediately.
We now develop some notation for analyzing real singular solutions of systems such as (3.2). Let ψ1, . . . , ψ2r be non-trivial polynomials in Z[x, y] of total degree at most k. When I, J ⊂{1, 2, . . . , 2r} with card(J ) = 2 card(I) and z, w ∈Z2r, define the Jacobian J(I, J ; ψ) = det à ∂ψj ∂zi (zi, wi) ∂ψj ∂wi(zi, wi) !
i∈I,j∈J .
Write Jd = {1, . . . , d}, and let I∗ d denote the set of all subsets of J2r of size d.
We will call the 4r-tuple of integers (z1, w1, . . . , z2r, w2r) highly singular for ψ if J(I, J2r; ψ) = 0 for each I ∈I∗ r. Also write di,j(z, w; ψ) = det à ∂ψi ∂z (z, w) ∂ψj ∂z (z, w) ∂ψi ∂w (z, w) ∂ψj ∂w (z, w) !
, and let Sr(P; ψ) denote the set of all integral 4r-tuples (z1, w1, . . . , z2r, w2r) with 1 ≤zi, wi ≤P that are highly singular for ψ.
Lemma 3.2.3. Suppose that ψ1, . . . , ψ2r satisfy the condition that d1,2 is non-trivial and degw(di,j) < degw(di′,j′) whenever i + j < i′ + j′. Then we have card(Sr(P; ψ)) ≪k P 3r−1.
Proof. Let T0(P; ψ) denote the set of integral 4r-tuples (z, w) with 1 ≤zi, wi ≤P and d1,2(zi, wi; ψ) = 0 (3.7) for i = 1, . . . , 2r. For a 4r-tuple counted by T0(P; ψ) and a given i, there are at most O(P) choices for zi and wi satisfying (3.7), since we have assumed that d1,2 is non-trivial, and it follows that card(T0(P; ψ)) ≪P 2r.
47 Now for 1 ≤d ≤r −1, we say that (z, w) ∈Td(P; ψ) if J(I, J2d; ψ) ̸= 0 (3.8) for some I ∈I∗ d but J(I ∪{i}, J2d+2; ψ) = 0 (3.9) for all i ∈J2r \ I. Consider a 4r-tuple counted by Td(P; ψ), where 1 ≤d ≤r −1.
There are O(1) choices for I and O(P 2d) choices for the zi and wi with i ∈I. Now we fix i ∈J2r \ I and expand the determinant in (3.9) using 2 × 2 blocks along the rows containing zi and wi. Then on using (3.8), together with our hypothesis on ψ, we see that the relation (3.9) is a non-trivial polynomial equation in the variables zi and wi and hence has O(P) solutions. Thus we have card(Td(P; ψ)) ≪P 2d+(2r−d) = P 2r+d and hence card(Sr(P; ψ)) ≤ r−1 X d=0 card(Td(P; ψ)) ≪P 3r−1, as desired.
Finally, we recall an estimate of Wooley for the number of integers in an interval with a given square-free kernel s0.
Lemma 3.2.4. Suppose that L is a positive real number and that r is a positive integer with log r ≪log L. Then for each ε > 0, one has card{y ≤L : s0(y) = s0(r)} ≪ε Lε.
Proof. This is Lemma 2.1 of Wooley .
3.3 The Fundamental Lemma For 0 ≤i ≤k, let ψi(z, w; c) be polynomials with integer coefficients in the vari-ables z, w, c1, . . . , cu and satisfying the conditions of Lemma 3.2.3. Further, suppose that Ci and C′ i satisfy 1 ≤C′ i ≤Ci ≪P, write ˜ C = Qu i=1 Ci, and let Di(c) be polyno-mials with total degrees bounded in terms of k such that Di(c) ̸= 0 for C′ i ≤ci ≤Ci.
Throughout the remainder of this chapter, ε, η, and γ will denote small positive numbers, whose values may change from statement to statement. Generally, η and γ will be chosen sufficiently small in terms of ε, and the implicit constants in our 48 analysis may depend at most on ε, η, γ, s, and k. Since our methods will involve only a finite number of steps, all implicit constants that arise remain under control, and the values assumed by η and γ throughout the arguments remain uniformly bounded away from zero.
When r ≤ £ k+1 2 ¤ , let Ss,r(P, Q, R; ψ) = Ss,r(P, Q, R; ψ; C, D; γ) be the number of solutions of the system r X n=1 ηn(ψi(zn, wn; c) −ψi(˜ zn, ˜ wn; c)) + Di(c) s X m=1 (xk−i m yi m −˜ xk−i m ˜ yi m) = 0 (0 ≤i ≤k) (3.10) with xm, ym, ˜ xm, ˜ ym ∈A(Q, R) (1 ≤m ≤s), (3.11) (xm, ym) ≤P γ and (˜ xm, ˜ ym) ≤P γ (1 ≤m ≤s), (3.12) 1 ≤zn, wn, ˜ zn, ˜ wn ≤P and ηn ∈{±1} (1 ≤n ≤r), (3.13) and C′ j ≤cj ≤Cj (1 ≤j ≤u).
(3.14) Further, write ˜ Ss,r(P, Q, R; ψ) for the number of solutions of (3.10) with (3.11), (3.12), (3.13), (3.14), and J2r(z, w; c) ̸= 0 and J2r(˜ z, ˜ w; c) ̸= 0, (3.15) where (recalling the notation of the previous section) we have put J2r(z, w; c) = J(Jr, J2r, ψ(z, w; c)).
Finally, let Ts,r(P, Q, R, θ; ψ) denote the number of solutions of r X n=1 ηn(ψi(zn, wn; c) −ψi(˜ zn, ˜ wn; c)) + Di(c)qk−ipi s X m=1 (uk−i m vi m −˜ uk−i m ˜ vi m) = 0 (0 ≤i ≤k) (3.16) with (3.13), (3.14), P θ < p, q ≤P θR and (q, p) ≤P γ, (3.17) 49 um, vm, ˜ um, ˜ vm ∈A(QP −θ, R) (1 ≤m ≤s), (3.18) (um, vm) ≤P γ and (˜ um, ˜ vm) ≤P γ (1 ≤m ≤s), (3.19) and (J2r(z, w; c), pq) = (J2r(˜ z, ˜ w; c), pq) = 1.
(3.20) Lemma 3.3.1. Given ε > 0, there exists a positive number η = η(ε, s, k) such that whenever R ≤P η one has Ss,r(P, Q, R; ψ) ≪˜ CP 3r−1Ss(Q, R) + ˜ CQ3sP 2r+sθ+ε + P (4s−2)θ+εTs,r(P, Q, R, θ; ψ).
Proof. Let S1 denote the number of solutions counted by Ss,r(P, Q, R; ψ) such that (z, w, ˜ z, ˜ w) is highly singular for ψ, and let S2 denote the number of solutions such that (z, w, ˜ z, ˜ w) is not highly singular for ψ, so that Ss,r(P, Q, R; ψ) = S1 + S2.
(i) Suppose that S1 ≥S2, so that Ss,r(P, Q, R; ψ) ≤2S1. By Lemma 3.2.3, we see that there are O(P 3r−1) permissible choices for z, w, ˜ z, and ˜ w. Now let fc(α; Q, R) = X x,y∈A(Q,R) (x,y)≤P γ e à k X i=0 αiDi(c)xk−iyi !
.
For a fixed choice of z, w, ˜ z, ˜ w, c, and η, the number of possible choices for x, y, ˜ x, and ˜ y is at most Z Tk+1 |fc(α; Q, R)|2sdα ≤Ss(Q, R), so we have S1 ≪P 3r−1 ˜ CSs(Q, R), which establishes the lemma in this case.
(ii) Suppose that S2 ≥S1, so that Ss,r(P, Q, R; ψ) ≤2S2. By rearranging vari-ables, we see that Ss,r(P, Q, R; ψ) ≪S3, where S3 denotes the number of solutions of (3.10) with (3.11), (3.12), (3.13), and (3.14), and J2r(z, w; c) ̸= 0. Then by using the Cauchy-Schwarz inequality as in the corresponding argument of Wooley to manipulate the underlying mean values, we see that Ss,r(P, Q, R; ψ) ≪S4, where S4 denotes the number of solutions of (3.10) with (3.11), (3.12), (3.13), (3.14), J2r(z, w; c) ̸= 0, and J2r(˜ z, ˜ w; c) ̸= 0.
50 We now further classify the solutions counted by S4. Write x D(L) y if there is some divisor d of x with d ≤L such that x/d has all of its prime divisors amongst those of y. Let S5 denote the number of solutions counted by S4 for which xj D(P θ) J2r(z, w; c) or ˜ xj D(P θ) J2r(˜ z, ˜ w; c) (3.21) or yj D(P θ) J2r(z, w; c) or ˜ yj D(P θ) J2r(˜ z, ˜ w; c) (3.22) for some j, and let S6 denote the number of solutions for which neither (3.21) nor (3.22) holds for any j. Then we have Ss,r(P, Q, R; ψ) ≪S5 + S6, and we divide into further cases.
(iii) Suppose that S5 ≥S6, and further suppose that (3.21) holds. Write S(z, w; c) = {x ∈A(Q, R) : x D(P θ) J2r(z, w; c)}, and let ˜ Hc,η(α; P, Q, R) = X z,w J2r(z,w;c)̸=0 X x∈S(z,w;c) y∈A(Q,R) (x,y)≤P γ e(Ξ(α; x, y, z, w; c, η)), where Ξ(α; x, y, z, w; c, η) = k X i=0 αi(Di(c)xk−iyi + η1ψi(z1, w1; c) + · · · + ηrψi(zr, wr; c)).
Then S5 ≪ X c,η,ω Z Tk+1 | ˜ Hc,η(α; P, Q, R) ˜ F ∗ c,ω(α; P)fc(α; Q, R)2s−1| dα, where ˜ F ∗ c,ω(α; P) = X z,w J2r(z,w;c)̸=0 e(Ξ(α; 0, 0, z, w; c, ω)).
By using the Cauchy-Schwarz inequality and considering the underlying Diophantine equations as in , we deduce that Ss,r(P, Q, R; ψ) ≪ X g,h,c V (g, h; c), 51 where V (g, h; c) denotes the number of solutions of the system r X n=1 ηn(ψi(zn, wn; c) −ψi(˜ zn, ˜ wn; c)) + Di(c) s−1 X m=1 (xk−i m yi m −˜ xk−i m ˜ yi m) = Di(c)((e˜ x)k−i˜ yi −(dx)k−iyi) (0 ≤i ≤k) with (3.11), (3.12), (3.13), (3.14), and J2r(z, w; c) ̸= 0, J2r(˜ z, ˜ w; c) ̸= 0, g|J2r(z, w; c), h|J2r(˜ z, ˜ w; c), 1 ≤d, e ≤P θ, x ≤Q/d, ˜ x ≤Q/e, y, ˜ y ≤Q, s0(x) = g, s0(˜ x) = h.
Write Gc,η,g(α; P) = X z,w g|J2r(z,w;c)̸=0 e(Ξ(α; 0, 0, z, w; c, η)) and Gc,η(α) = X g≤Q Gc,η,g(α; P) X d≤P θ X x≤Q/d s0(x)=g y≤Q e à k X i=0 αiDi(c)(dx)k−iyi !
.
Then Ss,r(P, Q, R; ψ) ≪ X c,η Z Tk+1 |Gc,η(α)2fc(α; Q, R)2s−2| dα.
(3.23) By Cauchy’s inequality, we have |Gc,η(α)|2 ≤H1,c,η(α)H2,c(α), (3.24) where H1,c,η(α) = X g≤Q |Gc,η,g(α; P)|2 and H2,c(α) = X g≤Q ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ X d≤P θ X x≤Q/d s0(x)=g X y≤Q e à k X i=0 αiDi(c)(dx)k−iyi !
¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 .
52 Now by interchanging the order of summation and using Cauchy’s inequality together with Lemma 3.2.4 as in , we obtain H2,c(α) = X g≤Q ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ X x,y≤Q s0(x)=g X d≤P θ d≤Q/x e à k X i=0 αiDi(c)(dx)k−iyi !
¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 ≪ X g≤Q Q1+ε X x,y≤Q s0(x)=g P θQ/x ≪ Q3P θ+ε.
(3.25) Thus an application of H¨ older’s inequality in (3.23) gives Ss,r ≪ Ã X c,η Z Tk+1|H1,c,η(α)fc(α)2s|dα !
1−1 s à X c,η Z Tk+1|H1,c,η(α)H2,c(α)s|dα !1 s ≪ Q3P θ+ε ÃX c,η X g≤Q Z Tk+1 |Gc,η,g(α; P)|2dα !1 s Ss,r(P, Q, R; ψ)1−1 s, where we have written fc(α) for fc(α; Q, R) and used a standard estimate for the divisor function. But for a fixed choice of c, η, ˜ z, and ˜ w, the Inverse Function The-orem, in combination with B´ ezout’s Theorem, shows that there are O(1) choices of z and w satisfying r X n=1 ηn(ψi(zn, wn; c) −ψi(˜ zn, ˜ wn; c)) = 0 (0 ≤i ≤k) with J2r(z, w; c) ̸= 0. Hence by another divisor estimate we see that X c,η X g≤Q Z Tk+1 |Gc,η,g(α; P)|2dα ≪˜ CP 2r+ε, and the result follows in the case where (3.21) holds. The case where (3.22) holds is handled in exactly the same manner.
(iv) Suppose that S6 ≥S5, and consider a solution counted by S6. For a given index j, let q and p denote the largest divisors of xj and yj, respectively, with (q, J2r(z, w; c)) = (p, J2r(z, w; c)) = 1.
Then, since neither (3.21) nor (3.22) holds, we have q > P θ and p > P θ. Thus we can find divisors qj of xj and pj of yj such that P θ < qj, pj ≤P θR and (qjpj, J2r(z, w; c)) = 1, and we proceed similarly with the ˜ xj and ˜ yj, except that 53 we replace J2r(z, w; c) by J2r(˜ z, ˜ w; c). Hence we see that S6 ≪V1, where V1 denotes the number of solutions of r X n=1 ηn(ψi(zn, wn; c) −ψi(˜ zn, ˜ wn; c)) + Di(c) s X j=1 ((qjuj)k−i(pjvj)i −(˜ qj˜ uj)k−i(˜ pj˜ vj)i) = 0 (0 ≤i ≤k).
with (3.13), (3.14), and for 1 ≤j ≤s P θ < qj, pj, ˜ qj, ˜ pj ≤P θR, (qj, pj), (˜ qj, ˜ pj) ≤P γ, (3.26) uj, vj, ˜ uj, ˜ vj ∈A(QP −θ, R), (uj, vj), (˜ uj, ˜ vj) ≤P γ, and (qjpj, J2r(z, w; c)) = (˜ qj ˜ pj, J2r(˜ z, ˜ w; c)) = 1.
Now write Fc,η,q(α; P, R) = X z,w (q,J2r(z,w;c))=1 e(Ξ(α; 0, 0, z, w; c, η)) and Fc,j(α) = fc(qjpjα; QP −θ, R)fc(−˜ qj˜ pjα; QP −θ, R), where qjpjα = (α0qk j , α1qk−1 j pj, . . . , αkpk j) and ˜ qj˜ pjα = (α0˜ qk j , α1˜ qk−1 j ˜ pj, . . . , αk˜ pk j).
Then we have V1 ≤ X c,η Z Tk+1 X q,p,˜ q,˜ p Fc,η,π(α; P, R)Fc,η,˜ π(−α; P, R) s Y i=1 Fc,j(α) dα, (3.27) where π = q1 · · · qsp1 · · · ps and ˜ π = ˜ q1 · · · ˜ qs˜ p1 · · · ˜ ps, and where the sum is over q, p, ˜ q, ˜ p satisfying (3.26). Let Xc,η,j(α) = ¯ ¯Fc,η,π(α; P, R)2fc(qjpjα; QP −θ, R)2s¯ ¯, 54 and let Yc,η,j(α) be the analogous function for the ˜ qj and ˜ pj. Then by (3.27) and two applications of H¨ older’s inequality (as in ), we obtain S6 ≪ X q,p,˜ q,˜ p s Y j=1 ÃX c,η Z Tk+1 Xc,η,j(α) dα !
1/2s ÃX c,η Z Tk+1 Yc,η,j(α) dα !
1/2s .
Now we observe that X c,η Z Tk+1 Xc,η,j(α) dα = W(P, Q, R, qj, pj), where W(P, Q, R, q, p) denotes the number of solutions of (3.16) with (3.13), (3.14), (3.18), (3.19), and (3.20). Thus we have S6 ≪ X q,p,˜ q,˜ p s Y j=1 W(P, Q, R, qj, pj)1/2s W(P, Q, R, ˜ qj, ˜ pj)1/2s, whence by H¨ older’s inequality S6 ≪ Ã X q,p,˜ q,˜ p 1 !
1−1/2s à X q,p,˜ q,˜ p s Y j=1 W(P, Q, R, qj, pj) W(P, Q, R, ˜ qj, ˜ pj) !
1/2s ≪ (P θR)4s−2 Ã 2s Y j=1 X q,p W(P, Q, R, qj, pj) !
1/2s ≪ (P θR)4s−2 Ts,r(P, Q, R, θ; ψ), and this completes the proof of the lemma.
The following modification of Lemma 3.3.1 may be more useful for smaller values of k.
Lemma 3.3.2. Given ε > 0, there exists a positive number η = η(ε, s, k) such that whenever R ≤P η one has Ss,r(P, Q, R; ψ) ≪˜ CP 3r−1Ss(Q, R) + Q3P θ+ε ˜ Ss−1,r(P, Q, R; ψ) + P (4s−2)θ+ε Ts,r(P, Q, R, θ; ψ).
Proof. The only change occurs in part (iii) of the proof, where the number of solutions counted by S5 is estimated. Substituting the bounds (3.24) and (3.25) into (3.23), we obtain Ss,r(P, Q, R; ψ) ≪Q3P θ+ε X c,η X g≤Q Z Tk+1 |Gc,η,g(α; P)2fc(α; Q, R)2s−2| dα, and the lemma follows on considering the underlying Diophantine equations and recalling a standard estimate for the divisor function.
55 Now let ˜ Ts,r(P, Q, R, θ; ψ) denote the number of solutions of (3.16) with (3.13), (3.14), (3.17), (3.18), (3.19) and also zn ≡˜ zn (mod qkpk) and wn ≡˜ wn (mod qkpk) (1 ≤n ≤r).
(3.28) Lemma 3.3.3. Given ε > 0, there exists a positive number γ0 = γ0(ε, s, k) such that whenever γ ≤γ0 one has Ts,r(P, Q, R, θ; ψ) ≪(P θR)2r(2r−1)+ε ˜ Ts,r(P, Q, R, θ; ψ).
Proof. When q and p satisfy (3.17), let Bq,p(u; c, η) denote the set of solutions (z, w) of the system of congruences Υi(z, w; c, η) ≡ui (mod qk−ipi) (0 ≤i ≤k) (3.29) with 1 ≤zn, wn ≤(qp)k and (qp, J2r(z, w; c)) = 1, where Υi(z, w; c, η) = r X n=1 ψi(zn, wn; c).
By Lemma 3.2.2 we have card(Bq,p(u; c, η)) ≪(pq)r(2r−1)+ε, on taking γ sufficiently small in terms of ε.
Now observe that for each solution counted by Ts,r(P, Q, R, θ; ψ) we have Υi(z, w; c, η) ≡Υi(˜ z, ˜ w; c, η) (mod qk−ipi), so for each i we can classify the solutions of (3.16) according to the common residue class modulo qk−ipi of Υi(z, w; c, η) and Υi(˜ z, ˜ w; c, η). Let Hq,p(α; z, w; c, η) = X x∈[1,P]r xn≡zn(qkpk) X y∈[1,P]r yn≡wn(qkpk) e à k X i=0 αiΥi(x, y; c, η) !
.
Then Ts,r(P, Q, R, θ; ψ) ≪ X q,p X c,η Z Tk+1 ˜ Hq,p(α; c, η)| ˜ fc,q,p(α; QP −θ, R)|2sdα, where ˜ Hq,p(α; c, η) = qk X u0=1 qk−1p X u1=1 · · · pk X uk=1 ¯ ¯ ¯ ¯ ¯ ¯ X (z,w)∈Bq,p(u;c,η) Hq,p(α; z, w; c, η) ¯ ¯ ¯ ¯ ¯ ¯ 2 56 and ˜ fc,q,p(α; L, R) = X x,y∈A(L,R) (x,y)≤P γ e à k X i=0 αiDi(c)(qx)k−i(py)i !
.
Now by Cauchy’s inequality, ˜ Hq,p(α; c, η) ≤ qk X u0=1 qk−1p X u1=1 · · · pk X uk=1 card(Bq,p(u; c, η)) X (z,w)∈Bq,p(u;c,η) |Hq,p(α; z, w; c, η)|2, and thus Ts,r(P, Q, R, θ; ψ) ≪ (P θR)2r(2r−1)+ε X q,p c,η X z,w 1≤zn≤qkpk 1≤wn≤qkpk Z Tk+1 |Hq,p|2| ˜ fc,q,p|2sdα ≪ (P θR)2r(2r−1)+ε ˜ Ts,r(P, Q, R, θ; ψ).
This completes the proof.
3.4 Efficient Differencing Define the difference operator ∆∗ j recursively by ∆∗ 1(f(x, y); h; g) = f(x + h, y + g) −f(x, y) and ∆∗ j+1(f(x, y); h1, . . . , hj+1; g1, . . . , gj+1) = ∆∗ 1(∆∗ j(f(x, y); h1, . . . , hj; g1, . . . , gj); hj+1; gj+1), with the convention that ∆∗ 0(f(x, y); h; g) = f(x, y).
Further, write ψi,j(z, w; h, g; m, n) = ∆∗ j(zk−iwi; h′ 1, . . . , h′ j; g′ 1, . . . , g′ j), where h′ i = hi(mini)k and g′ i = gi(mini)k, (3.30) 57 and put rj = ·k −j + 1 2 ¸ .
(3.31) Our first task is to show that the polynomials ψi,j satisfy the conditions of Lemma 3.2.3, so that the results of the previous section may be applied. We start by ex-pressing ∆∗ j in terms of the more familiar difference operators ∆j defined by ∆1(f(x); h) = f(x + h) −f(x) and ∆j+1(f(x); h1, . . . , hj+1) = ∆1(∆j(f(x); h1, . . . , hj); hj+1).
For simplicity, we introduce the functions χi,j(z, w; h; g) = ∆∗ j(zk−iwi; h1, . . . , hj; g1, . . . , gj) (3.32) and observe that ψi,j(z, w; h, g; m, n) = χi,j(z, w; h′, g′), where h′ and g′ are defined by (3.30). As in Section 3.2, we write Jd for the set {1, . . . , d}, and also write ˜ Ad for the set Jd \ A. When A = {i1, . . . , im} ⊂Jj with i1 < · · · < im, define q(i) m (w; g, A) = ∆m(wi; gi1, . . . , gim), (3.33) and when A is as above and B = {j1, . . . , jt} ⊂Jj with j1 < · · · < jt, define p(i) t (z; h, A, B) = ∆t((z + hi1 + · · · + him)k−i; hj1, . . . , hjt).
(3.34) Lemma 3.4.1. We have χi,j(z, w; h; g) = j X m=0 X A⊂Jj |A|=m p(i) j−m(z; h, A, ˜ Aj)q(i) m (w; g, A).
Proof. We fix i, h, and g and proceed by induction on j. For brevity, we write χi,j(z, w), qm(w; A), and pt(z; A, B) for the functions defined by (3.32), (3.33), and (3.34), respectively. For j = 0 we have χi,0(z, w) = zk−iwi = p0(z; ∅, ∅)q0(w; ∅).
58 Now assume the result holds for j −1. Then we have χi,j(z, w) = χi,j−1(z + hj, w + gj) −χi,j−1(z, w), so by the inductive hypothesis we obtain χi,j(z, w) = j−1 X m=0 X A⊂Jj−1 |A|=m θi,j(z, w; m; A), where θi,j = pj−1−m(z + hj; A, ˜ Aj−1)qm(w + gj; A) −pj−1−m(z; A, ˜ Aj−1)qm(w; A).
The above expression can be rewritten as θi,j = pj−m(z; A, ˜ Aj)qm(w; A) + pj−1−m(z + hj; A, ˜ Aj−1)qm+1(w; A ∪{j}), so we have χi,j = j−1 X m=0 µ X A⊂Jj−1 |A|=m pj−m(z; A; ˜ Aj)qm(w; A) + X A⊂Jj |A|=m+1 j∈A pj−(m+1)(z; A; ˜ Aj)qm+1(w; A) ¶ = j−1 X m=0 X A⊂Jj |A|=m j / ∈A pj−m(z; A; ˜ Aj)qm(w; A) + j X m=1 X A⊂Jj |A|=m j∈A pj−m(z; A; ˜ Aj)qm(w; A), and the lemma follows.
Now we show that the 2 × 2 Jacobians satisfy the condition imposed in Lemma 3.2.3.
Lemma 3.4.2. Suppose that 0 ≤j < k and i1 < i2 ≤k −j Then we have di1,i2(z, w; χj) = p(z)wi1+i2−1 + Oz(wi1+i2−2), where p(z) is a non-trivial polynomial of degree at most 2k.
Proof. When i < k −j, we have by Lemma 3.4.1 that ∂χi,j ∂z = ∂ ∂z ¡ ∆j(zk−i; h1, . . . , hj) ¢ wi + Oz(wi−1) and ∂χi,j ∂w = i∆j(zk−i; h1, . . . , hj)wi−1 + Oz(wi−2), 59 and we recall (see for example Exercise 2.1 of Vaughan ) that ∆j(zk; h1, . . . , hj) = k(k −1) · · · (k −j + 1)h1 · · · hjzk−j + O(zk−j−1).
Hence if i2 < k −j then we have di1,i2(z, w, χ) = p(z)wi1+i2−1 + Oz(wi1+i2−2), where the leading term of p(z) is (h1 · · · hj)2(k −i1)!(k −i2)!
(k −i1 −j)!(k −i2 −j)! ((k −i1 −j)i2 −(k −i2 −j)i1)z2k−i1−i2−2j−1, and the lemma follows in this case on noting that (k −i1 −j)i2 −(k −i2 −j)i1 = (k −j)(i2 −i1) ̸= 0.
Now if i = k −j we obtain from Lemma 3.4.1 that ∂χi,j ∂z = Oz(wi−1) and ∂χi,j ∂w = i(k −i)! h1 · · · hjwi−1 + Oz(wi−2).
Thus if i2 = k −j then we have di1,i2 = µi2(h1 · · · hj)2(k −i1)!(k −i2)!
(k −i1 −j −1)!
zk−i1−j−1 + O(zk−i1−j−2) ¶ wi1+i2−1 + Oz(wi1+i2−2), and this completes the proof.
We now consider the effect of substituting ψi,j(z, w; h, g; m, n) for ψi(z, w; c) in the analysis of Section 3.3. For 1 ≤j ≤k, suppose that 0 ≤φj ≤1/2k, and put Mj = P φj, Hj = PM −2k j , and Qj = P(M1 · · · Mj)−1.
Further, write ˜ Mj = j Y i=1 Mi and ˜ Hj = j Y i=1 Hi.
We replace (3.14) by the conditions 1 ≤hi, gi ≤Hi (1 ≤i ≤j), (3.35) 60 Mi < mi, ni ≤MiR, and (mi, ni) ≤P γ (1 ≤i ≤j), (3.36) and take Di(m, n) = j Y l=1 mk−i l ni l.
On replacing hi by hi(mini)k and gi by gi(mini)k in the above results, we see that ψ0,j, . . . , ψ2r−1,j satisfy the hypotheses of Lemma 3.2.3 whenever r ≤rj. Thus we may apply Lemma 3.3.1 to relate Ss,rj(P, Qj, R; ψj) to Ts,rj(P, Qj, R, φj+1; ψj). The following lemma then relates Ts,rj(P, Qj, R, φj+1; ψj) to Ss,rj+1(P, Qj, R; ψj+1) and hence allows us to repeat the differencing process.
Lemma 3.4.3. Suppose that r ≤2w and 0 ≤j < k. Then given ε > 0, there exists η = η(ε, s, k) such that whenever R ≤P η one has ˜ Ts,r(P, Qj, R, φj+1; ψj) ≪P (3−2kφj+1)r+ε ˜ H2 j ˜ M 2 j+1Ss(Qj+1, R) + P εH2r−2 j+1 ¡ ˜ H2 j+1 ˜ M 2 j+1Ss(Qj+1, R) ¢1−r/2w¡ Ss,w(P, Qj+1, R; ψj+1) ¢r/2w.
Proof. Write θ = φj+1, and define La,b,d(α; h, g; m, n) = X 1≤z≤P z≡a (d) X 1≤w≤P w≡b (d) e à k X i=0 αiψi,j(z, w; h, g; m, n) !
, Kd(α; h, g; m, n) = d X a=1 d X b=1 |La,b,d(α; h, g; m, n)|2, and gq,p(α; m, n) = X x,y∈A(Qj+1,R) (x,y)≤P γ e à k X i=0 αiDi(m, n)(qx)k−i(py)i !
.
Then on considering the underlying Diophantine equations, we have ˜ Ts,r ≍ X h,g,m,n X Mj+1<p,q≤Mj+1R (p,q)≤P γ Z Tk+1 Kqkpk(α; h, g; m, n)r|gq,p(α; m, n)|2sdα.
Let U0 be the number of solutions counted by ˜ Ts,r with zn = ˜ zn or wn = ˜ wn for some n, and let U1 be the number of solutions in which zn ̸= ˜ zn and wn ̸= ˜ wn for all n, so that ˜ Ts,r = U0 + U1.
61 First suppose that U0 ≥U1, so that ˜ Ts,r ≪U0. Then U0 ≪P 3−2kφj+1 X h,g,m,n X Mj+1<p,q≤Mj+1R Z Tk+1Kqkpk(α; h, g; m, n)r−1|gq,p(α; m, n)|2sdα, and by using H¨ older’s inequality twice as in , we find that ˜ Ts,r ≪P (3−2kφj+1)r+ε ˜ H2 j ˜ M 2 j+1Ss(Qj+1, R).
(3.37) Now suppose that U1 ≥U0, so that ˜ Ts,r ≪U1.
Note that for each solution counted by U1 we can write ˜ zn = zn + ˜ hnqkpk and ˜ wn = wn + ˜ gnqkpk for 1 ≤n ≤r, where ˜ hn, ˜ gn are integers satisfying 1 ≤|˜ hn|, |˜ gn| ≤Hj+1. Thus we see that U1 ≤ X η∈{±1}r U2(η), where U2(η) is the number of solutions of the system r X l=1 ηlψi,j+1(zl, wl; h, ˜ hl; g, ˜ gl; m, q; n, p) + Di(m, n)qk−ipi s X m=1 (uk−i m vi m −˜ uk−i m ˜ vi m) = 0 (0 ≤i ≤k) with z, w, u, v, ˜ u, ˜ v, h, g, m, n satisfying (3.13), (3.18), (3.19), (3.35), and (3.36), and with 1 ≤˜ hl, ˜ gl ≤Hj+1 (1 ≤l ≤r), Mj+1 < p, q ≤Mj+1R, and (q, p) ≤P γ.
On writing G(α; ˜ h, ˜ g; q, p) = X 1≤z,w≤P e à k X i=0 αiψi,j+1(z, w; h, ˜ h; g, ˜ g; m, q; n, p) !
, we have by H¨ older’s inequality that U2(η) ≪ X h,g,m,n X q,p Z Tk+1 ¯ ¯ ¯ ¯ ¯ ¯ X 1≤˜ g,˜ h≤Hj+1 G(α; ˜ h, ˜ g, q, p) ¯ ¯ ¯ ¯ ¯ ¯ r |gq,p(α; m, n)|2s dα ≪ H2r−2 j+1 X h,g,m,n X q,p,˜ h,˜ g Z Tk+1 |G(α; ˜ h, ˜ g, q, p)|r|gq,p(α; m, n)|2sdα.
62 Thus on using H¨ older’s inequality twice more and considering the underlying Dio-phantine equations, we see that U2(η) ≪ H2r−2 j+1 X h,g,m,n q,p,˜ h,˜ g µZ Tk+1 |G|2w|gq,p|2sdα ¶r/2w µZ Tk+1 |gq,p|2sdα ¶1−r/2w ≪ H2r−2 j+1 ÃX Z Tk+1 |G|2w|gq,p|2sdα !r/2w ÃX Z Tk+1 |gq,p|2sdα !1−r/2w ≪ H2r−2 j+1 µ Ss,w(P, Qj+1, R; ψj+1) ¶r/2wµ P ε ˜ H2 j+1 ˜ M 2 j+1Ss(Qj+1, R) ¶1−r/2w , and the lemma follows on combining this with (3.37).
In analogy with Lemma 4.2 of , one might hope to refine the above argument to allow the factor of P (3−2kφj+1)r in the first term of the estimate to be replaced by P 2r, but it is not clear that this can be achieved. As will be seen in Section 3.6, such an improvement would have a significant impact on the strength of our repeated efficient differencing procedure.
3.5 Mean Value Estimates Based on Single Differencing In this section, we consider estimates for Ss(P, R) arising from a single efficient difference, reserving the full power of the preceding analysis for Section 3.6.
Suppose that 0 < θ ≤1/2k, write r = r0 = £ k+1 2 ¤ , and put M = P θ, H = PM −2k, and Q = PM −1.
Further, let F(α; P) = X 1≤z,w≤P e(α0zk + α1zk−1w + · · · + αkwk), G(α; q, p) = X 1≤h,g≤H X 1≤z,w≤P e à k X i=0 αiψi,1(z, w; h, g; q, p) !
, gq,p(α; P, Q, R) = X x,y∈A(Q,R) (x,y)≤P γ e à k X i=0 αi(qx)k−i(py)i !
, and Ms,r(P, Q, R) = X M≤p,q≤MR Z Tk+1 ¯ ¯G(α; q, p)rgq,p(α; P, Q, R)2s¯ ¯ dα.
63 We say that λs is a permissible exponent if for every ε > 0 there exists η = η(ε, s, k) such that Ss(P, R) ≪ε P λs+ε whenever R ≤P η. Further, we recall that the exponent ∆s admissible if λs = 4s −k(k + 1) + ∆s is permissible.
Lemma 3.5.1. Let θ = 1/2k, and suppose that s ≥k2/(1 −θ). If ∆s ≤k(k + 1) is an admissible exponent, then the exponent ∆s+r = ∆s(1 −θ) is admissible.
Proof. By Lemmata 3.3.1 and 3.3.3, we have Ss,r(P, P, R; ψ0) ≪P 3r−1Ss(P, R) + P (3+θ)s+2r+ε + P εM 4s−2+2r(2r−1) ˜ Ts,r(P, P, R, θ; ψ0) (3.38) for γ sufficiently small, and by the argument of the proof of Lemma 3.4.3 we have ˜ Ts,r(P, P, R, θ; ψ0) ≪P (3−2kθ)r+εM 2Ss(Q, R) + Ms,r(P, Q, R).
(3.39) Since θ = 1/2k, we have H = 1, so by a trivial estimate we obtain Ms,r(P, Q, R) ≪M 2P 2r+εSs(Q, R).
Hence on recalling Lemma 3.2.1 and considering the underlying Diophantine equa-tions, we obtain from (3.38) and (3.39) that Ss+r(P, R) ≪P 2s+2r+ε + Ss,r(P, P, R; ψ0) ≪P 3r−1Ss(P, R) + P (3+θ)s+2r+ε + P 2r+εM 4s+2r(2r−1)Ss(Q, R).
(3.40) Thus, since λs = 4s −k(k + 1) + ∆s is permissible, we have Ss+r(P, R) ≪P Λ1+ε + P Λ2+ε + P Λ3+ε, where Λ1 = 4(s + r) −k(k + 1) −(r + 1) + ∆s, Λ2 = 4(s + r) −k(k + 1) −s(1 −θ) −2r + k(k + 1), and Λ3 = 4(s + r) −k(k + 1) + ∆s(1 −θ).
Now since r + 1 ≥k+1 2 and ∆s ≤k(k + 1), we have ∆sθ ≤r + 1 and hence Λ1 ≤Λ3.
Furthermore, since s(1 −θ) ≥k2 and 2r ≥k, we have Λ2 ≤Λ3. Therefore, the exponent ∆s+r = ∆s(1 −θ) is admissible, and this completes the proof.
64 Proof of Theorem 3.1. Let s1 be as in the statement of the theorem, and suppose that s ≥s1. Choose an integer t with s ≡t (mod r) and s1 −r < t ≤s1. Then since ∆t = k(k + 1) is trivially admissible, we find by repeated use of Lemma 3.5.1 that the exponent ∆s = k(k + 1) µ 1 −1 2k ¶ (s−t)/r ≤k(k + 1) µ 1 −1 2k ¶ (s−s1)/r is admissible, and this completes the proof.
3.6 Estimates Arising from Repeated Differencing In this section, we explore the possibility of obtaining improved mean value esti-mates by employing our efficient differencing procedure repeatedly. As we take more differences, we must reduce the number of variables taken in a complete interval, so that the difference polynomials ψj will satisfy the hypotheses of Lemma 3.2.3. This complicates the recursion for generating admissible exponents and therefore requires some additional notation. Recall the definition of rj from (3.31), and write Ωj = X 2rj<l≤k+1 (k −l + 1) = 1 2(k −2rj + 1)(k −2rj).
(3.41) For convenience, we also write r = r0 = £k+1 2 ¤ . Throughout this section, we will assume that k is taken to be sufficiently large.
Lemma 3.6.1. Suppose that u ≥k(k + 1) and that ∆u ≤k(k + 1) is an admissible exponent.
For any integer j with 1 ≤j ≤ √ k, define φ(j, s, J) recursively for s = u + lr and J = j, . . . , 2 by φ(j, s, j) = 1/2k, φ∗(j, s, J −1) = 1 4k + µ1 2 + 2ΩJ−1 −∆s−r 8krJ−1 ¶ φ(j, s, J), (3.42) and φ(j, s, J) = min(1/2k, φ∗(j, s, J)), where ∆s = ∆s−r(1 −θs) + r(2kθs −1) (3.43) and θs = min 1≤j≤k φ(j, s, 1).
Then ∆s is an admissible exponent for s = u + lr (l ∈N).
65 Proof. We start by noting that 0 < θs ≤1/2k and that θs is an increasing function of s. Now let j denote the least integer with φ(j, s + r, 1) = θs+r and write φJ = φ(j, s + r, J). As in the proof of , Theorem 6.1, we have φJ < 1/2k whenever J < j. In particular, it follows that whenever J < j we have 2ΩJ −∆s < 0 and φJ = φ∗(j, s + r, J). We claim that φJ ≤φJ+1 for J < j. By (3.42) and the above remarks, this is equivalent to φJ+1 µ1 2 + ∆s −2ΩJ 8krJ ¶ ≥1 4k, (3.44) and this is immediate when J = j−1, since ∆s−2Ωj−1 > 0 and φj = 1/2k. Assuming the claim holds for J, then we see from (3.42) that φJ µ1 2 + ∆s −2ΩJ−1 8krJ−1 ¶ µ1 2 + ∆s −2ΩJ 8krJ ¶ ≥1 4k µ1 2 + ∆s −2ΩJ−1 8krJ−1 ¶ , and it follows on using (3.41) that φJ µ1 2 + ∆s −2ΩJ−1 8krJ−1 ¶ ≥1 4k µ rJ rJ−1 ¶ 2rJ−1(4k + 1) + ∆s −k(k + 1) 2rJ(4k + 1) + ∆s −k(k + 1) .
Since ∆s ≤k(k + 1) and rJ ≤rJ−1, we see that (3.44) holds with J replaced by J −1, and our claim follows.
For 1 ≤i ≤j, we write Mi = P φi, Hi = PM −2k i , and Qi = P(M1 · · · Mi)−1, with the convention that Q0 = P. We prove the lemma by induction on l, the case l = 0 having been assumed. Suppose that ∆s is admissible, so that Ss(Q, R) ≪ Qλs+ε, where λs = 4s −k(k + 1) + ∆s. We show inductively that ˜ Ts,rJ(P, QJ, R, φJ+1; ψJ) ≪P (3−2kφJ+1)rJ+ε ˜ H2 J ˜ M 2 J+1Qλs J+1 (3.45) for J = j −1, . . . , 0. By Lemma 3.4.3 with j replaced by j −1, r = rj−1 and w = rj, we have that ˜ Ts,rj−1(P, Qj−1, R, φj; ψj−1) ≪P (3−2kφj)rj−1+ε ˜ H2 j−1 ˜ M 2 j Ss(Qj, R) + P εH 2rj−1−2 j ( ˜ H2 j ˜ M 2 j Ss(Qj, R))1−β(Ss,rj(P, Qj, R, ψj))β, where β = rj−1/(2rj). Then on making the trivial estimate Ss,rj(P, Qj, R; ψj) ≪P 4rj+ε ˜ H2 j ˜ M 2 j Ss(Qj, R) 66 and noting that φj = 1/2k and hence Hj = 1, we obtain ˜ Ts,rj−1(P, Qj−1, R, φj; ψj−1) ≪ P 2rj−1+ε ˜ H2 j−1 ˜ M 2 j Ss(Qj, R) ≪ P 2rj−1+ε ˜ H2 j−1 ˜ M 2 j Qλs j , on using the outer induction hypothesis. Thus (3.45) holds in the case J = j −1.
Now suppose that (3.45) holds for J. Then, for γ sufficiently small, we have by Lemmata 3.3.1 and 3.3.3 that Ss,rJ(P, QJ, R; ψJ) ≪P ε ˜ H2 J ˜ M 2 J ¡ P Λ1 + P Λ2 + P Λ3¢ , where Λ1 = 3rJ −1 + λs(1 −φ1 −· · · −φJ), (3.46) Λ2 = 3s(1 −φ1 −· · · −φJ) + sφJ+1 + 2rJ, (3.47) and Λ3 = (4s + 2rJ(2rJ −1))φJ+1 + (3 −2kφJ+1)rJ + λs(1 −φ1 −· · · −φJ+1).
(3.48) Now since J ≤ √ k, we have rJ ∼k/2, and it follows easily that Λ1 ≤Λ3 and Λ2 ≤Λ3 for s ≥k(k + 1) and k sufficiently large. Hence by Lemma 3.4.3 we have ˜ Ts,rJ−1(P, QJ−1, R, φJ; ψJ−1) ≪P (3−2kφJ)rJ−1+ε ˜ H2 J−1 ˜ M 2 JQλs J + P εH2rJ−1−2 J ¡ ˜ H2 J ˜ M 2 JQλs J ¢1−β′¡ P (3−2kφJ+1)rJ+εM 4s−2+2rJ(2rJ−1) J+1 ˜ H2 J ˜ M 2 J+1Qλs J+1 ¢β′ , where β′ = rJ−1/(2rJ). The second term here is ˜ H2 J−1 ˜ M 2 JQλs J P Λ+ε, where Λ = 2rJ−1(1 −2kφJ) + rJ−1 2rJ £ (3 −2kφJ+1)rJ + (4s + 2rJ(2rJ −1) −λs)φJ+1 ¤ .
By (3.41) and (3.42), we have (4s + 2rJ(2rJ −1) −λs)φJ+1 = (4krJ + 2ΩJ −∆s)φJ+1 = 8krJφJ −2rJ, and hence Λ = (5 2 −kφJ+1)rJ−1 ≤(3 −2kφJ)rJ−1 + rJ−1(kφJ −1 2) ≤(3 −2kφJ)rJ−1, 67 since φJ+1 ≥φJ and φJ ≤1/2k. Thus (3.45) holds with J replaced by J −1, so this completes the inner induction. Now we apply (3.45) with J = 0 to obtain ˜ Ts,r(P, P, R, φ1; ψ0) ≪P (3−2kφ1)r+2φ1+λs(1−φ1)+ε, whence by Lemmata 3.2.1, 3.3.1, and 3.3.3 we have (for γ sufficiently small) that Ss+r(P, R) ≪P 2s+ε + Ss,r(P, P, R; ψ0) ≪P Λ1+ε + P Λ2+ε + P Λ3+ε, where Λ1, Λ2, and Λ3 are given by (3.46), (3.47), and (3.48) with J = 0. Therefore, the exponent λs+r = 4(s + r) −k(k + 1) + ∆s(1 −θs+r) + r(2kθs+r −1) is permissible, and the desired conclusion holds with s replaced by s + r.
This completes the proof of the lemma.
Next we investigate the size of the admissible exponents supplied by Lemma 3.6.1.
Lemma 3.6.2. Suppose that s > k(k+1)+r and that ∆s−r is an admissible exponent satisfying (log k)2 < ∆s−r ≤2rk.
Write δs−r = ∆s−r/4rk, and define δs to be the unique positive solution of the equa-tion δs + log δs = δs−r + log δs−r −3 4k + 1 k(log k)3/2.
(3.49) Then the exponent ∆s = 4rkδs is admissible.
Proof. The proof is nearly identical to that of , Lemma 6.2. In view of (3.43), we may assume that 0 ≤∆s ≤2rk and hence that 0 ≤δs ≤1 2. By Lemma 3.6.1 with j = ·1 2(log k)1/4 ¸ + 1, (3.50) we see that the exponent ∆s = ∆s−r(1 −θ) + r(2kθ −1) = 4krδs−r −r + 2rk(1 −2δs−r)θ, (3.51) is admissible, where θ = θs = φ(j, s, 1). We note that for 1 ≤J < j one has ΩJ ≤1 2J(J + 1) < 1 2(log k)1/2, 68 so on writing φJ for φ∗(j, s, J) we have φJ−1 ≤1 4k + 1 2(1 −δ′)φJ, (3.52) where δ′ = ∆s−r −(log k)1/2 4kr ≥δs−r(1 −(log k)−3/2).
(3.53) An easy induction using (3.52) shows that φJ ≤ 1 2k(1 + δ′) Ã 1 + δ′ µ1 −δ′ 2 ¶j−J!
(1 ≤J ≤j), and therefore θ = φ1 ≤1 + 21−jδ′ 2k(1 + δ′).
Write L = (log k)−3/2. Since the expression on the right hand side of the above inequality is a decreasing function of δ′, we see from (3.50) and (3.53) that θ ≤1 + 21−jδs−r(1 −L) 2k(1 + δs−r(1 −L)) ≤1 + δs−rL + 21−jδs−r 2k(1 + δs−r) ≤1 + 2δs−rL 2k(1 + δs−r) for k sufficiently large. It now follows from (3.51) that ∆s 4rk ≤δs−r µ 1 − 3 2 −w 2k(1 + δs−r) ¶ , where w = (1 −2δs−r)(log k)−3/2.
Hence if δs is defined by (3.49), then since log(1 −x) ≤−x for 0 < x < 1, we have ∆s 4rk + log ∆s 4rk ≤ δs−r µ 1 − 3 2 −w 2k(1 + δs−r) ¶ + log δs−r − 3 2 −w 2k(1 + δs−r) ≤ δs−r + log δs−r −3 4k + 1 k(log k)3/2 = δs + log δs, so that δs ≥∆s/4rk, since δ + log δ is an increasing function of δ. It follows that 4rkδs is admissible, and this completes the proof of the lemma.
69 We are now fully equipped to prove Theorem 3.2.
Proof of Theorem 3.2. We first note that the theorem is trivial when 1 ≤s ≤s0.
Now when s > s0, define δs to be the unique positive solution of the equation δs + log δs = 1 −3(s −s0) 4rk + s −s0 rk(log k)3/2.
(3.54) We show by induction that the exponent ∆s = 4krδs is admissible whenever s0 < s ≤s1. First suppose that s0 < s ≤s0 + r, and observe that the exponent ∆s = k(k + 1) ≤2r(k + 1) is trivially admissible. Then we have ∆s 4rk ≤1 2 + 1 2k, and hence ∆s 4rk + log ∆s 4rk ≤3 4 + log 3 4 < 1 2 ≤1 −3 4k ≤δs + log δs for k ≥2. It it follows that the exponent 4rkδs is admissible, since δ + log δ is an increasing function of δ. Now suppose that ∆s−r = 4krδs−r is admissible, where s0 + r < s ≤s1. We have by (3.54) that δs−r ≤1 and δs−r + log δs−r ≥1 −3(s1 −s0) 4rk > 1 −log(4rk) + 2 log log k, from which it follows that δs−r > (log k)2 4rk .
Thus Lemma 3.6.2 shows that ∆s = 4rkγs is admissible, where γs is the unique positive solution of γs + log γs = δs−r + log δs−r −3 4k + 1 k(log k)3/2.
Applying (3.54) with s replaced by s −r now shows that γs + log γs = δs + log δs, whence γs = δs, and the induction is complete.
The theorem now follows immediately in the case where 1 ≤s ≤s1, since from (3.54) and the definition of s1 we see that log δs ≤2 −3(s −s0) 4rk for k sufficiently large.
70 Now suppose that s > s1, and let U denote the largest integer with s ≡U (mod r) and U ≤s1, so that U ≥s1 −r. Then the exponent ∆U = 4rke2−3(U−s0)/4rk < e4(log k)2 is admissible, and the theorem follows on applying Lemma 3.5.1 repeatedly.
We note that in the presence of the refined version of Lemma 3.4.3 discussed at the end of Section 3.4, we could replace the factor of r in the second term of (3.43) by 2r and the 3/4k term in (3.49) by 1/k. Hence we would obtain admissible exponents that decay like k2e−2s/k2 in many cases of interest.
CHAPTER IV Applications of the Mean Value Theorems 4.1 Overview The technical apparatus developed in the previous chapter, culminating in the mean value estimates of Theorems 3.1 and 3.2, allows us to consider a variety of Diophantine problems that would not otherwise be accessible. In this chapter, we are primarily concerned with problems involving forms of large degree, while the application of our methods to cubic forms is illustrated in Chapter 5.
In applications involving the circle method, the mean value estimates of Chapter 3 are of fundamental importance to the treatment of the minor arcs, but we also require Weyl estimates in order to proceed with the analysis. Fortunately, our mean value estimates themselves give rise to Weyl estimates by way of the large sieve inequality. Thus in Section 4.2 we will be able to prove Theorem 4.1. For µ > 0, define mµ to be the set of α ∈Rk+1 such that whenever ai ∈Z and q ∈N satisfy (a0, . . . , ak, q) = 1 and |qαi −ai| ≤P µ−kRk (0 ≤i ≤k) one has q > P µRk+1. Suppose that 0 < λ ≤1 2 and that ∆s denotes an admissible exponent. Then given ε > 0 there exists η = η(ε, k) such that whenever R ≤P η one has sup α∈mλ(k+1) |f(α; P, R)| ≪P 2−σ(λ)+ε, where σ(λ) = max 2s≥k+1 λ −(1 −λ)∆s 2s .
(4.1) The following special case of Theorem 4.1 will be sufficient for most of our appli-cations in this chapter.
71 72 Corollary 4.1.1. Given ε > 0, there exists η = η(ε, k) such that whenever R ≤P η one has sup α∈m1/2 |f(α; P, R)| ≪P 2−σ1(k)+ε, where σ1(k)−1 ∼28 3 k3 log k as k →∞.
The estimates contained in Theorem 4.1 have immediate applications to the prob-lem of giving localized bounds for the fractional parts of polynomials in two variables.
Thus in Section 4.3 we are able to establish the following result.
Theorem 4.2. Suppose that α ∈Rk+1, and let ∆s denote an admissible exponent.
Then given ε > 0, there exists N0 = N0(ε, k) such that whenever N > N0 one has min 1≤m,n≤N ||α0mk + α1mk−1n + · · · + αknk|| < N ε−τ(k), where τ(k) = max 2s≥k+1 1 −k∆s 2s(k + 1) + 1 + ∆s .
In particular, we have Corollary 4.2.1. Given α ∈Rk+1 and ε > 0, there exists N0 = N0(ε, k) such that whenever N > N0 one has min 1≤m,n≤N ||α0mk + α1mk−1n + · · · + αknk|| < N ε−ρ(k), where ρ(k)−1 ∼14 3 k3 log k as k →∞.
As remarked in the introduction, Theorem 4.2 and Corollary 4.2.1 are primarily of interest when both α0 and αk are non-zero, since one may otherwise obtain superior results by specializing one variable and applying single-variable methods.
Our remaining applications require the use of the Hardy-Littlewood method, so in Section 4.4 we take a brief detour to develop the necessary major arc approximations for our exponential sums.
In Section 4.5, we consider the multidimensional analogue of Waring’s problem discussed in the introduction. Let Ws(n, P) denote the number of solutions of the system of equations xk−j 1 yj 1 + · · · + xk−j s yj s = nj (0 ≤j ≤k) (4.2) with xi, yi ∈[1, P] ∩Z. The following result quantifies Theorem 6 by providing a lower bound for Ws(n, P).
73 Theorem 4.3. Suppose that s ≥14 3 k2 log k + 10 3 k2 log log k + O(k2), and fix real numbers µ0, . . . , µk with the property that the system ηk−j 1 ξj 1 + · · · + ηk−j s ξj s = µj (0 ≤j ≤k) (4.3) has a non-singular real solution with 0 < ηi, ξi < 1. Suppose also that the system (4.2) has a non-singular p-adic solution for all primes p. Then there exist positive numbers δ = δ(s, k, µ) and P0 = P0(s, k, µ) such that, whenever |nj −P kµj| < δP k (0 ≤j ≤k) (4.4) and P > P0, one has Ws(n, P) ≫P 2s−k(k+1).
Finally, in Section 4.6, we consider the problem of counting rational lines on the hypersurface defined by an additive equation. Let c1, . . . , cs be nonzero integers, and write Ns(P) for the number of solutions of the polynomial equation c1(x1t + y1)k + · · · + cs(xst + ys)k = 0 (4.5) with xi, yi ∈[−P, P] ∩Z.
Equivalently, by the binomial theorem, Ns(P) is the number of solutions of the system of equations c1xk−j 1 yj 1 + · · · + csxk−j s yj s = 0 (0 ≤j ≤k) (4.6) with xi, yi ∈[−P, P] ∩Z.
Theorem 4.4. Suppose that s ≥14 3 k2 log k + 10 3 k2 log log k + O(k2), and that the system of equations (4.6) has a non-singular real solution and a non-singular p-adic solution for all primes p. Then for P sufficiently large one has Ns(P) ≫P 2s−k(k+1).
Given a line ℓ: xt + y, we define the height of ℓby h(ℓ) = max(|xi|, |yi|). Among the solutions counted by Ns(P), we may of course have several that correspond to the same line, so Theorem 4.4 does not directly yield a lower bound for the number of lines on the hypersurface c1zk 1 + · · · + cszk s = 0 (4.7) 74 with h(ℓ) ≤P.
In Section 4.6, however, we will actually derive the estimate of Theorem 4.4 with the variables restricted to dyadic-type intervals and then show that in this situation the number of solutions of (4.6) corresponding to any particular line is at most O(1). Thus we will prove Theorem 4.5. Let Ls(P) denote the number of distinct rational lines ℓlying on the hypersurface (4.7) and satisfying h(ℓ) ≤P. Then, under the hypotheses of Theorem 4.4, one has Ls(P) ≫P 2s−k(k+1).
It is worth remarking that the p-adic solubility conditions imposed in the above theorems in fact need only be checked for finitely many primes p, as we will see in Sections 4.5 and 4.6 that primes sufficiently large in terms of k are dealt with unconditionally using exponential sums.
Throughout this chapter, any statement involving ε and R is taken to mean that for every ε > 0 there exists a positive number η = η(ε, s, k) such that the result holds whenever R ≤P η.
4.2 Weyl Estimates Here we obtain the estimates for smooth Weyl sums quoted in Theorem 4.1 by making simple modifications in the corresponding argument of Wooley . In the end, a standard application of the large sieve inequality shows that these estimates follow from the mean value estimates of Theorems 3.1 and 3.2. Let Cq(Q) = {x ∈Z ∩[1, Q] : s0(x)|s0(q)}, (4.8) write ψ(x, y; α) = k X i=0 αixk−iyi, (4.9) and define the exponential sum hr,v,v′(α; L, L′, R, R′; θ, θ′) = X u∈A(L,R) (u,r)=1 X u′∈A(L′,R′) (u′,r)=1 e(ψ(uv, u′v′; α) + θu + θ′u′).
Also, when π is a prime, we define a set of modified smooth numbers B(M, π, R) = {v ∈N : M < v ≤Mπ, π|v, and p|v ⇒π ≤p ≤R}.
(4.10) We have the following analogue of , Lemma 7.2.
75 Lemma 4.2.1. Suppose that α ∈Rk+1 and r ∈N. Then, whenever R ≤M < Q ≪P and R ≤M ′ < Q′ ≪P, we have X x∈A(Q,R) y∈A(Q′,R) (xy,r)=1 e(ψ(x, y; α)) ≪P ε max π,π′≤R π,π′prime sup θ,θ′∈[0,1] X v∈B(M,π,R) v′∈B(M′,π′,R) (vv′,r)=1 |hr,v,v′(α; T, T ′, π, π′; θ, θ′)| + E, where T = Q/M, T ′ = Q′/M ′, and E ≪Q′M + QM ′.
Proof. By Lemma 10.1 of Vaughan , we have X x∈A(Q,R) y∈A(Q′,R) (xy,r)=1 e(ψ(x, y; α)) = X M<x≤Q x∈A(Q,R) (x,r)=1 X M′<y≤Q′ y∈A(Q′,R) (y,r)=1 e(ψ(x, y; α)) + O(Q′M + QM ′) = X π,π′≤R π,π′prime (r,ππ′)=1 U(α; Q, Q′, M, M ′, R, r, π, π′) + O(Q′M + QM ′), where U(α; Q, Q′, M, M ′, R, r, π, π′) = X v∈B(M,π,R) (v,r)=1 X u∈A(Q/v,π) (u,r)=1 X v′∈B(M′,π′,R) (v′,r)=1 X u′∈A(Q′/v′,π′) (u′,r)=1 e(ψ(uv, u′v′; α)).
Now when v, v′ ≥M we can use orthogonality to write X u∈A(Q/v,π) u′∈A(Q′/v′,π′) (uu′,r)=1 e(ψ(uv, u′v′; α)) = Z 1 0 Z 1 0 hr,v,v′(θ, θ′) X x≤Q/v e(−θx) X x′≤Q′/v′ e(−θ′x′) dθ dθ′, where we have abbreviated hr,v,v′(α; T, T ′, π, π′; θ, θ′) by hr,v,v′(θ, θ′). Thus we see that U(α; Q, Q′, M, M ′, R, r, π, π′) ≪ Z 1 0 Z 1 0 X v∈B(M,π,R) v′∈B(M′,π′,R) (vv′,r)=1 |hr,v,v′(θ, θ′)| min(Q/M, ∥θ∥−1) min(Q′/M ′, ∥θ′∥−1) dθ dθ′, 76 and the lemma follows on noting that Z 1 0 min(X, ∥θ∥−1)dθ ≪1 + log X for X ≥1.
Theorem 4.1 is an easy consequence of the following lemma.
Lemma 4.2.2. Suppose that 0 < λ ≤1 2 and write M = P λ. Let j be an integer with 0 ≤j ≤k, and let α ∈Rk+1. Suppose that a ∈Z and q ∈N satisfy (a, q) = 1, |qαj −a| ≤1 2(MR)−k, q ≤2(MR)k, and either |qαj −a| > MP −k or q > MR.
Then whenever s is a natural number with 2s > max(j, k −j) and the exponent ∆s is admissible we have f(α; P, R)2s ≪P 4s+εM −1(P/M)∆s.
Proof. By Lemma 3.2.4, along with a standard estimate for the divisor function, we see that card(Cq(X)) ≪Xε whenever log q ≪log X, and it follows that f(α; P, R) = X d,e∈Cq(P)∩A(P,R) X x∈A(P/d,R) y∈A(P/e,R) (xy,q)=1 e(ψ(xd, ye; α)) ≪ P ε max d,e∈Cq(M/R) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ X x∈A(P/d,R) y∈A(P/e,R) (xy,q)=1 e(ψ(xd, ye; α)) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + P 1+ε(PR/M).
Thus by Lemma 4.2.1 there exist d, e ∈Cq(M/R), θ, θ′ ∈[0, 1] and primes π, π′ ≤R such that f(α; P, R) ≪P 2+εM −1 + P εg(α; d, e, π, π′, θ, θ′), (4.11) where g(α; d, e, π, π′, θ, θ′) = X v∈B(M/d,π,R) (v,q)=1 X v′∈B(M/e,π′,R) (v′,q)=1 |hr,vd,v′e(α; P/M, P/M, π, π′; θ, θ′)|.
Let J(q, v, d, e, h) denote the number of solutions of the congruence (vd)k−j(xe)j ≡ h (mod q) with 1 ≤x ≤q and (x, q) = 1. When (v, q) = 1, a solution x counted by J(q, v, d, e, h) satisfies dk−jejxj ≡h′ (mod q), and we then necessarily have 77 (h′, q)|dk−jej. In this instance, a simple application of the Chinese Remainder The-orem shows that J(q, v, d, e, h) ≪qεdk−jej.
Thus for any fixed v with (v, q) = 1, we may divide the integers v′ with M/e < v′ ≤MR/e and (v′, q) = 1 into L ≪qεdk−jej classes V1, . . . , VL such that, whenever v′ 1, v′ 2 ∈Vr and (vd)k−j(v′ 1e)j ≡(vd)k−j(v′ 2e)j (mod q), we have v′ 1 ≡v′ 2 (mod q).
Now put Q = P/M, and write cy for the number of solutions of the system s X i=1 uk−j i (u′ i)j = yj (0 ≤j ≤k) (4.12) with ui ∈A(Q, π) and u′ i ∈A(Q, π′) (1 ≤i ≤s) and (ui, q) = (u′ i, q) = 1 (1 ≤i ≤s).
Further, write g(α) for g(α; d, e, π, π′, θ, θ′). Then for some r with 1 ≤r ≤L we have by H¨ older’s inequality that |g(α)|2s ≪P εdk−jej(M 2R2/de)2s−1 X v∈B(M/d,π,R) (v,q)=1 X v′∈Vr ¯ ¯ ¯ ¯ ¯ X y cye(ψ(vd, v′e; αy)) ¯ ¯ ¯ ¯ ¯ 2 , where we have written αy = (α0y0, . . . , αkyk) and where the summation is over y with 1 ≤yi ≤sQk. Applying Cauchy’s inequality, we obtain |g(α)|2s ≪P εM 4s−2Qk2 X∗ y X v∈B(M/d,π,R) (v,q)=1 X v′∈Vr ¯ ¯ ¯ ¯ ¯ ¯ X yj cye(αj(vd)k−j(v′e)jyj) ¯ ¯ ¯ ¯ ¯ ¯ 2 , (4.13) where P∗denotes the sum over yi with i ̸= j.
We now show that the quantities αj(vd)k−j(v′e)j are well-spaced modulo 1 as v′ runs through the set Vr, and it is here that we use the “minor arc” conditions on αj imposed in the statement of the lemma. Fix v ∈B(M/d, π, R), and note that if v′ 1, v′ 2 ∈Vr and v′ 1 ̸≡v′ 2 (mod q) then since |qαj −a| ≤1 2(MR)−k we have ° °αj((vd)k−j(v′ 1e)j −(vd)k−j(v′ 2e)j) ° ° ≥ ° ° ° ° a q ¡ (vd)k−j(v′ 1e)j −(vd)k−j(v′ 2e)j¢° ° ° ° −1 2q ≥ 1 2q.
78 In particular, if q > MR/e, then the elements of Vr are distinct modulo q, so the αj(vd)k−j(v′e)j with v′ ∈Vr are spaced at least 1 2q−1 apart.
Thus it suffices to consider the case when v′ 1 and v′ 2 are distinct elements of Vr with v′ 1 ≡v′ 2 (mod q) and q ≤MR/e. In this case we have ° °αj((vd)k−j(v′ 1e)j −(vd)k−j(v′ 2e)j) ° ° = ° ° ° ° µ αj −a q ¶ (vd)k−jej((v′ 1)j −(v′ 2)j) ° ° ° ° = ¯ ¯ ¯ ¯αj −a q ¯ ¯ ¯ ¯ (vd)k−jej¯ ¯(v′ 1)j −(v′ 2)j¯ ¯.
Now since |qαj −a| > MP −k and v′ 1 −v′ 2 is a nonzero multiple of q, we get ° °αj((vd)k−j(v′ 1e)j −(vd)k−j(v′ 2e)j) ° ° ≥MP −k(vd)k−jej(v′ 1)j−1 ≥(P/M)−k, and thus on applying the large sieve inequality (see for example ) to (4.13) we obtain g(α; d, e, π, π′, θ, θ′)2s ≪P εM 4s−2(P/M)k2(q + (P/M)k) X v∈B(M/d,π,R) X y |cy|2.
But P y |cy|2 ≤Ss(P/M, R) and q ≤2(MR)k ≪(P/M)k so on recalling (4.11) we have f(α; P, R)2s ≪ P 4s+εM −2s + P εM 4s−1(P/M)k2(P/M)k(P/M)4s−k(k+1)+∆s ≪ P 4s+εM −1(P/M)∆s, as required.
Proof of Theorem 4.1. Suppose that α ∈mλ(k+1) and write M = P λ. By Dirichlet’s Theorem there exist bi ∈Z and qi ∈N with (bi, qi) = 1 such that |qiαi −bi| ≤1 2(MR)−k and qi ≤2(MR)k (0 ≤i ≤k).
If for some j we have either |αj −bj/qj| > q−1 j MP −k or qj > MR, then the desired conclusion follows from Lemma 4.2.2. Otherwise, write q = [q0, . . . , qk] and ai = biq/qi. Then (a0, . . . , ak, q) = 1, and for each i we have q ≤qi(MR)k ≤(MR)k+1 = P λ(k+1)Rk+1 and |αi −ai/q| ≤q−1(MR)kMP −k = q−1P λ(k+1)−kRk.
79 This contradicts the assumption that α ∈mλ(k+1) and hence completes the proof.
Proof of Corollary 4.1.1. We apply Theorem 4.1 with λ = 1 2(k+1). By (4.1), we have σ(λ) = max 2s≥k+1 1 −(2k + 1)∆s 4s(k + 1) .
Then on taking s = ·µ7 3 log 4rk + 2 log log k + 8 ¶ rk ¸ + 1 ∼7 3k2 log k, we have by Theorem 3.2 that the exponent ∆s = e4(log k)2e−(s−s1)/2rk ≤ 1 k(log k)1/3 is admissible. It follows that σ(λ) ≥ 1 + O((log k)−1/3) 28 3 k3(log k + O(log log k)) ∼ µ28 3 k3 log k ¶ −1 .
We remark that the proof of Lemma 4.2.2, with trivial modifications, may be applied to more general exponential sums of the form f(α; P, Q, R) X x∈A(P,R) X y∈A(Q,R) e(α0xk + α1xk−1y + · · · + αkyk), provided that P ≍Q, and hence Theorem 4.1 and Corollary 4.1.1 hold in this case as well. This observation will be useful in the analysis of Section 4.6.
4.3 Fractional Parts The following lemma allows us to deduce information about the distribution of fractional parts from the exponential sum estimates of Theorem 4.1.
Lemma 4.3.1. Suppose that g(m, n) is a non-negative function, L is a positive in-teger, and αm,n are real numbers. If L X l=1 ¯ ¯ ¯ ¯ ¯ N X m=1 N X n=1 g(m, n)e(lαm,n) ¯ ¯ ¯ ¯ ¯ < 1 6 N X m=1 N X n=1 g(m, n), then for any β there is a solution of ||αm,n + β|| < L−1 with 1 ≤m, n ≤N.
Proof. For each integer n with 1 ≤n ≤N 2, we can write n = qN + r for unique integers q = q(n) and r = r(n) satisfying 0 ≤q ≤N −1 and 1 ≤r ≤N. Now we apply Lemma 5 of Harman to the sequence α′ n = αq+1,r (1 ≤n ≤N 2).
80 The localized fractional parts estimates of Theorem 4.2 now follow in a routine manner.
Proof of Theorem 4.2. Fix α ∈Rk+1, let P be a large real number, and let ε be a real number with 0 < ε < τ. Further, write H1 = P τ−ε and T1(α) = X 1≤h≤H1 |f(hα; P, R)|, (4.14) where R = P η and η < η0(ε, k). We divide into cases.
(i) Suppose that there exist h, b, and q with 1 ≤h ≤H1, b ∈Zk+1, q ∈N, (q, b0, . . . , bk) = 1, and q ≤P 1−τRk+1 such that |qhαi −bi| ≤P 1−τ−kRk (0 ≤i ≤k).
Then for each i, we have ||αi(qh)k|| ≤|(qh)k−1(qhαi −bi)| < Hk−1 1 P −kτRk2+k−1 < 1 (k + 1)H1 for η sufficiently small. Thus on noting that qh ≤P we obtain min 1≤m,n≤P ||ψ(m, n; α)|| ≤||ψ(qh, qh; α)|| ≤ k X i=0 ||αi(qh)k|| < H−1 1 , which completes the proof in this case.
(ii) If the hypothesis of case (i) does not hold, then we may apply Theorem 4.1 with λ = 1−τ k+1 to obtain max 1≤h≤H1 |f(hα; P, R)| ≪P 2−σ(λ)+ε/2, and hence T1(α) ≪P 2+τ−σ(λ)−ε/2.
Now choose s so that τ = 1 −k∆s 2s(k + 1) + 1 + ∆s .
Then a simple calculation shows that τ = 1 −τ −(k + τ)∆s 2s(k + 1) ≤σ(λ), 81 whence T1(α) = o(P 2). The theorem now follows by applying Lemma 4.3.1 with g(m, n) = 1 if m, n ∈A(P, R), 0 otherwise, on recalling (see for example ) that card(A(P, R)) ≫P when R = P η.
Proof of Corollary 4.2.1. As in the proof of Corollary 4.1.1, we take s = ·µ7 3 log 4rk + 2 log log k + 8 ¶ rk ¸ + 1 and apply Theorem 3.2 to estimate ∆s. In the notation of Theorem 4.2, we find that τ(k) ≥1 + O((log k)−1/3) 2s(k + O(1)) ∼ µ14 3 k3 log k ¶ −1 , and this completes the proof.
4.4 Generating Function Asymptotics In this section, we derive the asymptotic formulas for our generating functions that will be required to handle the major arcs in our subsequent applications of the circle method.
As is now familiar in the applications of smooth numbers to additive number theory, one can only obtain asymptotics for the exponential sum f(α; P, R) on a very thin set of major arcs, so it is necessary to introduce sums over a complete interval in order to facilitate a pruning procedure. Thus we write F(α) = X 1≤x,y≤P e(α0xk + α1xk−1y + · · · + αkyk), and we also define S(q, a) = X 1≤x,y≤q e µa0xk + a1xk−1y + · · · + akyk q ¶ , v(β) = Z P 0 Z P 0 e(β0γk + β1γk−1ν + · · · + βkνk) dγ dν, (4.15) and V (α) = V (α; q, a) = q−2S(q, a)v(α −a/q).
82 Lemma 4.4.1. When αi = ai/q + βi for 0 ≤i ≤k, one has F(α) −V (α) ≪q2 + qP k+1 (|β0| + · · · + |βk|) .
Proof. On sorting the terms into arithmetic progressions modulo q, we have F(α) = q X r=1 q X s=1 e µa0rk + · · · + aksk q ¶ X 0≤i≤P −r q X 0≤j≤P −s q e(ψ(iq + r, jq + s; β)), where ψ(x, y; α) = k X i=0 αixk−iyi.
Thus on making the change of variables γ = qz + r and ν = qw + s in (4.15), we obtain F(α) −V (α) = X 1≤r,s≤q e µa0rk + · · · + aksk q ¶ X i,j i+1 Z i j+1 Z j H(z, w) dz dw + O(1) , where H(z, w) = H(z, w; r, s; i, j; β) = e(ψ(iq + r, jq + s; β)) −e(ψ(qz + r, qw + s; β)).
Using the mean value theorem, we find that H(z, w) ≪qP k−1 (|β0| + · · · + |βk|) when (z, w) ∈[i, i + 1] × [j, j + 1] and hence F(α) −V (α) ≪q2(1 + q−1P k+1 (|β0| + · · · + |βk|)), from which the lemma follows.
We now begin to analyze the sum f(α; P, R). First we record a partial summation lemma analogous to Lemma 2.6 of Vaughan .
Lemma 4.4.2. Let cm,n be arbitrary complex numbers, and suppose that F(x, y) has continuous partial derivatives on [0, X] × [0, Y ]. Then X m≤X n≤Y cm,nF(m, n) = X m≤X n≤Y cm,n ¡ F(X, n) + F(m, Y ) −F(X, Y ) ¢ + Z X 0 Z Y 0 ∂2 ∂γ∂ν F(γ, ν) X m≤γ n≤ν cm,n dν dγ.
83 Proof. Write Fγ(ν) = ∂ ∂γF(γ, ν). Then we have Fγ(n) = Fγ(Y ) − Z Y n ∂ ∂ν Fγ(ν) dν and F(m, n) = F(X, n) − Z X m Fγ(n) dγ.
Thus we can write F(m, n) = F(X, n) − Z X m Fγ(Y ) dγ + Z X m Z Y n ∂2 ∂γ∂ν F(γ, ν) dν dγ, and the lemma follows on summing over m and n and interchanging the order of integration and summation in the last term.
Using the well-known asymptotics for card(A(X, R)) in terms of Dickman’s ρ function, we can record the following lemma.
Lemma 4.4.3. Let τ be a fixed number, and suppose that R ≤m, n ≤Rτ. Then X x∈A(m,R) y∈A(n,R) 1 = ρ µlog m log R ¶ ρ µ log n log R ¶ mn + O µ mn log R ¶ .
Proof. By Lemma 5.3 of Vaughan , we have X x∈A(X,R) 1 = ρ µlog X log R ¶ X + O µ X log X ¶ whenever R ≤X ≤Rτ, and the result follows immediately.
Now let W be a parameter at our disposal, and write N(q, a) = {α ∈Tk+1 : |αi −ai/q| ≤WP −k (0 ≤i ≤k)} (4.16) whenever q ≤W and (q, a0, . . . , ak) = 1. Further, let R = P η, and write w(β) = Z P R Z P R ρ µ log γ log R ¶ ρ µ log ν log R ¶ e(β0γk + · · · + βkνk) dγ dν.
(4.17) Lemma 4.4.4. Suppose that α ∈N(q, a) with q ≤R, and write βi = αi −ai/q.
Then we have f(α; P, R) = q−2S(q, a)w(β) + O µq2P 2W 2 log P ¶ .
84 Proof. By arguing as in the proof of Vaughan , Lemma 5.4, we obtain X x∈A(m,R) x≡r (q) X y∈A(n,R) y≡s (q) 1 = 1 q2 X x∈A(m,R) y∈A(n,R) 1 + O µ P 2 log P ¶ whenever R ≤m, n ≤P, and hence by Lemma 4.4.3 we have X x∈A(m,R) y∈A(n,R) e µa0xk + · · · + akyk q ¶ = q−2S(q, a) X x∈A(m,R) y∈A(n,R) 1 + O µ q2P 2 log P ¶ = q−2S(q, a)ρ µlog m log R ¶ ρ µ log n log R ¶ mn + E1, where E1 ≪q2P 2/ log P. Now let B = A(P, R) × A(P, R), and write 1B for the characteristic function of B. Then by taking cx,y = e µa0xk + · · · + akyk q ¶ 1B(x, y) and F(x, y) = e(β0xk + · · · + βkyk) in Lemma 4.4.2 we find that f(α; P, R) = X 1≤x,y≤P cx,yF(x, y) = S0 −S1 + S2, (4.18) where S0 = X x,y∈A(P,R) e µa0xk + · · · + akyk q ¶¡ e(β0P k + · · · + βkyk) + e(β0xk + · · · + βkP k) ¢ , S1 = X x,y∈A(P,R) e µa0xk + · · · + akyk q ¶ e(β0P k + · · · + βkP k), and S2 = Z P 0 Z P 0 ∂2 ∂γ∂ν ¡ e(β0γk + · · · + βkνk) ¢ X x∈A(γ,R) y∈A(ν,R) e µa0xk + · · · + akyk q ¶ dν dγ.
From our observations above, we see immediately that S1 = q−2S(q, a)P 2ρ(1/η)2e(β0P k + · · · + βkP k) + O µ q2P 2 log P ¶ .
(4.19) 85 We next observe that, by equation (8.13) of Wooley , one has X x∈A(m,R) e µa0xk + · · · + akyk q ¶ = q−1S(q, a; y) m ρ µlog m log R ¶ + O µ qP log P ¶ , (4.20) where S(q, a; y) = X 1≤x≤q e µa0xk + a1xk−1y + · · · + akyk q ¶ .
If we write S0 = S3 + S4, then by (4.20) we have S3 = X x,y∈A(P,R) e µa0xk + · · · + akyk q ¶ e(β0P k + · · · + βkyk) = q−1ρ(1/η)P X y∈A(P,R) S(q, a; y)e(β0P k + · · · + βkyk) + O µ qP 2 log P ¶ , and then by partial summation S3 = q−1P ρ(1/η) T(P) e(β0P k + · · · + βkP k) −q−1P ρ(1/η) Z P R T(ν) ∂ ∂ν ¡ e(β0P k + · · · + βkνk) ¢ dν + O µ qP 2 log P ¶ , where T(ν) = X y∈A(ν,R) S(q, a; y).
But on using the obvious analogue of (4.20) we find that T(ν) = q−1S(q, a) νρ µ log ν log R ¶ + O µ q2P log P ¶ , and since α ∈N(q, a) have ∂ ∂ν ¡ e(β0P k + · · · + βkνk) ¢ ≪W/P.
Therefore we obtain S3 = Q ρ(1/η) P e(β0P k + · · · + βkP k) −Q I(P) + O µqP 2W log P ¶ , (4.21) where Q = q−2S(q, a)ρ(1/η)P and I(γ) = Z P R ν ρ µ log ν log R ¶ ∂ ∂ν ¡ e(β0γk + · · · + βkνk) ¢ dν.
86 Integration by parts yields I(γ) = ρ(1/η)P e(β0γk + · · · + βkP k) − Z P R e(β0γk + · · · + βkνk) ∂ ∂ν µ νρ µ log ν log R ¶¶ dν + O(R), but ∂ ∂ν µ νρ µ log ν log R ¶ ¶ = ρ µ log ν log R ¶ + 1 log R ρ′ µ log ν log R ¶ = ρ µ log ν log R ¶ + O µ 1 log P ¶ , since ρ′(x) ≪1. Thus we have I(γ) = ρ(1/η)P e(β0γk + · · · + βkP k) − Z P R e(β0γk + · · · + βkνk) ρ µ log ν log R ¶ dν + E2(γ), (4.22) where E2(γ) ≪P/ log P, so it follows from (4.21) that S3 = Q Z P R ρ µ log ν log R ¶ e(β0P k + · · · + βkνk) dν + O µqP 2W log P ¶ .
(4.23) Moreover, an identical argument shows that S4 = Q Z P R ρ µ log γ log R ¶ e(β0γk + · · · + βkP k) dγ + O µqP 2W log P ¶ .
(4.24) We now deal with S2. A simple calculation shows that ∂2 ∂γ∂ν ¡ e(β0γk + · · · + βkνk) ¢ ≪W 2/P 2.
when |βi| ≤WP −k, and it follows easily from the calculation at the beginning of the proof that S2 = Z P R Z P R ∂2 ∂γ∂ν ¡ e(β0γk + · · · + βkνk) ¢ q−2S(q, a) ρ µ log γ log R ¶ ρ µ log ν log R ¶ γν dγ dν + O µq2P 2W 2 log P ¶ .
After interchanging the order of differentiation and integration, we can write S2 = q−2S(q, a) Z P R γ ρ µ log γ log R ¶ I′(γ) dγ + O µq2P 2W 2 log P ¶ , 87 and on integrating by parts we get S2 = q−2S(q, a) µ P ρ(1/η) I(P) − Z P R I(γ) ρ µ log γ log R ¶ dγ ¶ + O µq2P 2W 2 log P ¶ .
Then from (4.22) we finally obtain S2 = q−2S(q, a)w(β) + E3, where E3 = q−2S(q, a)ρ(1/η)2P 2e(β0P k + · · · + βkP k) −Q Z P R ρ µ log ν log R ¶ e(β0P k + · · · + βkνk) dν −Q Z P R ρ µ log γ log R ¶ e(β0γk + · · · + βkP k) dγ + O µq2P 2W 2 log P ¶ , and the lemma follows on recalling (4.18), (4.19), (4.23), and (4.24).
4.5 A Multidimensional Analogue of Waring’s Problem Here we establish Theorem 4.3 by a fairly straightforward application of the Hardy-Littlewood method. Let P be a large positive number, and put R = P η, where η ≤η0(ε, k). Let F(α) be as in the previous section, and write f(α) = f(α; P, R).
Further, put s = t + 2u + v, and let Rs(n) = Z Tk+1 F(α)tf(α)2u+ve(−α · n) dα.
Then we have Ws(n, P) ≥Rs(n), so it suffices to obtain a lower bound for Rs(n).
We dissect Tk+1 into major and minor arcs as follows. Recalling the notation of Theorem 4.1, define m = m1/2 and M = Tk+1 \ m.
We take t = (k + 1)2, u = ·7 3k2 log k + 5 3k2 log log k + 6k2 ¸ , and v = · ∆u σ1(k) ¸ + 1, where ∆u is as in Theorem 3.2 and σ1(k) is as in Corollary 4.1.1. A simple calculation shows that v ≪k2, and hence s = 14 3 k2 log k + 10 3 k2 log log k + O(k2).
88 On applying the aforementioned theorem and corollary, we find that Z m |F(α)|t|f(α)|2u+vdα ≪ P 2t sup α∈m|f(α)|v Z Tk+1 |f(α)|2udα ≪ P 2s−k(k+1)−δ (4.25) for some δ > 0, since ∆u < vσ1(k). Thus it remains to deal with the major arcs.
When (q, a0, . . . , ak) = 1, define M(q, a) = {α ∈Tk+1 : |qαi −ai| ≤P 1/2−kRk (0 ≤i ≤k)}, (4.26) so that M = [ 1≤a0,...,ak≤q≤P 1/2Rk+1 (q,a0,...,ak)=1 M(q, a).
It is a simple exercise to show that the M(q, a) are pairwise disjoint. On recalling the notation of the previous section, we can record the following major arc approximation for F(α).
Lemma 4.5.1. Suppose that α ∈M(q, a), and write βi = αi −ai/q. Then one has F(α) −q−2S(q, a)v(β) ≪P 3/2+ε.
Proof. This follows immediately from Lemma 4.4.1, together with (4.26).
The following estimates for S(q, a), v(β), and w(β) are essentially immediate from the work of Arkhipov, Karatsuba, and Chubarikov .
Lemma 4.5.2. Whenever (q, a0, . . . , ak) = 1, we have S(q, a) ≪q2−1/k+ε.
Proof. This follows easily from , Lemma II.8, on recalling standard divisor function estimates.
Lemma 4.5.3. One has v(β) ≪P 2(1 + P k(|β0| + · · · + |βk|))−1/k and w(β) ≪P 2(1 + P k(|β0| + · · · + |βk|))−1/k.
89 Proof. The first estimate follows from , Lemma II.2, on making a change of vari-able, and the second follows in a similar manner (see the comment in the proof of , Lemma 8.6) on noting that ρ(log γ/ log R) ≍1 and is decreasing for R ≤γ ≤P.
We now use the information contained in the above lemmata to prune back to a very thin set of major arcs on which f(α) can be suitably approximated. Specifically, let W be a parameter at our disposal, and recall the definition of N(q, a) from the previous section. Further, let N = [ 1≤a0,...,ak≤q≤W (q,a0,...,ak)=1 N(q, a).
We have the following result, which is closely analogous to , Lemma 9.2.
Lemma 4.5.4. If t is an integer with t ≥(k + 1)2, then one has Z M |F(α)|t dα ≪P 2t−k(k+1) and Z M\N |F(α)|t dα ≪W −σP 2t−k(k+1) for some σ > 0.
Proof. When α ∈M(q, a), we have by Lemma 4.5.1 that F(α)t −V (α)t ≪ ¡ P (3/2+ε)¢t + P 3/2+ε|V (α)|t−1, (4.27) and the proof now follows the argument of Wooley , Lemma 9.2, employing Lemma 4.5.2 together with the estimate v(β) ≪P 2 k Y i=0 (1 + P k|βi|)−1/k(k+1), which is immediate from Lemma 4.5.3.
On making a trivial estimate for f(α), it follows directly from Lemma 4.5.4 that Z M\N |F(α)|t|f(α)|2u+v dα ≪W −σP 2s−k(k+1), (4.28) for some σ > 0, so it suffices to deal with the pruned major arcs N. When α ∈ N(q, a), we have by Lemma 4.4.4 that f(α)2u+v −W(α)2u+v ≪ µq2P 2W 2 log P ¶2u+v + q2P 2W 2 log P |W(α)|2u+v−1, 90 where W(α) = W(α; q, a) = q−2S(q, a)w(β) and βi = αi −ai/q.
On combining this with (4.27) and recalling the definition of N, we find that Z N F(α)tf(α)2u+ve(−α · n) dα = Z N V (α)tW(α)2u+ve(−α · n) dα + O(P 2s−k(k+1)(log P)−δ) for some δ > 0, provided that W is chosen to be a suitably small power of log P.
Now let S(q) = X 1≤a0,...,ak≤q (q,a0,...,ak)=1 (q−2S(q, a))se µ−a0n0 −· · · −aknk q ¶ , S(n, P) = X q≤W S(q), and S(n) = ∞ X q=1 S(q).
Notice that by Lemma 4.5.2 we have S(q) ≪qk+1−s/k+ε, whence S(n) ≪1 and S(n) −S(n, P) ≪P −δ for some δ > 0, provided that s ≥(k + 1)2. Further, let J(n, P) = Z B(P) v(β)tw(β)2u+ve(−β · n) dβ, where B(P) = [−WP −k, WP −k]k+1, and put J(n) = Z Rk+1 v(β)tw(β)2u+ve(−β · n) dβ.
Then when s ≥(k + 1)2, it follows easily from Lemmata 4.5.2 and 4.5.3 that J(n) ≪P 2s−k(k+1) 91 and X 1≤q≤P 1/2+ε |S(q)||J(n) −J(n, P)| ≪P 2s−k(k+1)(log P)−δ for some δ > 0. Combining these observations, we find that Z N F(α)tf(α)2u+ve(−α · n) dα = S(n)J(n) + O(P 2s−k(k+1)(log P)−δ) (4.29) for some δ > 0, again provided that W is a sufficiently small power of log P. The singular integral J(n) and the singular series S(n) require further analysis.
Lemma 4.5.5. Suppose that s ≥(k + 1)2, and fix real numbers µ0, . . . , µk with the property that the system (4.3) has a non-singular real solution with 0 < ηi, ξi < 1.
Then there exists a positive number δ′ = δ′(s, k, µ) such that, whenever |nj −P kµj| < δ′P k (0 ≤j ≤k) and P is sufficiently large, one has J(n) ≫P 2s−k(k+1).
Proof. After a change of variables, we have J(n) = P 2s−k(k+1) Z Rk+1 Z B T(γ, ν) e à k X j=0 βj(φj(γ, ν) −µj + δj) !
dγ dν dβ, where B = [0, 1]2t × [R/P, 1]4u+2v, T(γ, ν) = s Y i=t+1 ρ µlog Pγi log R ¶ ρ µlog Pνi log R ¶ , (4.30) φj(γ, ν) = γk−j 1 νj 1 + · · · + γk−j s νj s, and where |δj| ≤δ′ for each j. Notice that (η, ξ) is contained in B for P sufficiently large. Now let S(t0, . . . , tk) = {(γ, ν) ∈B : φj(γ, ν) −µj + δj = tj (0 ≤j ≤k)}, 92 so that J(n) = P 2s−k(k+1) Z Rk+1 Z C Z S(t0,...,tk) T(γ, ν) e(β0t0 + · · · + βktk) dS(t) dt dβ, where C ⊂Rk+1.
Since (η, ξ) ∈B, we see that C contains a neighborhood of (δ0, . . . , δk) and hence contains the origin when δ′ is sufficiently small. Thus after k +1 applications of Fourier’s Integral Theorem (see for example Davenport ) we obtain J(n) = P 2s−k(k+1) Z S(0) T(γ, ν) dS(0).
Now, for δ′ sufficiently small, the implicit function theorem shows that S(0) is a space of dimension 2s −k −1 with positive (2s −k −1)-dimensional measure, and the lemma follows on noting that T(γ, ν) ≫1 for R/P ≤γ, ν ≤1.
It remains to deal with p-adic solubility considerations and hence to obtain a lower bound for the singular series S(n).
Lemma 4.5.6. The function S(q) is multiplicative.
Proof. By , Lemma II.4, one has S(qr, a) = S(q, rk−1a)S(r, qk−1a) whenever (q, r) = 1, and the result now follows by a standard argument.
For each prime p, write σ(p) = ∞ X h=0 S(ph).
Whenever s ≥(k + 1)2 one finds using Lemmata 4.5.2 and 4.5.6 that S(n) = Y p σ(p) (4.31) and that there exists a constant C(k) such that 1 2 ≤ Y p>C(k) σ(p) ≤3 2.
(4.32) Hence it remains to deal with small primes. Let Mn(q) denote the number of solutions of the system of congruences xk−j 1 yj 1 + · · · + xk−j s yj s ≡nj (mod q) (0 ≤j ≤k).
(4.33) 93 Lemma 4.5.7. One has X d|q S(d) = qk+1−2sMn(q).
Proof. By the orthogonality of the additive characters modulo q, one has Mn(q) = 1 qk+1 q X r0=1 · · · q X rk=1 (S(q, r))s e (−(r · n)/q) .
Now on writing d = q/(q, r0, . . . , rk) and ai = rid/q we obtain Mn(q) = 1 qk+1 X d|q X 1≤a0,...,ak≤d (d,a0,...,ak)=1 (q/d)2s (S(d, a))s e (−(a · n)/d) , and the result follows.
We therefore have σ(p) = lim h→∞ph(k+1−2s)Mn(ph), (4.34) so to show that S(n) ≫1 it suffices to obtain a suitable lower bound for Mn(ph).
In order to deduce this from the existence of non-singular p-adic solutions to (4.2), we need a version of Hensel’s Lemma. In what follows, we write | · |p for the usual p-adic valuation, normalized so that |p|p = p−1.
Lemma 4.5.8. Let ψ1, . . . , ψr be polynomials in Zp[x1, . . . , xr] with Jacobian ∆(ψ; x), and suppose that a ∈Zr p satisfies |ψj(a)|p < |∆(ψ; a)|2 p (1 ≤j ≤r).
Then there exists a unique b ∈Zr p such that ψj(b) = 0 (1 ≤j ≤r) and |bi −ai|p < p−1|∆(ψ; a)|p (1 ≤i ≤r).
Proof. This is Proposition 5.20 of Greenberg with R = Zp.
Lemma 4.5.9. Suppose that the system (4.2) has a non-singular p-adic solution.
Then there exists an integer u = u(p) such that whenever h ≥u one has Mn(ph) ≥p(h−u)(2s−k−1).
94 Proof. We relabel the variables by writing (z1, . . . , z2s) = (x1, . . . , xs, y1, . . . , ys), and let a = (a1, . . . , a2s) be a non-singular p-adic solution of (4.2). Then there exist indices i0, . . . , ik such that ∆(ψ; ai0, . . . , aik) ̸= 0, so we can find an integer u such that |∆(ψ; ai0, . . . , aik)|2 p = p1−u > 0.
Now suppose that h ≥u. For i / ∈{i0, . . . , ik}, choose integers wi with wi ≡ai (mod pu), and write vi = ai for i = i0, . . . , ik and vi = wi otherwise. Then on writing ψj(z) = ψj(x, y) = xk−j 1 yj 1 + · · · + xk−j s yj s −nj for 0 ≤j ≤k, we see that ψj(v) ≡ψj(a) ≡0 (mod pu), and hence |ψj(v)|p ≤p−u < |∆(ψ; vi0, . . . , vik)|2 p.
Now if h ≥u then there are p(h−u)(2s−k−1) possible choices for the wi modulo ph.
Moreover, for any fixed choice we may regard ψj as a polynomial in the k + 1 vari-ables zi0, . . . , zik after substituting zi = wi on the remaining indices. Thus for each admissible choice of w we may apply Lemma 4.5.8 to obtain integers bi0, . . . , bik such that ψj(b; w) ≡0 (mod ph) for each j, whence the lemma follows.
Now by (4.34) and Lemma 4.5.9 we have σ(p) ≥pu(k+1−2s) for all primes p, so on combining this with (4.31) and (4.32) we see that S(n) ≫1. Hence the proof of Theorem 4.3 is complete upon recalling Lemma 4.5.5, together with (4.25), (4.28), and (4.29).
4.6 Lines on Additive Equations We now establish Theorems 4.4 and 4.5 by proceeding much as in the previous section. Before embarking on the circle method, however, we need to make some preliminary observations.
Lemma 4.6.1. Suppose that (x, y) ∈R2s is a solution of (4.6), and let a, b, c, and d be arbitrary real numbers. Then (ax + by, cx + dy) is also a solution.
95 Proof. For 0 ≤j ≤k, write Aj(x, y) = s X i=1 ci(axi + byi)k−j(cxi + dyi)j.
Then by the binomial theorem we have for each j that Aj(x, y) = s X i=1 ci k−j X r=0 µk −j r ¶ (axi)k−j−r(byi)r j X s=0 µj s ¶ (cxi)j−s(dyi)s = k−j X r=0 j X s=0 µk −j r ¶µj s ¶ ak−j−rbrcj−sds s X i=1 cixk−(r+s) i yr+s i , and the lemma follows.
Lemma 4.6.2. Suppose that the system of equations (4.6) has a non-singular real solution (η, ξ). Then we can find a non-singular real solution (η′, ξ′) such that η′ i and ξ′ i are nonzero for each i.
Proof. For 0 ≤j ≤k, let ψj(x, y) = c1xk−j 1 yj 1 + · · · + csxk−j s yj s, and write (z0, . . . , z2s−1) = (x1, . . . , x2s, y1, . . . , y2s). Then by rearranging variables, we may write the given real solution as (η, ξ) = (γ0, . . . , γ2s−1), where det µ∂ψj ∂zi (γ) ¶ 0≤i,j≤k ̸= 0.
Hence by using the Implicit Function Theorem as in the proof of , Lemma 6.2, we see that there exists a (2s −k −1)-dimensional neighborhood T0 of (γk+1, . . . , γ2s−1) and a function φ : T0 →Rk+1 such that γ = (φ(w), w) is a solution of (4.6) whenever w ∈T0. Thus by choosing w with |wi −γi| sufficiently small for k + 1 ≤i ≤2s −1, we may assume that γ is a non-singular solution whose last 2s −k −1 coordinates are nonzero. Moreover, a simple calculation shows that at most two of the remaining ηi and at most two of the remaining ξi are zero and that either ηi or ξi is nonzero for every i. In particular, when s ≥5, there is some i for which ηiξi ̸= 0. Now let b = min{|ηi/ξi| : ηiξi ̸= 0} and c = min{|ξi/ηi| : ηiξi ̸= 0}, and take b′ < 1 2b and c′ < 1 2c. Then by Lemma 4.6.1 we see that (η′, ξ′) is a solution of (4.6), where η′ = η + b′ξ and ξ′ = c′η + ξ, and it is easy to check that η′ i and ξ′ i are nonzero for each i. The non-singularity follows by continuity on choosing b′ and c′ sufficiently small.
96 By Lemma 4.6.2 we may henceforth suppose that the system (4.6) has a non-singular real solution (η, ξ) with ηi and ξi nonzero for all i, and by homogeneity we can re-scale to ensure that 0 < |ηi|, |ξi| < 1 2. For each i, write η+ i = ηi + 1 2|ηi| and η− i = ηi −1 2|ηi| and ξ+ i = ξi + 1 2|ξi| and ξ− i = ξi −1 2|ξi|.
Now let P be a large positive number, put R = P η with η ≤η0(ε, k), and let c1, . . . , cs be nonzero integers. Throughout this section, the implicit constants arising in our analysis may depend on c1, . . . , cs and on the real solution (η, ξ). We define the exponential sums Fi(α) = X η− i P<x≤η+ i P X ξ− i P cP 1/2Rk+1 and hence q > P 1/2Rk+1. Thus ciα ∈m1/2.
The second assertion now follows on recalling the remark at the end of Section 4.2 and noting that we may replace αj by −αj as needed so that our sums are over positive integers.
As in the previous section, we take t = (k + 1)2, u = ·7 3k2 log k + 5 3k2 log log k + 6k2 ¸ , and v = · ∆u σ1(k) ¸ + 1, where ∆u is as in Theorem 3.2 and σ1(k) is as in Corollary 4.1.1. Then by H¨ older’s inequality and a change of variables we obtain Z m |F(α)G(α)| dα ≪ P 2t+v(2−σ1+ε) t+2u Y i=t+1 µZ Tk+1 |fi(α)|2udα ¶1/2u ≪ P 2s−k(k+1)−δ (4.35) for some δ > 0, since ∆u < vσ1(k). Thus it remains to deal with the major arcs.
Recalling the notation of the previous section, we define Si(q, a) = S(q, cia), vi(β) = Z η+ i P η− i P Z ξ+ i P ξ− i P e(ci(β0γk + β1γk−1ν + · · · + βkνk)) dγ dν, 98 and Vi(α) = q−2Si(q, a)vi(α −a/q) (α ∈M(q, a)).
Further, we define the pruned major arcs N exactly as in the previous section, again with W a suitable power of log P, and write wi(β) = Z η+ i P η− i P Z ξ+ i P ξ− i P ρ µ log γ log R ¶ ρ µ log ν log R ¶ e(ci(β0γk + β1γk−1ν + · · · + βkνk)) dγ dν, and Wi(α) = q−2Si(q, a)wi(α −a/q) (α ∈N(q, a)).
The next several lemmas are simple adaptations of the corresponding results in the previous section.
Lemma 4.6.4. When α ∈M(q, a), one has Fi(α) −Vi(α) ≪P 3/2+ε, and when α ∈N(q, a), one has fi(α) −Wi(α) ≪q2P 2W 2 log P .
Proof. These estimates follow by making trivial modifications in the arguments of Lemmata 4.4.1 and 4.4.4, respectively.
Lemma 4.6.5. Whenever (q, a0, . . . , ak) = 1, we have Si(q, a) ≪q2−1/k+ε.
Proof. Put di = (q, ci). Then by Lemma 4.5.2 we have Si(q, a) = d2 i S(q/di, cia/di) ≪d1/k i q2−1/k+ε ≪c q2−1/k+ε, as required.
Lemma 4.6.6. One has vi(β) ≪P 2(1 + P k(|β0| + · · · + |βk|))−1/k and wi(β) ≪P 2(1 + P k(|β0| + · · · + |βk|))−1/k.
99 Proof. The argument is identical to the proof of Lemma 4.5.3.
Lemma 4.6.7. If t is an integer with t ≥(k + 1)2, then one has Z M |Fi(α)|t dα ≪P 2t−k(k+1) (4.36) and Z M\N |Fi(α)|t dα ≪W −σP 2t−k(k+1) (4.37) for some σ > 0.
Proof. The result follows as in Lemma 4.5.4 on using Lemmata 4.6.4, 4.6.5, and 4.6.6 in place of the corresponding results in the previous section.
Once again, Lemma 4.6.7, together with (4.35), allows us to focus attention on the pruned major arcs N. Let S(q) = X 1≤a0,...,ak≤q (q,a0,...,ak)=1 q−2s s Y i=1 Si(q, a), S(P) = X q≤X S(q), and S = ∞ X q=1 S(q).
Again we have S(q) ≪qk+1−s/k+ε, and hence S ≪1 and S −S(P) ≪P −δ for some δ > 0, provided that s ≥(k + 1)2. Further, let J(P) = Z B(P) t Y i=1 vi(β) s Y i=t+1 wi(β) dβ, where B(P) = [−WP −k, WP −k]k+1, and put J = Z Rk+1 t Y i=1 vi(β) s Y i=t+1 wi(β) dβ.
Then when s ≥(k + 1)2, we have by Lemmata 4.6.5 and 4.6.6 that J ≪P 2s−k(k+1) and X 1≤q≤cP 1/2+ε |S(q)||J −J(P)| ≪P 2s−k(k+1)(log P)−δ.
for some δ > 0. Thus, by employing standard arguments based on Lemmata 4.6.4, 4.6.5, and 4.6.6, we obtain Z N F(α)G(α) dα = SJ + O(P 2s−k(k+1)(log P)−δ) (4.38) for some δ > 0.
100 Lemma 4.6.8. Whenever s ≥(k + 1)2 and P is sufficiently large, one has J ≫P 2s−k(k+1).
Proof. By a change of variables, we find that J = P 2s−k(k+1) Z Rk+1 Z B T(γ, ν) e à k X j=0 βjφj(γ, ν) !
dγ dν dβ, where B = [η− 1 , η+ 1 ] × · · · × [η− s , η+ s ] × [ξ− 1 , ξ+ 1 ] × · · · × [ξ− s , ξ+ s ], φj(γ, ν) = c1γk−j 1 νj 1 + · · · + csγk−j s νj s, and where T(γ, ν) is as in (4.30). Now let S(t0, . . . , tk) = {(γ, ν) ∈B : φj(γ, ν) = tj (0 ≤j ≤k)}, so that J = P 2s−k(k+1) Z Rk+1 Z C Z S(t0,...,tk) T(γ, ν) e(β0t0 + · · · + βktk) dS(t) dt dβ, where C ⊂Rk+1. Since (η, ξ) ∈B, we see that C contains a neighborhood of the origin, whence after k + 1 applications of Fourier’s Integral Theorem we obtain J = P 2s−k(k+1) Z S(0) T(γ, ν) dS(0), and the result follows as in the proof of Lemma 4.5.5.
Lemma 4.6.9. The function S(q) is multiplicative.
Proof. This is identical to the proof of Lemma 4.5.6.
Whenever s ≥(k + 1)2 one finds using Lemma 4.6.9 that S = Y p σ(p), where σ(p) = ∞ X h=0 S(ph), and that there exists a constant C(k) such that 1 2 ≤ Y p>C(k) σ(p) ≤3 2.
Let M(q) denote the number of solutions of the system of congruences c1xk−j 1 yj 1 + · · · + csxk−j s yj s ≡0 (mod q) (0 ≤j ≤k).
101 Lemma 4.6.10. One has X d|q S(d) = qk+1−2sM(q).
Proof. This is identical to the proof of Lemma 4.5.7.
It follows that σ(p) = lim h→∞ph(k+1−2s)M(ph), so again to show that S ≫1 it suffices to show that M(ph) ≥p(h−u)(2s−k−1) for p ≤C(k), and this follows exactly as in the argument of Lemma 4.5.9. Hence the proof of Theorem 4.4 is complete on assembling (4.35), (4.37), and (4.38) and recalling Lemma 4.6.8.
In order to deduce Theorem 4.5, we need some additional observations.
Lemma 4.6.11. Let (x, y), (x′, y′) ∈Z2s be such that (x1, . . . , xs) = 1. Then xt + y and x′t + y′ parameterize the same line if and only if x′ = qx and y′ = y + rx for some integers q and r with q ̸= 0.
Proof. First suppose that x′ = qx and y′ = y + rx for some integers q and r with q ̸= 0. Then one has xt + y = x′ µt −r q ¶ + y′ and x′t + y′ = x(qt + r) + y, so the two lines are identical. Conversely, suppose that the two lines are the same.
By taking t = 0 on the line x′t+y′, we see that there exists t1 such that y′ = xt1+y, and then by taking t = 1 we find that there exists t2 such that x′ + y′ = xt2 + y and hence x′ = (t2 −t1)x. Moreover, the condition (x1, . . . , xs) = 1 implies that t1 and t2 are distinct integers, and this completes the proof.
Now let Rs(P, d) denote the number of solutions of (4.6) counted by Rs(P) for which (x1, . . . , xs) = d. The following estimate will be useful when d is large.
Lemma 4.6.12. One has Rs(P, d) ≪P 2s−k(k+1) d2 .
102 Proof. Consider a solution (x, y) counted by Rs(P, d). Since xs−1 and xs each have d as a divisor, the number of possible choices for xs−1, ys−1, xs, and ys is at most P 2(P/d)2. Given such a choice, the number of possibilities for the remaining variables is Z Tk+1 Ã t Y i=1 Fi(α) s−2 Y i=t+1 fi(α) !
e(α · m) dα, where mj = cs−1xk−j s−1yj s−1 + csxk−j s yj s, and thus Rs(P, d) ≪P 4 d2 Z Tk+1 t Y i=1 |Fi(α)| s−2 Y i=t+1 |fi(α)| dα.
The lemma now follows by dissecting Tk+1 into major and minor arcs and using (4.35) and (4.36).
We can now complete the proof of Theorem 4.5.
Proof of Theorem 4.5. Define an equivalence relation on the set of solutions to (4.6) by writing (x, y) ∼(x′, y′) whenever xt + y and x′t + y′ define the same line. We need a lower bound for the number of equivalence classes, so by Lemma 4.6.11 it suffices to estimate the number of solutions of (4.6) for which (x1, . . . , xs) = 1 and 1 ≤y1 ≤|x1|.
Let Ns(P, Q, d) be the number of solutions of (4.6) with x ∈BP, y ∈CQ, and (x1, . . . , xs) = d, where B = [η− 1 , η+ 1 ] × · · · × [η− s , η+ s ] and C = [ξ− 1 , ξ+ 1 ] × · · · × [ξ− s , ξ+ s ].
Then the solutions counted by Rs(P, d) are in bijective correspondence with a subset of those counted by Ns(P/d, P, 1). Moreover, Lemma 4.6.11 shows that two solutions (x, y) and (x′, y′) counted by Ns(P/d, P, 1) are equivalent if and only if x = x′ and y −y′ = rx for some integer r. Then since |x1| ≥P τd and |y1 −y′ 1| ≤P, where τ = 2/|η1|, we see that each equivalence class contains at most τd members counted by Ns(P/d, P, 1). Thus we see that Rs(P, d) ≤Ns(P/d, P, 1) ≤τd Ls(P), 103 and hence for any D we have that X d≤D Rs(P, d) ≤τD2Ls(P).
Thus by Lemma 4.6.12 there exist positive constants γ1 and γ2 such that γ1P 2s−k(k+1) ≤Rs(P) ≤τD2Ls(P) + X d>D µγ2P 2s−k(k+1) d2 ¶ for P sufficiently large, and hence we have Ls(P) ≥P 2s−k(k+1) τD2 ³ γ1 −γ2 D ´ .
The theorem now follows on taking D = 2γ2/γ1.
CHAPTER V The Density of Rational Lines on Cubic Hypersurfaces 5.1 Overview In this chapter, we refine the analysis of Chapters 3 and 4 in the case k = 3 by providing a more detailed consideration of the lower moments of the relevant exponential sums. In particular, we are able to establish diagonal behavior for the mean value S(P) = Z T4 |F(α)|6dα, where F(α) = X 1≤x,y≤P e(α0x3 + α1x2y + α2xy2 + α3y3), and this leads, via the iterative method, to improved estimates for higher moments.
While the assertion of diagonal behavior amounts to the estimate S(P) ≪P 6, we can actually establish a more precise result. Notice that, by orthogonality, S(P) is the number of solutions of the system of equations x3 1 + x3 2 + x3 3 = x3 4 + x3 5 + x3 6 x2 1y1 + x2 2y2 + x2 3y3 = x2 4y4 + x2 5y5 + x2 6y6 x1y2 1 + x2y2 2 + x3y2 3 = x4y2 4 + x5y2 5 + x6y2 6 y3 1 + y3 2 + y3 3 = y3 4 + y3 5 + y3 6 (5.1) with xi, yi ∈[1, P] ∩Z. Further, write T(P) for the number of “trivial” solutions, in which (x4, x5, x6) = σ(x1, x2, x3) and (y4, y5, y6) = σ(y1, y2, y3) for some permutation σ ∈S3. Then clearly one has T(P) = 6P 6 + O(P 4), and it transpires that almost all the solutions counted by S(P) are of this diagonal type. The following “paucity theorem” provides an upper bound for the number of non-diagonal solutions.
104 105 Theorem 5.1. For every ε > 0, one has S(P) −T(P) ≪ε P 6−11 192+ε, and hence in particular S(P) ∼6P 6.
Note that the second assertion of the theorem is immediate from the first in view of the above discussion. Section 5.2 is devoted to the proof of Theorem 5.1. In Section 5.3, we use this result, along with the techniques of its proof and the iterative method of Chapter 3, to obtain non-trivial estimates for higher moments. In Section 5.4, we establish Weyl estimates more suitable for small k than those of Theorem 4.1. Finally, in Section 5.5, we apply the circle method as in Sections 4.5 and 4.6 to obtain Theorem 5.2. Suppose that s ≥58 and let Ls(P) denote the number of distinct lines ℓon the hypersurface c1z3 1 + · · · + csz3 s = 0 with h(ℓ) ≤P. Then Ls(P) ≫P 2s−12 for P sufficiently large.
For comparison, we note that Wooley has demonstrated the existence of rational lines on arbitrary cubic hypersurfaces in at least 37 variables, whereas we require 58 variables in Theorem 5.2. In the additive situation we are considering, the existence of lines follows immediately from the theory of a single additive cubic equation (see R. Baker ), provided that s ≥14. Hence the significance of our result lies in the density estimate.
The existence of these “trivial” lines when s ≥14 is in fact key to our analysis, for they give rise to non-singular integer solutions of the system c1x3−j 1 yj 1 + · · · + csx3−j s yj s = 0 (0 ≤j ≤3) (5.2) and hence allow us to avoid imposing explicit local solubility hypotheses in Theorem 5.2. Unfortunately, the solutions arising in this way are singular for larger values of k and hence are of no use in the analysis leading to Theorem 8.
Local conditions also may present an obstacle to demonstrating the expected density of higher-dimensional linear spaces on a cubic hypersurface in a reasonable number of variables. While the results of Schmidt , could be applied to the analogues of (5.2), the number of variables required may in general be quite large.
The material of this chapter is contained in the author’s preliminary manuscript .
106 5.2 The Paucity Problem Our goal in this section is to establish Theorem 5.1. Before proceeding with the proof, we record for reference some of the key estimates we will use. The first of these is implicit in the work of Hooley on sums of four cubes.
Lemma 5.2.1. Let n be a non-zero integer, and let R(P) denote the number of integral solutions of the equation x3 1 + x3 2 + x3 3 + x3 4 = n with |xi| ≤P. Then one has R(P) ≪P 11/6+ε.
Proof. Clearly, we may focus attention on solutions in which at least two of the xi are non-zero. For any such solution x counted by R(P), we can find i and j such that xi and xj have the same parity and are not both zero. Now if xi +xj = 0 and xk and xl are the remaining two variables, then since n ̸= 0 we must have xk + xl ̸= 0, and if xk and xl do not have the same parity, then one of them has the same parity as xi and xj. Thus, after relabeling variables, we may assume that x1 ≡x2 (mod 2) and x1 ̸= −x2. This allows us to write x1 = r + s and x2 = r −s, where r and s are integers with r ̸= 0, and hence to consider solutions of the equation 2r(r2 + 3s2) = n −z3 −w3.
The argument is now identical to that of Hooley , the condition r ̸= 0 being essential to the consideration of congruences modulo divisors of r. The only change is that the upper bound of n1/3 for the moduli of r, z, and w is replaced throughout by P, and the sieving parameter ξ is now chosen to be P 1/6.
We also make use of some recent work of Heath-Brown on sums of two cubes.
Lemma 5.2.2. Let U(P) denote the number of integral solutions of the equation x3 1 + x3 2 = x3 3 + x3 4 with |xi| ≤P and x1 + x2 ̸= x3 + x4. Then one has U(P) ≪P 4/3+ε.
Proof. This is a special case of Heath-Brown , Theorem 1.
We remark that Hooley , using the Riemann hypothesis for varieties over finite fields, obtained a result of the above shape with the exponent 4/3 replaced by 5/3. Wooley later devised an elementary proof, and his ideas play a key role in Heath-Brown’s argument.
Finally, we recall a result on binary quadratic forms dating back to Estermann .
107 Lemma 5.2.3. Let a, b, and c be non-zero integers, and let Q(P) denote the number of integral solutions of the equation ax2 + by2 = c with 1 ≤x, y ≤P. Then one has Q(P) ≪|abcP|ε.
Proof. See (for example) Vaughan and Wooley , Lemma 3.5.
We are now ready to embark on the proof of Theorem 5.1. On writing h = x1−x4 and g = y1 −y4 and relabeling variables in (5.1), we see that S(P) is the number of solutions of the system of equations h(3x(x + h) + h2) = u3 1 + u3 2 −u3 3 −u3 4 (2hx + h2)y + g(x + h)2 = u2 1v1 + u2 2v2 −u2 3v3 −u2 4v4 (2gy + g2)x + h(y + g)2 = u1v2 1 + u2v2 2 −u3v2 3 −u4v2 4 g(3y(y + g) + g2) = v3 1 + v3 2 −v3 3 −v3 4 (5.3) with 1 ≤x, y, ui, vi ≤P and |h|, |g| < P.
(5.4) We shall estimate N(P) = S(P) −T(P) by dividing into several cases.
(i) Let N1 denote the number of solutions counted by N(P) for which h = g = 0, and consider a solution x, y, u, v counted by N1. Then one has (u1, u2, v1, v2) ̸= (u3, u4, v3, v4) and (u1, u2, v1, v2) ̸= (u4, u3, v4, v3), since otherwise the solution would be counted by T(P). If we have (u1, u2) = (u3, u4) and (v1, v2) = (v4, v3), then the second equation in (5.3) implies that either u1 = u2 or v1 = v2, whence the number of choices for u and v is O(P 3). Trivially, there are O(P 2) choices for x and y, so the total number of solutions is O(P 5), and the same analysis applies if (u1, u2) = (u4, u3) and (v1, v2) = (v3, v4). Otherwise, since ui and vi are positive, it follows that either u1 +u2 ̸= u3 +u4 or v1 +v2 ̸= v3 +v4, so Lemma 5.2.2 may be applied to estimate the number of choices for u or v (or possibly both).
On combining this with Hua’s Lemma, one sees that N1 ≪P 16/3+ε.
(ii) Let N2 denote the number of solutions counted by N(P) for which exactly one of h or g is zero. Suppose first that h = 0 and g ̸= 0. Then by Hua’s Lemma one has O(P 2+ε) choices for u, and by a trivial estimate there are O(P 2) choices for g and y.
Now for fixed non-zero g and y, we may apply Lemma 5.2.1 to deduce that there are 108 O(P 11/6+ε) choices of v satisfying the fourth equation of (5.3). Finally, since g ̸= 0, the second equation is a non-trivial polynomial in x and hence determines x to O(1).
By following a symmetric argument in the case where g = 0 and h ̸= 0, we find that N2 ≪P 35/6+ε.
(iii) Write d = (h, g), let β be a parameter at our disposal, and let N3 denote the number of solutions counted by N(P) for which hg ̸= 0 and |hg/d| ≤P 1+β. In this case, there are O(P 1+β) choices for the integer hg/d, of which d, h/d and g/d are all divisors. Thus by a standard estimate for the divisor function, we see that there are O(P 1+β+ε) choices for h and g. Trivially, there are O(P) choices for x, and then by Lemma 5.2.1 we have O(P 11/6+ε) choices for u. Now by taking a linear combination of the equations (5.3), with respective weights g3, −3g2h, 3gh2, and −h3, we find that any solution x, y, g, h, u, v satisfies (gu1 −hv1)3 + (gu2 −hv2)3 = (gu3 −hv3)3 + (gu4 −hv4)3, (5.5) and by applying Hua , Theorem 4, to the underlying mean value we find that, for fixed h, g, and u, there are O(P 2+ε) choices for v. Finally, y is determined to O(1) by a polynomial, whence N3 ≪P 35/6+β+ε.
(iv) For i = 1, . . . , 4, write Xi = gui −hvi, and let N4 denote the number of solutions counted by N(P) for which X1 + X2 = X3 + X4 and hg ̸= 0. The former condition, when combined with (5.5), implies that either X1 = X3, X2 = X3, or X1 = −X2. We may suppose that X1 = X3 and X2 = X4, so that g(u1 −u3) = h(v1 −v3) and g(u2 −u4) = h(v2 −v4), (5.6) the argument in the remaining two cases being identical. For convenience we write A = u1 −u3, B = u2 −u4, C = v1 −v3, and D = v2 −v4.
(5.7) Since h and g are non-zero, the first equation in (5.3) implies that either A or B is non-zero, and the fourth equation implies that either C or D is non-zero. Suppose that C ̸= 0 and D = 0. We first choose u2 = u4 and v2 = v4 in O(P 2) ways, and then by (5.3) we have (x + h)3 + u3 3 = x3 + u3 1 and (y + g)3 + v3 3 = y3 + v3 1.
Since solutions with x = u3 and y = v3 are counted by T(P), we may apply Lemma 5.2.2, together with Hua’s Lemma, to deduce that there are O(P 10/3+ε) choices for x, y, h, g, u1, u3, v1, and v3. The case where C = 0 and D ̸= 0 is identical.
109 It remains to consider solutions for which both C and D (and hence A and B) are non-zero. We first observe that, after substituting from (5.7) and completing the square, the first and fourth equations in (5.3) become h(3x(x + h) + h2) −1 4(A3 + B3) = 3A(u3 + 1 2A)2 + 3B(u4 + 1 2B)2 (5.8) and g(3y(y + g) + g2) −1 4(C3 + D3) = 3C(v3 + 1 2C)2 + 3D(v4 + 1 2D)2, (5.9) respectively. In view of (5.8) and (5.9), we further classify solutions according to whether h(3x(x + h) + h2) −1 4(A3 + B3) = 0 (5.10) or g(3y(y + g) + g2) −1 4(C3 + D3) = 0.
(5.11) If both (5.10) and (5.11) hold, then we start by selecting values for A and B from among O(P 2) possibilities, and (5.10) then determines h and x to O(P ε). Trivially, there are O(P) choices for g, and (5.6) then determines C and D to O(P ε), whence y is determined to O(1) by (5.11). Finally, u3 and v3 may be assigned in O(P 2) ways, and this choice determines u and v in light of (5.7), (5.8), and (5.9). Hence there are O(P 5+ε) solutions of this type.
If (5.10) holds but (5.11) does not, then we assign A, B, and u3 in O(P 3) ways, so that u is determined by (5.8). Then h and x are again determined up to O(P ε), and there are O(P 2) choices for y and g. This latter choice determines C and D to O(P ε) by (5.6), and we may apply Lemma 5.2.3 to (5.9), regarded as a binary quadratic equation in the variables v3 and v4. The case where (5.11) holds but (5.10) does not is exactly symmetric, so we see that there are O(P 5+ε) solutions of these two types.
Finally, if neither (5.10) nor (5.11) holds, then we fix h, C, and D in O(P 3) ways, from which g, A, and B are determined to O(P ε) by (5.6). There are O(P 2) possibilities for x and y, and Lemma 5.2.3 then shows that u and v are determined up to O(P ε) by (5.8) and (5.9). Thus we conclude that N4 ≪P 16/3+ε.
(v) Now let γ be a parameter at our disposal, write M = P 1+β, and let N5 be the number of solutions counted by N(P) in which hg ̸= 0, |hg/d| > M, X1 + X2 ̸= X3 + X4, (5.12) 110 and d = (h, g) ≤P γ. By symmetry, we may assume that |h| ≥|g|, the argument in the other case being identical. Write h′ = h/d and g′ = g/d, so that (h′, g′) = 1.
For any given d and |h′| ≥|g′|, we divide both sides of (5.5) by d3 and apply Lemma 5.2.2 to deduce that there are then O((|h′|P)4/3+ε) possible choices for X1, . . . , X4.
With Xi now fixed and (h′, g′) = 1, any two choices for ui must be congruent modulo h′, so one has O(P/|h′|) possibilities for each of u1, . . . , u4, and this determines v.
Since x and y are then determined by polynomials, we find that N5 ≪ X d≤P γ X 1≤g≤P/d X h≥max(g,M/gd) (hP)4/3+ε(P/h)4 ≪ P 16/3+ε X d≤P γ X g≤(M/d)1/2 X h≥M/gd h−8/3 + X g>(M/d)1/2 X h≥g h−8/3 ≪ P 16/3+ε X d≤P γ X g≤(M/d)1/2 (M/gd)−5/3 + X g>(M/d)1/2 g−5/3 ≪ P 16/3+ε X d≤P γ ¡ (M/d)−5/3(M/d)4/3 + (M/d)−1/3¢ ≪ P 16/3+εM −1/3 X d≤P γ d1/3 ≪P 16 3 −1 3(1+β)+ 4 3 γ+ε.
(vi) Finally, let N6 be the number of solutions counted by N(P) with (5.12) and d > P γ. In this case we use an affine slicing approach almost exactly as in Wooley . As before, we exploit the symmetry of our system to focus attention on solutions with |h| ≥|g|. On recalling (5.5), we have that X3 1 + X3 2 = X3 3 + X3 4 and X1 + X2 = X3 + X4 + H (5.13) for some integer H. For convenience, we write X′ i = Xi/d and H′ = H/d. For fixed h, g, and u, one has H′ = g′(u1 + u2 −u3 −u4) −h′(v1 + v2 −v3 −v4), which determines the residue class of H′ modulo h′. Furthermore, since |h′| ≥|g′|, one has |H′| ≤4|h′|P. Now from the equations (5.13), we find that (X1 + X2 −X3)3 −(X3 1 + X3 2 −X3 3) = (X4 + H)3 −X3 4, which simplifies to 3(X1 −X3)(X2 −X3)(X1 + X2) = H(3X2 4 + 3X4H + H2).
(5.14) 111 By (5.12), we have H ̸= 0, so after dividing both sides of (5.14) by d3 we see that at least one of X′ 1 −X′ 3, X′ 2 −X′ 3, or X′ 1 + X′ 2 has a divisor e ≫|H′|1/3 in common with H′. We suppose that e = (H′, X′ 1 −X′ 3) ≫|H′|1/3 and write X1 −X3 = deY , the analysis in the other two cases being identical.
Hence, for fixed d and e, there are O(|h′|P/e) choices for Y . Now, on substituting X4 = X1 + X2 −X3 −H and X3 = X1 −deY in (5.14), we obtain 3deY X2 1 −3(deY −H)X2 2 −3(deY )2X1 −3(deY −H)2X2 = (deY −H)3 −(deY )3, and after completing the square this becomes 3deY (X1 −1 2deY )2 −3(deY −H)(X2 + 1 2(deY −H))2 = 1 4(deY −H)3 −1 4(deY )3.
Since H ̸= 0, the quantities deY , deY −H, and (deY −H)3−(deY )3 are all non-zero, so Lemma 5.2.3 may be applied. Thus, for fixed d, e, H, and Y , the values of X1 and X2 are determined up to O(P ε), and this fixes X3 and X4. For fixed g, h, and u, this determines v, and y is then determined to O(1) by a polynomial. Thus we have N6 ≪ X d>P γ X h′,g′≤P/d X x,u X H′, e h′P 1+ε e .
We now divide the sum over H′ into dyadic intervals of the form 4h′P/2r+1 < |H′| ≤ 4h′P/2r. For fixed H′, a divisor estimate shows that there are O(P ε) possible choices for e, and for fixed h and x Lemma 5.2.1 shows that there are O(P 11/6+ε) choices for u. Thus on summing trivially over g′ and x we find that N6 ≪ ∞ X r=0 X d>P γ P d X h′≤P/d P · P 11/6+ε · P 1+ε 2r µ h′P 2r+1 ¶ −1/3 h′P 1+ε, on recalling that e ≫|H′|1/3 and that the choice of h, g, and u fixes the residue class of H′ modulo h′. Finally, we obtain N6 ≪ P 11/2+ε ∞ X r=0 2−2r/3 X d>P γ d−1 X h′≤P/d (h′)2/3 ≪ P 43/6+ε X d>P γ d−8/3 ≪P 43/6−5 3 γ+ε.
112 We now choose the value of γ so that N5 and N6 have the same order of magnitude.
Thus we set 16 3 −1 3 (1 + β) + 4 3γ = 43 6 −5 3γ, which yields γ = 11 18 + 1 9(1 + β). In view of our bound for N3, we choose β by setting 35 6 + β = 43 6 −5 3 µ11 18 + 1 9 (1 + β) ¶ , which gives β = 7 64 and γ = 47 64. The result of Theorem 5.1 now follows immediately on assembling the bounds for N1, . . . , N6 and noting that 1 6 −7 64 = 11 192.
5.3 Further Mean Value Estimates Here we use the result of the previous section to obtain estimates for higher moments, which will be required in our application of the Hardy-Littlewood method in Section 5.5. As usual, the sharpest estimates are obtained by considering solutions in which some of the variables have no large prime factors. Thus when P and R are positive integers, write A(P, R) = {n ∈[1, P] ∩Z : p|n, p prime ⇒p ≤R} for the set of R-smooth numbers up to P, and define the exponential sum f(α; P, R) = X x,y∈A(P,R) e(α0x3 + α1x2y + α2xy2 + α3y3).
It will also be useful to have some variables in a complete interval, so we define F(α; P) = X 1≤x,y≤P e(α0x3 + α1x2y + α2xy2 + α3y3).
When there is no danger of confusion, we shall write f(α) = f(α; P, R) and F(α) = F(α; P). Further, let Ss,r(P, R) = Z T4 |F(α)|2r|f(α)|2s dα.
We adopt the convention that any statement involving ε and R means that for each ε > 0 there exists η = η(ε) > 0 such that the assertion holds whenever R ≤P η. In this section, our implicit constants will depend at most on ε unless otherwise noted.
We start with an estimate for S3,2(P, R).
113 Lemma 5.3.1. One has S3,2(P, R) ≪P 12+ 1 20+ε.
Proof. Define the difference operator ∆∗ 1 by ∆∗ 1(f(x, y); h, g) = f(x + h, y + g) −f(x, y).
Then by Cauchy’s inequality, one has S3,2(P, R) = Z T4 ¯ ¯ ¯ ¯ ¯ X x,h X y,g e à 3 X i=0 αi∆∗ 1(x3−iyi; h, g) !¯ ¯ ¯ ¯ ¯ 2 |f(α)|6 dα ≤ P 2 X h,g Z T4 ¯ ¯ ¯ ¯ ¯ X x,y e à 3 X i=0 αi∆∗ 1(x3−iyi; h, g) !¯ ¯ ¯ ¯ ¯ 2 |f(α)|6 dα, and hence S3,2(P, R) ≤P 2U3,2(P, R), (5.15) where Us,2(P, R) denotes the number of solutions of the system 3h(x2 1 −x2 2 + h(x1 −x2)) = s X i=1 (u3 i −u3 s+i) h(2(x1y1 −x2y2) + h(y1 −y2)) + g(x2 1 −x2 2 + 2h(x1 −x2)) = s X i=1 (u2 i vi −u2 s+ivs+i) g(2(x1y1 −x2y2) + g(x1 −x2)) + h(y2 1 −y2 2 + 2g(y1 −y2)) = s X i=1 (uiv2 i −us+iv2 s+i) 3g(y2 1 −y2 2 + g(y1 −y2)) = s X i=1 (v3 i −v3 s+i) with 1 ≤xi, yi ≤P, ui, vi ∈A(P, R), and |h|, |g| < P.
(5.16) The argument is now very similar to the proof of Theorem 5.1 given in the previous section.
(i) Let U1 denote the number of solutions counted by U3,2(P, R) for which h = g = 0 or x1 −x2 = y1 −y2 = 0. In either case, there are O(P 4) choices for h, g, x, and y, and one sees that the number of choices for u and v is then Z T4 |f(α)|6 dα ≪P 6, 114 on recalling Theorem 5.1 and considering the underlying Diophantine equations.
Thus we have U1 ≪P 10.
(ii) Let U2 denote the number of solutions counted by U3,2(P, R) for which exactly one of h or g is zero. First suppose that h = 0, g ̸= 0, and y1 ̸= y2. Then by Vaughan , Theorem 4.4, one has O(P 13/4+ε) choices for u, and by a trivial estimate there are O(P 3) choices for g and y. Now for fixed g and y, again shows that there are O(P 13/4+ε) choices for v. Finally, since g(y1 −y2) ̸= 0, x1 and x2 are determined to O(P ε) by the second and third equations above. If instead y1 = y2, then one has O(P 2) choices for g and y, but O(P 1+ε) choices for x, so we get the same estimate in either case. The situation when g = 0 and h ̸= 0 is identical. Thus we see that U2 ≪P 19/2+ε.
(iii) Let U3 denote the number of solutions counted by U3,2(P, R) for which hg ̸= 0 and exactly one of x1 −x2 or y1 −y2 is zero. If x1 ̸= x2, and y1 = y2, then there are O(P 13/4+ε) choices for v and O(P 2) choices for h and g. Now, as in the previous section, we find that (gu1 −hv1)3 + (gu2 −hv2)3 + (gu3 −hv3)3 = (gu4 −hv4)3 + (gu5 −hv5)3 + (gu6 −hv6)3 (5.17) so by Hua , Theorem 4, there are O(P 7/2+ε) choices for u, and then x and y are determined to O(P 1+ε). The other case is identical, and thus U3 ≪P 39/4+ε.
(iv) Write d = (h, g), let β be a parameter at our disposal, and let U4 denote the number of solutions counted by U3,2(P, R) for which 0 < |hg/d| ≤P 5/4+β, x1 ̸= x2, and y1 ̸= y2.
In this case, there are O(P 5/4+β) choices for the integer hg/d, whence by a divisor estimate there are O(P 5/4+β+ε) choices for h and g. Trivially, there are O(P 2) choices for x, and then we have O(P 13/4+ε) choices for u. Thus by applying Hua to (5.17), we see that there are O(P 7/2+ε) choices for v, and then y is determined to O(P ε).
Hence U4 ≪P 10+β+ε.
(v) Now let γ be a parameter at our disposal, write M = P 5/4+β, and let U5 be the number of solutions counted by U3,2(P, R) in which hg ̸= 0, |hg/d| > M, x1 ̸= x2, y1 ̸= y2, (5.18) and d = (h, g) ≤P γ. As before, we assume that |h| ≥|g| and write h′ = h/d and g′ = g/d, so that (h′, g′) = 1. For any given d and |h′| ≥|g′|, we divide both sides of (5.17) by d3; then by Hua’s Lemma there are O((|h′|P)7/2+ε) possible choices for 115 X1, . . . , X6, where Xi = gui −hvi. With Xi now fixed and (h′, g′) = 1, any two choices for ui must be congruent modulo h′, so one has O(P/|h′|) possibilities for each of u1, . . . , u6, and this determines v. Since x and y are then determined to O(P ε), we find that U5 ≪ X d≤P γ X 1≤g≤P/d X h≥max(g,M/gd) (hP)7/2+ε(P/h)6 ≪ P 19/2+ε X d≤P γ X g≤(M/d)1/2 X h≥M/gd h−5/2 + X g>(M/d)1/2 X h≥g h−5/2 ≪ P 19/2+ε X d≤P γ X g≤(M/d)1/2 (M/gd)−3/2 + X g>(M/d)1/2 g−3/2 ≪ P 19/2+ε X d≤P γ ¡ (M/d)−3/2(M/d)5/4 + (M/d)−1/4¢ ≪ P 19/2+εM −1/4 X d≤P γ d1/4 ≪P 19 2 −1 4 ( 5 4 +β)+ 5 4 γ+ε.
(vi) Finally, let U6 be the number of solutions counted by U3,2(P, R) with (5.18) and d > P γ. Then for fixed d, there are (P/d)2 choices for h and g and P 2 choices for x. Now on recalling (5.17), we see that there are O(P 27/4+ε) choices for u and v, whence y is determined to O(P ε). Thus U6 ≪P 43/4+ε X d>P γ d−2 ≪P 43/4−γ+ε.
To optimize the results of (v) and (vi), we set 9 + 3 16 −1 4β + 5 4γ = 10 + 3 4 −γ, which gives γ = 25 36 + 1 9β. Now on recalling (iv), we choose β so that 10 + β = 43 4 −25 36 −1 9β, which gives β = 1 20. Hence U3,2(P, R) ≪P 10+ 1 20, and the lemma follows on recalling (5.15).
Before proceeding, we record an easy consequence of Lemma 5.3.1.
Lemma 5.3.2. One has S2,2(P, R) ≪P 9+ 1 40+ε.
116 Proof. By Theorem 5.1, we have Z T4 |F(α)|6dα ≪P 6.
Thus by applying the Cauchy-Schwarz inequality and Lemma 5.3.1 we obtain S2,2(P, R) ≪ µZ T4 |F(α)|6dα ¶1/2 µZ T4 |F(α)|2|f(α)|8dα ¶1/2 ≪P 9+ 1 40+ε on considering the underlying Diophantine equations.
We now proceed to estimate some higher moments.
Lemma 5.3.3. One has S4,2(P, R) ≪P 15+ 1 3 +ε.
Proof. By Cauchy’s inequality, we have S4,2(P, R) ≤P 2U4,2(P, R), and the estima-tion of U4,2(P, R) proceeds almost exactly as in the proof of Lemma 5.3.1. The only modifications are that we use Lemma 5.3.2 in place of Theorem 5.1 in the analysis of case (i) and we replace the 6th moment estimates of P 13/4+ε and P 7/2+ε by Hua’s 8th moment estimate of P 5+ε. Taking M = P 4/3 and γ = 2/3 produces identical bounds for the final three cases and hence gives an optimal result.
We remark that the argument of the preceding proof in fact shows that one may replace S4,2(P, R) by S2,4(P, R) in the statement of Lemma 5.3.3. Further, we note that tiny improvements in the exponents of the preceding lemmata may be achieved by using the results of Wooley , in place of Vaughan , but such improvements do not have significant consequences in the present application.
For higher moments, we apply the (single) efficient differencing procedure of Chapter 3. Although our methods always allow us to take a few variables rang-ing over a complete interval, we will often simplify by stating results for mean val-ues in which all the variables are smooth. Thus we adopt the notation of writing Ss(P, R) for Ss,0(P, R). Further, we say that ∆s is an admissible exponent if one has Ss(P, R) ≪P λs+ε, where λs = 4s −12 + ∆s, and in this situation we call λs a permissible exponent. To this point we have obtained the admissible exponents ∆3 = 6, ∆4 = 5 1 40, ∆5 = 4 1 20, and ∆6 = 3 1 3.
(5.19) The above method of generating admissible exponents becomes noticeably less effective when s > 6, since the maximum savings of P 3 in estimating the number of solutions of u3 1 + · · · + u3 t = u3 t+1 + · · · + u3 2t 117 is already achieved when t = 4. Since the results of Section 3.5 are directly applicable only for s ≥11, we will need the following lemma to work out admissible exponents when s lies in the intermediate range.
Lemma 5.3.4. One has Ss+2(P, R) ≪P 5Ss(P, R) + P 19/6+εSs−1,2(P, R) + P 2 3 s+6+εSs(P 5/6, R).
Proof. This follows on using Lemma 3.3.2 in the initial stages of the argument of Lemma 3.5.1.
We now apply Lemma 5.3.4 repeatedly to obtain admissible exponents ∆s for 7 ≤s ≤12. First of all, by using Lemma 5.3.3 and making a trivial estimate, we see that S5,2(P, R) ≪P 19+ 1 3 +ε, and using this together with Lemma 5.3.3 in Lemma 5.3.4 gives S8(P, R) ≪P 22+ 7 9 +ε.
(5.20) Now using the Cauchy-Schwarz inequality to interpolate between S4,2 and S8, we obtain S7(P, R) ≪(S4,2(P, R))1/2(S8(P, R))1/2 ≪P 19+ 1 18+ε.
(5.21) Putting (5.20) and (5.21) into Lemma 5.3.4 now yields S9(P, R) ≪P 26+ 59 108+ε, and this is used along with (5.20) to obtain S10(P, R) ≪P 30+ 17 54+ε.
Continuing the iteration, we find that S11(P, R) ≪P 34+ 79 648+ε and S12(P, R) ≪P 37+ 301 324+ε.
Thus we have the admissible exponents ∆7 = 3 1 18, ∆8 = 27 9, ∆9 = 2 59 108, ∆10 = 217 54, ∆11 = 2 79 648, ∆12 = 625 324.
(5.22) Further admissible exponents can now be read offfrom Lemma 3.5.1. Namely, if s ≥11 and ∆s is admissible then the exponent ∆s+2t = ∆s(5/6)t is also admissible.
118 5.4 Weyl Differencing Here we obtain estimates for the modulus of the exponential sum F(α) when at least one of the αj is badly approximated by rationals. In Section 4.2, estimates of this type were obtained for f(α) by using the large sieve to relate the modulus of the sum to known mean values. This treatment allowed us to obtain bounds of the form P 2−σ(k)+ε, where σ(k)−1 ≍k3 log k, and for large k this is substantially better than the exponential decay that results from Weyl differencing. For k = 3, however, we are much better offapplying a two-fold Weyl differencing procedure. For purposes of application, it is useful to consider the slightly more general exponential sum F(α; P, Q) = X 1≤x≤P X 1≤y≤Q e(α0x3 + α1x2y + α2xy2 + α3y3).
Lemma 5.4.1. Suppose that Q ≍P and that for some j there exist qj ∈N and aj ∈Z with (qj, aj) = 1 and |qjαj −aj| ≤q−1 j .
(5.23) Then one has |F(α; P, Q)| ≪P 2+ε(q−1 j + P −1 + qjP −3)1/4.
Proof. First suppose that (5.23) holds with j = 0. Then by Weyl’s inequality (see for instance Lemma 2.4 of Vaughan ) one has |F(α; P, Q)| ≤ X y≤Q ¯ ¯ ¯ ¯ ¯ X x≤P e(α0x3 + α1x2y + α2xy2) ¯ ¯ ¯ ¯ ¯ ≪ QP 1+ε(q−1 0 + P −1 + q0P −3)1/4, and the result follows. Note that if instead (5.23) holds with j = 3, then we obtain the same conclusion simply by interchanging the roles of x and y in the above argument.
Now suppose that (5.23) holds with j = 1. Then by Cauchy’s inequality we have |F(α)|2 ≤ P X y≤Q ¯ ¯ ¯ ¯ ¯ X x≤P e(α0x3 + α1x2y + α2xy2) ¯ ¯ ¯ ¯ ¯ 2 ≪ P X y≤Q Ã P + X x,h e(α0(3x2h + 3xh2 + h3) + α1y(2xh + h2) + α2y2h) !
≪ P 3 + P X y≤Q X x,h e(α0(3x2h + 3xh2 + h3) + α1y(2xh + h2) + α2y2h), 119 where the second sum is over x and h ̸= 0 with 1 ≤x ≤P and 1−x ≤h ≤P −x, and where we have abbreviated F(α; P, Q) by F(α). Then on using Cauchy’s inequality again we obtain |F(α)|4 ≪ P 6 + P 4 X x,h ¯ ¯ ¯ ¯ ¯ X y≤Q e(α0(3x2h + 3xh2 + h3) + α1y(2xh + h2) + α2y2h) ¯ ¯ ¯ ¯ ¯ 2 ≪ P 6 + P 4 X x,h à P + X y,g e(α1g(2xh + h2) + α2h(2yg + g2)) !
≪ P 7 + P 4 X x,h,y,g e(α1g(2xh + h2) + α2h(2yg + g2)) ≪ P 7 + P 4 X 1≤|h|,|g|≤P X y∈I(Q,g) ¯ ¯ ¯ ¯ ¯ ¯ X x∈I(P,h) e(2α1ghx) ¯ ¯ ¯ ¯ ¯ ¯ , where I(P, h) = [1, P] ∩[1 −h, P −h]. Thus on summing the geometric progression, recalling a standard divisor function estimate, and using Lemma 2.2 of , we find that |F(α; P, Q)|4 ≪ P 7 + QP 4 X 1≤|h|,|g|≤P min(P, ||2α1gh||−1) ≪ P 7 + QP 4+ε X n≤2P 2 min(P, ||α1n||−1) ≪ P 7 + QP 7+ε(q−1 1 + P −1 + q1P −3), whence |F(α; P, Q)| ≪P 2+ε(q−1 1 + P −1 + q1P −3)1/4.
Again, the same conclusion follows when (5.23) holds with j = 2 by repeating the argument with the roles of x and y reversed.
Next we record a consequence of the above lemma, which will be useful in our application of the circle method.
Lemma 5.4.2. Let α0, . . . , α3 be real numbers with the property that whenever there exist q ∈N and a0, . . . , a3 ∈Z with (q, a0, . . . , a3) = 1 and |qαj −aj| ≤P δ−3 one has q > P δ. Then whenever Q ≍P one has |F(α; P, Q)| ≪P 2−δ/16+ε.
Proof. Let α0, . . . , α3 be as in the statement of the lemma, and write ν = δ/4. For each j, Dirichlet’s Theorem allows us to find qj ∈N and bj ∈Z with (qj, bj) = 1 such 120 that |qjαj −bj| ≤P ν−3 and qj ≤P 3−ν. If qj > P ν for some j, then the conclusion follows from Lemma 5.4.1. Otherwise, put q = [q0, . . . , q3] and aj = bjq/qj. Then we have (q, a0, . . . , a3) = 1 and q ≤qjP 3ν for each j and hence |αj −aj/q| ≤q−1 j P ν−3 ≤q−1P δ−3 (0 ≤j ≤3) and q ≤P δ, contradicting the hypothesis of the lemma.
5.5 The Circle Method Now we are in a position to prove Theorem 5.2 by applying the circle method along the lines of Section 4.6. The following lemma provides us with non-singular local solutions to (5.2).
Lemma 5.5.1. If s ≥14 and c1, . . . , cs are non-zero integers, then the system (5.2) has a non-singular real solution and a non-singular p-adic solution for all primes p.
Proof. After setting y1 = · · · = y7 = 0, x8 = · · · = x14 = 0, and xi = yi = 0 (i > 14), the system (5.2) reduces to c1x3 1 + · · · + c7x3 7 = 0 and c8y3 8 + · · · + c14y3 14 = 0.
By Baker , each of these equations has a non-trivial integral solution; suppose that x and y are solutions with xI and yJ non-zero. Then on writing ψj(x, y) = c1x3−j 1 yj 1 + · · · + csx3−j s yj s (0 ≤j ≤3), we have det µ∂ψj ∂xI , ∂ψj ∂yI , ∂ψj ∂xJ , ∂ψj ∂yJ ¶ 0≤j≤3 = (3cIcJ)2x4 Iy4 J ̸= 0.
Thus (x, y) is a non-singular integer solution of (5.2), so it is non-singular in each local field as well.
By Lemmata 4.6.2 and 5.5.1, we may assume that the system (5.2) has a non-singular real solution (η, ξ) with 0 < |ηi|, |ξi| < 1 2 for i = 1, . . . , s. For each i, we write η+ i = ηi + 1 2|ηi| and η− i = ηi −1 2|ηi| 121 and ξ+ i = ξi + 1 2|ξi| and ξ− i = ξi −1 2|ξi|.
Now let P be a large positive number, put R = P η with η ≤η0(ε), and let c1, . . . , cs be nonzero integers. Throughout this section, the implicit constants arising in our analysis may depend on c1, . . . , cs and on the real solution (η, ξ). We define the exponential sums Fi(α) = X η− i P<x≤η+ i P X ξ− i P<y≤ξ+ i P e(ci(α0x3 + α1x2y + α2xy2 + α3y3)) and fi(α) = X η− i P<x≤η+ i P |x|∈A(P,R) X ξ− i P 0.
Proof. We use Lemma 4.4.1 in the argument of Lemma 4.6.7.
We are now finally in a position to derive a mean value estimate in which we obtain the full savings of P 12 over the trivial bound.
Lemma 5.5.3. One has Z T4 |F(α)|17|f(α)|44dα ≪P 110.
Proof. Dissect into major and minor arcs by taking δ = 3/4 and c = 1 in the above definitions. Then by Lemma 5.4.2 we have sup α∈m|F(α)| ≪P 2−3 64+ε, and by Lemma 3.5.1, together with (5.22), we see that the exponent ∆22 = 625 324 µ5 6 ¶ 5 < 51 64 is admissible. Thus by Lemma 5.5.2 and a change of variables we have Z T4 |F(α)|17|f(α)|44dα ≪ P 34−51 64+ε Z T4 |f(α)|44dα + P 88 Z M |F(α)|17dα ≪ P 110−51 64+∆22+ε + P 110, and the result follows.
We are now ready to derive the lower bound for Rs(P). Take t = 24, u = 16, and v ≥2, for a total of s ≥58 variables, and further let δ = 9/10 and c = max |ci|.
If α ∈m, then by the argument of Lemma 4.6.3 one sees that the hypotheses of Lemma 5.4.2 hold with α replaced by ciα and hence sup α∈m|Fi(α)| ≪P 2−δ/16+ε.
123 Moreover, the exponent ∆16 = 625 324 µ5 6 ¶ 2 < 1.34 is admissible, and it is clear by considering the underlying Diophantine equations that all the mean value estimates from Section 5.3 hold with f(α) replaced by fi(α).
Thus by H¨ older’s inequality we have Z m |F(α)G(α)| dα ≪ P 48+2v−3 2 δ+ε 56 Y i=25 µZ T4 |fi(α)|32dα ¶ 1/32 ≪ P 2s−12−27 20+∆16+ε ≪P 2s−12−τ for some τ > 0. Furthermore, after applying H¨ older’s inequality and making a change of variables, we have by Lemmata 5.5.2 and 5.5.3 that Z M\N |F(α)G(α)| dα ≪ P 2v µZ M\N |F(α)|48dα ¶ 1/2 µZ T4 |f(α)|64dα ¶ 1/2 ≪ P 2s−12W −σ/2, and so it suffices to deal with the pruned major arc N. But it follows immediately from the analysis of Section 4.6 that Z N F(α)G(α) dα ≫P 2s−12, provided that W is taken to be a suitably small power of log P. Finally, on recalling the notation of Section 4.6 we have by the argument of Lemma 4.6.12 that Rs(P, d) ≪P 2v dv Z T4 |F(α)| Ã 56 Y i=25 |fi(α)| !
dα ≪P 2s−12 d2 , since v ≥2. Thus on following through the corresponding argument in the proof of Theorem 4.5, we find that Ls(P) ≫P 2s−12, and this completes the proof of Theorem 5.2.
We remark that essentially the same analysis, modified along the lines of Section 4.5, may be applied to prove Theorem 7. In that argument, we may clearly take v = 0 and it follows that G∗ 1(3) ≤56.
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ABSTRACT Exponential Sums and Diophantine Problems by Scott T. Parsell Chair: Trevor D. Wooley This work is concerned with the theory of exponential sums and their application to various Diophantine problems. Particular attention is given to exponential sums over smooth numbers, i.e. numbers having no large prime factors.
As an application of the theory of exponential sums in a single variable, we consider pairs of Diophantine inequalities of different degrees. Specifically, we show that two additive forms, one cubic and one quadratic, with real coefficients in at least 13 variables and satisfying suitable conditions, take arbitrarily small values simultaneously at integer points. In fact, we obtain a quantitative version of this result, which indicates how rapidly the forms can be made to approach zero as the size of the variables increases. Moreover, we obtain a lower bound for the density of integer points at which these small values occur.
We then proceed to study double exponential sums over smooth numbers by developing a version of the Vaughan-Wooley iterative method. We obtain estimates for mean values of these exponential sums, and these estimates are then used within the fabric of the Hardy-Littlewood method to obtain a lower bound for the density of rational lines on the hypersurface defined by an additive equation. We show that one obtains the expected density provided that the number of variables is sufficiently large in terms of the degree and that certain natural local solubility hypotheses are satisfied. We also consider applications to a two-dimensional generalization of Waring’s problem and to fractional parts of polynomials in two variables.
Finally, we refine the above analysis in the case of a cubic hypersurface to show that the expected density of lines is obtained whenever the defining equation has at least 58 variables. In the process, we obtain a result on the paucity of non-trivial solutions to an associated system of Diophantine equations. |
9049 | https://pressbooks.pub/rothschildsintrotosociology/chapter/global-wealth-and-poverty/ | Global Wealth and Poverty – Rothschild’s Introduction to Sociology
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Contents
Preface
About the Book
Acknowledgments
Main Body
Chapter 1: Introduction to Sociology
What Is Sociology?
The History of Sociology
Theoretical Perspectives
Why Study Sociology?
Chapter 2: Introduction to Sociological Research
Approaches to Sociological Research
Research Methods
Ethical Concerns
Chapter 3: Studying Culture as Sociologists
What Is Culture?
Elements of Culture
Pop Culture, Subculture, and Cultural Change
Theoretical Perspectives on Culture
Chapter 4: Socialization and Society
Theories of Self-Development
Why Socialization Matters
Agents of Socialization
Socialization Across the Life Course
Chapter 5: Society and Social Interaction
Types of Societies
Social Constructions of Reality
Theoretical Perspectives on Society
Chapter 6: Introduction to Gender, Sex, and Sexuality
Sex and Gender
Gender
Sex and Sexuality
Chapter 7: Introduction to Social Stratification
Social Stratification and Mobility in the United States
Theoretical Perspectives on Social Stratification
Global Stratification and Inequality
Global Inequality
Global Stratification and Classification
Global Wealth and Poverty
Theoretical Perspectives on Global Stratification
Chapter 8: Introduction to Race and Ethnicity
Racial, Ethnic, and Minority Groups
Stereotypes, Prejudice, and Discrimination
Intergroup Relationships
Theories of Race and Ethnicity
Race and Ethnicity in the United States
Chapter 9: Introduction to Groups and Organizations
Types of Groups
Group Size and Structure
Formal Organizations
Chapter 10: Introduction to Media and Technology
Technology Today
Media and Technology in Society
Global Implications of Media and Technology
Theoretical Perspectives on Media and Technology
Chapter 11: Introduction to Marriage and Family
Theoretical Perspectives on Aging
What Is Marriage? What Is a Family?
Variations in Family Life
Challenges Families Face
Chapter 12: Introduction to Education
Education around the World
Theoretical Perspectives on Education
Issues in Education
Chapter 13: Introduction to Religion
The Sociological Approach to Religion
World Religions
Religion in the United States
Chapter 14: Introduction to Politics and Government
Power and Authority
Forms of Government
Politics in the United States
Theoretical Perspectives on Government and Power
Chapter 15: Economy and Work
Economic Systems
Globalization and the Economy
Work in the United States
Chapter 16: Introduction to Population, Urbanization, and the Environment
Demography and Population
Urbanization
The Environment and Society
Chapter 17: Introduction to Social Movements and Social Change
Collective Behavior
Social Movements
Social Change
Concluding Note
Rothschild’s Introduction to Sociology
Global Wealth and Poverty
Understand the differences between relative, absolute, and subjective poverty
Describe the economic situation of some of the world’s most impoverished areas
Explain the cyclical impact of the consequences of poverty
How poor is poor for these beggar children in Vietnam? (Photo courtesy of Augapfel/flickr)
What does it mean to be poor? Does it mean being a single mother with two kids in New York City, waiting for the next paycheck in order to buy groceries? Does it mean living with almost no furniture in your apartment because your income doesn’t allow for extras like beds or chairs? Or does it mean having to live with the distended bellies of the chronically malnourished throughout the peripheral nations of Sub-Saharan Africa and South Asia? Poverty has a thousand faces and a thousand gradations; there is no single definition that pulls together every part of the spectrum. You might feel you are poor if you can’t afford cable television or buy your own car. Every time you see a fellow student with a new laptop and smartphone you might feel that you, with your ten-year-old desktop computer, are barely keeping up. However, someone else might look at the clothes you wear and the calories you consume and consider you rich.
Types of Poverty
Social scientists define global poverty in different ways and take into account the complexities and the issues of relativism described above.
Relative poverty is a state of living where people can afford necessities but are unable to meet their society’s average standard of living.
Absolute poverty lack even the basic necessities, which typically include adequate food, clean water, safe housing, and access to healthcare.
Absolute poverty is defined by the World Bank (2014a) as when someone lives on less than $1.25 a day. According to the most recent estimates, in 2011, about 17 percent of people in the developing world lived at or below $1.25 a day, a decrease of 26 percent compared to ten years ago, and an overall decrease of 35 percent compared to twenty years ago. A shocking number of people––88 million––live in absolute poverty, and close to 3 billion people live on less than $2.50 a day (Shah 2011). If you were forced to live on $2.50 a day, how would you do it? What would you deem worthy of spending money on, and what could you do without? How would you manage the necessities—and how would you make up the gap between what you need to live and what you can afford?
Slums in India illustrate absolute poverty all too well. (Photo courtesy of Emmanuelle Dyan/flickr)
What do the driver of an unlicensed hack cab in New York, a piecework seamstress working from her home in Mumbai, and a street tortilla vendor in Mexico City have in common? They are all members of the underground economy, a loosely defined unregulated market unhindered by taxes, government permits, or human protections. Official statistics before the worldwide recession posit that the underground economy accounted for over 50 percent of nonagricultural work in Latin America; the figure went as high as 80 percent in parts of Asia and Africa (Chen 2001).The underground economy has never been viewed very positively by global economists. After all, its members don’t pay taxes, don’t take out loans to grow their businesses, and rarely earn enough to put money back into the economy in the form of consumer spending. Within the context of this recession, some see the underground economy as a key player in keeping people alive.
Global Feminization of Poverty
In some ways, the phrase “global feminization of poverty” says it all: around the world, women are bearing a disproportionate percentage of the burden of poverty. This means more women live in poor conditions, receive inadequate healthcare, bear the brunt of malnutrition and inadequate drinking water, and so on. Throughout the 1990s, data indicated that while overall poverty rates were rising, especially in peripheral nations, the rates of impoverishment increased for women nearly 20 percent more than for men (Mogadham 2005).
Why is this happening? While myriad variables affect women’s poverty, research specializing in this issue identifies three causes (Mogadham 2005):
The expansion in the number of female-headed households
The persistence and consequences of intra-household inequalities and biases against women
The implementation of neoliberal economic policies around the world
While women are living longer and healthier lives today compared to ten years ago, around the world many women are denied basic rights, particularly in the workplace. In peripheral nations, they accumulate fewer assets, farm less land, make less money, and face restricted civil rights and liberties. Women can stimulate the economic growth of peripheral nations, but they are often undereducated and lack access to credit needed to start small businesses.
Consequences of Poverty
For this child at a refugee camp in Ethiopia, poverty and malnutrition are a way of life. (Photo courtesy of DFID – UK Department for International Development/flickr)
The poor often experience inadequate healthcare, limited education, and the inaccessibility of birth control. But those born into these conditions are incredibly challenged in their efforts to break out since these consequences of poverty are also causes of poverty, perpetuating a cycle of disadvantage. According to sociologists Neckerman and Torche (2007) in their analysis of global inequality studies, the consequences of poverty are many. Neckerman and Torche have divided them into three areas.
poverty is difficult to reverse
it has effects on mental and physical health
it relates to a prevalence in crime(Fajnzylber, Lederman, and Loayza 2002).
Slavery refers to any situation in which people are sold, treated as property, or forced to work for little or no pay. Just as in the pre-Civil War United States, these humans are at the mercy of their employers. Chattel slavery, the form of slavery once practiced in the American South, occurs when one person owns another as property. Child slavery, which may include child prostitution, is a form of chattel slavery. In debt bondage, or bonded labor, the poor pledge themselves as servants in exchange for the cost of basic necessities like transportation, room, and board. In this scenario, people are paid less than they are charged for room and board. When travel is required, they can arrive in debt for their travel expenses and be unable to work their way free, since their wages do not allow them to ever get ahead.
The global watchdog group Anti-Slavery International recognizes other forms of slavery: human trafficking (in which people are moved away from their communities and forced to work against their will), child domestic work and child labor, and certain forms of servile marriage, in which women are little more than chattel slaves (Anti-Slavery International 2012).
Summary
When looking at the world’s poor, we first have to define the difference between relative poverty, absolute poverty, and subjective poverty. While those in relative poverty might not have enough to live at their country’s standard of living, those in absolute poverty do not have, or barely have, basic necessities such as food. Poverty has numerous negative consequences, from increased crime rates to a detrimental impact on physical and mental health.
Section Quiz
Slavery in the pre-Civil War U.S. South most closely resembled
chattel slavery
debt bondage
relative poverty
peonage
Answer: 1
Maya is a twelve-year-old girl living in Thailand. She is homeless, and often does not know where she will sleep or when she will eat. We might say that Maya lives in ___ poverty.
subjective
absolute
relative
global
Answer: 2
Mike, a college student, rents a studio apartment. He cannot afford a television and lives on cheap groceries like dried beans and ramen noodles. Since he does not have a regular job, he does not own a car. Mike is living in:
global poverty
absolute poverty
subjective poverty
relative poverty
Answer: 4
In a U.S. town, a mining company owns all the stores and most of the houses. It sells goods to the workers at inflated prices, offers house rentals for twice what a mortgage would be, and makes sure to always pay the workers less than needed to cover food and rent. Once the workers are in debt, they have no choice but to continue working for the company, since their skills will not transfer to a new position. This situation most closely resembles:
child slavery
chattel slavery
debt slavery
servile marriage
Answer: 3
Short Answer
Consider the concept of subjective poverty. Does it make sense that poverty is in the eye of the beholder? When you see a homeless person, is your reaction different if he or she is seemingly content versus begging? Why?
Think of people among your family, your friends, or your classmates who are relatively unequal in terms of wealth. What is their relationship like? What factors come into play?
Go to your campus bookstore or visit its web site. Find out who manufactures apparel and novelty items with your school’s insignias. In what countries are these produced? Conduct some research to determine how well your school adheres to the principles advocated by USAS.
Contemporary Research
Students often think that the United States is immune to the atrocity of human trafficking. Check out the following link to learn more about trafficking in the United States:
For more information about the ongoing practices of slavery in the modern world click here:
References
Anti-Slavery International. 2012. “What Is Modern Slavery?” Retrieved January 1, 2012 (
Barta, Patrick. 2009. “The Rise of the Underground.” Wall Street Journal, March 14. Retrieved January 1, 2012 (ttp://online.wsj.com/article/SB123698646833925567.html).
Buvinić, M. 1997. “Women in Poverty: A New Global Underclass.” Foreign Policy, Fall (108):1–7.
Chen, Martha. 2001. “Women in the Informal Sector: A Global Picture, the Global Movement.” The SAIS Review 21:71–82
Chronicle of Higher Education. 2006. “Nearly Nude Penn State Students Protest Sweatshop Labor.” March 26. Retrieved January 4, 2012 (
Fajnzylber, Pablo, Daniel Lederman, and Norman Loayza. 2002. “Inequality and Violent Crime.” Journal of Law and Economics 45:1–40.
International Labour Organization. 2012. “High Unemployment and Growing Inequality Fuel Social Unrest around the World.” Retrieved November 7, 2014 (
Neckerman, Kathryn, and Florencia Torche. 2007. “Inequality: Causes and Consequences.” Annual Review of Sociology 33:335–357.
Shah, Anup. 2011. “Poverty around the World.” Global Issues. Retrieved January 17, 2012 (
U.S. Department of State. 2011a. “Background Note: Argentina.” Retrieved January 3, 2012 (
U.S. Department of State. 2011b. “Background Note: China.” Retrieved January 3, 2012 (
U.S. Department of State. 2011c. “Background Note: Rwanda.” Retrieved January 3, 2012 (
USAS. 2009. “Mission, Vision and Organizing Philosophy.” August. Retrieved January 2, 2012 (
World Bank. 2013. “Middle East and North Africa.” Retrieved November 7, 2014 (
World Bank. 2014e. “Poverty Overview.” Retrieved November 7, 2014 (
GLOSSARY
absolute poverty the state where one is barely able, or unable, to afford basic necessities relative poverty the state of poverty where one is unable to live the lifestyle of the average person in the country underground economy an unregulated economy of labor and goods that operates outside of governance, regulatory systems, or human protections
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9050 | https://bswtogether.org.uk/medicines/documents/bsw-acne-prescribing-guidelines-clinical-management-in-primary-care-2/ | Adapted in Sept 2021 to incorporate NG198 ; a pproved at BSW APC October 2021. Reviewed June 2023 . Aklief added Nov 2023. Based on work from NHS Wilts CCG with Dr Deirdre Buckley, Co nsultant Dermatologist, RUH
Acne Vulgaris - Prescribing Guideline - Clinical Management in Primary Care
• Treatment i s determined by severity of acne and the extent to which it bothers the individual. Acne of any severity can cause psychological distress.
• Risk of scarring increases with severity/duration of acne and should prompt more intensive treatment .
• No treatment works quickly in acne. Advise patients to adhere to treatment; positive effects can take 6 -8 weeks to become noticeable.
• If childbearing potential : topical retinoids and oral tetracyclines are CONTRAINDICATED during and when planning a pregnancy. Use effective contracept ion or
choose alternative options. Also avoid both in breastfeeding and avoid tetracyclines in children (up to 12 years) .
• In pregnancy , recommended treatments are benzoyl pero xide (BPO) monotherapy [Acnecide ® SELF -PURCHASE ] OR fixed combination BPO /C lindamycin [ Duac ®].
• DO NOT USE MINOCYCLINE to treat acne ; it is associated with lupus erythematosus, hepatitis and pigmentation.
• DO NOT USE ANTIBIOTIC MONOTHERAPY topical or oral antibiotic OR a combination of topical and oral antibiotic; s ignificantly i ncreases the risk of resistance .
Offer a 12 -week course of ONE of the first -line o ptions in Table 1 overleaf . Consider severity of acne and pe rson's preferences. Use Table 1 to d iscuss
pros/ con s. Review at 12 weeks; assess for improvement and side effects. If inadequate response to a 12 -week course of 1 st line treatment option:
• If at review severity is mild -moderate : offer a second option from Table 1 . Consider referral if inadequate response to 2 different 12 -we ek courses .
• If at review severity is moderate to severe and initial treatment did not include an oral antibiotic : offer an option that includes an oral antibiotic.
• If at review severity is moderate to severe and initial treatment did include an oral antibiotic : consider referral.
• In people whose treatment incl udes oral antibiotics, if acne has completely cleared, consider stopping the antibiotic but continuing with topical treatment. If
acne has improved but not completely cleared, consider continuing the oral antibiotic, alongside topical treatment, for up to 12 more weeks.
Antimicrobial stewardship : Only continue a treatment option that includes antibiotic (topical/oral) for > 6 months in exceptional circumstances. Review at 12
weekly intervals and stop antibiotic as soon as poss ible.
Consider maintenance treatment in those with a history of frequent relapse. Consider fixed combination adapalene/BPO [ Epiduo ®] as maintenance treatment. If not
tolerated or contraindicated, consider topical monotherapy with adapalene [Differin® ], azelaic acid [Finacea ®] or benzoyl peroxide [Acnecide ® SELF -PURCHASE ]. Review
maintenance treatment in 12 weeks period to decide the need for continued. NG198 section 1.6 gives further advice on relapses .
Guideline and General Prescribing Points Updated in line with NICE Guidance NG198: Management of Acne Vulgaris published June ‘21
Skin care advice for all Encourage continued appropriate skincare at all points in the treatment pathway.
- Use a pH neutral or slightly acidic cleansing product twice daily (Self -purchase – not for prescribing on FP10)
- Avoid oil -based and comedogenic moistu rise rs and cosmetics. Avoid picking acne lesions to reduce risk of scarring
Treatment options and instructions for review Adapted in Sept 2021 to incorporate NG198 ; a pproved at BSW APC October 2021. Reviewed June 2023 . Aklief added Nov 2023. Based on work from NHS Wilts CCG with Dr Deirdre Buckley, Co nsultant Dermatologist, RUH
Acne Vulgaris - Prescribing Guideline - Clinical Management in Primary Care
TABLE 1
NICE state these two fixed combination products can be considered a first line option in acne of ANY SEVERITY
• Topical adapalene with topical benzoyl peroxide [ Epiduo® gel 2.5/0.1% or 2.5/0.3% Apply OD in evening ] Does not contain antibiotics. Avoid in
pregnancy, caution in breastfeeding. Can cause irritation/photosensitivity/bleaching hair and fabrics.
• Topical tretinoin with topical clindamycin [ Treclin® gel 0.025/1% Apply OD in evening ] Avoid in pregnancy and breastfeeding. Can cause irritation,
photosensitivity, bleaching hair and fabrics.
Mild to Moderate Acne Moderate to S evere Acne Very Severe Acne
•On the face and often mild truncal disease,
comedones present
•More extensive lesions or acne unresponsive to topical antibiotic
•Systemic treatments useful for truncal disease where topical application is difficult
•Facial lesions and widespread truncal disease .
•Nodule s & cysts present, scarring developing
TREATMENT: 12 weeks then review TREATMENT: 12 weeks then review TREATMENT: Specialist
Epiduo ® or Treclin ® applied OD in evening
(see notes above)
OR
Topical benzoyl peroxide with topical
clindamycin [Duac ® gel 1/3% or 1/5% Apply
OD in evening ].
Can be used with caution during pregnancy
and breastfeeding. Can cause skin irritation,
photosensitivity, bleaching of hair and
fabrics.
Epiduo ® – applied OD in evening (see notes above)
OR
Topical azelaic acid [Finacea® gel 15%] [Skinoren ® cream 20%] apply BD . An
alternative to BPO or to a topical retinoid. Useful for treating comedonal acne,
particularly of the face. It is less likely to cause local irritation than BPO .
PLUS, Systemic antibiotic therapy
Select one of the fol lowing oral antibiotics (do not use with topical antibiotic):
Suggested dosage schedules (in increasing cost order):
• Doxycycline capsules 100mg daily OR
• Lymecycline capsules 408mg daily (more expensive)
All tetracyclines can cause photosensitivity. And should be avoided in
pregnancy and breastfeeding and <12yrs.
For patients who cannot tolerate or have contraindications to oral lymecycline
or doxycycline, consider alternative antibiotics.
Oral Erythromycin 500mg BD OR
-
Oral Trimethoprim 300mg BD (off -label indication)
Systemic isotretinoin is indicated as
monotherapy and only available from
secondary care or community
dermatology clinics : REFER
If clinical judgement indicates oral
isotretinoin for acne may be needed in
future, be aware that it should not be
used unless adequate courses of
standard therapy with systemic
antibiotics and topical therapy have
been tried in line with MHRA guidance.
Ta ke into account when choosing initial
treatment option.
This primary care guidance does not
include prescribing info for oral
isotretinoin. Specialist prescribers can
access PILs/SPCs and risk minimisation
materials here .
Aklief ® (trifarotene 50microgram/g cream ) is a selective topical retinoid – added to BSW formulary with GREEN TLS Nov 2023.
Aklief comes in a pump dispenser and is a useful option in people ≥12years for acne of the trunk (truncal acne) when many
comedones, papules and pustules are present. Avoid in pregnancy and breastfeeding; can cause irritation, photosensitivity.
Apply thinly once a day, in the evening: One pump to cover the face (forehead, cheeks, nose and chin), Two pumps to cover the
upper trunk (reachable upper back, shoulders, and chest). Assess continued improvement after 3 months of treatment.
Not currently included in NICE NG198 but for info see Scottish Medicines Consortium assessment (Aug 2022).
Consider BPO monotherapy [ Acnecide® 5% gel ] SELF -PURCHASE if options in Table1 contra -indicated or if wishing to avoid topical retinoid or topical/oral antibiotic. Adapted in Sept 2021 to incorporate NG198 ; a pproved at BSW APC October 2021. Reviewed June 2023 . Aklief added Nov 2023. Based on work from NHS Wilts CCG with Dr Deirdre Buckley, Co nsultant Dermatologist, RUH
Acne Vulgaris - Prescribing Guideline - Clinical Management in Primary Care
To reduce the risk of skin irritation associated with topical BPO or retinoids , start with alternate day or short contact application (e.g. wash off after an hour). If
tolerated, progress to using standard application regimen.
Always combine systemic or oral antibiotics with topical anti -acne agents retinoids +/ - BPO to reduce resistance and improve outcome .
Oral Contraceptives
•If a person receiving treatment for acne wishes to use hormonal contraception, consider the combined oral contraceptive (COC) pill in preference to progestogen -only pill (POP).
POP can exacerbate acne.
•COC may be used in combination with topical treatments or systemic antibiotics. Recommendations are that no additional contra cep tive precautions are required when COCs are
used with antibacterials that do not induce liver enzymes (e.g. Doxycycline), unless diarrhoea or vomiting occur. Check indiv idu al Summary of Product Characteristics for the
patient's contraceptive and the chosen antibiotic for specific advice. These recommendations should be discussed with the pat ien t.
•Co -cyprindiol [Cyproterone acetate 2 mg/Ethinylestradiol 35 microgram] is licensed for moderate to severe acne. Cyproterone may increase the risk of venous thromboembolism
(VTE) .The risk of VTE with these medicines is 1.5 to 2 times higher than for COCs containing levonorgestrel and may be simil ar to the risk with contraceptives containing gestodene,
desogestrel or drospirenone. See MHRA DSU (references below). Cyproterone is also contraindicated in those with previous or current meningioma. See MHRA DSU (reference s
below). Co -cypindriol is only included in NG198 as a second line treatment option for people with polycystic ovary syndrome (PCOS) and a cne. If Co -cyprindiol is being used, the
need to continue treatment should be reviewed at 6 months, discussing continuation or alternative treatment options.
Reasons to consider referral to relevant specialist teams (See NG198 section 1.4 for more detail)
•Mild to mod acne that has not responded to 2 completed courses of treatment OR mod to severe acne that has not responded to pre vious treatment containing oral antibiotic.
•Severe nodulo -cystic acne /acne conglobata /acne with scarring / acne with persistent pigmentary changes.
•If acne (any severity) or related scarring is contributing to psychological distress or mental health disorder. Consider curr ent /past history of self -harm, suicidal ideation, severe
depressive or anxiety disorder, body dysmorphic disorder.
•If medication (including self -administered anabolic steroids) or a medical disorder (e.g. PCOS ) is likely to be contributing t o a person's acne . Also see NG198 section 1.5.29 for more
detail on treatment options for people with PCOS .
•Diagnostic difficulty (uncommon).
•Severe variant of acne such as acne fulminans - very rare severe inflammatory acne with fever, malaise and joint symptoms (very urgent referral) .
Referral Form - Please include list of all treatments used in referral letter and any concomitant other medication and information regarding oth er medical conditions
Useful Links and References
British Association of Dermatologists (BAD) Patient Information Leaflet on Acne (July 2020)
National Institute for Health and Care Excellence (2021). Overview | Acne vulgaris: management | Guidance | NICE . [online] www.nice.org.uk. Available at:
British National Formulary
Clinical Knowledge Summaries: Acne vulgaris (June 2021): -vulgaris
MHRA Cyproterone acetate with ethinylestradiol (co -cyprindiol): balance of benefits and risks remains positive (June 2013):
MHRA Cyproterone acetate: new advice to minimise riak of meningioma (June 2020): -safety -update/cyproterone -acetate -new -advice -to -minimise -risk -of -meningioma
MHRA Isotretinoin : Introduction of new safety measures Isotretinoin (Roaccutane▼): introduction of new safety measures, including additional oversight of the initiation of treatmen t for patients under 18 years of age - GOV.UK (www.gov.uk) Updated October 2023
Arowojolu AO, Gallo MF, Lopez LM, Grimes DA. Combined oral contraceptives pills for treatment of acne. Cochrane Database Syst em Reviews 2012
Yang Z, Zhang Y, Lazic Mosler E, Hu J, Li H, Zhang Y, Liu J, Zhang Q. Topical benzoyl peroxide for acne. Cochrane Dat abase of Systematic Reviews 2020
Liu H, Yu H, Xia J, Liu L, Liu GJ, Sang H, Peinemann F. Topical azelaic acid, salicylic acid, nicotinamide, sulphur, z inc and fruit acid (alpha ‐hydroxy acid) for acne. Cochrane Database of Systematic Reviews 2020
BMJ Management of acne vulgaris: summary of NICE guidance. BMJ 2021; 374 doi: INCLUDES VISUAL SUMMARY AND INFOGRAPHIC
Scottish Medicines Consortium (SMC) Product Update 2441 August 2022 - trifarotene 50 microgram/g cream (Aklief®) trifarotene (Aklief) (scottishmedicines.org.uk) |
9051 | https://www.thoughtco.com/activity-series-of-metals-603960 | Activity Series of Metals: Predicting Reactivity
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Periodic Table›
Activity Series of Metals: Predicting Reactivity
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By
Todd Helmenstine
Todd Helmenstine
Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
B.A., Physics and Mathematics, Hastings College
Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics.
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Updated on February 28, 2020
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Key Takeaways
The activity series helps predict metal reactions with water, acids, and in displacement reactions.
Metals at the top of the activity series are more reactive than those at the bottom.
The reactivity series can aid in predicting reactions in aqueous solutions at room temperature.
The activity series of metals is an empirical tool used to predict products in displacement reactions and reactivity of metals with water and acids in replacement reactions and ore extraction. It can be used to predict the products in similar reactions involving a different metal.
Exploring the Activity Series Chart
The activity series is a chart of metals listed in order of declining relative reactivity. The top metals are more reactive than the metals on the bottom. For example, both magnesium and zinc can react with hydrogen ions to displace H 2 from a solution by the reactions:
Mg(s) + 2 H+(aq) → H 2(g) + Mg 2+(aq)
Zn(s) + 2 H+(aq) → H 2(g) + Zn 2+(aq)
Both metals react with the hydrogen ions, but magnesium metal can also displace zinc ions in solution by the reaction:
Mg(s) + Zn 2+ → Zn(s) + Mg 2+
This shows magnesium is more reactive than zinc and both metals are more reactive than hydrogen. This third displacement reaction can be used for any metal that appears lower than itself on the table. The further apart the two metals appear, the more vigorous the reaction. Adding a metal like copper to zinc ions will not displace the zinc since copper appears lower than zinc on the table.
The first five elements are highly reactive metals that will react with cold water, hot water, and steam to form hydrogen gas and hydroxides.
The next four metals (magnesium through chromium) are active metals that will react with hot water or steam to form their oxides and hydrogen gas. All the oxides of these two groups of metals will resist reduction by H 2 gas.
The six metals from iron to lead will replace hydrogen from hydrochloric, sulfuric and nitric acids. Their oxides can be reduced by heating with hydrogen gas, carbon, and carbon monoxide.
All the metals from lithium to copper will combine readily with oxygen to form their oxides. The last five metals are found free in nature with little oxides. Their oxides form through alternate pathways and will readily decompose with heat.
The series chart below works remarkably well for reactions that occur at or near room temperatures and in aqueous solutions.
Activity Series of Metals
MetalSymbolReactivity
Lithium Li displaces H 2 gas from water, steam and acids and forms hydroxides
Potassium K
Strontium Sr
Calcium Ca
Sodium Na
Magnesium Mg displaces H 2 gas from steam and acids and forms hydroxides
Aluminum Al
Zinc Zn
Chromium Cr
Iron Fe displaces H 2 gas from acids only and forms hydroxides
Cadmium Cd
Cobalt Co
Nickel Ni
Tin Sn
Lead Pb
Hydrogen gas H 2 included for comparison
Antimony Sb combines with O 2 to form oxides and cannot displace H 2
Arsenic As
Bismuth Bi
Copper Cu
Mercury Hg found free in nature, oxides decompose with heating
Silver Ag
Palladium Pd
Platinum Pt
Gold Au
Sources
Greenwood, Norman N.; Earnshaw, Alan (1984). Chemistry of the Elements. Oxford: Pergamon Press. pp. 82–87. ISBN 0-08-022057-6.
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Helmenstine, Todd. "Activity Series of Metals: Predicting Reactivity." ThoughtCo, Apr. 8, 2025, thoughtco.com/activity-series-of-metals-603960.Helmenstine, Todd. (2025, April 8). Activity Series of Metals: Predicting Reactivity. Retrieved from Helmenstine, Todd. "Activity Series of Metals: Predicting Reactivity." ThoughtCo. (accessed September 28, 2025).
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9052 | https://brainly.com/question/11250906 | [FREE] What is the algebraic expression for "three more than the sum of a number and four"? - brainly.com
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What is the algebraic expression for "three more than the sum of a number and four"?
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Hey there!
"More than" could also mean "greater than"
So,
t h ree = 3
"More than" = >
S u m = +
N u mb er = p (We don't know the number so we substitute it as a variable)
F o u r = 4
Answer: 3>p+4
Good luck on your assignment and enjoy your day!
~L o v e Y o u rse l f F i rs t:)
Answered by Amphitrite1040 •4.6K answers•9M people helped
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The algebraic expression for "three more than the sum of a number and four" is x+7, where x is the unknown number. This is derived by first determining the sum of the number and four and then adding three to that sum. Hence, the expression captures the full statement mathematically.
Explanation
To create an algebraic expression for "three more than the sum of a number and four," we first need to define our variable. Let’s say the unknown number is represented by the variable x.
Sum of a number and four: This can be written as x+4.
Three more than that sum: To express "three more than" something, we add 3 to it. Therefore, we take the expression we found in step 1 and add 3 to it.
Putting it all together, we get:
Algebraic expression=(x+4)+3
We can simplify this expression further:
Algebraic expression=x+7
Thus, the final algebraic expression is x+7.
Examples & Evidence
If we consider 'x' as 2, then the expression x+7 would equal 9, which means three more than the sum of 2 and 4 is 9 (2 + 4 + 3 = 9).
This breakdown follows algebraic conventions for forming expressions based on written statements, which is a standard method in mathematics.
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Percentage Share
Number Problems
Simple Mathematics
Math Solving Problems
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5
How do I calculate the original number from a percentage?
·
To calculate the original number from a percentage, you can use the following steps:
Write the percentage as a decimal: Divide the percentage by 100. For example, if you have 25%, divide it by 100 to get 0.25.
Divide the given number by the decimal: Take the number you have and divide it by the decimal from step 1. This will give you the original number. For example, if you have 0.25 and the given number is 200, divide 200 by 0.25 to get the original number, which is 800.
Here's an example to illustrate the calculation:
Let's say you have a car that is sold at a 15% discount for 6,800.Tofindthe
To calculate the original number from a percentage, you can use the following steps:
Write the percentage as a decimal: Divide the percentage by 100. For example, if you have 25%, divide it by 100 to get 0.25.
Divide the given number by the decimal: Take the number you have and divide it by the decimal from step 1. This will give you the original number. For example, if you have 0.25 and the given number is 200, divide 200 by 0.25 to get the original number, which is 800.
Here's an example to illustrate the calculation:
Let's say you have a car that is sold at a 15% discount for 6,800.Tofindtheoriginalpriceofthecar,youcanfollowthesesteps:1.Writethepercentageasadecimal:156,800 by 0.15. The result is 45,333.33.Therefore,theoriginalpriceofthecarbeforethediscountwas45,333.33 .
Learn more:
1. Reverse Percentages: How to Calculate an Original Amount
2. Reverse Percentages Calculator Online
3. How to Calculate an Original Amount Using Reverse Percentages - mathsathome.com
Jeremy Goff
Studied Materials Science and Engineering at The University of Utah
·
Author has 1K answers and 1.8M answer views
·
5y
There are 3 parts to a percentage problem:
Initial amount
Percentage
Final amount
This looks like (initial amount)(number)%=(final amount)
You must have two pieces yo find the third.
So based upon the question, you must have something of the form “18 is 36% of what number?”
In that case, we go to the basic equation and plug things in.
Since we don't have the initial number, we can call it x.
Now we have all three parts… kind of.
x36%=18
Two problems arise, with one being the fact that we don't actually want “x” as an answer, so we need to find the numerical value which makes this equation true. Next we
There are 3 parts to a percentage problem:
Initial amount
Percentage
Final amount
This looks like (initial amount)(number)%=(final amount)
You must have two pieces yo find the third.
So based upon the question, you must have something of the form “18 is 36% of what number?”
In that case, we go to the basic equation and plug things in.
Since we don't have the initial number, we can call it x.
Now we have all three parts… kind of.
x36%=18
Two problems arise, with one being the fact that we don't actually want “x” as an answer, so we need to find the numerical value which makes this equation true. Next we also need to consider that we want 36% as a decimal or rational number, and not “36%” as written.
All percentages are just number%/100
So we have either 36/100, or .36 depending on how you want to work it.
Now our equation becomes (x)(.36)=18. The problem has been reduced to a simple single-step equation from algebra 1. X is our unknown, so we will leave it alone and divide by .36
This yields x=18/.36 which is 50, so x=50.
Just follow this process with whatever numbers you may have.
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Leslie Quarcu
Auto technician at Continental Blue Investment Ghana Limited (2023–present)
·
Upvoted by
Bobby Sue
, M.F.A Mathematics & Physics, Chemistry, Maths, and Computer Science, Yale University (2022)
·
Updated Apr 23
Related
How do you find the original number from a percentage increase?
Add the percentage increase to 100, divide 100 by it, then multiple by the final number to get the original number.
Example. Let's say 54 is increased by 50%. It gives us 81.
We are going to pretend we didn't know the original number was 54. All we know is that a number was increased by 50% to obtain 81.
So 50 + 100 = 150
100/150 81 = 54..
Easy right?
Thanks
PS: I appreciate the recognition. Didn't quite expect that.
Related questions
How do you find the original number from a percentage increase?
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Chris Robida
Former 40 Years Teaching Elem., Middle Sch. & College (1973–2016)
·
Author has 4.8K answers and 2.5M answer views
·
5y
To GET a % you’d move the decimal 2 places to the right (which is multiplying by 100)
If you start with a percentage, you do the opposite (divide by 100) by moving the decimal 2 places to the left to get the original number.
ex: 54% = .54
3.25 = 325%
Brandon Ripley
5y
To go from original number to the final result you need to multiply by the percentage, so to reverse engineer, divide by that same percentage. So if:
N 50% = 500 then
N = 500 / 50%
N = 1,000.
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Hilmar Zonneveld
Translator (1985–present)
·
Author has 58.5K answers and 19.1M answer views
·
3y
Related
How do you find the original number from a percentage increase?
This question - and many other questions related to percentages - is easier if you convert the percentage increase, to an increase or decrease factor.
As an example, let’s take a 10% increase. If you increase something by 10%, the corresponding factor is 1 + 10/100 = 1.1.
So, if you have an original amount of 300, and you increase by 10%, you multiply 300 x 1.1, to get 330. The simplification of the problem might not be that obvious here.
But if you only know the 330, you simply divide by the same factor, i.e., you divide by 1.1, to get the original, which in this example is 300.
Related questions
How do you figure out a percentage of a number?
What's the easiest way to calculate the percentage difference between two numbers?
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What's the formula for calculating percentage?
Laura Ragsdale
Former Driver for DoorDash
·
7y
Related
How do you calculate a reverse percentage? What is the formula?
Reverse Percentages are pretty easy to figure out, so I’m confused as to why every single answer you’ve received seems to be unimaginably complicated. The only thing you really need know is the basic reverse percentage formula. Don’t worry about figuring out Reverse Percentage Increases or Decreases just yet.
Basic Reverse Percentage Formula:
Example: You ask your friend for 25% of his apples, so he hands you 3 apples. You are trying to figure out how many apples he originally started with.
Calculation/Formula:
Number Provided ÷ Percent Value (as decimal) = Original Number
Result: 3 ÷ 0.25 = 12
Easy
Reverse Percentages are pretty easy to figure out, so I’m confused as to why every single answer you’ve received seems to be unimaginably complicated. The only thing you really need know is the basic reverse percentage formula. Don’t worry about figuring out Reverse Percentage Increases or Decreases just yet.
Basic Reverse Percentage Formula:
Example: You ask your friend for 25% of his apples, so he hands you 3 apples. You are trying to figure out how many apples he originally started with.
Calculation/Formula:
Number Provided ÷ Percent Value (as decimal) = Original Number
Result: 3 ÷ 0.25 = 12
Easy, right?
Now that you know the basics, calculating Reverse Percentage Increases and Decreases will be super easy!
Calculating Reverse Percentage Increase:
1 + Your Percent = Percent Value
Number Provided ÷ Percent Value = Original Number
Calculating Reverse Percentage Decrease:
1 - Your Percent = Percent Value
Number Provided ÷ Percent Value = Original Number
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Will Gates
Former Assistant chief cook and bottle washer.
·
Author has 2K answers and 1.8M answer views
·
4y
Mathematicians were very crafty in naming a percentage as exactly what it is:
per (divided by) cent (100)
So, for example, 36 % = 36/100 = 0.36
Hans Davidsson
Rust Developer/programmer (2020–present)
·
Author has 135 answers and 126.8K answer views
·
5y
Related
How do you find the original number from a percentage increase?
If
x ( 1 + ( y / 100) ) = z
and
y is not -100
then
x = z / ( 1 + ( y / 100 ) )
If you for example have 28 and add 25% of 28 to it, then you have 35.
To get the original 28 back, using 25 and 35, you have to calculate
35 / ( 1 + ( 25 / 100 ) )
which is equal to
3500 / ( 100 + 25 )
which is equal to
3500 / 125
which is 28.
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Janet Clarke
Former Programmer at Ministry of Defence of the United Kingdom (1972–1983)
·
Author has 544 answers and 128.7K answer views
·
1y
Originally Answered: Can you explain how to calculate a percentage from a number and how to find the original number given the percentage and other information?
·
Percentage is out of 100. If you divide a number by 100 you have 1%. Multiply that 1% by the % you require. eg to find 60% of £90 90 divided by 100 = 0.9, then 0.9 x 60 = 54 so the answer is £54.
If you started off with £54 and knew it was 60% then divide 54 by 60 to get 0.9 then multiply by 100 to get £90.
With a calculator it could be one sum 54 /60 x 100 = 90
Calvin L.
18, Mathematics & Statistics major
·
Author has 10K answers and 2.1M answer views
·
3y
Related
How do you find the original number from a percentage increase?
Let's say you have a number x which is increased 40% to a value of 150.
Let's express that mathematically.
x×(140%)=150⟺1.4x=150
∴x≈107.14
Since x is 100%, adding 40% to it makes the new value 140% or 1.4 of x.
Then, you can divide the new value by the new percentage (or 1+%increase in decimal) to get the original x.
Samarth Golchha
Curious Thinker!
·
Author has 52 answers and 186.8K answer views
·
9y
Related
How do you calculate a reverse percentage? What is the formula?
Sometimes a question will ask you to work backwards and find the original price of something after the price has increased.
If you are given a quantity after a percentage increase or decrease, and you need to find the original amount, use this method:
Example 1
A radio sells for £63, after a 40% increase in the cost price. Find the cost price.
Solution
Start with the original amount as 100%.
Cost price = 100%
We are told the selling price is a 40% in the cost price.
So the selling price is 100% + 40% = 140% of the cost price.
We know that the selling price is £63, so 140% = £63.
Now calculate
Sometimes a question will ask you to work backwards and find the original price of something after the price has increased.
If you are given a quantity after a percentage increase or decrease, and you need to find the original amount, use this method:
Example 1
A radio sells for £63, after a 40% increase in the cost price. Find the cost price.
Solution
Start with the original amount as 100%.
Cost price = 100%
We are told the selling price is a 40% in the cost price.
So the selling price is 100% + 40% = 140% of the cost price.
We know that the selling price is £63, so 140% = £63.
Now calculate 1%:
140% = £63
1% = £63/140
1% = £0.45
The cost price is 100%, so multiply £0.45 by 100.
Cost price = 0.45 × 100 = £45.
Shadab Ahamed
Works at Fiserv (company)
·
5y
Related
How do you find the original number from a percentage increase?
(100×increase amount)/percentage
Original: 28
Percentage: 25%
Increase: 7
Original = (100×7)/25 = 700/25 = 28
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Truth Table
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A truth table is defined as a canonical representation of a Boolean function that exhaustively enumerates every truth assignment and its corresponding function value. It is particularly effective for functions with a limited number of input variables, allowing for constant-time basic Boolean operations through parallel bitwise operations.
AI generated definition based on: Electronic Design Automation, 2009
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Outline
1. Introduction to Truth Tables in Computer Science
2. Logical Foundations and Boolean Algebra
3. Truth Tables in Digital Logic Design
4. Truth Tables in Formal Methods and Verification
5. Conclusion
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1. Introduction to Truth Tables in Computer Science
A truth table is a tabular representation of a Boolean function that exhaustively enumerates every possible combination of input variables and lists the corresponding output value for each assignment. 1 In computer science, truth tables are foundational tools used to describe the behavior of digital logic circuits and analyze Boolean algebraic expressions. 2 Each row of a truth table corresponds to a unique combination of input values, and the output column specifies the result of the Boolean function for that combination. Truth tables serve as canonical representations of Boolean functions, meaning that two Boolean functions are equivalent if and only if they have identical truth tables; this property is essential for logic synthesis and verification applications. For a Boolean function with n input variables, the truth table contains 2ⁿ rows, which leads to exponential growth in size as the number of inputs increases, making truth tables impractical for functions with many variables. 3
2. Logical Foundations and Boolean Algebra
Truth tables provide an exhaustive enumeration of the mapping of a Boolean function, listing every possible truth assignment alongside its corresponding function value, and serve as canonical representations of Boolean functions; two Boolean functions are equivalent if and only if they have the same truth table, which is a property crucial for logic synthesis and verification applications. 1 Standard logical operators such as AND, OR, and NOT are defined by their respective truth tables, which specify the output for every combination of input values; for example, the AND operator yields true only when all inputs are true, while the OR operator yields true if at least one input is true, and the NOT operator inverts the input value. 4 5 The sum-of-products (SOP) form, also known as disjunctive normal form, is a Boolean formula consisting of disjunctions (sums) of conjunctions of literals. In propositional logic, truth tables establish the semantics of logical connectives by determining the truth value of compound statements from the truth values of their components, and for a formula with n variables, the truth table has 2^n rows, each corresponding to a possible interpretation.
3. Truth Tables in Digital Logic Design
A truth table is the simplest way to represent a digital system of combinational nature, specifying the output for every possible combination of inputs to a logic circuit or gate. 6 7 For basic gates, the truth table lists all possible input combinations and the corresponding output. For example, an AND gate produces a TRUE output only when all inputs are TRUE, while an OR gate produces a TRUE output if at least one input is TRUE. 8 The NOT gate inverts its single input, and the NAND and NOR gates combine AND/OR operations with inversion. The XOR gate outputs TRUE if an odd number of inputs are TRUE, and the XNOR gate outputs TRUE when all inputs are equal.
A truth table is a useful way of representing a combinational logic circuit, and can be used to design the circuit needed to achieve a desired function. 9 The number of rows in a truth table grows exponentially with the number of inputs, requiring 2ⁿ rows for n inputs, which can become impractical for large circuits. 6 7 For multi-input gates, such as three-input NOR or four-input AND gates, the truth table explicitly lists all possible input combinations and their outputs. 8
In logic synthesis and minimization, truth tables serve as the basis for methods like Karnaugh maps and Quine-McCluskey, which help reduce the complexity of logic circuits. 10 1 Truth tables are also used in the implementation of programmable logic devices, such as programmable logic arrays (PLAs), read-only memories (ROMs), random-access memories (RAMs), and field-programmable gate arrays (FPGAs), where the table is programmed to define the logic function. 11 Any gate circuit can be analyzed by drawing up a truth table, or a set of truth tables, by using lettering to identify all inputs, outputs, and intermediate points, and drawing up the truth table in stages, starting with all possible combinations at the inputs. 7
4. Truth Tables in Formal Methods and Verification
Truth tables are canonical representations of Boolean functions, providing an exhaustive enumeration of all possible truth assignments and their corresponding function values, which is essential for logic synthesis and verification in computer science. 1 12 In equivalence checking, a truth table serves as a canonical representation for Boolean functions, allowing equivalence to be determined by direct comparison of the canonical representations of two circuits. 13 Truth tables are used to determine the truth value of statements in propositional logic by evaluating all possible combinations of input values, which can be systematically represented in a truth table. 2 4Automated theorem proving uses truth tables to verify inference rules and to evaluate all possible variable assignments, which can quickly become a satisfiability problem with exponential time complexity as the number of variables increases. 14 3 The exponential growth in the size of truth tables, known as state space explosion, presents a challenge for formal methods, but techniques such as recognizing sparseness and using more compact representations help mitigate this issue.
5. Conclusion
Truth tables provide an exhaustive enumeration of the mapping of a Boolean function, listing every possible truth assignment and the corresponding function value, and serve as canonical representations of Boolean functions, ensuring that two Boolean functions are equivalent if and only if they have the same truth table. 1 This canonicity is important in applications such as logic synthesis and verification. Truth tables are effective for representing functions with a small number of input variables, typically no more than 5 or 6 for modern computers with word sizes of 32 or 64 bits, and allow basic Boolean operations over small functions to be performed in constant time using parallel bitwise operations. However, the size of a truth table grows exponentially with the number of input variables, requiring 2ⁿ rows for n variables, which makes them impractical for functions with many inputs. 3 In practice, more compact representations can be generated by exploiting the sparseness of actual data, as it is often unnecessary to repeat rows with identical or zero results.
References
Reference 1 Book Chapter Logic Synthesis in a NutshellJiang J.H., Devadas S. Electronic Design Automation, 2009 pp 299-404 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... By Shannon expansion , every Boolean function f can be decomposed with respect to some variable x i as (6.12) f = x i f x i ∨ ¬ x i f ¬ x i Note that the variable x i needs not be a functional support variable of f . 6.2.3 Boolean function representation Below we discuss different ways of representing Boolean functions. 6.2.3.1 Truth table The mapping of a Boolean function can be exhaustively enumerated with a truth table , where every truth assignment has a corresponding function value listed. Example 6.7 Figure 6.2 shows the truth table of the majority function f ( x 1 , x 2 , x 3 ), which valuates to true if and only if at least two of the variables { x 1 , x 2 , x 3 } valuate to true. Truth tables are canonical representations of Boolean functions. That is, two Boolean functions are equivalent if and only if they have the same truth table. Canonicity is an important property that may be useful in many applications of logic synthesis and verification. For practical implementation, a truth table is effective in representing functions with a few input variables (often no more than 5 or 6 variables for modern computers having a word size 32 or 64 bits). By storing a truth table as a computer word, basic Boolean operations over two small functions can be done in constant time by parallel bitwise operation over their truth tables. Truth tables however are impractical to represent functions with many input variables. 6.2.3.2 SOP Sum-of-products (SOP), or disjunctive normal form (DNF) as it is called in computer science, is a special form of Boolean formulas consisting of disjunctions (sums) of conjunctions of literals (product terms or cubes). It is a flat structure corresponding to a two-level circuit representation (the first level of AND gates and the second level of an OR gate). In two-level logic minimization, the set of product terms ( i.e. , cubes) of an SOP representation of a Boolean function is called a cover of the Boolean function. A Boolean function may have many different covers, and a cover uniquely determines a Boolean function. ..." What are related quotes? Related quote(s)2 / 2 "... Truth table The mapping of a Boolean function can be exhaustively enumerated with a truth table , where every truth assignment has a corresponding function value listed. Example 6.7 Figure 6.2 shows the truth table of the majority function f ( x 1 , x 2 , x 3 ), which valuates to true if and only if at least two of the variables { x 1 , x 2 , x 3 } valuate to true. Truth tables are canonical representations of Boolean functions. That is, two Boolean functions are equivalent if and only if they have the same truth table. Canonicity is an important property that may be useful in many applications of logic synthesis and verification. For practical implementation, a truth table is effective in representing functions with a few input variables (often no more than 5 or 6 variables for modern computers having a word size 32 or 64 bits). By storing a truth table as a computer word, basic Boolean operations over two small functions can be done in constant time by parallel bitwise operation over their truth tables. Truth tables however are impractical to represent functions with many input variables. ..." What are related quotes?
Reference 2 Book Chapter Computer EngineeringPhilip Kosky, Robert Balmer, William Keat, George Wise Exploring Engineering, 2013 pp 161-183 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Truth tables It is often convenient to summarize the results of a logic analysis for all possible combinations of the values of the input variables of an “if … then” statement. This can be done with a truth table . It is simply a table with columns representing variables and rows representing combinations of variable values. The variables for the “if” conditions start from the left, and their rows can be filled in systematically to include all possible combinations of inputs. The column at the far right represents “then.” Its value can be computed for each possible input combination. Example 8.4 Consider the condition in Example 8.2: open the throttle if the speed is below the set speed and the set speed is not above the speed limit. The “if” conditions are A = “speed is below the speed limit” and B = “set speed is above the speed limit.” The “then” condition is X = “open the throttle.” The truth table is set up as shown here. A B B ′ X = A B ′ Need: All 16 entries to the truth table. Know: Negation operator ′ and the AND operator . How: Fill in all possible binary combinations of statements A and B . Solve: One convenient way of making sure you insert all the possible input values is to “count” in binary from all 0s at the top to all 1s at the bottom. (If you're not already able to count in binary, this is explained on the next page or so.) In this example, the top line on the input side represents the binary number 00 (equal to decimal 0), the second line is 01 (decimal 1), the third is 10 (decimal 2), and the fourth is 11 (decimal 3). 6 6 If the input had three columns, A , B , and C , you would count from 0 to 7 in binary (000 to 111). If four columns, the 16 entries would count from 0 to 15 in binary , and so on. ..." What are related quotes?
Reference 3 Book Chapter Basic formal verification algorithmsErik Seligman, Tom Schubert, M V Achutha Kiran Kumar Formal Verification, 2015 pp 23-47 View PDFView chapterShow related quote(s) Related quote(s)1 / 3 "... We can use truth tables to define functions or systems with multiple stimulus arguments and multiple results. For example, the truth table below describes a 1-bit ADD operation ( Table 2.1 ). The size of truth tables grows quickly as stimuli are added: with each additional input column, we double the number of rows for the number of new possible stimulus values. Defining a MUX operation, for example, requires twice as many rows as the 2-bit ADD ( Table 2.2 ): This means that a truth table for n inputs will have 2 n rows; as we discussed in the last chapter, this exponential growth quickly makes it very difficult to analyze with simple tools and techniques. However, many FV algorithms take advantage of a key insight that mitigates this problem: in real life, such a truth table can often be quite sparse in terms of information content, since many rows produce 0 values, and many sets of rows produce the same value. As an example, consider developing a specification for a simple vending machine truth table that determines whether 35 cents has been deposited. (Yes, one of the authors may be revealing their age by hypothesizing such a vending machine, but bear with us for now.) We’ll limit our specification to require that the machine only accepts exact change consisting of nickels (5 cents), dimes (10 cents), or quarters (25 cents). Legitimate inputs to the vending machine that should be accepted are: One quarter, one dime One quarter, two nickels Three dimes, one nickel Two dimes, three nickels One dime, five nickels Seven nickels For these six cases, our truth table result should be 1, and otherwise it should be 0. To keep track of the possible inputs at any time we will need six bits: 1 bit to track how many quarters have been entered (0 or 1), 2 bits to track how many dimes have been entered (0–3), and 3 bits to track how many nickels have been entered (0–7). To write out a complete truth table, we will need 64 rows (2 6 ), yet only six rows will produce a 1 result. ..." What are related quotes? Related quote(s)2 / 3 "... Building Truth Tables Intelligently As we mentioned in the beginning of this chapter, the truth table in Table 2.1 described a 1-bit add operation, and contained four rows. Suppose we want to expand this ADD operation to allow more input variables. With each additional input column, we double the number of rows for the number of new possible stimulus values, leading to exponential growth in the size of a basic truth table: we will require 2 n rows to handle n variables. However, as we saw in the vending machine example, we can take advantage of the sparseness of the actual data to generate much more compact representations of truth tables. By recognizing the fact that we do not need to repeat rows that represent identical or zero results, we can use significantly fewer rows in practice than the naïve exponential calculation would suggest. 1 ..." What are related quotes? Related quote(s)3 / 3 "... Formalizing Operation Definitions Within a requirements or implementation specification, we use Boolean variables for the inputs, outputs, and state elements that may retain different values over time. We will initially assume that these variables can take on binary values ( 0 or 1 ), though it can be useful to add a third unknown value, “ X ,” to indicate we don’t know whether the value is 0 or 1 . Conventionally, 0 is also interpreted as representing low/off/false and 1 as representing high/on/true. Operations on values or combinations of values can be described in terms of truth tables where the results (outputs and next states) are a function of stimulus arguments (inputs and current states). Since the reader is presumed to be a design or validation engineer proficient in SystemVerilog, we will assume you have a basic understanding of the common Boolean operations: AND (&&), OR (||), NOT (!), and their various negations and variants. Building Truth Tables Intelligently As we mentioned in the beginning of this chapter, the truth table in Table 2.1 described a 1-bit add operation, and contained four rows. Suppose we want to expand this ADD operation to allow more input variables. With each additional input column, we double the number of rows for the number of new possible stimulus values, leading to exponential growth in the size of a basic truth table: we will require 2 n rows to handle n variables. However, as we saw in the vending machine example, we can take advantage of the sparseness of the actual data to generate much more compact representations of truth tables. By recognizing the fact that we do not need to repeat rows that represent identical or zero results, we can use significantly fewer rows in practice than the naïve exponential calculation would suggest. Adding Sequential Logic Truth tables can also capture systems with sequential logic, where the model can store state information that changes over time, by adding input stimulus and output result columns to represent the current and next states of a variable. Though two variables are used, it is convenient to give them the same name with an added suffix tick mark (‘) to the successor state variable. 1 ..." What are related quotes?
Reference 4 Book Chapter The Propositional CalculusNils J. Nilsson Artificial Intelligence: A New Synthesis, 1998 pp 217-230 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... 13.5.2 The Propositional Truth Table Given the values of atoms under some interpretation, we can use a truth table to compute a value for any wff under that same interpretation. The truth table establishes the semantics (meanings) of the propositional connectives. Let ω 1 and ω 2 be wffs; then the truth table rules are (ω 1 ∧ ω 2 ) has value True if both ω 1 and ω 2 have value True ; otherwise, it has value False. (ω 1 ∨ ω 2 ) has value True if one or both of ω 1 or ω 2 has value True ; otherwise, it has value False. ¬ω 1 has value True if ω 1 has value False ; otherwise, it has value False. The semantics of ⊃ is defined in terms of ∨ and ¬. Specifically, (ω 1 ⊃ ω 2 ) is an alternative and equivalent form of (¬ω 1 ∨ ω 2 ). These are called truth table rules because they are usually presented in tabular form, as they are in Table 13.1 . We can use the truth table to compute the value of any wff given the values of the constituent atoms in the wff. As an example of the use of the truth table to compute the value of a wff, suppose ρ has value False, Q has value False, and R has value True. In this interpretation, what is the truth value of [(P ⊃ Q) ⊃ R] ⊃ P? Working from the “inside out,” we first compute the value of P ⊃ Q to be True ; then we see that (P ⊃ Q) ⊃ R is True. Finally, since ρ is False, the value of the entire expression must be False. If an agent describes its world using n features (corresponding to propositions), and these features are represented in the agent's model of the world by a corresponding set of n atoms, then there are 2 n different ways its world can be—so far as the agent can discern—because there are 2 n ways in which the n atoms can have values of True and False. Each of the ways the world can be corresponds to an interpretation. 1 ..." What are related quotes?
Reference 5 Book Chapter The Relational ParadigmTom Johnston Bitemporal Data, 2014 pp 43-58 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Connectives The usual set of statement connectives used in propositional logic are formal equivalents of what is expressed in ordinary language by “and”, “or”, “not”, “if…then” and “if and only if”. The usual symbols for these connectives are, respectively, “∧”, “∨”, “~”, “→” and “↔”. Equivalent symbols for these connectives are, respectively, “AND”, “OR”, “NOT”, “IF/THEN” and “IFF”. In standard SQL, only the first three connectives are used. The other two can easily be defined in terms of the first three. In fact, based on either AND and NOT, or on OR and NOT, all three other connectives can be defined. 1,2 1 “AND” designates the formalization of the “and” of ordinary English, and so too for “OR” and “or”, and “NOT” and “not”. Thus, these three strings of capital letters are not English words; they are equivalent strings for, respectively, “∧”, “∨” and “~”. 2 There is even a single operator in terms of which all the others may be defined. It is called the Sheffer stroke. But it is so terse and non-intuitive a connective that it is almost never used in solving problems in logic. AND The logical AND operator applies between two statements, and forms a compound statement called a conjunction . It says that if either or both of a pair of component statements is false, then their conjunction is false, and otherwise is true. To put it another way, the conjunction (X∧Y) is true is when both its components are true, and otherwise is false. A truth table contains one row for every assignment of true (T) and false (F) to the component statements of a compound statement, and then shows the truth-value assignment to the compound statement, given those component assignments. The truth table for conjunction – the formalization of “and” – is this: Figure 4.1 The Truth Table for AND. OR The logical OR operator applies between two statements, and forms a compound statement called a disjunction . 1 ..." What are related quotes?
Reference 6 Book Chapter Fault diagnosis and fault toleranceAfaq Ahmad, Sayyid Samir Al Busaidi System Assurances, 2022 pp 307-321 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... Functional modeling at the logic level The simplest way to represent a digital system of combinational nature is by its Truth table . An example of a typical Truth table of function F ( x 1, x 2, x 3 ) is shown in Table 17.1 . Assuming binary input values, a digital system realizing a function F ( x 1, x 2, x 3 ,…, x n ) of n variables require a table with 2 n entries. The data structure representing a Truth table is, therefore, an array V of dimension 2 n . The table is arranged with all possible input combinations of n -bit binary representations of decimal values varying from 0 to 2 n − 1. From the perspective of testing of digital circuit, a typical example of Primitive cubes model representation of Table 17.1 is given in Table 17.2 . In Table 17.2 , d represents do not care, i.e., the input may be 0 or 1. Sequential circuit in a digital system is a finite state sequential function. The function can be modeled as a sequential machine which is represented as State tables and Flow tables (State diagram) [7,8] . In digital system modeling, also, exists Binary Decision Diagrams technique [9,10] which is basically a graph model representation of the function of a digital circuit. Depicted in Fig. 17.2 is the binary decision diagrams Boolean function F = ( x ′ y ′) p . q . 1 ..." What are related quotes? Related quote(s)2 / 2 "... Digital systems modeling Digital systems design requires rigorous modeling and simulation analysis that eliminates design risks so as to avoid potential harm to consumers. Since digital systems designer is responsible for the integrity of the system design as a whole, it must encompass the interconnectedness of the various aspects of the system. Hence, modeling of a digital system plays an important role in the system’s design, synthesis fabrication, analysis, and testing. Mainly the modeling belong to three of the categories namely. (1) Behavioral, (2) Functional, and (3) Structural. A behavioral model is basically defined by a functional model but it also associates with timing relationships. Modeling of a digital system plays an important role in system design, synthesis fabrication, analysis, and testing. Mainly the functional level modeling can further be classified as follows: 17.2.1 Functional modeling at the logic level The simplest way to represent a digital system of combinational nature is by its Truth table . An example of a typical Truth table of function F ( x 1, x 2, x 3 ) is shown in Table 17.1 . Assuming binary input values, a digital system realizing a function F ( x 1, x 2, x 3 ,…, x n ) of n variables require a table with 2 n entries. The data structure representing a Truth table is, therefore, an array V of dimension 2 n . The table is arranged with all possible input combinations of n -bit binary representations of decimal values varying from 0 to 2 n − 1. From the perspective of testing of digital circuit, a typical example of Primitive cubes model representation of Table 17.1 is given in Table 17.2 . In Table 17.2 , d represents do not care, i.e., the input may be 0 or 1. Sequential circuit in a digital system is a finite state sequential function. The function can be modeled as a sequential machine which is represented as State tables and Flow tables (State diagram) [7,8] . In digital system modeling, also, exists Binary Decision Diagrams technique [9,10] which is basically a graph model representation of the function of a digital circuit. 1 ..." What are related quotes?
Reference 7 Book Chapter Gating and Logic CircuitsIan Sinclair Electronics Simplified, 2011 pp 185-201 View PDFView chapterShow related quote(s) Related quote(s)1 / 6 "... Publisher Summary Logic circuits exist to carry out a set of logic actions such as are used for controls for washing machines, tape-recorder drives, computer disk drives, security systems, and a host of industrial control actions. Simple arithmetic actions can also be carried out using logic circuits. All logic actions, however complicated, can be analyzed into simple actions that are called AND, OR, and NOT, so that circuits, called gates, which carry out such actions, are the basis of logical circuits. Any gate circuit can be analyzed by drawing up a truth table, or a set of truth tables. A truth table is a simple way of showing what the output of a gate or gate circuit will be for each and every possible combination of inputs. The alternative is to use Boolean algebra, a process that is greatly simplified by computer programs that carry out an analysis of gate circuits. The Boolean algebra alternative method relies on using lettering to identify all inputs, outputs, and intermediate points, and drawing up the truth table in stages, starting with all possible combinations at the inputs. An arithmetic circuit is a set of gates with a separate set of inputs for each number that has to be processed. The gates are connected to carry out an arithmetic action and the outputs of the gate circuit are the digits of the result (addition, subtraction, multiplication, or division). Encoders, decoders, multiplexers, and demultiplexers are used to change the way that digital codes are carried. 1 ..." What are related quotes? Related quote(s)2 / 6 "... The more useful method for professional use is called Boolean algebra but, like other mathematical methods, it is beyond the scope of this book. There are computer programs that will analyze the action of a gate circuit using Boolean algebra. Figure 10.2 shows truth tables for the basic two-input AND, OR, and NOT gates. Of these, the NOT gate is a simple one, with just one input and one output. Its action is that of a logic inverter. If the input is 0, then the output is not0, which is 1. If the input is 1, then the output is not1, which is 0. The other two gate types permit more than one input, and the examples show two inputs, the most common number. The action of the AND gate is to give a 1 output only when both inputs are at 1, and a 0 output for any other combination. The action of the OR gate is to give a 0 output when each input is 0, but a 1 for any other combination of inputs. The same arguments apply to gates with more than two inputs. Summary A truth table is a simple but clumsy way of showing what the output of a gate or gate circuit will be for each and every possible combination of inputs. The alternative is to use Boolean algebra, a process that is greatly simplified by computer programs that carry out an analysis of gate circuits. NAND and NOR Gates Two particularly useful gate types can be made by combining the action of an inverter with that of the AND and OR gates. The combination of NOT and AND gives the NAND gate, whose symbols and truth table (for two inputs) are shown in Figure 10.3 . The action of this gate is that the output is 0 only when all of its inputs are at 1, which is the action of the AND gate followed by an inverter. The combined action of the OR gate and a following inverter gives the NOR gate, whose symbols and truth table are shown in Figure 10.4 . The output of this gate will be at logic 0 when any one (or more) of its inputs is at logic 1. There is one further gate that is often used and which is called exclusive-OR (XOR). 1 ..." What are related quotes? Related quote(s)3 / 6 "... NAND and NOR Gates Two particularly useful gate types can be made by combining the action of an inverter with that of the AND and OR gates. The combination of NOT and AND gives the NAND gate, whose symbols and truth table (for two inputs) are shown in Figure 10.3 . The action of this gate is that the output is 0 only when all of its inputs are at 1, which is the action of the AND gate followed by an inverter. The combined action of the OR gate and a following inverter gives the NOR gate, whose symbols and truth table are shown in Figure 10.4 . The output of this gate will be at logic 0 when any one (or more) of its inputs is at logic 1. There is one further gate that is often used and which is called exclusive-OR (XOR). This action ( Figure 10.5 ) is closer to what we normally mean by the word ‘or’ (meaning one or the other but not both), and the output is 1 if either input is 1, but not when both inputs are zero or both are 1. The diagram also shows that the XOR gate is equivalent to the action of a circuit made using an OR, AND, and NAND gate combination. Note If you would like to read further about gates and other logic circuits, with details of the more advanced methods such as Boolean algebra, take a look at the book Digital Logic Gates and Flip-Flops , from PC Publishing. Summary A truth table is a simple way of expressing the action of a logic circuit, and the standard gates called OR, AND, and NOT can all be illustrated in this way. Gates in IC form often consist of the NAND and NOR type of gates, equivalent to an AND or OR, respectively, followed by NOT. These inverting gates are easier to produce and the action is often more useful than that of the simpler AND or OR type. The XOR gate is another useful type which gives an action closer to the normal meaning of OR, as ‘one or the other but not both’. 1 ..." What are related quotes? Related quote(s)4 / 6 "... A gate circuit diagram will consist of a large number of gate symbols with joining lines so that the output of a gate will be connected to one or more other gate inputs. Provided that we know what each basic type of gate does, we can analyze the action of complete gate circuits. In this book we are concerned more with block diagrams than with gate circuits, but some knowledge of gate circuits is useful, and in any case, these are closer in spirit to block diagrams than to circuit diagrams. Summary Logic circuits exist to carry out a set of logic actions such as are used for controls for washing machines, tape-recorder drives, computer disk drives, security systems, and a host of industrial control actions. Simple arithmetic actions can also be carried out using logic circuits. All logic actions, however complicated, can be analyzed into simple actions that are called AND, OR, and NOT, so that circuits, called gates, which carry out such actions, are the basis of logical circuits. Note Remember that ICs are classed as medium-scale integration (MSI), large-scale integration (LSI), very large-scale integration (VLSI), etc., by the number of equivalent gate circuits on a single chip. Definition The action of any gate can be expressed in a truth table . This is just a table that shows all the possible inputs to the gate, and the output for each set of inputs. Remember that each input can be 0 or 1 only, so that each input contributes two possible outputs. The total number of outputs is equal to 2 n , where n is the number of inputs. For example, if there are four inputs to a gate, then the number of possibilities is 2 4 = 16, and its truth table will consist of 16 lines. For a lot of truth tables, there is only one output that is different from the rest, and it is easier to remember which one this is than to try to remember the whole of a truth table. Note Truth tables are the simplest way of showing what a small-scale gate or gate circuit does, but they are impossibly clumsy when we try to use them on complex gates with a large number of inputs. 1 ..." What are related quotes? Related quote(s)5 / 6 "... This action ( Figure 10.5 ) is closer to what we normally mean by the word ‘or’ (meaning one or the other but not both), and the output is 1 if either input is 1, but not when both inputs are zero or both are 1. The diagram also shows that the XOR gate is equivalent to the action of a circuit made using an OR, AND, and NAND gate combination. Note If you would like to read further about gates and other logic circuits, with details of the more advanced methods such as Boolean algebra, take a look at the book Digital Logic Gates and Flip-Flops , from PC Publishing. Summary A truth table is a simple way of expressing the action of a logic circuit, and the standard gates called OR, AND, and NOT can all be illustrated in this way. Gates in IC form often consist of the NAND and NOR type of gates, equivalent to an AND or OR, respectively, followed by NOT. These inverting gates are easier to produce and the action is often more useful than that of the simpler AND or OR type. The XOR gate is another useful type which gives an action closer to the normal meaning of OR, as ‘one or the other but not both’. Analyzing Gate Systems A circuit that has been made up by connecting several standard gates together, which has several inputs and an output, can be analyzed to find what its action is. This analysis can be done by drawing up truth tables, or by a method called Boolean algebra. The truth-table method is simpler, but more tedious than the Boolean algebra method, which is not dealt with in this book. The method of analysis by truth table can be summarized in a few rules. 1. Letter each input to the circuit (A, B, C) and also each point where the output of one gate is connected to the input of another gate, using different letters for each point. Label the final output as Q. 2. Draw up a blank truth table, using one column for each letter that has been allocated, and with 2 n rows, where n is the number of signal inputs to the circuit. 3. Write in every possible combination of inputs. 1 ..." What are related quotes? Related quote(s)6 / 6 "... Analyzing Gate Systems A circuit that has been made up by connecting several standard gates together, which has several inputs and an output, can be analyzed to find what its action is. This analysis can be done by drawing up truth tables, or by a method called Boolean algebra. The truth-table method is simpler, but more tedious than the Boolean algebra method, which is not dealt with in this book. The method of analysis by truth table can be summarized in a few rules. 1. Letter each input to the circuit (A, B, C) and also each point where the output of one gate is connected to the input of another gate, using different letters for each point. Label the final output as Q. 2. Draw up a blank truth table, using one column for each letter that has been allocated, and with 2 n rows, where n is the number of signal inputs to the circuit. 3. Write in every possible combination of inputs. This is most easily done by starting with 0000 and continuing in the form of a binary count up to an input which consists entirely of 1s. 4. Knowing the truth tables for the standard gates, write in the logic states (0 or 1) for the outputs of the gates at the inputs in each line of the truth table. 5. The first set of outputs will now be the inputs for the next set of gates, so that their outputs can be written into the truth table. 6. Continue in this way until the truth table has been completed. As an example, Figure 10.6 shows a logic diagram for an electronic combination lock. This is a simple design, with four main inputs, and therefore 16 combinations, ignoring the unlock input E. The lock is arranged so that only the correct combination of inputs will open the lock, and any other combination will cause an alarm to sound, so that it cannot be solved by trying each possible combination. The inputs, A, B, C, and D are from switches which are to be set in the pattern needed to open the lock. When these switches have been set, pressing the button E will cause the door to unlock (Q = 1) if the combination is correct, or cause the alarm to sound (X = 1) if the combination is incorrect. 1 ..." What are related quotes?
Reference 8 Book Chapter From Zero to OneSarah L. Harris, David Harris Digital Design and Computer Architecture, 2022 pp 1-50 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... Multiple-Input Gates Many Boolean functions of three or more inputs exist. The most common are AND, OR, XOR, NAND, NOR, and XNOR. An N -input AND gate produces a TRUE output when all N inputs are TRUE. An N -input OR gate produces a TRUE output when at least one input is TRUE. Example 1.16 Three-Input NOR Gate Figure 1.19 shows the symbol and Boolean equation for a three-input NOR gate. Complete the truth table. Solution Figure 1.20 shows the truth table. The output is TRUE only if none of the inputs are TRUE. Example 1.17 Four-Input And Gate Figure 1.21 shows the symbol and Boolean equation for a four-input AND gate. Create a truth table. Solution Figure 1.22 shows the truth table. The output is TRUE only if all of the inputs are TRUE. 1 ..." What are related quotes? Related quote(s)2 / 2 "... The XOR operation is indicated by ⊕, a plus sign with a circle around it. Any gate can be followed by a bubble to invert its operation. The NAND gate performs NOT AND. Its output is TRUE unless both inputs are TRUE. The NOR gate performs NOT OR. Its output is TRUE if neither A nor B is TRUE. An N -input XOR gate is sometimes called a parity gate and produces a TRUE output if an odd number of inputs are TRUE. As with two-input gates, the input combinations in the truth table are listed in counting order. A silly way to remember the OR symbol is that its input side is curved like Pacman’s mouth, so the gate is hungry and willing to eat any TRUE inputs it can find! Example 1.15 XNOR Gate Figure 1.17 shows the symbol and Boolean equation for a two-input XNOR (pronounced ex-NOR) gate that performs the inverse of an XOR. Complete the truth table. Solution Figure 1.18 shows the truth table. The XNOR output is TRUE if both inputs are FALSE or both inputs are TRUE. The two-input XNOR gate is sometimes called an equality gate because its output is TRUE when the inputs are equal. 1.5.6 Multiple-Input Gates Many Boolean functions of three or more inputs exist. The most common are AND, OR, XOR, NAND, NOR, and XNOR. An N -input AND gate produces a TRUE output when all N inputs are TRUE. An N -input OR gate produces a TRUE output when at least one input is TRUE. Example 1.16 Three-Input NOR Gate Figure 1.19 shows the symbol and Boolean equation for a three-input NOR gate. Complete the truth table. Solution Figure 1.20 shows the truth table. The output is TRUE only if none of the inputs are TRUE. Example 1.17 Four-Input And Gate Figure 1.21 shows the symbol and Boolean equation for a four-input AND gate. Create a truth table. Solution Figure 1.22 shows the truth table. The output is TRUE only if all of the inputs are TRUE. 1 ..." What are related quotes?
Reference 9 Book Chapter Digital Control SystemsParr E.A. Electrical Engineer's Reference Book, 2003 pp 14-1 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Combinational logic 14.3.1 Introduction Combinational logic is based around the block diagram of Figure 14.21 (a) . Such systems have several inputs and one, or more, outputs. The output states are uniquely defined for each and every combination of inputs and the ‘block’ does not contain any device such as storage, timers or counters. We therefore have n inputs I 1 to I n and Z outputs Q 1 to Q z . In systems with multiple outputs it is usually easier to consider each separately as Figure 14.21 (b) , allowing us to consider the circuit as Z blocks, each different but represented by Figure 14.21 (c) . The number of possible input states depends on the number of inputs: For two inputs there are four input combinations For three inputs there are eight input combinations For four inputs there are sixteen input combinations and so on. Not all of these may be needed. There are frequently only a certain number of input combinations that may occur because of physical restrictions elsewhere in the system. The design of combinational logic systems first involves examining all the input states that can occur and defining the output states that must occur for each and every input state. A logic design to achieve this is then constructed from the gates described in Section 14.1.3 . In many systems the design can be done in an intuitive manner, but the rest of this section describes more formal design procedures. Few real life systems need pure combinational logic, most need storage and similar dynamic functions. Such systems can be analysed and designed considering them as smaller subsystems linked together. The design of dynamic systems is discussed in Section 14.8 . 14.3.2 Truth tables A truth table is a useful way of representing a combinational logic circuit, and can be used to design the circuit needed to achieve a desired function. Suppose we have three contacts monitoring some event (overpressure in a chemical reactor for example) and we wish to construct a majority vote circuit. 1 ..." What are related quotes?
Reference 10 Book Chapter IntroductionStroud C.E., Wang L.T., Chang Y.-W. Electronic Design Automation, 2009 pp 1-38 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... The actual digital logic continued to be hand-designed by use of state diagrams, state tables, Karnaugh maps, and a few simple CAD tools for logic minimization. This limited the complexity of ASICs in terms of the number of gates or transistors that could be correctly designed and verified by a typical designer. In the early 1980s, an ASIC with more than 100,000 transistors was considered to be near the upper limit for a single designer. By the late 1980s, the limit was significantly increased as a result of multi-designer teams working on a single IC and as a result of advances in EDA capabilities, particularly in the area of logic synthesis. Currently, the largest ICs exceed 1 billion transistors [ Naffziger 2006 ; Stackhouse 2008 ]. One of the earliest approaches to EDA in terms of combinational logic synthesis was for implementing programmable logic arrays (PLAs) in very large-scale integrated (VLSI) circuits in the late 1970s [ Mead 1980 ]. Any combinational logic function can be expressed as Boolean logic equations, sum-of-products (SOP) or product-of-sums (POS) expressions, and truth tables or Karnaugh maps. There are other representations, but these three are illustrated for the example circuit in Figure 1.6 and are important for understanding the implementation of PLAs and other programmable logic. We can program the truth table onto a read-only memory (ROM) with eight words and two bits/word and then use the ROM address lines as the three input signals ( A, B, C ) and the ROM outputs as the output signals ( X, Y ). Similarly, we can also write the truth table into a random-access memory (RAM) with eight words and two bits/word and then disable the write enable to the RAM and use the address lines as the inputs. Note that this is the same thing as the ROM, except that we can reprogram the logic function by rewriting the RAM; this also forms the basis for combinational logic implementations in FPGAs. Another option for implementing a truth table is the PLA. 1 ..." What are related quotes? Related quote(s)2 / 2 "... Currently, the largest ICs exceed 1 billion transistors [ Naffziger 2006 ; Stackhouse 2008 ]. One of the earliest approaches to EDA in terms of combinational logic synthesis was for implementing programmable logic arrays (PLAs) in very large-scale integrated (VLSI) circuits in the late 1970s [ Mead 1980 ]. Any combinational logic function can be expressed as Boolean logic equations, sum-of-products (SOP) or product-of-sums (POS) expressions, and truth tables or Karnaugh maps. There are other representations, but these three are illustrated for the example circuit in Figure 1.6 and are important for understanding the implementation of PLAs and other programmable logic. We can program the truth table onto a read-only memory (ROM) with eight words and two bits/word and then use the ROM address lines as the three input signals ( A, B, C ) and the ROM outputs as the output signals ( X, Y ). Similarly, we can also write the truth table into a random-access memory (RAM) with eight words and two bits/word and then disable the write enable to the RAM and use the address lines as the inputs. Note that this is the same thing as the ROM, except that we can reprogram the logic function by rewriting the RAM; this also forms the basis for combinational logic implementations in FPGAs. Another option for implementing a truth table is the PLA. In the connection array in Figure 1.6e , only three product terms produce logic 1 at the output signals. The PLA allows implementing only those three product terms and not the other five, which is much smaller than either the ROM or RAM implementation. Any SOP can be implemented as a 2-level AND-OR or NAND-NAND logic function. Any SOP can also be implemented as a 2-level NOR-NOR logic function if we invert the inputs and the output as illustrated in Figure 1.7a . Note that AB' is a shared product term and allows us to share an AND gate in the gate-level implementation. 1 ..." What are related quotes?
Reference 11 Book Chapter Computer Hardware for Industrial ControlPeng Zhang Industrial Control Technology, 2008 pp 187-257 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Logic cells are also known as logic array blocks, logic elements, and configurable logic blocks. Lookup tables or truth tables are used to implement a single logic function by storing the correct output logic state in a memory location that corresponds to each particular combination of input variables. Field-programmable gate arrays are available with many logic families. Transistor-transistor logic and related technologies such as Fairchild advanced Schottky use transistors as digital switches. By contrast, emittercoupled logic uses transistors to steer current through gates that compute logical functions. Another logic family, CMOS, uses a combination of P-type and N-type metal-oxide-semiconductor field effect transistors to implement logic gates and other digital circuits. Logic families for field-programmable gate arrays include crossbar switch technology, gallium arsenide, integrated injection logic, and silicon on sapphire. Gunning with transceiver logic and gunning with transceiver logic plus are also available. Field-programmable gate arrays are available in a variety of IC package types and with different numbers of pins and flip-flops. Basic IC package types for field-programmable gate arrays include ball grid array, quad flat package, single in-line package, and dual in-line package. Many packaging variants are available. 2.3.3.2 FPGA Architecture The typical basic architecture consists of an array of logic blocks and routing channels. Multiple I/O pads may fit into the height of one row or the width of one column. Generally, all the routing channels have the same width (number of wires). Figure 2.18 illustrates the FPGA structure. (1) FPGA logic block. An application circuit must be mapped into an FPGA with adequate resources. The typical FPGA logic block consists of a 4-input lookup table (LUT) and a flip-flop, as shown in Fig. 2.19 . There is only one output in each logic block, which can be either the registered or the unregistered LUT output. As shown in Figure 2.20 , the logic block has four inputs for the LUT and a clock input. 1 ..." What are related quotes?
Reference 12 Book Chapter Basic formal verification algorithmsErik Seligman, Tom Schubert, M V Achutha Kiran Kumar Formal Verification, 2023 pp 25-49 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... A simple vending machine example Before we get into too much detail on the general techniques, we begin with a small example showing how a clever data structure can be used to transform an apparently exponential complexity problem into something much more tractable. Boolean logic is usually represented through truth tables, which describe expected output values of a system in terms of each of their inputs. We can use truth tables to define functions or systems with multiple stimulus arguments and multiple results. For example, the truth table below describes a 1-bit ADD operation ( Table 2.1 ). The size of truth tables grows quickly as stimuli are added: with each additional input column, we double the number of rows for the number of new possible stimulus values. Defining a MUX operation, for example, requires twice as many rows as the 2-bit ADD ( Table 2.2 ). This means that a truth table for n inputs will have 2 n rows; as we discussed in the last chapter, this exponential growth quickly makes it very difficult to analyze with simple tools and techniques. However, many FV algorithms take advantage of a key insight that mitigates this problem: in real life, such a truth table can often be quite sparse in terms of information content, since many rows produce 0 values, and many sets of rows produce the same value. As an example, consider developing a specification for a simple vending machine truth table that determines whether 35 cents have been deposited (Yes, one of the authors may be revealing their age by hypothesizing such a vending machine, but bear with us for now.). We'll limit our specification to require that the machine only accepts exact change consisting of nickels (5 cents), dimes (10 cents), or quarters (25 cents). Legitimate inputs to the vending machine that should be accepted are: one quarter, one dime; one quarter, two nickels; three dimes, one nickel; two dimes, three nickels; one dime, five nickels; seven nickels. For these six cases, our truth table result should be 1, and otherwise it should be 1 ..." What are related quotes?
Reference 13 Book Chapter Functional VerificationWen H.P., Wang L.-C., Cheng K.T. Electronic Design Automation, 2009 pp 513-573 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... Checking based on functional equivalence The first step of functional CEC is to translate the combinational circuits into a canonical representation . A representation of a Boolean function is canonical if the representation for each function is unique and independent of the implementation of the function. A truth table is one example of a canonical representation for Boolean functions. Equivalence can be determined by directly comparing the two canonical representations. Among all canonical representations, the reduced ordered binary decision diagram (OBDD), introduced in Chapter 4 , is the most prevalent, because OBDD yields a more compact representation than other representations. The CEC problem can be resolved by building the OBDDs for the outputs of the circuits on the basis of their primary inputs. Two circuits are equivalent if the OBDDs from each pair of corresponding outputs are graphically isomorphic. 1 ..." What are related quotes? Related quote(s)2 / 2 "... A representation of a Boolean function is canonical if the representation for each function is unique and independent of the implementation of the function. A truth table is one example of a canonical representation for Boolean functions. Equivalence can be determined by directly comparing the two canonical representations. Among all canonical representations, the reduced ordered binary decision diagram (OBDD), introduced in Chapter 4 , is the most prevalent, because OBDD yields a more compact representation than other representations. The CEC problem can be resolved by building the OBDDs for the outputs of the circuits on the basis of their primary inputs. Two circuits are equivalent if the OBDDs from each pair of corresponding outputs are graphically isomorphic. 9.5.1.2 Checking based on structural search A structural search approach checks to see whether any vector exists at primary inputs that would cause a mismatch between the two circuits at their primary outputs. If no such input vector can be found, the two circuits are proven equivalent. The satisfiability (SAT) solvers, introduced in Chapter 4 , can be used as the structural search engine for checking equivalence. A SAT solver can be used to check if an assignment at PIs exists to satisfy a 1 at the miter's output. An UNSAT answer from the solver proves the equivalence of the two circuits. An ATPG tool developed for generating manufacturing tests for stuck-at faults can also be used for checking structural equivalence. As illustrated in Figure 9.24 , if the stuck-at-0 fault at the XOR output is proven a redundant fault by an ATPG tool, the two circuits are equivalent. A thorough treatment of ATPG techniques will be provided in Chapter 14 . For complex circuits, directly applying SAT solving at the miter's output signal may result in an exponential number of backtracks, which makes the approach inefficient. Structural similarity between the two circuits under checking can be explored to improve its efficiency, which attempts to solve the structural equivalence problem by incrementally solving a sequence of easier sub-problems [ Brand 1993 ; Kunz 1993 ; Goldberg 2000 ; Huang 2000 ]. 1 ..." What are related quotes?
Reference 14 Handbook Chapter Machine learning algorithms, applications, and practices in data scienceYeturu K. Handbook of Statistics, 2020 pp 81-206 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Automated theorem proof Given a knowledgebase of truth statements and a query truth statement, the task is to prove or disprove. There are two ways of accomplishing this—(i) Forward proof method and (ii) Backward proof method. 8.4.3.1 Forward chaining A set of rules are provided to the automated theorem proof algorithm. The algorithm systematically expands the knowledgebase by combinations of the pieces of knowledge to produce derived knowledge. Some of the rules of inference is shown in Table 13 . Here, → denotes implication, ∼ denotes negation, ∨ or denotes logical OR, or ∧ and comma, denotes logical AND, [] denote grouping. The rules are verifiable via truth table . There can be more rules based on combinations in truth table. 8.4.3.2 Incompleteness of the forward chaining The rules of inference may not be sufficient for proving a given query in the forward method. New rules need to be amended for each special query that cannot be addressed. For instance, the following rule is not possible to deduce from a given of rules ( Table 13 ). [ B ∧ D , A ∨ C ] → ( A ∧ B ) ∨ ( C ∧ D ) In order to consider this, one can augment this as new rule. However, it poses a sense of uncertainty for being applicable to all the queries. 8.4.3.3 Backward chaining In backward chaining, the query ( or goal ) is considered along side the knowledgebase and deductions are evaluated. One approach is, to augment the query to the knowledgebase and to prove the whole group to be true for some assignment of values to the variables. However, this reduces to evaluating for all possible values and identifying a particular combination of variable assignments leading to truth. The scenario quickly becomes satisfiability problem which has exponential time complexity . Another way to prove is a query is deducible is by a method of contraction which is presented next. 1 ..." What are related quotes?
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Reference 1 Book Chapter Logic Synthesis in a NutshellJiang J.H., Devadas S. Electronic Design Automation, 2009 pp 299-404 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... By Shannon expansion , every Boolean function f can be decomposed with respect to some variable x i as (6.12) f = x i f x i ∨ ¬ x i f ¬ x i Note that the variable x i needs not be a functional support variable of f . 6.2.3 Boolean function representation Below we discuss different ways of representing Boolean functions. 6.2.3.1 Truth table The mapping of a Boolean function can be exhaustively enumerated with a truth table , where every truth assignment has a corresponding function value listed. Example 6.7 Figure 6.2 shows the truth table of the majority function f ( x 1 , x 2 , x 3 ), which valuates to true if and only if at least two of the variables { x 1 , x 2 , x 3 } valuate to true. Truth tables are canonical representations of Boolean functions. That is, two Boolean functions are equivalent if and only if they have the same truth table. Canonicity is an important property that may be useful in many applications of logic synthesis and verification. For practical implementation, a truth table is effective in representing functions with a few input variables (often no more than 5 or 6 variables for modern computers having a word size 32 or 64 bits). By storing a truth table as a computer word, basic Boolean operations over two small functions can be done in constant time by parallel bitwise operation over their truth tables. Truth tables however are impractical to represent functions with many input variables. 6.2.3.2 SOP Sum-of-products (SOP), or disjunctive normal form (DNF) as it is called in computer science, is a special form of Boolean formulas consisting of disjunctions (sums) of conjunctions of literals (product terms or cubes). It is a flat structure corresponding to a two-level circuit representation (the first level of AND gates and the second level of an OR gate). In two-level logic minimization, the set of product terms ( i.e. , cubes) of an SOP representation of a Boolean function is called a cover of the Boolean function. A Boolean function may have many different covers, and a cover uniquely determines a Boolean function. ..." What are related quotes? Related quote(s)2 / 2 "... Truth table The mapping of a Boolean function can be exhaustively enumerated with a truth table , where every truth assignment has a corresponding function value listed. Example 6.7 Figure 6.2 shows the truth table of the majority function f ( x 1 , x 2 , x 3 ), which valuates to true if and only if at least two of the variables { x 1 , x 2 , x 3 } valuate to true. Truth tables are canonical representations of Boolean functions. That is, two Boolean functions are equivalent if and only if they have the same truth table. Canonicity is an important property that may be useful in many applications of logic synthesis and verification. For practical implementation, a truth table is effective in representing functions with a few input variables (often no more than 5 or 6 variables for modern computers having a word size 32 or 64 bits). By storing a truth table as a computer word, basic Boolean operations over two small functions can be done in constant time by parallel bitwise operation over their truth tables. Truth tables however are impractical to represent functions with many input variables. ..." What are related quotes?
Reference 2 Book Chapter Computer EngineeringPhilip Kosky, Robert Balmer, William Keat, George Wise Exploring Engineering, 2013 pp 161-183 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Truth tables It is often convenient to summarize the results of a logic analysis for all possible combinations of the values of the input variables of an “if … then” statement. This can be done with a truth table . It is simply a table with columns representing variables and rows representing combinations of variable values. The variables for the “if” conditions start from the left, and their rows can be filled in systematically to include all possible combinations of inputs. The column at the far right represents “then.” Its value can be computed for each possible input combination. Example 8.4 Consider the condition in Example 8.2: open the throttle if the speed is below the set speed and the set speed is not above the speed limit. The “if” conditions are A = “speed is below the speed limit” and B = “set speed is above the speed limit.” The “then” condition is X = “open the throttle.” The truth table is set up as shown here. A B B ′ X = A B ′ Need: All 16 entries to the truth table. Know: Negation operator ′ and the AND operator . How: Fill in all possible binary combinations of statements A and B . Solve: One convenient way of making sure you insert all the possible input values is to “count” in binary from all 0s at the top to all 1s at the bottom. (If you're not already able to count in binary, this is explained on the next page or so.) In this example, the top line on the input side represents the binary number 00 (equal to decimal 0), the second line is 01 (decimal 1), the third is 10 (decimal 2), and the fourth is 11 (decimal 3). 6 6 If the input had three columns, A , B , and C , you would count from 0 to 7 in binary (000 to 111). If four columns, the 16 entries would count from 0 to 15 in binary , and so on. ..." What are related quotes?
Reference 3 Book Chapter Basic formal verification algorithmsErik Seligman, Tom Schubert, M V Achutha Kiran Kumar Formal Verification, 2015 pp 23-47 View PDFView chapterShow related quote(s) Related quote(s)1 / 3 "... We can use truth tables to define functions or systems with multiple stimulus arguments and multiple results. For example, the truth table below describes a 1-bit ADD operation ( Table 2.1 ). The size of truth tables grows quickly as stimuli are added: with each additional input column, we double the number of rows for the number of new possible stimulus values. Defining a MUX operation, for example, requires twice as many rows as the 2-bit ADD ( Table 2.2 ): This means that a truth table for n inputs will have 2 n rows; as we discussed in the last chapter, this exponential growth quickly makes it very difficult to analyze with simple tools and techniques. However, many FV algorithms take advantage of a key insight that mitigates this problem: in real life, such a truth table can often be quite sparse in terms of information content, since many rows produce 0 values, and many sets of rows produce the same value. As an example, consider developing a specification for a simple vending machine truth table that determines whether 35 cents has been deposited. (Yes, one of the authors may be revealing their age by hypothesizing such a vending machine, but bear with us for now.) We’ll limit our specification to require that the machine only accepts exact change consisting of nickels (5 cents), dimes (10 cents), or quarters (25 cents). Legitimate inputs to the vending machine that should be accepted are: One quarter, one dime One quarter, two nickels Three dimes, one nickel Two dimes, three nickels One dime, five nickels Seven nickels For these six cases, our truth table result should be 1, and otherwise it should be 0. To keep track of the possible inputs at any time we will need six bits: 1 bit to track how many quarters have been entered (0 or 1), 2 bits to track how many dimes have been entered (0–3), and 3 bits to track how many nickels have been entered (0–7). To write out a complete truth table, we will need 64 rows (2 6 ), yet only six rows will produce a 1 result. ..." What are related quotes? Related quote(s)2 / 3 "... Building Truth Tables Intelligently As we mentioned in the beginning of this chapter, the truth table in Table 2.1 described a 1-bit add operation, and contained four rows. Suppose we want to expand this ADD operation to allow more input variables. With each additional input column, we double the number of rows for the number of new possible stimulus values, leading to exponential growth in the size of a basic truth table: we will require 2 n rows to handle n variables. However, as we saw in the vending machine example, we can take advantage of the sparseness of the actual data to generate much more compact representations of truth tables. By recognizing the fact that we do not need to repeat rows that represent identical or zero results, we can use significantly fewer rows in practice than the naïve exponential calculation would suggest. 1 ..." What are related quotes? Related quote(s)3 / 3 "... Formalizing Operation Definitions Within a requirements or implementation specification, we use Boolean variables for the inputs, outputs, and state elements that may retain different values over time. We will initially assume that these variables can take on binary values ( 0 or 1 ), though it can be useful to add a third unknown value, “ X ,” to indicate we don’t know whether the value is 0 or 1 . Conventionally, 0 is also interpreted as representing low/off/false and 1 as representing high/on/true. Operations on values or combinations of values can be described in terms of truth tables where the results (outputs and next states) are a function of stimulus arguments (inputs and current states). Since the reader is presumed to be a design or validation engineer proficient in SystemVerilog, we will assume you have a basic understanding of the common Boolean operations: AND (&&), OR (||), NOT (!), and their various negations and variants. Building Truth Tables Intelligently As we mentioned in the beginning of this chapter, the truth table in Table 2.1 described a 1-bit add operation, and contained four rows. Suppose we want to expand this ADD operation to allow more input variables. With each additional input column, we double the number of rows for the number of new possible stimulus values, leading to exponential growth in the size of a basic truth table: we will require 2 n rows to handle n variables. However, as we saw in the vending machine example, we can take advantage of the sparseness of the actual data to generate much more compact representations of truth tables. By recognizing the fact that we do not need to repeat rows that represent identical or zero results, we can use significantly fewer rows in practice than the naïve exponential calculation would suggest. Adding Sequential Logic Truth tables can also capture systems with sequential logic, where the model can store state information that changes over time, by adding input stimulus and output result columns to represent the current and next states of a variable. Though two variables are used, it is convenient to give them the same name with an added suffix tick mark (‘) to the successor state variable. 1 ..." What are related quotes?
Reference 4 Book Chapter The Propositional CalculusNils J. Nilsson Artificial Intelligence: A New Synthesis, 1998 pp 217-230 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... 13.5.2 The Propositional Truth Table Given the values of atoms under some interpretation, we can use a truth table to compute a value for any wff under that same interpretation. The truth table establishes the semantics (meanings) of the propositional connectives. Let ω 1 and ω 2 be wffs; then the truth table rules are (ω 1 ∧ ω 2 ) has value True if both ω 1 and ω 2 have value True ; otherwise, it has value False. (ω 1 ∨ ω 2 ) has value True if one or both of ω 1 or ω 2 has value True ; otherwise, it has value False. ¬ω 1 has value True if ω 1 has value False ; otherwise, it has value False. The semantics of ⊃ is defined in terms of ∨ and ¬. Specifically, (ω 1 ⊃ ω 2 ) is an alternative and equivalent form of (¬ω 1 ∨ ω 2 ). These are called truth table rules because they are usually presented in tabular form, as they are in Table 13.1 . We can use the truth table to compute the value of any wff given the values of the constituent atoms in the wff. As an example of the use of the truth table to compute the value of a wff, suppose ρ has value False, Q has value False, and R has value True. In this interpretation, what is the truth value of [(P ⊃ Q) ⊃ R] ⊃ P? Working from the “inside out,” we first compute the value of P ⊃ Q to be True ; then we see that (P ⊃ Q) ⊃ R is True. Finally, since ρ is False, the value of the entire expression must be False. If an agent describes its world using n features (corresponding to propositions), and these features are represented in the agent's model of the world by a corresponding set of n atoms, then there are 2 n different ways its world can be—so far as the agent can discern—because there are 2 n ways in which the n atoms can have values of True and False. Each of the ways the world can be corresponds to an interpretation. 1 ..." What are related quotes?
Reference 5 Book Chapter The Relational ParadigmTom Johnston Bitemporal Data, 2014 pp 43-58 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Connectives The usual set of statement connectives used in propositional logic are formal equivalents of what is expressed in ordinary language by “and”, “or”, “not”, “if…then” and “if and only if”. The usual symbols for these connectives are, respectively, “∧”, “∨”, “~”, “→” and “↔”. Equivalent symbols for these connectives are, respectively, “AND”, “OR”, “NOT”, “IF/THEN” and “IFF”. In standard SQL, only the first three connectives are used. The other two can easily be defined in terms of the first three. In fact, based on either AND and NOT, or on OR and NOT, all three other connectives can be defined. 1,2 1 “AND” designates the formalization of the “and” of ordinary English, and so too for “OR” and “or”, and “NOT” and “not”. Thus, these three strings of capital letters are not English words; they are equivalent strings for, respectively, “∧”, “∨” and “~”. 2 There is even a single operator in terms of which all the others may be defined. It is called the Sheffer stroke. But it is so terse and non-intuitive a connective that it is almost never used in solving problems in logic. AND The logical AND operator applies between two statements, and forms a compound statement called a conjunction . It says that if either or both of a pair of component statements is false, then their conjunction is false, and otherwise is true. To put it another way, the conjunction (X∧Y) is true is when both its components are true, and otherwise is false. A truth table contains one row for every assignment of true (T) and false (F) to the component statements of a compound statement, and then shows the truth-value assignment to the compound statement, given those component assignments. The truth table for conjunction – the formalization of “and” – is this: Figure 4.1 The Truth Table for AND. OR The logical OR operator applies between two statements, and forms a compound statement called a disjunction . 1 ..." What are related quotes?
Reference 6 Book Chapter Fault diagnosis and fault toleranceAfaq Ahmad, Sayyid Samir Al Busaidi System Assurances, 2022 pp 307-321 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... Functional modeling at the logic level The simplest way to represent a digital system of combinational nature is by its Truth table . An example of a typical Truth table of function F ( x 1, x 2, x 3 ) is shown in Table 17.1 . Assuming binary input values, a digital system realizing a function F ( x 1, x 2, x 3 ,…, x n ) of n variables require a table with 2 n entries. The data structure representing a Truth table is, therefore, an array V of dimension 2 n . The table is arranged with all possible input combinations of n -bit binary representations of decimal values varying from 0 to 2 n − 1. From the perspective of testing of digital circuit, a typical example of Primitive cubes model representation of Table 17.1 is given in Table 17.2 . In Table 17.2 , d represents do not care, i.e., the input may be 0 or 1. Sequential circuit in a digital system is a finite state sequential function. The function can be modeled as a sequential machine which is represented as State tables and Flow tables (State diagram) [7,8] . In digital system modeling, also, exists Binary Decision Diagrams technique [9,10] which is basically a graph model representation of the function of a digital circuit. Depicted in Fig. 17.2 is the binary decision diagrams Boolean function F = ( x ′ y ′) p . q . 1 ..." What are related quotes? Related quote(s)2 / 2 "... Digital systems modeling Digital systems design requires rigorous modeling and simulation analysis that eliminates design risks so as to avoid potential harm to consumers. Since digital systems designer is responsible for the integrity of the system design as a whole, it must encompass the interconnectedness of the various aspects of the system. Hence, modeling of a digital system plays an important role in the system’s design, synthesis fabrication, analysis, and testing. Mainly the modeling belong to three of the categories namely. (1) Behavioral, (2) Functional, and (3) Structural. A behavioral model is basically defined by a functional model but it also associates with timing relationships. Modeling of a digital system plays an important role in system design, synthesis fabrication, analysis, and testing. Mainly the functional level modeling can further be classified as follows: 17.2.1 Functional modeling at the logic level The simplest way to represent a digital system of combinational nature is by its Truth table . An example of a typical Truth table of function F ( x 1, x 2, x 3 ) is shown in Table 17.1 . Assuming binary input values, a digital system realizing a function F ( x 1, x 2, x 3 ,…, x n ) of n variables require a table with 2 n entries. The data structure representing a Truth table is, therefore, an array V of dimension 2 n . The table is arranged with all possible input combinations of n -bit binary representations of decimal values varying from 0 to 2 n − 1. From the perspective of testing of digital circuit, a typical example of Primitive cubes model representation of Table 17.1 is given in Table 17.2 . In Table 17.2 , d represents do not care, i.e., the input may be 0 or 1. Sequential circuit in a digital system is a finite state sequential function. The function can be modeled as a sequential machine which is represented as State tables and Flow tables (State diagram) [7,8] . In digital system modeling, also, exists Binary Decision Diagrams technique [9,10] which is basically a graph model representation of the function of a digital circuit. 1 ..." What are related quotes?
Reference 7 Book Chapter Gating and Logic CircuitsIan Sinclair Electronics Simplified, 2011 pp 185-201 View PDFView chapterShow related quote(s) Related quote(s)1 / 6 "... Publisher Summary Logic circuits exist to carry out a set of logic actions such as are used for controls for washing machines, tape-recorder drives, computer disk drives, security systems, and a host of industrial control actions. Simple arithmetic actions can also be carried out using logic circuits. All logic actions, however complicated, can be analyzed into simple actions that are called AND, OR, and NOT, so that circuits, called gates, which carry out such actions, are the basis of logical circuits. Any gate circuit can be analyzed by drawing up a truth table, or a set of truth tables. A truth table is a simple way of showing what the output of a gate or gate circuit will be for each and every possible combination of inputs. The alternative is to use Boolean algebra, a process that is greatly simplified by computer programs that carry out an analysis of gate circuits. The Boolean algebra alternative method relies on using lettering to identify all inputs, outputs, and intermediate points, and drawing up the truth table in stages, starting with all possible combinations at the inputs. An arithmetic circuit is a set of gates with a separate set of inputs for each number that has to be processed. The gates are connected to carry out an arithmetic action and the outputs of the gate circuit are the digits of the result (addition, subtraction, multiplication, or division). Encoders, decoders, multiplexers, and demultiplexers are used to change the way that digital codes are carried. 1 ..." What are related quotes? Related quote(s)2 / 6 "... The more useful method for professional use is called Boolean algebra but, like other mathematical methods, it is beyond the scope of this book. There are computer programs that will analyze the action of a gate circuit using Boolean algebra. Figure 10.2 shows truth tables for the basic two-input AND, OR, and NOT gates. Of these, the NOT gate is a simple one, with just one input and one output. Its action is that of a logic inverter. If the input is 0, then the output is not0, which is 1. If the input is 1, then the output is not1, which is 0. The other two gate types permit more than one input, and the examples show two inputs, the most common number. The action of the AND gate is to give a 1 output only when both inputs are at 1, and a 0 output for any other combination. The action of the OR gate is to give a 0 output when each input is 0, but a 1 for any other combination of inputs. The same arguments apply to gates with more than two inputs. Summary A truth table is a simple but clumsy way of showing what the output of a gate or gate circuit will be for each and every possible combination of inputs. The alternative is to use Boolean algebra, a process that is greatly simplified by computer programs that carry out an analysis of gate circuits. NAND and NOR Gates Two particularly useful gate types can be made by combining the action of an inverter with that of the AND and OR gates. The combination of NOT and AND gives the NAND gate, whose symbols and truth table (for two inputs) are shown in Figure 10.3 . The action of this gate is that the output is 0 only when all of its inputs are at 1, which is the action of the AND gate followed by an inverter. The combined action of the OR gate and a following inverter gives the NOR gate, whose symbols and truth table are shown in Figure 10.4 . The output of this gate will be at logic 0 when any one (or more) of its inputs is at logic 1. There is one further gate that is often used and which is called exclusive-OR (XOR). 1 ..." What are related quotes? Related quote(s)3 / 6 "... NAND and NOR Gates Two particularly useful gate types can be made by combining the action of an inverter with that of the AND and OR gates. The combination of NOT and AND gives the NAND gate, whose symbols and truth table (for two inputs) are shown in Figure 10.3 . The action of this gate is that the output is 0 only when all of its inputs are at 1, which is the action of the AND gate followed by an inverter. The combined action of the OR gate and a following inverter gives the NOR gate, whose symbols and truth table are shown in Figure 10.4 . The output of this gate will be at logic 0 when any one (or more) of its inputs is at logic 1. There is one further gate that is often used and which is called exclusive-OR (XOR). This action ( Figure 10.5 ) is closer to what we normally mean by the word ‘or’ (meaning one or the other but not both), and the output is 1 if either input is 1, but not when both inputs are zero or both are 1. The diagram also shows that the XOR gate is equivalent to the action of a circuit made using an OR, AND, and NAND gate combination. Note If you would like to read further about gates and other logic circuits, with details of the more advanced methods such as Boolean algebra, take a look at the book Digital Logic Gates and Flip-Flops , from PC Publishing. Summary A truth table is a simple way of expressing the action of a logic circuit, and the standard gates called OR, AND, and NOT can all be illustrated in this way. Gates in IC form often consist of the NAND and NOR type of gates, equivalent to an AND or OR, respectively, followed by NOT. These inverting gates are easier to produce and the action is often more useful than that of the simpler AND or OR type. The XOR gate is another useful type which gives an action closer to the normal meaning of OR, as ‘one or the other but not both’. 1 ..." What are related quotes? Related quote(s)4 / 6 "... A gate circuit diagram will consist of a large number of gate symbols with joining lines so that the output of a gate will be connected to one or more other gate inputs. Provided that we know what each basic type of gate does, we can analyze the action of complete gate circuits. In this book we are concerned more with block diagrams than with gate circuits, but some knowledge of gate circuits is useful, and in any case, these are closer in spirit to block diagrams than to circuit diagrams. Summary Logic circuits exist to carry out a set of logic actions such as are used for controls for washing machines, tape-recorder drives, computer disk drives, security systems, and a host of industrial control actions. Simple arithmetic actions can also be carried out using logic circuits. All logic actions, however complicated, can be analyzed into simple actions that are called AND, OR, and NOT, so that circuits, called gates, which carry out such actions, are the basis of logical circuits. Note Remember that ICs are classed as medium-scale integration (MSI), large-scale integration (LSI), very large-scale integration (VLSI), etc., by the number of equivalent gate circuits on a single chip. Definition The action of any gate can be expressed in a truth table . This is just a table that shows all the possible inputs to the gate, and the output for each set of inputs. Remember that each input can be 0 or 1 only, so that each input contributes two possible outputs. The total number of outputs is equal to 2 n , where n is the number of inputs. For example, if there are four inputs to a gate, then the number of possibilities is 2 4 = 16, and its truth table will consist of 16 lines. For a lot of truth tables, there is only one output that is different from the rest, and it is easier to remember which one this is than to try to remember the whole of a truth table. Note Truth tables are the simplest way of showing what a small-scale gate or gate circuit does, but they are impossibly clumsy when we try to use them on complex gates with a large number of inputs. 1 ..." What are related quotes? Related quote(s)5 / 6 "... This action ( Figure 10.5 ) is closer to what we normally mean by the word ‘or’ (meaning one or the other but not both), and the output is 1 if either input is 1, but not when both inputs are zero or both are 1. The diagram also shows that the XOR gate is equivalent to the action of a circuit made using an OR, AND, and NAND gate combination. Note If you would like to read further about gates and other logic circuits, with details of the more advanced methods such as Boolean algebra, take a look at the book Digital Logic Gates and Flip-Flops , from PC Publishing. Summary A truth table is a simple way of expressing the action of a logic circuit, and the standard gates called OR, AND, and NOT can all be illustrated in this way. Gates in IC form often consist of the NAND and NOR type of gates, equivalent to an AND or OR, respectively, followed by NOT. These inverting gates are easier to produce and the action is often more useful than that of the simpler AND or OR type. The XOR gate is another useful type which gives an action closer to the normal meaning of OR, as ‘one or the other but not both’. Analyzing Gate Systems A circuit that has been made up by connecting several standard gates together, which has several inputs and an output, can be analyzed to find what its action is. This analysis can be done by drawing up truth tables, or by a method called Boolean algebra. The truth-table method is simpler, but more tedious than the Boolean algebra method, which is not dealt with in this book. The method of analysis by truth table can be summarized in a few rules. 1. Letter each input to the circuit (A, B, C) and also each point where the output of one gate is connected to the input of another gate, using different letters for each point. Label the final output as Q. 2. Draw up a blank truth table, using one column for each letter that has been allocated, and with 2 n rows, where n is the number of signal inputs to the circuit. 3. Write in every possible combination of inputs. 1 ..." What are related quotes? Related quote(s)6 / 6 "... Analyzing Gate Systems A circuit that has been made up by connecting several standard gates together, which has several inputs and an output, can be analyzed to find what its action is. This analysis can be done by drawing up truth tables, or by a method called Boolean algebra. The truth-table method is simpler, but more tedious than the Boolean algebra method, which is not dealt with in this book. The method of analysis by truth table can be summarized in a few rules. 1. Letter each input to the circuit (A, B, C) and also each point where the output of one gate is connected to the input of another gate, using different letters for each point. Label the final output as Q. 2. Draw up a blank truth table, using one column for each letter that has been allocated, and with 2 n rows, where n is the number of signal inputs to the circuit. 3. Write in every possible combination of inputs. This is most easily done by starting with 0000 and continuing in the form of a binary count up to an input which consists entirely of 1s. 4. Knowing the truth tables for the standard gates, write in the logic states (0 or 1) for the outputs of the gates at the inputs in each line of the truth table. 5. The first set of outputs will now be the inputs for the next set of gates, so that their outputs can be written into the truth table. 6. Continue in this way until the truth table has been completed. As an example, Figure 10.6 shows a logic diagram for an electronic combination lock. This is a simple design, with four main inputs, and therefore 16 combinations, ignoring the unlock input E. The lock is arranged so that only the correct combination of inputs will open the lock, and any other combination will cause an alarm to sound, so that it cannot be solved by trying each possible combination. The inputs, A, B, C, and D are from switches which are to be set in the pattern needed to open the lock. When these switches have been set, pressing the button E will cause the door to unlock (Q = 1) if the combination is correct, or cause the alarm to sound (X = 1) if the combination is incorrect. 1 ..." What are related quotes?
Reference 8 Book Chapter From Zero to OneSarah L. Harris, David Harris Digital Design and Computer Architecture, 2022 pp 1-50 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... Multiple-Input Gates Many Boolean functions of three or more inputs exist. The most common are AND, OR, XOR, NAND, NOR, and XNOR. An N -input AND gate produces a TRUE output when all N inputs are TRUE. An N -input OR gate produces a TRUE output when at least one input is TRUE. Example 1.16 Three-Input NOR Gate Figure 1.19 shows the symbol and Boolean equation for a three-input NOR gate. Complete the truth table. Solution Figure 1.20 shows the truth table. The output is TRUE only if none of the inputs are TRUE. Example 1.17 Four-Input And Gate Figure 1.21 shows the symbol and Boolean equation for a four-input AND gate. Create a truth table. Solution Figure 1.22 shows the truth table. The output is TRUE only if all of the inputs are TRUE. 1 ..." What are related quotes? Related quote(s)2 / 2 "... The XOR operation is indicated by ⊕, a plus sign with a circle around it. Any gate can be followed by a bubble to invert its operation. The NAND gate performs NOT AND. Its output is TRUE unless both inputs are TRUE. The NOR gate performs NOT OR. Its output is TRUE if neither A nor B is TRUE. An N -input XOR gate is sometimes called a parity gate and produces a TRUE output if an odd number of inputs are TRUE. As with two-input gates, the input combinations in the truth table are listed in counting order. A silly way to remember the OR symbol is that its input side is curved like Pacman’s mouth, so the gate is hungry and willing to eat any TRUE inputs it can find! Example 1.15 XNOR Gate Figure 1.17 shows the symbol and Boolean equation for a two-input XNOR (pronounced ex-NOR) gate that performs the inverse of an XOR. Complete the truth table. Solution Figure 1.18 shows the truth table. The XNOR output is TRUE if both inputs are FALSE or both inputs are TRUE. The two-input XNOR gate is sometimes called an equality gate because its output is TRUE when the inputs are equal. 1.5.6 Multiple-Input Gates Many Boolean functions of three or more inputs exist. The most common are AND, OR, XOR, NAND, NOR, and XNOR. An N -input AND gate produces a TRUE output when all N inputs are TRUE. An N -input OR gate produces a TRUE output when at least one input is TRUE. Example 1.16 Three-Input NOR Gate Figure 1.19 shows the symbol and Boolean equation for a three-input NOR gate. Complete the truth table. Solution Figure 1.20 shows the truth table. The output is TRUE only if none of the inputs are TRUE. Example 1.17 Four-Input And Gate Figure 1.21 shows the symbol and Boolean equation for a four-input AND gate. Create a truth table. Solution Figure 1.22 shows the truth table. The output is TRUE only if all of the inputs are TRUE. 1 ..." What are related quotes?
Reference 9 Book Chapter Digital Control SystemsParr E.A. Electrical Engineer's Reference Book, 2003 pp 14-1 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Combinational logic 14.3.1 Introduction Combinational logic is based around the block diagram of Figure 14.21 (a) . Such systems have several inputs and one, or more, outputs. The output states are uniquely defined for each and every combination of inputs and the ‘block’ does not contain any device such as storage, timers or counters. We therefore have n inputs I 1 to I n and Z outputs Q 1 to Q z . In systems with multiple outputs it is usually easier to consider each separately as Figure 14.21 (b) , allowing us to consider the circuit as Z blocks, each different but represented by Figure 14.21 (c) . The number of possible input states depends on the number of inputs: For two inputs there are four input combinations For three inputs there are eight input combinations For four inputs there are sixteen input combinations and so on. Not all of these may be needed. There are frequently only a certain number of input combinations that may occur because of physical restrictions elsewhere in the system. The design of combinational logic systems first involves examining all the input states that can occur and defining the output states that must occur for each and every input state. A logic design to achieve this is then constructed from the gates described in Section 14.1.3 . In many systems the design can be done in an intuitive manner, but the rest of this section describes more formal design procedures. Few real life systems need pure combinational logic, most need storage and similar dynamic functions. Such systems can be analysed and designed considering them as smaller subsystems linked together. The design of dynamic systems is discussed in Section 14.8 . 14.3.2 Truth tables A truth table is a useful way of representing a combinational logic circuit, and can be used to design the circuit needed to achieve a desired function. Suppose we have three contacts monitoring some event (overpressure in a chemical reactor for example) and we wish to construct a majority vote circuit. 1 ..." What are related quotes?
Reference 10 Book Chapter IntroductionStroud C.E., Wang L.T., Chang Y.-W. Electronic Design Automation, 2009 pp 1-38 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... The actual digital logic continued to be hand-designed by use of state diagrams, state tables, Karnaugh maps, and a few simple CAD tools for logic minimization. This limited the complexity of ASICs in terms of the number of gates or transistors that could be correctly designed and verified by a typical designer. In the early 1980s, an ASIC with more than 100,000 transistors was considered to be near the upper limit for a single designer. By the late 1980s, the limit was significantly increased as a result of multi-designer teams working on a single IC and as a result of advances in EDA capabilities, particularly in the area of logic synthesis. Currently, the largest ICs exceed 1 billion transistors [ Naffziger 2006 ; Stackhouse 2008 ]. One of the earliest approaches to EDA in terms of combinational logic synthesis was for implementing programmable logic arrays (PLAs) in very large-scale integrated (VLSI) circuits in the late 1970s [ Mead 1980 ]. Any combinational logic function can be expressed as Boolean logic equations, sum-of-products (SOP) or product-of-sums (POS) expressions, and truth tables or Karnaugh maps. There are other representations, but these three are illustrated for the example circuit in Figure 1.6 and are important for understanding the implementation of PLAs and other programmable logic. We can program the truth table onto a read-only memory (ROM) with eight words and two bits/word and then use the ROM address lines as the three input signals ( A, B, C ) and the ROM outputs as the output signals ( X, Y ). Similarly, we can also write the truth table into a random-access memory (RAM) with eight words and two bits/word and then disable the write enable to the RAM and use the address lines as the inputs. Note that this is the same thing as the ROM, except that we can reprogram the logic function by rewriting the RAM; this also forms the basis for combinational logic implementations in FPGAs. Another option for implementing a truth table is the PLA. 1 ..." What are related quotes? Related quote(s)2 / 2 "... Currently, the largest ICs exceed 1 billion transistors [ Naffziger 2006 ; Stackhouse 2008 ]. One of the earliest approaches to EDA in terms of combinational logic synthesis was for implementing programmable logic arrays (PLAs) in very large-scale integrated (VLSI) circuits in the late 1970s [ Mead 1980 ]. Any combinational logic function can be expressed as Boolean logic equations, sum-of-products (SOP) or product-of-sums (POS) expressions, and truth tables or Karnaugh maps. There are other representations, but these three are illustrated for the example circuit in Figure 1.6 and are important for understanding the implementation of PLAs and other programmable logic. We can program the truth table onto a read-only memory (ROM) with eight words and two bits/word and then use the ROM address lines as the three input signals ( A, B, C ) and the ROM outputs as the output signals ( X, Y ). Similarly, we can also write the truth table into a random-access memory (RAM) with eight words and two bits/word and then disable the write enable to the RAM and use the address lines as the inputs. Note that this is the same thing as the ROM, except that we can reprogram the logic function by rewriting the RAM; this also forms the basis for combinational logic implementations in FPGAs. Another option for implementing a truth table is the PLA. In the connection array in Figure 1.6e , only three product terms produce logic 1 at the output signals. The PLA allows implementing only those three product terms and not the other five, which is much smaller than either the ROM or RAM implementation. Any SOP can be implemented as a 2-level AND-OR or NAND-NAND logic function. Any SOP can also be implemented as a 2-level NOR-NOR logic function if we invert the inputs and the output as illustrated in Figure 1.7a . Note that AB' is a shared product term and allows us to share an AND gate in the gate-level implementation. 1 ..." What are related quotes?
Reference 11 Book Chapter Computer Hardware for Industrial ControlPeng Zhang Industrial Control Technology, 2008 pp 187-257 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Logic cells are also known as logic array blocks, logic elements, and configurable logic blocks. Lookup tables or truth tables are used to implement a single logic function by storing the correct output logic state in a memory location that corresponds to each particular combination of input variables. Field-programmable gate arrays are available with many logic families. Transistor-transistor logic and related technologies such as Fairchild advanced Schottky use transistors as digital switches. By contrast, emittercoupled logic uses transistors to steer current through gates that compute logical functions. Another logic family, CMOS, uses a combination of P-type and N-type metal-oxide-semiconductor field effect transistors to implement logic gates and other digital circuits. Logic families for field-programmable gate arrays include crossbar switch technology, gallium arsenide, integrated injection logic, and silicon on sapphire. Gunning with transceiver logic and gunning with transceiver logic plus are also available. Field-programmable gate arrays are available in a variety of IC package types and with different numbers of pins and flip-flops. Basic IC package types for field-programmable gate arrays include ball grid array, quad flat package, single in-line package, and dual in-line package. Many packaging variants are available. 2.3.3.2 FPGA Architecture The typical basic architecture consists of an array of logic blocks and routing channels. Multiple I/O pads may fit into the height of one row or the width of one column. Generally, all the routing channels have the same width (number of wires). Figure 2.18 illustrates the FPGA structure. (1) FPGA logic block. An application circuit must be mapped into an FPGA with adequate resources. The typical FPGA logic block consists of a 4-input lookup table (LUT) and a flip-flop, as shown in Fig. 2.19 . There is only one output in each logic block, which can be either the registered or the unregistered LUT output. As shown in Figure 2.20 , the logic block has four inputs for the LUT and a clock input. 1 ..." What are related quotes?
Reference 12 Book Chapter Basic formal verification algorithmsErik Seligman, Tom Schubert, M V Achutha Kiran Kumar Formal Verification, 2023 pp 25-49 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... A simple vending machine example Before we get into too much detail on the general techniques, we begin with a small example showing how a clever data structure can be used to transform an apparently exponential complexity problem into something much more tractable. Boolean logic is usually represented through truth tables, which describe expected output values of a system in terms of each of their inputs. We can use truth tables to define functions or systems with multiple stimulus arguments and multiple results. For example, the truth table below describes a 1-bit ADD operation ( Table 2.1 ). The size of truth tables grows quickly as stimuli are added: with each additional input column, we double the number of rows for the number of new possible stimulus values. Defining a MUX operation, for example, requires twice as many rows as the 2-bit ADD ( Table 2.2 ). This means that a truth table for n inputs will have 2 n rows; as we discussed in the last chapter, this exponential growth quickly makes it very difficult to analyze with simple tools and techniques. However, many FV algorithms take advantage of a key insight that mitigates this problem: in real life, such a truth table can often be quite sparse in terms of information content, since many rows produce 0 values, and many sets of rows produce the same value. As an example, consider developing a specification for a simple vending machine truth table that determines whether 35 cents have been deposited (Yes, one of the authors may be revealing their age by hypothesizing such a vending machine, but bear with us for now.). We'll limit our specification to require that the machine only accepts exact change consisting of nickels (5 cents), dimes (10 cents), or quarters (25 cents). Legitimate inputs to the vending machine that should be accepted are: one quarter, one dime; one quarter, two nickels; three dimes, one nickel; two dimes, three nickels; one dime, five nickels; seven nickels. For these six cases, our truth table result should be 1, and otherwise it should be 1 ..." What are related quotes?
Reference 13 Book Chapter Functional VerificationWen H.P., Wang L.-C., Cheng K.T. Electronic Design Automation, 2009 pp 513-573 View PDFView chapterShow related quote(s) Related quote(s)1 / 2 "... Checking based on functional equivalence The first step of functional CEC is to translate the combinational circuits into a canonical representation . A representation of a Boolean function is canonical if the representation for each function is unique and independent of the implementation of the function. A truth table is one example of a canonical representation for Boolean functions. Equivalence can be determined by directly comparing the two canonical representations. Among all canonical representations, the reduced ordered binary decision diagram (OBDD), introduced in Chapter 4 , is the most prevalent, because OBDD yields a more compact representation than other representations. The CEC problem can be resolved by building the OBDDs for the outputs of the circuits on the basis of their primary inputs. Two circuits are equivalent if the OBDDs from each pair of corresponding outputs are graphically isomorphic. 1 ..." What are related quotes? Related quote(s)2 / 2 "... A representation of a Boolean function is canonical if the representation for each function is unique and independent of the implementation of the function. A truth table is one example of a canonical representation for Boolean functions. Equivalence can be determined by directly comparing the two canonical representations. Among all canonical representations, the reduced ordered binary decision diagram (OBDD), introduced in Chapter 4 , is the most prevalent, because OBDD yields a more compact representation than other representations. The CEC problem can be resolved by building the OBDDs for the outputs of the circuits on the basis of their primary inputs. Two circuits are equivalent if the OBDDs from each pair of corresponding outputs are graphically isomorphic. 9.5.1.2 Checking based on structural search A structural search approach checks to see whether any vector exists at primary inputs that would cause a mismatch between the two circuits at their primary outputs. If no such input vector can be found, the two circuits are proven equivalent. The satisfiability (SAT) solvers, introduced in Chapter 4 , can be used as the structural search engine for checking equivalence. A SAT solver can be used to check if an assignment at PIs exists to satisfy a 1 at the miter's output. An UNSAT answer from the solver proves the equivalence of the two circuits. An ATPG tool developed for generating manufacturing tests for stuck-at faults can also be used for checking structural equivalence. As illustrated in Figure 9.24 , if the stuck-at-0 fault at the XOR output is proven a redundant fault by an ATPG tool, the two circuits are equivalent. A thorough treatment of ATPG techniques will be provided in Chapter 14 . For complex circuits, directly applying SAT solving at the miter's output signal may result in an exponential number of backtracks, which makes the approach inefficient. Structural similarity between the two circuits under checking can be explored to improve its efficiency, which attempts to solve the structural equivalence problem by incrementally solving a sequence of easier sub-problems [ Brand 1993 ; Kunz 1993 ; Goldberg 2000 ; Huang 2000 ]. 1 ..." What are related quotes?
Reference 14 Handbook Chapter Machine learning algorithms, applications, and practices in data scienceYeturu K. Handbook of Statistics, 2020 pp 81-206 View PDFView chapterShow related quote(s) Related quote(s)1 / 1 "... Automated theorem proof Given a knowledgebase of truth statements and a query truth statement, the task is to prove or disprove. There are two ways of accomplishing this—(i) Forward proof method and (ii) Backward proof method. 8.4.3.1 Forward chaining A set of rules are provided to the automated theorem proof algorithm. The algorithm systematically expands the knowledgebase by combinations of the pieces of knowledge to produce derived knowledge. Some of the rules of inference is shown in Table 13 . Here, → denotes implication, ∼ denotes negation, ∨ or denotes logical OR, or ∧ and comma, denotes logical AND, [] denote grouping. The rules are verifiable via truth table . There can be more rules based on combinations in truth table. 8.4.3.2 Incompleteness of the forward chaining The rules of inference may not be sufficient for proving a given query in the forward method. New rules need to be amended for each special query that cannot be addressed. For instance, the following rule is not possible to deduce from a given of rules ( Table 13 ). [ B ∧ D , A ∨ C ] → ( A ∧ B ) ∨ ( C ∧ D ) In order to consider this, one can augment this as new rule. However, it poses a sense of uncertainty for being applicable to all the queries. 8.4.3.3 Backward chaining In backward chaining, the query ( or goal ) is considered along side the knowledgebase and deductions are evaluated. One approach is, to augment the query to the knowledgebase and to prove the whole group to be true for some assignment of values to the variables. However, this reduces to evaluating for all possible values and identifying a particular combination of variable assignments leading to truth. The scenario quickly becomes satisfiability problem which has exponential time complexity . Another way to prove is a query is deducible is by a method of contraction which is presented next. 1 ..." What are related quotes?
Related topics (10)
Boolean Algebra
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Topic summary
Outline
1. Introduction to Truth Tables in Computer Science
2. Logical Foundations and Boolean Algebra
3. Truth Tables in Digital Logic Design
4. Truth Tables in Formal Methods and Verification
5. Conclusion
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What is Titration?
A titration is a chemical analysis in which a researcher determines the concentration of a chemical solution (analyte or titrand) by adding a measured volume of standard solution (titrant) until the produced reaction reaches the equivalence point. Titrations are generally used to find the unknown concentration of an acid or base in solution.
Traditional Titration Procedure
To perform a titration, clamp an empty burette above a beaker or an Erlenmeyer flask. Carefully fill the burette with a titrant of known concentration, and fill the flask with a known volume of the analyte of unknown concentration. Add a magnetic stir bar to the analyte solution to assist with mixing. Then, adjust the stopcock on the burette, and slowly drip the titrant into the analyte.
Add a pH Sensor, chemical indicator, or both to the analyte solution to assist with tracking the titration’s progress. During the titration, the pH changes as the titrant is added. Plotting the analyte pH against the titrant volume produces a graph known as a titration curve.
The shape of the titration curve is analyzed to determine the equivalence point of the reaction. The total titration volume is determined by reading the change in titrant volume from the burette at the equivalence point.
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Modern Titration Procedure
If you’ve ever used the methods described above, then you likely know how tricky acid-base titrations can be. One moment your analyte looks clear, and the next it’s bright fuchsia. For this reason, many modern acid-base titration setups utilize both a pH Sensor and a drop counter so that the titration curve can be plotted and monitored in real time. With this method, the experimenter uses a live plot to predict the equivalence point before it occurs and adjusts the drop rate as it nears for greater precision.
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Simply align the drop window with the burette, connect the sensor to your device, and begin monitoring the titration in real time. A built-in IR filter protects the drop window from light interference for reliable accuracy, regardless of drop size or pacing.
When used with a Wireless pH Sensor, the Drop Counter generates a live titration curve as the reaction progresses, making it easier to predict both the equivalence point and the endpoint. Titrant volumes can be reported in drops or mL.
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Chemical Indicators
A chemical indicator is a substance that visually indicates the presence or absence of a chemical compound, usually by changing color. Most chemical indicators are weak organic acids or bases that exhibit different colors in their protonated and deprotonated states.
Prior to starting a titration, add a few drops of phenolphthalein into the analyte solution. If the solution is acidic or neutral, the phenolphthalein will appear colorless.
Chemical indicators are most often used to visually track the pH of the analyte during a titration. A change in color signals the endpoint of a reaction, or the moment when the reaction is complete.
Common titration indicators include:
Phenolphthalein
Bromothymol Blue
Methyl Red
Universal Indicator
Methyl Orange
Phenolphthalein
Phenolphthalein is a widely used indicator. It is naturally colorless but turns pink when present in a basic solution with a pH of about 8.3 to 10. Above a pH of 9, the fuschia color darkens remarkably; above a pH of 10, the solution becomes colorless once more.
Phenolphthalein is commonly used in acid-base reactions where the analyte neutralizes the mixture, leading to neutrality—the moment just after all the acid and base have reacted. After neutralization is reached, even one or two additional drops of titrant will raise the pH to between 8 and 10, causing the analyte solution to turn pink. Since the indicator changes color almost immediately after neutralization, it serves as a great benchmark for identifying the equivalence point.
Other visual indicators include bromothymol blue, which turns blue between a pH of 6 and 8, or methyl red, which turns reddish-orange between pH 4 and pH 6.
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What Causes an Indicator to Change Color During a Titration?
Indicator color change is caused by changes in chemical structure that occur as the indicator binds or releases a proton in solution. You can learn more about the relationship between chemical structure, light, and color in our guide, "What is Spectroscopy?"
Indicators are generally used in low concentrations so they do not significantly alter a reaction’s equivalence point. Typically, the indicator should be a lower concentration than the acid or base being titrated, so it doesn’t interfere with the chemical reaction.
The indicator used depends on the type of titration being performed. For example, there are four types of titrations that involve acids and bases—and they all have different equivalence points. We will discuss equivalence points and titration curves in more detail in the next section.
How to Find the Equivalence Point
The equivalence point of a reaction occurs when the number of moles of the titrant is chemically equivalent to the number of moles of the analyte.
The equivalence point happens immediately before the chemical indicator changes color, as the color change signifies the endpoint of the reaction. A single titration, such as a polyprotic acid-base titration, can have multiple equivalence points but only one endpoint, depending on the indicator involved.
Steps to find the equivalence point:
Monitor the pH of the analyte with a pH probe.
Monitor the reaction visually with an indicator.
Analyze the titration curve.
We can calculate the equivalence point for an acid-base titration using just a few values. Recall that M=m o l V and that at equivalence there are equal moles of both hydrogen and hydroxide ions.
M o l e H+=M o l e O H−w h e r e M o l e H+=H M o l e H n A f o r d i p r o t i c(2 x)o r t r i p r o t i c(3 x)a c i d s
M H+V a c i d=M O H−V b a s e
We can use stoichiometry to determine the amount of hydroxide ions (OH-) required to neutralize the hydrogen ions (H+). In the neutralization equations above, the n subscript comes from the H+ and OH– subscripts included in the balanced chemical equation. By measuring the exact volume of the reagent required to reach the equivalence point, it is possible to calculate the molarity of the unknown solution. So the above equations become:
M H+=M O H−×V b a s e V a c i d
Titration Curves
A titration curve graphically represents the changes in pH and volume that occur during a titration, making it easier to identify the equivalence point of a reaction. It is a plot of the pH of the analyte versus the volume (mL) of titrant added during the reaction.
The shape of a titration curve is used to track the reaction’s progress. As the equivalence point nears, the curve slopes upward, and the pH changes more drastically. After the equivalence point, the endpoint is reached, and the curve begins to flatline. For an acid-base titration, the shape of the titration curve depends on the type of acid and base used, their concentrations, and the volume of titrant added. Keep reading to learn how to interpret common types of titration curves.
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Acid-Base Titrations
The objective of an acid-base titration is to solve for the unknown concentration of an acid or base by performing an acid-base reaction. Acid-base titrations have a wide range of applications in a variety of industries, ranging from pharmaceuticals to food quality assurance. There are four types of acid-base titrations: Strong acid-strong base, weak acid-strong base, strong acid-weak base, and weak acid-weak base. In the next section, we’ll explain the different reactions and how to interpret their corresponding titration curves.
Strong Acid-Strong Base Titrations
When titrating a strong acid with a strong base, the reactants are highly dissociated, and the pH changes rapidly as the equivalence point nears.
The equivalence point marks the moment at which just enough titrant has been added to neutralize the analyte solution. In a strong acid-strong base titration, the equivalence point of the reaction occurs when the analyte solution reaches a pH of 7.
Reactions between strong acids and strong bases are also known as neutralization reactions, as both species fully dissociate to form the neutral products of salt and water, as shown below.
In the curve below, the strong base, NaOH, titrates the strong acid, HCl. As the total amount of added NaOH increases, the pH of the analyte solution also increases.
N a O H(a q)+H C l(a q)→N a C l(a q)+H 2 O(l)
Notice how the curve transitions from strongly acidic (below pH 2) to strongly basic (above pH 11). As the hydroxide (OH–) ions neutralize the hydronium (H+) ions, the curve takes a vertical shape. It then begins to flatten as the number of free hydronium ions decreases and the analyte pH becomes basic.
In any acid-base titration, the equivalence point occurs at the midpoint of the vertical portion of the titration curve.
Strong Base-Strong Acid Titrations
When the solutions have opposite characteristics (basic analyte/acidic titrant), the titration curve is inverted; meaning the curve begins high (basic) and slopes downward toward the low pH range (acidic) as the reaction progresses.
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Strong Acid-Strong Base Example | Titration of HCl with NaOH
To better understand strong acid-strong base titrations, let’s consider an example. Suppose we want to determine the concentration of hydrochloric acid (HCl) using sodium hydroxide (NaOH) as a titrant. We know that sodium hydroxide is a strong base and hydrochloric acid is a strong acid.
First we would obtain a known volume of hydrochloric acid and titrate it with sodium hydroxide until we reach the equivalence point. This point is determined using a visual indicator, a pH probe, or both. For the titration of a strong acid and strong base, the equivalence point happens at a pH of 7, so we could use Bromothymol blue or Phenolphthalein as the indicator.
After the titration, we can use the balanced chemical equation plus the known volume and concentration of the sodium hydroxide to calculate the unknown concentration of the hydrochloric acid.
Calculating the Unknown Molarity
Imagine the titration involves titrating 25 mL of HCl with 0.10 M NaOH. We want to determine the molarity of the HCl in the 25 mL solution.
We titrate the HCl with the NaOH and find that it takes 55 mL of NaOH to reach the equivalence point.
Now we can calculate the moles of NaOH used in the titration by converting the volume to liters and multiplying it by the known molarity.
(55 m L N a O H)(L 1000 m L)(0.10 m o l N a O H L N a O H)=0.0055 m o l e s N a O H
In the balanced equation below, we can see that the reaction between HCl and NaOH has a 1:1 mole ratio. Therefore, we know that there will be equal moles of HCl and NaOH at the equivalence point. We can now find the molarity (M) of HCl by dividing the number of moles by the known volume of HCl.
H C l+N a O H→H 2 O+N a C l
(25 m L H C l)(L 1000 m L)=0.025 L H C l
(0.0055 m o l H C l 0.025 L H C l)=0.22 M H C l
When the mole ratio is 1:1 for an acid-base titration no differentiation is needed to compare acid concentration and H+ concentration. We can calculate the unknown concentration using the equation below, where M is molarity (M) and V is volume:
M A V A=M B V B
(M A)(0.025 L H C l)=(0.10 M N a O H)(0.055 L N a O H)
M A=0.22 M H C l
Weak Acid-Strong Base Titrations
In a titration between a weak acid and a strong base, only the strong base dissociates fully. Once the equivalence point is reached, the excess hydroxide ions accumulate in the analyte solution, and the titration curve shifts toward strongly basic. For these types of reactions, the analyte solution is considered basic at the equivalence point (pH > 7).
The initial pH of the analyte in a weak acid-strong base titration is higher than the initial pH of the analyte in a strong acid-strong base titration, because weak acids only partially dissociate in solution.
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Strong Acid-Weak Base Titrations
In a titration between a strong acid and a weak base, only the strong acid dissociates fully. At the equivalence point, the analyte solution contains a slight excess of hydronium ions, which keep the pH of the analyte solution below 7.
In a strong acid-weak base titration, the starting pH of the analyte is lower than the initial pH of the analyte in a strong acid-strong base titration. This is because weak bases dissociate less than strong acids.
Weak Acid-Weak Base Titration | Titration of Acetic Acid with Ammonium Hydroxide
In a titration between a weak acid and a weak base, both the analyte and the titrant dissociate partially. At the equivalence point, the analyte approaches neutralization (pH = 7). For these reactions, the initial pH of the analyte generally ranges from three to eleven. The titration curve resembles a compressed version of the traditional “S” shape, and the equivalence point is indicated by the smaller but still steep rise in the curve.
For example, consider the following equation between acetic acid (a weak acid) and ammonium hydroxide (a weak base):
C H 3 C O O H(a q)+N H 4 O H(a q)⇌C H 3 C O O N H 4(a q)+H 2 O(l)
In a weak acid-weak base titration, neither the analyte nor the titrant dissolve fully in solution. The resulting curve is a compressed version of the strong acid-strong base titration curve.
Comparing Titration Curves
The plot below shows the titration curves produced from two different titrations: a weak acid-strong base titration and a strong acid-strong base titration.
Reaction 1 (Strong Base-Weak Acid): The blue line is the titration curve produced when the strong base, sodium hydroxide (NaOH), titrates the weak acid, acetic acid (CH 3 COOH).
C H 3 C O O H(a q)+N a O H(a q)⇌N a+C H 3 C O O–(a q)+H 3 O+(a q)
Reaction 2 (Strong Base-Strong Acid): The red line is the titration curve produced when the strong base, ammonium chloride (NH 4 OH), titrates the strong acid, hydrochloric acid (HCl).
H C l(a q)+N H 4 O H(a q)→N H 4 C l(a q)+H 2 O(l)
A strong base-weak acid titration (blue) is plotted against a strong base-strong acid titration (red).
In the plot above, the analyte solution in Reaction 1 begins at a higher pH than the analyte used in Reaction 2. Weak acids generally have pH values of about 7. Reaction 1 uses the weak acid, acetic acid, which only partially dissociates. In contrast, Reaction 2 utilizes the highly dissociative, strong acid, hydrochloric acid, as the analyte. Reaction 2 uses a more dissociative acid, and thus, has a higher concentration of hydronium ions in the analyte solution, giving it a lower pH.
Because the analyte in Reaction 1 doesn’t fully dissociate, the addition of the highly dissociative strong base, sodium hydroxide, quickly neutralizes the acid in solution. At the equivalence point for Reaction 1, the pH of the analyte is greater than seven. This is true for all titrations that involve a weak acid and a strong base.
In Reaction 2, both the analyte solution and the titrant fully dissociate because it is a strong acid-strong base reaction. At the equivalence point, the analyte solution contains chemically equal amounts of hydroxide and hydronium ions, and the pH is near 7.
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Polyprotic Acid-Base Titration Reactions ----------------------------------------
While most acid-base reactions have a single equivalence point, polyprotic acids and bases produce reactions with multiple equivalence points. A polyprotic acid can donate more than one proton in solution, while a polyprotic base can accept more than one proton in solution.
When used in a titration, the protons in a polyprotic acid ionize in succession, resulting in multiple equivalence points. A similar effect is seen when using polyprotic bases, which accept protons in succession. On a titration curve, the multiple equivalence points occur at the middle of each vertical section.
Polyprotic acid titrations have multiple equivalence points, as shown in the graphs above. The equivalence points are located in the middle of each vertical section.
Examples of polyprotic acids include the diprotic acid, sulfurous acid (H 2 SO 3), which donates up to two protons per molecule and the triprotic acid, phosphoric acid (H 3 PO 4), which has three dissociable protons.
Examples of polyprotic bases include the diprotic base, sulfate ion (SO 4 2-), which accepts up to two ionized protons, and the triprotic base, phosphate ion (PO 4 3-), which accepts up to three ionized protons.
Other Types of Titrations
Acid-base titrations are just one type of titration used by chemists to determine the unknown characteristics of a solution. Other types of titrations include redox titrations, conductometric titrations, precipitation titrations, complexometric titrations, and non-aqueous titrations.
Discover more than a dozen standards-based titration labs—all available for free from the PASCO Experiment Library.
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Redox Titrations
A redox titration utilizes reduction and oxidation reactions to determine the concentration of a known analyte. Whereas acid-base titrations follow the exchange of protons, redox reactions follow the exchange of electrons. In a reduction reaction, electrons are gained. In an oxidation reaction, electrons are lost; the other added substance would gain these lost electrons and become reduced. Reduction and oxidation reactions always occur in pairs, hence the term “redox reaction.”
In the following redox titration, iron (II) is oxidized, and chromium (VI) is reduced. Fe 2+ loses an electron and becomes Fe 3+.
Cr 2 O 7 2- accepts electrons from the hydrogen ions, producing water and Cr 3+.
6 F e 2++C r 2 O 7 2–+14 H+→6 F e 3++2 C r 3++7 H 2 O
Redox titrations are often measured using a special probe, called an oxidation reduction potential probe (ORP Probe). ORP probes measure the oxidative and reductive potential of an analyte solution, enabling redox reactions to be tracked in real time.
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Oxidation Reduction Potential Probe
The Oxidation Reduction Potential Probe connects to the Wireless pH Sensor and allows students to determine the ability of a species in a solution to act as an oxidizing agent or reducing agent during redox reactions. Use this probe to monitor solutions during oxidation-reduction titrations, perform water quality studies, and study the effects of water chlorination. This probe is not a standalone sensor. It connects to and requires an amplifier.
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Conductometric Titrations
In a conductometric titration, the concentration of the analyte solution is determined by tracking the exchange of ions during the reaction. To do so, the conductivity of the analyte is measured throughout the procedure, usually with a conductivity probe such as the Wireless Conductivity Sensor.
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Why is Conductivity Used in Titrations?
Conductivity is a measurement of an ion’s capacity to conduct electricity in solution, also known as its ionic or electrolytic conductivity. During a titration, the reagents dissociate into ions, each of which has a specific ionic conductivity. When one ion replaces another, the conductivity of the analyte solution changes, making it possible to estimate the reaction’s progress.
The conductivity of a solution depends on the concentrations of cations and anions present and the temperature of the solution. To a smaller extent, conductivity also depends on an ion’s charge and mobility. When larger, less mobile ions, such as a sodium ion, replace a mobile ion, such as a hydrogen ion, the conductivity of the solution decreases. This effect is larger in more viscous fluids.
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The Wireless Conductivity Sensor measures the electrical conductivity of an aqueous solution. It is ideal for investigating the properties of solutions, including total dissolved solids (TDS) for water quality inquiry. Because it is temperature compensated, calibrations are less frequent and can be applied across a range of temperatures. With a range of 0 to 20,000 µS/cm, this sensor can be utilized for chemical, biological, and environmental studies.
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Covalent and Ionic Compounds
Covalent and ionic compounds make excellent reactants for conductometric titrations because they exhibit different ionic conductivities. A covalent compound cannot conduct electricity in its solid, liquid, or molten state. In contrast, an ionic compound can conduct electricity, so long as it is dissolved in solution or used in its molten form, and its ions are free to move.
Common covalent compounds include:
Water (H 2 O)
Carbon monoxide (CO)
Carbon dioxide (CO 2)
Ammonia (NH 3)
Methane (CH 4)
Common ionic compounds include:
Sodium chloride (NaCl)
Sodium carbonate (Na 2 CO 3)
Calcium carbonate (CaCO 3)
Lithium chloride (LiCl)
Calcium Iodide (CaI 2)
Precipitation Conductometric Titration
In a precipitation conductometric titration, an electrolyte titrates another electrolyte, forming an insoluble product called a precipitate. Precipitation titrations follow the general solubility rules.
During a precipitation titration, the conductivity of the analyte solution usually drops with the loss of ions as they form precipitate. The conductivity continues to drop until the endpoint is reached. After the endpoint, adding additional titrant adds more ions to solution. The conductivity rises until it matches the conductivity of the titrant, and then it remains constant.
One example is the precipitation reaction between sodium carbonate Na 2 CO 3 and calcium chloride CaCl 2.
N a 2 C O 3+C a C l 2→N a C l+C a C O 3
Conductivity of the analyte solution is plotted against the volume of added titrant for precipitation conductometric titrations.
When sodium carbonate titrates calcium chloride, the calcium (Ca 2+) ions and the carbonate (CO 3 2-) ions precipitate out and (Na+) replaces them. On the titration curve, the portion of the curve sloping downward indicates this exchange. The reaction equivalence point occurs once all the ions have been exchanged, indicated by the lowest point on the curve. The reaction begins to reverse when (Na 2 CO 3) is added beyond the endpoint, increasing the conductivity. The curve will eventually level out and become a flat line once the ion levels of the solution are equal to the ion levels of the titrant.
When a strong base, such a NaOH, is used in a conductometric titration, the conductivity of an analyte with two acids—one weak acid and one strong acid—can be determined.
This graph shows the conductivity change when the strong base NaOH is added to a solution of both strong and weak acids.
The strong acid is neutralized first, as indicated by the steep downward curve as ions form water and salt. However, the weak acid neutralization encourages more acid to dissociate until the acid is neutralized. After this equivalence, conductivity will rise quickly due to the added strong base. The conductivity due to the conjugate weak acid will no longer change as no more is being produced and is shown as a straight line in the graph.
Non-Aqueous Titrations
In a non-aqueous titration, a solvent other than water dissolves the analyte substance. The amphoteric properties of water make it a tough competitor for very weak acids and bases, which is why non-aqueous titrations exclude it from the reaction.
Non-aqueous acid-base titrations are used when:
the species of interest is not soluble in water
the sample contains fats or oils
the sample contains a mixture of acids or bases that must be identified by titration
both the titrant and the analyte have very weak basic or very weak acidic properties
Commonly used non-aqueous solvents include:
ethanol
methanol
isopropanol
toluene
glacial acetic acid
Other Types of Titrations
Titration is a widely used analytical technique in chemistry to determine the concentration of a substance in a sample. There are many different types of titration methods, each with its own unique advantages and limitations. Learn more about the different methods of titration in our blog, Beyond Acid-Base: Other Titration Types.
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Authored by: Roger Palmer2025-07-31
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9056 | http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap4.pdf | Bessel Functions of the First and Second Kind Reading Problems Outline Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 Modified Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Kelvin’s Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Hankel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Orthogonality of Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33 Assigned Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .40 1 Background Bessel functions are named for Friedrich Wilhelm Bessel (1784 - 1846), however, Daniel Bernoulli is generally credited with being the first to introduce the concept of Bessels func-tions in 1732. He used the function of zero order as a solution to the problem of an oscillating chain suspended at one end. In 1764 Leonhard Euler employed Bessel functions of both zero and integral orders in an analysis of vibrations of a stretched membrane, an investigation which was further developed by Lord Rayleigh in 1878, where he demonstrated that Bessels functions are particular cases of Laplaces functions.
Bessel, while receiving named credit for these functions, did not incorporate them into his work as an astronomer until 1817. The Bessel function was the result of Bessels study of a problem of Kepler for determining the motion of three bodies moving under mutual gravita-tion. In 1824, he incorporated Bessel functions in a study of planetary perturbations where the Bessel functions appear as coefficients in a series expansion of the indirect perturbation of a planet, that is the motion of the Sun caused by the perturbing body. It was likely Lagrange’s work on elliptical orbits that first suggested to Bessel to work on the Bessel functions.
The notation Jz,n was first used by Hansen9 (1843) and subsequently by Schlomilch10 (1857) and later modified to Jn(2z) by Watson (1922).
Subsequent studies of Bessel functions included the works of Mathews11 in 1895, “A treatise on Bessel functions and their applications to physics” written in collaboration with Andrew Gray. It was the first major treatise on Bessel functions in English and covered topics such as applications of Bessel functions to electricity, hydrodynamics and diffraction. In 1922, Watson first published his comprehensive examination of Bessel functions “A Treatise on the Theory of Bessel Functions” 12.
9Hansen, P.A. “Ermittelung der absoluten Strungen in Ellipsen von beliebiger Excentricitt und Neigung, I.” Schriften der Sternwarte Seeberg. Gotha, 1843.
10Schlmilch, O.X. “Ueber die Bessel’schen Function.” Z. fr Math. u. Phys. 2, 137-165, 1857.
11George Ballard Mathews, “A Treatise on Bessel Functions and Their Applications to Physics,” 1895 12G. N. Watson , “A Treatise on the Theory of Bessel Functions,” Cambridge University Press, 1922.
2 Definitions 1. Bessel Equation The second order differential equation given as x2 d2y dx2 + xdy dx + (x2 −ν2)y = 0 is known as Bessel’s equation. Where the solution to Bessel’s equation yields Bessel functions of the first and second kind as follows: y = A Jν(x) + B Yν(x) where A and B are arbitrary constants. While Bessel functions are often presented in text books and tables in the form of integer order, i.e. ν = 0, 1, 2, . . . , in fact they are defined for all real values of −∞< ν < ∞.
2. Bessel Functions a) First Kind: Jν(x) in the solution to Bessel’s equation is referred to as a Bessel function of the first kind.
b) Second Kind: Yν(x) in the solution to Bessel’s equation is referred to as a Bessel function of the second kind or sometimes the Weber function or the Neumann function.
b) Third Kind: The Hankel function or Bessel function of the third kind can be written as H(1) ν (x) = Jν(x) + iYν(x) x > 0 H(2) ν (x) = Jν(x) −iYν(x) x > 0 Because of the linear independence of the Bessel function of the first and second kind, the Hankel functions provide an alternative pair of solutions to the Bessel differential equation.
3 3. Modified Bessel Equation By letting x = i x (where i = √−1) in the Bessel equation we can obtain the modified Bessel equation of order ν, given as x2 d2y dx2 + xdy dx −(x2 + ν2)y = 0 The solution to the modified Bessel equation yields modified Bessel functions of the first and second kind as follows: y = C Iν(x) + D Kν(x) x > 0 4. Modified Bessel Functions a) First Kind: Iν(x) in the solution to the modified Bessel’s equation is referred to as a modified Bessel function of the first kind.
b) Second Kind: Kν(x) in the solution to the modified Bessel’s equation is re-ferred to as a modified Bessel function of the second kind or sometimes the Weber function or the Neumann function.
5. Kelvin’s Functions A more general form of Bessel’s modified equation can be written as x2 d2y dx2 + xdy dx −(β2x2 + ν2)y = 0 where β is an arbitrary constant and the solutions is now y = C Iν(βx) + D Kν(βx) If we let β2 = i where i = √ −1 4 and we note Iν(x) = i−νJν(ix) = Jν(i3/2x) then the solution is written as y = C Jν(i3/2x) + DKν(i1/2x) The Kelvin functions are obtained from the real and imaginary portions of this solution as follows: berν = Re Jν(i3/2x) beiν = Im Jν(i3/2x) Jν(i3/2x) = berν x + i bei x kerν = Re i−νKν(i1/2x) keiν = Im i−νKν(i1/2x) i−νKν(i1/2x) = kerν x + i kei x 5 Theory Bessel Functions Bessel’s differential equation, given as x2 d2y dx2 + xdy dx + (x2 −ν2)y = 0 is often encountered when solving boundary value problems, such as separable solutions to Laplace’s equation or the Helmholtz equation, especially when working in cylindrical or spherical coordinates. The constant ν, determines the order of the Bessel functions found in the solution to Bessel’s differential equation and can take on any real numbered value. For cylindrical problems the order of the Bessel function is an integer value (ν = n) while for spherical problems the order is of half integer value (ν = n + 1/2).
Since Bessel’s differential equation is a second-order equation, there must be two linearly independent solutions. Typically the general solution is given as: y = AJν(x) + BYν(x) where the special functions Jν(x) and Yν(x) are: 1. Bessel functions of the first kind, Jν(x), which are finite at x = 0 for all real values of ν 2. Bessel functions of the second kind, Yν(x), (also known as Weber or Neumann func-tions) which are singular at x = 0 The Bessel function of the first kind of order ν can be be determined using an infinite power series expansion as follows: Jν(x) = ∞ k=0 (−1)k(x/2)ν+2k k!Γ(ν + k + 1) = 1 Γ(1 + ν) x 2 ν 1 − (x/2)2 1(1 + ν) 1 − (x/2)2 2(2 + ν) 1 − (x/2)2 3(3 + ν) (1 −· · · 6 0 2 4 6 8 10 12 14 x -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Jnx J0 J1 J2 Figure 4.1: Plot of the Bessel Functions of the First Kind, Integer Order or by noting that Γ(ν + k + 1) = (ν + k)!, we can write Jν(x) = ∞ k=0 (−1)k(x/2)ν+2k k!(ν + k)!
Bessel Functions of the first kind of order 0, 1, 2 are shown in Fig. 4.1.
The Bessel function of the second kind, Yν(x) is sometimes referred to as a Weber function or a Neumann function (which can be denoted as Nν(x)). It is related to the Bessel function of the first kind as follows: Yν(x) = Jν(x) cos(νπ) −J−ν(x) sin(νπ) where we take the limit ν →n for integer values of ν.
For integer order ν, Jν, J−ν are not linearly independent: J−ν(x) = (−1)νJν(x) Y−ν(x) = (−1)νYν(x) 7 in which case Yν is needed to provide the second linearly independent solution of Bessel’s equation. In contrast, for non-integer orders, Jν and J−ν are linearly independent and Yν is redundant.
The Bessel function of the second kind of order ν can be expressed in terms of the Bessel function of the first kind as follows: Yν(x) = 2 πJν(x) ln x 2 + γ −1 π ν−1 k=0 (ν −k −1)!
k!
x 2 2k−ν + + 1 π ∞ k=0 (−1)k−1 1 + 1 2 + · · · + 1 k + 1 + 1 2 + · · · + 1 k + ν k!(k + ν)!
x 2 2k+ν Bessel Functions of the second kind of order 0, 1, 2 are shown in Fig. 4.2.
0 2 4 6 8 10 12 14 x -1.5 -1 -0.5 0 0.5 1 Ynx Y0 Y1 Y2 Figure 4.2: Plot of the Bessel Functions of the Second Kind, Integer Order 8 Relations Satisfied by the Bessel Function Recurrence Formulas Bessel functions of higher order be expressed by Bessel functions of lower orders for all real values of ν.
Jν+1(x) = 2ν x Jν(x) −Jν−1(x) Yν+1(x) = 2ν x Yν(x) −Yν−1(x) J ′ ν+1(x) = 1 2 [Jν−1(x) −Jν+1(x)] Y ′ ν+1(x) = 1 2 [Yν−1(x) −Yν+1(x)] J ′ ν(x) = Jν−1(x) −ν xJν(x) Y ′ ν(x) = Yν−1(x) −ν xYν(x) J ′ ν(x) = ν xJν(x) −Jν+1(x) Y ′ ν(x) = ν xYν(x) −Yν+1(x) d dx [xνJν(x)] = xνJν−1(x) d dx [xνYν(x)] = xνYν−1(x) d dx x−νJν(x) = −x−νJν+1(x) d dx x−νYν(x) = −x−νYν+1(x) Integral Forms of Bessel Functions for Integer Orders n = 0, 1, 2, 3, . . .
First Kind Jn(x) = 1 π π 0 cos(nθ −x sin θ) dθ = 1 π π 0 cos(x sin θ −nθ) dθ J0(x) = 1 π π 0 cos(x sin θ) dθ = 1 π π 0 cos(x cos θ) dθ J1(x) = 1 π π 0 cos(θ −x sin θ) dθ = 1 π π 0 cos(x sin θ −θ) dθ = 1 π π 0 cos θ sin(x cos θ) dθ 9 from Bowman, pg. 57 J0(x) = 2 π π/2 0 cos(x sin θ) dθ = 2 π π/2 0 cos(x cos θ) dθ Second Kind for Integer Orders n = 0, 1, 2, 3, . . .
Yn(x) = − 2(x/2)−n √πΓ 1 2 −n ∞ 1 cos(xt) dt (t2 −1)n+1/2 x > 0 Yn(x) = 1 π π 0 sin(x sin θ −nθ) dθ −1 π π 0 ent + e−nt cos(nπ) exp(−x sinh t) dt x > 0 Y0(x) = 4 π2 π/2 0 cos(x cos θ) γ + ln(2x sin2 θ) dθ x > 0 Y0(x) = −2 π ∞ 0 cos(x cosh t) dt x > 0 Approximations Polynomial Approximation of Bessel Functions For x ≥2 one can use the following approximation based upon asymptotic expansions: Jn(x) = 2 πx 1/2 [Pn(x) cos u −Qn(x) sin u] where u ≡x −(2n + 1)π 4 and the polynomials Pn(x) and Qn(x) are given by 10 Pn(x) = 1 −(4n2 −12)(4n2 −32) 2 · 1(8x)2 1 −(4n2 −52)(4n2 −72) 4 · 3(8x)2 1 −(4n2 −92)(4n2 −112) 6 · 5(8x)2 (1 −· · · ) and Qn(x) = 4n2 −12 1!(8x) 1 −(4n2 −32)(4n2 −52) 3 · 2(8x)2 1 −(4n2 −72)(4n2 −92) 5 · 4(8x)2 (1 −· · · ) The general form of these terms can be written as Pn(x) = (4n2 −(4k −3)2) (4n2 −(4k −1)2) 2k(2k −1)(8x)2 k = 1, 2, 3 . . .
Qn(x) = (4n2 −(4k −1)2) (4n2 −(4k + 1)2) 2k(2k + 1)(8x)2 k = 1, 2, 3 . . .
For n = 0 sin u = 1 √ 2 (sin x −cos x) cos u = 1 √ 2 (sin x + cos x) J0(x) = 1 √πx [P0(x)(sin x + cos x) −Q0(x)(sin x −cos x)] 11 or J0(x) = 2 πx 1/2 P0(x) cos x −π 4 −Q0(x) sin x −π 4 P0(x) = 1 −12 · 32 2!(8x)2 1 − 52 · 72 4 · 3(8x)2 1 − 92 · 112 6 · 5(8x)2 (1 −· · · ) Q0(x) = −12 8x 1 − 32 · 52 3 · 2(8x)2 1 − 72 · 92 5 · 4(8x)2 (1 −· · · ) For n = 1 sin u = 1 √ 2 (sin x + cos x) cos u = 1 √ 2 (sin x −cos x) J1(x) = 1 √πx [P1(x)(sin x −cos x) −Q1(x)(sin x + cos x)] or J1(x) = 2 πx 1/2 P1(x) cos x −3π 4 −Q1(x) sin x −3π 4 P1(x) = 1 + 3 · 5 2 · 1(8x)2 1 − 21 · 45 4 · 3(8x)2 1 −77 · 117 6 · 5(8x)2 (1 −· · · ) Q1(x) = 3 8x 1 − 35 2 · 1(8x)2 1 − 45 · 77 5 · 4(8x)2 (1 −· · · ) 12 Asymptotic Approximation of Bessel Functions Large Values of x Y0(x) = 2 πx 1/2 [P0(x) sin(x −π/4) + Q0(x) cos(x −π/4)] Y1(x) = 2 πx 1/2 [P1(x) sin(x −3π/4) + Q1(x) cos(x −3π/4)] where the polynomials have been defined earlier.
Power Series Expansion of Bessel Functions First Kind, Positive Order Jν(x) = ∞ k=0 (−1)k(x/2)ν+2k k!Γ(ν + k + 1) = 1 Γ(1 + ν) x 2 ν 1 − (x/2)2 1(1 + ν) 1 − (x/2)2 2(2 + ν) 1 − (x/2)2 3(3 + ν) (1 −· · · The General Term can be written as Zk = −Y k(k + ν) k = 1, 2, 3, . . .
Y = (x/2)2 where B0 = 1 B + 1 = Z1 · B0 B2 = Z2 · B1 .
.
.
Bk = Zk · Bk−1 13 The approximation can be written as Jν(x) = (x/2)ν Γ(1 + ν U k=0 Bk First Kind, Negative Order J−ν(x) = ∞ k=0 (−1)k(x/2)2k−ν k!Γ(k + 1 −ν) = 1 Γ(1 −ν) x 2 −ν 1 − (x/2)2 1(1 −ν) 1 − (x/2)2 2(2 −ν) 1 − (x/2)2 3(3 −ν) (1 −· · · The General Term can be written as Zk = −Y k(k −ν) k = 1, 2, 3, . . .
Y = (x/2)2 where B0 = 1 B + 1 = Z1 · B0 B2 = Z2 · B1 .
.
.
Bk = Zk · Bk−1 The approximation can be written as J−ν(x) = (x/2)−ν Γ(1 −ν) U k=0 Bk where U is some arbitrary value for the upper limit of the summation.
14 Second Kind, Positive Order Yν(x) = Jν(x) cos νπ −J−ν(x) sin νπ , ν ̸= 0, 1, 2, . . .
Roots of Bessel Functions First Kind, Order Zero, J0(x) = 0 This equation has an infinite set of positive roots x1 < x2 < x3 . . . < xn < xn+1 . . .
Note: xn+1 −xn →π as n →∞ The roots of J0(x) can be computed approximately by Stokes’s approximation which was developed for large n xn = α 4 1 + 2 α2 −62 3α4 + 15116 15α6 −12554474 105α8 + 8368654292 315α10 −. . .
with α = π(4n −1).
An approximation for small n is xn = α 4 1 + 2 α2 −62 3α4 + 7558 15α6 First Kind, Order One, J1(x) = 0 This equation has an infinite set of positive, non-zero roots x1 < x2 < x3 < . . . xn < xn+1 < . . .
15 Note: xn+1 −xn →π as n →∞ These roots can also be computed using Stoke’s approximation which was developed for large n.
xn = β 4 1 + 6 β2 + 6 β4 −4716 5β6 + 3902418 35β8 −895167324 35β10 + . . .
with β = π(4n + 1).
An approximation for small n is xn = β 4 1 −6 β2 + 6 β4 −4716 10β6 The roots of the transcendental equation xnJ1(xn) −BJ0(xn) = 0 with 0 ≤B < ∞are infinite in number and they can be computed accurately and efficiently using the Newton-Raphson method. Thus the (i + 1)th iteration is given by xi+1 n = xi n −xi nJ1(xi n) −BJ0(xi n) xi nJ0(xi n) + BJ1(xi n) Accurate polynomial approximations of the Bessel functions J0(·) and J1(·) may be em-ployed. To accelerate the convergence of the Newton-Raphson method, the first value for the (n + 1)th root can be related to the converged value of the nth root plus π.
Aside: Fisher-Yovanovich modified the Stoke’s approximation for roots of J0(x) = 0 and J1(x) = 0. It is based on taking the arithmetic average of the first three and four term expressions For Bi →∞roots are solutions of J0(x) = 0 δn,∞= α 4 1 + 2 α2 −62 3α4 + 15116 30α6 16 with α = π(4n −1).
For Bi →0 roots are solutions of J1(x) = 0 δn,0 = β 4 1 −6 β2 + 6 β4 −4716 10β6 with β = π(4n + 1).
Potential Applications 1. problems involving electric fields, vibrations, heat conduction, optical diffraction plus others involving cylindrical or spherical symmetry 2. transient heat conduction in a thin wall 3. steady heat flow in a circular cylinder of finite length 17 Modified Bessel Functions Bessel’s equation and its solution is valid for complex arguments of x. Through a simple change of variable in Bessel’s equation, from x to ix (where i = √−1), we obtain the modified Bessel’s equation as follows: x2 d2y dx2 + xdy dx + ((ix)2 −ν2)y = 0 or equivalently x2 d2y dx2 + xdy dx −(x2 + ν2)y = 0 The last equation is the so-called modified Bessel equation of order ν. Its solution is y = AJν(ix) + BYν(ix) x > 0 or y = CIν(x) + DKν(x) x > 0 and Iν(x) and Kν(x) are the modified Bessel functions of the first and second kind of order ν.
Unlike the ordinary Bessel functions, which are oscillating, Iν(x) and Kν(x) are exponen-tially growing and decaying functions as shown in Figs. 4.3 and 4.4.
It should be noted that the modified Bessel function of the First Kind of order 0 has a value of 1 at x = 0 while for all other orders of ν > 0 the value of the modified Bessel function is 0 at x = 0. The modified Bessel function of the Second Kind diverges for all orders at x = 0.
18 0 0.5 1 1.5 2 2.5 3 x 0.5 1 1.5 2 2.5 Ix I0 I1 I2 Figure 4.3: Plot of the Modified Bessel Functions of the First Kind, Integer Order 0 0.5 1 1.5 2 2.5 3 x 0.5 1 1.5 2 2.5 Kx K0 K1 K2 Figure 4.4: Plot of the Modified Bessel Functions of the Second Kind, Integer Order 19 Relations Satisfied by the Modified Bessel Function Recurrence Formulas Bessel functions of higher order can be expressed by Bessel functions of lower orders for all real values of ν.
Iν+1(x) = Iν−1(x) −2ν x Iν(x) Kν+1(x) = Kν−1(x) + 2ν x Kν(x) I′ ν(x) = 1 2 [Iν−1(x) + Iν+1(x)] K′ ν(x) = −1 2 [Kν−1(x) + Kν+1(x)] I′ ν(x) = Iν−1(x) −ν xIν(x) K′ ν(x) = −Kν−1(x) −ν xKν(x) I′ ν(x) = ν xIν(x) + Iν+1(x) K′ ν(x) = ν xKν(x) −Kν+1(x) d dx [xνIν(x)] = xνIν−1(x) d dx [xνKν(x)] = −xνKν−1(x) d dx x−νIν(x) = x−νIν+1(x) d dx x−νKν(x) = −x−νKν+1(x) Integral Forms of Modified Bessel Functions for Integer Orders n = 0, 1, 2, 3, . . .
First Kind In(x) = 1 π π 0 cos(nθ) exp(x cos θ) dθ I0(x) = 1 π π 0 exp(x cos θ) dθ I1(x) = 1 π π 0 cos(θ) exp(x cos θ) dθ 20 Alternate Integral Representation of I0(x) and I1(x) I0(x) = 1 π π 0 cosh(x cos θ) dθ I1(x) = dI0(x) dx = 1 π π 0 sinh(x cos θ) cos θ dθ Second Kind Note: These are also valid for non-integer values of Kv(x).
Kn(x) = √π(x/2)n Γ n + 1 2 ∞ 0 sinh2n t exp(−x cosh t) dt x > 0 Kn(x) = ∞ 0 cosh(nt) exp(−x cosh t) dt x > 0 K0(x) = ∞ 0 exp(−x cosh t) dt x > 0 K1(x) = ∞ 0 cosh t exp(−x cosh t) dt x > 0 21 Approximations Asymptotic Approximation of Modified Bessel Functions for Large Values of x In(x) = ex √ 2πx 1 −4n2 −12 1(8x) 1 −(4n2 −32) 2(8x) 1 −(4n2 −52) 3(8x) (1 −. . . ) I0(x) = ex √ 2πx 1 + 1 8x 1 + 9 2(8x) 1 + 25 3(8x) (1 + . . . ) I1(x) = ex √ 2πx 1 −3 8x 1 + 5 2(8x) 1 + 21 3(8x) (1 + . . . ) The general term can be written as −4n2 −(2k −1)2 k(8x) Power Series Expansion of Modified Bessel Functions First Kind, Positive Order Iν(x) = ∞ k=0 (x/2)ν+2k k!Γ(ν + k + 1) = 1 Γ(1 + ν) x 2 ν 1 + (x/2)2 1(1 + ν) 1 + (x/2)2 2(2 + ν) 1 + (x/2)2 3(3 + ν) (1 + · · · The General Term can be written as Zk = Y k(k + ν) k = 1, 2, 3, . . .
Y = (x/2)2 22 where B0 = 1 B + 1 = Z1 · B0 B2 = Z2 · B1 .
.
.
Bk = Zk · Bk−1 The approximation can be written as Iν(x) = (x/2)ν Γ(1 + ν) U k=0 Bk where U is some arbitrary value for the upper limit of the summation.
First Kind, Negative Order I−ν(x) = ∞ k=0 (x/2)2k−ν k!Γ(k + 1 −ν) = 1 Γ(1 −ν) x 2 −ν 1 + (x/2)2 1(1 −ν) 1 + (x/2)2 2(2 −ν) 1 + (x/2)2 3(3 −ν) (1 + · · · The General Term can be written as Zk = Y k(k −ν) k = 1, 2, 3, . . .
Y = (x/2)2 23 where B0 = 1 B + 1 = Z1 · B0 B2 = Z2 · B1 .
.
.
Bk = Zk · Bk−1 The approximation can be written as I−ν(x) = (x/2)−ν Γ(1 −ν) U k=0 Bk where U is some arbitrary value for the upper limit of the summation.
Second Kind, Positive Order Kν(x) = π[I−ν(x) −Iν(x)] 2 sin νπ Alternate Forms of Power Series Expansion for Modified Bessel Functions First Kind In(x) = zn n!
1 + z2 1(n + 1) 1 + z2 2(n + 2) 1 + z2 3(n + 3) 1 + z2 4(n + 4) (1 + . . . ) I0(x) = 1 + z2 1 + z2 2 · 2 1 + z2 3 · 3 1 + z2 4 · 4 1 + z2 5 · 5 (1 + . . . ) I1(x) = z 1 + z2 1 · 2 1 + z2 2 · 3 1 + z2 3 · 4 1 + z2 4 · 5 (1 + . . . ) 24 Second Kind When n is a positive integer Kn(x) = π 2x 1/2 e−x 1 + (4n2 −12) 1(8x) 1 + (4n2 −32) 2(8x) 1 + (4n2 −52) 3(8x) (1 + . . . ) K0(x) = π 2x 1/2 e−x 1 −1 8x 1 − 9 2(8x) 1 − 25 3(8x) K1(x) = π 2x 1/2 e−x 1 + 3 8x 1 − 5 2(8x) 1 − 21 3(8x) Series expansions based upon the trapezoidal rule applied to certain forms of the integral representation of the Bessel functions can be developed for any desired accuracy. Several expansions are given below.
For x ≤12, 8 decimal place accuracy is obtained by 15J0(x) = cos x + 2 7 j=1 cos(x cos jπ/15) 15J1(x) = sin x + 2 7 j=1 sin(x cos jπ/15) cos(jπ/15) For x ≤20, 8 decimal place accuracy is obtained by 15I0(x) = cosh x + 2 7 j=1 cosh(x cos jπ/15) 15I1(x) = sinh x + 2 7 j=1 sinh(x cos jπ/15) cos(jπ/15) 25 For x ≥0.1, 8 decimal place accuracy is obtained by 4K0(x) = e−x + e−x cosh 6 + 2 11 j=1 exp[−x cosh(j/2)] 4K1(x) = e−x + cosh 6e−x cosh 6 + 2 11 j=1 exp[−x cosh(j/2) cosh(j/2)] Potential Applications 1. displacement of a vibrating membrane 2. heat conduction in an annular fin of rectangular cross section attached to a circular base 26 Kelvin’s Functions Consider the differential equation x2 d2y dx2 + xdy dx −(ik2x2 + n2)y = 0 i = √ −1 This is of the form of Bessel’s modified equation x2 d2y dx2 + xdy dx −(β2x2 + n2)y = 0 i = √ −1 with β2 = ik2. Since the general solution of Bessel’s modified equation is y = AIn(βx) + BKn(βx) the general solution of the given equation can be expressed as y = AIn( √ i kx) + BKn( √ i kx) Also, since In(x) = i−nJn(ix) ⇒ inIn(x) = Jn(ix) we may take the independent solutions of the given equation as y = AJn(i3/2 kx) + BKn(i1/2 kx) when x is real, Jn(i3/2x) and Kn(i1/2x) are not necessarily real. We obtain real functions 27 by the following definitions: bern = Re Jn(i3/2x) bein = Im Jn(i3/2x) Jn(i3/2x) = bern x + i bei x kern = Re i−nKn(i1/2x) kein = Im i−nKn(i1/2x) i−nKn(i1/2x) = kern x + i kei x It is, however, customary to omit the subscript from the latter definitions when the order n is zero and to write simply J0(i3/2x) = ber x + i bei x K0(i1/2x) = ker x + i kei x The complex function ber x + i bei x is often expressed in terms of its modulus and its amplitude: ber x + i bei x = M0(x)eiθ0(x) where M0(x) = [(ber x)2 + (bei x)2]1/2, θ0 = arc tan bei x ber x Similarly we can write bern x + i bern x = Mn(x)eiθn(x) where 28 Mn(x) = [(bern x)2 + (bein x)2]1/2, θn = arc tan bein x bern x Kelvin’s Functions ber x and bei x The equation for I0(t) is t2d2y dt2 + tdy dt −t2y = 0 Set t = x √ i and the equation becomes x2 d2y dx2 + xdy dx −i x2y = 0 with the solutions I0(x √ i) and K0(x √ i). The ber and bei functions are defined as follows.
Since I0(t) = 1 + t 2 2 + (t/2)2 (2!)2 + (t/2)6 (3!)2 + · · · we have real and imaginary parts in I0(x √ i) = 1 −(x/2)4 (2!)2 + (x/2)8 (4!)2 −· · · +i (x/2)2 −(x/2)6 (3!)2 + (x/2)10 (5!)2 −· · · = ber x + i bei x Equating real and imaginary parts we have ber x = 1 −(x/2)4 (2!)2 + (x/2)8 (4!)2 −· · · bei x = (x/2)2 −(x/2)6 (3!)2 + (x/2)10 (5!)2 −· · · 29 Both ber x and bei x are real for real x, and it can be seen that both series are absolutely convergent for all values of x. Among the more obvious properties are ber 0 = 1 bei 0 = 0 and x 0 x ber x dx = x bei′ x, x 0 x bei x dx = −x ber′ x In a similar manner the functions ker x and kei x are defined to be respectively the real and imaginary parts of the complex function K0(x √ i), namely ker x + i kei x = K0(x √ i) From the definition of K0(x) we can see that ker x = −[ln(x/2) + δ] ber x + π 4 bei x + ∞ r=1 (−1)r(x/2)4r [(2r)!]2 φ(2r) and kei x = −[ln(x/2) + δ] bei x −π 4 ber x + ∞ r=1 (−1)r(x/2)4r+2 [(2r + 1)!]2 φ(2r + 1) where φ(r) = r s=1 1 s 30 Potential Applications 1. calculation of the current distribution within a cylindrical conductor 2. electrodynamics of a conducting cylinder 31 Hankel Functions We can define two new linearly dependent functions H(1) n (x) = Jn(x) + iYn(x) x > 0 H(2) n (x) = Jn(x) −iYn(x) x > 0 which are obviously solutions of Bessel’s equation and therefore the general solution can be written as y = AH(1) n (x) + BH(2) n (x) where A and B are arbitrary constants. The functions H(1) n (x) and H(2) n (x) are called Hankel’s Bessel functions of the third kind. Both are, of course, infinite at x = 0, their usefulness is connected with their behavior for large values of x.
Since Hankel functions are linear combinations of Jn and Yn, they satisfy the same recurrence relationships.
32 Orthogonality of Bessel Functions Let u = Jn(λx) and v = Jn(µx) with λ ̸= µ be two solutions of the Bessel equations x2u′′ + xu′ + (λ2x2 −n2)u = 0 and x2v′′ + xv′ + (µ2x2 −n2)v = 0 where the primes denote differentiation with respect to x.
Multiplying the first equation by v and the second by u, and subtracting, we obtain x2(vu′′ −uv′′) + x(vu′ −uv′) = (µ2 −λ2)x2uv Division by x gives x(vu′′ −uv′′) + (vu′ −uv′) = (µ2 −λ2)xuv or d dx[x(vu′ −uv′)] = (µ2 −λ2)xuv Then by integration and omitting the constant of integration we have (µ2 −λ2) xuv dx = x(vu′ −uv′) or making the substitutions for u, u′, v and v′ we have (µ2 −λ2) xJn(λx) Jn(µx) dx = x[Jn(µx) J ′ n(λx) −Jn(λx) J ′ n(µx)] 33 This integral is the so-called Lommel integral. The right hand side vanishes at the lower limit zero. It also vanishes at some arbitrary upper limit, say x = b, provided Jn(µb) = 0 = Jn(λb) or J ′ n(λb) = 0 = J ′ n(µb) In the first case this means that µb and λb are two roots of Jn(x) = 0, and in the second case it means that µb and λb are two roots of J ′ n(x) = 0. In either case we have the following orthogonality property b a xJn(λx)Jn(µx) dx = 0 This property is useful in Bessel-Fourier expansions of some arbitrary function f(x) over the finite interval 0 ≤x ≤b. Further the functions Jn(λx) and Jn(µx) are said to be orthogonal in the interval 0 ≤x ≤b with respect to the weight function x.
As µ →λ, it can be shown by the use of L’Hopital’s rule that x 0 xJ 2 n(λx) dx = x2 2 J ′ n(λx) 2 + 1 − n2 (λx)2 [Jn(λx)]2 where J ′ n(λx) = dJn(r) dr with r = λx 34 Assigned Problems Problem Set for Bessel Equations and Functions 1. By means of power series, asymptotic expansions, polynomial approximations or Ta-bles, compute to 6 decimal places the following Bessel functions: a) J0(x) e) Y0(x) b) J1(y) f) Y1(y) c) I0(z) g) K0(z) d) I1(z) h) K1(z) given x = 3.83171 y = 2.40482 z = 1.75755 2. Compute to 6 decimal places the first six roots xn of the transcendental equation x J1(x) −B J0(x) = 0 B ≥0 when B = 0.1, 1.0, 10, and 100.
3. Compute to 4 decimal places the coefficients An(n = 1, 2, 3, 4, 5, 6) given An = 2B (x2 n + B2) J0(xn) for B = 0.1, 1.0, 10, and 100. The xn are the roots found in Problem 2.
35 4. Compute to 4 decimal places the coefficients Bn(n = 1, 2, 3, 4) given Bn = 2AnJ1(xn) xn for B = 0.1, 1.0, 10, and 100. The xn are the roots found in Problem 2 and the An are the coefficients found in Problem 3.
5. The fin efficiency of a longitudinal fin of convex parabolic profile is given as η = 1 γ I2/3(4√γ/3) I−2/3(4√γ/3) Compute η for γ = 3.178.
6. The fin efficiency of a longitudinal fin of triangular profile is given by η = 1 γ I1(2√γ) I0(2√γ) Compute η for γ = 0.5, 1.0, 2.0, 3.0, and 4.0.
7. The fin efficiency of a radial fin of rectangular profile is given by η = 2ρ x(1 −ρ2) I1(x)K1(ρx) −K1(x)I1(ρx) I0(ρx)K1(x) + I1(x)K0(ρx) Compute the efficiency for x = 2.24, ρx = 0.894.
Ans: η = 0.537 8. Show that i) J ′ 0(x) = −J1(x) ii) d dx [xJ1(x)] = xJ0(x) 36 9. Given the function f(x) = xJ1(x) −B J0(x) with B = constant ≥0 determine f ′, f ′′, and f ′′′. Reduce all expressions to functions of J0(x), J1(x) and B only.
10. Show that i) x2J0(x) dx = x2J1(x) + xJ0(x) − J0(x) dx ii) x3J0(x) dx = x(x2 −4) J1(x) + 2x2J0(x) 11. Show that i) 1 0 xJ0(λx) dx = 1 λJ1(λ) ii) 1 0 x3J0(λx) dx = λ2 −4 λ3 J1(λ) + 2 λ2 J0(λ) 12. If δ is any root of the equation J0(x) = 0, show that i) 1 0 J1(δx) dx = 1 δ ii) δ 0 J1(x) dx = 1 13. If δ(> 0) is a root of the equation J1(x) = 0 show that 1 0 xJ0(δx) dx = 0 37 14. Given the Fourier-Bessel expansion of f(x) of zero order over the interval 0 ≤x ≤1 f(x) = A1 J0(δ1x) + A2 J0(δ2x) + A3 J0(δ3x) + . . .
where δn are the roots of the equation J0(x) = 0. Determine the coefficients An when f(x) = 1 −x2.
15. Show that over the interval 0 ≤x ≤1 x = 2 ∞ n=1 J1(δn) δn J2(δn) where δn are the positive roots of J1(x) = 0.
16. Obtain the solution to the following second order ordinary differential equations: i) y′′ + xy = 0 ii) y′′ + 4x2y = 0 iii) y′′ + e2xy = 0 Hint: let u = ex iv) xy′′ + y′ + k2y = 0 k > 0 v) x2y′′ + x2y′ + 1 4y = 0 vi) y′′ + 1 xy′ − 1 + 4 x2 y = 0 vii) xy′′ + 2y′ + xy = 0 17. Obtain the solution for the following problem: xy′′ + y′ −m2by = 0 0 ≤x ≤b, m > 0 with y(0) ̸= ∞ and y(b) = y0 38 18. Obtain the solution for the following problem: x2y′′ + 2xy′ −m2xy = 0 0 ≤x ≤b, m > 0 with y(0) ̸= ∞ and y(b) = y0 19. Show that I3/2(x) = 2 πx 1/2 cosh x −sinh x x I−3/2(x) = 2 πx 1/2 sinh x −cosh x x 39 Selected References 1. Abramowitz, M. and Stegun, I.A., Handbook of Mathematical Functions, Dover, New York, 1965.
2. Bowman, F., Introduction to Bessel Functions, Dover New York, 1958.
3. Gray A, Mathews, G. B. and MacRobert, T. M., A Treatise on Bessel Functions, Basingstoke MacMillan, 1953.
4. Luke, Y. L., Integrals of Bessel Functions, McGraw-Hill, New York, 1962.
5. Magnus W, Oberhettinger, F. and Soni, R. P. Formulas and Theorems for the Special Functions of Mathematical Physics, Springer, Berlin, 1966.
6. McLachlan, N.W., Bessel Functions for Engineers, 2nd Edition, Oxford University Press, London, 1955.
7. Rainville, E.D., Special Functions, MacMillan, New York, 1960.
8. Relton, F.E., Applied Bessel Functions, Blackie, London, 1946.
9. Sneddon, I.N., Special Functions of Mathematical Physics and Chemistry, 2nd Edi-tion, Oliver and Boyd, Edinburgh, 1961.
10. Watson, G.N., A Treatise on the Theory of Bessel Functions, 2nd Edition, Cambridge University Press, London, 1931.
11. Wheelon, A. D., Tables of Summable Series and Integrals Involving Bessel Functions, Holden-Day, San Francisco, 1968.
40 |
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Place Value Names
| Place Value Name | Number | Scientific Count Format |
---
| Units (Ones) | 1 | 1 |
| Tens | 10 | 1e+1 |
| Hundreds | 100 | 1e+2 |
| Thousands | 1,000 | 1e+3 |
| Ten thousands | 10,000 | 1e+4 |
| Hundred thousands | 100,000 | 1e+5 |
| Millions | 1,000,000 | 1e+6 |
| Ten millions | 10,000,000 | 1e+7 |
| Hundred millions | 100,000,000 | 1e+8 |
| Billions | 1,000,000,000 | 1e+9 |
| Ten billions | 10,000,000,000 | 1e+10 |
| Hundred billions | 100,000,000,000 | 1e+11 |
| Trillions | 1,000,000,000,000 | 1e+12 |
| Ten trillions | 10,000,000,000,000 | 1e+13 |
| Hundred trillions | 100,000,000,000,000 | 1e+14 |
| Quadrillions | 1,000,000,000,000,000 | 1e+15 |
| Ten quadrillions | 10,000,000,000,000,000 | 1e+16 |
| Hundred quadrillions | 100,000,000,000,000,000 | 1e+17 |
| Quintillions | 1,000,000,000,000,000,000 | 1e+18 |
| Ten quintillions | 10,000,000,000,000,000,000 | 1e+19 |
| Hundred quintillions | 100,000,000,000,000,000,000 | 1e+20 |
| Sextillions | 1,000,000,000,000,000,000,000 | 1e+21 |
| Ten sextillions | 10,000,000,000,000,000,000,000 | 1e+22 |
| Hundred sextillions | 100,000,000,000,000,000,000,000 | 1e+23 |
| Septillions | 1,000,000,000,000,000,000,000,000 | 1e+24 |
| Ten septillions | 10,000,000,000,000,000,000,000,000 | 1e+25 |
| Hundred septillions | 100,000,000,000,000,000,000,000,000 | 1e+26 |
| Octillions | 1,000,000,000,000,000,000,000,000,000 | 1e+27 |
| Ten octillions | 10,000,000,000,000,000,000,000,000,000 | 1e+28 |
| Hundred octillions | 100,000,000,000,000,000,000,000,000,000 | 1e+29 |
| Decimals (Tenths) | 0.1 | 1e-1 |
| Hundredths | 0.01 | 1e-2 |
| Thousandths | 0.001 | 1e-3 |
| Ten thousandths | 0.0001 | 1e-4 |
| Hundred thousandths | 0.00001 | 1e-5 |
| Millionths | 0.000001 | 1e-6 |
| Ten millionths | 0.0000001 | 1e-7 |
| Hundred millionths | 0.00000001 | 1e-8 |
| Billionths | 0.000000001 | 1e-9 |
| Ten billionths | 0.0000000001 | 1e-10 |
| Hundred billionths | 0.00000000001 | 1e-11 |
| Trillionths | 0.000000000001 | 1e-12 |
| Ten trillionths | 0.0000000000001 | 1e-13 |
| Hundred trillionths | 0.00000000000001 | 1e-14 |
| Quadrillionths | 0.000000000000001 | 1e-15 |
| Ten quadrillionths | 0.0000000000000001 | 1e-16 |
| Hundred quadrillionths | 0.00000000000000001 | 1e-17 |
| Quintillionths | 0.000000000000000001 | 1e-18 |
| Ten quintillionths | 0.0000000000000000001 | 1e-19 |
| Hundred quintillionths | 0.00000000000000000001 | 1e-20 |
| Sextillionths | 0.000000000000000000001 | 1e-21 |
| Ten sextillionths | 0.0000000000000000000001 | 1e-22 |
| Hundred sextillionths | 0.00000000000000000000001 | 1e-23 |
| Septillionths | 0.000000000000000000000001 | 1e-24 |
| Ten septillionths | 0.0000000000000000000000001 | 1e-25 |
| Hundred septillionths | 0.00000000000000000000000001 | 1e-26 |
| Octillionths | 0.000000000000000000000000001 | 1e-27 |
| Ten octillionths | 0.0000000000000000000000000001 | 1e-28 |
| Hundred octillionths | 0.00000000000000000000000000001 | 1e-29 |
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Open Access Case Report
A Non-Secreting Pituitary Adenoma That Changed to a Prolactinoma
by
Cristina Santiago-Vazquez
Cristina Santiago-Vazquez
SciProfilesScilitPreprints.orgGoogle Scholar
1,
Nuria Palacios-Paino
Nuria Palacios-Paino
SciProfilesScilitPreprints.orgGoogle Scholar
1 and
Fernando Cordido
Fernando Cordido
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1,2,3,
1
Servicio Endocrinología y Nutrición, Hospital Universitario A Coruña, 15006 A Coruña, Spain
2
Grupo Fisiopatoloxía Endócrina, Nutricional e Médica (FENM), Facultad de Ciencias de la Salud, Universidade da Coruña, 15006 A Coruña, Spain
3
Instituto de Investigación Biomedica (INIBIC), Centro de Investigaciones Científicas avanzadas (CICA), Universidade da Coruña, 15006 A Coruña, Spain
Author to whom correspondence should be addressed.
Clin. Pract.2024, 14(4), 1310-1318;
Submission received: 23 May 2024 / Revised: 18 June 2024 / Accepted: 3 July 2024 / Published: 4 July 2024
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Abstract
Pituitary adenomas (PAs) are the third most common brain tumors in adults right after meningiomas and gliomas. Taking into account their hormonal activity in vivo, they can be divided in functioning PAs, which secrete hormones, and nonfunctioning pituitary adenomas (NFPAs), which are not associated with increased hormone secretion. We present the case of a man diagnosed with pituitary apoplexy. A transsphenoidal surgery was performed with subtotal removal of the mass. Pituitary hormones were measured before and after the procedure on several occasions, showing always normal PRL values, so he was diagnosed with a clinically NFPA. Two years later, the patient noticed a visual deficit. A new magnetic resonance imaging study was performed, showing adenomatous recurrence, and the patient underwent a new surgery. After this, hormonal evaluation revealed high levels of PRL on several occasions. After treatment with cabergoline was started, PRL levels normalized, the visual deficit improved, and there was a slight adenoma reduction. This case report represents an exception to the paradigm that in the presence of a macroadenoma and normal PRL levels (avoiding the “hook effect”), a prolactinoma can be discarded. Moreover, it stresses the importance of comprehensive, regular, and lifelong surveillance of patients with NFPAs and the close monitoring of serum PRL.
Keywords:
pituitary adenomas; non-secreting pituitary adenoma; prolactinomas; macroprolactinomas
1. Introduction
Pituitary adenomas (PAs) are the most common neoplasms of the sellar region, and taking into account their hormonal activity in vivo, they can be divided in functioning PAs, which secrete pituitary hormones autonomously, and nonfunctioning pituitary adenomas (NFPAs), which are not associated with increased hormone secretion . PAs are present in approximately 10% of people in the general population on imaging studies or at autopsy. The great majority of these are microadenomas . Clinically nonfunctioning PAs are present in approximately 1 in 1100 individuals in the general population. Of these, 48% are macroadenomas [3,4]. Macroadenomas can cause mass effect, consisting of visual field defects, headache, or hypopituitarism. People with hypopituitarism, including GH deficiency, due to PA have an approximately two-fold increase in mortality compared with the general population [5,6]. NFPAs account for approximately 30% of PAs that come to clinical attention and cause symptoms, and they arise from cells of gonadotroph lineage in about 80% of patients . Prolactin-secreting PA (prolactinomas) account for approximately 53% of PAs and arise from cells of lactotroph lineage Compared with men, women are approximately ten times more likely to develop prolactinomas and are diagnosed at a younger age; however, the tumors tend to be smaller . Serum prolactin (PRL) should be measured to evaluate for prolactinoma . These tumors secrete PRL in proportion to their size. A PA with a circulating PRL greater than 250 ng/mL is considered a prolactinoma. Other causes of an elevated serum PRL level, with values below 250 ng/mL, include drugs, pregnancy, kidney failure, liver failure, polycystic ovary syndrome, and primary hypothyroidism. Alternatively, PRL levels may be elevated due to a “stalk effect”. PRL secretion is normally inhibited by hypothalamic dopamine. Patients with an NFPA compressing the pituitary stalk may develop hyperprolactinemia due to disruption of dopamine outflow through the pituitary stalk (stalk effect) . In patients with stalk effect, serum prolactin is generally below 150 ng/mL . Artifactually low values of PRL occur when a very high serum PRL level saturates both the capture and signal antibodies used in immunoradiometric and chemiluminescent assays, preventing the binding of the two antibodies. This phenomenon is called the “hook effect”. Most authors agree that in the presence of a macroadenoma and a confirmed normal PRL level (after avoiding the “hook effect”), a prolactinoma can be discarded .
We present the clinical case of an initially non-secreting pituitary adenoma that changed to a prolactinoma, an entity that, as far as we know, has not been previously described in clinical practice.
2. Materials and Methods
The patient is a 53-year-old man whose most prominent personal background consisted in a cured hepatitis C virus infection and gastroesophageal reflux. He was initially evaluated in July 2018 in the Emergency Room for a headache that he reported to be more intense than usual, plus vomiting and fever of three days of evolution. He did not notice any visual deficit nor any other neurological symptom. At the physical examination, he presented a score of 15 on the Glasgow Coma Scale (GCS), cranial pairs examination was normal, and the perimetry test was also apparently normal.
A computed tomography (CT) scan was performed, revealing a pituitary macroadenoma with cystic–hemorrhagic degeneration and compression of the optic chiasm. To complete the study, magnetic resonance imaging (MRI) was requested. MRI showed a macroadenoma (20 × 20 × 18 mm) with signs of pituitary apoplexy and an invasion of not only the optic chiasm but also the hypothalamus and partially the left cavernous sinus (Figure 1).
Furthermore, in laboratory studies, no hormone hypersecretion was found (T4, TSH, FSH, LH, PRL, cortisol, and testosterone were measured). At that moment, the PRL value was 22.1 ng/mL. Regarding biochemical determinations, mild hyponatremia (Na 131 mEq/L) was noted. At this time, corticosteroid therapy was started at stress doses (hydrocortisone at 100 mg every 8 h intravenously). The patient underwent transsphenoidal surgery, and subtotal removal was achieved with glandular remnants on the bottom of the sellar region extended to the right cavernous sinus. The postoperative histological examination showed immunohistochemical expression of CAM 5.2, PRL, and ACTH.
The days after the surgery, mild hyponatremia was still present, with levels around 131–133 mEq/L, with no neurological symptoms associated. The level of PRL five days after surgery was 9.5 ng/mL. The thyroid and gonadotropic axes were also evaluated in this determination, showing a hormonal deficit of both axes at the central level. The patient was discharged with diagnoses of non-secretory pituitary macroadenoma with data of pituitary apoplexy, panhypopituitarism, and mild asymptomatic hyponatremia compatible with excessive fluid intake. Home treatment was hydrocortisone and levothyroxine.
For the next year, PRL was measured several times, obtaining normal or minimally elevated values repeatedly: 15.8 ng/mL (August 2018), 27.6 ng/mL (January 2019), 71.2 ng/mL (July 2019). Dilution of the serum of the patient was performed to avoid the “hook effect”.
Almost two years after the surgical procedure, the patient reported that he had noticed a progressive visual deficit in his right eye for three weeks. A perimetry test was performed revealing a nasal defect in the right eye. A new MRI scan was requested (June 2020) (Figure 2), showing adenomatous recurrence with right optoquiasthmatic compressive effect, so the patient underwent a new transsphenoidal surgery.
A probable right suprasellar adenomatous rest was evidenced in the post-surgery CT scan. This time, the postoperative histological examination showed fragments of extensively fibrous connective tissue without evidence of viable tumor cellularity, as well as respiratory submucosal glands without histological alterations.
At that time, the patient was sent to our pituitary unit, and PRL was measured again after the new procedure, showing a relevant increase in the PRL level (565.3 ng/mL). The laboratory test was repeated, this time obtaining a PRL level of 681.9 ng/mL. A dynamic PRL release test was also performed, and the results were consistent with the isolated determination (PRL: 791.8 ng; PRL 30′: 702.4 ng/mL; PRL 60′: 667 ng/mL).
3. Results
3.1. Final Diagnosis
According to laboratory studies and imaging tests, the final diagnosis was a macroprolactinoma.
3.2. Treatment
At the time, treatment with a dopaminergic agonist (cabergoline) was started (October 2020).
3.3. Outcome and Follow-Up
Only one month later, PRL levels were markedly decreased (0.4 ng/mL) (Table 1) and in the following perimetry test, a clear improvement in the visual deficit was shown (visual field index of 98% in the left eye and visual field index of 92% in the right eye).
On the other hand, in terms of tumor size reduction, the benefit was more subtle: the adenomatous rest moderately reduced its volume, particularly in its cranial part, so it was no longer in contact with the gyrus rectus and the overlying subcallosal area (Figure 3).
The patient is currently on cabergoline treatment at a dose of 0.5 mg twice a week, with a good clinical evolution, maintaining a normal prolactin level as well as normal visual acuity.
4. Discussion
We present a not previously described clinical case of a pituitary macroadenoma with initial repeatedly normal PRL levels that during clinical evolution changed to markedly increased circulating PRL levels, compatible with macroprolactinoma with a good clinical response to dopaminergic agonists.
In the presence of a pituitary macroadenoma, markedly increased PRL makes the diagnosis of prolactinoma [10,11]. In men, prolactinomas are usually large and invasive, with signs and symptoms of hypogonadism and mass effects being the most frequent clinical features . Additionally, according to the last World Health Organization classification , lactotroph tumors in men have a high probability of recurrence. In the diagnostic workup for prolactinoma, other causes of hyperprolactinemia should be excluded. A number of physiologic (stress, exercise, pregnancy, and breastfeeding) and pathologic conditions (cirrhosis, chronic kidney failure, primary hypothyroidism, compression of the pituitary stalk by a non-PRL-secreting pituitary tumor or different parasellar mass, and infiltration of the hypothalamus), as well as several drugs (antipsychotics, antidepressants, dopamine receptor blockers, dopamine synthesis inhibitors, and oral contraceptives) can induce symptomatic PRL level increase . Attention should be paid to medical history, concomitant medications, and biochemical assessment . In our patient, on initial clinical presentation, the PRL values were repeatedly normal, and the minimally elevated PRL values found on one occasion could have been due to compression of the pituitary stalk.
Diagnostic workup for prolactinomas may be complicated by several challenges. Assay errors, macroprolactinemia, and high-dose “hook effect” are all possible reasons for false-positive or false-negative PRL measurements . The “hook effect” is a possible explanation for the presence of artifactually low values of prolactin in the presence of a pituitary macroadenoma. Caution should be exercised in interpreting slightly increased serum PRL concentrations in the presence of a macroadenoma because of possible artifactually low values due to the “hook effect” [15,16,17]. This effect occurs when a very high PRL value, for example 4000 ng/mL or more, saturates both the capture and signal antibodies used in immunoradiometric and chemiluminescent assays, preventing the binding of the two antibodies in a “sandwich.” The result is an apparent PRL value that is only slightly increased, suggesting that the macroadenoma is an NFPA. The artifact can be avoided by repeating the analytical determination by performing dilution of serum . In the present case, we performed dilution of the serum of the studied patient to avoid this “hook effect”.
NFPAs that completely changed their phenotype have been reported. Fang et al. reported a patient with an initial NFPA. Several years later, the patient developed Cushing’s syndrome due to ACTH-dependent increased cortisol secretion, and the tumor extirpated during the fourth intervention showed marked immunopositivity for ACTH. To explain this event, some hypotheses have been formulated. The possibility that more than one tumor could be involved has been suggested in the differential diagnosis of patients with transformed-phenotype PAs . Multiple or double adenoma are presented synchronously (although sometimes asynchronously). Most of them are clinically silent, small, and discovered incidentally . Plurihormonal collision tumors from different lineages have been reported ; nevertheless, these pathological findings have not been described in cases reported as changing phenotype. Instead, multiple asynchronous PAs could be a plausible consideration in patients with transformed PAs. It is possible to assume a complex and multicausal mechanism. It has been proposed that a combination of gene alterations and several stimuli (drugs, pituitary surgery, or radiotherapy) may contribute to an accumulation of genetic changes, which may result in the functional variation observed in PAs . Some PAs originate in an uncommitted stem cell, which can differentiate into two separate cell types. Also, one cell type can “trans-differentiate” into another cell type as a result of subsequent mutations during tumor progression. Starting from this concept, Dessimoz et al. suggested the possible “trans-differentiation” into another cell type as an explanation for phenotype change in some PAs .
Silent PAs are tumors with positive staining for pituitary hormones or their transcription factors in the absence of hormonal hypersecretion. Silent somatotroph adenomas and silent lactotroph adenomas are rare . On the contrary, PRL-producing pituitary adenomas without elevated PRL were found in a study in 12.1% of pituitary adenomas, and most patients were resistant to dopamine agonist treatment . NFPAs are not associated with hormone hypersecretion, but sometimes, they even present slightly elevated PRL values . Increased pressure on the pituitary stalk is probably the main mechanism for moderately increased PRL in NFPAs, although other factors, like young age, female sex, and decreased thyroid function, could participate . An important aspect to be considered is that clinical or subclinical hemorrhage or infarction could modify the tumor hormone secretion and induce the phenotype change . This could be a possibility in PAs with changing phenotype from functioning to nonfunctioning. However, it seems unlikely to explain the reverse transformation or cases with transformed phenotype . These aspects highlight the rarity of the present case.
Pituitary apoplexy may be the presenting feature of NFPAs or prolactinomas . Between 2% and 12% of patients with all types of adenoma experience apoplexy, and the diagnosis of pituitary tumor was unknown at the time of apoplexy in more than three out of four cases . Pituitary hemorrhage in prolactinomas has an overall prevalence rate of 6.8% and is significantly higher in macroprolactinomas (20.3%) compared with microprolactinomas (3.1%). These data show that incidental hemorrhage in prolactinomas is not uncommon. The present clinical case presentation was consistent with a clear pituitary apoplexy.
In our view, the most plausible explanation for the present case is that the pituitary apoplexy with hemorrhage of the macroadenoma at its initial presentation, with necrosis of the tumor lactotroph cells, provoked the normalization of the circulating PRL levels, which were previously elevated. Subsequently, as tumor regrowth slowly developed, marked hyperprolactinemia appeared. Thus, according to this explanation, no tumor transformation occurred. However, from a clinical point of view, the management of pituitary adenomas with markedly increased prolactin (prolactinoma) and with normal prolactin (NFPA) is completely different. We cannot completely rule out the possibility of an NFPA that completely changed its pathological and clinical phenotype to a prolactinoma or the presence of double adenomas not detected on pathological examination, as previously discussed.
This case report represents an exception to the paradigm that in the presence of a macroadenoma and normal PRL levels (avoiding the “hook effect”), a prolactinoma can be discarded. Additionally, this case report shows the need for ongoing surveillance for these tumor types and the need for clinical suspicion for a different hormonal type. Moreover, it stresses the importance of comprehensive, regular, and lifelong vigilance of patients with NFPAs and the close monitoring of serum PRL concentrations.
Another secondary aspect that could be considered in relation to the medical treatment of PAs with dopamine agonists is cabergoline treatment. which has been employed as treatment for NFPAs [29,30,31], the present case highlights the possibility that some of the good results of such treatment for NFPAs [29,30] could be due to the inclusion of macroprolactinoma with normal circulating prolactin levels in some of those studies.
5. Conclusions
This case represents an exception to the widely accepted paradigm that in the presence of a macroadenoma and a confirmed normal PRL level (after avoiding the “hook effect”), a prolactinoma can be discarded. Moreover, it stresses the importance of comprehensive, regular, and lifelong surveillance of patients with NFPAs and the close monitoring and control of serum PRL concentrations.
Author Contributions
Conceptualization: F.C.; Methodology: C.S.-V., N.P.-P. and F.C.; Investigation, C.S.-V., N.P.-P. and F.C.; Writing—Original Draft Preparation, C.S.-V., N.P.-P. and F.C.; Writing—Review and Editing, C.S.-V., N.P.-P. and F.C.; Supervision, F.C. All authors have read and agreed to the published version of the manuscript.
Funding
This research received no external funding.
Institutional Review Board Statement
Not applicable. The study was conducted according to the guidelines of the Declaration of Helsinki.
Informed Consent Statement
Written informed consent was obtained from the patient to publish this paper.
Data Availability Statement
The data used during the current study are available from the corresponding author upon a reasonable request.
Conflicts of Interest
The authors declare no conflicts of interest.
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Figure 1. MRI, July 2018. Macroadenoma (20 × 20 × 18 mm) with signs of pituitary apoplexy and an invasion of not only the optic chiasm but also the hypothalamus and partially the left cavernous sinus. The tumor is delineated in all imaging studies. MRI, magnetic resonance imaging.
Figure 1. MRI, July 2018. Macroadenoma (20 × 20 × 18 mm) with signs of pituitary apoplexy and an invasion of not only the optic chiasm but also the hypothalamus and partially the left cavernous sinus. The tumor is delineated in all imaging studies. MRI, magnetic resonance imaging.
Figure 2. MRI, June 2020. Adenomatous recurrence with right optoquiasthmatic compressive effect, MRI, magnetic resonance imaging.
Figure 2. MRI, June 2020. Adenomatous recurrence with right optoquiasthmatic compressive effect, MRI, magnetic resonance imaging.
Figure 3. MRI, June 2021. The adenomatous rest reduced its volume, particularly in its cranial part, so it was no longer in contact with the gyrus rectus and the overlying subcallosal area. MRI, magnetic resonance imaging.
Figure 3. MRI, June 2021. The adenomatous rest reduced its volume, particularly in its cranial part, so it was no longer in contact with the gyrus rectus and the overlying subcallosal area. MRI, magnetic resonance imaging.
Table 1. Prolactin levels, radiological evolution, and clinical management.
Table 1. Prolactin levels, radiological evolution, and clinical management.
| Date | July 2018 | January 2019 | July 2019 | June 2020 | October 2020 | December 2020 | June 2021 |
:---: :---: :---: :---: |
| Prolactin (ng/mL) | 22.1 | 27.6 | 71.2 | | 681.9 | 0.4 | 1.8 |
| MRI | Macroadenoma (20 × 20 × 18 mm) with signs of pituitary apoplexy. Invasion of the optic chiasm, hypothalamus, and the left cavernous sinus. | 5 × 2 mm nodular structure that could be related to remnant glandular tissue. Persistent left deviation of the pituitary stalk. | | Tumor growth (16 × 14 × 11 mm). Significant compression of the optic nerve and the optic chiasm. | | Discrete decrease in volume of the tumor (13 × 12 × 8 mm). No compression of the optic nerve and the chiasm. | Volume reduction (13 × 10 × 6 mm). The structure is no longer in contact with the gyrus rectus and the subcallosal area. |
| Treatment | First surgical procedure | | | Second surgical procedure | Cabergoline started | | |
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Santiago-Vazquez, C.; Palacios-Paino, N.; Cordido, F. A Non-Secreting Pituitary Adenoma That Changed to a Prolactinoma. Clin. Pract.2024, 14, 1310-1318.
AMA Style
Santiago-Vazquez C, Palacios-Paino N, Cordido F. A Non-Secreting Pituitary Adenoma That Changed to a Prolactinoma. Clinics and Practice. 2024; 14(4):1310-1318.
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Santiago-Vazquez, Cristina, Nuria Palacios-Paino, and Fernando Cordido. 2024. "A Non-Secreting Pituitary Adenoma That Changed to a Prolactinoma" Clinics and Practice 14, no. 4: 1310-1318.
APA Style
Santiago-Vazquez, C., Palacios-Paino, N., & Cordido, F. (2024). A Non-Secreting Pituitary Adenoma That Changed to a Prolactinoma. Clinics and Practice, 14(4), 1310-1318.
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Santiago-Vazquez, C.; Palacios-Paino, N.; Cordido, F. A Non-Secreting Pituitary Adenoma That Changed to a Prolactinoma. Clin. Pract.2024, 14, 1310-1318.
AMA Style
Santiago-Vazquez C, Palacios-Paino N, Cordido F. A Non-Secreting Pituitary Adenoma That Changed to a Prolactinoma. Clinics and Practice. 2024; 14(4):1310-1318.
Chicago/Turabian Style
Santiago-Vazquez, Cristina, Nuria Palacios-Paino, and Fernando Cordido. 2024. "A Non-Secreting Pituitary Adenoma That Changed to a Prolactinoma" Clinics and Practice 14, no. 4: 1310-1318.
APA Style
Santiago-Vazquez, C., Palacios-Paino, N., & Cordido, F. (2024). A Non-Secreting Pituitary Adenoma That Changed to a Prolactinoma. Clinics and Practice, 14(4), 1310-1318.
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9059 | https://www.bbc.co.uk/bitesize/guides/zyt8qty/revision/2 | Sharing in a given ratio - Ratio - WJEC - GCSE Maths Numeracy (WJEC) Revision - BBC Bitesize
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Ratio - WJEC Sharing in a given ratio
Ratios show the relationship between two values. They may be in direct proportion and increase as the other increases, or they can be in inverse proportion; as one increases the other decreases.
Part ofMaths Numeracy (WJEC)Number
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Understanding ratio
Sharing in a given ratio
Direct proportion - Intermediate and Higher tier
Inverse proportion - Intermediate and Higher tier
Example for Intermediate and Higher tier
Further example for Higher tier
Sharing in a given ratio
A ratio can also be used to share a quantity into parts.
Example 1
Figure caption,
CREDIT: Getty/yorkfoto
Rebeckah and Amy share £280 in the ratio 5:2. How much money will they each receive?
1. Add up the ratio to find the total number of parts:
5 + 2 = 7 parts
2. Divide the total amount by the number of parts:
£280 ÷ 7 = £40
Each part is worth £40
3. Multiply by the ratio to find each person’s share:
5 × £40 = £200 (Rebeckah’s share)
2 × £40 = £80 (Amy’s share)
4. Check these add up to the original amount:
£200 + £80 = £280
Example 2
Figure caption,
CREDIT: Paul Bradbury
David and Paul share the cost of a football season ticket in the ratio 3:7. If David pays £105, how much does the season ticket cost in full?
David pays three parts = £105: £105 ÷ 3 = £35 = 1 part
The cost had been split into 3 + 7 = 10 parts
£35 × 10 = £350
The total cost of the ticket is £350.
Question
An alloy consists of magnesium, zinc and copper in the ratio 5:3:4. If an alloy compound weighs 720 g, how much of this would be magnesium, zinc and copper?
Show answer Hide answer
1. Add the total number of parts:
5 + 3 + 4 = 12 parts
2. Find out how much one part is:
720 g ÷ 12 = 60 g
3. Multiply each ratio by 60:
60 g × 5 = 300 g of magnesium
60 g × 3 = 180 g of zinc
60 g × 4 = 240 g of copper
4. Check they add up to 720 g:
300 + 180 + 240 = 720 g
Question
To make pink paint I mix red and white paint in the ratio 4:5. If I use 600 ml of red paint, how many litres of paint will I make?
Show answer Hide answer
Red paint = 4 parts = 600 ml
600 ml ÷ 4 = 150 ml = One part
Total parts = 4 + 5 = 9
150 ml × 9 = 1,350 ml of paint
Convert to litres by ÷ 1,000: 1,350 ÷ 1,000 = 1.35 litres of paint made
Next page Direct proportion - Intermediate and Higher tier
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9060 | https://scienceready.com.au/pages/standing-wave-in-pipes?srsltid=AfmBOortixLykj_6730bHRu0FWqUULnGFK1m-wRExLFXf2sqhqcX7K-V | Standing Waves in Pipes: Everything Explained –HSC Physics – Science Ready
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Standing Wave in Pipes
This topic is part of the HSC Physics course under the section Sound Waves.
HSC Physics Syllabus
investigate and model the behaviour of standing waves on strings and/or in pipes to relate quantitatively the fundamental and harmonic frequencies of the waves that are produced to the physical characteristics (eg length, mass, tension, wave velocity) of the medium (ACSPH072)
Standing Waves in Closed/Open Pipes Explained
What are Standing Waves?
Contrary to progressive waves, standing waves, or stationary waves, don't travel or propagate. They form when two waves of the same frequency interfere while moving in opposite directions. The resultant wave appears to "stand still", that is, it continues to oscillate but does not propagate from one place to another.
The following video shows the formation of a standing wave (black) from two progressive waves of the same frequency travelling in the opposite direction. Note that the standing wave oscillates, but does not seem to move to the left nor the right.
Waves do not need to have the same amplitude to produce a standing wave. However, when their amplitudes are equal, the resultant standing wave is the most pronounced, and this is more commonly seen.
Antinodes and nodes of a standing waves are the result ofsuperpositionof progressive waves.
Constructive Interference: At some points, the two waves add up to produce areas of maximum displacement, known as antinodes.
Destructive Interference: At other points, they cancel each other out, resulting in areas of no displacement, called nodes.
This alternating pattern of nodes and antinodes is a hallmark of standing waves.
How are Standing Waves Formed?
Standing waves are typically formed due to the interference of two waves with the same frequency, travelling in opposite directions within a bounded or confined medium.
This interference can be a result of:
Reflection: When a wave reflects off a boundary or barrier and interferes with incoming waves of the same kind. For example, a wave moving down a string reflects back upon reaching a fixed end, creating a standing wave.
The production of standing waves using sound in pipes and tubes is discussed separately here.
Standing Waves on Strings
Standing waves can be produced by the reflection of a progressive wave in a string system that is either fixed or free to move at one end. The frequency of such a progressive wave must be specific to produce the standing wave for reasons we discussed above.
Standing wave on strings is discussed separately here
Standing Waves in Pipes
Pipe with two open ends
The lowest frequency required to produce a standing wave is thefundamental frequency,f 0 f 0, also called the first harmonic.
Frequencies required to form the standing wave are integral multiples of the fundamental frequency:
f=n f 0 where n = 1, 2, 3...f=n f 0 where n = 1, 2, 3...
The frequency twice the fundamental frequency is called thesecond harmonic. This is also the first frequency after the fundamental frequency that can produce a standing wave, hence its alternative name,first overtone.
The next frequency (three times the fundamental) is called the third harmonic and second overtone.
The wavelength of the standing wave decreases as its frequency increases such that:
λ=2 n L where n = 1, 2, 3...λ=2 n L where n = 1, 2, 3...
where L is the length of the pipe.
Standing waves form in a pipe with both ends open such that an antinode (point at which maximum displacement occurs) if present at each open end. The number of nodes (points at which there are no oscillation) also corresponds to the harmonic number (n). For example, a second harmonic standing wave has two nodes.
Pipe with one closed end
Frequencies required to form a standing wave in this system are given by
f=n f 0 where n = 1, 3, 5...f=n f 0 where n = 1, 3, 5...
Note that n must be an odd integer when one end of the pipe is closed.
The first frequency after the fundamental frequency is the third harmonic, which is three times the fundamental frequency.
This also means there are no second (n= 2), fourth (n = 4) harmonics for standing waves in this system.
The wavelength of standing waves is given by
λ=4 n L where n = 1, 3, 5...λ=4 n L where n = 1, 3, 5...
where L is the length of the pipe.
In a pipe with one open and one closed end, an antinode again occurs at the open end, and a node occurs at the closed end.
Example
The fundamental frequency of a standing wave in an open pipe is 300 Hz. Calculate:
(a) The length of the pipe
(b) The frequency of the 3rd harmonic
(c) The length required to produce the same fundamental frequency if the pipe is closed.
Solution to part (a):
The wavelength is given by
λ=v f λ=v f
λ=340 300=1.13 m λ=340 300=1.13 m
The relationship between wavelength and the length of the open pipe for the first harmonic is given by
λ=2 L λ=2 L
L=0.57 m L=0.57 m
Solution to part(b):
In an open pipe, the frequency of the third harmonic is three times the fundamental frequency:
f=3×300=900 Hz f=3×300=900 Hz
Solution to part (c):
The relationship between wavelength and length of a closed pipe for the fundamental frequency is given by
l a m b d a=4 L l a m b d a=4 L
L=0.283 m L=0.283 m
RETURN TO MODULE 3:Waves and Thermodynamics
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9061 | https://www.johndcook.com/blog/2009/01/09/log-concave-functions/ | Log concave functions
Log concave functions have some very interesting and useful properties. I’ll list some of these shortly after a three definitions.
A function is convex if the line segment joining two points on the graph lies above the graph. In symbols, f(x) is convex if for every t between 0 and 1, and for every x and y,
f(tx + (1 − t)y) ≤ t f(x) + (1 − t) f(y).
A function f(x) is concave if −f(x) is convex. Equivalently, the line connecting two points on the graph lies below the graph. In symbols, reverse the direction of the inequality for convexity.
A function f(x) is log concave if log( f(x) ) is concave.
The basic properties of convex functions are obvious. It’s easy to show that the sum of two convex functions is convex, the maximum of two convex functions is convex, etc. It would seem that log concave functions would be unremarkable because they are so simply related to convex functions. But they have some surprising properties.
The Laplace transform of a non-negative function is a log convex.
The running average of a log concave function is also log concave.
If a random variable has a log concave PDF (probability density function), then the CDF (cumulative distribution function) is also log concave.
The convolution of two log concave functions is log concave.
If two independent random variables have log-concave PDFs, then their sum also has a log concave PDF.
A few properties of log concave functions do follow easily from analogous properties of convex and concave functions.
The product of log concave functions is log concave.
A function f(x) is log concave if and only if for every t between 0 and 1 and for every x and y,
f(tx + (1 − t)y) ≥ f(x)t f(y)1−t
A smooth function f(x) is log concave if and only if the inequality f ∇2 f ≤ ∇ f ⋅ ∇ f holds everywhere.
Here are a few specific functions that are log concave.
The exponential function eax
The inverse logit function ex/(ex + 1)
The function ∏ xi / ∑ xi
The function xa for a ≥ 0
Determinant is a log concave function on positive definite matrices
For more, see Stephen Boyd’s book Convex Optimization.
10 thoughts on “Log concave functions”
gb
Great post.
One minor issue – I haven’t seen the function exp(x)/(1+exp(x)) called the inverse logic function before (which might simply be ignorance on my part).
(I have seen it called the inverse logit function or the logistic function. )
However, given “logit” and “logic” only differ by one character, I did wonder if it might have been a simple typo.
2. John
Thanks! You were right: “logic” was a typo for “logit.” I’ve corrected the post.
3. seh chun
Hi,
This post is great and useful. About a point you have put, saying “The running average of a log concave function is also log concave”. May I know is there any book / reference that I can find more information about that?
Thanks.
4. John
seh chun: For a reference, see the book by Stephen Boyd mentioned at the bottom on the post.
5. KW
John,
A concave function could be increasing or decreasing fast in one segment of the curve and slow in other segments, while the upper half of a circle seems increasing and decreasing constantly. My question is how to interpret this in a mathematics form.
Thanks.
KW
6. Brooks
Thanks! I have a question about “log concave”.
Assume Q(x) is a log concave and it denotes the CDF.
I can’t be sure that “p+(1-2p)Q(x)” is also a log concave function, if given p (0<p<1).
Can you give some analysis for me , please?
7. John
I doubt your conjecture holds. Let f(x) = p + (1 – 2p)Q(x) and test whether f(x) f”(x) ≤ (f'(x))^2 for all p. I believe the inequality does not hold for p = 3/8.
I only did a hurried calculation and so I may have made errors, but I believe that approach to resolving the question will work.
8. Brooks
Hi,
Q(x) =erf(x);
I find a intersting result. When p=1-2p, f =p+(1-2p)Q(x) will be a log concave function. The inequality always holds.
%%%%%
clear
clc
close all
x=[-10:0.01:10];
y=1/3+1/3.erf(x);
f1= 1/3 1/(sqrt(2pi)) exp(-x.^2/2);
f2= -1/3 1/(sqrt(2pi)) exp(-x.^2/2) . x;
L=f2.y;
R=f1.^2;
boot=(R>=L); %% (if R>=L, the value will be set to 1, or set to 0)
plot(-10:0.01:10,L,’+’,-10:0.01:10,R,’’,-10:0.01:10,boot,’s’)
%%%
Thanks.
9. wen
Hi,
Thanks! I have a question: is the sum of infinitely many log concave functions a log concave function?
Thanks!
10. If you would like a guide through Steven Boyd’s book, lectures from the two courses he teaches on convex optimization are available on youtube and other course material is on the Stanford web site |
9062 | https://www.shmoop.com/study-guides/systems-equations-inequalities/three-d-graphs-help.html | Graphing Linear Equations in Three Dimensions at a Glance
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Graphing Linear Equations in Three Dimensions at a Glance
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The x and y variables of a system of linear equations match up with the x- and y-axes on a graph. There are two variables for two dimensions. When we add in a third variable, z, that means we travel into The Third Dimension. Wooooaoooo.
Linear equations in two dimensions are represented on a graph as lines. Guess what linear equations with three variables look like on a graph? They look like planes—and not the kind with pilots.
To graph linear equations in three variables, we need to use a 3D coordinate system like the one below.
To make a 3D coordinate system, first draw a normal 2D coordinate plane like usual. Then add in a brand-spankin'-new z-axis through the origin—only it's popping out of the page in brilliant, stereoscopic 3D.
Think of it like this; x and y are lying flat, like a piece of paper on a table, and z is sticking straight up. That's much better than poking a stick through our paper to represent the z-axis. It's hard to turn in homework that way.
Sample Problem
Plot the point (-3, 1, 4).
We need to draw the coordinate system we'll be placing this point in. A point without a system to put it in is like a star without a sky. If it's not in the sky, that means it's here on Earth. That would get very hot, very quickly.
To start off, we draw our 2D plane as usual, labeling them y and z. Then, the x-axis is drawn like the line y = x. It's just easier to draw it this way, flat on a page, to give the impression of three dimensions.
For our purposes here, the positive part of the x-axis is in the lower left corner, in Quadrant III. The negative part of the x-axis is in the upper-right corner, Quadrant I. Check out the numbers; they don't lie.
Next comes putting the point on the graph. Think of the x- and y-axes as being flat on the page, and the z-axis as straight up, off the page. It can't actually grab you, though, don't worry. It might take a while to see it at first, but practice makes better.
Find the (y, z) points first, like we would x and y before, and then shift the point "forward" or "backwards" by x.
This point might look to you like it is in a weird spot. Remember, we're trying to cram three dimensions into two. A bit of awkwardness is to be expected. Again, looking at lots of different graphs will help.
Sample Problem
Graph 4 x – 3 y + 6 z = 24.
This equation has three variables, so it represents the three-dimensional equivalent of a line: a plane. To make the graph, we'll need to make three interceptions, er, find three intercepts. Sorry, we've got football on the brain.
To find the x-intercept, plug y = 0 and z = 0 into the equation. Yep, both of them are 0 at the same time. Remember what an intercept actually is: it's the point where the equation crosses that particular axis.
4 x – 3(0) + 6(0) = 24
4 x = 24
x = 6
We can represent the x-intercept as an ordered triplet, (6, 0, 0). There are too many variables for it to be an ordered pair anymore. sniff They grow so fast at this age.
To find the y-intercept, we plug x = 0 and z = 0 into the equation.
4(0) – 3 y + 6(0) = 24
-3 y = 24
y = -8
This is (0, -8, 0) to its friends.
Now to find the z-intercept, the new kid on the block. Plug in x = 0 and y = 0:
4(0) – 3(0) + 6 z = 24
6 z = 24
z = 4
We want to plot the three intercepts: (6, 0, 0), (0, -8, 0), and (0, 0, 4). First, though, we need to draw the coordinate plane we'll be placing them on. Remember that x and y have been shuffled around.
Now for our intercepts. For x, it will be 6 points in the positive direction; that's "towards" us, along what would be y = x in a 2D graph. The y-intercept is hanging out 8 to the left, which is negative for it now. The z-intercept lies 4 point straight up.
At this point, we connect the intercepts to form a triangular plane. Pretend it's a "test plane" from Area 51. Don't expect anybody to confirm its existence.
We've drawn this like a triangle, but don't be fooled. This represents just one part of the whole plane, which actually extends on and on far past where we got lazy and stopped drawing. It passes through our triangle, though.
Graphing in three dimensions is an exercise in optical illusions and imagination. And headaches, at least at first. Because paper is two-dimensional, we can't properly depict a 3D object on it. We have to trick our brains into seeing the image.
Sample Problem
Graph -10 x + 5 y – 10 z = 20
Even though we have more variables, finding the intercepts of an equation is still a straightforward affair. Set everything but the variable of interest equal to 0.
For x we have:
-10 x + 5(0) – 10(0) = 20
-10 x = 20
x = -2
The y-intercept has x and z sitting on the sidelines.
-10(0) + 5 y – 10(0) = 20
5 y = 20
y = 4
With x and y at 0, it's up to the z-intercept to get to solve the case and save the day.
-10(0) + 5(0) –10 z = 20
-10 z = 20
z = -2
We have the points (-2, 0, 0), (0, 4, 0), and (0, 0, -2). Put 'em on the board. One point is 2 towards us, another is 4 to the right, and the last is 2 down from the origin.
Finally, connect the intercepts to form a triangular plane. Airplane noises optional.
We personally think the airplane noises were crucial to our success.
Sample Problem
Graph 3 x + 3 y – 6 z = 18
We start off by catching our three intercepts. Use any tools at our disposal: comically oversized nets, microphones, or setting the other two variables equal to 0.
To find the x-intercept, plug y = 0 and z = 0 into the equation.
3 x + 3(0) – 6(0) = 18
3 x = 18
x = 6
To find the y-intercept, plug x = 0 and z = 0 into the equation.
3(0) + 3 y – 6(0) = 18
3 y = 18
y = 6
To find the z-intercept, plug x = 0 and y = 0 into the equation.
3(0) + 3(0) – 6 z = 18
-6 z = 18
z = -3
Next, plot the three intercepts: (6, 0, 0), (0, 6, 0), and (0, 0, -3).
Finally, connect the intercepts to form a triangular plane.
Call the captain, we've found the plane. It's still in one piece, even.
Examples
Exercises
Prev
1
2
Next
Example 1
Graph 6 x + 3 y – 4 z = 12.
The x-intercept is where the plane passes through the x-axis, and so y = 0 and z = 0.
6 x + 3(0) – 4(0) = 12
6 x = 12
x = 2
Then we find the y-intercept:
6(0) + 3 y – 4(0) = 12
3 y = 12
y = 4
And finally the z-intercept.
6(0) + 3(0) – 4 z= 12
-4 z = 12
z = -3
The intercepts are (2, 0, 0), (0, 4, 0), and (0, 0, -3). Connect them up inside the graphing cube.
Show Next Step
Example 2
Graph -2 x – 2 y + 14 z = 14.
The x-intercept:
-2 x – 2(0) + 14(0) = 14
-2 x = 14
x = -7
The y-intercept:
-2(0) – 2 y + 14(0) = 14
-2 y = 14
y = -7
The z-intercept:
-2(0) – 2(0) + 14 z = 14
14 z = 14
z = 1
Now we have three intercepts: (-7, 0, 0), (0, -7, 0), and (0, 0, 1). Graph them, and then get out of Dodge.
Show Next Step
Prev
1
2
3
4
5
Next
Exercise 1
Graph 5 x + 4 y – 10 z = 20.
Show AnswerGimme a Hint
Hint
The intercepts are (4, 0, 0), (0, 5, 0), and (0, 0, -2).
Answer
Exercise 2
Graph 6 x – 4 y + 12 z = 36.
Show AnswerGimme a Hint
Hint
The intercepts are (6, 0, 0), (0, -9, 0), and (0, 0, 3).
Answer
Exercise 3
Graph x + y +z = 10.
Show AnswerGimme a Hint
Hint
All of the intercepts are at 10.
Answer
Exercise 4
Graph -3 x – 3 y – 7 z = -21.
Show AnswerGimme a Hint
Hint
The intercepts are at (7, 0, 0), (0, 7, 0), and (0, 0, 3).
Answer
Exercise 5
Graph 2 x + 6 y– 2 z = 18.
Show AnswerGimme a Hint
Hint
The intercepts are (9, 0, 0), (0, 3, 0), and (0, 0, -9).
Answer
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W hy's T his F unny?
CLOSE |
9063 | https://www.quora.com/Given-four-letters-a-b-c-d-how-many-combinations-can-you-make-and-what-would-the-formula-be | Something went wrong. Wait a moment and try again.
Combinations Ma
Solution Counting
Combinatorics of Words
Permutations and Co
Combinations of Events
Algorithms, Combinatorics
Combinatorics C
5
Given four letters (a, b, c, d), how many combinations can you make, and what would the formula be?
·
The formula to calculate the number of combinations that can be made with 4 letters (a, b, c, d) is:
n! / (k! (n-k)!)
Where:
- n is the total number of elements (in this case, 4 letters)
- k is the number of elements being combined at a time (in this case, the number of letters in each combination)
- ! represents the factorial operation
To find the total number of combinations:
- When k = 1 (single letter combinations), the formula is: 4! / (1! (4-1)!) = 4
- When k = 2 (2-letter combinations), the formula is: 4! / (2! (4-2)!) = 6
- When k = 3 (3-letter combinations), the formula is: 4! / (
The formula to calculate the number of combinations that can be made with 4 letters (a, b, c, d) is:
n! / (k! (n-k)!)
Where:
- n is the total number of elements (in this case, 4 letters)
- k is the number of elements being combined at a time (in this case, the number of letters in each combination)
- ! represents the factorial operation
To find the total number of combinations:
- When k = 1 (single letter combinations), the formula is: 4! / (1! (4-1)!) = 4
- When k = 2 (2-letter combinations), the formula is: 4! / (2! (4-2)!) = 6
- When k = 3 (3-letter combinations), the formula is: 4! / (3! (4-3)!) = 4
- When k = 4 (4-letter combinations), the formula is: 4! / (4! (4-4)!) = 1
Adding up all the combinations:
4 + 6 + 4 + 1 = 15
Therefore, with 4 letters (a, b, c, d), you can make a total of 15 different combinations.
Ellis Cave
40+ years as an Electrical Engineer
·
Author has 7.9K answers and 4.3M answer views
·
6y
Using the J programming language:
The total number of combinations of four items is 1, as combinations are not order specific.
<<<>>
However if you want to find the number of permutations of the four characters ‘abcd’ this can be found by taking the factorial of the number of items:
4! = 1234 = 24
or using J:
!4 - Factorial 4
24
However if you want to see all the permutations, the formula is:
a=.'abcd'{~ perm 4
24
The answer is still 24 combinations, but now we have stored all the permutations in a, so we can list all the permutations:
{a
│abcd│abdc│acbd│acdb│adbc│adcb│bacd│badc│bcad│bcda│bdac│bdca│cab
Using the J programming language:
The total number of combinations of four items is 1, as combinations are not order specific.
<<<>>
However if you want to find the number of permutations of the four characters ‘abcd’ this can be found by taking the factorial of the number of items:
4! = 1234 = 24
or using J:
!4 - Factorial 4
24
However if you want to see all the permutations, the formula is:
a=.'abcd'{~ perm 4
24
The answer is still 24 combinations, but now we have stored all the permutations in a, so we can list all the permutations:
{a
│abcd│abdc│acbd│acdb│adbc│adcb│bacd│badc│bcad│bcda│bdac│bdca│cabd│cadb│cbad│cbda│cdab│cdba│dabc│dacb│dbac│dbca│dcab│dcba│
<<<<>>>>
How it works:
perm is a J verb (a function implemented using several J primitives that together generate permutation indices:
perm =. i.@! A. i.
perm 4 generates the permutation indices for 4 items:
{perm 4
│0 1 2 3│0 1 3 2│0 2 1 3│0 2 3 1│0 3 1 2│0 3 2 1│1 0 2 3│1 0 3 2│1 2 0 3│1 2 3 0│1 3 0 2│1 3 2 0│2 0 1 3│2 0 3 1│2 1 0 3│2 1 3 0│2 3 0 1│2 3 1 0│3 0 1 2│3 0 2 1│3 1 0 2│3 1 2 0│3 2 0 1│3 2 1 0│
{~ uses the permutation indices to select characters from the string ‘abcd’ which is then stored in a and displayed:
{a=.'abcd'{~ perm 4
│abcd│abdc│acbd│acdb│adbc│adcb│bacd│badc│bcad│bcda│bdac│bdca│cabd│cadb│cbad│cbda│cdab│cdba│dabc│dacb│dbac│dbca│dcab│dcba│
a counts the number of items in a:
a
24
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Chris McClinch
Senior IT Management Consultant (2012–present)
·
Author has 2.8K answers and 4M answer views
·
6y
If I can repeat letters, I can make 256 combinations (4^4).
If they need to be unique, I can make 24 combinations (4!).
Ellis Retting
Former Paralegal (1997–2010)
·
Author has 157 answers and 422.4K answer views
·
7y
Originally Answered: How many different 4 letter combinations can be created from the letters A,B, C, and D?
·
Your question implies there cannot be replacement of the individual letters, that is, the same letter cannot appear more than once. If this is the case, it is 24. This is because you have a set of 4, and all items will be used.
That makes it 4!= 4 x 3 x 2 x 1 = 24.
The possible combinations, without using a letter more than once, are:
ABCD BCDA CDAB DBAC
ABDC BCAD CDBA DBCA
ACDB BDAC CABD DCBA
ACBD BDCA CADB DCAB
ADBC BACD CBAD DABC
ABDC BADC CBDA DACB
If you are use the letter more than once, it is 44 = 256
Matias Relyea (贺彦龙）
Devoted Learner of Mathematics
·
Author has 205 answers and 347.6K answer views
·
5y
A permutation describes the total number of possible arrangements of a set of variables or numbers:
For example, find as many arrangements of 1,2,3 and 4 as possible. First, you would use the permutation formula (n!)/(n-r)!=nPr to determine the number of permutations. First, n represents the number of values you are attempting to arrange, and r is the number of values that n goes in to.
For this example, n=4 and r=4. Therefore, the final answer would be
(4!)/(4–4)!=(4×3×2×1)/0!=(4×3×2×1)/1=24 (as the definition of 0! is 1)
Therefore, this set contains 24 unique arrangements of 1,2,3, and 4. Determ
A permutation describes the total number of possible arrangements of a set of variables or numbers:
For example, find as many arrangements of 1,2,3 and 4 as possible. First, you would use the permutation formula (n!)/(n-r)!=nPr to determine the number of permutations. First, n represents the number of values you are attempting to arrange, and r is the number of values that n goes in to.
For this example, n=4 and r=4. Therefore, the final answer would be
(4!)/(4–4)!=(4×3×2×1)/0!=(4×3×2×1)/1=24 (as the definition of 0! is 1)
Therefore, this set contains 24 unique arrangements of 1,2,3, and 4. Determining the number of permutations manually while guessing is less efficient but still effective as it will take alot of unnecessary time away.
However, if you encounter a problem that states that you need to arrange variables a,b,c, and d if you arrange it with two variables in each set, then you must understand that n=4 and r=2.
This question is completely different and will be brought up in the description of the combination.
The answer is
(4!)/(4–2)!=(4×3×2×1)/(2×1)=4×3=12
A combination has a completely different definition despite it's similarities to the permutation formula. It is defined by (n!)/[r!(n-r)!]=nCr and determines the number of different arrangements of a set not dependent on the order of the numbers or variables.
For example, determine the number of combinations of the set 1,2,3, and 4.
Immediately we can determine that there is one combination as there is not difference between 1,2,3,4 and 3,1,2,4. However,
Using math, n=4 and r=4 so
(4!)/[4!(4–4)!]=(4×3×2×1)/4!×0!)=(4×3×2×1)/(4×3×2×1)=1
If we were to complicate this and state that we need to determine the number of combinations of a,b,c, and d if eachs set contains two variables, the answer wouldn't be as obvious.
Therefore,
(4!)/[4!(4–2)!]=(4×3×2×1)/(4×3×2×1×2×1)=2
Hope this helps.
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Robert Nichols
Author has 5K answers and 15.5M answer views
·
6y
Originally Answered: How many different 4 letter combinations can be created from the letters A,B, C, and D?
·
How many different 4 letter combinations can be created from the letters A,B, C, and D?
You did not say, Can you use the same letter more than once?
Additionally you use the word “Combination” combination implies that order does not matter. ABCD is the same as ACBD when counting combinations.
So if you really do mean combinations there is only 1 combination without using repeats, since ABCD = ABDC = ACBD = … = DCBA
Combinations with repeats becomes a trial and error for me: 35
AAAA, AAAB, AAAC, AAAD, AABB, AACC, AADD, AABC, AABD, AACD, ABBB, ACCC, ADDD, ABBC, ABBD, ACCB, ACCD, ADDB, ADDC, ABCD, BBB
How many different 4 letter combinations can be created from the letters A,B, C, and D?
You did not say, Can you use the same letter more than once?
Additionally you use the word “Combination” combination implies that order does not matter. ABCD is the same as ACBD when counting combinations.
So if you really do mean combinations there is only 1 combination without using repeats, since ABCD = ABDC = ACBD = … = DCBA
Combinations with repeats becomes a trial and error for me: 35
AAAA, AAAB, AAAC, AAAD, AABB, AACC, AADD, AABC, AABD, AACD, ABBB, ACCC, ADDD, ABBC, ABBD, ACCB, ACCD, ADDB, ADDC, ABCD, BBBB, BBBC, BBBD, BBCC, BBDD, BBCD, BCCC, BDDD, BCCD, BCDD, CCCC, CCCD, CCDD, CDDD, DDDD
The number of permutations with no repeats is 4 x 3 x 2 x 1 = 24 Permutations. This is because you have 4 options for the first letter, but only 3 options for the 2nd letter, 2 choices for the third letter, and finally only 1 choice for the last letter.
The number of Permutations if you allow repeats is 4 x 4 x 4 x 4 = 256
Related questions
How many different 4 letter combinations can be created from the letters A,B, C, and D?
How many combinations can you make with ABC?
How many 4-letter combinations are there?
How many combinations of 4 colors are there?
How many combinations of 4 objects are taken two at a time?
Alberto Cid
M.S.E. in Telecommunications Engineering & Data Transmission, Technical University of Madrid (Graduated 2008)
·
Author has 2K answers and 3.8M answer views
·
5y
Originally Answered: How many different 4 letter combinations can be created from the letters A,B, C, and D?
·
The question is a bit confusing / ambiguous / unclear …
but Thao Dao and Andrei Ciobotaru asked me to answer this.
OK.
I will give my answer.
This question is not very clear because of several reasons:
On the one hand, the word “combination” has several meanings.
In the context of combinatorics or mathematics the word “combination” means “a selection of elements (or items) from a collection (or set) such that the order of selection does not matter.
This meaning is very similar to the first meaning from the dictionary: “the mixture you get when two or more things are combined”. For exam
Footnotes
COMBINATION | significado, definición en el Cambridge English Dictionary
Combination - Wikipedia
The question is a bit confusing / ambiguous / unclear …
but Thao Dao and Andrei Ciobotaru asked me to answer this.
OK.
I will give my answer.
This question is not very clear because of several reasons:
On the one hand, the word “combination” has several meanings.
In the context of combinatorics or mathematics the word “combination” means “a selection of elements (or items) from a collection (or set) such that the order of selection does not matter.
This meaning is very similar to the first meaning from the dictionary: “the mixture you get when two or more things are combined”. For example, when you combine rum with lemon juice you have a combination… which is basically the same as the combination of lemon juice with rum. That is, in most usual meaning of the word combination the order doesn’t matter.
That concept would be very similar to the concept of subset, except in the case where you select the same element (or item) more than once… which is called a “multiset” (order does not matter, but you can have the same element repeated several times)… and these are called “combinations with repetition”.
But the dictionary gives another meaning, too. It says a combination can also be
“a sequence of numbers or letters used to open a combination lock.”
And, also, the common use of letters is to make “words”, that is, sequences of letters where the order matters.
So, the question is confusing because it mixes together the word combination that usually has a preferred meaning with letters, which usually have a very different purpose.
On the other hand, the question doesn’t say whether the letters can be repeated or not.
Order does matter …
I take this as the first option because the most common use of letters is to make words, where order DOES matter.
1.1 With repetition
→ “[mathematical] N-tuples of m-sets”
→ or “[mathematical] permutations with repetition”
(this second name can be a bit confusing)
When you make words you can repeat letters, right??
(example: the word ABAC has 2 letters A … the letter A is repeated)
In this case it’s the number of 4-tuples of 4-sets, … of sets with 4 elements.
Or simply, words of length 4 from an alphabet of 4
44=4∗4∗4∗4=28= = 256 words with 4 letters each from {A, B, C, D} .
You would have from AAAA to DDDD
I guess this is the intended sense of the question.
1.2 Without repetition → “[mathematical] permutations”
Then it would be permutations of 4 elements:
P4 = 4! = 4321 = 24 orderings
The first could be ABCD (in alphabetical order)
And the last would be DCBA.
Order doesn’t matter → “[mathematical] combinations”
2.1 Combinations
If you have 4 elements and you get 4 and order doesn’t matter, then there’s only one way to do it: all the four elements.
That is, {A, B, C, D} is the only 1 combination.
or “1 subset of 4 letters from {A, B, C, D}”
(44)=1
2.2 Combinations with repetition:
((nk))=(n+k−1k).
In this case :
((44))=(74)=7∗6∗53∗2=35.
Then:
35 multisets of 4 elements, taking 4 at a time.
In this last case one multiset would be AAAA.
But the multiset AAAB would be the same as BAAA, ABAA and AABA.
The answer by Robert Nichols gives all the 35 different multisets.
I hope my answer is clear enough.
Footnotes
COMBINATION | significado, definición en el Cambridge English Dictionary
Combination - Wikipedia
Tuple - Wikipedia
Permutation - Wikipedia
Permutation - Wikipedia
Combination - Wikipedia
Combination - Wikipedia
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Deb P. Choudhury
Former Professor at University of Allahabad
·
Author has 10K answers and 8M answer views
·
6y
Non-repeating combinations (as distinct from permutations) are just subsets of the given set. A set with n elements, has C(n,r) subsets having r elements. This symbol equals
n!/[r!(n-r)] if n >/= r, and 0 if n < r.
The number of all combinations is equal to the total number of subsets which is 2^n, because C(n,0) + C(n,1) + ……C(n,n) = 2^n.
For your question, n=4, and the number of all combinations is 2^4 = 16.
Jerry Weinberg
Studied Business (college major)
·
Author has 61 answers and 23.8K answer views
·
6y
There would be 24 combinations using 4!
4! Is the same as 4x3x2x1. These numbers represent the four letters. We can’t multiply letters so we use numbers. 24
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Glenn Redd
B.S. in Mathematics, Kennesaw State University (Graduated 2016)
·
Author has 1.9K answers and 3.7M answer views
·
6y
Originally Answered: How many different 4 letter combinations can be created from the letters A,B, C, and D?
·
Assuming order is important (ABCD is not the same as DCAB):
44=256 if you can reuse letters (ie - AAAB is acceptable).
4!=24 if each letter can only be used once.
Related questions
How many combinations are in a 4-digit code?
How many combinations are there in a 4 letters + 4 digits code (example: AAAA0000)?
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Tante Tagamolila
Plant Design & Operations
·
Author has 90 answers and 49.8K answer views
·
4y
If you are asking how many unique combinations/sequencies you can make out fd the 4 letters, it is 4!
Raymond Beck
Former Former Army Sergeant
·
Author has 10.2K answers and 1.2M answer views
·
May 17
4! = 4x3x2= 24
for 2 letters, a and b, 2!=2, a,b and b,a
for 3 letters a,b and c 3!=3x2 = 6, abc, acb, bac, bca, cab, cba
Options Trader (2005–present)
·
Author has 119 answers and 158K answer views
·
6y
Originally Answered: How many different 4 letter combinations can be created from the letters A,B, C, and D?
·
For combinations order does not matter.
If each letter used once then 4C4 = 1
For permutations order does matter.
If each letter is used once then 4P4 = 24
Related questions
How many different 4 letter combinations can be created from the letters A,B, C, and D?
How many combinations can you make with ABC?
How many 4-letter combinations are there?
How many combinations of 4 colors are there?
How many combinations of 4 objects are taken two at a time?
How many combinations are in a 4-digit code?
How many combinations are there in a 4 letters + 4 digits code (example: AAAA0000)?
How many different combinations can I make out of 24 letters?
How many combinations you can make with 1, 2, 3, and 4 with no repeating digits?
How many 4 character combinations can possibly be made with letters and numbers only?
How many different combinations are there of four letters (ABCD) where none of the letters are duplicated?
Given the set of four letters, {A, B, C, D}, how many possibilities are there for selecting any two letters where order is important?
How many combinations are there to a 4-digit lock?
Which 4 letters have the most word combinations possible?
What is the number of combinations of three letters, say a, b, and c?
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9064 | https://www.ck12.org/flexi/geometry/angle-pairs-and-segments-proofs/lesspgreater%22for-every-line-lessmathgreaterlless-by-mathgreater-and-for-every-point-lessmathgreatermathrmpless-by-mathgreater-not-lying-on-a-given-line-lessmathgreaterlless-by-mathgreater-there-exists-a-unique-line-lessmat-2a07f11/ | Flexi answers - "For every line @$\begin{align}l\end{align}@$ and for every point @$\begin{align}\mathrm{P}\end{align}@$ not lying on a given line @$\begin{align}l\end{align}@$ , there exists a unique line @$\begin{align}m\end{align}@$ passing through @$\begin{align}\mathrm{P}\end{align}@$ and parallel to @$\begin{align}l\end{align}@$ " is known as Playfair's axiom. | CK-12 Foundation
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Geometry
Angle Pairs and Segments Proofs
Question
"For every line l and for every point P not lying on a given line l , there exists a unique line m passing through P and parallel to l " is known as Playfair's axiom.
Flexi Says:
Yes, that's correct! Playfair's axiom is a fundamental concept in Euclidean geometry. It states that through any point not on a given line, there is exactly one line parallel to the given line. This axiom is an alternative to Euclid's fifth postulate and is used to define parallel lines in a plane.
Analogy / Example
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The sketch shows two parallel lines cut by a transversal. Angle 1 is 58°. What is the measure of angle 7?Are corresponding angles supplementary?If two lines intersect, prove that the vertically opposite angles are equal.
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By messaging Flexi, you agree to our Terms and Privacy Policy |
9065 | https://doodlelearning.com/maths/skills/numbers/rounding-numbers-to-nearest-100-and-1000-overview | Skip to content
Round Numbers to the Nearest 100 and 1000 (Up to and Beyond 1000)
To round a number to the nearest 100, check the tens digit to decide whether to round up or round down.
If the tens and ones are 50 or more, round up.
If they are 49 or less, round down.
Example => 4149 to the nearest 100 is 4100, but 4150 to the nearest hundred is 4200. To round a number to the nearest 1000, check the hundreds digit to decide whether to round up or round down. If the hundreds, tens and ones are 500 or more, round up. If they are 499 or less, round down.
Example => 16,500 to the nearest 1000 is 17,000, but 16,499 to the nearest thousand is 16,000.
Now try a few practice questions!
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9066 | https://agenda.infn.it/event/36245/contributions/200455/attachments/105865/148925/Presentazione%2030.pptx%20-%20Google%20Slides.pdf | Presentazione 30 .PPTX File Edit View Insert Format Slide Arrange Tools Help Menus Background Layout Theme Transition 1 LA VELOCITA' DELLA LUCE Joshua Grumetti, Oliver Rossi 4ICL Emma Novarini, Arianna Masi 3DLS Liceo delle Scienze Applicate ITIS G. Cardano di Pavia Lab2Go 2022/2023 2 TEORIA DELLA VELOCITA' DELLA LUCE • Per velocità della luce si intende la velocità di propagazione di un’onda elettromagnetica e di una particella libera senza massa.
• Il suo valore nel vuoto cosmico è pari a 299 792 458 metri al secondo (m/s).
• Per convenzione si indica con la lettera “c” e si approssima il suo valore a 3 · 10⁸ m/s per facilitare i calcoli associati.
• La velocità della luce è la massima raggiungibile in natura.
• Nessun corpo è in grado di raggiungerla e superarla perché, con l’applicazione di una forza la sua velocità aumenterà, ma occorrerà sempre più energia per aumentarne ancora la velocità, dato che sempre più energia si trasforma in massa: l’unico risultato sarebbe un grande aumento della massa con una velocità praticamente inalterata.
3 TEORIA DEL FOTODIODO Il fotodiodo è un diodo caratterizzato da una giunzione P-N, un cristallo a semiconduttore, formato da due zone o regioni, la zona P e la zona N adiacenti l'una con l'altra e di materiali semiconduttori diversi: la prima è formata da silicio drogato con boro ed ha eccedenza di elettroni; la seconda, invece, è formata da silicio con una piccolissima quantità di fosforo ad eccedenza di lacune, cioè portatrici di carica elettrica, le quali contribuiscono al passaggio della corrente elettrica nei semiconduttori 4 TEORIA DEL FOTODIODO Il fotodiodo è costituito da semiconduttori perché essi consentono di conservare la carica elettrica, mentre usando altri materiali, la carica elettrica potrebbe disperdersi.
Da essi dipende l'energia minima che il fotone dovrà possedere per poter generare la fotocorrente.
Semiconduttori più usati nei fotodiodi: Silicio, Germanio, Arseniuro di indio gallio, Solfuro di piombo.
5 EFFETTO FOTOELETTRICO L'effetto fotoelettrico è un fenomeno che consiste nell'emissione, da parte di un atomo, di elettroni quando viene investito da radiazione elettromagnetica avente una determinata energia. Gli elettroni espulsi nel fenomeno dell'effetto fotoelettrico sono solitamente trattenuti da una certa energia e per espellerli all'esterno, occorre, investire l’atomo con una radiazione elettromagnetica avente una energia almeno uguale all'energia che li trattiene. E = h · v 6 EFFETTO FOTOELETTRICO • La frequenza di tale radiazione viene detta frequenza critica v0, ed è caratteristica di ogni atomo: • Utilizzando una luce di frequenza v < v0, anche se molto intensa, non si verifica alcun effetto.
• Utilizzando una luce di frequenza v > v0, gli elettroni emessi mostrano una energia cinetica tanto più grande quanto maggiore è la frequenza v. 7 PROCEDURA SPERIMENTALE ●Il trasmettitore (fotodiodo rosso modulato a 60 MHz) e il ricevitore (fotodiodo, amplificatore, mescolatore e parte collegata alla rete) sono in due custodie separate, unite tra loro da un cavo coassiale lungo 6 m.
●Attraverso questo cavo, l'emettitore viene alimentato con la corrente proveniente dalla sezione di rete del ricevitore.
●Questo segnale elettrico di 60MHz arriva al trasmettitore che lo converte in un segnale ottico tramite il fotodiodo incorporato.
●Il segnale ottico viene inviato nuovamente alla scatola rossa, che diventa ricevitore, con un segnale di 59,9 MHz e viene trasformato in elettrico.
●Esso viene poi mescolato con un altro segnale elettrico di 60 MHz, generato internamente dalla scatola rossa, per ottenere un segnale a bassa frequenza di 100 kHz, leggibile e supportato dall’oscilloscopio scolastico che può leggere segnali fino a 100 kHz, per questo i segnali sono stati ridotti a una frequenza minore.
●Sul display dell’oscilloscopio si vede l’onda del segnale ottico in trasmissione.
8 PROCEDURA SPERIMENTALE Descriveremo i passaggi della procedura sperimentale per misurare la velocità della luce in due diversi mezzi: aria e vetro.
L'obiettivo è quello di calcolare la velocità della luce ottenendo due misure: -il periodo del segnale ottico -la differenza di tempo tra il segnale ottico e quello ottenuto miscelando i due segnali elettrici nella scatola rossa 9 PROCEDURA SPERIMENTALE La misurazione deve essere avviata dopo circa mezz’ora dall’accensione del trasmettitore di luce e del trasmettitore, perchè per ottenere una precisione soddisfacente del risultato è necessaria la loro stabilità termica.
10 STRUMENTO DELLA LEYBOLD Il trasmettitore e il ricevitore sono dei fotodiodi e si occupano della conversione del segnale: -la scatola rossa emette un segnale elettrico (trasmesso ad un cavo adatto per avere un errore impercettibile) -il segnale arriva al trasmettitore - il trasmettitore trasforma il segnale elettrico in segnale ottico e lo trasmette alla scatola rossa -la scatola rossa agisce come ricevitore e trasforma di nuovo il segnale da ottico ad elettrico -il segnale ottico viene trasmesso all’oscilloscopio.
11 STRUMENTO DELLA LEYBOLD Il tubo con due finestre finali consente di misurare la velocità della luce tenendo conto dell’indice di rifrazione in acqua, determinato con l’aiuto del trasmettitore e del ricevitore di luce.
12 CONCLUSIONI DELL'ESPERIENZA 13 Velocità della luce nell’aria Il valore della velocità della luce nell’aria, calcolato con i dati sperimentali, è c = 2,8 · 10⁸ m/s, molto vicino al valore teorico originale: c = 3,00 · 10⁸ m/s.
Velocità della luce nel vetro Il valore della velocità della luce nel vetro acrilico, calcolato con i dati sperimentali è cvetro = 2,60 · 10⁸ m/s , molto vicino al valore teorico calcolato con l’indice di rifrazione: cvetro = 2 · 10⁸ m/s.
1 1 2 3 4 5 6 7 8 9 10 11 12 13 1 1 2 3 4 5 6 7 Velocità della luce nell’aria Il valore della velocità della luce nell’aria, calcolato con i dati sperimentali, è c = 2,8 · 10⁸ m/s, molto vicino al valore teorico originale: c = 3,00 · 10⁸ m/s.
Velocità della luce nel vetro Il valore della velocità della luce nel vetro acrilico, calcolato con i dati sperimentali è cvetro = 2,60 · 10⁸ m/s , molto vicino al valore teorico calcolato con l’indice di rifrazione: cvetro = 2 · 10⁸ m/s.
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9067 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7616719/ | GTP before ATP: The energy currency at the origin of genes - PMC
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Biochim Biophys Acta Bioenerg
. Author manuscript; available in PMC: 2024 Oct 21.
Published in final edited form as: Biochim Biophys Acta Bioenerg. 2024 Sep 24;1866(1):149514. doi: 10.1016/j.bbabio.2024.149514
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GTP before ATP: The energy currency at the origin of genes
Natalia Mrnjavac
Natalia Mrnjavac
1 Institute of Molecular Evolution, Faculty of Mathematics and Natural Sciences, Heinrich Heine University Düsseldorf, 40225 Düsseldorf, Germany
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1, William F Martin
William F Martin
1 Institute of Molecular Evolution, Faculty of Mathematics and Natural Sciences, Heinrich Heine University Düsseldorf, 40225 Düsseldorf, Germany
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1,
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1 Institute of Molecular Evolution, Faculty of Mathematics and Natural Sciences, Heinrich Heine University Düsseldorf, 40225 Düsseldorf, Germany
Corresponding author. bill@hhu.de (W.F. Martin).
Issue date 2024 Sep 24.
This is an open access article under the CC BY license (CC BY license).
PMC Copyright notice
PMCID: PMC7616719 EMSID: EMS199412 PMID: 39326542
The publisher's version of this article is available at Biochim Biophys Acta Bioenerg
Abstract
Life is an exergonic chemical reaction. Many individual reactions in metabolism entail slightly endergonic steps that are coupled to free energy release, typically as ATP hydrolysis, in order to go forward. ATP is almost always supplied by the rotor-stator ATP synthase, which harnesses chemiosmotic ion gradients. Because the ATP synthase is a protein, it arose after the ribosome did. What was the energy currency of metabolism before the origin of the ATP synthase and how (and why) did ATP come to be the universal energy currency? About 27 % of a cell’s energy budget is consumed as GTP during translation. The universality of GTP-dependence in ribosome function indicates that GTP was the ancestral energy currency of protein synthesis. The use of GTP in translation and ATP in small molecule synthesis are conserved across all lineages, representing energetic compartments that arose in the last universal common ancestor, LUCA. And what came before GTP? Recent findings indicate that the energy supporting the origin of LUCA’s metabolism stemmed from H 2-dependent CO 2 reduction along routes that strongly resemble the reactions and transition metal catalysts of the acetyl-CoA pathway.
Keywords: ATP synthase, Ribosome, Origin of translation, Origin of life, Acyl phosphates, Acetyl-CoA pathway
1. Introduction
Thoughts on biochemical evolution invariably lead to questions concerning the nature of energy conservation in early evolution as well as the chemical carriers of metabolic energy (energy currencies) that helped metabolism (and life) evolve. All prokaryotic groups are united by chemiosmotic ATP synthesis in bioenergetic membranes. Bioenergetic membranes in prokaryotes are diverse in composition [2,3] yet uniform in function. They present myriad pathways by which cells couple exergonic reactions to the pumping of protons or Na+ ions across a membrane from the inside of the cell to the outside, generating an ion gradient that sets the biological dynamo, the rotor-stator ATP synthase (termed in the following the ATPase for short) , in motion . That principle is conserved in all cells (except for a few highly derived parasites). The ATPases of bacteria and archaea are, like the ribosome, clearly homologous [6,7], hence the harnessing of ion gradients goes back to the common ancestor of bacteria and archaea , which by modern accounts was the last universal common ancestor LUCA . Bioenergetic membranes in eukaryotes are localized in mitochondria and chloroplasts, which arose over a billion years ago via endosymbiosis from their α-proteobacterial and cyanobacterial ancestors. In this paper, our focus is the very early evolution of bioenergetics, going back to the time before, during, and after the ATPase arose, homing in on the nature of energetics in LUCA with a focus on two main questions: 1) What energy currency (or currencies) preceded ATP in bioenergetic evolution? 2) How (and why) did ATP come to be the universal energy currency?
Though energetic properties can help to reconstruct intermediate stages in the transition from inanimate chemical reactions to the earliest living systems, early evolution carries the general caveat that the further back in time we go, the more hazy the contours of early bioenergetic processes become. But there are some robust constraints on these questions, as we will show. In general, the cell bears witness to its own ancient history. Its chemistry holds insights into early bioenergetic evolution, if we look at the right cells, and from the right angle.
2. Energy currencies
When we hear the term “energy currency” we rightly think about ATP , synthesized either by substrate-level phosphorylation using organophosphorus compounds as phosphoryl donors for phosphorylation of ADP or by the ATP synthase using ion gradients , ion gradients being an energy currency in their own right . In the context of energy currencies, some might also think of acyl phosphates , thioesters , or GTP , which is used in many reactions of central metabolism. Some will think of pyrophosphate , although the role of pyrophosphate in metabolism is not that of an energy currency, it is instead a mediator of irreversibility , as Kornberg explained. Some might think of carbamoyl phosphate or acyl anilides like formyltetrahydrofolate . Others might think of phosphagens like creatine , although phosphagens are storage forms for ATP. Still others might think of reduced coenzymes and electron carriers such as NADH, FADH 2, reduced ferredoxins or cytochromes. Some might consider highly exergonic reactions, such as the formation of aromatic compounds from aliphatic starting materials, for example as in the biosynthesis of pyridoxine phosphate from 4-phospho-hydroxy-threonine and 1-deoxyxylulose-5-phosphate , which is far more exergonic than ATP hydrolysis because of water eliminations that generate aromaticity (hence stability) in the pyridoxine phosphate product . Yet aromatic formation is not used by cells for substrate-level phosphorylation, nor can it be readily coupled to slightly endergonic reactions in order to improve their thermodynamic favorability, whereas ATP hydrolysis can.
When it comes to sources and ‘currencies’ of biochemical energy, few people might think of exergonic reactions of CO 2 with H 2. But according to one theory for the origin of metabolism (and life), that is the energy source that got the chemistry of life off the ground [24,25]. For example, the synthesis of pyruvate—the arguably most central compound in metabolism —from H 2 and CO 2 is a close to equilibrium (hence reversible) reaction under physiological conditions , but is exergonic in the direction of pyruvate synthesis by roughly −57 kJ/mol under laboratory conditions that simulate H 2-producing hydrothermal vents . A series of recent papers has shown that the synthesis of formate, acetate, pyruvate and other biologically relevant organics from H 2 and CO 2 using transition metal catalysts—with or without various inorganic support materials—is facile under the conditions of serpentinizing hydrothermal vents [27–31]. The energy to drive those reactions forward does not reside in ATP hydrolysis or any other coupled reaction, but resides instead in the redox reactions of H 2 with CO 2 themselves, reactions in which the equilibrium lies on the side of reduced carbon compounds (provided the absence of strong oxidants).
The key to making such reactions occur at relevant rates in water are catalysts: solid-state, zero-valent transition metal catalysts, mainly Fe, Ni, and Co. Those are the same catalysts that acetogens (an ancient bacterial lineage [33,34]) and methanogens (an ancient archaeal lineage [35,36]) use in the acetyl-CoA pathway, the most ancient pathway of CO 2 fixation [24,27,37]: In the acetyl-CoA pathway, the metals are not zero-valent but coordinated as divalent ions by S, C, O, and N in active-site residues of enzymes and cofactors that synthesize pyruvate from H 2 and CO 2 [38,39]. Under some conditions, native metals in water can convert H 2 and CO 2 to pyruvate at concentrations up to 200 μM , equaling the physiological concentration of pyruvate in growing acetogenic cells . In some experiments, magnetite (Fe 3 O 4) and greigite (Fe 3 S 4) also provide effective catalysis for H 2-dependent CO 2 reduction , but it cannot currently be excluded that a portion of the Fe 2+ in those catalysts was reduced to Fe 0 (as the active catalyst) by H 2 under alkaline conditions during the experiment.
3. Energy currencies require an energy source
The transition metal-catalyzed reactions of H 2 with CO 2 to metabolically relevant organics require neither phosphate-containing compounds, nor ion gradients, nor cofactors, thioesters, RNA, peptides or other organic compounds, although they do require alkaline pH, which shifts the equilibrium in the reaction H 2 → 2e− + 2H+ to the right by proton removal [41–43]. This renders H 2 a sufficiently strong reductant to reduce CO 2 without the help of reduced ferredoxin or flavin-based electron bifurcation . The reactions run all by themselves in water, on the metal surface, driven by the exergonic nature of H 2-dependent CO 2 reduction. The reactions are highly specific in terms of products, which are mainly formate, acetate, and pyruvate, the products of the acetyl-CoA pathway [24,45] of CO 2 fixation in acetogens and methanogens. They go forward in water without enzymes, with nothing more than a piece of solid-state metal as catalyst [27–31] whereby Fe 0 can, in some conditions, serve as both catalyst and reductant . Such findings suggest that the bedrock-primordial reactions of metabolism started from gasses reacting on metals, which themselves are deposited in H 2-producing hydrothermal environments . The reactions are fueled by pure redox energy: H 2-dependent CO 2 reduction on metals, reactions that differ only in mechanism, not in product spectrum, from the energy-releasing redox reactions of the acetyl-CoA pathway that acetogens (bacteria) and methanogens (archaea) still use today to fix CO 2 and simultaneously generate ion gradients for ATP synthesis [48,49].
Even if metabolism really did start by such H 2-dependent reactions (which is possible), this still leaves a massive energetic gap between H 2-dependent organic synthesis catalyzed by solid-state metals in water on the one hand, versus enzymatic ATP synthesis via an exquisitely complex rotor-stator ATP synthase [5,6] that harnesses an ion gradient across membranes on the other. Because that gap is large, the intermediate steps are difficult to reconstruct. But there had to be a starting point.
Recent findings indicate that solid-state transition metals are good candidates for the first catalysts for an environmental origin of metabolism [27–31,46,50]. These solid-state, zero-valent metal catalysts cannot be incorporated into enzymes (hence free-living cells) during early evolution, although zero-valent metals are thought to arise as intermediate states in some enzymatic reaction mechanisms of metalloenzymes involving nickel . The metals are catalysts. In H 2-dependent reactions, they catalyze (i) the reduction of CO 2 to the intermediates and end products of the acetyl-CoA pathway [27–31], (ii) the synthesis of amino acids from 2-oxoacids and NH 3 , (iii) the synthesis of long chain hydrocarbons from HCO 3− , (iv) the reduction of NAD+ to NADH , and (v) the reduction of Clostridium pasteurianum ferredoxin as an evolutionary precursor to flavin-based electron bifurcation . In each case, the energy source is H 2, the catalyst is Ni 0, Fe 0, Co 0, or their alloys, and the reaction takes place under the conditions of serpentinizing hydrothermal vents. These findings are in generally good agreement with the predictions of some theories about the origin of metabolism [17,37,53,54] and the nature of LUCA , though the newly discovered role of native metals as catalysts [46,55] departs fundamentally from earlier ideas concerning a catalytic role for FeS minerals in early evolution .
4. Ion gradients and phosphates
Today the exergonic reaction of H 2-dependent CO 2 reduction in acetogens and methanogens is coupled to ion pumping [48,49]. At some point in early evolution, there had to be conversions of the redox energy in CO 2 reduction into soluble chemical currencies of bioenergetic utility. Acyl phosphates are widely discussed as a likely intermediate state in the evolution of energy conservation [9,17,57] because they are the most common organophosphate used for ATP syntheses via substrate-level phosphorylation , and because they are formed during the process of H 2-dependent CO 2 reduction in acetogens and under some conditions in methanogens .
Are acyl phosphates a ‘missing link’ intermediate between H 2 and ATP in bioenergetic evolution? Probably, but they are not the only one. The search for intermediates in early bioenergetic evolution includes identification of possible alternative energy currencies that could substitute for ATP based on free energy of hydrolysis (for example Δ G 0′ = −43 kJ⋅mol−1 for acetyl phosphate vs. Δ G 0′ = −32 kJ⋅mol−1 for ATP) . But the biochemical reactions of cells themselves can also inform on early energetic evolution, if properties are sufficiently conserved to be identifiable in modern metabolism.
ATP synthesis is the main bioenergetic chemical reaction that cells perform and the ATPase is the main source of chemical energy in almost all cells, certainly in all autotrophs. However, the ATPase is a protein that is produced by the ribosome (Fig. 1) meaning that, beyond all doubt, the ribosome preceded the ATP synthase in evolution. More generally, the ribosome preceded the origin of all protein-coding genes we know today, because they are all synthesized on ribosomes using aminoacyl-tRNAs and the universal genetic code . That the ribosome is ancient is not news to anyone, but the bioenergetic environment surrounding its origin is not a well-charted space . Protein synthesis at the ribosome consumes roughly 75 % of the ATP produced by any modern prokaryotic cell, but protein synthesis arose at a time before there was an ATPase. The foregoing context, admittedly long but perhaps necessary, brings us to our first question: What was that energy currency at the origin of translation?
Fig. 1.
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The ribosome (left) and the ATPase (right) drawn to scale. The ribosome is from E. coli (PDB ID: 4YBB) , as is the ATPase (PDB ID: 5T4O) . During translation, the mRNA is threaded through the ‘donut hole’ in the ribosome. Images prepared using Visual Molecular Dynamics version 1.9.3. Lipids surrounding the F o subunit drawn for illustration.
5. Energy currencies before the ATPase arose
A world without the ATPase is hard to imagine but it had to exist. The ATP synthase is a protein. At the time of its origin, it entered in a world in which protein synthesis, hence the genetic code and the ribosome, were already present and producing proteins with the help of energy that was not supplied by the ATPase. From that it follows that at the time that the ATPase arose, there was already something that we can call the biosynthetic core in place, because the specific amino acids of which the ATP synthase consists had to be in place, as did the bases of which RNA consists and the code itself (mediated by aminoacyl-tRNA synthetases [61–63] and tRNA-rRNA interactions), as well as the co-factors required to fix CO 2 and make the amino acids, bases, and co-factors involved in protein synthesis. Such a universal biosynthetic core was postulated by Kluyver 100 years ago on the basis of very sparse data suggesting the central role of redox reactions in carbon metabolism . Morowitz pursued the idea of universality in metabolism based on the notion that the reverse citric acid cycle is the most ancient biochemical pathway although it turned out not to occur in archaea ; the acetyl-CoA pathway is the only CO 2 assimilation route that occurs in both bacteria and archaea . Today we know that, as it concerns core biosynthesis, Kluyver was right. There is a conserved metabolic core that starts from H 2 and CO 2 and spans bacteria and archaea [23–27,37]. All cells use the same 20 amino acids in their proteins, plus selenocysteine and pyrrolysine in some lineages. All cells use the same nucleoside phosphates in their nucleic acids, with many chemical modifications in rRNA and hundreds of chemical modifications in tRNA . And all cells use a common set of about 18 cofactors with some lineage-specific cofactors, for example the Ni-containing tetrapyrrole F 430 plus specialized thiol-containing cofactors coenzyme M and coenzyme B in methanogens .
The conserved core of microbial metabolism that supplies the building blocks of life is surprisingly small. It consists of only about 400 reactions . Most, but not all of the enzymes that catalyze those 400 reactions are homologous across all lineages, although the chemical reactions that are catalyzed and the chemical intermediates are conserved. This suggests that parts of the chemical network might be older than the enzymes that catalyze the reactions , which makes biochemical sense: Enzymes do not perform feats of magic, they accelerate reactions that tend to occur anyway. The universality of the biosynthetic core is, like the universality of the genetic code, a source of strong evidence for the single origin of life.
We can thus be sure that the ribosome was working when the ATPase arose. It is furthermore likely that there was a functioning biosynthetic network, probably an autocatalytic network , supplying the metabolic products required for translation (possibly with some help from non-enzymatic reactions catalyzed by the environment [27,50,74,75]. Note that our question is not the origin of the ribosome or translation [58,76–78], which we take in this paper as a given, because the ATPase is a protein. Our question is the energy currency at the time that the ATPase arose, which required translation in order to occur. Translation required the ribosome and the code, but not the ATPase.
That is not to say that the biosynthetic core arose entirely in the absence of high energy phosphorus compounds. It is possible to computationally construct networks of organic compounds that can arise without the participation of phosphorus . But metabolism in cells, also in LUCA, requires thermodynamic impetus in order to go forward [80–82]. Of the 400 reactions in the biosynthetic core, 77 of them (19 %) are enzymatically coupled to the hydrolysis of ATP so that these otherwise slightly endergonic reactions can go forward in the biosynthetic direction. Acyl phosphates, which have a higher free energy of hydrolysis than ATP and which generate ATP in most substrate-level phosphorylations , could have energetically substituted for ATP in primordial versions of those reactions, as could other primitive energy currencies, in principle, as long as the free energy release was sufficient to drive the coupled reaction forward.
If we look for energy-releasing reactions in the biosynthetic core (Table 1), we find ATP hydrolysis, but also many reaction types that are exergonic by more than −20 kJ/mol without ATP hydrolysis. In addition, the energy demand for biochemical reactions also depends on environmental conditions. For example, the reduction of ferredoxin with H 2 and metallic iron, the long-sought evolutionary precursor to flavinbased electron bifurcation (a bioenergetic and evolutionary problem in its own right , but not the topic of this paper) proceeds readily at alkaline pH as in serpentinizing hydrothermal systems, but not at acidic pH, because of the pH-dependence of H 2 oxidation. The midpoint potential of the reaction H 2 → 2e− + 2H+ (E 0′ = −414 mV) becomes more negative with increasing alkalinity, because of the pulling effect that OH− exerts by removing protons from the right-hand side of the reaction, leading to highly reducing conditions in serpentinizing hydrothermal vents [34,41] and making H 2-dependent reductions more exergonic under alkaline conditions .
Table 1. Energy-releasing reactions in the biosynthetic core.a.
| Reaction | Number | Δ G [kJ·mol −1] | Reference |
:--- :--- |
| Aromatic formation | 31 | − 60 to −150 | |
| Pyruvate formation from H 2 + CO 2 | 1 | − 57 | |
| Acyl phosphate hydrolyses | 4 | − 45 | |
| ATP hydrolyses | 77 | − 32 | |
| Acyl thioester hydrolyses | 14 | − 32 | |
| Pterin-dependent alkyl transfers | 4 | − 30 | |
| Reductions | 44 | − 28b | |
| Folate-dependent acyl transfers | 2 | − 26 | |
| SAM-dependent alkyl transfers | 10 | − 24 | |
| Decarboxylations | 30 | − 20 | |
| Ring-forming reactions | 35 | −10 to − 25 | |
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a
The ~400 reactions for synthesis of amino acids, bases and cofactors from H 2, CO 2, NH 3, H 2 S and P i (plus salts) are compiled and listed in . Values of Δ G for conditions given in the corresponding references, usually Δ G°′ (1 M reactants, 25 °C).
b
Average for 44 NAD(P)H-, ferredoxin- and formate-dependent reductions of organic compounds in the biosynthetic core .
6. The environment of metabolic and ribosomal origin
Where do the components come from that are required to synthesize an ATPase on a ribosome? In metabolism, they come from the biosynthetic core (and ATP synthesis as the source of energy). Roughly 98 % of the reactions in the biosynthetic core are exergonic under the highly reducing, non-equilibrium and hot (~80 °C) environmental conditions of serpentinizing hydrothermal vents . The H 2-dependent reduction of CO 2 to pyruvate was discussed above. Recent work by the groups of Joseph Moran and Harun Tüysüz show that Ni 0 and H 2 plus ammonia will convert seven different biological 2-oxo acids (including pyruvate) into the corresponding amino acids (glycine, alanine, aspartate, glutamate, valine, leucine and isoleucine) in water at room temperature at yields in the range of 6−51 % . The same conditions without ammonium promote the synthesis of biological 2-oxo acids . Those findings converge seamlessly with H 2-dependent CO 2 reduction to pyruvate (a 2-oxo acid) under similar hydrothermal conditions [27–31]. The similarities between the chemical conditions of hydrothermal vents and the enzymatic reactions of anaerobic microbial cells, first proposed by John Baross [53,89] are turning out to be surprisingly robust [42,90,91]. By contrast, the cyanide-based chemistry that classical schools have maintained to be ancestral to metabolism since Oro’s 1960 synthesis of adenine from cyanide , though presenting elegant organic syntheses, often works without catalysts altogether (in utter contrast to biochemical reactions) and has therefore never been compatible with, nor intersected, the chemistry (and catalysts) used by cells [43,50]. The chemistry of serpentinizing hydrothermal vents is relevant in an early evolution context because it is energetically conducive to the formation of the biosynthetic core, a reaction network forming the ABC compounds required for translation, a prerequisite for the origins of an ATP synthase.
How might a primordial ATPase have operated? It requires an ion gradient and a hydrophobic layer into which the F o subunit can insert. Serpentinization generates pH gradients with ocean water at vents [94,95]. Both the bottom-up (chemical synthetic) [27–31,43,46,50] and the top-down (inference from modern cells) [9,53,94] approaches to the ecological context of primordial ATPase function suggest that LUCA and the biosynthetic core arose in a serpentinizing (H 2-producing) hydrothermal vent , and that the natural pH gradient between the hydrothermal effluent (pH 9−11) [94,95] and the Hadean ocean (pH 6.5) could power an ATP synthase [17,97]. This requires the existence of lipids or other hydrophobic compounds that can form a protonpermeability barrier. The synthesis of such compounds is efficiently performed by native cobalt (Co 0 is naturally deposited in serpentinizing systems ) in the presence of H 2 and CO 2 under hydrothermal vent conditions . That would provide contours, context, and an energy source for ATP synthase function, but does not answer our question: What energy currency supported translation before the ATP synthase arose and became functional to supply ATP? Having attempted to justify our premises about the origin of metabolism at a H 2-producing hydrothermal vent, we have a suggestion.
7. GTP before ATP
Although ATP is the main energy currency for biosynthesis today, GTP is the main energy currency of ribosome biogenesis and function [69,98]. How much and what kind of energy does protein synthesis consume? By dry weight, cells are about 50–60 % protein and about 20 % RNA, with variation depending on growth conditions . Using E. coli as a proxy for a typical prokaryotic cell , protein synthesis consumes about 75 % of the total ATP budget (Table 2), with ~55 % of ATP consumption attributable to the formation of peptide bonds by the ribosome, 4 % attributable to amino acid synthesis and 16 % attributable to the synthesis of tRNA, mRNA and rRNA, whereby almost all of the RNA in a cell is rRNA.
Table 2. ATP costs per cell in E. coli grown on glucose and NH 4.
| Polymer | Gram per gram of cellsa | ATP required per gram of cells [mol· 10 4] | Proportion of ATP cost per cell [%] |
:--- :--- |
| Protein | 0.52 | | 59.1 |
| Amino acid | | 14 | |
| synthesis | | | |
| Polymerization | | 191 | |
| RNA | 0.16 | | 16.4 |
| NMP synthesis | | 34 | |
| Polymerization | | 9 | |
| mRNA turnover | | 14 | |
| Import of salts | | 52 | 14.9 |
| DNA | 0.03 | 11 | 3.2 |
| Lipid | 0.09 | 1 | 0.3 |
| Polysaccharideb | 0.17 | 21 | 6.1 |
| Solutesc | 0.04 | – | – |
| Total | 1.01 | 347 | 100 |
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a
Values originally from and tabulated in [14,99].
b
Stouthamer’s polysaccharide value includes 10.3 % hexose, consistent with variation in glycogen content across studies .
c
Solutes (metabolites) not reported in , but in other studies . The high cost of protein synthesis comes from 4 ATP equivalents expended per peptide bond synthesized at the ribosome: the PP i-producing step at aminoacyl-tRNA synthesis (counted as two ATP equivalents) and the two GTP-consuming steps at translation.
The main energetic cost in cells is peptide bond formation at the ribosome. It consumes 4 ATP equivalents per peptide bond formed. Half of the energetic costs for translation are incurred in the cytosol in reactions catalyzed by 20 aminoacyl-tRNA synthetases (AARS) [61–63]. AARSs activate the amino acids to aminoacyl-adenylates by transferring an AMP residue from ATP to the corresponding amino acid with the concurrent release of PP i, then transferring the aminoacyl moiety to the corresponding tRNA molecule to form the aminoacyl-tRNA for translation. The two-step AARS reaction generates AMP and PP i, whereby PP i is immediately hydrolyzed to two P i by ubiquitous pyrophosphatases, making the process of aminoacyl-tRNA synthesis, hence translation, irreversible (the PP i substrate for the back reaction is removed).
Protein synthesis at the ribosome starts with the formation of the mRNA-30S subunit (SSU) complex and the binding of the initiator tRNA with the help of the initiation factors IF1, IF2 and IF3. The initiation factor IF2 is a large GTPase that hydrolyzes GTP upon arrival of the 50S subunit (LSU) to form the 70S initiation complex [58,101]. That GTP-dependent step primes the ribosome for protein synthesis, but it occurs only once, followed by many subsequent peptide bond formations per translated mRNA molecule. The energetic cost of translation is incurred at the elongation and the translocation phase, which require two further GTPases, the elongation factors EF-Tu and EF-G. EF-Tu is a GTPase that binds aminoacyl-tRNAs, delivers them to the ribosome and strengthens their interaction with it. Upon proper codon-anticodon recognition, conformational changes in the ribosome are transmitted to EF-Tu, GTP is hydrolyzed, followed by release of EF-Tu-GDP [102,103]. Peptide bond formation at the peptidyl transferase center proceeds without GTP hydrolysis , but the mRNA must translocate by one codon relative to the ribosome for the next peptide bond to form. This is achieved by conformational changes induced by EF-G, another large GTPase. It binds the ribosome and hydrolyzes GTP to induce translocation of the ribosome relative to the mRNA, releasing GDP and P i in the process . The cost and energy currency of elongation and translocation at the ribosome is two GTP per peptide bond formed.
Thus, AARS in the cytosol consumes one ATP per aminoacyl-tRNA synthesized, but produces PP i which is hydrolyzed, such that the cost is two ATP equivalents per amino acid (two high-energy bonds), while translation at the ribosome consumes two GTP per peptide bond. GTP is generated from GDP and ATP by ubiquitous nucleoside diphosphate kinases . Broken down across the total ATP pool, roughly 34 % of the E. coli energy budget is expended for irreversible reactions (mostly to make translation irreversible), roughly 38 % is expended for biosynthesis, while 27 % of the E. coli energy budget is converted to GTP for translation (Fig. 2).
Fig. 2.
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Biosynthetic energy budget of E. coli with a focus on protein synthesis. About 28 % of the biosynthetic ATP in E. coli is converted into GTP for the elongation and translocation steps of translation. About 35 % of the ATP is dedicated to irreversible processes. GTP use is highlighted in green, irreversible processes are highlighted in yellow. (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)
Because the GTPases EF-Tu and EF-G are universally conserved in all cells , the process of translation is GTP-dependent in all cells. If the E. coli budget in Fig. 2 is a roughly valid proxy, all lineages of pro-karyotes have maintained a GTP-dependent translation process that is resupplied with energy via the ATPase since the origin of the latter. Common sense (and Occam’s razor) have it that traits that are present in all members of a group trace to the common ancestor. That means that translation was GTP-dependent at the origin of translation, hence at the time that the ATPase arose. The ancient role of GTP in ribosome biology is corroborated by its role in ribosome biogenesis [106–111].
Ribosome biogenesis reveals a predominance of small GTPases (Fig. 3). Small GTPases contain a structurally conserved GTP-binding domain and are widely distributed among bacteria and archaea where they typically function by inducing conformational changes in target proteins upon GTP hydrolysis. This allows small GTPases to modulate diverse functions such as ribosome binding, tRNA binding, Fe 2+ transport and Ni 2+ transport, but most importantly here, ribosome biogenesis and tRNA modification. tRNA modifications are essential for the proper operation of the genetic code, as nucleic acid interactions modulated by post-transcriptional base modifications are important for accurate translation, and require the presence of various modified bases in tRNA . Based on universal conservation, we can infer that LUCA had ribosomes that translated in a GTP-dependent manner and both large and the small GTPase families essential to ribosome biogenesis trace to LUCA. Functions associated with prokaryotic ribosome function and biogenesis are conspicuously GTP-dependent.
Fig. 3.
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GTP is the energy currency of the ribosome (PDB ID: 7K00 ). The list of GTPases is from references [106–111]. The figure was created using The PyMOL Molecular Graphics System, version 2.5.4, Schrödinger, LLC.
In addition to being the energy currency for the initiation, elongation and translocation steps of ribosomal translation, GTP is also the energy currency for cotranslational insertion of proteins into membranes via the signal recognition particle and a variant of the universally conserved Sec system [115,116]. Protein membrane targeting and secretion via the SRP pathway is mediated by the interaction of the signal recognition particle SRP, an RNA−GTPase ribonucleoprotein complex, and the SRP receptor, a second large GTPase [115,117]. This insertional mechanism is required for ATPase function.
Though acyl phosphates were probably crucial at the onset of metabolic reactions [17,57], a number of ancient core biochemical functions are also GTP-dependent in some organisms: the succinyl CoA synthetase step in the TCA cycle , the phosphoenolpyruvate carboxykinase reaction , or the phosphate pyrophosphokinase step that initiates nucleotide synthesis . In some cases GTP forms the carbon backbone of cofactors, such as in the synthesis of the ironguanylylpyridinol (FeGP) cofactor of methanogen Fe-hydrogenase , and pterin synthesis . Folate and methanopterin are pterins essential to the acetyl-CoA pathway of acetogens and methanogens, perhaps the oldest biochemical pathway known [24–27,37]. These pterins are biosynthetically derived from the carbon backbone of GTP, as are FAD, F 420 (an archaeal homologue of FAD) and the molybdopterin cofactor (MoCo). Pterin synthesis starts with the GTP cyclohydrolase I reaction that includes opening of both the guanine and the ribose ring of GTP, followed by chain rearrangement and formate elimination to form the pteridine backbone . ATP is not used in the same way for cofactor biosyntheses, it is a more specialized energy currency, though often covalently bound to cofactors as a kind of biochemical handle. The foregoing suggests that GTP was integrated into CO 2 fixation chemistry as a carbon backbone for folate and pterin synthesis to enable the acetyl-CoA pathway (which requires no ATP hydrolysis in the pathway to pyruvate in methanogens) . This implies GTP was readily available at the time enzymatic pterin biosynthesis came about.
The list of 400 reactions that comprise the biosynthetic core of LUCA’s metabolism reveals 11 that consume or produce GTP. In addition to the examples just mentioned (pterin synthesis: folate, MoCo, methanopterin, F 420, and flavins), there are GTP-dependent enzymes in cobalamin synthesis (required in the acetyl-CoA pathway by the corrinoid iron-sulfur methyltransferase CoFeS), but one example is curious. The last step in the synthesis of AMP is GTP-dependent. Adenylosuccinate synthase (EC 6.3.4.4) catalyzes the GTP-dependent transfer of an aspartate moiety to the carbonyl carbon of inosine monophosphate (IMP) to yield adenylosuccinate, which eliminates fumarate to yield AMP. This mirrors the last step of GMP synthesis catalyzed by GMP synthase (EC 6.3.5.2), which converts the carbonyl at C2 of xanthylate to an amino group via ATP-dependent adenylation and PP i formation, rendering the reaction irreversible, as in the case of amino acid activation by aminoacyl-tRNA synthetases .
Independent of ancient reactions in core metabolism, the universality of GTP as the source of energy in translation suggests that GTP was the energy currency of translation in the environment where the ribosome arose. The ribosome gave rise to the ATPase, therefore the main energy currency before the origin of the ATP synthase appears, in the most direct inference, to have been GTP.
8. A relict of GTP predominance in rRNA
The aminoacylation of tRNA performed by AARS enzymes is generally ATP-dependent, though there are two reports of archaeal AARS, for aspartyl-tRNA and phenylalanyl-tRNA synthesis from two different hyperthermophilic species, that accept GTP in addition to ATP for the synthesis of the aminoacyl-tRNA [123,124]. A critic might conjure a list of reasons why GTP is an evolutionary replacement for an ancestrally ATP-dependent translation process. Yet were that so, then ATP would have been the universal energy currency at the origin of the ribosome, only to have been replaced (for some unknown reason) by GTP in LUCA, which is possible, though not the inference of choice. If the ribosome and ancestral rRNAs arose at a time when ATP was the main energy currency, then one would expect A to be the most common base in rRNA, simply because it was, by ATP-first reasoning, the most commonly available triphosphate in the system (as a substrate for rRNA synthesis).
It is therefore all the more noteworthy that G, not A, predominates in rRNA, both in 16S and in 23S rRNA sequences, both in bacteria and in archaea (Fig. 4). This is not a G + C effect (thermostability of GC basepairing in folding) because G predominates over C, nor is it a purine effect, because G also predominates over A. The excess of G is highly significant. Seen in light of GTP-dependent ribosome biogenesis and function, the excess of G in rRNA appears to be a conserved trait that traces to the ancestral ribosomal RNA of the ancestral ribosome. It could reflect a frozen accident, a predominance of GTP over other NTPs as substrates for primordial rRNA synthesis in the environment where the ribosome and translation arose.
Fig. 4.
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Frequency of the nucleobases (as read from gene sequences, not including modifications) in rRNAs of the ribosomal small subunit (SSU), i.e.16S, and large subunit (LSU), i.e. 23S, in bacteria and archaea. The number of rRNA sequences, taken from the SILVA database , is indicated underneath the corresponding bar chart for each of the four cases. The distribution of base frequencies is highly nonrandom (p< 10−300, Smirnov-Komolgorov test).
9. The rise to prominence of ATP
The ribosome had to exist and function before the ATPase came into existence. The energy to power the origin of the ribosome and the origin of the first protein-coding genes either came from some chemiosmotic coupling mechanisms for which there is no trace in modern cells, or more likely (from the present perspective) it came from acyl phosphates (or other organophosphates) and was used for substrate-level phosphorylation of GDP. This is not far-fetched, as it is known that acetate kinase, the enzyme catalyzing the substrate-level phosphorylation reaction in the acetyl-CoA pathway, starting from acetyl-phosphate, accepts GTP . But caveats are always in order, also when it comes to acyl phosphates. We point to recent findings by Bernhard Schink and colleagues who isolated an enzyme from the bacterium Phosphitispora fastidiosa that uses the acetyl-CoA pathway for CO 2 fixation, but derives all of its electrons from phosphite (HPO 3 2−) rather than from H 2. The protein they isolated, AMP-dependent phosphite dehydrogenase, catalyzes a very unusual enzymatic reaction:
AMP+NAD++phosphite→ADP+NADH
which is a completely unique, redox-dependent substrate-level phosphorylation involving only four cofactors and phosphite. The reaction is powered by the extremely negative midpoint potential of electron release in the phosphite to phosphate oxidation reaction with E o′ = -690 mV . Two molecules of ADP are then converted by the cell to ATP and AMP via a ubiquitous and well-known enzyme, adenylate kinase. Given the presence of phosphite in serpentinized minerals , this is a possible (and maximally compact) ancient route of energy conservation: phosphorylation, NADH synthesis and ATP synthesis in one. It may very well be a primordial energy source, as phosphite has been reported in the kinds of rocks that generate H 2 . Such environments might have been more widespread early in evolution , but today almost all cells use the normal route of membrane-associated redox reactions, proton pumping, and ATP synthesis via the ATPase.
If GTP was the energy currency at the origin of translation, as the ribosome itself would indicate, why change a running system? Why did ATP become the universal energy currency, supplanting GTP in apparently all core biosynthetic and most other processes, except ribosome biogenesis, translation, and GTP-dependent signaling? A recent report proposed that ATP rose to prominence because acetyl phosphate phosphorylates ADP better than other nucleoside diphosphates in substrate-level phosphorylation . However, subsequent work by the Moran group showed that ADP was, in fact, not the better substrate at all, rather that ADP (in a complex with metal ions) serves as a catalyst that promotes the phosphorylation of all NDPs more or less equally , including ADP itself.
The simplest suggestion as to what force or process could have driven the rise of ATP to the status of the universal energy currency, in our view, is that the first rotor-stator ATP synthase had from its inception, or soon developed, a pronounced substrate specificity for ADP over other competing substrates, such that its main product was ATP rather than some other possible energy currency. This property appears to be conserved in the enzyme of all cells today. In that sense, the identity of ATP as the universal energy currency of biosynthesis (but not translation, where it apparently could not displace GTP) would be a true frozen accident, of no burgeoning functional significance in comparison to other possible energy currencies (including GTP), a chance consequence of active site conformation in the F 1 subunit, but of enormous utility to a world of evolving proteins. Just as the ribosome has never lost its dependence upon GTP, the ATP synthase has never altered its sub-strate in any known lineage to phosphorylate GDP or other NDPs. Evolutionary experiments that altered the main supply of chemical energy in the cell surely took place over 4 billion years, but were strongly counterselected, even though nucleotide diphosphate kinases can readily interconvert NTPs.
If the ancestral ion gradient that powered the first ATPase was a geochemical pH gradient generated by serpentinization , then the ATPase would be turning and churning out ATP day and night, 365 days/year, 1000 years per millennium, putting a very constant and very high energy charge (ATP/ADP ratio) on the contents of compartments within which it was synthesized and functional. The massive difference to all modern energy metabolism would be that no energy was needed from metabolic reactions in order to generate the ion gradient. That would have led to direct and lasting impact on early protein evolution, freeing H 2-dependent CO 2 reduction from the burden of generating acyl phosphates while “accelerating biochemical innovations, energetically financing gene inventions, and the selective pressure on evolving proteins to adapt to a new energy currency […]. ATP-binding domains are so prevalent in genomes, not because ATP is a constituent of RNA, but because it became the most popular energy currency. The ATPase transduced a geochemically generated ion gradient into usable chemical energy, and since the energy was free, the means to harness it as ATP ‘just’ required a suitable protein for the job, a complicated protein, but a protein” . In other words, with an ATPase powered by a geochemical ion gradient, ATP was the energy currency in free supply (independent of exergonic CO 2 fixation reactions), hence any newly evolved protein that was involved in reactions that benefited from added thermodynamic drive by coupling to the hydrolysis of phosphoanhydride bonds would readily accommodate ATP as the energy currency in a Darwinian sense. This would explain the prevalence of ATP usage in enzymes of metabolism . The paucity of ATP hydrolysis in the synthesis and function of the ribosome itself suggests that GTP was the ribosome’s ancestral energy currency and was never displaced in that function.
10. An ancient energy currency dichotomy
The foregoing suggests a path of early bioenergetic evolution along the lines of the summary in Fig. 5. The advent of ATP as a second major energy currency after GTP could mark the onset of an energy dichotomy in early evolution: GTP at the origin of translation and ATP following the origin of the ATPase and at the origin of enzymes that arose subsequently. It is possible that the 77 enzymes of the biosynthetic core that use ATP today could have used currencies other than ATP before the time of LUCA and that they adapted to the new and abundant energy currency supply via incorporation of ATP-binding domains. However, neither the ribosome (GTP) nor the ATP synthase (ATP) have altered their substrate specificity during evolution. Such complete conservation is best explained as preservation of the ancestral state and the operation of very strict functional constraints over billions of years across all lineages. A few enzymes that accept either ATP or GTP are known [120,133,134], it is possible that systematic searches might reveal many more.
Fig. 5.
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GTP before ATP. The tendency of the ribosome to arise on GTP during ontogeny and to function during translation with GTP suggests that GTP was the energy currency of translation before ATP came to be the universal energy currency. The origin of ATP as a second major energy currency (the energy currency dichotomy) is likely the result of the substrate specificity of the ancestral ATP synthase. The small white arrows pointing from ATP to GTP indicate enzymatic conversion of ATP to GTP to supply energy for the ribosome in modern metabolism. See also references [136, 137].
The commitment of roughly a quarter of the cell’s energy supply to GTP for translation reflects a kind of metabolic compartmentation within the cell that has persisted in all lineages throughout all of evolution: GTP for protein synthesis at the ribosome, ATP for the biosynthesis of components needed to synthesize a ribosome. The free energy released by hydrolysis of ATP and GTP is the same, there is no energetic reason to prefer GTP over ATP at the ribosome (or the converse in biosynthesis). Its conservation is probably an unerasable relict of the substrate specificity of the GTPases that run the ribosome and power its biogenesis. In that sense, bioenergetic evolution was marked by two frozen accidents: GTP for translation and ATP for all else.
That would supply the answer to the question of how and why ATP rose to prominence as the universal energy currency. It was the substrate specificity of the ancestral ATPase and the unlimited supply of geochemical ion gradients that resulted in an unstoppably high ATP/ ADP ratio, providing obvious advantage to reactions that required coupling to the hydrolysis of phosphoanhydride bonds in ATP to go forward. If ATP was the currency in supply, enzymes would evolve, where possible, to adapt their demand accordingly, either by modifications of the NTP binding site or by recombinative incorporation of ATP binding domains . But seen in the larger context, the very beginning of biochemical evolution started long before proteins, beginning with CO 2 fixation with energy from H 2 which required neither ATP nor GTP .
Our focus is the energy currency at the origin of the ribosome, not the origin of the ribosome itself, though newer findings suggest that the process of ribosome origin followed a path from simple to complex , possibly beginning with a ca. 70 nucleotide long hyperconserved RNA structure identified at the heart of the ribosome, termed the protoribosome [76,77], that i) encompasses the peptidyl transferase site and ii) is able to perform the peptidyl transferase reaction in vitro [77,78]. We offer that it is not fully accurate to refer to ATP as the “universal” energy currency. ATP is the universal energy currency of small molecule synthesis, but in the overall energy budget of cells, the biochemical power of growth is shared: ATP enjoys a 73 % majority in biosynthesis, but in 4 billion years has been unable to displace a 27 % minority of GTP use at translation. The data suggest that both fractions are ultimately the result of frozen accidents that occurred during early biochemical evolution in LUCA.
Acknowledgements
We thank Katharina Trost and Michael Knopp for calculating the nucleotide frequencies in rRNA from data base entries underlying the preparation of Fig. 4 and Andrea Alexa for help in preparing the manuscript. We thank Ute Römling for helpful discussions. This work was supported by the European Research Council (ERC) under the Horizon 2020 research and innovation program (grant 101018894 to WFM), the Volkswagen Foundation (grant 96742 to WFM) and the Deutsche Forschungsgemeinschaft (grant MA 1426/21-1 to WFM).
Footnotes
CRediT authorship contribution statement
Natalia Mrnjavac: Writing − review & editing, Writing − original draft, Visualization, Investigation, Conceptualization. William F. Martin: Writing − review & editing, Writing − original draft, Visualization, Supervision, Investigation, Funding acquisition, Conceptualization.
Declaration of competing interest
The authors declare no competing interests.
Data availability
Data will be made available on request.
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9068 | https://math.fandom.com/wiki/Convergent_sequence | Convergent sequence
A Convergent Sequence is a sequence which becomes arbitrarily close to a specific value, called its "limit". The sequence may or may not take the value of the limit.
The formal definition of a sequence,
⟨
x
n
⟩
{\displaystyle \langle x_n\rangle}
, tending to a limit
L
{\displaystyle L}
is:
∀
ε
0
,
∃
N
∈
N
{\displaystyle \forall\varepsilon>0,\exists N\in\N}
such that
∀
n
≥
N
,
|
x
n
−
L
|
<
ϵ
{\displaystyle \forall n\ge N,|x_n-L|<\epsilon}
.
It may be written
lim
n
→
∞
x
n
=
L
{\displaystyle \lim_{n\to\infty}x_n=L}
, or
x
n
→
n
→
∞
L
{\displaystyle x_n\xrightarrow[n\to\infty]{}L}
.
Proving that a sequence converges from the definition requires knowledge of what the limit is. Therefore we introduce the notion of Cauchy sequences, which is an equivalent condition for convergence (by the General Principle of Convergence).
Examples[]
n
{\displaystyle n}
to the power of 0 (starting with
n
1
{\displaystyle n=1}
) converges to 1.
n
0
→
1
{\displaystyle n^0\to1}
The
n
{\displaystyle n}
-th root of
x
0
{\displaystyle x>0}
converges to 1.
x
n
→
1
{\displaystyle \sqrt[n]{x}\to1}
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9069 | https://www.stuartschool.org/uploaded/parents/back_to_school/2022-2023_Files/2022_Upper_School_Summer/Algebra_2_Summer_Assignment_2022.pdf | CPT Review 4/17/01 1 Algebra Review 1. Evaluate the expression 2 2 3 b a + when a = -3 and b = -4. A) 2 1 − B) 2 17 − C) 2 1 D) 2 17 2. Simplify: 4 6 5 3 − • + A) 17 B) 29 C) 16 D) 30 3. Simplify: 5 2 2 2 6 + • − A) 40 B) 18 C) 34 D) 12 4. Evaluate: x z y x − − 6 3 if x = 2, y = 8, and z = –2. A) 7 1 B) 7 1 − C) 5 1 D) 5 1 − CPT Review 4/17/01 2 5. Simplify: ( ) 4 2 30 14 − − A) –2 B) 2 C) 2 11 D) 2 11 − 6. Use the distributive property to simplify. ( ) − − + 3 10 x x A) − + 4 30 x B) − − 4 30 x C) − + 2 30 x D) − − 2 30 x 7. Simplify: 8y – 2 – 3(y – 4) A) 11y – 6 B) 5y – 6 C) 5y – 14 D) 5y + 10 8. Write the fraction in lowest terms: 2 4 2 3 24 36 c ab bc a A) 3 2 2 3 a b B) 2 3 3 2 a b C) 3 2 2 3 b a D) 3 2 3 2 b a CPT Review 4/17/01 3 9. Solve for x: 3 1 6 ( ) x + = − A) −2 B) −3 C) 1 D) 7 3 10. Add the polynomials: b a b a 7 5 3 2 − + + A) b a 4 7 − B) b a 4 7 + C) ab 3 D) b a 10 7 − 11. Subtract the polynomials: ( ) ( ) 2 2 3 11 4 9 2 2 + − − + − x x x x A) 9 2 6 2 + −x x B) 13 2 6 2 + − x x C) 9 6 6 2 + −x x D) 13 6 6 2 + −x x 12. ( )( )= + − + 4 2 2 2 x x x A) 8 3 + x B) 8 8 4 2 3 + − + x x x C) 8 8 4 2 3 + + − x x x D) 8 8 3 + + x x 13. The difference of twice a number and six is four times the number. Find an equation to solve for the number. A) 2x – 6 = 4 B) 2x – 6 = 4x C) 2x + 6 = 4x D) 2x – 6 = x + 4 14. Expand: ( ) 2 3 2 − x A) 9 4 2 − x B) 9 4 2 + x C) 9 12 2 2 + − x x D) 9 12 4 2 + − x x CPT Review 4/17/01 4 15. Which of the following numbers is the smallest? A) 4 3 − B) 2 3 − C) 1 − D) 3 2 − 16. Which of the following is the largest? A) 2 5 − B) 5 2 − C) 5 2− − D) 5 2 2 5 − + − 17. Solve: ( ) 8 5 3 − ≤ − x x A) 2 7 ≤ x B) 7 2 ≤ x C) 1 − ≤ x D) 1 − ≥ x 18. A flower-bed is in the shape of a triangle with one side twice the length of the shortest side and the third side 15 feet longer than the shortest side. If the perimeter is 100 feet and if x represents the length of the shortest side, find an equation to solve for the lengths of the three sides. A) x + 2x + x + 15 = 100 B) x + 15 = 2x C) x + 15 + 2x = 100 D) x + 2x = x + 115 19. If John has $50 more money than Mary and you choose to represent John’s amount of money as X how should you represent Mary’s amount of money in terms of X? A) X + $50 B) X - $50 C) $50 – X D) $50·X CPT Review 4/17/01 5 20. Multiply: ( ) 3 5 3 2 2 − −x x x A) x x x 6 5 6 2 3 − − B) 3 5 6 3 − −x x C) x x x 3 10 6 2 3 − − D) x x x 6 10 6 2 3 − − 21. Divide: m m m m 7 7 28 14 8 2 + − A) m m m 7 28 2 8 + − B) 1 4 2 7 + −m m C) 7 4 2 m m − D) m m m + − 8 2 4 2 22. Factor completely: 2 3 4 4 20 12 x x x + − One factor is: A) 4 4x B) 1 3 + x C) 1 − x D) 1 5 3 2 + −x x 23. Factor completely: 36 12 2 + − x x One factor is: A) x 6 B) 6 − x C) 12 − x D) 3 + x 24. Factor completely: 21 14 7 2 − + x x One factor is: A) x 7 B) 1 + x C) 3 − x D) 3 + x 25. Solve: 0 10 3 2 = − −x x One solution is: A) x = 10 B) x = 1 C) x = –2 D) x = 2 CPT Review 4/17/01 6 26. Solve: 0 5 2 2 = −x x The solutions are: A) x = 0 B) x = 0, x = 5 C) x = 0, x = –5 D) x = 0, 2 5 = x 27. Simplify and reduce: 18 9 12 3 2 + − x x A) 6 2 − x B) 3 2 3 − x C) 3 2 + x D) 3 2 − x 28. Given the equation –2x + 3y = 12, find the missing value in the ordered pair (–3,) A) –6 B) –2 C) 6 D) 2 29. What are the coordinates of the x-intercept in the graph below? -2 -1 0 1 2 3 2 1 Y X A) ( ) 2 , 1 − B) ( ) 2 , 0 − C) ( ) 0 , 2 − D) ( ) 1 , 0 CPT Review 4/17/01 7 30. Graph the line 3x + y = 6. A) B) C) D) CPT Review 4/17/01 8 31. Solve and simplify if possible: 3 9 3 2 − − − x x x = A) x x 3 3 − − B) -1 C) 3 + x D) 3 − x 32. Solve the following system of equations for the y-value: 13 2 2 7 2 = + = + y x y x A) 2 1 = y B) 6 = y C) 5 = y D) 4 13 = y 33. = − 2 3 2 6 5 a a A) 2 6 3 a B) a 3 3 C) 2 6 1 a D) 2 6 4 5 a a − 34. b b b 3 3 9 2 − = A) 1 9 2 − b B) 1 3 − b C) 1 − b D) b 9 CPT Review 4/17/01 9 35. The DoBee.Com Corporation has 5 more than three times as many female as male supervisors. If “x” represents the number of male supervisors write an expression that would represent the total number of female supervisors in terms of “x”. A) x + 5 B) 3x + 5 C) 4x + 5 D) 9x 36. Which of the following is not an equivalent statement? A) + − = − 5 2 5 2 25 4 2 b b b B) ( ) 3 3 3 2 2 5 − = − x x x x C) ( ) 7 3 4 x x = D) 4 4 1 x x = − 37. Charles needs enough fencing to enclose a rectangular garden with a perimeter of 140 feet. If the width of his garden is to be 30 feet, write the equation that can be used to solve for the length of the garden. A) x + 30 = 140 B) 2x + 30 = 140 C) 2x + 60 = 140 D) 140 – x = 60 38. Which one of the following ordered pairs is NOT a solution for the equation: 3x + y = 12? A) 2 1 10 , 2 1 B) (6,2) C) (12, -24) D) (2,6) 39. For what value(s) of x will each expression be undefined? x x x x 2 2 4 4 6 + + + − A) 6 B) -2 C) -2 and 3 D) 2 and -3 CPT Review 4/17/01 10 40. Which of the following is a factor of both expressions? 5 3 2 5 4 2 2 − + − + x x x x A) ( ) 5 + x B) ( ) 5 − x C) ( ) 1 − x D) ( ) 3 − x Solutions Elementary Algebra Sample Items 1. B Evaluate the expression 2 2 3 b a + when a = -3 and b = -4 ( ) ( ) 2 17 2 8 9 2 4 2 3 3 − − + − − + − 2. B Simplify: 4 6 5 3 − • + 29 4 33 4 30 3 4 6 5 3 = − − + − • + 3. C Simplify: 5 2 2 2 6 + • − 34 32 2 32 4 6 32 2 2 6 2 2 2 6 5 = + + − + • − + • − CPT Review 4/17/01 11 4. A x z y x − − 6 3 if x = 2, y = 8, and z = –2. ( ) ( ) 1/7 14 2 12-2 8 6 8 2 3 = − − − − 6 −2 − 5. B Simplify: ( ) 4 2 30 14 − − ( ) 2 8 16 4 2 30 14 = − − − − 6. C Use the distributive property to simplify. ( ) − − + 3 10 x x ( ) 30 2 30 3 10 3 + − + + − + − − x x x x x 7. D Simplify: 8y – 2 – 3(y – 4) ( ) 10 5 12 3 2 8 4 3 2 8 + + − − − − − y y y y y 8. C Write the fraction in lowest terms: 2 4 2 3 24 36 c ab bc a 3 2 3 0 2 1 4 2 2 1 3 2 4 2 3 2 3 2 3 2 3 24 36 b a b c a b c a c ab bc a − − − CPT Review 4/17/01 12 9. B Solve for x:3 1 6 ( ) x + = − ( ) ( ) ( ) 3 3 9 3 3 9 3 3 6 3 3 3 6 3 3 6 1 3 − = − = − = − − = − + − = + − = + x x x x x x 10. A Add the polynomials: b a b a 7 5 3 2 − + + b a b a b a b a 4 7 ) 7 3 ( ) 5 2 ( 7 5 3 2 − − + + − + + 11. A Subtract the polynomials: ( ) ( ) 2 2 3 11 4 9 2 2 + − − + − x x x x ( ) ( ) 9 2 6 2 2 3 11 4 9 2 2 3 11 4 9 2 2 2 2 + − − + − + − + − − + − x x x x x x x x x x 12. A ( )( ) 4 2 2 2 + − + x x x ( )( ) 8 8 0 0 8 4 2 4 2 4 2 2 x 3 2 3 2 2 3 2 + + + + + − + + − + − + x x x x x x x x x x x 13. B The difference of twice a number and six is four times the number. Find an equation to solve for the number. x x 4 6 2 Therefore equal means is subtract, means Difference = − CPT Review 4/17/01 13 14. D Expand: ( ) 2 3 2 − x ( ) ( )( ) 9 12 4 9 6 6 4 3 2 3 2 3 2 2 2 2 + − + − − − − − x x x x x x x x 15. B Which of the following numbers is the smallest? Make all of the numbers have common denominators and compare. 12 9 4 3 − = − numerator smallest the is 18 - since number smallest the is This 12 18 2 3 → − = − 12 12 1 − = − 12 8 3 2 − = − 16. C Which of the following is the largest? Solve each absolute value and compare results. Remember absolute value is always positive. 3 3 2 5 = = − 3 3 5 2 = − = − 7 7 5 2 = − = − − ? This is the largest 6 3 3 3 3 5 2 2 5 = + = − + = − + − 17. A Solve: ( ) 8 5 3 − ≤ − x x ( ) 2 7 2 7 2 2 7 2 15 8 15 15 2 8 15 2 8 15 3 8 15 3 8 5 3 ≤ ≤ ≤ + − ≤ + − − ≤ − − − ≤ − − − ≤ − − ≤ − x x x x x x x x x x x x x CPT Review 4/17/01 14 18. A A flower-bed is in the shape of a triangle with one side twice the length of the shortest side and the third side 15 feet longer than the shortest side. If the perimeter is 100 feet and if x represents the length of the shortest side, find an equation to solve for the lengths of the three sides. 2x x+15 x x+2x+x+15=100 19. B If John has $50 more money than Mary and you choose to represent John’s amount of money as X how should you represent Mary’s amount of money in terms of X? If John has $50 more than Mary than Mary must have $50 less than John. Therefore if John is represented by X, Mary would be represented by X-$50. 20. D Multiply: ( ) 3 5 3 2 2 − −x x x ( ) x x x x x x 6 10 6 3 5 3 2 2 3 2 − − − − 21. B Divide: m m m m 7 7 28 14 8 2 + − 1 4 2 7 7 7 28 7 14 7 7 28 14 7 8 2 8 2 + − = + − = + − m m m m m m m m m m m m 22. D Factor completely: 2 3 4 4 20 12 x x x + − . One factor is: ( ) 1 5 3 4 4 20 12 2 2 2 3 4 + − + − x x x x x x Therefore the correct listed factor is 1 5 3 2 + −x x CPT Review 4/17/01 15 23. B Factor completely: 36 12 2 + − x x . One factor is: ( )( ) 6 6 36 12 2 − − + − x x x x Therefore the correct listed factor is 6 − x . 24. D Factor completely: 21 14 7 2 − + x x . One factor is: ( ) ( )( ) 1 3 7 3 2 7 21 14 7 2 2 − + − + − + x x x x x x Therefore the correct listed factor is 3 + x . 25. C Solve: 0 10 3 2 = − −x x One solution is: 5 or x 2 0 5 or 0 2 0 ) 5 )( 2 ( 0 10 3 2 = − = = − = + = − + = − − x x x x x x x Therefore the correct listed solution is –2. 26. D Solve: 0 5 2 2 = −x x . The solutions are: ( ) 2 5 x 5 2x 0 5 2 or 0 0 5 2 0 5 2 = = = − = = − = − x x x x x x Therefore your answers are 2 5 and 0 . 27. D Factor and reduce: 18 9 12 3 2 + − x x ( ) ( ) ( )( ) ( ) ( ) 3 2 2 9 2 2 3 2 9 4 3 18 9 12 3 2 2 − = + − + = + − = + − x x x x x x x x CPT Review 4/17/01 16 28. D Given the equation –2x + 3y = 12, find the missing value in the ordered pair (–3,). Substitute the “-3” into the equation for x and solve for y. 2 6 3 12 3 6 12 3 ) 3 ( 2 12 3 2 = = = + = + − − = + − y y y y y x 29. C The x-axis is the horizontal axis. The line intersects the x-axis at ( ) 0 , 2 − . 30. C Using the intercept method to graph a line, substitute “0” in for the x value and solve for y. Then substitute “0” in for the y value and solve for x. Then plot these point and draw the line. ( ) ( ) 0 , 2 and 6 , 0 are intercepts 2 x 6 6 3x 6 0 6 0 3x 6 ) 0 ( 3 = = = = + = + = + y y y 31. C Solve and simplify if possible: 3 9 3 2 − − − x x x ( )( ) 3 3 3 3 3 9 3 9 3 2 2 + = − − + = − − = − − − x x x x x x x x x 32. A Solve the following system of equations for the y-value: 13 2 2 7 2 = + = + y x y x ( ) 13 2y 2x 13 2 2 7 2y x 2 - 2 -by multiply 7 2 = + = + = + = + y x y x Now add the equations vertically 2 1 y 2 1 2 -2y - -1 2y - 13 2 2 14 4 2 = − − = = = + − = − − y x y x CPT Review 4/17/01 17 33. D = − 2 3 2 6 5 a a First find common denominators for your fractions. 2 2 2 2 6 4 5 numerators the subtract Now 6 4 2 2 3 2 and 6 5 6 5 a a a x a a a a a x a − = = 34. B b b b 3 3 9 2 − Factor the numerator ( ) b b b 3 1 3 3 − . Now reduce. Answer 3b-1 35. B The DoBee.Com Corporation has 5 more than three times as many female as male supervisors. If “x” represents the number of male supervisors write an expression that would represent the total number of female supervisors in terms of “x”. The statement “five more than” indicates adding “5” to the product of 3x. Answer ? 5 3 + x 36. C Which of the following is not an equivalent statement? + − = − 5 2 5 2 25 4 2 b b b This is the correct factored form of the given. ( ) 3 3 3 2 2 5 − = − x x x x This is the correct factored form of the given. ( ) 7 3 4 x x = This is incorrect. ( ) 12 3 4 x x = 4 4 1 x x = − This is correct 37. C Charles needs enough fencing to enclose a rectangular garden with a perimeter of 140 feet. If the width of his garden is to be 30 feet, write the equation that can be used to solve for the length of the garden. 2 lengths + 2 widths = perimeter Let the unknown length be “x” ( ) 140 60 2 140 30 2 2 = + = + x x 38. B Which one of the following ordered pairs is NOT a solution for the equation: 3x + y = 12? The solution is ( ) 2 , 6 because when substituted into the equation your solution is incorrect. ( ) 12 2 6 3 ≠ + CPT Review 4/17/01 18 39. D For what value(s) of x will each expression be undefined? x x x x 2 2 4 4 6 + + + − An expression is undefined when the denominator equals zero. To solve set the denominator equal to zero and solve for x. ( )( ) 2 or 3 0 2 -or x 0 3 0 2 3 0 6 2 = − = = = + = − + = − + x x x x x x x 40. C Which of the following is a factor of both equations? 5 3 2 5 4 2 2 − + − + x x x x Factor both equations and compare results ( )( ) 1 5 5 4 2 − + − + x x x x ( )( ) 1 5 2 5 3 2 2 − + − + x x x x The matching factor is (x-1) |
9070 | https://www.flippingphysics.com/uploads/2/1/1/0/21103672/0402_lecture_notes_-_work_done_by_a_spring_-_deriving_elastic_potential_energy.pdf | 0402 Lecture Notes - Work done by a Spring - Deriving Elastic Potential Energy.docx page 1 of 1 Flipping Physics Lecture Notes: Work done by a Spring – Deriving Elastic Potential Energy When we place a 1.00 kg mass on a spring, the end of the spring moves from the 34.4 cm mark to the 30.2 cm mark and comes to rest. (a) What is the spring constant of the spring? (b) What is the work done by the spring on the mass? Where x is the displacement from equilibrium position in Hooke’s Law. Draw the free body diagram: For part (b), there are two equations for work: Which we are not going to use because the force of a spring is not constant. Which we are going to use because we can use this when the force is not constant. I need to point out that the work done by the spring on the mass equals the negative of the elastic potential energy stored in the spring. The spring does negative work on the mass because the force of the spring and displacement of the mass are in opposite directions. And that energy is now stored in the spring as elastic potential energy. |
9071 | https://fiveable.me/heat-mass-transfer/unit-2/one-dimensional-steady-state-conduction/study-guide/EXtv7APlgslOsqHt | printables
❤️🔥Heat and Mass Transfer Unit 2 Review
2.1 One-Dimensional Steady-State Conduction
❤️🔥Heat and Mass Transfer Unit 2 Review
2.1 One-Dimensional Steady-State Conduction
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
APA
❤️🔥Heat and Mass Transfer
Unit & Topic Study Guides
2.1 One-Dimensional Steady-State Conduction
2.2 Thermal Conductivity and Thermal Resistance
2.3 Heat Diffusion Equation
2.4 Multidimensional and Unsteady Conduction
2.5 Numerical Methods for Conduction Problems
One-dimensional steady-state conduction is the foundation of heat transfer analysis. It covers how heat moves through materials in a straight line when conditions don't change over time. This concept is key to understanding more complex heat transfer scenarios.
Fourier's law, thermal conductivity, and boundary conditions are crucial in this topic. We'll explore how these factors affect heat flow and temperature distribution in various systems, from simple walls to multi-layer structures with heat generation.
Heat Flux and Temperature Distribution
Fourier's Law and One-Dimensional Heat Flux
Fourier's law states that the heat flux is proportional to the negative temperature gradient, with the proportionality constant being the thermal conductivity of the material
The one-dimensional form of Fourier's law is $q" = -k(dT/dx)$, where $q"$ is the heat flux, $k$ is the thermal conductivity, and $dT/dx$ is the temperature gradient
The heat transfer rate can be calculated using $Q = -kA(dT/dx)$, where $A$ is the cross-sectional area perpendicular to the direction of heat transfer
Fourier's law is used to calculate heat flux and temperature distribution in one-dimensional steady-state conduction problems (plane wall, cylindrical, or spherical systems)
Steady-State Conduction and Boundary Conditions
For steady-state conduction, the temperature distribution is linear in a plane wall, cylindrical, or spherical system with constant thermal conductivity and no internal heat generation
Boundary conditions, such as specified temperature or heat flux, are necessary to determine the temperature distribution and heat transfer rate in a system
Examples of boundary conditions include constant surface temperature (isothermal), constant heat flux, convection, and radiation
Boundary conditions are used to solve the heat diffusion equation and determine the temperature profile and heat transfer rate in a system
Thermal Conductivity of Materials
Definition and Dependence on Material Properties
Thermal conductivity is a material property that quantifies the ability of a substance to conduct heat
The thermal conductivity of a material depends on its composition, structure, and temperature
Materials with high thermal conductivity, such as metals (copper, aluminum), efficiently transfer heat, while materials with low thermal conductivity, such as insulators (fiberglass, polyurethane foam), resist heat transfer
Thermal conductivity is an important consideration in the selection of materials for various applications, such as insulation, heat exchangers, and electronic devices
Temperature Dependence and Data
The temperature dependence of thermal conductivity can be described by empirical equations or tabulated data for specific materials
Thermal conductivity data is essential for accurate modeling and calculation of heat transfer in various applications
Examples of temperature-dependent thermal conductivity include increasing thermal conductivity with temperature for metals and decreasing thermal conductivity with temperature for gases
Thermal conductivity data can be found in handbooks, databases, or obtained through experimental measurements
Conduction with Heat Generation
Heat Generation and Its Effects
Heat generation within a material can occur due to chemical reactions, nuclear reactions, or electrical resistance heating (Joule heating)
The presence of heat generation affects the temperature distribution and heat transfer in a system, leading to a nonlinear temperature profile
The heat diffusion equation with heat generation is $d/dx(k(dT/dx)) + \dot{q} = 0$, where $\dot{q}$ is the volumetric heat generation rate
Examples of systems with heat generation include nuclear fuel rods, electrical heating elements, and exothermic chemical reactions
Solving Conduction Problems with Heat Generation
Analytical solutions for conduction problems with heat generation are available for simple geometries and boundary conditions
Numerical methods, such as finite difference or finite element methods, are often employed to solve complex conduction problems with heat generation
Variable thermal conductivity, which depends on temperature, also results in a nonlinear temperature distribution
Examples of analytical solutions include the temperature distribution in a plane wall with uniform heat generation and constant surface temperatures
Multi-Layer Conduction with Resistance
Multilayer Systems and Thermal Resistance
Multilayer systems consist of two or more materials with different thermal conductivities in series (composite walls, insulated pipes)
The temperature distribution in a multilayer system is piecewise linear, with discontinuities at the interfaces between layers
The overall heat transfer rate in a multilayer system can be determined using the concept of thermal resistance, which is the reciprocal of the product of thermal conductivity and area for each layer
The effective thermal conductivity of a multilayer system can be calculated using the sum of the individual thermal resistances
Thermal Contact Resistance
Thermal contact resistance occurs at the interface between two materials due to surface roughness, air gaps, or imperfect contact
Thermal contact resistance is modeled as an additional thermal resistance in series with the layers, and its value depends on factors such as surface finish, contact pressure, and interstitial materials
Examples of thermal contact resistance include the interface between a heat sink and a microprocessor, or between two mating surfaces in a heat exchanger
Thermal contact resistance can be minimized by applying thermal interface materials (thermal grease, thermal pads) or increasing contact pressure |
9072 | https://www.convertunits.com/from/g/cm3/to/kg/m3 | Convert gram/cubic centimetre to kilogram/cubic metre
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The SI derived unit for density is the kilogram/cubic meter.
1 g/cm3 is equal to 1000 kilogram/cubic meter.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between grams/cubic centimeter and kilograms/cubic meter.
Type in your own numbers in the form to convert the units!
Quick conversion chart of g/cm3 to kg/m3
1 g/cm3 to kg/m3 = 1000 kg/m3
2 g/cm3 to kg/m3 = 2000 kg/m3
3 g/cm3 to kg/m3 = 3000 kg/m3
4 g/cm3 to kg/m3 = 4000 kg/m3
5 g/cm3 to kg/m3 = 5000 kg/m3
6 g/cm3 to kg/m3 = 6000 kg/m3
7 g/cm3 to kg/m3 = 7000 kg/m3
8 g/cm3 to kg/m3 = 8000 kg/m3
9 g/cm3 to kg/m3 = 9000 kg/m3
10 g/cm3 to kg/m3 = 10000 kg/m3
Want other units?
You can do the reverse unit conversion from
kg/m3 to g/cm3, or enter any two units below:
Common density conversions
g/cm3 to tonne/milliliter
g/cm3 to grain/cubic foot
g/cm3 to ton/gallon
g/cm3 to megagram/milliliter
g/cm3 to grain/gallon
g/cm3 to ounce/cubic foot
g/cm3 to slug/cubic foot
g/cm3 to gram/milliliter
g/cm3 to pound/cubic foot
g/cm3 to gram/cubic decimeter
Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! |
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Adding Fractions – Definition With Examples
Home » Math Vocabulary » Adding Fractions – Definition With Examples
What is “Adding Fractions”?
How To Add Fractions?
Formulas For Adding Fractions
Solved Examples On Adding Fractions
Practice Problems On Adding Fractions
Frequently Asked Questions On Adding Fractions
What is “Adding Fractions”?
Adding fractions can be discussed using two cases:
Adding fractions with like denominators
Adding fractions with unlike denominators
A fraction represents part of a whole. A fraction has two parts: a numerator and a denominator. A numerator represents the parts taken from a whole, written at the top; the denominator represents the number of equal parts the whole is divided into, written at the bottom.
For example, $\frac{5}{6},\; \frac{4}{5},\; \frac{3}{10},\; \frac{8}{9}$, and $\frac{1}{3}$. In the given image, the whole cake is cut into 8 equal pieces. If you take out 3 pieces, they represent the fraction $\frac{3}{8}$.
The addition of fractions is the method of finding the sum of two fractions. To add two fractions, we first make their denominators same (by LCM method or by rationalization) and then add the numerators, keeping the new denominator common.
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Adding Fractions: Definition
Adding fractions refers to finding the sum of two or more fractions with same or different denominators.
Adding fractions with the same denominator is simple. We add the numerators and keep the denominator the same.When the denominators are different, we first have to make the denominators same.
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How To Add Fractions?
Let us look at the steps for adding two fractions.
Step 1: Check whether the given fractions are like or unlike fractions. Find out whether the denominators are the same.
Step 2: If the denominator is the same, find the sum of the numerators and place the sum over the common denominator. Simplify the final fraction if required.
Example: $\frac{2}{5} + \frac{1}{5} = \frac{2 + 1}{5} = \frac{3}{5}$
Step 3: If the denominators are different, the fractions are unlike fractions. We will have to convert them into like fractions by making the denominators the same. We will learn different methods to do so later in the article.
What Are Like Fractions and Unlike Fractions?
The fractions that have the same denominator are called “like fractions.”
Example: $\frac{2}{5},\; \frac{1}{5},\; \frac{3}{5},\; \frac{4}{5}$
The fractions that have different denominators are called “unlike fractions.”
Example: $\frac{2}{5},\; \frac{1}{4},\; \frac{1}{2},\; \frac{3}{7}$
Adding Fractions with Like Denominators
Adding fractions is the easiest when the denominators of the fractions are the same. Hence, if denominators of two or more fractions are the same, we can directly add their numerators, and keep the denominator the same.
Example 1: Add the fractions: $\frac{3}{8}$ and $\frac{1}{8}$.
Since the denominators are like, therefore we can add the numerators directly.
$\frac{3}{8} + \frac{1}{8} = \frac{3 + 1}{8} = \frac{4}{8}$
Now, simplify the fraction $\frac{4}{8} = \frac{1}{2}$
Hence, the sum of $\frac{3}{8}$ and $\frac{1}{8}$ is $\frac{1}{2}$.
Example 2: $\frac{1}{5} + \frac{2}{5} = \frac{3}{5}$
Example 3: $\frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1$
Adding Fractions with Unlike Denominators
Now that we have learnt how to add like fractions, let us move on to the addition of unlike fractions. When we add two or more fractions with different denominators, then we cannot add the numerators directly.
Let’s discuss methods we can use to convert the unlike fractions into like fractions.
LCM Method
Let’s solve $\frac{2}{5} + \frac{3}{7}$ using the LCM method.
Here, we are adding two fractions with different denominators.
Find the LCM of the denominators, 5 and 7.
LCM of 5 and $7 = 35$
Multiply 25 by 77 so that the denominator becomes 35.
$\frac{2}{5} \times \frac{7}{7} = \frac{14}{35}$
Now, multiply $\frac{3}{7}$ by $\frac{5}{5}$.
$\frac{3}{7} \times \frac{5}{5} = \frac{15}{35}$
The fractions $\frac{14}{35}$ and $\frac{15}{35}$ are like fractions.
Now, add the numerators and keep 35 as the denominator.
$\frac{14}{35} + \frac{15}{35} = \frac{29}{35}$
We cannot simplify this further.
Hence, the sum of $\frac{2}{5} + \frac{3}{7} = \frac{29}{35}$.
Cross Multiplication Method
This method works in all cases but when the numerators and denominators of the two unlike fractions are small numbers, the cross multiplication method is convenient to use.
Example: $\frac{3}{7} + \frac{2}{5}$
Adding simple fractions like these is much easier with the cross multiplication method.
Multiply the numerator of the first fraction with the denominator of the second fraction.
$(3 \times 5) = 15$
Multiply the denominator of the first fraction with the numerator of the second fraction.
$(7 \times 2) = 14$
Add these products together and write the sum as the numerator. Multiply the denominators and write the product as the denominator of the final answer.
$15 + 14 = 29$
$7 \times 5 = 35$
Thus, $\frac{3}{7} + \frac{2}{5} = \frac{29}{35}$
We can write all the steps together as follows:
$\frac{3}{7} + \frac{2}{5} = \frac{(3 \times 5) + (7 \times 2)}{7 \times 5} = \frac{15 + 14}{35} = \frac{29}{35}$
Adding unlike fractions can be represented visually as follows:
Note that when the denominators of two numbers are co-prime (HCF $= 1$, the denominators that do not have common factors, except 1), we can use the cross multiplication method.
Adding Fractions with Co-prime Denominators
Co-prime or relatively prime numbers are numbers that have only 1 as a common factor. How to add fractions with co-prime denominators? Let’s understand with an example.
Add the fractions $\frac{5}{7}$ and $\frac{3}{4}$.
We see that the denominators 7 and 4 are co-prime because they have only one common factor which is 1. Such fractions can be added easily using the cross multiplication method!
$\frac{5}{7} + \frac{3}{4} = \frac{(5 \times 4 ) + ( 3 \times 7)}{7 \times 4} = \frac{20 + 21}{28} = \frac{41}{28}$
Addition of a Fraction and a Whole Number
Let us learn how to add a fraction and a whole number with the below mentioned steps:
Write the given whole number in the form of a fraction (for example $5 = \frac{5}{1}$).
Make the denominators the same.
Add the fractions.
Simplify the sum.
For example: Add \frac{5}{2} + 3$
Here, $\frac{5}{2}$ is a fraction and 3 is a whole number.
We can write 3 as $\frac{3}{1}$.
Now after making the denominators same, we get;
$\frac{5}{2}$ and $\frac{3}{1} \times \frac{2}{2} = \frac{6}{2}$
Now, add $\frac{5}{2}$ and $\frac{6}{2}$.
$\frac{5}{2} + \frac{6}{2} = \frac{11}{2} = 5\frac{1}{2}$
Therefore, the sum of $\frac{5}{2}$ and 3 is $5\frac{1}{2}$.
Adding Mixed Fractions
A mixed fraction is a fraction, which consists of a whole number and a fraction. To add two mixed fractions, first we can convert them into improper fractions and then add them together.
Let’s look at an example to understand it better.
Example: Add : $3\frac{1}{4} + 2\frac{3}{4}$
First, convert the given mixed fractions into improper fractions as shown below.
$3\frac{1}{4} = \frac{(4 \times 3) + 1}{4} = \frac{13}{4}$
$2\frac{3}{4} = \frac{(4 \times 2) + 3}{4} = \frac{11}{4}$
The fractions are like fractions.
$\frac{13}{4} + \frac{11}{4} = \frac{13 + 11}{4} = \frac{24}{4} = 6$
Hence, the sum of $3\frac{1}{4}$ and $2\frac{3}{4}$ is 6.
Adding Fractions with Variables
If the numerators of the given fractions have a variable coefficient, we can still add the fractions using the similar steps.
Example 1: Add $\frac{x}{5} + \frac{2x}{5}$, where “x” is the variable.
Solution: Let us learn how to add these fractions using the following steps:
Step 1: The given fractions, $\frac{x}{5} + \frac{2x}{5}$ are like fractions since they have the same denominator and we can see that “x” is common.
Step 2: We can take out the common factor and rewrite it as: $\frac{x}{5} + \frac{2x}{5} = (\frac{1}{5} + \frac{2}{5})x = \frac{3x}{5}$
Step 3: Therefore, $\frac{x}{5} + \frac{2x}{5} = \frac{3x}{5}$
Example 2: Add unlike fractions: $\frac{x}{3} + \frac{x}{5}$
Solution: Let us learn how to add the fractions using the following steps.
We know that the given fractions, $\frac{x}{3} + \frac{x}{5}$ are unlike fractions, we will take the LCM of the denominators and convert them into like fractions.
Next, we need to take the common variable out and rewrite it as follows:
LCM $(3,\; 5) = 15$
$\frac{x}{3} = \frac{x}{3} \times \frac{5}{5} = \frac{5x}{15}$
$\frac{x}{5} = \frac{x}{5} \times \frac{3}{3} = \frac{3x}{15}$
We got two fractions with common denominators, $\frac{5x}{15} + \frac{3x}{15} = \frac{5x + 3x}{15} = \frac{8x}{15}$
Therefore, the sum of $\frac{x}{2} + \frac{x}{3} = \frac{8x}{15}$
Note: Different variables, like “x” and “y” are treated as unlike terms.
Formulas For Adding Fractions
Let’s summarize the formulas we discussed for adding fractions.
Adding fractions with like denominators: $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$
Adding fractions with unlike denominators: $\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$
Adding a fraction and a whole number: $w + \frac{a}{b} = \frac{a + wb}{b}$
Fun Fact!
Subtracting fractions also uses the same set of rules as addition of fractions, only the operation is different.
Conclusion
In this article, we learned how to add fractions. We learned how to add like fractions, unlike fractions, mixed fractions, fractions and whole number addition, etc. Now let us solve a few examples on adding fractions along with practice problems to better understand the concepts.
Solved Examples On Adding Fractions
1. What is the sum of $\frac{5}{30}$ and $\frac{10}{30}$?
Solution:
Since, the denominators of the given fractions are equal, the fractions are like fractions.
Add the numerators.
$\frac{5}{30} + \frac{10}{30} = \frac{5 + 10}{30} = \frac{15}{30}$
In the end, we have to simplify the fraction.
$\frac{15}{30} = \frac{1}{2}$
Therefore, the sum of $\frac{5}{30}$ and $\frac{10}{30}$ is $\frac{1}{2}$.
2. What is the sum of $\frac{7}{10}$ and $\frac{4}{15}$?
Solution:
First, we have to check if the given fractions are like fractions or unlike fractions. Since, the denominators of the given fractions are not equal, hence, the fractions are unlike fractions.
Now, let us make the denominator equal by taking the LCM of 10 and 15.
The LCM of 10 and 15 is equal to 30.
Now let us make both the denominators equal.
$\frac{7}{10} \times \frac{3}{3} = \frac{21}{30}$ and,
$\frac{4}{15} \times \frac{2}{2} = \frac{8}{30}$
Now, let us add both the fractions
$\frac{21}{30} + \frac{8}{30} = \frac{21 + 8}{30} = \frac{29}{30}$.
Therefore the sum of $\frac{7}{10}$ and $\frac{4}{15}$ is equal to $\frac{29}{30}$.
3. What is the sum of $\frac{4}{5}$ and 7?
Solution:
Here, $\frac{4}{5}$ is a fraction and 7 is a whole number.
We can write 7 as $\frac{7}{1}$.
Now, after making the denominators same, we get:
$\frac{7}{1} \times \frac{5}{5} = \frac{35}{5}$
Now, add $\frac{4}{5}$ and $\frac{35}{5}$
$\frac{4}{5} + \frac{35}{5} = \frac{39}{5}$
Therefore, the sum of 45 and 7 is 395.
4. What is the sum of $3\frac{2}{5}$ and $6\frac{3}{7}$ ?
Solution:
First, convert the given mixed fractions to improper fractions.
$3\frac{2}{5} = \frac{17}{5}$
$6\frac{3}{7} = \frac{45}{7}$
Make the denominators the same by taking LCM.
The LCM of 5 and 7 is 35.
$\frac{17}{5} \times \frac{7}{7} = \frac{119}{35}$
$\frac{45}{7} \times \frac{5}{5} = \frac{225}{35}$
Now, add the numerators.
$\frac{119}{35} + \frac{225}{35} = \frac{119 + 225}{35} = \frac{344}{35}$
Hence, the sum of $3\frac{2}{5}$ and $6\frac{3}{7}$ is $\frac{344}{35}$.
5. Add the fractions $\frac{y}{5}$ and $\frac{3y}{5}$.
Solution:
Let us look at the steps to add the fractions:
Step 1: The given fractions, $\frac{y}{5}$ + $\frac{3y}{5}$ are like fractions since they have the same denominator and we can see that “y” is common.
Step 2: We can take out the common factor and rewrite it as: $\frac{y}{5} + \frac{3y}{5} = (\frac{1}{5} + \frac{3}{5})y = \frac{4y}{5}$
Step 3: Therefore, the sum of $\frac{y}{5} + \frac{3y}{5} = \frac{4y}{5}$
Practice Problems On Adding Fractions
Adding Fractions - Definition With Examples
Attend this quiz & Test your knowledge.
1
What is the sum of $\frac{17}{20}$ and $\frac{3}{20}$?
1
2
3
4
CorrectIncorrect
Correct answer is: 1$\frac{17}{20} + \frac{3}{20} = \frac{17 + 3}{20} = \frac{20}{20} = 1$. Therefore, the sum of $\frac{17}{20}$ and $\frac{3}{20}$ is 1.
2
What is the sum of $\frac{12}{21}$ and $\frac{3}{7}$?
1
2
3
4
CorrectIncorrect
Correct answer is: 1$\frac{12}{21} \times \frac{1}{1} = \frac{12}{21}$ and $\frac{3}{7} \times \frac{3}{3} = \frac{9}{21}$ Now let us add both the fractions. $\frac{12}{21} + \frac{9}{21} = \frac{12 + 9}{21} = \frac{21}{21} = 1$
3
What is the sum of $\frac{9}{10}$ and 5?
CorrectIncorrect
Correct answer is: $\frac{59}{10}$Here, $\frac{9}{10}$ is a fraction and 5 is a whole number. We can write 5 as $\frac{5}{1}$. Now after making the denominators same, we get; $\frac{9}{10}$ and $\frac{5}{1} \times \frac{10}{10} = \frac{50}{10}$ Now, add $\frac{9}{10}$ and $\frac{50}{10}$ $\frac{9}{10} + \frac{50}{10} = \frac{59}{10}$ Therefore, the sum of $\frac{9}{10}$ and 5 is $\frac{59}{10}$.
4
What is the sum of $4\frac{2}{3}$ and $7\frac{1}{3}$ ?
CorrectIncorrect
Correct answer is: 12First, convert the given mixed fractions to improper fractions. $4\frac{2}{3} = \frac{14}{3}$ $7\frac{1}{3} = \frac{22}{3}$ We see that the fractions are like fractions. $\frac{14}{3} + \frac{22}{3} = \frac{14 + 22}{3} = \frac{36}{3}$ This is equal to 12 after simplification.
5
What is the result after adding the fractions $\frac{y}{10}$ and $\frac{4y}{5}$?
CorrectIncorrect
Correct answer is: $\frac{9y}{10}$The given fractions, $\frac{y}{10} + \frac{4y}{5}$ are unlike fractions since they have different denominators and we can see that “y” is common. Here, $\frac{4y}{5} \times \frac{2}{2} = \frac{8y}{10}$ $\frac{y}{10} + \frac{4y}{5} = (\frac{1}{10} + \frac{8}{10} ) y = \frac{9y}{10}$
Frequently Asked Questions On Adding Fractions
What are equivalent fractions?
Fractions that have different numerators and denominators but still represent the same value are called equivalent fractions. Example: $\frac{1}{2},\; \frac{2}{4}$
What is a proper fraction?
Proper fractions are fractions in which the numerator is less than the denominator. The decimal value of a proper fraction is always less than 1.
How do we multiply fractions?
To multiply any two fractions, we first multiply the numerators, then multiply the denominators. Finally, we simplify the resultant fraction. For example, $\frac{3}{5} \times \frac{7}{9} = \frac{21}{45} = \frac{7}{15}$
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Question: Suppose that ac + bd = 0. Show that vectors [a, b] and [c, d] are perpendicular. The number ac + bd is called the dot product of the vectors [a, b] and [c, d].
Suppose that ac + bd = 0. Show that vectors [a, b] and [c, d]
are perpendicular. The number ac + bd is called the dot product of
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Given that and recognizing that is the dot product of the vectors and , what follows is the next step in the solution.
When the dot product is zero, it implies that , hence working with this definition of the dot product is crucial for the proof.
Solution 1 :-
Here we're given ac+bd = 0
And we also know that ac+bd is the dot product of the
vectors [a, b] and [c, d]
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Bound of derivative of Lipschitz continuous function
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I have a function f(x)=g(x)T x, where f:R n→R, g(x):R n→R n, x∈R n. Furthermore assume g(0)=0, ‖g(x)‖ is bounded and also the partial derivative ‖∂g(x)/∂x i‖ is bounded for all x∈R n.
Since g(x) is bounded, it holds that ‖f(x)‖≤‖g(x)‖max‖x‖ and if ‖g(x)‖max=:L, then L can be considered as a Lipschitz constant.
Now, with this information, I am looking for an upper bound for ∂f(x)/∂x i. It is possible to find a bound ‖∂f(x)/∂x i‖≤˜L independent of x?
Since ∂f(x)/∂x i=(∂g(x)/∂x i)x+g(x)T e i it is clear that ‖∂f(x)/∂x i‖≤‖(∂g(x)/∂x i)‖max‖x‖+L.
I am wondering if I could obtain a better result. I considered using the mean value theorem, but at this point I got stuck.
Can anybody give me a hint how to say from this something about the bound of ‖∂f(x)/∂x i‖? Also, if I have an error in the derivation please let me know.
lipschitz-functions
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edited Aug 28, 2018 at 21:22
Michael Hardy
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asked Aug 28, 2018 at 19:36
CarlosCarlos
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I heavily edited your question, feel free to reject the edit. Also, could you provide some more context on this question? Is it related to some exercise or a greater result you need for some specific purpose?Mefitico –Mefitico 2018-08-28 19:49:46 +00:00 Commented Aug 28, 2018 at 19:49
@Mefitico: It is not an excercise itself, it is used in a Lyapunov stability proof. In what point is the MVT wrong?Carlos –Carlos 2018-08-28 20:00:12 +00:00 Commented Aug 28, 2018 at 20:00
1 @Carlos: You wrote (∂f(ξ)/∂x)x=g(x)x, which is unclear if intended to be a gradient times a vector (∇f(ξ)T x) (that would be correct) or a single partial derivative (∂f/∂x i) which is a scalar, times a vector (x, which indeed would be wrong). Maybe "wrong" was a strong word, but this is misleading notation.Mefitico –Mefitico 2018-08-28 20:06:42 +00:00 Commented Aug 28, 2018 at 20:06
@Mefitico: Ok, I agree. Thanks for editing. For simple functions like f=arctan(x)x with x∈R, a bound |f′|≤˜L, can be found. However, with ˜L>|arctan(x)|max=π/2. Actually ˜L=π/2+1/2. Do you have a hint regarding the more general case?Carlos –Carlos 2018-08-28 20:12:44 +00:00 Commented Aug 28, 2018 at 20:12
Can't really think of any good hints, but see my answer: I have a felling no such bound could be found, and attempted proving by counter-example. Maybe adding a few more assumptions would do your trick.Mefitico –Mefitico 2018-08-28 20:31:45 +00:00 Commented Aug 28, 2018 at 20:31
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We have
∂f∂x i(x)=lim t→0 f(x+t⋅e i)−f(x)t=lim t→0 g(x+t⋅e i)T(x+t⋅e i)−g(x)T x t=lim t→0(g(x+t⋅e i)T−g(x)T)x t+g(x+t⋅e i)T e i=(lim t→0 g(x+t⋅e i)−g(x)t)T x+g(x)T e i=(∂g∂x i(x))T x+g(x)T e i
Since g is bounded, g(x)T e i also is. However, it not enough to know that ∂g∂x i is bounded to conclude that the term (∂g∂x i(x))T x also is. It would seem one needs some decay condition on ∂g∂x i(x) as ‖x‖ increases.
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answered Aug 28, 2018 at 20:39
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Let k be any constant vector and g(x)=k sin(||x||)
Then g(x) is bounded with g(0)=0. The partial derivative is given by: ∂g∂x i=k i x i||x||cos(||x||)
So also bounded. Thus satisfying your conditions.
However even for the one-dimensional case with k=1: f(x)=x sin(|x|)
f′(x)=sin(|x|)+sign(x)x cos(|x|)
We have that f′(x) is unbounded. Granted that this function is not continuously differentiable on 0. I'm thus suspecting that at least more conditions should be given so that there could exist the requested bound.
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edited Aug 28, 2018 at 20:37
answered Aug 28, 2018 at 20:17
MefiticoMefitico
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Thanks for your answer. Now I also assume that further conditions may be necessary, like a decay condition, as Fimpellizieri mentioned in his answer. Also the sine fuction in your counterexample is not decaying, rather it is varying fast as x tends to infinity.Carlos –Carlos 2018-08-28 20:51:55 +00:00 Commented Aug 28, 2018 at 20:51
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9076 | https://www.partmfg.com/thermal-conductivity-of-aluminum-aluminum-grades-and-alternative-choices/ | Skip to content
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Thermal Conductivity of Aluminum, Aluminum Grades And Alternative Choices
This guide covers the thermal conductivity of aluminum and aluminum grades for the most accurate experimental readings.
Aluminum is usually known for its lightweight properties or its electrical conductivity. Surprisingly, it also ranks high among the best thermally conductive metals. This explains why it is a popular choice for many heat-related applications.
When choosing a metal for such applications, it’s almost impossible to find the metal in its pure elemental state.
Aluminum is usually mixed with alloying metals to enhance its properties, and for your project calculations, you should usually use aluminum’s accurate thermal conductivity.
This guide covers the thermal conductivity of aluminum and aluminum grades for the most accurate experimental readings.
What Is Thermal Conductivity?
Thermal conductivity is the property of a material that allows it to transfer heat.
Materials measuring high in thermal conductivity can transfer heat faster to another object and those low in thermal conductivity take longer to transfer the same amount of heat.
The symbol for thermal conductivity is k or
The SI units for thermal conductivity are Wm-1K-1.
Thermally conductive metals also absorb heat more readily from the environment. Their temperature changes more readily than for example wood.
Thermal Conductivity of Aluminum
The thermal conductivity of aluminum is 237 W/mK. Aluminum ranks 7th in terms of best thermal conductivity behind gold and silver but –is roughly 400 times cheaper than silver.
This makes it a popular choice for applications like heat exchangers and radiators.
Heat Capacity of Aluminum
There is another similar concept to thermal conductivity. Heat capacity is the amount of energy needed by a substance to produce a change of 1 unit in its temperature.
In simpler terms, the energy in Joules (J) needed to increase the temperature of a metal, water, or even a gas by 1°C.
The heat capacity of aluminum is 0.90 J/g°C. For reference, water has a heat capacity of 4.184 J/g°C. So, it would take about 4 times the amount of heat energy to change the temperature of water by 1°C.
How is Thermal Conductivity Measured?
Thermal conductivity or thermal resistance can be measured in many ways. Some involve a static analysis while others are time dependent.
Hot plate method
This process involves using a hot plate at a fixed temperature and a cold plate to initiate heat flow. A guarded or insulated hot plate is placed on top with the cold plate at the bottom and sensors at different heights between them.
This allows the observer to get a temperature gradient to use in the given formula.
k=QdAT
Where Q is the heat flow rate
d is the sample thickness
A is the cross-sectional area
Thermal Conductivity Table For Aluminum And Other Metals
| | |
--- |
| Material | Thermal Conductivity (W/m K) |
| Diamond | 2,200 |
| Silver | 406 |
| Copper | 398 |
| Gold | 315 |
| Aluminum Nitride | 321 |
| Aluminum | 237 |
| Tungsten | 175 |
| Magnesium Oxide | 48.4 |
| Steel | 45 |
| Silicon | 139.4 |
| Lead | 34.7 |
| Glass | 1.38 |
| Brass | 111 |
| Mercury | 8.3 |
| Iron | 73 |
| Brick | 1.0 |
| Concrete | 1.95 |
| Cork | 0.38 |
| Styrofoam | 0.04 |
Table Of Thermal Conductivity Of Materials
Thermal Conductivity of Aluminum Alloys
It is extremely rare to use pure aluminum for any heat application. Most conductive applications involve highly pure aluminum grades like alloy 1050, 3003, or 6063.
Aluminum grades are based on the alloying elements. The most popular aluminum grade 6061 has a thermal conductivity of 154 W/m K. Compared to 200 W/m K of aluminum grade 6063, this is much lower however it is suitable for motorbike engine components and automotive parts that require the strength that comes from the Si alloy.
Thermal Conductivity Table for Aluminum Alloys
| | |
--- |
| Aluminum alloys | Thermal Conductivity (W/m K) |
| 1050 | 227 |
| 1100 | 222 |
| 3003 | 193 |
| 3004 | 163 |
| 3105 | 171 |
| 4045 | 171 |
| 5052 | 138 |
| 5083 | 117 |
| 6063 | 200 |
| 6061 | 154 |
| 2024 | 121 |
| 7075 | 130 |
Table of Thermal Conductivities of Aluminum Alloys
Thermal Conductivity of 6061 T6 Aluminum
Aluminum 6061 T6 is a popular choice for heat exchangers and has a thermal conductivity value of 152 W/m K.
Copper VS Aluminum In Terms Of Thermal Conductivity
Copper and aluminum both rank high for their thermal conductivities. However, you will likely choose a single metal to manufacture your product. So which one should you choose?
Copper conducts heat almost twice as fast as aluminum and can transfer it more readily. Copper dissipates heat better than aluminum. For applications involving removing heat from a system, such as heatsinks and HVACs, copper is the better choice.
However, there are other factors to consider
Cost: Copper is more expensive than aluminum and for large-scale production this can be a concern.
Weight: Aluminum is less dense and lighter than copper, making it a better choice for lightweight applications like aerospace parts.
Corrosion Resistance: Aluminum offers natural protection against rust while copper can oxidize over time.
Conclusion
Aluminum is one of the best heat conductors and has a high thermal conductivity, k value among metals. It is suitable for many specialized applications like heatsinks in electronics and general case applications like evaporators and condensers.
Even in electronics, aluminum is used not just for its electrical conductivity but also for heat dissipation. In LED lighting systems aluminum dissipates heat.
Selecting an Aluminum Grade for Heat Applications | Reliable Aluminum Manufacturing Partner
When considering aluminum for its thermal applications. It’s best to select a grade that offers the right stiffness, strength, and thermal conductivity.
Aluminum 1050 is great for aluminum busbars and heatsinks
Aluminum 3003 has a thermal conductivity of 160 W/m K, smaller than 1xxx series but offers great corrosion resistance making it suitable for HVAC systems
6061 aluminum is great for LED fixtures
Part MFG offers high-quality aluminum manufacturing and our engineering team can help you source the right aluminum grade for your project.
FAQ
How hot can aluminum foil get?
Aluminum foil does not produce smoke or burn, it can reach temperatures of 660°C before melting.
Does aluminum absorb heat?
Aluminum is one of the best heat conductors with a thermal conductivity of 237 W/m K and it is good at absorbing heat and readily transferring it to other places.
Does aluminum reflect heat?
Aluminum has a shiny surface that reflects light and Infrared rays incoming from another heat source.
What is the thermal conductivity of aluminum foil?
Aluminum foil has a thermal conductivity of 235 W/m K and a melting point of 660°C.
What is the thermal conductivity of aluminum vs stainless steel?
Aluminum has a thermal conductivity in the range of 117 – 237 with the most common value around 235 W/m K whereas stainless steel has a poor thermal conductivity of around 15 W/ m K.
Table of Contents
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9077 | https://www.bmj.com/content/bmj/281/6235/277.full.pdf | BRITISH MEDICAL JOURNAL 26 JULY 1980 277 MEDICAL PRACTICE Consensus in Medicine Febrile seizures: long-term management of children with fever-associated seizures SUMMARY OF AN NIH CONSENSUS STATEMENT A consensus development conference on febrile seizures was held at the National Institutes of Health on 19-21 May 1980.
The purpose of the conference was to bring together practising physicians, research scientists, consumers, and others in an effort to reach general agreement on the risks of sequelae in children with febrile seizures and to compare them with the potential risks and benefits of prophylaxis with anticonvulsants.
A consensus developmental panel met to consider, together with members of the audience, the questions presented below.
The members of the panel represented the disciplines concerned with treating and evaluating the management of children with febrile seizures. They were nominated to the panel by the American Academy of Pediatrics, the American Academy of Neurology, the American Academy of Family Physicians, and the Child Neurological Society. After hearing formal presenta-tions of evidence on the various questions and subsequent discussion the panel debated the issues. This summary is the result of these deliberations.
The Conference was sponsored by the National Institute of Neurological and Communicative Disorders and Stroke, in co-operation with the National Institute of Child Health and Human Development and with the assistance of the Office for Medical Applications of Research, NIH.
Members of the consensus development panel were: Edwin L Kendig, jun (chairperson), professor of paediatrics, Virginia Commonwealth Uni-versity, Medical College of Virginia, Richmond, Virginia; Paul R Dyken, professor and chief, paediatric neurology, Medical College of Georgia, Augusta, Georgia; Nyrma Hernandez, deputy executive director for program services, Epilepsy Foundation of America, Washington, DC; Marion C McKee, clinical professor of paediatrics, University of Vermont, Burlington, Vermont; Hart de C Peterson, associate attending neurologist and paedia-trician, New York Hospital-Cornell Medical Centre, New York; Nelson G Richards, clinical associate professor of neurology, Medical College of Virginia, Richmond, Virginia; Robert H Shackelford, clinical professor of family practice, Duke University, Bowman Gray, University of North Carolina and East Carolina University Medical Schools, Mount Olive, North Carolina; Elbert E Smith, paediatrician, Kansas City, Missouri; B J Wilder, professor of neurology and neuroscience, University of Florida College of Medicine, Gainesville, Florida; Janet Wittes, associate professor, Depart-ment of Mathematical Sciences, Hunter College of the City University of New York, New York.
(1) What is febrile seizure?
A febrile seizure is an event in infancy or childhood, usually occurring between 3 months and 5 years of age, associated with fever but without -evidence of intracranial infection or defined cause. Seizures with fever in children who have suffered a previous non-febrile seizure are excluded. Febrile seizures are to be distinguished from epilepsy, which is characterised by recurrent non-febrile seizures.
(2) What are the risks facing the child who has a febrile seizure?
Children who suffer a febrile seizure generally enjoy normal health after the episode. They are, however, at some risk in several respects. Thirty to forty per cent of children who have one febrile seizure and who do not receive prophylactic therapy will experience a second. The seizure itself is frightening and frequently may be emotionally disturbing to the family. During a seizure there is, additionally, a minimal chance of physical injury. The occurrence of a second or subsequent febrile seizure does not, in itself, greatly change the risk of epilepsy.
A small percentage of children who have had a febrile seizure may have non-febrile seizures-that is, epilepsy. Significant risk factors separate children at high risk of developing non-febrile seizures from those at low risk. The high-risk group, for which one study showed a 13% incidence of epilepsy, is characterised by the presence of at least two of the following risk factors: a family history of non-febrile seizures, abnormal neurological or developmental status before febrile seizure, A cerebral seizure is an abnormal, sudden, excessive electrical discharge of neurons (grey matter) which propagates down the neuronal processes (white matter) to affect an end organ in a clinically measurable fashion.
Protected by copyright, including for uses related to text and data mining, AI training, and similar technologies. .
by guest on 2 September 2025 Downloaded from 26 July 1980. 10.1136/bmj.281.6235.277 on Br Med J: first published as 278 an atypical febrile seizure, such as a prolonged or focal seizure.
Only 2% to 3% of children who have none or one of the above risk factors subsequently develop non-febrile seizures.
The child with febrile seizures may have a neurological deficit, such as mental retardation, motor deficit, and sensory and perceptual abnormalities. It is generally believed that when present these central nervous system (CNS) deficits have usually antedated the febrile seizures. There is no convincing experimental or epidemiological evidence that these deficits reflect neurological injury occurring at the time of the febrile seizure.
(3) What can chronic or intermittent prophylaxis accomplish in reducing those risks?
The only risk shown to be affected by treatment is that of recurrence of a febrile seizure. Many studies show that the risk of recurrence of febrile seizures can be reduced by the continuous daily administration of phenobarbitone at appro-priate dosage to achieve minimum therapeutic blood levels (about 15 mg/l).
Preliminary evidence suggests that the frequency of febrile seizures may also be reduced by the prompt administration of diazepam per rectum, or possibly other agents, at the onset of acute febrile illness. There is no evidence to suggest that phenytoin is effective in prophylaxis of febrile seizures.
Maintenance therapy-While prolonged administration of phenobarbitone is an accepted method ofreducing the frequency of febrile seizures, the long-term effects of such management are still poorly understood and require investigation. Valproic acid has been reported to be effective; however, hepatic toxicity after the prolonged use of valproic acid has been reported. Thus liver function should be closely monitored when this drug is used.
Intermittent therapy during febrile episodes-Intermittent use of phenobarbitone administered orally at the usual recom-mended dosage (3-5 mg/kg) has been shown to be ineffective in promoting therapeutic blood concentrations. When administered in suppositories diazepam is apparently absorbed rapidly enough to provide immediate protection from subsequent seizures in a high percentage of febrile children.
There is no evidence to support the concept that prolonged therapy with anticonvulsants prevents the development of epilepsy or significant neurological deficits.
(4) What are the potential risks of prophylaxis, using the available forms of therapy?
The potential risks of continuous prophylaxis are those predictable side effects, toxic manifestations, or idiosyncratic reactions that may be peculiar to the anticonvulsant drug selected for therapy.
Drugs currently administered on a long-term basis to prevent febrile seizures are phenobarbitone, diazepam, valproic acid, and, to a lesser extent, phenytoin and carbamazepine. Pheno-barbitone is by far the most commonly used agent.
Phenobarbitone-Side effects and toxic reactions are reported in up to 40% of infants or children receiving phenobarbitone.
These reactions are usually of the following types: Behavioural changes-hyperactivity to extreme irritability and, rarely, somnolence.
Sleep pattern disturbances-prolonged nocturnal awakenings.
Not yet available in the United States.
BRITISH MEDICAL JOURNAL 26 JULY 1980 Interference with higher cortical or cognitive functions for example, (a) defects in short-term memory formation; (b) inattentiveness or decreased attention span; (c) defects in general comprehension.
Disturbances in behaviour and patterns of sleep are not predictable. They are the cause for discontinuation of therapy in up to 25% of patients. Animal and tissue culture studies, which are not yet conclusive in regard to humans, raise questions about more serious effects of phenobarbitone and certain other anticonvulsant agents on the developing nervous system.
Phenobarbitone is capable of enzyme induction and may therefore interact with other drugs. The induction of more rapid metabolism of acetaminophen (a major antipyretic) may increase the risks of hepatotoxicity. Phenobarbitone has been used intermittently during febrile episodes to prevent re-currence of febrile seizures. In less than loading doses this method is ineffective. When given in loading doses (10-15 mg/kg) phenobarbitone produces somnolence, lethargy, and other behavioural manifestations.
Valproic acid has been shown to be an effective prophylactic agent in preventing recurrent febrile seizures. Few serious side effects and toxic reactions occur; however, gastrointestinal upset, toxic hepatitis, pancreatitis, and other side effects have been reported. Behavioural side effects are rare and disturbances in higher cortical function- have not been reported. Liver function studies should be monitored periodically in patients on prolonged valproic acid therapy.
Diazepam, given by rectum, has been reported to be effective in the prophylaxis of seizures during febrile episodes. Side effects and toxic reactions have been rare, though sedation has been observed.
Other agents-One report found that antipyretic therapy was not effective in preventing febrile seizures. Acetaminophen and aspirin both produce relatively prompt lowering of temperature and both are relatively free of side effects and toxic reactions.
(5) What is a rational approach to management of children with febrile seizures? Which children should be considered for prophylaxis?
A rational approach to the management of febrile seizures should take into account that the long-term prognosis is excellent, that prophylaxis reduces the risk of subsequent febrile seizures, and that there is no evidence that prophylaxis reduces the risk of subsequent non-febrile seizures.
Initial investigations in a case of febrile seizure should include a complete history and complete paediatric and neuro-logical examination, including characterisation of the febrile illness, degree of increase in temperature, and complete description of the febrile seizure. If a CNS infection is suspected a lumbar puncture is indicated.
The role of the electroencephalogram (EEG) in the in-vestigation of febrile seizures remains controversial. Abnormal EEGs do not reliably predict the development of epilepsy in patients with febrile convulsions. Other studies, such as a complete blood count; measurement of concentrations of serum electrolytes, calcium, and glucose in the blood; and skull radiography or computed tomography (CT) of the brain, are rarely useful in the uncomplicated febrile seizure.
Febrile convulsions rarely presage complex partial seizures or other forms of epilepsy and are generally benign and self-limiting.
Anticonvulsant prophylaxis in therapeutic levels may be considered under any of the following conditions: In the presence of abnormal neurological development-Protected by copyright, including for uses related to text and data mining, AI training, and similar technologies. .
by guest on 2 September 2025 Downloaded from 26 July 1980. 10.1136/bmj.281.6235.277 on Br Med J: first published as BRITISH MEDICAL JOURNAL 26 JULY 1980 279 for example, cerebral palsy syndromes, mental retardation, microcephaly.
When a febrile seizure is: longer than 15 minutes, focal, or followed by transient or persistent neurological abnor-malities.
If there is a history of non-febrile seizures of genetic origin in a parent or sibling.
The physician may elect in certain selected cases to provide anticonvulsant treatment when a patient has multiple febrile seizures or when seizures occur in an infant under the age of 12 months.
Once anticonvulsant prophylaxis is started, it is usually continued for at least two years or one year after the last seizure, whichever is the longer. Treatment should be discontinued slowly over one to two months.
Parents and others who are responsible for the care of young children play a key role in the prevention and management of febrile seizures.
Family education and counselling should address: the relatively benign nature of febrile szizures; the recognition of and management of fever; the use of antipyretic agents; medication and compliance; side effects of medication; first aid for a seizure; and when and how to seek emergency assistance, if needed.
Educational materials may be an effective means to com-plement the physician's efforts toward family education.
Since nurses and allied health professionals, health educators, and social workers are important in family education and counselling, they should receive adequate information in the management of febrile seizures.
Efforts should also be directed to disseminating this know-ledge to the public, including day care centres, through mass media and other means.
(6) Are further clinical, experimental, or epidemiological studies necessary to help in answering these questions?
Certain important questions remain unanswered. Studies are needed to address these issues: Determination of risk factors predicting an initial febrile seizure.
Continuation of the present efforts to follow into adulthood children who have had a history of febrile seizures. This follow-up should include exploration of any association between febrile seizures and learning disorders, epilepsy, behaviour aberrations, intellectual development, and changes in the EEG.
Continued efforts to clarify the role of anticonvulsant treatment in febrile convulsions. Can evidence be obtained as to whether anticonvulsant treatment can: (a) alter the probability of developing epilepsy at some later date ?
(b) alter the probability of development of other neurological sequelae-for example, a lowered intelligence quotient ?
(c) alter psychological and behavioural characteristics ?
Pharmacological studies: (a) What anticonvulsants are safe and effective in the chronic prophylaxis of febrile convulsions ? What is the proper dosage and the therapeutic blood level ?
(b) What anticonvulsants are safe and effective for short-term prophylaxis of febrile convulsions ? What is the proper dosage and the therapeutic blood level ?
(c) What are the long-term risks of the use of the above anticonvulsants ?
(d) Does phenobarbitone enhance the possible carcinogenic effects of other drugs or chemical agents in humans ?
Controlled study of antipyretic measures with the onset of febrile illness as a means of decreasing the risk of recurrence of febrile seizures.
Continuation of experiments of the effects of drugs on brain growth and development, using animal and tissue culture techniques.
Continuation of animal experiments to clarify the effects of single recurrent seizures on brain maturation and development, using both experimental animal models and kindling techniques.
A woman in her 60s was treated 10 years ago in Australia for thyro-toxicosis with radioactive iodine. She developed severe thyroid deficiency, and was subsequently stabilised on thyroxine sodium, 100 F±g thrice daily.
Are thyroid function tests of any value now or is "how she feels" the main indication whether she is taking adequate dosage or not ?
The most acceptable replacement dose of thyroxine (T4) is now considered to be the lowest dosage that will maintain a normal circulating concentration of serum thyrotrophin (TSH). Although replacement doses of T4 of 300 ,ug daily were conventionally accepted at the time of this patient's hypothyroidism, using the above-mentioned criteria, it has been shown that such doses are often physiologically excessive and that symptomatic relief and biochemical normality may be restored in most patients with 100-150 FLg of T4 daily.'' 2Since mild hyperthyroidism at this age may cause or exacer-bate cardiac arrhythmias or congestive heart failure, I would recommend decreasing her dose to 150 Wug of T4 daily. In view of the long half-life of thyroxine this could be given as a single dose and for psychological reasons in a step-wise fashion. To determine the adequacy of T4 replacement, clinical assessment, including testing the Achilles reflex, is essential, as dosage may need adjusting on the basis of clinical findings before available laboratory results.
Clinically euthyroid individuals with normal thyroid hormone concentrations may, however, have subclinical hypothyroidism as shown only by a raised TSH concentration. This suggests that an individual's hormonal requirements may be more finely regulated than the "normal" range for the population- may suggest. Therefore, for this patient, a clinical and hormonal assessment including serum TSH in one or two months' time, allowing her to equilibrate on her adjusted T4 dosage, would be indicated. Once a maintenance dose is established, a similar assessment every one or two years is sufficient.
Evered DC, Young ET, Ormston BJ, et al. Treatment of hypothyroidism: a reappraisal of thyroxine therapy. Br MedJ 1973;iii:131-4.
2Stock JM, Martin MD, Oppenheimer JH. Replacement dosage of 1-thyroxine in hypothyroidism. N EnglJ Med 1974;290:529-33.
Is microwave radiation suitable for sterilising surgical instruments ?
Microwaves in some conditions are merely a convenient way of heating a substance. Some materials, when put into a microwave field, absorb energy from the field and become hot. In preparing food it is generally the water in the food that is the major energy absorber; the advantage is that the food is heated throughout its mass and not merely through its surface-that is, the inside cooks as soon as the outside. It is not safe to put metal objects into microwave ovens. Furthermore, there would be no advantage in applying heat to a surgical instrument by expensive microwaves rather than by a conventional electric heating element and boiling water or steam.
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by guest on 2 September 2025 Downloaded from 26 July 1980. 10.1136/bmj.281.6235.277 on Br Med J: first published as |
9078 | https://www.sigmaaldrich.com/US/en/substance/1propanol601071238?srsltid=AfmBOoqMYuo0HykVzczF1TbpxCHW89pWlKob_w9rEPmq7XzrY_qaCB5d | 1-Propanol - Propyl alcohol, 1-Propanol
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1-Propanol
Synonym(s):
Propyl alcohol, 1-Propanol
Linear Formula:
CH 3 CH 2 CH 2 OH
CAS Number:
71-23-8
Molecular Weight:
60.10
EC Number:
200-746-9
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9079 | https://goldbook.iupac.org/terms/view/D01801 | Alphabetical Index
Additional Indexes
Physical ConstantsUnits of MeasurePhysical QuantitiesSI PrefixesRing IndexGeneral FormulaeExact FormulaeSource DocumentsTerms by IUPAC DivisionTerms by Organization
Version 5.0.0 (12318 Terms)
DOI: 10.1351/goldbook
Jan Kaiser - Content Editor Stuart J. Chalk - Technical Editor Joint Subcommitteeon the IUPAC Gold Book
dissociation
The separation of a molecular entity into two or more molecular entities (or any similar separation within a polyatomic molecular entity). Examples include unimolecular heterolysis and homolysis, and the separation of the constituents of an ion pair into free ions.
The separation of the constituents of any aggregate of molecular entities.
In both senses dissociation is the reverse of association.
Source: PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1107 [Terms] [Paper]
Citation: 'dissociation' in IUPAC Compendium of Chemical Terminology, 5th ed. International Union of Pure and Applied Chemistry; 2025. Online version 5.0.0, 2025. RIS BibTex EndNote
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9080 | https://www.sciencedirect.com/science/article/abs/pii/S0263822322002124 | Compression and energy absorption characteristics of additively manufactured reticulated tubes filled with spherical reticulated shells under axial crushing - ScienceDirect
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Composite Structures
Volume 288, 15 May 2022, 115415
Compression and energy absorption characteristics of additively manufactured reticulated tubes filled with spherical reticulated shells under axial crushing
Author links open overlay panel Fei Wu, Yating Chen, Shunqiu Zhao, Zhengrong Zhang, Shiwei Zheng
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Abstract
In this study, the concept of energy absorption and collision avoidance of metal tubes in the transportation field is combined with a spherical reticulated shell structure with excellent vibration absorption and impact resistance in the construction field. New energy absorption structures—reticulated tubes (RTs), equator skeleton spherical reticulated shells (ESs), truss spherical reticulated shells (TRs), and thin-walled sphere with round holes (TSHs), sphere-filled RTs, and sphere-filled thin-walled tubes—are proposed. For comparison, aluminium alloy thin-walled tubes and hollow spheres (HSs) are also included in the study. The HS, TSH, ES, and TR-filled RTs can produce more prominent filling effect than the corresponding aluminium alloy tubes and significantly improve the carrying capacity, energy absorption, crushing force efficiency, and specific energy absorption. Specifically, the filled RTs not only reduce the impact effect more notably and keep the initial impact peak almost unchanged but also significantly increase the energy absorption energy per unit mass compared to the filled aluminium alloy tubes. Thus, the HS, TSH, ES, and TR-filled RTs can be new types of lightweight structures which can meet the requirements of light weight, high energy absorption, and low impact effect.
Introduction
Energy conservation and environmental protection and safety have become the three major research themes of the modern automobile industry. Under the general trends of energy conservation and emission reduction, there is development towards high-speed cars and low-fuel consumption or new energy vehicles with long range. Therefore, vehicle lightweighting has become inevitable. Thus, the research on the development of vehicle collision safety protection will also become increasingly challenging , , , . The bumper absorbing box, which is used for automobile low speed collision to effectively absorb collision energy, is a research hotpot in automobile passive safety field. It is an important energy absorbing metal thin-walled device in automobile bumper system, which is installed between the beam and the frame rail. At present, the research focus is mainly on new lightweight energy-absorbing materials and structures, such as polygonal tubes , taper/gradient/ corrugated tubes , , , surface defect tubes , , , , multi-cellular composite tubes , , , , , , , , , lightweight material-filled tubes , , , , foam aluminium/honeycomb aluminium and its composite structures, lattice/truss structures , , , metal fibre/wire porous structures , and bionic structures .
The research on energy absorption of metal thin-walled circular tubes is the most mature, and all subsequent studies are based on it. Studies on the energy absorption characteristics of circular, polygonal, and prism-shaped structures under axial compression show that the energy absorption per unit mass of circular tubes is the largest, whereas that of polygonal tubes with few edges is small. Compared to a straight tube, for a taper tube, the energy absorption capacity is increased by approximately 120%, and the initial impact peak is also increased . A mesh tube has a certain bearing capacity and absorbs energy in a low initial peak state. A mesh tube prepared by the author using an ultra-fine soft wire rope as the matrix has high energy absorption efficiency and no initial peak value. However, because of its low bearing capacity, it cannot be used as an independent energy absorption component , . Compared to traditional tubes, expansion honeycomb hole tubes and lattice hole thin-walled tubes with a negative Poisson's ratio effect , not only have higher specific energy absorption but also better structural damping performance during axial collision . These researches mainly improve the axial energy absorption characteristics of metal thin-walled tubes by changing the section and wall shapes of tubes.
It is very limited to improve the energy absorption characteristics of thin-walled metal tubes only by changing their section and wall shapes. The composite tubes of metal thin-walled tubes come into being, which mainly include multicellular tubes, lightweight material-filled tubes and laminated tubes. The inner space of such a tube is divided into several multicellular structures with independent spaces separated by metal thin panels, which can significantly improve the energy absorption capacity. The energy absorption per unit mass of a square straight multicellular tube is increased by approximately 220% compared to that of an empty tube . A corrugated multicellular tube has a lower initial peak value, higher energy absorption and specific energy absorption, and more controllable failure mode than a multicellular straight tube. Although foamed and honeycomb aluminium have low carrying capacity, filling them into thin-walled tubes can effectively improve the energy absorption capacity of the tubes . When aluminium alloy thin-walled tubes are filled with metal thin-walled balls with high energy absorption efficiency (0.6–0.8), the energy absorption efficiency of the tubes is increased from 0.6 to 0.7–0.8, and their carrying capacity is increased by approximately 300%. When carbon fibre tubes are laminated by aluminium alloy tubes , both the energy absorption capacity and initial peak value are significantly increased, and the brittle fracture failure mode of the carbon fibre tubes is not conducive for their application. The energy absorption characteristics of a three-layer laminated structure of a crochet-sintered mesh tube formed with internal and external mesh tubes and metal thin-walled tubes lead to improved carrying capacity, deformation stability, crushing force efficiency, and crashworthiness . As a compound tube with spacings, its folds have low mutual constraints along the radial direction during deformation . Shen et al. filled aluminium foam between tubes to study their energy absorption characteristics. The structure of bellows sandwiched between tubes has a good interface effect, and the energy absorption of the resulting tube is approximately 35.6% higher than that of its components . Wang et al. combined a Koch-curve metal thin-walled tube with a metal thin-walled tube to form a laminated tube, which presented a remarkable synergistic effect and effectively enhanced the energy absorption capacity.
In addition to the traditional metal structures mentioned above, some porous materials and braided structures have been used for energy absorption. Aluminium foam, aluminium honeycomb, and their composite structures were initially studied extensively. Aluminium foam is light weight and has high specific strength, a long flat energy-absorbing plastic plateau, and almost no initial peak, which suggests that its interaction process with collision objects is long and the initial impact effect is weak. Honeycomb aluminium develops from an initial hexahedral structure to corrugated honeycomb and subsequently to auxetic honeycomb . Compared with the traditional hexagonal honeycomb structure, its specific energy absorption can be improved at a low initial peak and the compressive strength can be increased by approximately 300%. Kang et al. developed a braided equipment independently to form a periodic wire cell-like porous structure for energy absorption and collision avoidance application. Carbon fibre structures braided by three-dimensional four- and five-directional methods have strong impact resistance; however, their brittle fracture characteristics affect their applications .
At present, the research focus is mainly on finding a method to improve the energy absorption capacity by changing the complexity of energy-absorbing materials or structures. This generally leads to a common problem: the increase in the energy absorption capacity is accompanied by an increase in the initial peak value. When a structure absorbs more impact energy, the impact effect of the structure is also enhanced, which indirectly increases the impact damage to the protected object. Finding a method to weaken the impact vibration of a structure is also a research direction that cannot be ignored. In 2019 and 2020, Guo et al. and Lee et al. reported on auxetic lattice honeycomb tubes and negative Poisson's ratio effect lattice hole thin-walled tubes, respectively, focusing not only on high specific energy absorption but also on the structural damping characteristics. A spherical reticulated shell structure, which is commonly used in the construction field, is prominently examined for its practical applications and theoretically studied in regards to its seismic resistance, wind resistance, explosion resistance, and impact resistance. By combining theoretical analysis, numerical simulation, and experimental research, numerous studies have been conducted on the mechanical behaviour, failure mode, and failure mechanism of reticulated shell structures under strong earthquake and impact loads. Studies have shown that such structures have excellent vibration reduction and impact resistance as well as good structural characteristics of dissipative vibrations and shock waves , , , , . This paper innovatively proposes combining lightweight reticulated shells used in buildings and metal tubes employed for energy absorption and collision avoidance, to achieve energy absorption and protection. The objective is to realize the dissipation of impact vibration based on the structure. Focusing on the energy absorption capacity and the dissipative capacity of a structure to the impact effect is of high research relevance and significance.
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Section snippets
Structure and sample fabrication
Fig. 1 shows the components of the filled composite structures used in this study. Reticulated tubes (Fig. 1a) with an outer diameter of approximately 30 mm and a height of 50 mm were manufactured by selective laser melting (SLM). Their three-dimensional model was obtained by the graphical reconstruction of a crochet-sintered mesh tube prepared in a previous study by X-ray tomography (Fig. 2a). Truss spherical reticulated shells (Fig. 1b), equator skeleton spherical reticulated shells (
Mechanical behavior of empty T and additively manufactured RT
Fig. 6a shows the load–displacement curves obtained when 50-mm-high tubes are compressed downward by 35 mm. The thin-walled tube (hereafter T) curve presents three distinct deformation stages. First, the initial peak force is reached in the elastic deformation stage; subsequently the load decreases sharply and enters a wave-type plastic deformation stage, and finally the load sharply increases. The initial peak force of the reticulate tube (hereafter RT) is approximately one third of that of
Conclusion
By experimental studies, the quasi-static and dynamic compression mechanical behaviour and energy absorption characteristics of spheres, RTs, aluminium alloy Ts, sphere-filled RTs, and sphere-filled Ts were investigated. The research conclusions are as follows:
(1) About compression mechanical behaviour
The four different types of spheres showed different load–displacement curves and deformation modes. The load–displacement curves of T, RT and their composite tubes filled with spheres all
CRediT authorship contribution statement
Fei Wu: Project administration, Conceptualization, Methodology, Investigation, Writing – original draft, Writing – review & editing, Funding acquisition. Yating Chen: Formal analysis, Investigation, Data curation. Shunqiu Zhao: Validation, Resources, Visualization. Zhengrong Zhang: Supervision, Conceptualization, Writing – review & editing, Funding acquisition. Shiwei Zhen: Formal analysis, Data curation.
Declaration of Competing Interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgements
The authors gratefully acknowledge the financial support from the National Natural Science Foundation of China (Grant No. 51705085 and No. 52175294) and the State Engineering Research Center for Metallic Materials Net-shape Processing, South China University of Technology (Grant No. 2020011).
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A review of the crashworthiness performance of energy absorbing composite structure within the context of materials, manufacturing and maintenance for sustainability
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A review of recent research on bio-inspired structures and materials for energy absorption applications
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Modeling the deformation of thin-walled circular tubes filled with metallic foam under two lateral loading patterns
2024, Structures Citation Excerpt :
Filling light-weight foams or multi-cell elements into thin-walled tubes has been popular to enhance load-carrying and energy-absorbing capacity. Extensive research has been conducted on the axial compression and lateral bending characteristics [29–33] and bending behaviors of circular tubes filled with metallic foam, as they show promise as effective energy absorbers. When the foam filled tubes as energy-absorbers are crushed axially, a high peak force generally arises which may cause serious injure to the protected objects. Show abstract Thin-walled circular tubes with metallic foam fillers are widely used in various engineering fields. This work examines the plastic behavior of thin-walled circular tubes filled with metallic foam under two different lateral loading patterns, by using both analytical and numerical methods. Two types of loading indenters, including a line load and a rigid plane, are considered. There have been very few studies on the lateral compression of foam-filled circular tubes crushed by a line load or a rigid plane. The analytical models assume that the tubular wall material is rigid and ideally plastic, and that the metallic foam fillers maintain a constant resistance known as plateau stress. Using the principle of virtual velocities, we derive succinct explicit solutions for the crushing forces as well as deformation characteristics in regard with the radius of the tubes, the flow stress of the tubular wall and the plateau stress of the filled foam under the two loading conditions. Finite element analysis is performed using the ABAQUS2023/Explicit code to model aluminum alloy circular tubes filled with aluminum foam. A comparison and analysis of the deformation characteristics of the tubular top generator and foam cross-sections are conducted. The proposed analytical crushing forces and deformation properties obtained herein agree well with the numerical simulation results and outperform previous theoretical models, which may serve as valuable formulations for the engineering application.
### A novel cylindrical mechanical metamaterial: Design, fabrication, and compressive properties
2024, Structures Show abstract Mechanical metamaterials exhibit fascinating deformation characteristics and exceptional mechanical properties due to their distinctive unit cell structure and the repetitive development of these units. This research presents the design of a novel cylindrical metamaterial featuring a concave hexagonal structure. This design incorporates an auxetic structure within the outer circular tube in order to enhance radial shrinkage and improve bearing capacity and stability by leveraging the negative Poisson's ratio effect. The specimens were fabricated using 3D printing technology, and their mechanical properties and uniaxial compression deformation characteristics were subsequently investigated through experiments. Additionally, the accuracy of the finite element model was verified by comparing the simulated predictions with the experimental results. The specimens differ in terms of their internal and external structures. The external structures vary based on the combination of single cells and interlayers, resulting in three distinct types. The internal structures differ based on the angle of inward depression. The mechanical properties and deformation characteristics of specimens composed of different internal and external structures were analyzed, revealing that the cylindrical structure can possess adjustable stiffness, high stability, and compressive capacity.
### Crushing behavior of multi-layer lattice-web reinforced double-braced composite cylinders under lateral compression and impact loading
2023, Thin Walled Structures Citation Excerpt :
Steel fenders have been more commonly used in China due to its better designability and cushioning performance [10,11], but the application of steel fenders has been limited by their weaknesses such as poor corrosion resistance and high maintenance requirements . Fiber reinforced composite structures have drawn increasing attention to transport infrastructure and its anti-collision structures due to their high strength to weight ratios, great cushioning performance and excellent corrosion resistance [13–18]. Foam-filled composite sandwich panels consisting of glass fiber reinforced polymer (GFRP) face sheets and polyurethane (PU) foam core were investigated through quasi-static compression tests and the results demonstrated that the foam material made an essential contribution to the energy absorption of the panels . Show abstract This paper reports on the crushing behavior of several novel multi-layer lattice-web reinforced double-braced composite cylinders composed of glass fiber reinforced polymer (GFRP) face sheets, GFRP lattice webs, polyurethane (PU) foam core and steel bars. A series of quasi-static lateral compression and low-velocity impact experiments were carried out to investigate the feasibility of the proposed cylinders. All the experimental specimens were manufactured using a vacuum assisted resin infusion process (VARIP) method. The bearing capacity and energy absorption performance of the composite cylinders can be significantly improved with the enhancement of multi-layer lattice-web layout and use of bracing. Among the proposed three types of lattice-web layouts, the double-layer dislocated lattice-web layout made the composite cylinder exhibit the greatest specific energy absorption (SEA) performance and good impact resistance property and can be chosen as an optimal configuration. Furthermore, numerical models were established using LS-DYNA software to simulate the large deformation of the composite cylinders with double-layer dislocated lattice-web layout. Based on the numerical models, parametric analysis was carried out to discuss the effects of various parameters on the crushing behavior of the composite cylinders. The bearing capacity and impact resistance property can be generally improved with the increase of GFRP thickness or radial lattice-web height. Additionally, using stronger foam material or smaller inclination of bracing can increase the absorbed energy in PU foam but the GFRP material always makes an essential contribution to the energy absorption of the composite cylinders.
### A hybrid architectural metamaterial combing plate lattice and hollow-truss lattice with advanced mechanical performances
2023, Additive Manufacturing Citation Excerpt :
The quasi-static compressive test was introduced to obtain the crushing behavior of specimens based on the universal testing machine SUNS 5305. The specimen was located on the support plate, and the pressing plate crushed to the specimen at a constant velocity of 2 mm/min [56,57], as shown in Fig. 4B. The sensor upon the pressing plate recorded the force and displacement, and a camera captured the deformation process of specimens. Show abstract Inspired by the hollow-tube architecture and chambers found in bamboo, we propose a design concept for mechanical metamaterials based on the hybridization of plate lattice and hollow-truss lattice (HPHL) to enhance the mechanical performance of lattice metamaterials in this paper. The design feasibility of HPHL is demonstrated through sample manufacturing using the laser powder bed fusion technique. Quasi-static compressive tests are conducted to investigate the crushing behavior of HPHL. Compared with the typical plate lattice, the HPHL exhibits a more effective crushing region, a higher mean crushing stress and a larger specific energy absorption (SEA). The SEA of HPHL is improved by 53.18% compared with that of the pure plate lattice. Based on the verified numerical simulations, the crushing mechanisms of HPHL are investigated through examining the deformation process and the stress-strain curve. It is found that the incorporation of hollow-truss lattices effectively delays the densification of the plate lattice and provides a more homogeneous and efficient stress distribution in HPHL, leading to higher crushing stress and higher energy absorption efficiency. Moreover, the effects of various design parameters, such as the hollow-truss diameter d T, hollow-truss thickness t T and plate thickness t P, on the mechanical behavior of HPHL are further explored to identify the optimal HPHL and realize the mechanical customization. The bionic hybrid design strategy proposed in this paper paves a new promising route to improve the mechanical performance of lattice metamaterials.
### Mechanical properties and energy absorption of composite bio-inspired multi-cell tubes
2023, Thin Walled Structures Show abstract New composite bio-inspired multi-cell metal tubes—lotus root inspired structures (LR-IS), lotus root-horsetail inspired structures (LR-HS1 and LR-HS3) and lotus root-honeycomb inspired structures (LR-HC1 and LR-HC2) — were proposed and fabricated by selective laser melting. Their mechanical behaviors and energy absorption characteristics were investigated under the axial, radial and three-bending crushing via experiments and numerical simulation. Their crushing force efficiency values ranged from 0.8–1.0, which were much higher than that (0.536) of metal thin-walled hollow tubes (HT). The specific energy absorption values in axial, radial and three-point bending directions were 27.7–81.0 J/g, 1.5–12.3 J/g and 0.8–10.7 J/g, respectively, which were approximately 2-13 times of HT. Compared with the HT, the structural strength, rigidity, energy absorption, crushing force efficiency and specific energy absorption of the bionic composite metal tubes were greatly improved in axial, three-point bending and radial directions. Their axial directions were the main directions of energy absorption and collision avoidance, and the lateral directions had sufficient load bearing capacity. These bionic multi-cell tubes have the potential to be used as lightweight, rigid, high strength, impact-resistant, and energy-absorbing independent energy absorbers.
### Properties of 3D Double-Arrow Negative Poisson's Ratio Lattice Structure under Quasi-Static Compression and Low-Speed Impact
2022, Advances in Materials Science and Engineering
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9081 | http://beta.chem.uw.edu.pl/chemanal/PDFs/1994/CHAN1994V0039P00623.pdf | Chem. Anal. (ll'Orsaw), 39, 623 (1994) '-Formulation of Generalized Equations for Redox Titration Curves by T. Michalowski and A. Lesiak Faculty ofChemistry, Jagiellonian University, 30-060Krakow; Poland Key words: potentiometry, redox titration curves The paper provides several examples of rigorous formulation of-redox titration curves.
All available physico-chemical. data .of a system considered. are employed, and no simplifying assumptions are made. A new, generalized notation of the electron balance and an iterative computer program were applied for this purpose. The approach proposed enable: a) to construct titration curves and the plots of log[Xi] vs. fraction titrated ct> for different species Xi, b}to formulate the reactions proceeding at any stage of the titration together with their efficiencies and c) to locate characteristic points on the titration curves.
W praey przedstawiono kilka przykladow formulowanla krzywych miareczkowania redoks z wykorzystaniem wszystkich dostepnych danyeh fizykochemicznych zwiqza-nych z rozwazanym ukladem, bez stosowania jakiehkolwiek zalozeri upraszczajaeych.
Zastosowano w tym celu nowy, uogolnlony zapisbilansuelektronowego i komputerowy programiteracyjny. Z3proponowany sposob opracowywania uklad6wredoks umozli-win: a}konstruowanie krzywych miareczkowania i wykresow zaleznosci 10g[Xi] od ulamka zmiareczkowania .. ¢l dla r6znych skladnik6w Xi, b) sformulowaniereakcji zachodzacych w dowolnym momencie miareczkowania wraz z ich efektywnosciami oraz c) umiejscowieniecharakterystycznych punktow na krzywej miareczkowania.
Titration curves for non-redox systems are usually obtained on the basis ofcharge and concentration balances . The proton balance instead of the charge balance is sometimes applied, see e.g. . The proton balance was formulated according to ex-change phenomena ; however, it can also be obtained from a combination of the charge and concentration balances [1, 3]. The latter procedure of the proton balance formulation is especially useful for complicated, multicomponent systems .
624 T.MichalowskiandA. Lesiak In order to construct a redox titration curve, the set of equations should be completed by the electron balance. The electron balance, applied for this purpose, was previously obtained according to the electron exchange phenomena [4-6]. Such formulation of the electron balance is, in general, very cumbersome, when applied to more complex systems met, as a rule, to redox titrimetric analysis.
A more useful and convenient form of the electron balance, named the "electron pre-balance", has been recently proposed by Michalowski [7-9]. It enables to avoid some simplifying assumptions ascertained elsewhere [10, 11]. Moreover, it appears to be an indispensable tool for the systems containing more than two elements participating in the redox equilibria and the systems with polynuclear complexes formed.
The stoichiometry of a redox reactionis "coded" in the due electron balance. It usually happens that the predominating reaction is accompanied by side reactions.
The influence of side reactions depends frequently on the conditions of analysis.
The redox equilibria involve, in general, all species formed also in (de)protona-tion, (de)complexation or precipitation reactions. Thus the electron-, charge- and concentration balancestogether with an appropriate set of equations for the equili-brium constants arc required for the quantitative description of redox systems. The equilibrium constants involving some species in a system considered (an aqueous medium) are, among others: the ionic product of water (Kw) , stability constants of complexes (10 and standard potentials (Eo). All known physico-chemical data related to the system should be involved there. Such thermodynamic image should be completed with the kinetic data which impose some restric-tions on a system in question.
A notation of any (properly written) chemical reaction fulfils the laws of the matter and charge conservation. These laws are then expressed by the equalities for the numbers of individual elements and for the charges of species on both sides of the reaction. In general, the reaction thus formulated does not take into account its efficiency. The notation, although comprehensible when formulated for the equili-brium constants, is not a basis for indicating a priori the main reaction and the efficiencies of the accompanying reactions which result from an addition of the reagent(s). These reactions can be formulated, however, using the dependences of 10g[Xil vs. fraction titrated plotted for different species Xi. By this way it is possible to state, for example, the changes in the scheme of the HBrO dispro-por-tionation after an addition ofa NaOH solution . The reactions occurring in a system are sometimes entirely changed by some component(s) introduced there.
Accurate a priori results of titrimctric analyses can be ensured by the well-estab-lished theory of titration curves. It should be confessedly stated, however.that the theory of redox titration curves, related to the systems of analytical applicability, was not elaborated satisfactorily as yet. The computational aspect should not make a hindrance in it nowadays, and the state of matter should be perceived in methodology applied.
The methodology provided in the present paper enables to involve all the species existing in a system considered, Any analytical prescription is then reproducible and the simplifying assumptions arc thus avoided. The calculations can be made for the titrand-titrant systems of any degree of complexity. Interfering action of some Equations/or redox titration curves 625 species can be thus defined a priori. This provides a basis for choosing the best conditions for the titrimctric analysis and defining the equivalence point(s) accurately, with no necessity to refer to the notation of the redox reaction considered. The "equivalentweight" and the"normal concentration" are the redundant concepts only.
An analogous viewpoint has been already presented in acid-base [12, 13], complex- .
ometric , and precipitation titrations proceeding in the non-redox systems.
In all instances, none of the reactions proceeding in the system is distinguished [7, 8, 12].
This approach was a basis for an unconventional unified concept of the equivalent weight [12, 13], common for all kinds of potentiometric titrations, in contradistinc-tion.tothe one recommended by the IUPAC .
The resolution of complicated chemical systems requires the application of an iterative computer program. It should be confessedly stated, however, that potential possibilities inherent in computational technique are cxploited in avcry limited degree in classical analysis, particularly in theoreticalapproach to redox systems.
The present paper concerns Iive redox systems: 1) Fe2++ Mn04", 2) Fe2++ Ce4+, 3)V2++ Mn04", 4) y2++ Cr20~- (all considered by de Levie ) and 5) a more com-plex system where iodate is titrated with ascorbic acid in the presence of some accompanying species.
Equilibrium constants In the systems 1-4 we apply pKw=14 for the ionic product of water and the following stability constants (IogK): HS04" (1.8), MnOH+ (3.4), MnOH2+ (14.2), MnS04 (2.28), FcOH+ (4.5), FeOH2+ (11.0), Fe(OH)~ (21.7), Fe2(OH)~+ (25.1), FCS04 (2.3), FeSO! (4.18), Fe(S04)2 (7.4), CeOH2+ (5.0),. CeOH3+(13.3), Ce2(OH)~+ (40.3), Cez(OH)j+ (53.7), CeSO! (1.63), Ce(S04)2 (2.34), CC(S04)3-(3.08), CeSO~+ (3.5), Ce(S04)2 (8.0), CC(S04)~- (10.4), CrOH2+ (10.2), Cr(OH)!
(18.3), HCr02 (14.4), CrS04 (1.3), CrSO! (1.76), HCrZ07 (1.64), H2Cr207 (0.24), HCr04 (6.52), H2Cr04 (7.27), VOH2+ (11.1), V(OH)~ (21.6); VOS04 (2.48), VOzS04" (1.8); Other equilibrium constants (IogK): VO~-+H+=HVO~- (13.0), HVO~-+ H+= HzV04" (7.88), HzVO~-+H+ =H3V04 (3.4), HV~- = V03"+OH-(-6), VOi+H+=HV03 (3.7), 2HV~- =VZO~-+H20 (1.68), Vz0t" + H~ =HV20~ (9.73), HVzO~- + H+=HzVzO~-(8.78), 2VO~ + 3H20=H3V207 + 3H+ (-3.18), 3HV04- = V30ij-+30H- (-10.4), V02++H20=VOOH++H+ (-4.8); Eo: 1.507 for Mn04,H+/Mn2+; 1.743 for MnO~-,H+lMn2+; 1.509 for Mn3+/M112+; 0.771 for Fe3+/Fe2+; 1.70 for Cc4+/Ce3+; 1.333 for Cr20~-,H+/Cr3+; 1.477 for CrO~-, H+/Cr3+; 0.945 for CrOJ-,H+/Cr02"; -0.407 for cr3+/cr1+; -0.255 for V3+/V2+; 0.337 for V02+, H+/V3+; 1.(J04for VO~+, H+N02+;-0.044 for V02+/VO+; 1.256 for VO~-,H+NO+.
Further equilibrium constants, especially those related to the system. 5, are involved in the relations given in Table 1.
626 1:Michalowski andA. Lesiak -----------Presentationofthe data The potentiometric titration curves obtained on the basis of the equilibrium data areplotted in coordinates (,E), where E is the emf of the system and =CVI(CoVo) is the fraction titrated ; Vo=100 ml was assumed in all calculations. Thcequili-briumconstant (Ke) of a redox reactionis expressed by the formula IogKe =zAEo (1) wherezis the number of electrons involved in the reaction,Eois the standard potential of the reaction, A:.F'/(Rl1nlO)=16.92 at T=298 K. TIle related curves (a, Figs. 1-4) were compared with ones found according to a simplified model presented in (curves Il in Figs. .1-4).
Redox systems 1. Fe2+-Mn04" system. The electron pre-balance [7, 8] related to the system obtained after an addition ofV ml of C mol 1-1 KMl104into '10 mI of the acidified (Ca mol 1-1 H2S04) solution of FeS04 (Co moll-I) is (Zl=26 for Fe and Z2=25 for Mll): (Zl-2){[Fe2+]+[FeOH+]+[FeS0 4]}+(Zr -3){[Fe3+]+[FeOHz+ ]+[Fe(OHH]+ 2[Fe2(OH)1+]+[FeSO!]+[Fe(S04)2] + (Zz-7)[Mn041 + (Z2-6)[MnO~-]+ (Zz-3){[Mn3+]+[MnOH 2+ ]) + (Zz-2){[Mnz+]+[MnOH+]+[MnS04])-{(Zl-2)CoVo + (Zz--"7)CV}/(Vo+V) =0 (2) Equation (2) together with concentration balances for Fe, Mn, SOa-[Fez+]+[FeOH+]+[FeS041+[Fe3+]+[FcOH2+]+[Fe(OH)j:J+2[Fe2(OH)i+]+ [FeSO!]+[Fe(S04)2] - CoVoIeVo;'-V) =0 (3) [Mn04]+[M110~-]+[Mn3+]+[MnOHz+]+[Mn2'l-]+[M1l0H+]+[MnS04]+ CV/(Vo+V)=0 (4) [HS04]+[SO~-]+[FeS04]+[FcSOj]+2[Fe(S04)2]+[MnS04]-(Co+Ca)VoI(Vo+V)=0 (5) and the charge balance 2[Fe2+]+[FeOH+]+3[Fe3+]+2[FeOH2+]+[Fe(OH)2]+4[Fc2(OH)~+]+[FcS04] [Fe(S04)2]-[M1l04]~2[MnO~-]+3[Mn3+]+2[MnOH2+]+2rMl1z+]+[MnOHn+ CV/(Vo+ V)-[HS04]-2[SO~-]+[H+]-[OH-] =0 (6) form a set of five independent equations with six variables: [Fez+] (or [Fc3+ ] after the equivalent point), [Mn2+], [SOa-], [H+], E, V. Setting the successive values for V, we find the related values of other variables according to iteration procedure. The points (<1>, E) thus obtained form the titration curve (Fig. 1a) for the system in question.
Equationsforred()x titrationcurves 627 1.5 ~--------------, a 1.1 "0 Dr ....
~j ••L;••• ·.~ 1.3, 0.9 0.7 0.5 0.3 0.1 0.3 +---~-~--r-""'--~---; 0.0 Figure 1. a) TIle titration curves (<< - our, p- simplified) and b) the pH l~S. H+]+[FeSO4]}+(Z1-3){[Fe3+]+[FeOHZ+]+[Fe{OH)!l+ 2[Fe2(OH)~+]+[F~SO!]+[Fc(S04)2]} + (Zz-3){[Cc3+]+[CeOH2+]+[CeSO!1+ [Ce(S04>:2]+[Ce(S04)~-]}·+ (Z2-4){[Ce4+]+[CeOH~+]+2[Cez(OH)j+l+ 2[Ce2(OH)~+]+[CeSO~+]+[Ce(S04)2]+[Ce(S04)~-]}-((Zl-:z)COVo+(Zz-4)CV}/(V o+V)= o (7) Equation (7) together with the charge and concentration balances enable to get the titration curve which is plotted in Fig. 2.
Ut;J pl.
..
. b 0.•.•• ·.Cll ~~•...' ••..·L·'U 1.7 a 1.5 >-1.3 ~ .....
1.1 td :::J s:l 0.9 (l) +.> Q Poe 0.7 0.5 0.3 , 0.0 0.5 1.'0 1.5 2.'0 Figure 2. a) The titration curves (<<- our, j3 - simplified) and b) the pH vs. ~ relationship for the Fe2+Ce4+ system; Co=O.Ol, C=O.l, Ca=O.l, C~=O.5 mol 1-1 628 T. MichalowskiandA. Lesiak 3. V2+-Mn04 system. The electron pre-balance related to the system obtained after an addition of Vml of C mol 1-1 KMn04 into Vo ml orco mol 1-1 VS04 acidified with H2S04 (Ca mol lrl) has the form (Zl=23 for V,Z2=25 for Mn): (Zl-2)[V2+]+(Z 1-3){[V3+]+[VO+]+[VOH2+]+[V(OH)t]} + (Zl-4){[V02+]+ [VOOH+]+2[(VOOH)~+]+[VOS04]} + (Zl-5){[VO~]+[VO~l+[HVOr-]+[H2V04]+ [H3 V04]+[V°3]+[HVOj]+[V02S04]+2([V20t]+[HV20r ]+[H2V20¥-]+[H3V2°7"])+ 3[V30~l} + (Z2-7)[Mn0 4]+(Z2-6)[MnOr-]+(Z2-3){[Mn3+]+[MnOH2+]}+ (Z2-2){[Mn2+]+[MnOH+]+[MnS04H - {(Zl-2)CoVO+(Z2-7)CV}/(Vo+V) = 0 (8) The balance (8) together with the charge and concentration balances enable the related titration curve to be plotted (Fig. 3).
1.4 a > 1.0 ~ - 0.6 os :a c G> ~ 0.2 0 p..
-0.2 -0.6 0.2 0.0 uoa ..
'.llI IX ,...
~u ......
0.4-0.6 0.8 t Figure 3. a) The titration curves (a - our, ~ - simplified) and b) the pH vs. ~ relationship plotted for V2+-Mn04" system; Co=O.Ol, C=O.04, Ca=O.l mol 1-1 4. V2+-Cr20~- system. The electron pre-balance related to the system obtained after an addition of Vml ofC mol 1-1 K2Cr207 into Vo ml ofComol 1-1 VS04acidified with H2S04 (Ca) mol 1-1 has the form (Zl=23 for V, Zz=24 for Cr): (Zl-2)[V2+]+(Zl-3){[V3+]+[VO+]+[VOH2+]+[V(OH)~]) + (Zl-4){[V02+]+ [VOOH+]+2[(VOOH)~+]+[VOS04]) + (Zl-5){[VO~]+[VO~-]+[HVO~-]+[H2V04 ]+ [H3 V04]+[V03]+[HV03]+[V02S04]+2([V20t]+[HV20t]+[H2V20r-]+[H3V2°7"])+ 3[V30~-]} + (Zz-6){2([H2Crz07]+[HCr207]+[Cr209-])+[H2Cr04]+[HCr04]+ [CrarH + (Z2-3){[Cc3+]+[CrOHZ+]+[Cr(OH)~]+[cr02]+[HCr02]+[CrSOS]+ (Z2-2){(Cc2+]+[CrS04]) - {(Zl-2)CoVo+2(Zz-6)CV}/(Vo+V) = 0 (9) The balance (9) together with the charge and concentration balances are the ba-sis for the construction of the titration curve presented in Fig. 4.
Equationsfor redox titration curves 629 1.4-a > 1.0 f;I:I ..
0.6 ....
~ e ~ 0.2 0 Pot -0.2 -0.6 0.0 0.;2 0.4 0.6· t Figure 4. a) The titration curves (a - our, J3 - simplified) and b) the pH vs.~relationshipdue to V2+-Cr20~- system;Co=0.012, C=0.04, <::.=0.1 mol 1-1 5. IOj-ascorbicacid system. The.electron-, charge- and concentration balances will be fOfIl1ulated for ascorbinometric titration of iodate according 10 the prescrip-tion taken from the Erdey and Svehla's monograph . This determination seemed to be the most interesting among those quoted in there.
The physico-chemical data [18-21] applied in the calculations reported below are involved in 36 independent relations [Xl= l(y,[y]q (10) and collected in Table 1. Dehydroascorbic acid (C6H606) is considered as unproto-lysable species in aqueous media.
Table 1. Expressions forp related to different X and Y [Eq.(1)]; pD =-log[D], D = H+, 1-,CI-; A=F/(Rl1n10)=16.92 for T=298 K; £-r~ox potential X Y q p X Y q P C(}I~6 C6Hs06 1 2A(E-o.39)+2pH CI03 cr 1 6A(E-l.4S)+6pH C(}Is06 C6H706 1 4.21-pH CI04 CI"" 1 8A(E-1.38)+8pH C6HS06 C6H60~-1 IS.78-2pH 1 2CI-12 1 O.2-pCl 12 1-2 2A(£-o.621 ) 1 2 ICI 2 -2A(£-1.10S)+2pCl h 1-3 2A(£-o.545) I(:J2 ICI 1 2.2-pCl 10-I-I 2A(£-o.49)+2pl1-28 11+ on-1 -14 HIO 10-I 10.6-pIl H~e03 HSe03 -1 2.62-;>1-1 103 I-t 6A(E-1.08)+6pII HZ Se03 SeOr-I 1O.94-2pH 1-1103 1°3 1 0.79-pH HgOH+ Hi+ 1 -3.7+pH HsI06 I-I M(£-1.24)+7pl-l Hg(OH)2 Hi+ I -6.3+2pl-l 630 T.Michalowski and A. Lesiak Table 1 continued H4I06 HsI06 1 -3.3+pH HgCl+ H~+ 1 6.74-pCI H3IOt r 1 M(E-Q.37)+9pH-126 HgCh Hg2+ 1 13.22-2pCI Ch CI-2 2A(E-l.359) HgCli Hg2+ 1 14.07-3pCI CIO-CI-1 2A(E-Q.88)+2pH-28 HgCl~-Hg2+ 1 15.07-4pCI HCIO CIO-I 7.3-pH HgI+ Hg2+ 1 12.87-pl CI02 cr l 4A(E-Q.77)+4pH-56 HgI2 n,i+ 1 23.8-2pl HCI02 CI02 1 1.97-pH Hgl:3 Hg2+ 1 27.6-3pI CI02 CI-I 5A(E-l.S)+4pH HgI~-Hg2+ 1 29.8-4pI Let us assume that Vo=100 mI ofthe solution containing KI03 (Co=O.Ol mol I-I), HCI (Ca=O .02 mol I'"), HZSe03 (Cse=O .02 mol I'") (as the catalyst) and HgCl2 (CHg=O.07 mot r ') is titrated with Vml of C=O.l mol 1-1 ascorbic acid. The electron pre-balance written fora monophase system, under the assumption that chlorine species participate in electron transitions whereas mercury docs not, is (ZI+1)[I-]+(3Z1+1)[I3"1+2([IZ1+a[I2(s)])+(Zl-1)([HIO]+[IO-])+ (Zl-5)(HI03]+[I03])+(Zl-7)([H5I06]+[H4I06]+[H3IOg-])+(Zl+1)[HgI+]+ 2(Zt+l)[HgI2]+3(Zl+1)[HgI3]+4(Zl+1)[HgI~-]+Z2([C6H806]+[C6H7061+ [C6H60t])+(Zz-2)[C6H606]+(Z3+1)[CI-]+ZZ3[Clz]+(Z3-1)([HCIO]+[CIO-])+ (Z3-3)([HCI02]+[CI0Z])+(Z3-4)[CIOz]+(Z3--5)[CI03]+(Z3-1)[CI04]+ (Zl+Z3)[ICI]+(21+2Z3+1)[IClz l+(2Zt+Z3+1)[IzCI-] -(Zt-5)CoVO+Z2CV+(Z3+1)CaVo}/(Vo+V) =0 (11) The charge and concentration balances have the form: CoVo/(Va+ V)+[H+J-[OH-]-[I-]-[I3]-[IO-]-[I03]--[H4I06]-2(H3IO~-] [CI-]-[ICl zl-[I2CI]+2[Hg 2+]+[HgOH+]+[HgCI+]--[HgCI3]-2[HgCI~-]+ [HgI+]-[HgI3]-2[HgI~-]-[C6H706]-2[C6H60~]-[HSe03]-2[Seo~-]-[CIO-]-[CIOz]-[CI03]-[CI04"1 = 0 (12) [I-]+3[I31+2([Iz]+a[I2(s)])+[HIO]+[IO-]+[HI03]+[I03"]+[H51061+[H4I061+ [H3IO~-]+[ICI1+[IC121+2[IzCl-]+[HgI+]+2[HgIz]+3[HgI3]+4[HgIa-] -CoVo/(Vo+V) :: 0 (13) [C6H806]+[C6H706]+[C6H60~-]+[C6H606]-CV/(Vo+V)=0 (14) Equatio11$Jar red~ (itr(ltiolf,.curves 631 [Cl-]+2[Clz]+[HCIO]+[CIO-]+ [HCIOz]+[CI02]+[CIOz]+[CI03 ]+[CI04]+ [ICl]+2[ICI2]+[IzCl-]+[HgCl+]+2[HgClz]+3[HgCI3 ]+4[HgClt]-(2CHg+CJVoI(Vo+V)=0 (15) 4 4 [Hg2+]+[HgOH+]+[Hg(0H)z]+~ [HgClf2- i)+] +~[HgIf2- i)+] i-1 i-I CHgVoI(Vo + V) =0 (17) 7independent balances (11)-(17) together with the relationships given in Table 1 in-volve 7 independent variables chosen for optimization, e.g., the variables Xi (i=1,...,7) expressed by equations: [HgI+] :: lo-XI, [HgClz] =lo-x2, [HZSeZ0 3]=10-x3, [C6H606] =10~x4, .[Cl] =lO-xs, pH=X6' E =x7 Other choice of independent variables is also admissible. It is advantageous (from computational viewpoint), however, to choose concentrations with relatively high values during titration or at defined stage of the titration. For this purpose, a change of variable(s) after passing an equivalence point may prove to be desirable.
The curves thus obtained (Figs. 5-8) are related to the system with (CHg= 0.07 mol 1-1, system A) or without (CHg=O, system B) mercuryrll) chloride in the titrand (with other data unchanged).
1.1 .11' .....'.B.B' :> 0.9 ~ ..
-as =' 0.7 g .s 0 ~ 0.5 0.3 0 1 .1 ICII B' B .....',f.f· Figure S.Graphs ofE vs. relationships plotted for Vo=100, Co=O.OI, C=O.I, Cse=O.02,Ca=O.02 mol l"!
and: CHg=O (curve A), CHg=O.07 (curve B); ct»=CV/(CoVo). Curves A', B' are plotted under assumption that iodine does not precipitate in the defined ct» range. For further details iieetext 632 T. Michalowski andA. Lesiak 1.95 == 1.85 ~ 1.75 1.65 o 1 2 3 Figure 6. pH vs. ct> relationships: A - for the system A, B - for the system R Curves A', B' are plotted under assumption that iodine docs not precipitate in the defined range ..-------.--.--..------~--,~"-, 1.
-5 0.,-------------,----"1 E r == --III -10 1,-, 0 ' ....
-" ,.
fir , "..
-15 I .......
· , lHIO I· I• -20 I 0 1 2 3 ~ 0 IO~·· B -,,---.-----,".' ol.i -5 ---------i~-----------~ -10 -~---...._;;;...--.
~ I.
-10-~15 -200 1 2 3 ~ Figure 7. Graphs of 10g[X;] vs. relationship for some iodine species Xi: A - related to the curve A, B - related to the curve B ill Fig. 5 Equations for redox titration curves 633 0..------------------.
2 1 --10 -15 -20+------~-----_r--4I--+--; o 0 ..------------........----, ~ :; -10 -15 B 2 1 -20+--...----,r------..-~'--~--+--...; o Figure 8. Graphsof 10g[Xill's. relationships for some species Xi indicated at the corresponding lines: A - related to the curve A, B - related to the curve B in Fig. 5; H2LR, HLR-, LR2- and LO -abbreviations for C6Hg06, C6H706, C6H60~- and C61I606, respectively DISCUSSION The constancy of pH during titration is not fulfilled in the systems considered, even at greater buffer capacities, see Figs. Ib, 2b, 3b, 4b and 6. pH affects the redox potentials and the related titration curves are put out of shape under this simplifying assumption.
Fe3+ and Ce4+ form relatively stable sulfate complexes. In this respect, the simplification of the problem as was made in , is then incorrect. For this reason, "electron balances" given in are not valid.
The following comments concern ascorbinometric titration of iodate, taken as the most interesting and demonstrative example. The problem is considered in the aspect of an influence of some factors on the shape of the related titration curves. It enables 634 T.MichtilowskiandA. Lesiak to settle the best apriori conditions of the determination. The results obtained there provide the following conclusions: 1. The first inflection point in the curve A disappears after addidion of an excess of HgCl2 (curve B); £=0.904 at <1>=2.5 (V=25), £=0.483 at <1>=3 (V=30) (curve A) and £=0.653 at <1>=3 (V=30) (curve B).
2. Curves A and B (Fig. 5) were plotted allowing for the limited solubility ofsolid iodine, 12(s)' in water s=1.33xl0-3 mol 1-1 [8-!_20]. To plot the related part of the titration curve (V-intervals: ac for curve A, bd for curve B, Fig. 5) it was assumed that =S and [Iz(s)] was taken as the variable [a=1 in Eqs. (11), (13)]; V values limiting the ac and bd intervals are: 7.82 (a), 26.91 (c), 8.51 (b), 28.22 (d). The curves A and B nearly overlap with the curves A', B' which were plotted under. the assumption that solid iodine is not formed there [a=O in Eqs. (11),(13)].
3. Curve A in Fig.6 is a broken line, very interesting from analytical viewpoint.
Its maximum and minimum values coincide exactly with the corresponding inflection points on the curve A in Fig. 5. The reactions related to the first, ascending part of the curve A are: 2103 + 5C6Hs06 + 2H+ = Iz(s) + 5C6~06 + 6HzO 2103 + 5C6HS06 + r + 2H+ = 13' + 5C6~06 + 6HzO (18) (19) whereas the descending part of the curve corresponds to the reactions where protons are evolved Iz + C6HS06 = 21- + C6H606 + 2H+ 13 + C6HS06 = 31- + C6H606 + 2H+ (20) Further growth of pH is affected by the dilution effect. Such conclusions resulted from an analysis ofthe graphs in Fig. 7.
4. Reduction ofiodate in the system B (with HgClzinvloved) proceeds according to the scheme described by equations (18)-(20). Concentrations of 1- and 13' are low there.
5. The concentrations of the oxidized forms of chloride are negligibly small in comparison with [CI-]; concentration of Clz (the oxidized chlorine species of the highestconcentration) is smaller than 10-14 mol 1-1. Chlorine could then be excluded as the participant in the redox equilibria.
6. HgClz is present in great stoichiometric excess compared with the iodide concentration there. Taking into account the predominating species, the reaction HgCI2+21- = Hglz+2CI- instead of Hgz++4I- = HgI~- [17, 22] should be considered there (Fig. 8). Hglzis the predominating species among mercuric iodide .complexes.
Equations[artedoxtitration curves 1.6 .,.......................................--..........----........................................., 6:lS E, Y 1.5 1.4 1.3 t,.~,1J 0.22 0.26.
0.30 4> Figure 9. Fragments of hypothetical titration cUrves plotted for (K5hK52) pairs: 1 (104, 10'); 2 - (103, 106); 3 - (102•5, 105)~ 4 .... (102,104) ; 5 .... (104,0); 6 .... (103,0); 7 (102,0); 8 - (0,0) 1. The. cenccutratioas of heptavalcut .ioclitte species, I.e. Hsl0 6, H410 C; and H3IO~- are below 10-20 mol I'".
8.The shape ofthe the titrationcurves A and13 in Fig. 5isin exquisite accordance with the ones found experimentally [17, 2~J. Thereare<minot, however, differences.
Namely, the theoreticalcurves in Fig. 5fall abruptly in the immediatevicinity of the initial point (V=O) of the titration. At V=O, E=1.152 for curves A and B; at V=O.Ol ml, E=1.072for AalldE=1.068 forB.The sccondexperiU1CI1falpointoltthe experfmental curve ShOWIl in [22} is far distinct fromV=O andthc two points Wereconnected bya rounded line. Moreover, the mistake made by the authors itt paper (which erroneously recalculated the potentials against the normal hydrogen electrodewhile the calomel electrode was applied in the titratioll)waS revealed by this way.
1. The electron balance notation. Zi values in the electron balances can be chosen arbitrarily; it results from the fact tlmt any llnearcombination of the electron and concentration balances fulfil all properties of the electron balance [8). For example, the atomic numbers of the elements participating iti the electron transitions were applied above.
2. The eledronpre-balance and electron balance, A linear combinatlou of the electron.pre-balance and the coneerttratioll balances provides the ;electron balance' equation, in the meaning similar to one already known in the literature, e.g., [4). For example, the linear combination-of equations (2)-=(4) gives {[Fe2+)+[FeOH+]+[FeS04]}-{5[Ml104]+4[MI10~1+[Mn3+]+[MnOH 2+]} = (CoVo-5CV)/(Vo+V) (21) 636 T. Michalowski and A. Lesiak Equation (21) is much more completed than the one obtained from the electron balance involving only the species participating in the redox reaction notation, see e.g., ; all attainable physico-chemical knowledge on the system considered is involved there. There is no necessity to consider separately the parts of the titration curves located before and after the related inflection points.
3. Complexes presumed. The literature provides no information concerning the sulfate complexes of the Mn3+ ions. For this purpose, some simulated titration curves were plotted for different stability constants values (K3Ddue to the complexes Mn(S04)t(3 - 2;) presumed (i=l, 2). The inflection point, visible in Fig. 9 at =0.25, and related to a hypothetical reaction 4MnS04+ Mn04'+ HS04' + 2H+ =5Mn(S04)2 + 4H20, disappears at lower K31 and K32 values assumed in the simulating procedure.
Comparing the constructed curves with the one obtained experimentally and the one plotted under the assumption that K31=K32=0 (curve 8) we concluded that the complexes suggested do not exist at all or their stability constants arc small.
4. Kinetic aspect. The calculations based on the equilibrium constants do not involve the kinetics. For example, it is known that from thermodynamic viewpoint Mn04' should decompose water and O2 is evolved. Atthe.same time Mn04' should be reduced into Mn02 with great efficiency, which results from the equation log{[Mn02]/[Mn04']} = 23.4 - pH - 0.7510gp(02) where P(02) [arm] is a partial pressure of 02' In reality, Mn04' and H20 form a metastable system. For this reason, H20 does not enter the electron balances in the systems 1 and 3 considered above.
5. Some computntlonal details. The iteration procedure of the computer pro-gram MINRES which employs the Levenberg-Marquardt algorithm was applied in the calculations. Denoting the set of n independent variables involved in a system by "" • " x, x = [x(l),...,x(n)], the related balances were written in general formfi(x) = 0 (i=l,..., m).
For Fe2++ Mn04' system, ~ is represented by 5 variables: Fe2+, fMn2+], 2 " " [H+], [S04-], E; Eqs. (2)-(6) form a set ofm=5 equations. Optimum values for x, Xopt, minimize the sum of the squares m SS = LfJi (~)]2 ;=1 The error of the fit is expressed as the square root of SS, er = VSS.
A computer IBM PC 386 was used in calculations. The optimal values of the variables found for the defined volume ~. have been applied as the initial values for the consecutive volume Vj+1• The volumes considered were refined in close vicinity of the equivalence point(s). The values er < 10-15 were obtained in all instances.
6. Software. Full, very ingenious and versatile computer program has been elaborated. It enables any redox or non-redox, mono- and two-phase systems to be easily resolved, not only from the titration curves viewpoint.
Equations for redox titration curves REFERENCES 637 1. Michalowski 1:, Talanta, 39, 1127 (1992).
2. Olivieri A. C.,J. Chem. Educ., 61,229 (1990).
3. Michalowski 1:,J. Chem. Educ., 69,858 (1992).
4. Butler J. N., Ionic Equilibrium.A MathematicalApproach, Addison-Wesley, Reading, MA 1964, p,396.
5. Anfalt 1: and Jagner D., Anal. Chim. Acta, 41,57 (1969).
6. Machado A. A. S. C.,J. Chem. Educ., 53, 305 (1976).
7. Michalowski 1:,J. Chem. Educ., in print.
8. Michalowski 1: and LesiakA.,J. Chem. Educ.; in print.
9. Michalowski T. and Lesiak A., Euroanalysis VIII, Book of Abstracts, I)J35, PM 10, Edinburgh 1993.
10. de MouraD. R.,J. Chem.Edu~., 61,226 (1990).
11.de Levie R.,J. Electroanal.Chem., 323, 347 (1992).
12. Michalowski T., Chem.Anal. (WarsaK1, 26,799 (1981).
13. Michalowski 1:, Zesz. Nauk. UJ, PnChem., 32, 15(1989).
14. Koscielniak P. and Michalowski T., Chem. Anal. (WarsaK'~, 31,313 (1986).
15.Michalowski T. andStepak R.,Anal. Chim.Acta, 112,207 (1985).
16. West 1: S.,PureAppl. Chem., 50, 325 (1978).
17. Erdey L and Svehla .G.,Ascorbinometric Tltrations, Akad. Kiado, Budapest 1973.
18.lnczedy J.,AnalyticaIApplic(llions ofComplex Equilibria, lIorwood, Chichester 1976.
19. Lurie Yu.Yu., Guide-J)ook on AnalyticalChemistry, Khimia, Moscow 1989 (in Russian).
20.Guide-Book/or Chemist, (Nikolsky B. P., Ed.), Vol. 3,1964 (in Russian).
21. Charlot G., Les reactions chimiques en solution; L'analyse qualitative minemle, Masson, Paris 1969.
22. Erdey L, Bodor E. and Buzas II., Z. Anal.Chellf., 134,22 (1951/52).
23. Treatise on AnalyticalChemistry (Kolthoff I. M., Elving P. J. and Sandell E. B., Eds.), Part I, Vol.1, p.670, Intersc, Enc., Inc., New YorkJ959.
ReceivedJanuary 1994 AcceptedJune 1994 |
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2019 AMC 10A Problems/Problem 20
The following problem is from both the 2019 AMC 10A #20 and 2019 AMC 12A #16, so both problems redirect to this page.
Contents
1 Problem
2 Solutions
2.1 Solution 1
2.2 Solution 2 (Pigeonhole)
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5
2.6 Solution 6
2.7 Solution 7
3 Video Solutions
3.1 Video Solution by OmegaLearn
3.2 Video Solution 1
3.3 Video Solution 2
3.4 Video Solution 3
3.5 Video Solution 4
4 See Also
Problem
The numbers are randomly placed into the squares of a grid. Each square gets one number, and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?
Solutions
Solution 1
Note that odd sums can only be formed by or so we focus on placing the evens: we need to have each even be with another even in each row/column. Because there are only odd numbers overall, we can only have row and column. It can be seen that there are ways to do this. There are then ways to permute the odd numbers, and ways to permute the even numbers, thus giving the answer as .
~Petallstorm (Minor edits by JeffersonJ and monkey_land)
Solution 2 (Pigeonhole)
By the Pigeonhole Principle, there must be at least one row with or more odd numbers in it. Therefore, that row must contain odd numbers in order to have an odd sum. The same thing can be done with the columns. Thus we simply have to choose one row and one column to be filled with odd numbers, so the number of valid odd/even configurations (without regard to which particular odd and even numbers are placed where) is . The denominator will be , the total number of ways we could choose which of the squares will contain an even number. Hence the answer is
The Pigeonhole Principle isn't really necessary here: After noting from the first solution that any row that contains evens must contain two evens, the result follows that the four evens must form the corners of a rectangle.
~Petallstorm
Solution 3
Note that there are 5 odds and 4 evens, and for three numbers to sum an odd number, either 1 or three must be odd. Hence, one column must be all odd and one row must be all odd. First, we choose a row, for which there are three choices and within the row P(5,3). There are 3! ways to order the remaining odds and 4! ways to order the evens. The total possible ways is 9!.
~Petallstorm
Solution 4
Note that the odd sums are only formed by, or any permutation of . When looking at a 3x3 box, we realize that there must always be one column that is odd and one row that is odd. Calculating the probability of one permutation of this we get:
Odd - Odd - Odd -
Odd - Even Even
Odd - Even Even
Now, there are 9 ways you can get this particular permutation (3 choices for all odd row, 3 choices for all odd column), so multiplying the result by , we get:
~petallstorm
Solution 5
To get an odd sum we need or we need . Note that there are ways to arrange the numbers. Also notice that there are even numbers and odd numbers from through .
The number of ways to arrange the even numbers is and the number of ways to arrange the odd numbers is . Now a common strategy is to deal with the category with fewer numbers so this suggests for us to look at even numbers and try something out.
Since we have even numbers we need to split two evens to one row and two evens to another row. Note that there are ways to choose the rows, and the to choose where the two even numbers go in that row. The odd numbers go wherever there is a spot left which means we don't need to care about them. So we totally have ways to place the even numbers.
Don't forget, we still need to arrange these even and odd numbers and like mentioned above there are and ways respectively.
Therefore our answer is simply
Solution 6
We need to know the number of successful outcomes/Total outcomes. We start by the total number of ways to place the odd numbers, which is , which is because there are odd number between and . Note we need odd number and even numbers OR odd numbers in each row to satisfy our conditions. We have types of arrangements for the number of successful outcomes, we could have all the even numbers on the corners, which is case, We could have the odd numbers arranged like a T, which has cases (you could twist them around), and you have the even numbers in a by square, and you have cases for that. We have successful cases and total cases, so the probability is = which is
~Arcticturn
Solution 7
Since we only care about the parity of the sum of our rows and columns, let's take everything modulo . We can see that there will be odd numbers, and even numbers. Testing a few configurations, we realize that if we have one complete row and column of , the sum of all rows and columns will be odd. We find distinct configurations: We can see that rotating the first configuration by and , will give us cases for this configuration. Similarly, the second configuration has cases by the same rotations. The third configuration only has case because its rotational symmetry is . There are ways to choose the odd and even numbers for all of the configurations, so our answer is .
We can also realize that there are places for the row and column of to overlap, resulting in cases for all of the possible configurations.
~kn07
Video Solutions
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution 1
Education, the Study of Everything
Video Solution 2
Video Solution 3
Video Solution 4
See Also
2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19Followed by
Problem 21
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 10 Problems and Solutions
2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15Followed by
Problem 17
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
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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9083 | https://www.merriam-webster.com/thesaurus/rudimentary | RUDIMENTARY Synonyms: 99 Similar and Opposite Words | Merriam-Webster Thesaurus
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Est. 1828
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as in basic
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Synonyms of rudimentary
rudimentary
adjective
ˌrü-də-ˈmen-tə-rē
Definition of rudimentary
1
as in primitive
belonging to or characteristic of an early level of skill or development rudimentary shelters built by prehistoric peoples
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advanced
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Recent Examples of rudimentary These pioneers used rudimentary plows, but when steel plows became available farmers were able to plant and harvest much more food per acre to feed a growing population.—Mary Ann Grossmann, Twin Cities, 31 May 2025 In short, DSLRs are hampered by a relatively narrow area of interest for autofocus, and only a few have rudimentary face detection support through the viewfinder.—PC Magazine, 20 May 2025 Only a rudimentary monument on the side of a road stands to mark the site.—Jay Cheshes, Smithsonian Magazine, 20 May 2025 And no, this doesn't mean turning the office into a place full of rudimentary and shallow perks.—Julian Hayes Ii, Forbes.com, 26 May 2025 See All Example Sentences for rudimentary
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About nineteen minutes in, after a meditative string of A-flats, a halting procession of some two hundred and seventy-five chords begins—permutations of eight basictypes, containing up to twelve notes. — Alex Ross, New Yorker, 23 June 2025
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9084 | https://crypto.stackexchange.com/questions/75408/efficient-function-algorithm-method-to-do-modular-exponentiation | rsa - Efficient function/algorithm/method to do modular exponentiation - Cryptography Stack Exchange
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Efficient function/algorithm/method to do modular exponentiation
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I was working on this project where I needed an RSA key, and I wondered if there was and more efficient way of doing g a mod n g a mod n other than calculating g a g a and then finding the remainder when you divide?
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edited Oct 30, 2019 at 19:10
kelalaka
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asked Oct 30, 2019 at 13:12
Ömer Enes ÖzmenÖmer Enes Özmen
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Efficient is not sufficient in cryptography. You also need secure computation. Consider a standard repeated squaring implementation in Python;
```
def fast_power(base, power):
result = 1
while power > 0:
# If power is odd
if power % 2 == 1:
result = (result base) % MOD
# Divide the power by 2
power = power // 2
# Multiply base to itself
base = (base base) % MOD
return result
```
The if condition is an attack point for side-channel attacks. One can measure power usage and determine exponent bits. The below image or similar research images can show the idea of the attack.
If you are signing with RSA your private exponent can be revealed. Squaring can be implemented faster than multiplication, in that case, it can also be exploited. Therefore, you need constant timing to prevent this. The python library contains power modulus pow(a,e,n) which is based on repeated squaring and it is not secure. The below version of the repeated squaring method always calculates the (result base) % MOD and discards it when it is not necessary. This may help mitigate the attack, however, this can be still traced if there is an instrument with enough resolution. Also, during compilation, make sure that this mitigation is not removed by compilers. Compilers, like Cray's CC can discard unused calculations as an optimization by default. The below algorithm has given a name as square and always multiply algorithm.
```
def constant_time_power(base, power):
result = 1
while power > 0:
# else to prevent leak
if power % 2 == 1:
result = (result base) % MOD
else:
result_temp = (result base) % MOD //will be discarded.
# Divide the power by 2
power = power // 2
# Multiply base to itself
base = (base base) % MOD
return result
```
As mentioned above code, the above can be optimized by compilers with dead code elimination technique since the code contains unused code. Some compilers have pragma directives to disable local optimization. Instead of relying on compiler directives and their bugs, changes, etc., it is possible to use a constant-time conditional copy. The below show this;
```
def constant_time_power_with_conditional_copy(base, power):
result = 1
while power > 0:
# constant-time conditional copy
sBit = power%2==1
result = ((result base) % MOD)sBit+(1-sBit)result
# Divide the power by 2
power = power // 2
# Multiply base to itself
base = (base base) % MOD
return result
```
The famous GMP library has a secure version mpn_sec_powm of their standard mpn_powm.
This function is designed to take the same time and have the same cache access patterns for any two same-size arguments, assuming that function arguments are placed at the same position and that the machine state is identical upon function entry. This function is intended for cryptographic purposes, where resilience to side-channel attacks is desired.
There are other modular multiplications like Barret or Montgomery. They have also constant-time implementations. For example for Montgomery based
2009 Fast and Constant-Time Implementation of Modular Exponentiation by Gopal et al.
Gmpy now has powmod_sec(), see the update and use at own your risk.
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edited Apr 13, 2023 at 6:29
answered Oct 30, 2019 at 19:08
kelalakakelalaka
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3
4 Note that this is not a secure sample code because a smart compiler will notice that power_temp is never used and that its defining RHS doesn't have side-effects as per the language specification and will optimize it away. A better solution would be to remove the if and add a constant time conditional copy in the next line.SEJPM –SEJPM 2019-10-31 18:49:18 +00:00 Commented Oct 31, 2019 at 18:49
@SEJPM I'll look into that.kelalaka –kelalaka 2019-10-31 18:56:11 +00:00 Commented Oct 31, 2019 at 18:56
2 I think the basic python interpreter doesn't do dead-store / dead-code elimination. But I suppose it's not a great idea to make such an implicit assumption about the interpreter / compiler and "just providing a secure example" might be a better plan. Also to counter power-analysis attacks multiplicative blinding may be necessary.SEJPM –SEJPM 2019-10-31 19:19:21 +00:00 Commented Oct 31, 2019 at 19:19
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Yes. You don't need to wait until the end of the computation to compute the remainder, you can do that in each step of the exponentiation; this way, the largest numbers you'll need to handle are twice the size of n.
There are many algorithms to compute the exponentiation itself, the simplest is square-and-multiply.
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edited Oct 30, 2019 at 15:33
answered Oct 30, 2019 at 13:19
ConradoConrado
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Linked
23How can I understand whether my C implementation is constant-time or not (i.e. resistant to timing attacks)
3Is there an easy way to make textbook RSA secure enough so it can be used in real life?
3Timing attacks on homomorphic encryption
2Is an if-else calling same function(s) vulnerable to side channel attacks
2How do I try a timing attack on a public-key implementation on a general-purpose computer?
1Side channel attack - execution time depends on the size of the key
2Is "double-and-add-always" a good mitigation against side channel attacks?
1Is there a benefit in cryptography to return 0 in square and mutliply algorithm when a modulo is 0?
0What efficiency can we expect from an average consumer computer for the RSA algorithm?
-1Can the public key be derived from the private key?
See more linked questions
Related
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4Solve a Modular Exponentiation
4Efficient algorithm for finding Sophie Germain primes
2How to solve polynomial modular equation to create a correct decryption algorithm
2Montgomery Multiplication with CRT
2How to handle modular arithmetic with regard to two's-complement negative numbers?
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6Efficient multiplication modulo a square
0Reverse modular exponentiation: finding the base given the exponenent, the modulus semi‑prime and the result…
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9085 | https://covetdentist.com/variations-in-tooth-anatomy-in-dogs/ | Variations in Tooth Anatomy in Dogs
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Variations in Tooth Anatomy in Dogs
15 Mar Variations in Tooth Anatomy in Dogs
Posted at 18:20h in Dog Dentistry by New Wine Digital
Not every dog and cat have teeth that are just like a picture in a textbook. Some animals are born with extra teeth, missing teeth, malformed teeth, extra roots and/or teeth growing in the wrong place. Sometimes these abnormal teeth cause problems, but sometimes they are just incidental findings that illustrate how unique each pet is. Our post will discuss the variations in tooth anatomy in dogs.
Adult Dog Teeth
Dogs have a great variety of shapes in their 42 teeth. They have teeth with single roots (canines, incisors, first premolars and third molars), double roots (upper second and third premolars, lower premolars and first and second molars) and triple roots (upper fourth premolars and molars). Cats have only 30 teeth and the only triple rooted teeth are the upper fourth premolars. All the rest are single or double rooted.
Cats rarely have abnormalities in tooth shape and number, but sometimes dogs can be a little more creative. It is not unusual to find a triple rooted upper third premolar in dogs like this one. These supernumerary roots are usually bilateral, meaning if present on one side, the same tooth on the other side is likely to have it as well.
Extra roots are generally not a problem unless the tooth needs to be extracted. In that case, the veterinarian needs to be sure to get all three roots.
Puppy Teeth
The puppy in the photo below came to visit us because he had extra teeth growing on the inside of his lower canines. Extra teeth are called “supernumerary” teeth. And although it is obvious that something is different about this puppy, we did not know if those extra teeth were a second set of canine teeth or incisors (the smaller teeth between the canines).
Fortunately, a dental radiograph could easily tell the difference. The extra teeth were supernumerary third incisors. Some dogs have room for extra teeth. If the extra teeth don’t cause any crowding and don’t interfere with the mouth closing, they can stay in place. In this puppy’s situation, the extra teeth were very crowded with the canines. Crowded teeth are predisposed to periodontal disease, so the extra incisors were extracted. With regular home and professional dental care, Cooper’s remaining teeth should stay healthy.
Even though incisors look pretty small, their roots are quite long. These teeth had to be extracted very carefully to avoid damage to all the other teeth nearby.
Diagnostic Imaging in Fort Collins
There’s something fascinating about variations in normal tooth anatomy. These cases illustrate the importance of diagnostic imaging such as dental radiographs and cone beam CT. At Animal Dental Care & Oral Surgery in Fort Collins, every tooth in every patient is carefully examined to determine if it is healthy and to document any abnormalities. If you think your pet has a “special” tooth, give us a call to schedule an exam!
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dental extraction, dog incisor, incisor extraction, incisor removal, incisors, tooth anatomy, variations in normal tooth anatomy
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Why do tangent curves have the same derivative?
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Suppose I have two level sets and I choose two curves, one from each set. Let's suppose these two curves touch at a point. Why will their derivatives be the same?
calculus
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asked Oct 23, 2015 at 2:44
Stan ShunpikeStan Shunpike
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That makes a lot of sense Stan Shunpike –Stan Shunpike 2015-10-23 03:38:20 +00:00 Commented Oct 23, 2015 at 3:38
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Here's an intuitive picture: Recall the interpretation of the tangent at a point p p on a curve as the direction in which you move on changing p p slightly.
If the tangent to curve A A at p p intersects curve B B, and curves A A and B B coincide at p p, that means that A A must move into B B and cut it at two distinct points: p p and something close to p p, and the two can't 'touch' at p p, only intersect. So their tangents coincide at p p. The derivative is simply the slope of the tangent, and must also be identical.
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answered Oct 23, 2015 at 3:19
stochasticboy321stochasticboy321
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The derivative of a single variable function f(x)f(x) evaluated at some point x=a x=a is the slope of the tangent line at that point.
m t a n g e n t=f′(a)m t a n g e n t=f′(a)
Note the emphasis on the tangent line. There's only one such line. So if two curves are tangent to one another at x=a x=a, then they share that same tangent line. Thus they have the same derivative at that point.
We can generalize this to any set of level curves.
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answered Oct 23, 2015 at 3:19
zahbazzahbaz
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9087 | https://faculty.washington.edu/seattle/physics227/reading/reading-24-25.pdf | Chapter 9 Angular Momentum Quantum Mechanical Angular Momentum Operators Classical angular momentum is a vector quantity denoted ⃗ L = ⃗ r X ⃗ p.
A common mnemonic to calculate the components is ⃗ L = ¯ ¯ ¯ ¯ ¯ ¯ ˆ i ˆ j ˆ k x y z px py pz ¯ ¯ ¯ ¯ ¯ ¯ = ¡ ypz −zpy ¢ˆ i + ¡ zpx −xpz ¢ˆ j + ¡ xpy −ypx ¢ˆ j = Lxˆ i + Lyˆ j + Lzˆ j.
Let’s focus on one component of angular momentum, say Lx = ypz −zpy.
On the right side of the equation are two components of position and two components of linear momentum.
Quantum mechanically, all four quantities are operators. Since the product of two operators is an operator, and the difference of operators is another operator, we expect the components of angular momentum to be operators. In other words, quantum mechanically Lx = YPz −ZPy, Ly = ZPx −XPz, Lz = X Py −YPx.
These are the components. Angular momentum is the vector sum of the components. The sum of operators is another operator, so angular momentum is an operator. We have not encountered an operator like this one, however, this operator is comparable to a vector sum of operators; it is essentially a ket with operator components. We might write ¯ ¯ L > = Lx Ly Lz = YPz −ZPy ZPx −X Pz X Py −YPx .
(9 −1) A word of caution concerning common notation—this is usually written just L, and the ket/vector nature of quantum mechanical angular momentum is not explicitly written but implied.
Equation (9-1) is in abstract Hilbert space and is completely devoid of a representation. We will want to pick a basis to perform a calculation. In position space, for instance X →x, Y →y, and Z →z, and Px →−i¯ h ∂ ∂x, Py →−i¯ h ∂ ∂y , and Pz →−i¯ h ∂ ∂z .
Equation (9–1) in position space would then be written ¯ ¯ L > = −i¯ hy ∂ ∂z + i¯ hz ∂ ∂y −i¯ hz ∂ ∂x + i¯ hx ∂ ∂z −i¯ hx ∂ ∂y + i¯ hy ∂ ∂x .
(9 −2) The operator nature of the components promise difficulty, because unlike their classical analogs which are scalars, the angular momentum operators do not commute.
300 Example 9–1: Show the components of angular momentum in position space do not commute.
Let the commutator of any two components, say £ Lx, Ly ¤ , act on the function x.
This means £ Lx, Ly ¤ x = ¡ Lx Ly −Ly Lx ¢ x → µ −i¯ hy ∂ ∂z + i¯ hz ∂ ∂y ¶ µ −i¯ hz ∂ ∂x + i¯ hx ∂ ∂z ¶ x − µ −i¯ hz ∂ ∂x + i¯ hx ∂ ∂z ¶ µ −i¯ hy ∂ ∂z + i¯ hz ∂ ∂y ¶ x = µ −i¯ hy ∂ ∂z + i¯ hz ∂ ∂y ¶ ¡ −i¯ hz ¢ − µ −i¯ hz ∂ ∂x + i¯ hx ∂ ∂z ¶ ¡ 0 ¢ = ³¡ −i¯ h ¢2y ´ = −¯ h2y ̸= 0, therefore Lx and Ly do not commute. Using functions which are simply appropriate posi-tion space components, other components of angular momentum can be shown not to commute similarly.
Example 9–2: What is equation (9–1) in the momentum basis?
In momentum space, the operators are X →i¯ h ∂ ∂px , Y →i¯ h ∂ ∂py , and Z →i¯ h ∂ ∂pz , and Px →px, Py →py, and Pz →pz.
Equation (9–1) in momentum space would be written ¯ ¯ L > = i¯ h ∂ ∂py pz −i¯ h ∂ ∂pz py i¯ h ∂ ∂pz px −i¯ h ∂ ∂px pz i¯ h ∂ ∂px py −i¯ h ∂ ∂py px .
Canonical Commutation Relations in Three Dimensions We indicated in equation (9–3) the fundamental canonical commutator is £ X , P ¤ = i¯ h.
This is fine when working in one dimension, however, descriptions of angular momentum are generally three dimensional. The generalization to three dimensions2,3 is £ Xi, Xj ¤ = 0, (9 −3) 2 Cohen-Tannoudji, Quantum Mechanics (John Wiley & Sons, New York, 1977), pp 149 – 151.
3 Sakurai, Modern Quantum Mechanics (Addison–Wesley Publishing Company, Reading, Mas-sachusetts; 1994), pp 44 – 51.
301 which means any position component commutes with any other position component including itself, £ Pi, Pj ¤ = 0, (9 −4) which means any linear momentum component commutes with any other linear momentum com-ponent including itself, £ Xi, Pj ¤ = i¯ hδi,j, (9 −5) and the meaning of this equation requires some discussion. This means a position component will commute with an unlike component of linear momentum, £ X, Py ¤ = £ X , Pz ¤ = £ Y, Px ¤ = £ Y, Pz ¤ = £ Z, Px ¤ = £ Z, Py ¤ = 0, but a position component and a like component of linear momentum are canonical commutators, i.e., £ Xx, Px ¤ = £ Y, Py ¤ = £ Z, Pz ¤ = i¯ h.
Commutator Algebra In order to use the canonical commutators of equations (9–3) through (9–5), we need to develop some relations for commutators in excess of those discussed in chapter 3. For any operators A, B, and C, the relations below, some of which we have used previously, may be a useful list.
£ A, A ¤ = 0 £ A, B ¤ = − £ B, A ¤ £ A, c ¤ = 0, for any scalar c, £ A, c B ¤ = c £ A, B ¤ , for any scalar c, £ A + B, C ¤ = £ A, C ¤ + £ B, C ¤ £ A, B C ¤ = £ A, B ¤ C + B £ A, C ¤ (9 −6) h A, £ B, C ¤i + h B, £ C, A ¤i + h C, £ A, B ¤i = 0.
You may have encountered relations similar to these in classical mechanics where the brackets are Poisson brackets. In particular, the last relation is known as the Jacobi identity. We are interested in quantum mechanical commutators and there are two important differences. Classical mechanics is concerned with quantities which are intrinsically real and are of finite dimension. Quantum mechanics is concerned with quantitites which are intrinsically complex and are generally of infinite dimension. Equation (9–6) is a relation we want to develop further.
Example 9–3: Prove equation (9–6).
£ A, B C ¤ = A B C −B C A = A B C −B A C + B A C −B C A = ¡ A B −B A ¢ C + B ¡ A C −C A ¢ = £ A, B ¤ C + B £ A, C ¤ , where we have added zero, in the form −B A C + B A C, in the second line.
302 Example 9–4: Develop a relation for £ A B, C ¤ in terms of commutators of individual operators.
£ A B, C ¤ = A B C −C A B = A B C −A C B + A C B −C A B = A ¡ B C −C B ¢ + ¡ A C −C A ¢ B = A £ B, C ¤ + £ A, C ¤ B.
Example 9–5: Develop a relation for £ A B, C D ¤ in terms of commutators of individual operators.
Using the result of example 9–3, £ A B, C D ¤ = £ A B, C ¤ D + C £ A B, D ¤ , and using the result of example 9–4 on both of the commutators on the right, £ A B, C D ¤ = ³ A £ B, C ¤ + £ A, C ¤ B ´ D + C ³ A £ B, D ¤ + £ A, D ¤ B ´ = A £ B, C ¤ D + £ A, C ¤ B D + C A £ B, D ¤ + C £ A, D ¤ B, which is the desired result.
Angular Momentum Commutation Relations Given the relations of equations (9–3) through (9–5), it follows that £ Lx, Ly ¤ = i¯ h Lz, £ Ly, Lz ¤ = i¯ h Lx, and £ Lz, Lx ¤ = i¯ h Ly.
(9 −7) Example 9–6: Show £ Lx, Ly ¤ = i¯ h Lz.
£ Lx, Ly ¤ = £ Y Pz −Z Py, Z Px −X Pz ¤ = ³ Y Pz −Z Py ´³ Z Px −X Pz ´ − ³ Z Px −X Pz ´³ Y Pz −Z Py ´ = Y PzZ Px −Y PzX Pz −Z PyZ Px +Z PyX Pz −Z PxY Pz +Z PxZ Py +X PzY Pz −X PzZ Py = ³ Y PzZ Px−Z PxY Pz ´ + ³ Z PyX Pz−X PzZ Py ´ + ³ Z PxZ Py −Z PyZ Px ´ + ³ X PzY Pz −Y PzX Pz ´ = £ Y Pz, Z Px ¤ + £ Z Py, X Pz ¤ + £ Z Px, Z Py ¤ + £ X Pz, Y Pz ¤ .
Using the result of example 9–5, the plan is to express these commutators in terms of individual operators, and then evaluate those using the commutation relations of equations (9–3) through (9– 5). In example 9–5, one commutator of the products of two operators turns into four commutators.
Since we start with four commutators of the products of two operators, we are going to get 16 303 commutators in terms of individual operators. The good news is 14 of them are zero from equations (9–3), (9–4), and (9–5), so will be struck.
£ Lx, Ly ¤ = Y £ Pz, Z ¤ Px + £ Y, Z ¤ Pz Px Á + Z Y £ Pz, Px ¤ Á + Z £ Y, Px ¤ Pz Á + Z £ Py, X ¤ Pz Á + £ Z, X ¤ Py Pz Á + X Z £ Py, Pz ¤ Á + X £ Z, Pz ¤ Py + Z £ Px, Z ¤ Py Á + £ Z, Z ¤ Px Py Á + Z Z £ Px, Py ¤ Á + Z £ Z, Py ¤ Px Á + X £ Pz, Y ¤ Pz Á + £ X , Y ¤ Pz Pz Á + Y X £ Pz, Pz ¤ Á + Y £ X, Pz ¤ Pz Á = Y £ Pz, Z ¤ Px + X £ Z, Pz ¤ Py = Y ¡ −i¯ h ¢ Px + X ¡ i¯ h ¢ Py = i¯ h ¡ X Py −Y Px ¢ = i¯ hLz.
The other two relations, £ Ly, Lz ¤ = i¯ h Lx and £ Lz, Lx ¤ = i¯ h Ly can be calculated using similar procedures.
A Representation of Angular Momentum Operators We would like to have matrix operators for the angular momentum operators Lx, Ly, and Lz.
In the form Lx, Ly, and Lz, these are abstract operators in an infinite dimensional Hilbert space. Remember from chapter 2 that a subspace is a specific subset of a general complex linear vector space. In this case, we are going to find relations in a subspace C3 of an infinite dimensional Hilbert space. The idea is to find three 3 X 3 matrix operators that satisfy relations (9–7), which are £ Lx, Ly ¤ = i¯ h Lz, £ Ly, Lz ¤ = i¯ h Lx, and £ Lz, Lx ¤ = i¯ h Ly.
One such group of objects is Lx = 1 √ 2 0 1 0 1 0 1 0 1 0 ¯ h, Ly = 1 √ 2 0 −i 0 i 0 −i 0 i 0 ¯ h, Lz = 1 0 0 0 0 0 0 0 −1 ¯ h.
(9 −8) You have seen these matrices in chapters 2 and 3. In addition to illustrating some of the math-ematical operations of those chapters, they were used when appropriate there, so you may have a degree of familiarity with them here. There are other ways to express these matrices in C3.
Relations (9–8) are dominantly the most popular. Since the three operators do not commute, we arbitrarily have selected a basis for one of them, and then expressed the other two in that basis.
Notice Lz is diagonal. That means the basis selected is natural for Lz.
The terminology usually used is the operators in equations (9–8) are in the Lz basis.
We could have selected a basis which makes Lx or Ly, and expressed the other two in terms of the natural basis for Lx or Ly.
If we had done that, the operators are different than 304 those seen in relations (9–8). The mathematics of this is not important at the moment, but it is important that you understand there are other self consistent ways to express these operators as 3 X 3 matrices.
Example 9–7: Show £ Lx, Ly ¤ = i¯ h Lz using relations (9–8).
£ Lx, Ly ¤ = 1 √ 2 0 1 0 1 0 1 0 1 0 ¯ h 1 √ 2 0 −i 0 i 0 −i 0 i 0 ¯ h −1 √ 2 0 −i 0 i 0 −i 0 i 0 ¯ h 1 √ 2 0 1 0 1 0 1 0 1 0 ¯ h = ¯ h2 2 i 0 −i 0 −i + i 0 i 0 −i −¯ h2 2 −i 0 −i 0 i −i 0 i 0 i = ¯ h2 2 i + i 0 −i + i 0 0 0 i −i 0 −i −i = ¯ h2 2 2i 0 0 0 0 0 0 0 −2i = i¯ h 1 0 0 0 0 0 0 0 −1 ¯ h = i¯ h Lz.
Again, the other two relations can be calculated using similar procedures. In fact, the arith-metic for the other two relations is simpler. Why would this be so? ...Because Lz is a diagonal operator.
Remember L is comparable to a vector sum of the three component operators, so in vec-tor/matrix notation would look like ¯ ¯L > = Lx Ly Lz = 1 √ 2 0 1 0 1 0 1 0 1 0 ¯ h 1 √ 2 0 −i 0 i 0 −i 0 i 0 ¯ h 1 0 0 0 0 0 0 0 −1 ¯ h .
Again, this operator will normally be denoted just L.
The L operator is a different sort of object than the component operators. It is a different object in a different space. Yet, we would like a way to address angular momentum with a 3 X 3 matrix which is in the same subspace as the components. We can do this if we use L2.
This operator is L2 = 2¯ h2I = 2¯ h2 1 0 0 0 1 0 0 0 1 .
(9 −9) 305 Example 9–8: Show L2 = 2¯ h2I.
L2 = →⟨1 √ 2 0 1 0 1 0 1 0 1 0 ¯ h, 1 √ 2 0 −i 0 i 0 −i 0 i 0 ¯ h, 1 0 0 0 0 0 0 0 −1 ¯ h ¯ ¯ ¯ ¯ ¯ 1 √ 2 0 1 0 1 0 1 0 1 0 ¯ h 1 √ 2 0 −i 0 i 0 −i 0 i 0 ¯ h 1 0 0 0 0 0 0 0 −1 ¯ h ⟩ = 1 √ 2 0 1 0 1 0 1 0 1 0 ¯ h 1 √ 2 0 1 0 1 0 1 0 1 0 ¯ h + 1 √ 2 0 −i 0 i 0 −i 0 i 0 ¯ h 1 √ 2 0 −i 0 i 0 −i 0 i 0 ¯ h + 1 0 0 0 0 0 0 0 −1 ¯ h 1 0 0 0 0 0 0 0 −1 ¯ h = 1 2 1 0 1 0 1 + 1 0 1 0 1 ¯ h2 + 1 2 1 0 −1 0 1 + 1 0 −1 0 1 ¯ h2 + 1 0 0 0 0 0 0 0 1 ¯ h2 = 1/2 0 1/2 0 1 0 1/2 0 1/2 ¯ h2 + 1/2 0 −1/2 0 1 0 −1/2 0 1/2 ¯ h2 + 1 0 0 0 0 0 0 0 1 ¯ h2 = 1 0 0 0 2 0 0 0 1 ¯ h2 + 1 0 0 0 0 0 0 0 1 ¯ h2 = 2 0 0 0 2 0 0 0 2 ¯ h2 = 2¯ h2I.
Complete Set of Commuting Observables ...A Discussion about Operators which do not Commute....
The intent of this section is to appreciate non–commutivity from a new perspective, and explain “what can be done about it” if the non–commuting operators represent physical quanti-ties we want to measure. The following toy example is adapted from Quantum Mechanics and Experience4.
We want two operators which do not commute. We are deliberately using simple operators in an effort to focus on principles.
In a two dimensional linear vector space, the property of “hardness” is modelled Hard = µ 1 0 0 −1 ¶ 4 Albert, Quantum Mechanics and Experience (Harvard University Press, Cambridge, Mas-sachusetts, 1992), pp 30–33.
306 and has eigenvalues of ±1 and eigenvectors |1>hard = µ 1 0 ¶ and | −1>hard = µ 0 1 ¶ .
Let’s also consider the “color” operator, Color = µ 0 1 1 0 ¶ with eigenvalues of ±1 and eigenvectors |1>color = 1 √ 2 µ 1 1 ¶ and | −1>color = 1 √ 2 µ 1 −1 ¶ .
Note that a “hardness” or “color” eigenvector is a superposition of the eigenvectors of the other property, i.e., |1>hard = 1 √ 2 |1>color + 1 √ 2 | −1>color | −1>hard = 1 √ 2 |1>color − 1 √ 2 | −1>color |1>color = 1 √ 2 |1>hard + 1 √ 2 | −1>hard | −1>color = 1 √ 2 |1>hard − 1 √ 2 | −1 >hard Hardness is a superposition of color states and color is a superposition of hardness states. That is the foundation of incompatibility, or non–commutivity. Each measurable state is a linear combi-nation or superposition of the measurable states of the other property. To disturb one property is to disturb both properties.
Also in chapter 3, we indicated if two Hermitian operators commute, there exists a basis of common eigenvectors. Conversely, if they do not commute, there is no basis of common eigenvec-tors. We conclude there is no common eigenbasis for the “hardness” and “color” operators.
This is exactly the status of the three angular momentum component operators, except there are three vice two operators which do not commute with one another. None of the component operators commutes with any other. There is no common basis of eigenvectors between any two, so can be no common eigenbasis between all three.
Back to the hardness and color operators. If we can find an operator with which both commute, say the two dimensional identity operator I, we can ascertain the eigenstate of the system. If we measure an eigenvalue of 1 for color, the eigenstate is proportional to µ 1 1 ¶ , were we to operate on this with the identity operator, the eigenstate of system is either µ 1 0 ¶ or µ 0 1 ¶ .
If we then measure with the hardness operator, the eigenvalue will be 1 if the state was µ 1 0 ¶ , or −1 if the state was µ 0 1 ¶ .
We have effectively removed the indeterminacy of the system by including 307 I.
If we measure either “hardness” or “color,” and then operate with the identity, we attain a distinct, unique unit vector. There are two complete sets of commuting operators possible, I and Hard, or I and Color.
The eigenvalues, indicated in the ket, and eigenvectors for the three angular momentum component operators are | − √ 2> = 1 2 1 − √ 2 1 , |0> = 1 √ 2 1 0 −1 , | √ 2> = 1 2 1 √ 2 1 , for Lx, | − √ 2> = 1 2 1 −i √ 2 −1 , |0> = 1 √ 2 1 0 1 , | √ 2> = 1 2 1 i √ 2 −1 , for Ly, and | −1> = 0 0 1 , |0> = 0 1 0 , |1> = 1 0 0 , for Lz.
Notice like the nonsense operators hardness and color, none of the angular momen-tum component operators commute and none of the eigenvectors correspond. Also comparable, L2 is proportional to the identity operator, except in three dimensions. We can do something similar to the “hardness, color” case to remove the indeterminacy. It must be similar and not the same...because we need a fourth operator with which the three non–commuting component angular momentum operators all commute, and any one of the angular momentum components to form a complete set of commuting observables. We choose L2, which commutes with all three component operators, and Lz, which is the conventional choice of components.
The requirement for a complete set of commuting observables is equivalent to removing or lifting a degeneracy. The idea is closely related to the discussion at the end of example 3–33.
If you comprehend the idea behind that discussion, you have the basic principle of this discussion.
Also, “complete” here means all possiblities are clear, i.e., that any degeneracy is removed.
This is the same word but a different context than “span the space” as the word was used in chapter 2. Both uses are conventional and meaning is ascertained only by usage, so do not be confused by its use in both contexts.
Precurser to the Hydrogen Atom The Hamiltonian for a spherically symmetric potential commutes with L2 and the three component angular momentum operators. So H, L2, and one of the three component angular momentum operators, conventially Lz, is a complete set of commuting observables for a spherically symmetric potential.
We will use a Hamiltonian with a Coulomb potential for the hydrogen atom. The Coulomb potential is rotationally invariant, or spherically symmetric. We have indicated H, L2, and Lz form a complete set of commuting observables for such a system. You may be familiar with the principal quantum number n, the angular momentum quantum number l, and the magnetic quantum number m.
We will find there is a correspondence between these two sets of three quantities, which is n comes from application of H, l comes from application of L2, and m 308 comes from application of Lz.
A significant portion of the reason to address angular momentum and explain the concept of a complete set of commuting observables now is for use in the next chapter on the hydrogen atom.
Ladder Operators for Angular Momentum We are going to address angular momentum, like the SHO, from both a linear algebra and a differential equation perspective. We are going to assume rotational invariance, or spherical symmetry, so we have H, L2, and Lz as a complete set of commuting observables. We will address linear algebra arguments first.
And we will work only with the components and L2, saving the Hamiltonian for the next chapter.
The four angular momentum operators are related as L2 = L2 x + L2 y + L2 z ⇒ L2 −L2 z = L2 x + L2 y.
The sum of the two components L2 x + L2 y would appear to factor ¡ Lx + iLy ¢¡ Lx −iLy ¢ , and they would if the factors were scalars, but they are operators which do not commute, so this is not factoring. Just like the SHO, it is a good mnemonic, nevertheless.
Example 9–9: Show L2 x + L2 y ̸= ¡ Lx + iLy ¢¡ Lx −iLy ¢ .
¡ Lx + iLy ¢¡ Lx −iLy ¢ = L2 x −iLxLy + iLyLx + L2 y = L2 x + L2 y −i ¡ LxLy −LyLx ¢ = L2 x + L2 y −i £ Lx, Ly ¤ = L2 x + L2 y −i ¡ i¯ hLz ¢ = L2 x + L2 y + ¯ hLz ̸= L2 x + L2 y, where the expression in the next to last line is a significant intermediate result, and we will have reason to refer to it.
Like the SHO, the idea is to take advantage of the commutation relations of equations (9–7).
We will use the notation L+ = Lx + iLy, and L−= Lx −iLy, (9 −12) which together are often denoted L±.
We need commutators for L±, which are £ L2, L± ¤ = 0, (9 −13) £ Lz, L± ¤ = ±¯ h L±.
(9 −14) Example 9–10: Show £ L2, L+ ¤ = 0.
£ L2, L+ ¤ = £ L2, Lx + iLy ¤ = £ L2, Lx ¤ + i £ L2, Ly ¤ = 0 + i(0) = 0.
309 Example 9–11: Show £ Lz, L+ ¤ = ¯ h L+.
£ Lz, L+ ¤ = £ Lz, Lx + iLy ¤ = £ Lz, Lx ¤ + i £ Lz, Ly ¤ = i¯ hLy + i ¡ −i¯ hLx ¢ = ¯ h ¡ Lx + iLy ¢ = ¯ hL+.
We will proceed essentially as we did the the raising and lowering operators of the SHO. Since L2 and Lz commute, they share a common eigenbasis.
Example 9–12: Show L2 and Lz commute.
£ L2, Lz ¤ = £ L2 x + L2 y + L2 z, Lz ¤ = £ L2 x, Lz ¤ + £ L2 y, Lz ¤ + £ L2 z, Lz ¤ Á = £ Lx Lx, Lz ¤ + £ Ly Ly, Lz ¤ = Lx £ Lx, Lz ¤ + £ Lx, Lz ¤ Lx + Ly £ Ly, Lz ¤ + £ Ly, Lz ¤ Ly = Lx ¡ −i¯ hLy ¢ + ¡ −i¯ hLy ¢ Lx + Ly ¡ i¯ hLx ¢ + ¡ i¯ hLx ¢ Ly = ¡ −i¯ hLx Ly + i¯ hLx Ly ¢ + ¡ −i¯ hLy Lx + i¯ hLy Lx ¢ = 0, where we have used the results of example 9–4 and two of equations (9–7) in the reduction.
We assume L2 and Lz will have different eigenvalues when they operate on the same basis vector, so we need two indices for each basis vector. The first index is the eigenvalue for L2, we will use α for the eigenvalue, and the second index is the eigenvalue for Lz, denoted by β.
If we had a third commuting operator, for instance H which we will add in the next chapter, we would need three eigenvalues to uniquely identify each ket. Here we are considering two commuting operators, so we need two indices representing the eigenvalues of the two commuting operators.
Considering just L2 and Lz here, the form of the eigenvalue equations must be L2¯ ¯α, β> = α ¯ ¯α, β>, (9 −15) Lz ¯ ¯α, β> = β ¯ ¯α, β>, (9 −16) where ¯ ¯α, β> is the eigenstate, α is the eigenvalue of L2, and β is the eigenvalue of Lz.
Equation (9–14)/example 9–11 give us £ Lz, L+ ¤ = Lz L+ −L+ Lz = ¯ h L+ ⇒ Lz L+ = L+ Lz + ¯ h L+.
Using this in equation (9–16), Lz L+ ¯ ¯α, β> = ¡ L+ Lz + ¯ h L+ ¢¯ ¯α, β> = L+ Lz ¯ ¯α, β> +¯ h L+ ¯ ¯α, β> = L+ β ¯ ¯α, β> +¯ h L+ ¯ ¯α, β> (9 −17) = ¡ β + ¯ h ¢ L+ ¯ ¯α, β> .
310 Summarizing, Lz ¡ L+ ¯ ¯α, β> ¢ = ¡ β + ¯ h ¢¡ L+ ¯ ¯α, β> ¢ , which means L+ ¯ ¯α, β> is itself an eigenvector of Lz with eigenvalue ¡ β + ¯ h ¢ .
The effect of L+ is to increase the eigenvalue of Lz by the amount ¯ h, so it is called the raising operator.
Note that it raises only the eigenvalue of Lz.
A better name would be the raising operator for Lz, but the convention is when angular momentum is being discussed is to refer simply to the raising operator, and you need to know it applies only to Lz.
Were we to calculate similarly, we would find L− ¯ ¯α, β> is itself an eigenvector of Lz with eigenvalue ¡ β −¯ h ¢ .
The effect of L−is to decrease the eigenvalue by the amount ¯ h, so it is called the lowering operator. Again, the convention when angular momentum is being discussed is to refer to the lowering operator without reference to Lz.
Example 9–13: Show ¯ ¯α, β> is an eigenvector of L2. Equation (9–13) yields £ L2, L+ ¤ = L2 L+ −L+ L2 = 0 ⇒ L2 L+ = L+ L2.
Then L2 L+ ¯ ¯α, β> = L+ L2¯ ¯α, β> = L+ α ¯ ¯α, β> = α L+ ¯ ¯α, β>, or summarizing L2¡ L+ ¯ ¯α, β> ¢ = α ¡ L+ ¯ ¯α, β> ¢ , so L+ ¯ ¯α, β> is itself an eigenvector of L2 with eigenvalue α.
Similarly, L− ¯ ¯α, β> is itself an eigenvector of L2 with eigenvalue α.
It is important that L+ ¯ ¯α, β> is itself an eigenvector of L2, but be sure to notice that the raising/lowering operator has no effect on the eigenvalue of L2.
The eigenvalue of L2 acting on an eigenstate is α.
The eigenvalue of L2 acting on a combination of the raising/lowering operator and an eigenstate is still α.
Eigenvalue Solution for the Square of Orbital Angular Momentum Recalling the relation between the four angular momentum operators, L2 −L2 z = L2 x + L2 y, we are going use the eigenvalue equations and apply these operators to the generic eigenstate, i.e., ¡ L2 −L2 z ¢¯ ¯α, β> = L2¯ ¯α, β> −L2 z ¯ ¯α, β> = α ¯ ¯α, β> −Lzβ ¯ ¯α, β> = α ¯ ¯α, β> −β2¯ ¯α, β> = ¡ α −β2¢¯ ¯α, β> .
Forming an adjoint eigenstate and a braket, <α, β ¯ ¯L2 −L2 z ¯ ¯α, β> = <α, β ¯ ¯L2 x + L2 y ¯ ¯α, β> (9 −18) = <α, β ¯ ¯α −β2¯ ¯α, β> = ¡ α −β2¢ <α, β ¯ ¯α, β> (9 −19) = α −β2 ≥0, (9 −20) 311 where we have assumed orthonormality of eigenstates in equation (9–19). The condition that the difference in equation (9–20) is non–negative is from the fact the braket is expressible in terms of a sum of L2 x and L2 y, as seen in equation (9–18). Both Lx and Ly are Hermitian, so their eigenvalues are real. The sum of the squares of the eigenvalues, corresponding to operations by L2 x and L2 y in equation (9–18), must be non–negative. In mathematical vernacular, L2 x and L2 y are positive definite.
Equation (9–20) is equivalent to α ≥β2, which means β is bounded for a given value of α.
Therefore there is an eigenstate |α, βmax> which cannot be raised, and another eigenstate |α, βmin> which cannot be lowered. In other words, we have a ladder which has a bottom, like the SHO, and a top, unlike the SHO. In a calculation similar to example 9–9, L−L+ = L2 x + L2 y −¯ h = L2 −L2 z −¯ hLz, so L−L+|α, βmax> = ⃗ 0 ⇒ ¡ L2 −L2 z −¯ hLz ¢ |α, βmax> = 0 (9 −21) ⇒ L2|α, βmax> −L2 z|α, βmax> −¯ hLz|α, βmax> = 0 ⇒ α|α, βmax> −β2 max|α, βmax> −¯ h βmax|α, βmax> = 0 ⇒ ¡ α −β2 max −¯ h βmax ¢ |α, βmax> = 0 ⇒ α −β2 max −¯ h βmax = 0 ⇒ α = β2 max + ¯ h βmax.
(9 −22) Similarly, L+L−|α, βmin> = ⃗ 0 ⇒ α = β2 max −¯ h βmax.
(9 −23) Equating equations (9–22) and (9–23), we get β2 max + ¯ hβmax −β2 min + ¯ hβmin = 0.
This is quadratic in both βmax and βmin, and to solve the equation, we will use the quadratic formula to solve for βmax, or βmax = −1 2¯ h ± 1 2 q ¯ h2 −4(−β2 min + ¯ h βmin) = −1 2¯ h ± 1 2 q 4β2 min −4¯ h βmin + ¯ h2 = −1 2¯ h ± 1 2 p (2βmin −¯ h)2 = −1 2¯ h ± 1 2(2βmin −¯ h) ⇒ βmax = −βmin, βmin −¯ h.
(9 −24) The case βmax = −βmin is the maximum sep-aration case. It gives us the top and bottom of the ladder. We assume the rungs of the ladder are 312 separated by ¯ h, because that is the amount of change indicated by the raising and lowering op-erators. The picture corresponds to figure 9–1. If there is other than minimum separation, say there are n steps between the bottom and top rungs of the ladder, there is a total separation of n¯ h between the bottom and the top. From figure 9–1 we expect 2βmax = n¯ h ⇒ βmax = n¯ h 2 .
Using this in equation (9–22), α = βmax ¡ βmax + ¯ h ¢ = n¯ h 2 µn¯ h 2 + ¯ h ¶ = ¯ h2 ³n 2 ´ ³n 2 + 1 ´ .
We are going to re–label, letting j = n/2, so α = ¯ h2 j ¡ j + 1 ¢ .
(9 −25) Wait a minute....
The fact j = ¯ h/2 vice just ¯ h does not appear consistent with the assumption that the rungs of the ladder are separated by ¯ h...and it isn’t. It appears the rungs of the ladder are separated by ¯ h/2 vice ¯ h.
What has occurred is that we have actually solved a more general problem than intended.
Because of symmetry, the linear algebra arguments have given us the solution for total angular momentum. Total angular momentum is ⃗ J = ⃗ L + ⃗ S, (9 −26) where ⃗ L is orbital angular momentum, ⃗ S is spin angular momentum or just spin. We posed the problem for orbital angular momentum, but because total angular momentum and spin obey analogous commutation relations to orbital angular momentum, we arrive at the solution for total angular momentum. Equations (9–7) indicated components of orbital angular momentum do not commute, £ Lx, Ly ¤ = i¯ h Lz, £ Ly, Lz ¤ = i¯ h Lx, and £ Lz, Lx ¤ = i¯ h Ly, and for the ladder operator solution, we formed L± = Lx ± iLy.
The commutation relations among the components of total angular momentum and spin angular momentum are exactly the same, i.e., £ Jx, Jy ¤ = i¯ h Jz, £ Jy, Jz ¤ = i¯ h Jx, and £ Jz, Jx ¤ = i¯ h Jy, and £ Sx, Sy ¤ = i¯ h Sz, £ Sy, Sz ¤ = i¯ h Sx, and £ Sz, Sx ¤ = i¯ h Sy.
313 If we had started out with J± = Jx ± iJy, or S± = Sx ± iSy, we would have come out with exactly the same result. In fact, this is the problem we solved, except using the symbol L vice J or S.
We will reinforce in chapter 13 that spin can have half integral values, or values of multiples of ¯ h/2.
Since spin can be half integral, values of total angular momentum can be half integral.
When we use symbols such as L2 and Li, we get the information contained in the commutation relations, independent of whatever symbols we choose.
Had we used explicit representations, such as equations (9–8) and (9–9), we would get the same information, however, limited by the representation. In that case, only integral values would be possible, though the form of the result analogous to equation (9–25) would remain the same. Using l as the quantum number for orbital angular momentum, the eigenvalue for orbital angular momentum squared is α = ¯ h2 l(l + 1).
(9 −27) A comment about the picture and notation is appropriate. The first impression is that this is similar to classical mechanics. The earth orbits the sun and has orbital angular momentum in that regard, and also spins on its axis so has spin angular momentum as well. It is tempting to apply this picture to a quantum mechanical system, say an electron in a hydrogen atom. It simply does not apply. The electron is not a small ball spinning on its axis as it orbits the proton.
Per the first chapter, an electron is not a particle, it is not a wave, it is an electron. There is no classical analogy for an electron, and many of the manifestations of quantum mechanical angular momentum are similarly not classical analogs.
Equation (9–26) says the total is the sum of the parts, but it is an operator equation which in Dirac notation is |J > = |L> + |S>.
Since earlier development was in this form, it may be useful to assist you to realize that each of these three operators has three components which are also each operators. Equation (9-26) is standard notation nevertheless.
Eigenvalue Solution for the Z Component of Orbital Angular Momentum We have calculated the eigenvalue of L2, but still need to find the eigenvalue of Lz.
We know one of the possible eigenvalues of Lz is zero from the last of equations (9–8), the explicit representations, regardless of the eigenstate. We have also calculated Lz ¡ L+ ¯ ¯α, β> ¢ = ¡ β + ¯ h ¢¡ L+ ¯ ¯α, β> ¢ .
If we start with an eigenstate that has the z component of angular momentum equal to zero, Lz ¡ L+ ¯ ¯α, 0> ¢ = ¡ 0 + ¯ h ¢¡ L+ ¯ ¯α, 0> ¢ = ¯ h ¡ L+ ¯ ¯α, 0> ¢ , so ¯ h is the next eigenvalue. Using ¯ h as the eigenvalue, Lz ¡ L+ ¯ ¯α, ¯ h> ¢ = ¡ ¯ h + ¯ h ¢¡ L+ ¯ ¯α, ¯ h> ¢ = 2¯ h ¡ L+ ¯ ¯α, ¯ h> ¢ , so 2¯ h is the next eigenvalue. If we use this as an eigenvalue, Lz ¡ L+ ¯ ¯α, 2¯ h> ¢ = ¡ ¯ h + 2¯ h ¢¡ L+ ¯ ¯α, 2¯ h> ¢ = 3¯ h ¡ L+ ¯ ¯α, 2¯ h> ¢ , 314 and 3¯ h is the next eigenvalue up the ladder. We can continue, and will attain integral values of ¯ h.
But we cannot continue forever, because we determined β is bounded by the eigenvalue of L2.
What is the maximum value? We go back to figure 9–1 and the result from this figure is βmax = n¯ h 2 , where we want only integral values for the orbital angular momentum, so this becomes βmax = l¯ h.
Were we to do the same calculation with the lowering operator, that is Lz ¡ L− ¯ ¯α, 0> ¢ = −¯ h ¡ L− ¯ ¯α, 0> ¢ , we step down the ladder in increments of −¯ h until we get to βmin.
Remember βmin also has a minimum, which is of the same magnitude but negative or βmin = −l¯ h.
So we have eigenvalues which climb to l¯ h and drop to −l¯ h in integral increments of ¯ h.
The eigenvalue of the z component of angular momentum is just an integer times ¯ h, from minimum to maximum values. The symbol conventionally used to denote this integer is m, so Lz ¯ ¯α, β> = m¯ h ¯ ¯α, β>, −l < m < l is the eigenvalue/eigenvector equation for the z component of angular momentum. The quantum number m, occasionally denoted ml, is known as the magnetic quantum number.
Eigenvalue/Eigenvector Equations for Orbital Angular Momentum If we use l vice α to denote the state of total angular momentum, realizing l itself is not an eigenvalue of L2, and m to denote the state of the z component of angular momentum, realizing the eigenvalue of Lz is actually m¯ h, the eigenvalue/eigenvector equations for L2 and Lz are L2¯ ¯l, m> = ¯ h2 l(l + 1) ¯ ¯l, m>, (9 −28) Lz ¯ ¯l, m> = m¯ h ¯ ¯l, m>, −l < m < l (9 −29) which is the conventional form of the two eigenvalue/eigenvector equations for L2 and Lz.
315 Angular Momentum Eigenvalue Picture for Eigenstates What is |l, m>?
It is an eigenstate of the commuting operators L2 and Lz.
The quantum numbers l and m are not eigenvalues. The corresponding eigenvalues are ¯ h2l(l + 1) and m¯ h.
Were we to use eigenvalues in the ket, the eigenstate would look like |¯ h2l(l + 1), m¯ h>.
But just l and m uniquely identify the state, and that is more economical, so only the quantum numbers are conventionally used. This is essentially the same sort of convenient shorthand used to denote an eigenstate of a SHO |n>, vice using the eigenvalue ¯ ¯¡ n + 1 2 ¢ ¯ hω>.
Only one quantum number is needed to uniquely identify an eigenstate of a SHO, but two are needed to uniquely identify an eigenstate of angular momentum. Because the angular momentum component operators do not commute, a complete set of commuting observables are needed. Each of the component operators commutes with L2, so we use it and one other, which is Lz chosen by convention. One quantum number is needed for each operator in the complete set. Multiple quantum numbers used to identify a ket denote a complete set of commuting observables is needed.
Remember that a system is assumed to exist in a linear combination of all possible eigenstates until we measure. If we measure, what are the possible outcomes? Possible outcomes are the eigenvalues. For a given value of the orbital angular momentum quantum number, the magnetic quantum number can assume integer values ranging from −l to l.
The simplest case is l = 0 ⇒ m = 0 is the only possible value of the magnetic quantum number. The possible outcomes of a measurement of such a system are eigenvalues of ¯ h2¡ 0 ¢¡ 0 + 1 ¢ = 0¯ h2 or just 0 for L2, and m¯ h = ¡ 0 ¢ ¯ h or just 0 for Lz as well, corresponding to figure 9–2.a.
Figure 9 −2.a. l = 0.
Figure 9 −2.b. l = 1.
Figure 9 −2.c. l = 2.
If we somehow knew l = 1, which could be ascertained by a measurement of ¯ h2¡ 1 ¢¡ 1 + 1 ¢ = 2¯ h2 for L2, the possible values of the magnetic quantum number are m = −1, 0, or 1, so the eigenvalues which could be measured are −¯ h, 0, or ¯ h for Lz, per figure 9–2.b. If we measured ¯ h2¡ 2 ¢¡ 2+1 ¢ = 6¯ h2 for L2, we would know we had l = 2, and the possible values of the magnetic quantum number are m = −2, −1, 0, 1, or 2, so the eigenvalues which could be measured are −2¯ h, −¯ h, 0, ¯ h, or 2¯ h for Lz, per figure 9–2.c. Though the magnetic quantum number is bounded by the orbital angular momentum quantum number, the orbital angular momentum quantum number is not bounded, so we can continue indefinitely. Notice there are 2l +1 possible values of m for every value of l.
A semi–classical diagram is often used. A simple interpretation of |l, m> is that it is a vector quantized in length of ¯ ¯L ¯ ¯ → ¯ ¯L ¯ ¯ = √ L2 = ¯ h p l(l + 1).
316 This vector has values for which the z component is also quantized in units of m¯ h.
These Figure 9 −3. Semi −Classical Picture for l = 2.
features are illustrated in figure 9–3 for l = 2.
The vectors are free to rotate around the z axis at any azimuthal angle φ, but are fixed at polar angles θ determined by the fact the projection on the z axis must be −2¯ h, −¯ h, 0, ¯ h, or 2¯ h.
Notice there is no information concerning the x or y components other than the square of their sum is fixed. We could express this for |ψ(t)> = |l, m> by stating the projection on the xy plane will be cos(ωt) or sin(ωt).
In such a case we can determine x and y component expectation values from symmetry alone, i.e., = 0, = 0.
Finally, what fixes any axis in space? And how do we know which axis is the z axis? The answer is we must introduce some asymmetry. Without an asymmetry of some sort, the axes and their labels are arbitrary. The practical assymmetry to introduce is a magnetic field, and that will establish a component quantization axis which will be the z axis.
Eigenvalue/Eigenvector Equations for the Raising and Lowering Operators Using quantum number notation, the fact L+ ¯ ¯α, β > is and eigenstate of Lz would be written LzL+ ¯ ¯l, m> = ¡ m¯ h + ¯ h ¢ L+ ¯ ¯l, m> = ¡ m + 1 ¢ ¯ hL+ ¯ ¯l, m> = γLz ¯ ¯l, m + 1>, where γ is a proportionality constant. Then LzL+ ¯ ¯l, m> = Lzγ ¯ ¯l, m + 1> ⇒ L+ ¯ ¯l, m> = γ ¯ ¯l, m + 1>, is the eigenvalue/eigenvector equation for the raising operator, where γ is evidently the eigenvalue, and the eigenvector is raised by one element of quantization in the z component. This means if the z component of the state on which the raising operator acts is m¯ h, the new state has a z component of m¯ h + ¯ h = (m + 1)¯ h, and thus the index m + 1 is used in the new eigenket. We 317 want to solve for γ and have an equation analogous to the forms of equations (9–28) and (9–29).
Forming the adjoint equation, = \ .
Though we did it for βmax, the maximum eigenvalue of Lz, the algebra leading to equation (9–22) remains the same for any β, any eigenvalue of Lz, so we know L−L+ = α −β2 −¯ hβ = ¯ h2 l(l + 1) −m2¯ h2 −m¯ h2.
Using this in the braket, \ = \ ⇒ ¯ h2³ l(l + 1) −m(m + 1) ´ \ = ¯ ¯γ∗γ ¯ ¯ \ ⇒ ¯ h2³ l(l + 1) −m(m + 1) ´ = ¯ ¯γ ¯ ¯2 (9 −30) ⇒ γ = p l(l + 1) −m(m + 1) ¯ h, where we used the orthonormality of eigenstates to arrive at equation (9–30). The eigenvalue/ eigenvector equation is then L+ ¯ ¯l, m> = p l(l + 1) −m(m + 1) ¯ h ¯ ¯l, m + 1> .
Were we to do the similar calculation for L−, we find L− ¯ ¯l, m> = p l(l + 1) −m(m −1) ¯ h ¯ ¯l, m −1> .
These are most often expressed as one relation, L± ¯ ¯l, m> = p l(l + 1) −m(m ± 1) ¯ h ¯ ¯l, m ± 1> .
(9 −31) Example 9–14: For the eigenstate ¯ ¯l, m> = ¯ ¯3, m>, what measurements are possible for L2 and Lz?
The only measurements that are possible are the eigenvalues.
From equation (9–28), the eigenvalue of L2 is ¯ h2 l(l + 1) = ¯ h2 3(3 + 1) = 12¯ h2.
For l = 3, the possible eigenvalues of Lz can range from −3¯ h to 3¯ h in increments of ¯ h.
Explicitly, the measurements that are possible for Lz for the eigenstate ¯ ¯3, m> are −3¯ h, −2¯ h, −¯ h, 0, ¯ h, 2¯ h, or 3¯ h.
Example 9–15: What are L+ and L−operating on the eigenstate ¯ ¯2, −1>?
318 Using equation (9–31), L+ ¯ ¯2, −1> = p 2(2 + 1) −(−1)((−1) + 1) ¯ h ¯ ¯2, −1 + 1> = p 2(3) −(−1)(0) ¯ h ¯ ¯2, 0> = √ 6 ¯ h ¯ ¯2, 0> .
L− ¯ ¯2, −1> = p 2(2 + 1) −(−1)((−1) −1) ¯ h ¯ ¯2, −1 −1> = p 2(3) −(−1)(−2) ¯ h ¯ ¯2, −2> = √ 6 −2 ¯ h ¯ ¯2, −2> = √ 4 ¯ h ¯ ¯2, −2> = 2¯ h ¯ ¯2, −2> .
Possibilities, Probabilities, Expectation Value, Uncertainty, and Time Dependence Examples 9–16 through 9–21 are intended to interface, apply, and extend calculations de-veloped previously to eigenstates of angular momentum. As indicated earlier, a state vector will be a linear combination of eigenstates, which this development should reinforce. Examples 9–16 through 9–21 all refer to the t = 0 state vector ¯ ¯ψ(t = 0)> = A ¡¯ ¯2, 1> +3 ¯ ¯1, −1> ¢ (9 −32) is is a linear combination of two eigenstates.
Example 9–16: Normalize the state vector of equation (9–32).
There are two eigenstates, so we can work in a two dimensional subspace. We can model the first eigenstate µ 1 0 ¶ and the second µ 0 1 ¶ .
Then the state vector can be written ¯ ¯ψ(0)> = A ·µ 1 0 ¶ + 3 µ 0 1 ¶¸ = A µ 1 3 ¶ .
Another way to look at it is the state vector is two dimensional with one part the first eigenstate and three parts the second eigenstate. This technique makes the normalization calculation, and a number of others, particularly simple.
¡ 1, 3 ¢ A∗A µ 1 3 ¶ = ¯ ¯A ¯ ¯2¡ 1 + 9 ¢ = 10 ¯ ¯A ¯ ¯2 = 1 ⇒ A = 1 √ 10 ⇒ ¯ ¯ψ(0)> = 1 √ 10 µ 1 3 ¶ = 1 √ 10 ¡¯ ¯2, 1> +3 ¯ ¯1, −1> ¢ .
Example 9–17: What are the possibilities and probabilities of a measurement of L2?
The possibilities are the eigenvalues. There are two eigenstates, each with its own eigenvalue.
If we measure and put the system into the first eigenstate, we measure the state corresponding to the quantum number l = 2, which has the eigenvalue ¯ h2 l(l + 1) = ¯ h2 2(2 + 1) = 6¯ h2.
If 319 we measure and place the state vector into the second eigenstate corresponding to the quantum number l = 1, the eigenvalue measured is ¯ h2 l(l + 1) = ¯ h2 1(1 + 1) = 2¯ h2.
Since the state function is normalized, P ¡ L2 = 6¯ h2¢ = ¯ ¯ <ψ ¯ ¯ψi> ¯ ¯2 = ¯ ¯ ¯ ¯ 1 √ 10 ¡ 1, 3 ¢ µ 1 0 ¶¯ ¯ ¯ ¯ 2 = 1 10 |1 + 0|2 = 1 10.
P ¡ L2 = 2¯ h2¢ = ¯ ¯ <ψ ¯ ¯ψi> ¯ ¯2 = ¯ ¯ ¯ ¯ 1 √ 10 ¡ 1, 3 ¢ µ 0 1 ¶¯ ¯ ¯ ¯ 2 = 1 10 |0 + 3|2 = 9 10.
Example 9–18: What are the possibilities and probabilities of a measurement of Lz?
For exactly the same reasons, the possible results of a measurement are m = 1 ⇒ ¯ h is the first eigenvalue and m = −1 ⇒ −¯ h is the second possible eigenvalue. Using exactly the same math, P ¡ Lz = ¯ h ¢ = 1 10, P ¡ Lz = −¯ h ¢ = 9 10.
Example 9–19: What is the expectation value of L2?
= X P(αi)αi = 1 106¯ h2 + 9 102¯ h2 = 6 10¯ h2 + 18 10¯ h2 = 24 10¯ h2 = 2.4¯ h2.
Example 9–20: What is the uncertainty of L2?
△L2 = qX P(αi) ¡ αi− ¢2 = · 1 10 ¡ 6¯ h2 −2.4¯ h2¢2 + 9 10 ¡ 2¯ h2 −2.4¯ h2¢2 ¸1/2 = ¯ h2 · 1 10 ¡ 3.6 ¢2 + 9 10 ¡ −0.4 ¢2 ¸1/2 = ¯ h2£ 1.296 + 0.144 ¤1/2 = ¯ h2√ 1.44 = 1.2¯ h2.
Example 9–21: What is the time dependent state vector?
¯ ¯ψ(t)> = X |j> e−iEjt/¯ h = µ 1 0 ¶ ¡ 1, 0 ¢ 1 √ 10 µ 1 3 ¶ e−iE1t/¯ h + µ 0 1 ¶ ¡ 0, 1 ¢ 1 √ 10 µ 1 3 ¶ e−iE2t/¯ h = 1 √ 10 µ 1 0 ¶ e−iE1t/¯ h + 3 √ 10 µ 0 1 ¶ e−iE2t/¯ h = 1 √ 10 ¯ ¯2, 1> e−iE1t/¯ h + 3 √ 10 ¯ ¯1, −1> e−iE2t/¯ h which is as far as we can go with the given information. We need a specific system and an energy operator, a Hamiltonian, to attain specific Ei.
320 Angular Momentum Operators in Spherical Coordinates The conservation of angular momentum, or rotational invariance, implies circular or spherical symmetry. We want to examine spherical symmetry, because spherical symmetry is often a rea-sonable assumption for simple physical systems. We will assume a hydrogen atom is spherically symmetric, for instance. Remember in spherical coordinates, x = r sin θ cos φ, r = ¡ x2 + y2 + z2¢1/2 y = r sin θ sin φ, θ = tan−1 ³p x2 + y2/z ´ z = r cos θ, φ = tan−1 (y/x) .
From these it follows that position space representations in spherical coordinates are Lx = i¯ h µ sin φ ∂ ∂θ + cos φ cot θ ∂ ∂φ ¶ , Ly = i¯ h µ −cos φ ∂ ∂θ + sin φ cot θ ∂ ∂φ ¶ , Lz = −i¯ h ∂ ∂φ, (9 −32) L2 = −¯ h2 µ ∂2 ∂θ2 + 1 tan θ ∂ ∂θ + 1 sin2 θ ∂2 ∂φ2 ¶ , (9 −33) L± = ±¯ h e±iφ µ ∂ ∂θ ± i cot θ ∂ ∂φ ¶ .
(9 −34) Example 9–22: Derive equation (9–32).
From equation (9–2), Lz = i¯ h µ −x ∂ ∂y + y ∂ ∂x ¶ .
We can develop the desired partial differentials from the relation between azimuthal angle and position coordinates, or φ = tan−1 (y/x) ⇒ y = x tan φ ⇒ ∂y ∂φ = x ∂ ¡ tan φ ¢ = x sec2 φ = x cos2 φ ⇒ ∂y = x ∂φ cos2 φ.
The same relation gives us x = y tan φ = y cos φ sin φ = y cos φ sin−1 φ ⇒ ∂x ∂φ = y ¡ −sin φ sin−1 φ + cos φ (−1) sin−2 φ cos φ ¢ 321 = −y µ 1 + cos2 φ sin2 φ ¶ = −y µsin2 φ + cos2 φ sin2 φ ¶ = − y sin2 φ ⇒ ∂x = −y ∂φ sin2 φ.
Using the partial differentials in the Cartesian formulation for the z component of angular momentum, Lz = i¯ h µ −x cos2 ∂ x ∂φ + y µ −sin2 φ ∂x y∂φ ¶¶ = −i¯ h ¡ cos2 φ + sin2 φ ¢ ∂ ∂φ = −i¯ h ∂ ∂φ.
Example 9–23: Given the spherical coordinate representations of Lx and Ly, show equation (9–34) is true for L+.
L+ = Lx + iLy = i¯ h µ sin φ ∂ ∂θ + cos φ cot θ ∂ ∂φ ¶ + i · i¯ h µ −cos φ ∂ ∂θ + sin φ cot θ ∂ ∂φ ¶¸ = ¯ h · i sin φ ∂ ∂θ + i cos φ cot θ ∂ ∂φ + cos φ ∂ ∂θ −sin φ cot θ ∂ ∂φ ¸ = ¯ h · (cos φ + i sin φ) ∂ ∂θ + (i cos φ −sin φ) cot θ ∂ ∂φ ¸ = ¯ h · (cos φ + i sin φ) ∂ ∂θ + i (cos φ + i sin φ) cot θ ∂ ∂φ ¸ = ¯ h ·¡ eiφ¢ ∂ ∂θ + i ¡ eiφ¢ cot θ ∂ ∂φ ¸ = ¯ h2 eiφ µ ∂ ∂θ + i cot θ ∂ ∂φ ¶ .
An outline of the derivations of the all components and square of angular momentum in spher-ical coordinates is included in Ziock5. These calculations can be “messy” by practical standards.
Special Functions Used for the Hydrogen Atom Two special functions are particularly useful in describing a hydrogen atom assumed to have spherical symmetry. These are spherical harmonics and Associated Laguerre functions. The plan will be to separate the Schrodinger equation into radial and angular equations. The solutions to the radial equation can be expressed in terms of associated Laguerre polynomials, which we will examine in the next chapter. The solutions to the angular equation can be expressed in terms of spherical harmonic functions, which we will examine in the next section. Spherical harmonics are closely related to a third special function, Legendre functions. They are so closely related, the spherical harmonics can be expressed in terms of associated Legendre polynomials.
5 Ziock Basic Quantum Mechanics (John Wiley & Sons, New York, 1969), pp. 91–94 322 The name spherical harmonic comes from the geometry the functions naturally describe, spheri-cal, and the fact any solution of Laplace’s equation is known as harmonic. Picture a ball. The surface may be smooth, which is likely the first picture you form. Put a rubber band around the center, and you get a minima at the center and bulges, or maxima, in the top and bottom half. Put rub-ber bands on the circumference, like lines of lon-gitude, and you get a different pattern of maxima and minima. We could imagine other, more com-plex patterns of maxima and minima. When these maxima and minima are symmetric with respect to an origin, the center of the ball, Legendre func-tions, associated Legendre functions, and spherical harmonics provide useful descriptions.
Properties that makes these special functions particularly useful is they are orthogonal and complete. Any set that is orthogonal can be made orthonormal. We have used orthonormality in a number of calculations, and the property of orthonormality continues to be a practical necessity.
They are also complete in the sense any phenomenon can be described by an appropriate linear combination. Other complete sets of orthonormal functions we have encountered are sines and cosines for the square well, and Hermite polynomials for the SHO. A set of complete, orthonormal functions is equivalent to a linear vector space; these special functions are different manifestations of a complex linear vector space.
Spherical Harmonics The ket ¯ ¯l, m> is an eigenstate of the commuting operators L2 and Lz, but it is an abstract eigenstate. That ¯ ¯l, m> is abstract is irrelevant for the eigenvalues, since eigenvalues are properties of the operators. We would, however, like a representation useful for description for the eigenvectors. Per chapter 4, we can form an inner product with an abstract vector to attain a representation. Using a guided choice, the angles of spherical coordinate system will yield an appropriate representation. Just as = g(x), we will write <θ, φ ¯ ¯l, m> = Yl,m(θ, φ).
The functions of polar and azimuthal angles, Yl,m(θ, φ), are the spherical harmonics.
The spherical harmonics are related so strongly to the geometry of the current problem, they can be derived from the spherical coordinate form of the eigenvalue/eigenvector equation (9–29), Lz ¯ ¯l, m> = m¯ h ¯ ¯l, m>, and use of the raising/lowering operator equation (9–31).
Using the spherical coordinate system form of the operator and the functional forms of the eigenstates, equation (9–29) is −i¯ h ∂ ∂φYl,m(θ, φ) = m¯ h Yl,m(θ, φ).
We are going to assume the spherical harmonics are separable, that they can be expressed as a product of a function of θ and a second function of φ, or Yl,m(θ, φ) = fl,m(θ) gl,m(φ).
323 Using this in the differential equation, −i¯ h ∂ ∂φfl,m(θ) gl,m(φ) = m¯ h fl,m(θ) gl,m(φ) ⇒ −i fl,m(θ) ∂ ∂φgl,m(φ) = mfl,m(θ) gl,m(φ) ⇒ −i ∂ ∂φgl,m(φ) = m gl,m(φ) ⇒ ∂gl,m(φ) gl,m(φ) = im ∂φ ⇒ ln gl,m(φ) = imφ ⇒ gl,m(φ) = eimφ.
Notice the exponential has no dependence on l, so we can write gm(φ) = eimφ, (9 −35) which is the azimuthal dependence.
Remember that there is a top and bottom to the ladder for a given l.
The top of the ladder is at m = l.
If we act on an eigenstate on the top of the ladder, we get zero, meaning L+ ¯ ¯l, l> = 0, Using the spherical coordinate forms of the raising operator and separated eigenstate including equation (9–35), this is ¯ h eiφ · ∂ ∂θ + i cot θ ∂ ∂φ ¸ fl,l(θ) eilφ = 0 ⇒ eilφ ∂ ∂θfl,l(θ) + i fl,l(θ) cot θ ¡ il ¢ eilφ = 0 ⇒ ∂ ∂θfl,l(θ) −l fl,l(θ) cot θ = 0.
The solution to this is fl,l(θ) = A ¡ sin θ ¢l.
To see that it is a solution, ∂ ∂θfl,l(θ) = ∂ ∂θA ¡ sin θ ¢l = A l ¡ sin θ ¢l−1 cos θ, and substituting this in the differential equation, A l ¡ sin θ ¢l−1 cos θ −l h A ¡ sin θ ¢li cos θ sin θ = A l ¡ sin θ ¢l−1 · cos θ −sin θcos θ sin θ ¸ = 0.
So the unnormalized form of the m = l spherical harmonics is Yl,m(θ, φ) = A ¡ sin θ ¢leimφ.
(9 −36) Example 9–24 derives Y1,1(θ, φ) starting with equation (9–36).
324 So how do we get the spherical harmonics for which m ̸= l?
The answer is to attain a Yl,l(θ, φ) and operate on it with the lowering operator. Example 9–25 derives Y1,0(θ, φ) in this manner.
One comment before we proceed. The spherical harmonics of equation (9–36) can be made orthonormal, so we need to calculate the normalization constants, A for each Yl,m(θ, φ).
Having selected a representation, this is most easily approached by the appropriate form of integration. The appropriate form of integration for spherical angles is with respect to solid angle, dΩ= sin θdθdφ, or Z Y ∗ l,m(θ, φ) Yl,m(θ, φ) dΩ= Z 2π 0 dφ Z π 0 dθ sin θ ¯ ¯Yl,m(θ, φ) ¯ ¯2 = 1, which will also be illustrated in examples 9–24 and 9–25. These and other special functions are addressed in most mathematical physics texts including Arken6 and Mathews and Walker7.
A list of the first few spherical harmonics is Y0,0(θ, φ) = 1 4π Y2,0(θ, φ) = r 5 16π ¡ 3 cos2 θ −1 ¢ Y1,±1(θ, φ) = r 3 8π sin θ e±iφ Y3,±3(θ, φ) = r 35 64π sin3 θ e±3iφ Y1,0(θ, φ) = r 3 4π cos θ Y3,±2(θ, φ) = r 105 32π sin2 θ cos θ e±2iφ Y2,±2(θ, φ) = r 15 32π sin2 θ e±2iφ Y3,±1(θ, φ) = r 21 64π sin θ ³ 5 cos2 θ −1 ´ e±iφ Y2,±1(θ, φ) = r 15 8π sin θ cos θ e±iφ Y3,0(θ, φ) = r 7 16π ³ 5 cos3 θ −3 cos θ ´ Table 9 −1. The First Sixteen Spherical Harmonic Functions.
A few comments about the list are appropriate.
First, notice the symmetry about m = 0.
For example, Y2,1 and Y2,−1 are exactly the same except for the sign of the argument of the exponential. Second, notice the Yl,0 are independent of φ.
When m = 0, the spherical harmonic functions are constant with respect to azimuthal angle. Next, per the previous sentences, it is common to refer to spherical harmonic functions without explicitly indicating that the arguments are polar and azimuthal angles. Finally, and most significantly, some texts will use a negative sign leading the spherical harmonic functions for which m < 0.
This is a different choice of phase.
We will use the convention denoted in table 9–1, where all spherical harmonics are positive. Used consistently, either choice is reasonable and both choices have advantages and disadvantages.
Figure 9–2 illustrates the functional form of the first 16 spherical harmonic functions. Note that the radial coordinate has not yet been addressed. Angular distribution is all that is being 6 Arfken Mathematical Methods for Physicists (Academic Press, New York, 1970), 2nd ed., chapters 9–13.
7 Mathews and Walker Mathematical Methods of Physics (The Benjamin/Cummings Publishing Co., Menlo Park, California, 1970), 2nd ed., chapter 7.
325 illustrated. The radial coordinate will be examined in the next chapter. The size of any of the individual pictures in figure 9–2 is arbitrary; they could be very large or very small. We assume a radius of one unit to draw the sketches. In other words, you can look at the smooth sphere of Y0,0 as having radius one unit, and the relative sizes of other spherical harmonic functions are comparable on the same radial scale.
Figure 9 −2. Illustrations of the First Sixteen Spherical Harmonic Functions.
326 There is a technique here we want to exploit when we address radial functions. The spher-ical harmonics are orthonormal so are normalized. The figures represent spherical harmonics of magnitude one, multiplied by one, so remain orthonormal. We want the radial functions to be orthonormal, or individually to have magnitude one. Just as we have assumed a one unit radius to draw the figures here, if we multiply two quantities of magnitude one, we attain a product of magnitude one. If the angular function and radial function are individually normalized, the product function will be normalized as well.
Example 9–24: Show Yl,l = A ¡ sin θ ¢leimφ yields the normalized Y1,1 of table 9–1.
Y1,1 = A ¡ sin θ ¢1ei(1)φ = A sin θ eiφ.
To normalize this, 1 = Z (Y1,1)∗Y1,1 dΩ= Z A∗sin θ e−iφ A sin θ eiφ dΩ = ¯ ¯A ¯ ¯2 Z sin2 θ e0 dΩ= ¯ ¯A ¯ ¯2 Z 2π 0 dφ Z π 0 dθ sin2 θ sin θ = ¯ ¯A ¯ ¯2 Z π 0 dθ sin3 θ Z 2π 0 dφ = 2π ¯ ¯A ¯ ¯2 Z π 0 dθ sin3 θ = 2π ¯ ¯A ¯ ¯2 · −1 3 cos θ ¡ sin2 θ + 2 ¢¸π 0 = 2π 3 ¯ ¯A ¯ ¯2 · cos θ ¡ sin2 θ + 2 ¢¸0 π = 2π 3 ¯ ¯A ¯ ¯2 · cos(0) ³ sin2(0) Á + 2 ´ −cos(π) ³ sin2(π) Á + 2 ´¸ = 2π 3 ¯ ¯A ¯ ¯2h (1)(2) −(−1)(2) i = 2π 3 ¯ ¯A ¯ ¯2£ 4 ¤ ⇒ 8π 3 ¯ ¯A ¯ ¯2 = 1 ⇒ A = r 3 8π ⇒ Y1,1 = r 3 8π sin θ eiφ, which is identical to Y1,1 in table 9–1.
Example 9–25: Derive Y1,0 from the result of the previous example using the lowering operator.
A lowering operator acting on an abstract eigenstate is L− ¯ ¯l, m> = B ¯ ¯l, m −1>, where B is a proportionality constant. Using the spherical angle representation on the eigenstate Y1,1, this eigenvalue/eigenvector equation is −¯ h e−iφ µ ∂ ∂θ −i cot θ ∂ ∂φ ¶ Y1,1 = B Y1,0, where B is the eigenvalue. Using the unnormalized form of Y1,1, we have B Y1,0 = −¯ h e−iφ µ ∂ ∂θ −i cot θ ∂ ∂φ ¶ A sin θ eiφ = −A¯ h e−iφ µ eiφ ∂ ∂θ sin θ −i cot θ sin θ ∂ ∂φeiφ ¶ = −A¯ h e−iφ µ eiφ cos θ −icos θ sin θ sin θ(i)eiφ ¶ = −A¯ h ¡ cos θ + cos θ ¢ = −2A¯ h ¡ cos θ ¢ 327 ⇒ Y1,0 = C cos θ, where all constants have been combined to form C, which becomes simply a normalization constant. We normalize this using the same procedure as the previous example, 1 = Z C∗cos θ C cos θ dΩ= ¯ ¯C ¯ ¯2 Z 2π 0 dφ Z π 0 dθ cos2 θ sin θ = 2π ¯ ¯C ¯ ¯2 Z π 0 dθ cos2 θ sin θ = 2π ¯ ¯C ¯ ¯2h −cos3 θ 3 iπ 0 = 2π 3 ¯ ¯C ¯ ¯2h cos3 θ i0 π = 2π 3 ¯ ¯C ¯ ¯2h cos3(0) −cos3(π) i = 2π 3 ¯ ¯C ¯ ¯2h 1 −(−1) i = 2π 3 ¯ ¯C ¯ ¯2£ 2 ¤ ⇒ 4π 3 ¯ ¯C ¯ ¯2 = 1 ⇒ C = r 3 4π ⇒ Y1,0 = r 3 4π cos θ, which is identical to Y1,0 as listed in table 9–1.
Generating Function for Spherical Harmonics A generating functions for higher index spherical harmonics is Yl,m(θ, φ) = (−1)m s (2l + 1)(l −m)!
4π(l + m)!
Pl,m(cos θ) eimφ, m ≥0, and Yl,−m(θ, φ) = Y ∗ l,m(θ, φ), m < 0, where the Pl,m(cos θ) are associated Legendre polynomials. Associated Legendre polynomials can be generated from Legendre polynomials using Pl,m(u) = (−1)mp (1 −u2)m dm dum Pl(u), where the Pl(u) are Legendre polynomials. Legendre polynomials can be generated using Pl(u) = (−1)l 2ll!
dl dul (1 −u2)l.
Notice the generating function for spherical harmonics contains the restriction m ≥0.
Our strategy to attain spherical harmonics with m < 0 will be to form them from the adjoint of the corresponding spherical harmonic with m > 0 as indicated. The advantage of this strategy is we do not need to consider associated Legendre polynomials with m < 0, though those also have meaning and can be attained using Pl,−m(u) = (l −m)!
(l + m)! Pl,m(u), in our phase scheme.
328 |
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9090 | https://spsweb.fltops.jpl.nasa.gov/portaldataops/mpg/MPG_Docs/Source%20Docs/CelestialMechanics-2-Conic%20sections.pdf | 1 CHAPTER 2 CONIC SECTIONS 2.1 Introduction A particle moving under the influence of an inverse square force moves in an orbit that is a conic section; that is to say an ellipse, a parabola or a hyperbola. We shall prove this from dynamical principles in a later chapter. In this chapter we review the geometry of the conic sections. We start off, however, with a brief review (eight equation-packed pages) of the geometry of the straight line. 2.2 The Straight Line It might be thought that there is rather a limited amount that could be written about the geometry of a straight line. We can manage a few equations here, however, (there are 35 in this section on the Straight Line) and shall return for more on the subject in Chapter 4. Most readers will be familiar with the equation y = mx + c 2.2.1 for a straight line. The slope (or gradient) of the line, which is the tangent of the angle that it makes with the x-axis, is m, and the intercept on the y-axis is c. There are various other forms that may be of use, such as x x y y 0 0 1 + = 2.2.2 y y x x y y x x − − = − − 1 1 2 1 2 1 2.2.3 which can also be written x y 1 x1 y1 1 = 0 2.2.4 x2 y2 1 x cos θ + y sin θ = p 2.2.5 2 The four forms are illustrated in figure II.1. A straight line can also be written in the form Ax + By + C = 0. 2.2.6 If C = 0, the line passes through the origin. If C ≠ 0, no information is lost, and some arithmetic and algebra are saved, if we divide the equation by C and re-write it in the form ax + by = 1. 2.2.7 x y x y c slope = m 2.2.1 x y x0 y0 2.2.2 x y (x2 , y2 ) ( x1 , y1 ) 2.2.3 or 4 α p 2.2.5 FIGURE II.1 • • 3 Let P (x , y) be a point on the line and let P0 (x0 , y0 ) be a point in the plane not necessarily on the line. It is of interest to find the perpendicular distance between P0 and the line. Let S be the square of the distance between P0 and P. Then ( ) ( ) 2 0 2 0 y y x x S − + − = 2.2.8 We can express this in terms of the single variable x by substitution for y from equation 2.2.7. Differentiation of S with respect to x will then show that S is least for ( ) 2 2 0 0 b a ay bx b a x + − + = 2.2.9 The corresponding value for y, found from equations 2.2.7 and 2.2.9, is ( ).
2 2 0 0 b a bx ay a b y + − + = 2.2.10 The point P described by equations 2.2.9 and 2.2.10 is the closest point to P0 on the line. The perpendicular distance of P from the line is p = √S or p ax by a b = − − + 1 0 0 2 2 . 2.2.11 This is positive if P0 is on the same side of the line as the origin, and negative if it is on the opposite side. If the perpendicular distances of two points from the line, as calculated from equation 2.2.11, are of opposite signs, they are on opposite sides of the line. If p = 0, or indeed if the numerator of 2.2.11 is zero, the point P0 (x0 , y0 ) is, of course, on the line. Let A(x1 , y1 ), B(x2 , y2 ) and C(x3 , y3 ) be three points in the plane. What is the area of the triangle ABC? One way to answer this is suggested by figure II.2. A(x1 , y1) C(x3 , y3) B(x2 , y2) A' B' C' FIGURE II.2 4 We see that area of triangle ABC = area of trapezium A'ACC' (see comment) + area of trapezium C'CBB' − area of trapezium A'ABB'. ( )( ) ( )( ) ( )( ) 1 2 1 2 2 1 3 2 3 2 2 1 1 3 1 3 2 1 y y x x y y x x y y x x + − − + − + + − = ( ) ( ) ( ) [ ] 2 1 3 1 3 2 3 2 1 2 1 y y x y y x y y x − + − + − = x1 x2 x3 = 2 1 y1 y2 y3 2.2.12 1 1 1 Since writing this section I have become aware of a difference in U.S./British usages of the word "trapezium". Apparently in British usage, "trapezium" means a quadrilateral with two parallel sides. In U.S. usage, a trapezium means a quadrilateral with no parallel sides, while a quadrilateral with two parallel sides is a "trapezoid". As with many words, either British or U.S. usages may be heard in Canada. In the above derivation, I intended the British usage. What is to be learned from this is that we must always take care to make ourselves clearly understood when using such ambiguous words, and not to assume that the reader will interpret them the way we intend. The reader might like to work through an alternative method, using results that we have obtained earlier. The same result will be obtained. In case the algebra proves a little tedious, it may be found easier to work through a numerical example, such as: calculate the area of the triangle ABC, where A, B, C are the points (2,3), (7,4), (5,6) respectively. In the second method, we note that the area of a triangle is 1 2 × base × height. Thus, if we can find the length of the side BC, and the perpendicular distance of A from BC, we can do it. The first is easy: (BC)2 ( ) ( ) .
2 2 3 2 2 3 y y x x − + − = 2.2.13 To find the second, we can easily write down the equation to the line BC from equation 2.2.3, and then re-write it in the form 2.2.7. Then equation 2.2.11 enables us to find the perpendicular distance of A from BC, and the rest is easy. If the determinant in equation 2.2.12 is zero, the area of the triangle is zero. This means that the three points are collinear. The angle between two lines y m x c = + 1 1 2.2.14 and y m x c = + 2 2 2.2.15 5 is easily found by recalling that the angles that they make with the x-axis are tan-1 m1 and tan-1 m2 together with the elementary trigonometry formula tan( ) (tan tan ) / ( tan tan ).
A B A B A B − = − + 1 It is then clear that the tangent of the angle between the two lines is m m m m 2 1 1 2 1 − + . 2.2.16 The two lines are at right angles to each other if m1 m2 = −1 2.2.17 The line that bisects the angle between the lines is the locus of points that are equidistant from the two lines. For example, consider the two lines −2x + 5y = 1 2.2.18 30x − 10y = 1 2.2.19 Making use of equation 2.2.11, we see that a point (x , y) is equidistant from these two lines if 1 2 5 29 1 30 10 1000 + − = ± − + x y x y . 2.2.20 The significance of the ± will become apparent shortly. The + and − choices result, respectively, in −8.568x + 8.079y = 1 2.2.21 and 2.656x + 2.817y = 1. 2.2.22 The two continuous lines in figure II.3 are the lines 2.2.18 and 2.2.19. There are two bisectors, represented by equations 2.2.21 and 2.2.22, shown as dotted lines in the figure, and they are at right angles to each other. The choice of the + sign in equation 2.2.20 (which in this case results in equation 2.2.21, the bisector in figure II.3 with the positive slope) gives the bisector of the sector that contains the origin. An equation of the form ax2 + 2hxy + by2 = 0 2.2.23 can be factored into two linear factors with no constant term, and it therefore represents two lines intersecting at the origin. It is left as an exercise to determine the angles that the two lines make with each other and with the x axis, and to show that the lines 6 0 2 2 = − − + y xy h b a x 2.2.24 are the bisectors of 2.2.23 and are perpendicular to each other. Given the equations to three straight lines, can we find the area of the triangle bounded by them? To find a general algebraic expression might be a bit tedious, though the reader might like to try it, but a numerical example is straightforward. For example, consider the lines x y − + = 5 12 0, 2.2.25 3 4 9 0 x y + − = , 2.2.26 3 3 0 x y − − = . 2.2.27 By solving the equations in pairs, it is soon found that they intersect at the points (−0.15789, 2.36842), (1.4, 1.2) and (1.92857, 2.78571). Application of equation 2.2.12 then gives the area as 1.544. The triangle is drawn in figure II.4. Measure any side and the corresponding height with a ruler and see if the area is indeed about 1.54. FIGURE II.3 y x 7 But now consider the three lines x y − + = 5 12 0, 2.2.28 3 4 9 0 x y + − = , 2.2.29 3 23 54 0 x y + − = . 2.2.30 By solving the equations in pairs, it will be found that all three lines intersect at the same point (please do this), and the area of the triangle is, of course, zero. Any one of these equations is, in fact, a linear combination of the other two. You should draw these three lines accurately on graph paper (or by computer). In general, if three lines are A x B y C 1 1 1 0 + + = 2.2.31 A x B y C 2 2 2 0 + + = 2.2.32 A x B y C 3 3 3 0 + + = 2.2.33 they will be concurrent at a single point if -0.5 0 0.5 1 1.5 2 2.5 1 1.5 2 2.5 3 3.5 x y FIGURE II.4 8 0 3 3 3 2 2 2 1 1 1 = C B A C B A C B A . 2.2.34 Thus the determinant in equation 2.2.12 provides a test of whether three points are collinear, and the determinant in equation 2.2.34 provides a test of whether three lines are concurrent. Finally - at least for the present chapter - there may be rare occasion to write the equation of a straight line in polar coordinates. It should be evident from figure II.5 that the equations ( ) ( ) θ − α = α − θ = csc or csc p r p r 2.2.35 describe a straight line passing at a distance p from the pole and making an angle α with the initial line. If p = 0, the polar equation is merely θ = α. p α θ r p r α θ FIGURE II.5 9 2.3 The Ellipse An ellipse is a figure that can be drawn by sticking two pins in a sheet of paper, tying a length of string to the pins, stretching the string taut with a pencil, and drawing the figure that results. During this process, the sum of the two distances from pencil to one pin and from pencil to the other pin remains constant and equal to the length of the string. This method of drawing an ellipse provides us with a formal definition, which we shall adopt in this chapter, of an ellipse, namely: An ellipse is the locus of a point that moves such that the sum of its distances from two fixed points called the foci is constant (see figure II.6). We shall call the sum of these two distances (i.e the length of the string) 2a.. The ratio of the distance between the foci to length of the string is called the eccentricity e of the ellipse, so that the distance between the foci is 2ae, and e is a number between 0 and 1. The longest axis of the ellipse is its major axis, and a little bit of thought will show that its length is equal to the length of the string; that is, 2a. The shortest axis is the minor axis, and its length is usually denoted by 2b. The eccentricity is related to the ratio b/a in a manner that we shall shortly discuss. The ratio η = (a − b)/a is called the ellipticity if the ellipse. It is merely an alternative measure of the noncircularity. It is related to the eccentricity, and we shall obtain that relation shortly, too. Until then, Figure II.7 shows pictorially the relation between the two. FIGURE II.6 • • 10 We shall use our definition of an ellipse to obtain its equation in rectangular coordinates. We shall place the two foci on the x-axis at coordinates (−ae, 0) and (ae, 0) (see figure II.8). The definition requires that PF1 + PF2 = 2a. That is: ( ) [ ] ( ) [ ] , 2 2 1 2 1 2 2 2 2 a y ae x y ae x = + − + + + 2.3.1 and this is the equation to the ellipse. The reader should be able, after a little bit of slightly awkward algebra, to show that this can be written more conveniently as 0 1 η 0 e 1 FIGURE II.7 P(x , y) ae ae FIGURE II.8 A B F1 l F2 11 ( ) .
1 1 2 2 2 2 2 = − + e a y a x 2.3.2 By putting x = 0, it is seen that the ellipse intersects the y-axis at ± − a e 1 2 and therefore that a e 1 2 − is equal to the semi minor axis b. Thus we have the familiar equation to the ellipse x a y b 2 2 2 2 1 + = 2.3.3 as well as the important relation between a, b and e: b2 = a2 (1 - e2 ) 2.3.4 The reader can also now derive the relation between ellipticity η and eccentricity e: ( ).
1 1 2 e − − = η 2.3.5 This can also be written ( ) η − η = 2 2 e 2.3.6 or ( ) .
1 1 2 2 = − η + e 2.3.7 This shows, incidentally, that the graph of η versus e, which we have drawn in figure II.7, is part of a circle of radius 1 centred at e = 0, η = 1. In figures II.9 I have drawn ellipses of eccentricities 0.1 to 0.9 in steps of 0.1, and in figure II.10 I have drawn ellipses of ellipticities 0.1 to 0.9 in steps of 0.1. You may find that ellipticity gives you a better idea than eccentricity of the noncircularity of an ellipse. For an exercise, you should draw in the positions of the foci of each of these ellipses, and decide whether eccentricity or ellipticity gives you a better idea of the "ex-centricity" of the foci. Note that the eccentricities of the orbits of Mars and Mercury are, respectively, about 0.1 and 0.2 (these are the most eccentric of the planetary orbits except for comet-like Pluto), and it is difficult for the eye to see that they depart at all from circles - although, when the foci are drawn, it is obvious that the foci are "ex-centric". 12 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 FIGURE II.9 The number inside each ellipse is its eccentricity. 13 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 FIGURE II.10 The figure inside or below each ellipse is its ellipticity. 14 In the theory of planetary orbits, the Sun will be at one focus. Let us suppose it to be at F2 (see figure II.8). In that case the distance F2 B is the perihelion distance q, and is equal to ( ).
1 e a q − = 2.3.8 The distance F2 A is the aphelion distance Q (pronounced ap-helion by some and affelion by others − and both have defensible positions), and it is equal to ( ).
1 e a Q + = 2.3.9 A line parallel to the minor axis and passing through a focus is called a latus rectum (plural: latera recta). The length of a semi latus rectum is commonly denoted by l (sometimes by p). Its length is obtained by putting x = ae in the equation to the ellipse, and it will be readily found that ( ).
1 2 e a l − = 2.3.10 The length of the semi latus rectum is an important quantity in orbit theory. It will be found, for example, that the energy of a planet is closely related to the semi major axis a of its orbit, while its angular momentum is closely related to the semi latus rectum. The circle whose diameter is the major axis of the ellipse is called the eccentric circle or, preferably, the auxiliary circle (figure II.11). Its equation is x y a 2 2 2 + = . 2.3.11 FIGURE II.11 v E F • P' P r 15 In orbit theory the angle v (denoted by f by some authors) is called the true anomaly of a planet in its orbit. The angle E is called the eccentric anomaly, and it is important to find a relation between them. We first note that, if the eccentric anomaly is E, the abscissas of P' and of P are each a E cos . The ordinate of P' is a E sin . By putting x a E = cos in the equation to the ellipse, we immediately find that the ordinate of P is b sin E. Several deductions follow. One is that any point whose abscissa and ordinate are of the form x a E y a E = = cos , sin 2.3.12 is on an ellipse of semi major axis a and semi minor axis b. These two equations can be regarded as parametric equations to the ellipse. They can be used to describe an ellipse just as readily as x a y b 2 2 2 2 1 + = 2.3.13 and indeed this equation is the E-eliminant of the parametric equations. The ratio PM/P'M for any line perpendicular to the major axis is b/a. Consequently the area of the ellipse is b/a times the area of the auxiliary circle; and since the area of the auxiliary circle is πa2 , it follows that the area of the ellipse is πab. In figure II.11, the distance r is called the radius vector (plural radii vectores), and from the theorem of Pythagoras its length is given by ( ) .
cos sin 2 2 2 2 2 e E a E b r − + = 2.3.14 On substituting 1 2 −cos E for sin2 E and a2 (1 - e2 ) for b2 , we soon find that r = a (1 − e cos E ) 2.3.15 It then follows immediately that the desired relation between v and E is cos cos cos .
v = − − E e e E 1 2.3.16 From trigonometric identities, this can also be written E e E e cos 1 sin 1 sin 2 − − = v 2.3.17a 16 or E E e cos 1 sin 1 tan 2 − − = v 2.3.17b or .
tan 1 1 tan 2 1 2 1 E e e − + = v 2.3.17c There are a number of miscellaneous geometric properties of an ellipse, some, but not necessarily all, of which may prove to be of use in orbital calculations. We describe some of them in what follows. Tangents to an ellipse. Find where the straight line y = mx + c intersects the ellipse .
1 2 2 2 2 = + b y a x 2.3.18 The answer to this question is to be found by substituting mx + c for y in the equation to the ellipse. After some rearrangement, a quadratic equation in x results: ( ) ( ) .
0 2 2 2 2 2 2 2 2 2 = − + + + b c a cmx a x b m a 2.3.19 If this equation has two real roots, the roots are the x-coordinates of the two points where the line intersects the ellipse. If it has no real roots, the line misses the ellipse. If it has two coincident real roots, the line is tangent to the ellipse. The condition for this is found by setting the discriminant of the quadratic equation to zero, from which it is found that c a m b 2 2 2 2 = + . 2.3.20 Thus a straight line of the form y mx a m b = ± + 2 2 2 2.3.21 is tangent to the ellipse. Figure II.12 shows several such lines, for a = 2b and slopes (tan-1 m) of 0o to 180o in steps of 10o 17 Director Circle. The equation we have just derived for a tangent to the ellipse can be rearranged to read ( ) .
0 2 2 2 2 2 2 = − + + − y b mx x a m 2.3.22 Now the product of the slopes of two lines that are at right angles to each other is −1 (equation 2.2.17). Therefore, if we replace m in the above equation by −1/m we shall obtain another tangent to the ellipse, at right angles to the first one. The equation to this second tangent becomes (after multiplication throughout by m ) ( ) .
0 2 2 2 2 2 2 = − + − − x a mx y b m 2.3.23 If we eliminate m from these two equations, we shall obtain an equation in x and y that describes the point where the two perpendicular tangents meet; that is, the equation that will describe a curve that is the locus of the point of intersection of two perpendicular tangents. It turns out that this curve is a circle of radius a b 2 2 + , and it is called the director circle. It is easier than it might first appear to eliminate m from the equations. We merely have to add the equations 2.3.22 and 2.3.23: ( ) .
0 2 2 2 2 2 2 2 2 2 = − − + + − − + y x b a y x b a m 2.3.24 For real m, this can only be if x y a b 2 2 2 2 + = + , 2.3.25 which is the required locus of the director circle of radius a b 2 2 + . It is illustrated in figure II.13. We shall now derive an equation to the line that is tangent to the ellipse at the point (x1 , y1 ). Let (x1 , y1 ) = (a cos E1 , b sin E1 ) and (x2 , y2 ) = (a cos E2 , b sin E2 ) be two points on the ellipse. The line joining these two points is ( ) ( ) 1 2 1 2 1 1 cos cos sin sin cos sin E E a E E b E a x E b y − − = − − ( ) ( ) ( ) ( ) 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 sin sin 2 sin cos 2 E E E E a E E E E b − + − − + = ( ) ( ).
sin cos 1 2 2 1 1 2 2 1 E E a E E b + + − = 2.3.26 18 Now let E2 approach E1 , eventually coinciding with it. The resulting equation y b E x a E b E a E − − = − sin cos cos sin , 2.3.27 in which we no longer distinguish between E1 and E2, is the equation of the straight line that is tangent to the ellipse at (a cos E , b sin E ). This can be written x E a y E b cos sin + = 1 2.3.28 FIGURE II.13 19 or, in terms of (x1 , y1 ), x x a y y b 1 2 1 2 1 + = , 2.3.29 which is the tangent to the ellipse at (x1 , y1 ). An interesting property of a tangent to an ellipse, the proof of which I leave to the reader, is that F1P and F2 P make equal angles with the tangent at P. If the inside of the ellipse were a reflecting mirror and a point source of light were to be placed at F1 , it would be imaged at F2 . (Have a look at figure II.6 or II.8.) This has had an interesting medical application. A patient has a kidney stone. The patient is asked to lie in an elliptical bath, with the kidney stone at F2 . A small explosion is detonated at F1; the explosive sound wave emanating from F1 is focused as an implosion at F2 and the kidney stone at F2 is shattered. Don't try this at home. Directrices. The two lines x a e = ± / are called the directrices (singular directrix) of the ellipse (figure II.14). FIGURE II.14 F2 P N 20 The ellipse has the property that, for any point P on the ellipse, the ratio of the distance PF2 to a focus to the distance PN to a directrix is constant and is equal to the eccentricity of the ellipse. Indeed, this property is sometimes used as the definition of an ellipse, and all the equations and properties that we have so far derived can be deduced from such a definition. We, however, adopted a different definition, and the focus-directrix property must be derived. This is straightforward, for, (recalling that the abcissa of F2 is ae) we see from figure II.14 that the square of the desired ratio is ( ) ( ) .
/ 2 2 2 x e a y ae x − + − 2.3.30 On substitution of ( ) ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 2 2 1 1 1 1 x a e e a b a x a x − − = − − = − 2.3.31 for y2 , the above expression is seen to reduce to e2 . Another interesting property of the focus and directrix, although a property probably with not much application to orbit theory, is that if the tangent to an ellipse at a point P intersects the directrix at Q, then P and Q subtend a right angle at the focus. (See figure II.15). Thus the tangent at P = (x1 , y1 ) is x x a y y b 1 2 1 2 1 + = 2.3.32 FIGURE II.15 F2 P Q 21 and it is straightforward to show that it intersects the directrix x = a/e at the point .
1 , 1 1 2 −ae x y b e a The coordinates of the focus F2 are (ae, 0). The slope of the line PF2 is ( ) 1 1 / y ae x − and the slope of the line QF2 is ( ).
1 1 1 2 ae e a ae x y b − − It is easy to show that the product of these two slopes is −1, and hence that PF2 and QF2 are at right angles. Conjugate Diameters. The left hand of figure II.16 shows a circle and two perpendicular diameters. The right hand figure shows what the circle would look like when viewed at some oblique angle. The circle has become an ellipse, and the diameters are no longer perpendicular. The diameters are called conjugate diameters of the ellipse. One is conjugate to the other, and the other is conjugate to the one. They have the property - or the definition - that each bisects all chords parallel to the other, because this property of bisection, which is obviously held by the perpendicular diameters of the circle, is unaltered in projection. FIGURE II.16 22 It is easy to draw two conjugate diameters of an ellipse of eccentricity e either by making use of this last-mentioned property or by noting that that the product of the slopes of two conjugate diameters is e2 − 1. The proof of this is left for the enjoyment of the reader. A Ladder Problem. No book on elementary applied mathematics is complete without a ladder problem. A ladder of length a + b leans against a smooth vertical wall and a smooth horizontal floor. A particular rung is at a distance a from the top of the ladder and b from the bottom of the ladder. Show that, when the ladder slips, the rung describes an ellipse. (This result will suggest another way of drawing an ellipse.) See figure II.17. If you have not done this problem after one minute, here is a hint. Let the angle that the ladder makes with the floor at any time be E. That is the end of the hint. The reader may be aware that some of the geometrical properties that we have discussed in the last few paragraphs are more of recreational interest and may not have much direct application in the theory of orbits. In the next subsection we return to properties and equations that are very relevant to orbital theory - perhaps the most important of all for the orbit computer to understand. 23 Polar Equation to the Ellipse. We shall obtain the equation in polar coordinates to an ellipse whose focus is the pole of the polar coordinates and whose major axis is the initial line (θ = 0o ) of the polar coordinates. In figure II.18 we have indicated the angle θ of polar coordinates, and it may occur to the reader that we have previously used the symbol v for this angle and called it the true anomaly. Indeed at present, v and θ are identical, but a little later we shall distinguish between them. From our definition of the ellipse, s = 2a − r, and so s a ar r 2 2 2 4 4 = − + . 2.3.33 From the cosine formulae for a plane triangle, s2 = 4a2e2 + r2 + 4aer cos θ. 2.3.34 On equating these expressions we soon obtain ( ) ( ).
cos 1 1 2 θ + = − e r e a 2.3.35 The left hand side is equal to the semi latus rectum l, and so we arrive at the polar equation to the ellipse, focus as pole, major axis as initial line: r l e = + 1 cos .
θ 2.3.36 If the major axis is inclined at an angle ω to the initial line (figure II.19 ), the equation becomes ( ) .
cos 1 cos 1 v e l e l r + = ω − θ + = 2.3.37 r θ FIGURE II.18 s 2ae 24 The distinction between θ and v is now evident. θ is the angle of polar coordinates, ω is the angle between the major axis and the initial line (ω will be referred to in orbital theory as the "argument of perihelion"), and v, the true anomaly, is the angle between the radius vector and the initial line. v ω θ = v + ω FIGURE II.19 25 2.4 The Parabola We define a parabola as the locus of a point that moves such that its distance from a fixed straight line called the directrix is equal to its distance from a fixed point called the focus. Unlike the ellipse, a parabola has only one focus and one directrix. However, comparison of this definition with the focus - directrix property of the ellipse (which can also be used to define the ellipse) shows that the parabola can be regarded as a limiting form of an ellipse with eccentricity equal to unity. We shall find the equation to a parabola whose directrix is the line y = −q and whose focus is the point (q , 0). Figure II.20 shows the parabola. F is the focus and O is the origin of the coordinate system. The vertex of the parabola is at the origin. In an orbital context, for example, the orbit of a comet moving around the Sun in parabolic orbit, the Sun would be at the focus F, and the distance between vertex and focus would be the perihelion distance, for which the symbol q is traditionally used in orbit theory. From figure II.20, it is evident that the definition of the parabola (PF = PN) requires that ( ) ( ) , 2 2 2 q x y q x + = + − 2.4.1 from which y2 = 4qx, 2.4.2 26 which is the equation to the parabola. The reader should also try sketching the parabolas ( ) ( ).
3 4 2 , 4 , 4 , 4 2 2 2 2 − = − − = = − = x q y qy x qy x qx y The line parallel to the y-axis and passing through the focus is the latus rectum. Substitution of x = q into y 2 = 4ax shows that the latus rectum intersects the parabola at the two points (q , ±2q), and that the length l of the semi latus rectum is 2q. The equations x = qt2 , y = 2qt 2.4.3 are the parametric equations to the parabola, for y2 = 4qx results from the elimination of t between them. In other words, if t is any variable, then any point that satisfies these two equations lies on the parabola. Most readers will know that if a particle is moving with constant speed in one direction and constant acceleration at right angles to that direction, as with a ball projected in a uniform gravitational field or an electron moving in a uniform electric field, the path is a parabola. In the constant speed direction the distance is proportional to the time, and in the constant acceleration direction, the distance is proportional to the square of the time, and hence the path is a parabola. Tangents to a Parabola. Where does the straight line y = mx + c intersect the parabola y2 = 4qx? The answer is found by substituting mx + c for y to obtain, after rearrangement, ( ) .
0 2 2 2 2 2 = + − + c x q mc x m 2.4.4 The line is tangent if the discriminant is zero, which leads to c = q/m. 2.4.5 Thus a straight line of the form y = mx + q/m 2.4.6 is tangent to the parabola. Figure II.22 (sorry - there is no figure II.21!) illustrates this for several lines, the slopes of each differing by 5o from the next. 27 We shall now derive an equation to the line that is tangent to the parabola at the point (x1 , y1 ). Let ( , ) ( , ) x y qt qt 1 1 1 2 1 2 = be a point on the parabola, and Let ( , ) ( , ) x y qt qt 2 2 2 2 2 2 = be another point on the parabola. The line joining these two points is ( ) .
2 ) ( 2 2 1 2 2 1 2 2 1 2 2 1 1 t t t t q t t q qt x qt y + = − − = − − 2.4.7 Now let t2 approach t1 , eventually coinciding with it. Putting t t t = = 2 1 in the last equation results, after simplification, in ty = x + qt2 , 2.4.8 being the equation to the tangent at (qt2 , 2qt ). Multiply by 2q: 28 2qty = 2q(x + qt2 ) 2.4.9 and it is seen that the equation to the tangent at (x1 , y1 ) is y1y = 2q (x1 + x). 2.4.10 There are a number of interesting geometric properties, some of which are given here. For example, if a tangent to the parabola at a point P meets the directrix at Q, then, just as for the ellipse, P and Q subtend a right angle at the focus (figure II.23). The proof is similar to that given for the ellipse, and is left for the reader. The reader will recall that perpendicular tangents to an ellipse meet on the director circle. The analogous theorem vis-à-vis the parabola is that perpendicular tangents meet on the directrix. This is also illustrated in figure II.23. The theorem is not specially important in orbit theory, and the proof is also left to the reader. Let PG be the normal to the parabola at point P, meeting the axis at G (figure II.24). We shall call the length GH the subnormal. A curious property is that the length of GH is always equal to l, the length of the semi latus rectum (which in figure II.24 is of length 2 − i.e. the ordinate where x = 1), irrespective of the position of P. This proof again is left to the reader. 29 The following two geometrical properties, while not having immediate applications to orbit theory, certainly have applications to astronomy. α β Q 30 The tangent at P makes an angle α with the x-axis, and PF makes an angle β with the x-axis (figure II.25). We shall show that β = 2α and deduce an interesting consequence. The equation to the tangent (see equation 2.4.8) is ty = x + qt2 , which shows that tan α = 1/t. 2.4.11 The coordinates of P and F are, respectively, ( ) ( ) 0 , and 2 , 2 a qt qt , and so, from the triangle PFH, we find. tan .
β = − 2 1 2 t t 2.4.12 Let τ = 1/t, then tan α = τ and tan β = 2τ/(1 - τ 2 ), which shows that β = 2α . This also shows that triangle JFP is isosceles, with the angles at J and P each being α. This can also be shown as follows. From the equation ty = x + qt2 , we see that J is the point (−qt2 , 0), so that JF = q (t2 + 1). From triangle PFH, we see that (PF)2 = ( ) ( ) .
1 1 4 2 2 2 2 2 2 2 2 + − − + t q t q t q 2.4.13 Therefore PF = JF. 2.4.14 Either way, since the triangle JPF is isosceles, it follows that QP and PF make the same angle α to the tangent. If the parabola is a cross section of a telescopic mirror, any ray of light coming in parallel to the axis will be focussed at F, so that a paraboloidal mirror, used on-axis, does not suffer from spherical aberration. (This property holds, of course, only for light parallel to the axis of the paraboloid, so that a paraboloidal mirror, without some sort of correction, gives good images over only a narrow field of view.) Now consider what happens when you stir a cup of tea. The surface takes up a shape that looks as though it might resemble the parabola y = x2/(4q) - see figure II.26: $ R mω2x mg θ 31 Suppose the liquid is circulating at angular speed ω. A tea leaf floating on the surface is in equilibrium (in the rotating reference frame) under three forces: its weight mg, the centrifugal force mω2x and the normal reaction R. The normal to the surface makes an angle θ with the vertical (and the tangent makes an angle θ with the horizontal) given by tan .
θ ω = 2x g 2.4.15 But the slope of the parabola y = x2/(4q) is x/(2q), so that the surface is indeed a parabola with semi latus rectum 2q = g/ω2 . This phenomenon has been used in Canada to make a successful large telescope (diameter 6 m ) in which the mirror is a spinning disc of mercury that takes up a perfectly paraboloidal shape. Another example is the spin casting method that has been successfully used for the production of large, solid glass paraboloidal telescope mirrors. In this process, the furnace is rotated about a vertical axis while the molten glass cools and eventually solidifies into the required paraboloidal effect. Exercise. The 6.5 metre diameter mirrors for the twin Magellan telescopes at Las as n ~ Campa , Chile, have a focal ratio f/1.25. They were made by the technique of spin casting at The University of Arizona's Mirror Laboratory.. At what speed would the furnace have had to be rotated in order to achieve the desired focal ratio? (Answer = 7.4 rpm.) Notice that f/1.25 is quite a deep paraboloid. If this mirror had been made by traditional grinding from a solid disc, what volume of material would have had to be removed to make the desired paraboloid? (Answer - a whopping 5.4 cubic metres, or about 12 tons!) Polar equation to the Parabola. As with the ellipse, we choose the focus as pole and the axis of the parabola as initial line. We shall orient the parabola so that the vertex is toward the right, as in figure II.27. We recall the focus-directrix property, FP = PN. Also, from the definition of the directrix, FO=OM = q, so that FM = 2q = l, the length of the semi latus rectum. It is therefore immediately evident from figure II.27 that r r q l cos , θ+ = = 2 so that the polar equation to the parabola is r l = + 1 cos .
θ 2.4.16 32 This is the same as the polar equation to the ellipse (equation 2.3.36), with e = 1 for the parabola. I have given different derivations for the ellipse and for the parabola; the reader might like to interchange the two approaches and develop equation 2.3.36 in the same manner as we have developed equation 2.4.16. When we discuss the hyperbola, I shall ask you to show that its polar equation is also the same as 2.3.36. In other words, equation 2.3.36 is the equation to a conic section, and it represents an ellipse, parabola or hyperbola according to whether e <1, e = 1 or e >1. θ θ 33 2.5 The Hyperbola A hyperbola is the locus of a point that moves such that the difference between its distances from two fixed points called the foci is constant. We shall call the difference between these two distances 2a and the distance between the foci 2ae, where e is the eccentricity of the hyperbola, and is a number greater than 1. See figure II.28. For example, in a Young's double-slit interference experiment, the mth bright fringe is located at a point on the screen such that the path difference for the rays from the two slits is m wavelengths. As the screen is moved forward or backwards, this relation continues to hold for the mth bright fringe, whose locus between the slits and the screen is therefore a hyperbola. The "Decca" system of radar navigation, first used at the D-Day landings in the Second World War and decommissioned only as late as 2000 on account of being rendered obsolete by the "GPS" (Global Positioning Satellite) system, depended on this property of the hyperbola. (Since writing this, part of the Decca system has been re-commissioned as a back-up in case of problems with GPS.) Two radar transmitters some distance apart would simultaneously transmit radar pulses. A ship would receive the two signals separated by a short time interval, depending on the difference between the distances from the ship to the two transmitters. This placed the ship on a particular hyperbola. The ship would also listen in to another pair of transmitters, and this would place the ship on a second hyperbola. This then placed the ship at one of the four points where the two hyperbolas intersected. It would usually be obvious which of the four points was the correct one, but any ambiguity could be resolved by the signals from a third pair of transmitters. In figure II.28, the coordinates of F1 and F2 are, respectively, (−ae , 0) and (ae , 0). 34 The condition PF1 − PF2 = 2a requires that ( ) [ ] ( ) [ ] , 2 2 / 1 2 2 2 / 1 2 2 a y ae x y ae x = + − − + + 2.5.1 and this is the equation to the hyperbola. After some arrangement, this can be written ( ) , 1 1 2 2 2 2 2 = − − e a y a x 2.5.2 which is a more familiar form for the equation to the hyperbola. Let us define a length b by b2 = a2 (e2 − 1). 2.5.3 The equation then becomes x a y b 2 2 2 2 1 − = , 2.5.4 which is the most familiar form for the equation to a hyperbola. ____ Problem. When a meteor streaks across the sky, it can be tracked by radar. The radar instrumentation can determine the range (distance) of the meteoroid as a function of time. Show that, if the meteoroid is moving at constant speed (a questionable assumption, because it must be decelerating, but perhaps we can assume the decrease in speed is negligible during the course of the observation), and if the range r is plotted against the time, the graph will be a hyperbola. Show also that, if r2 is plotted against t, the graph will be a parabola of the form r a t bt c 2 2 = + + , where a V b V t c V t r = = − = + 2 2 0 2 0 2 0 2 2 , , , V = speed of the meteoroid, t0 = time of closest approach, r0 = distance of closest approach. Radar observation of a meteor yields the following range-time data: t (s) r (km) 0.0 101.4 0.1 103.0 0.2 105.8 35 0.3 107.8 0.4 111.1 0.5 112.6 0.6 116.7 0.7 119.3 0.8 123.8 0.9 126.4 1.0 130.6 1.1 133.3 1.2 138.1 1.3 141.3 Assume that the velocity of the meteor is constant. Determine i. The time of closest approach (to 0.01 s) ii. The distance of closest approach (to 0.1 km) iii. The speed (to 1.0 km s-1) If you wish, just use the three asterisked data to determine a, b and c. If you are more energetic, use all the data, and determine a, b and c by least squares, and the probable errors of V, t0 and r0. ____ The distance between the two vertices of the hyperbola is its transverse axis, and the length of the semi transverse axis is a − but what is the geometric meaning of the length b? This is discussed below in the next subsection (on the conjugate hyperbola) and again in a later section on the impact parameter. The lines perpendicular to the x-axis and passing through the foci are the two latera recta. Since the foci are at (±ae , 0), the points where the latera recta intersect the hyperbola can be found by putting x = ae into the equation to the hyperbola, and it is then found that the length l of a semi latus rectum is l = a (e2 − 1). 2.5.5 The Conjugate Hyperbola. The equation x a y b 2 2 2 2 1 − = − 2.5.6 is the equation to the conjugate hyperbola. The conjugate hyperbola is drawn dashed in figure II.28, and it is seen that the geometric meaning of b is that it is the length of the semi transverse axis of the conjugate hyperbola. It is a simple matter to show that the eccentricity of the conjugate hyperbola is e2/(e2 − 1). 36 The Asymptotes. The lines y bx a = ± 2.5.7 are the asymptotes of the hyperbola. This can also be written x a y b 2 2 2 2 0 − = . 2.5.8 Thus x a y b c 2 2 2 2 − = 2.5.9 is the hyperbola, the asymptotes, or the conjugate hyperbola, if c = +1, 0 or −1 respectively. The asymptotes are drawn as dotted lines in figure II.28. The semi angle ψ between the asymptotes is given by tan ψ = b/a. 2.5.10 The Directrices. The lines y a e = ± / are the directrices, and, as with the ellipse (and with a similar proof), the hyperbola has the property that the ratio of the distance PF2 to a focus to the distance PN to the directrix is constant and is equal to the eccentricity of the hyperbola. This ratio (i.e. the eccentricity) is less than one for the ellipse, equal to one for the parabola, and greater than one for the hyperbola. It is not a property that will be of great importance for our purposes, but is worth mentioning because it is a property that is sometimes used to define a hyperbola. I leave it to the reader to draw the directrices in their correct positions in figure II.28. Parametric Equations to the Hyperbola. The reader will recall that the point (a cos E, b sin E) is on the ellipse ( ) 1 / ) / ( 2 2 2 2 = + b y a x and that this is evident because this equation is the E-eliminant of x = a cos E and y = b sin E. The angle E has a geometric interpretation as the eccentric anomaly. Likewise, recalling the relation cosh2 φ − sinh2 φ = 1 , it will be evident that ( ) 1 / ) / ( 2 2 2 2 = − b y a x can also be obtained as the φ−eliminant of the equations x a y b = = cosh , sinh φ φ 2.5.11 37 These two equations are therefore the parametric equations to the hyperbola, and any point satisfying these two equations lies on the hyperbola. The variable φ is not an angle, and has no geometric interpretation analogous to the eccentric anomaly of an ellipse. The equations x = a sec E , y = b tan E 2.5.12 can also be used as parametric equations to the hyperbola, on account of the trigonometric identity 1 + tan2 E = sec2 E. In that case, the angle E does have a geometric interpretation (albeit not a particularly interesting one) in relation to the auxiliary circle, which is the circle of radius a centred at the origin. The meaning of the angle should be evident from figure II.29, in which E is the eccentric angle corresponding to the point P. Impact Parameter. A particle travelling very fast under the action of an inverse square attractive force (such as an interstellar meteoroid or comet - if there are such things - passing by the Sun, or an electron in the vicinity of a positively charged atomic nucleus) will move in a hyperbolic path. We prove this in a later chapter, as well as discussing the necessary speed. We may imagine the particle initially approaching from a great distance along the asymptote at the bottom right hand corner of figure II.30. As it approaches the focus, it no longer moves along the asymptote but along an arm of the hyperbola. 38 The distance K2 F2 , which is the distance by which the particle would have missed F2 in the absence of an attractive force, is commonly called the impact parameter. Likewise, if the force had been a repulsive force (for example, suppose the moving particle were a positively charged particle and there were a centre of repulsion at F1, F1K1 would be the impact parameter. Clearly, F1K1 and F2K2 are equal in length. The symbol that is often used in scattering theory, whether in celestial mechanics or in particle physics, is b - but is this b the same b that goes into the equation to the hyperbola and which is equal to the semi major axis of the conjugate hyperbola? OF2 = ae, and therefore K2F2 = ae sin ψ. This, in conjunction with tan ψ = b/a and b2 = a2 (e2 − 1), will soon show that the impact parameter is indeed the same b that we are familiar with, and that b is therefore a very suitable symbol to use for impact parameter. Tangents to the Hyperbola. Using the same arguments as for the ellipse, the reader should easily find that lines of the form y mx a m b = ± − 2 2 2 2.5.13 F2 K2 F1 K1 O ψ 39 are tangent to the hyperbola. This is illustrated in figure II.33 for a hyperbola with b = a /2, with tangents drawn with slopes 30o to 150o in steps of 5o. (The asymptotes have ψ = 26o 34'.) (Sorry, but there are no figures II.31 or II.32 - computer problems!) Likewise, from similar arguments used for the ellipse, the tangent to the hyperbola at the point (x ,y) is found to be .
1 2 1 2 1 = − b y y a x x 2.5.14 Director Circle. As for the ellipse, and with a similar derivation, the locus of the points of intersection of perpendicular tangents is a circle, called the director circle, which is of radius √(a2 − b2 ). This is not of particular importance for our purposes, but the reader who is interested might like to prove this by the same method as was done for the director circle of the ellipse, and might like to try drawing the circle and some tangents. If b > a, that is to say if ψ > 45o and the angle between the asymptotes is greater than 90o , the director circle is not real and it is of course not possible to draw perpendicular tangents. Rectangular Hyperbola. If the angle between the asymptotes is 90o , the hyperbola is called a rectangular hyperbola. For such a hyperbola, b = a, the eccentricity is √2, the director circle is a point, namely the origin, and perpendicular tangents can be drawn only from the asymptotes. 40 The equation to a rectangular hyperbola is x y a 2 2 2 − = 2.5.15 and the asymptotes are at 45o to the x axis. Let Ox', Oy' be a set of axes at 45o to the x axis. (That is to say, they are the asymptotes of the rectangular hyperbola.) Then the equation to the rectangular hyperbola referred to its asymptotes as coordinate axes is found by the substitutions − = ' ' 45 cos 45 sin 45 sin 45 cos o o o o y x y x 2.5.16 into x2 − y2 = a2 . This results in the equation x y a c ' ' , = = 1 2 2 2 where , 2 / a c = 2.5.17 for the equation to the rectangular hyperbola referred to its asymptotes as coordinate axes. The geometric interpretation of c is shown in figure II.32, which is drawn for c = 1, and we have called the coordinate axes x and y. The length of the semi transverse axis is c 2. 41 The simple equation y = 1/x is a rectangular hyperbola and indeed it is this equation that is shown in figure II.32. It is left to the reader to show that the parametric equations to the rectangular hyperbola xy = c2 (we have dropped the primes) are x = ct, y = c/t, that lines of the form y mx c m = ± − 2 are tangent to xy = c2 (figure II.35, drawn with slopes from 90o to 180o in steps of 5o), and that the tangent at (x1 , y1 ) is x1y + y1x = 2c . Polar equation to the hyperbola. We found the polar equations to the ellipse and the parabola in different ways. Now go back and look at both methods and use either (or both) to show that the polar equation to the hyperbola (focus as pole) is r l e = + 1 cos .
θ 2.5.18 This is the polar equation to any conic section - which one being determined solely by the value of e. You should also ask yourself what is represented by the equation r l e = − 1 cos .
θ 2.5.19 42 Try sketching it for different values of e. 2.6 Conic Sections. We have so far defined an ellipse, a parabola and a hyperbola without any reference to a cone. Many readers will know that a plane section of a cone is either an ellipse, a parabola or a hyperbola, depending on whether the angle that the plane makes with the base of the cone is less than, equal to or greater than the angle that the generator of the cone makes with its base. However, given the definitions of the ellipse, parabola and hyperbola that we have given, proof is required that they are in fact conic sections. It might also be mentioned at this point that a plane section of a circular cylinder is also an ellipse. Also, of course, if the plane is parallel with the base of the cone, or perpendicular to the axis of the cylinder, the ellipse reduces to a circle. K F1 F2 P Q1 Q2 C2 C1 FIGURE II.36 43 A simple and remarkable proof can be given in the classical Euclidean "Given. Required. Construction. Proof. Q.E.D." style. Given: A cone and a plane such that the angle that the plane makes with the base of the cone is less than the angle that the generator of the cone makes with its base, and the plane cuts the cone in a closed curve K. Figure II.36. Required: To prove that K is an ellipse. Construction: Construct a sphere above the plane, which touches the cone internally in the circle C1 and the plane at the point F1 . Construct also a sphere below the plane, which touches the cone internally in the circle C2 and the plane at the point F2 . Join a point P on the curve K to F1 and to F2 . Draw the generator that passes through the point P and which intersects C1 at Q1 and C2 at Q2 . Proof: PF1 = PQ1 (Tangents to a sphere from an external point.) PF2 = PQ2 (Tangents to a sphere from an external point.) ∴ PF1 + PF2 = PQ1 + PQ2 = Q1 Q2 and Q1Q2 is independent of the position of P, since it is the distance between the circles C1 and C2 measured along a generator. ∴ K is an ellipse and F1 and F2 are its foci. Q.E.D. A similar argument will show that a plane section of a cylinder is also an ellipse. The reader can also devise drawings that will show that a plane section of a cone parallel to a generator is a parabola, and that a plane steeper than a generator cuts the cone in a hyperbola. The drawings are easiest to do with paper, pencil, compass and ruler, and will require some ingenuity. While I have seen the above proof for an ellipse in several books, I have not seen the corresponding proofs for a parabola and a hyperbola, but they can indeed be done, and the reader should find it an interesting challenge. If the reader can use a computer to make the drawings and can do better than my poor effort with figure II.36, s/he is pretty good with a computer, which is a sign of a misspent youth. 2.7 The General Conic Section The equation x a y b 2 2 2 2 1 + = 2.7.1 44 represents an ellipse whose major axis is along the x axis and whose centre is at the origin of coordinates. But what if its centre is not at the origin, and if the major axis is at some skew angle to the x axis? What will be the equation that represents such an ellipse? Figure II.37. If the centre is translated from the origin to the point (p, q), the equation that represents the ellipse will be found by replacing x by x − p and y by y − q. If the major axis is inclined at an angle θ to the x axis, the equation that represents the ellipse will be found by replacing x by x y cos sin θ θ + and y by −x sin θ + y cos θ . In any case, if the ellipse is translated or rotated or both, x and y will each be replaced by linear expressions in x and y, and the resulting equation will have at most terms in x2, y2, xy, x, y and a constant. The same is true of a parabola or a hyperbola. Thus, any of these three curves will be represented by an equation of the form ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. 2.7.2 (The coefficients a and b are not the semi major and semi minor axes.) The apparently random notation for the coefficients arises because these figures are plane sections of three-dimensional surfaces (the ellipsoid, paraboloid and hyperboloid) which are described by terms involving the coordinate z as well as x and y. The customary notation for these three-dimensional surfaces is very systematic, but when the terms in z are removed for the two- dimensional case, the apparently random notation a, b, c, f, g, h remains. In any case, the above equation can be divided through by the constant term without loss of generality, so that the equation to an ellipse, parabola or hyperbola can be written, if preferred, as ax2 + 2hxy + by2 + 2gx + 2fy + 1 = 0. 2.7.3 Is the converse true? That is, does an equation of this form always necessarily represent an ellipse, parabola or hyperbola? Not quite. For example, 6 17 12 0 2 2 x xy y x y + − − − + = 2.7.4 . (p , q ) y x θ FIGURE II.37 45 represents two straight lines (it can be factored into two linear terms - try it), while 2 4 4 4 4 0 2 2 x xy y x − + − + = 2.7.5 is satisfied only by a single point. (Find it.) However, a plane section of a cone can be two lines or a single point, so perhaps we can now ask whether the general second degree equation must always represent a conic section. The answer is: close, but not quite. For example, 4 12 9 14 21 6 0 2 2 x xy y x y + + + + + = 2.7.6 represents two parallel straight lines, while x y x y 2 2 3 4 15 0 + + + + = 2.7.7 cannot be satisfied by any real (x,y). However, a plane can intersect a cylinder in two parallel straight lines, or a single straight line, or not at all. Therefore, if we stretch the definition of a cone somewhat to include a cylinder as a special limiting case, then we can say that the general second degree equation in x and y does indeed always represent a conic section. Is there any means by which one can tell by a glance at a particular second degree equation, for example 8 10 3 2 4 2 0 2 2 x xy y x y + − − − − = , 2.7.8 what type of conic section is represented? The answer is yes, and this one happens to be a hyperbola. The discrimination is made by examining the properties of the determinant ∆= a h g h b f g f c 2.7.9 I have devised a table after the design of the dichotomous tables commonly used by taxonomists in biology, in which the user is confronted by a couplet (or sometimes triplet) of alternatives, and is then directed to the next part of the table. I shall spare the reader the derivation of the table; instead, I shall describe its use. In the table, I have used the symbol a to mean the cofactor of a in the determinant, h the cofactor of h, g the cofactor of g, etc. Explicitly, these are 46 , 2 f bc a − = 2.7.10 , 2 g ca b − = 2.7.11 , 2 h ab c − = 2.7.12 , af gh f − = 2.7.13 bg hf g − = 2.7.14 and .
ch fg h − = 2.7.15 The first column labels the choices that the user is asked to make. At the beginning, there are two choices to make, 1 and 1' The second column says what these choices are, and the fourth column says where to go next. Thus, if the determinant is zero, go to 2; otherwise, go to 5. If there is an asterisk in column 4, you are finished. Column 3 says what sort of a conic section you have arrived at, and column 5 gives an example. No matter what type the conic section is, the coordinates of its centre are ( ) c f c g / , / and the angle θ that its major or transverse axis makes with the x axis is given by tan .
2 2 θ = − h a b 2.7.16 Thus if x is first replaced with x g c + / and y with y f c + / , and then the new x is replaced with x y cos sin θ θ − and the new y with x y sin cos , θ θ + the equation will take the familiar form of a conic section with its major or transverse axis coincident with the x axis and its centre at the origin. Any of its properties, such as the eccentricity, can then be deduced from the familiar equations. You should try this with equation 2.7.8. 47 Key to the Conic Sections 1 ∆=0 2 1' ∆≠0 5 2 0 > c Point 0 2 2 2 2 2 2 = + − + − x y xy x 2' 0 = c 3 2" 0 < c Two nonparallel straight lines 4 3 0 = h Straight line 0 9 6 12 4 4 2 2 = + + + + + y x y xy x 3' 0 ≠ h Two parallel straight lines 0 6 21 14 9 12 4 2 2 = + + + + + y x y xy x 4 0 = + b a Two perpendicular straight lines 0 2 8 6 5 6 2 2 = − + + − + y x y xy x 4' 0 ≠ + b a Two straight lines, neither parallel nor perpendicular 0 12 13 34 6 2 2 = − + + − − y x y xy x 5 0 > c 6 5' 0 = c Parabola 0 19 101 18 4 12 9 2 2 = + − − + − y x y xy x 5" 0 < c 8 6 0 > ∆ a Nothing 0 15 4 3 2 2 = + + + + y x y x 6' 0 < ∆ a 7 7 a=b and h=0 Circle 0 9 8 6 2 2 = + − − + y x y x 7' Not so Ellipse 0 71 58 44 11 4 14 2 2 = + − − + − y x y xy x 8 a+b=0 Rectangular hyperbola 0 100 28 10 7 48 7 2 2 = + − + − − y x y xy x 8' a+b≠0 Hyperbola (not Rectangular) 0 2 4 2 3 10 8 2 2 = − − − − + y x y xy x 48 2.8 Fitting a Conic Section Through Five Points. Figure II.38 shows the five points A(1,8), B(4,9), C(5,2), D(7,6), E(8,4). Problem: Draw a conic section through the five points. The first thing to notice is that, since a conic section is of the form ax2 + 2hxy + by2 + 2gx + 2fy + 1 = 0 2.8.1 (remember that there is no loss of generality by taking the constant term to be 1), five points are necessary and sufficient to define a conic section uniquely. One and only one conic section can be drawn through these five points. We merely have to determine the five coefficients. The most direct (but not the fastest or most efficient) way to do this is to substitute each of the (x , y) pairs into the equation in turn, thus obtaining five linear equations in the five coefficients. There is a better way. We write down the equations for the straight lines AB, CD, AC and BD. Let us call these equations α = 0, β = 0, γ = 0 and δ = 0 respectively (figure II.39). Then αβ = 0 is the equation that represents the two straight lines AB and CD, and γδ = 0 is the equation that represents the two straight lines AC and BD. The equation αβ λγδ + = 0, where λ is an arbitrary constant, is a second degree equation that represents any conic section that passes through the points A, B, C and D. By inserting the coordinates of E in this equation, we can find 49 the value of that forces the equation to go through all five points. This model of unclarity will become clear on following an actual calculation for the five points of the present example. The four straight lines are α = 0: x − 3y + 23 = 0 β = 0: 2x − y − 8 = 0 γ = 0: 3x + y − 19 = 0 δ = 0: x + y − 13 = 0 The two ;pairs of lines are αβ = − + + + − = 0 2 7 3 38 184 0 2 2 : x xy y x y γδ = + + − − + = 0 3 5 2 58 45 247 0 2 2 : x xy y x y and the family of conic sections that passes through A, B, C and D is α=0 β=0 γ=0 δ=0 50 αβ + λγδ = 0: ( ) ( ) ( ) ( ) ( ) .
2 3 7 5 3 2 38 58 1 45 184 247 0 2 2 + − − + + + − + − − + = λ λ λ λ λ λ x xy y y Now substitute x = 8, y = 4 to force the conic section to pass through the point E. This results in the value λ = 76 13 . The equation to the conic section passing through all five points is therefore 508x2 + 578xy + 382y2 − 7828x − 6814y + 32760 = 0 We can, if desired, divide this equation by 2 (since all coefficients are even), or by 32760 (to make the constant term equal to 1) but, to make the analysis that is to follow easier, I choose to leave the equation in the above form, so that the constants f, g and h remain integers. The constants have the values a = 508, b = 382, c = 32760, f = −3407, g = −3914, h = 289 and the cofactors have the values a = 906 671, b = 1 322 684, c = 110 535 f = 599 610, g = 510 525, h = 867 358 Let us consult the dichotomous table. The value of the determinant is ∆= + + aa bb cc (or hh bb ff + + , or gg ff cc + + ; try all three sums to check for arithmetic mistakes). It comes to ∆= −419 939 520, so we proceed to option 5. c > 0,so we proceed to option 6. a and ∆ have opposite signs, so we proceed to 7. a does not equal b, nor is h equal to zero. Therefore we have an ellipse. It is drawn in figure II.40. The centre of the ellipse is at (4.619, 5.425), and its major axis is inclined at an angle 128o 51' to the x-axis. If we now substitute x + 4.619 for x and y + 5.425 for y, and then substitute x cos 128o 51' + y sin 1280 51' for the new value of x and −x sin 128o 51' + y cos 128o 51' for the new value of y, the equation will assume its the familiar form for an ellipse referred to its axes as coordinate axes and its centre as origin. 51 2.9 Fitting a Conic Section Through n Points What is the best ellipse passing near to the following 16 points? ( 1,50) (11,58) (20,63) (30,60) (42,59) (48,52) (54,46) (61,42) (61,19) (45,12) (35,10) (25,13) (17,17) (14,22) ( 5,29) ( 3,43) This is answered by substituting each point (x , y) in turn in the equation ax2 + 2hxy + by2 +2gx + 2fy + 1 = 0, 2.9.1 thus obtaining 16 equations in the coefficients a, h, b, g, f. (The constant term can be taken to be unity.) These are the equations of condition. The five normal equations can then be set up and solved to give those values for the coefficients that will result in the sum of the squares of the residuals being least, and it is in that sense that the "best" ellipse results. The details of the method are given in the chapter on numerical methods. The actual solution for the points given above is left as an exercise for the energetic. It might be thought that we are now well on the way to doing some real orbital theory. After all, suppose that we have several positions of a planet in orbit around the Sun, or several positions of 52 the secondary component of a visual binary star with respect to its primary component; we can now fit an ellipse through these positions. However, in a real orbital situation we have some additional information as well as an additional constraint. The additional information is that, for each position, we also have a time. The constraint is that the orbit that we deduce must obey Kepler's second law of planetary motion - namely, that the radius vector sweeps out equal areas in equal times. We shall have to await Part II before we get around actually to computing orbits. |
9091 | https://www.georgebrown.ca/sites/default/files/uploadedfiles/tlc/_documents/Common_Factoring.pdf | Common Factoring Tutoring and Learning Centre, George Brown College 2014 www.georgebrown.ca/tlc Introduction to Factoring Factoring is the OPPOSITE operation of expanding algebraic expressions. To factor an algebraic expression means to find the numbers, monomials, binomials or polynomials that multiplied together result in the given algebraic expression. Common Factoring Step 1: Find the greatest common factor of all terms in the algebraic expression. Consider the numbers and variables making up each term. Step 2: Write the common factor in front of the brackets. In brackets, write the algebraic expression resulting from dividing EACH term by the common factor. Example 1: Factor 16xy2 + 20x2y – 4x3y2. Step 1: Find the greatest common factor of 16xy2, 20x2y and 4x3y2. Look at the numbers 16, 20 and 4. The greatest common factor of these numbers is 4, since all of these numbers can be divided by 4 evenly. 16 ÷ 4 = 4 20 ÷ 4 = 5 4 ÷ 4 = 1 Look at the variables in each term, xy2, x2y and x3y2. The greatest common factor of these variables is xy since all of these terms can be divided by xy evenly. xy2 ÷ xy = y x2y ÷ xy = x x3y2 ÷ xy = x2y (Note: the terms are divided according to exponent rules). Therefore, the greatest common factor is 4xy. Step 2: Write the greatest common factor in front of brackets. Determine the algebraic expression in brackets by dividing each term in the given algebraic expression by the greatest common factor. 4xy (16xy2 + 20x2y – 4x3y2) 4xy 4xy 4xy = 4xy (4y + 5x – x2y) Common Factoring Tutoring and Learning Centre, George Brown College 2014 www.georgebrown.ca/tlc Step 3 (Optional): Double check the answer by expanding 4xy(4y + 5x – x2y). The answer should be the algebraic expression that was given in the question. 4xy(4y + 5x – x2y) = 4xy(4y) + 4xy(5x) + 4xy(–x2y) = 16xy2 + 20x2y – 4x3y2 Example 2: Factor –144p5q – 12p3 −6p. Step 1: Find the greatest common factor of –144p5q, –12p3 and −6p. The greatest common factor of –144, –12 and −6 is −6. –144 ÷ (−6) = 24 –12 ÷ (−6) = 2 −6 ÷ (−6) = 1 The greatest common factor of p5q, p3 and p is p. p5q ÷ p = p4q p3 ÷ p = p2 p ÷ p = 1 Therefore, the greatest common factor is −6p. Step 2: Write the greatest common factor in front of brackets. Determine the algebraic expression in brackets by dividing each term in the given algebraic expression by the greatest common factor. −6p (–144p5q – 12p3 − 6p) −6p −6p −6p = −6p (24p4q + 2p2 + 1) Practice questions: 1. Factor the following algebraic expressions. Check your work by expanding. a) 24k3s2 + 12k2s5 – 48k2s3 b) 10d4c3 + 50d3c4 – 40d4c2 + 5d3c2 c) 12x3 + 6x – 2y d) 81c4 + c2 + 10d e) 36q3r + 90rq2 + 81rq f) 3x3 – 6x2 + 20y – 4y2 Answers: 1. a) 12k2s2 (2k + s3 –4s) b) 5d3c2(2dc + 10c2 –8d +1) c) 2(6x3 + 3x –y) d) c2 (81c2 + 1) + 10d e) 9qr(4q2 +10q +9) f) 3x2 (x – 2) + 4y(5–y) |
9092 | https://www.youtube.com/watch?v=5qcHb_WFUWQ | Magic Square 4 x 4... can you fill in the blanks CORRECTLY?
Math Puzzles
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Welcome to our exciting new math puzzle series! In this episode, we're diving into the intriguing challenge of 4 x 4 Magic Squares. The goal is simple yet challenging: fill in numbers 1-16 so that every row, column, and long diagonal add up to the same number.
00:00 Introduction
00:05 Question
00:30 Explanation
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Let's put our math skills to the test and see how creative we can get with just four 6s! Happy puzzling!
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Transcript:
Introduction [Music] Question what's up everybody and welcome back to another math puzzle in this one the question is as follows find the missing numbers 1 through 16 for each Square so that every row column and diagonal have the same sum so this is a super fun magic square question hit that pause button and see if you can figure it out when you're ready hit play and I'll give you the explanation Explanation so the first thing we want to do here is figure out what that sum is going to be that's going to be consistent throughout every row column and diagonal and if you'll notice every row column and diagonal and when I say diagonal I mean just the long diagonals it's going to consist of four numbers so the way to figure out what these sums are going to be is we're going to add up all the numbers 1 through 16 and divide by four so as we line all these up now we can add them together first 1 through 9 adds up to 45 and then here we're going to do it in little chunks these three are going to add up to 33 these two are going to add up to 27 and these two are going to add up to 31 so now we add these together 27 + 33 is 60 uh 45 + 31 is 76 and we add these together and we get a nice 136 136 divided by 4 we can do that mentally by dividing by two two times that's what I like to do so 136 divid 2 is 68 divided 2 again is 34 so 34 four is the magic number now since we know everything that's got to add up to 34 and we're using each number only once we can cross off the numbers that are already in here so now we know that these numbers are off the table let's plug and chug and see if we can figure out what the remaining spaces should be I like to start with this row because we have three out of four numbers so if I already know that 6 + 13 is 19 + 3 is 22 right if we're trying to get to that magic number of 34 we need a remaining 12 right here now the next place I'm going to look is this row right here with three and nine because I know that three and nine add up to 12 so to get to 34 I'm missing 22 so if we're trying to get up to 22 and two spots I I don't have many options right I could say 16 and six but six is gone I could say 15 and seven those are possible viable options I could also say 14 and 8 and after that we don't have many options so we're going to say it's one of those two it's either 14 and 8 or 15 and 7 another place that's worthwhile looking is these two two inside spots because I already have 10 and 13 and those are adding up to 23 so to get to 34 I only need 11 right so two spaces that are filled by 11 obviously these big dogs are out pretty much my only options are seven and four right there's no other remaining combination so we know we've got 7 4 74 now this is pretty interesting because I know for a fact Seven's got to be one of these two spaces meaning this is out this is out so now I know it's got to be 74 7 4 and 148 148 we'll figure out which number goes exactly where as we flesh more of this out as we go because I have these limited already like so I'm going to move 14 or excuse me 4 7 14 and 8 so now the remaining numbers we have for the remaining spots are 1 5 15 and 16 so looking at this first row knowing that we already got 6 and 11 we're going to need 17 more right 17 + 17 is 34 of the remaining values the only thing that's going to give me 17 is 16 and 1 one so we know that we got 16 or one here 16 or one here and of course that lets me cross those off and then last but not least we got 15 and five and it makes total sense that they got to go here and here because 15 and 5 add up to 20 12 and 2 add up to 14 20 + 14 is again 34 so the first place I'm going to look to make some choices is I'm going to look at these values right here the reason why is because we've already got nine and two down here that are adding up to 11 meaning that I need these two spots to add up to 23 the only possible way I can see that is with 16 and 7 so that means one is out here four is out here and if four is out here we got to have four here if 16 is over here we got to have one here now if I know that this is adding to 11 right to get to 34 I need 23 of the options listed below 14 and 5 won't work 14 and 15 won't work but 8 and 15 will do the job so we've got eight and 15 and if I got eight here that means I got to have 14 here if I have 15 here I got to have five here so now here we have our final solution with all of the magic squares filled in and you can see any row column or long diagonal it's going to have that equivalent sum of that magic number 34 I hope you guys enjoyed this math puzzle and if you did please click that like button if you want to see more math puzzles on the regular make sure to click subscribe thank you guys so much for joining see you in the next video take it easy [Music] |
9093 | https://www.ijmedicine.com/index.php/ijam/article/view/487 | Screening of syphilis using rapid plasma reagin test in apparently healthy population in Madhya Pradesh: a 6 years study | International Journal of Advances in Medicine
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Vol. 1 No. 3 (2014): October-December 2014/
Original Research Articles
Screening of syphilis using rapid plasma reagin test in apparently healthy population in Madhya Pradesh: a 6 years study
Authors
Savita Bharat Jain Department of Microbiology, Gajra Raja Medical College, Gwalior, Madhya Pradesh, India
K. P. Ranjan Department of Microbiology, Gajra Raja Medical College, Gwalior, Madhya Pradesh, India
Shashi Gandhi Department of Microbiology, Gajra Raja Medical College, Gwalior, Madhya Pradesh, India
Keywords:
Syphilis, Rapid plasma reagin, Syphilis in apparently healthy population
Abstract
Background: To analyze the prevalence of syphilis in apparently healthy population in Madhya Pradesh during last 6 years.
Methods: Serum samples were collected from Departments of Obstetrics and Gynecology, Skin and Venereal diseases, Medicine, Surgery, etc., of the period of 6 years. All the serum samples were screened for syphilis by rapid plasma reagin (RPR) test using standard methods.
Results: Out of 21,742 samples, 133(0.62%) serum samples were found to be positive. The highest seroprevalence was observed in the group aged 25-31 years (0.89%) and lowest in 18-24 years (0.31%).
Conclusion: RPR testing in this populations group plays a special value in the diagnosis, treatment and prevention of syphilis.
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References
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Willey JM, Sherwood LM, Woolverton CJ. Prescott, Harley and Klein’s Microbiology. 7th Edition. Boston: McGraw Hill Companies Inc.; 2008: 976-7.
Nagi AM, Allah HAW, Khalid OM. Seroprevalence of syphilis among pregnant women in the Tri-capital, Khartoum, Sudan. Res J Med Sci. 2008;3:48-52.
Olokoba AB, Olokoba LB, Salawu FK, Danburam A, Desalu OO, Badung LH, et al. Syphilis in voluntary blood donours in North-Eastern, Nigeria. Eur J Sci Res. 2009;31(3):235-40.
Lukehart SA. Syphilis. In: Fauci AS, Braunwald E, Kasper DL, Hauser S, Longo D, Jameson J, et al., editors. Harrison’s Principles of Internal Medicine. 17th Edition. New York: McGraw Hill, 2008: 1038 45.
Ophori EA, Atanunu O, Johnny EJ, Adu M. Seroprevalence of syphilis in apparently healthy students from a tertiary institution in Benin City, Nigeria. Jpn J Infect Dis. 2010;63(6):437-9.
Genç M, Ledger WJ. Syphilis in pregnancy. Sex Transm Infect. 2000;76(2):73-9.
Van Dyck E, Van de Velden L, Ndoye I, Piot P, Meheus A. Evaluation of the rapid plasma reagin “teardrop” card test for screening of syphilis in field conditions. Sex Transm Dis. 1993;20(4):194-7.
World Health Organization. Treponemal Infections. Technical Reports Series 674. Geneva: World Health Organization; 1999.
Young H. Treponema: serological tests for syphilis. In: Collee JG, Duguid JP, Fraser Ag, Marmion BP, editors. Mackie and McCartney Practical Medical Microbiology. 14th Edition. London: Churchill Livingstone; 2007: 549-58.
Hook EW III. Syphilis. In: Goldman L, Schafer AI, editors. Goldman’s Cecil Medicine. 24th Edition, Chapter 327. Philadelphia, PA: Saunders Elsevier; 2011.
Koneman EW. Spirochetal infections. Koneman’s Textbook of Diagnostic Microbiology. 6th Edition. Philadelphia: Lippincott William & Wilkins; 2001: 126-33.
Pramod JR. Sexually Transmitted Diseases (STDs): Principles of Practical oral Medicine and Patients Evaluation. New Delhi: CBS Publications and Distributors; 2004: 308-11.
Larsen SA, Hambie EA, Pettit DE, Perryman MW, Kraus SJ. Specificity, sensitivity, and reproducibility among the fluorescent treponemal antibody-absorption test, the microhemagglutination assay for Treponema pallidum antibodies, and the hemagglutination treponemal test for syphilis. J Clin Microbiol. 1981;14(4):441-5.
Johnson RC, Hughes CA. Spirochetes. Clinical and Pathogenic Microbiology. 2nd Edition. St. Louis: Mosby; 1987: 529-40.
Cunningham FG, Leveno KJ, Bloom SL, Hauth JC, Gilstrop LC, Wenstrom KD. Sexually transmitted diseases. In: Cunningham FG, Leveno KJ, Bloom SL, Hauth JC, Gilstrop LC, Wenstrom KD, editors. Williams Obstetrics. 22nd Edition. United States: McGraw Hill; 2001: 1301-25.
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Jain, S. B., Ranjan, K. P., & Gandhi, S. (2017). Screening of syphilis using rapid plasma reagin test in apparently healthy population in Madhya Pradesh: a 6 years study. International Journal of Advances in Medicine, 1(3), 238–240. Retrieved from
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9094 | https://pubmed.ncbi.nlm.nih.gov/23608817/ | Specificities of lung adenocarcinoma in women who have never smoked - PubMed
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Multicenter Study
J Thorac Oncol
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. 2013 Jul;8(7):923-9.
doi: 10.1097/JTO.0b013e3182904dfb.
Specificities of lung adenocarcinoma in women who have never smoked
Julien Mazières1,Isabelle Rouquette,Benoit Lepage,Julie Milia,Laurent Brouchet,Nicolas Guibert,Michèle Beau-Faller,Pierre Validire,Paul Hofman,Pierre Fouret
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Affiliation
1 Unité d'Oncologie Thoracique, Service de Pneumologie, Hôpital Larrey-CHU de Toulouse, Université de Toulouse III (Paul Sabatier), Cedex 9, France. mazieres.j@chu-toulouse.fr
PMID: 23608817
DOI: 10.1097/JTO.0b013e3182904dfb
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Multicenter Study
Specificities of lung adenocarcinoma in women who have never smoked
Julien Mazières et al. J Thorac Oncol.2013 Jul.
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. 2013 Jul;8(7):923-9.
doi: 10.1097/JTO.0b013e3182904dfb.
Authors
Julien Mazières1,Isabelle Rouquette,Benoit Lepage,Julie Milia,Laurent Brouchet,Nicolas Guibert,Michèle Beau-Faller,Pierre Validire,Paul Hofman,Pierre Fouret
Affiliation
1 Unité d'Oncologie Thoracique, Service de Pneumologie, Hôpital Larrey-CHU de Toulouse, Université de Toulouse III (Paul Sabatier), Cedex 9, France. mazieres.j@chu-toulouse.fr
PMID: 23608817
DOI: 10.1097/JTO.0b013e3182904dfb
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Introduction: No clear data are available on the high rate of tobacco-independent lung cancer in women. We hypothesize that genetic events or hormonal factors may be partly involved.
Methods: We aimed to compare clinical, pathological, and biological characteristics of lung cancer in two cohorts of women: smokers and never-smokers. A total of 140 women (63 never-smokers and 77 former/current smokers) with adenocarcinoma, were included in this study.
Results: The never-smokers were characterized by a higher age (67 versus 58.7 years; p < 0.0001) and a higher frequency of lepidic features (60.3% versus 37.7%; p = 0.008) compared with smokers. We observed differential genetic alteration repartition in women according to their tobacco status: 50.8% of never-smokers displayed an epidermal growth factor receptor (EGFR) mutation versus 10.4% of smokers (p < 0.001). In contrast, K-Ras was more frequently mutated in smokers (33.8%) than in never-smokers (9.5%; p = 0.001). We also observed a higher percentage of estrogen receptors (ER) α expression (p = 0.03; and p = 0.008 with two different antibodies) in patients who never smoked when compared with smokers. There was no significant difference in ERβ and progesterone receptors between the groups. Finally, ERα expression was correlated with the presence of an EGFR mutation.
Conclusions: This study suggests that when lung cancer occurs in women who have never smoked, it is more frequently associated with an EGFR mutation and ERα expression, with a correlation between both markers. These findings underline the possibility of treating women who have never smoked by targeting both hormonal factors and genetic abnormalities.
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Differences in driver genes between smoking-related and non-smoking-related lung cancer in the Chinese population.Gou LY, Niu FY, Wu YL, Zhong WZ.Gou LY, et al.Cancer. 2015 Sep 1;121 Suppl 17:3069-79. doi: 10.1002/cncr.29531.Cancer. 2015.PMID: 26331813 Review.
Differential expression of biomarkers in men and women.Planchard D, Loriot Y, Goubar A, Commo F, Soria JC.Planchard D, et al.Semin Oncol. 2009 Dec;36(6):553-65. doi: 10.1053/j.seminoncol.2009.09.004.Semin Oncol. 2009.PMID: 19995647 Review.
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9096 | https://www.thejcdp.com/doi/10.5005/jp-journals-10024-1240 | T Shankar et al
842 JAYPEE
ORIGINAL RESEARCH
A Comparative Evaluation of the Dimensional Accuracy of Heat Polymerized Acrylic Resin Denture Base Clamped by the Conventional Method and by New-press Technique and Cured by Long Curing Cycle: An
in vitro Study T Shankar, Snigdha Gowd, Syed Tauqheer Ahmed, V Vinod, M Vikas Goud, N Venugopal Rao
10.5005/jp-journals-10024-1240
ABSTRACT
Aim : The aim of this study is to evaluate the dimensional accuracy of heat polymerized acrylic resin denture base clamped by the conventional method and by new-press technique and cured by long curing cycle.
Materials and methods: In this study, a total of 60 standardized maxillary record bases were fabricated with seven reference points as follows: • Point A: Incisive papilla, • Point B and C: Canine region on either side • Point E and G: Midpoint of tuberosities on either side • Point F: Midpoint of the line joining the two tuberosities • Point D: Midpoint between the line joining A and F
Group A : Ten maxillary record bases were fabricated by conventional clamping method and cured by long curing cycle.
Group A ’: Ten maxillary record bases were fabricated by New Press or RS tension clamping method and cured by long curing cycle. The distances between the reference points, i.e. A-B, A-C, A-D, D-F, B-E, C-G, E-F, F-G, B-D, D-G, C-D, D-E of all three thermoplastic denture base plates were measured and recorded with the help of travelling microscope and were used for comparison with the measured and recorded readings of processed acrylic denture bases. The data obtained was analyzed by using the one-way analysis of variance and HSD Multiple Comparison Test.
Results : The overall results of the study indicate that among all the denture bases cured by the two clamping systems and the long curing cycle, group A‘ were the most dimensionally stable, followed by control group A.
Conclusion : The study concluded that the denture bases fabricated by the New Press method using the long curing cycle would produce the most dimensionally stable denture bases.
Keywords: Long-curing cycle, New-press, Tension plate clamping.
How to cite this article: Shankar T, Gowd S, Ahmed ST, Vinod V, Goud MV, Rao NV. A Comparative Evaluation of the Dimensional Accuracy of Heat Polymerized Acrylic Resin Denture Base Clamped by the Conventional Method and by New-press Technique and Cured by Long Curing Cycle: An in vitro Study. J Contemp Dent Pract 2012;13(6):842-849.
Source of support: Nil
Conflict of interest: None declared
INTRODUCTION
Polymethylmethacrylate (PMMA) has become the most used material for denture bases since its introduction in 1937. It has excellent esthetic properties, adequate strength, low water sorption, low solubility, lack of toxicity and can be constructed by simple molding and processing technique. 1 A certain lack of dimensional accuracy as of dimensional change in denture base, has been accepted as one of its prime disadvantages. This dimensional change in the denture base can be influenced by polymerization shrinkage, the resin flasking method and time-temperature correlation during the polymerization procedure. 2,3 When the acrylic resin flasking procedure is performed under routine laboratory conditions, the flask is placed in traditional clamps after final pressing in hydraulic press. This condition may lead to release of residual internal stresses from the acrylic resin dough before polymerization. The release of residual internal stress along with polymerization shrinkage, thermal contraction during flask cooling and strain accompanying stress release during deflasking may cause diminished adaptation of the denture to the tissues. 4Rafael and Saide introduced an improved processing technique (New-Press Technique) with the help of RS tension plate clamping, which proved to decreases the magnitude of this gap. 7A Comparative Evaluation of the Dimensional Accuracy of Heat Polymerized Acrylic Resin Denture Base Clamped
The Journal of Contemporary Dental Practice, November-December 2012;13(6):842-849
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JCDP
Earlier studies, 7 measured the adaptability of the bases with the cast, only in the posterior palatal region and the fact being that the dimensional changes in the denture base happens overall, a study only in one area would not justify the results. The dimensional accuracy of the denture bases processed by different clamping methods, in different areas and in different directions has not been extensively studied. The purpose of this study was to investigate the dimensional accuracy of denture bases processed with the conventional heat-polymerizing acrylic resin and the New-Press Technique (Rafael and Saide) by means of long-curing cycle in different areas and different directions.
MATERIALS AND METHODS
The materials and methods used in this study have been described in the following order: 1. Heat polymerizing acrylic resin (DPI Heat Cure) TM ,Dental products of India Ltd., Batch No 6103). 2. Thermoplastic sheet 5 × 5 ”, 4 mm thickness (Ultradent) for making standard denture base plate (Fig. 1). 3. Transparent thermoplastic sheet 5 × 5 ”, 2 mm thick for making transfer template (Fig. 2). 4. Metallic Edentulous maxillary cast made up of Aluminum (Fig. 3). 5. Petroleum jelly (Medi Soft).
Fabrication of Thermoplastic Denture Base Plate
A 4-mm thick thermoplastic sheet was vacuum pressed over the metallic edentulous maxillary cast (Fig. 4). One thermoplastic denture base plate (SP1) was fabricated (Fig. 5).
Fabrication of Thermoplastic Transfer Template
A 2-mm thick transparent thermoplastic sheet was vacuum pressed over the thermoplastic denture base plate placed over the metallic edentulous maxillary cast (Fig. 6).
Fabrication of RS Tension Plates
The assembly consisted of 2 iron plates, each of dimensions 150 × 40 × 8 mm with two screws (9 mm dia) soldered into the centre of each end of one plate (Fig. 7), and was named the lower plate. Two corresponding holes (10 mm dia) were drilled on another plate which formed the upper plate. During the definitive flask closure in the hydraulic bench press, the screws of the lower plate were fit into the holes of upper plate and after obtaining an adequate pressure of 1200 lbs, the screw nuts were strongly tightened. This procedure maintained constant metal to metal contact on the flask halves (Fig. 8).
Marking of Points on the Thermoplastic Denture base Plate and Transfer Template
Seven reference points were marked on the thermoplastic denture base plate SP1 (Fig. 9) with the help of 0.5 mm tip permanent marking pen. • Point A: Incisive papilla. • Point B and C: Canine region on either side. • Point E and G: Midpoint of tuberosity on either side • Point F: Midpoint of the line joining the two tuberosity • Point D: Midpoint between the line joining A and F. The transparent thermoplastic transfer template was placed over the marked thermoplastic denture base plate and both were together kept on the metallic maxillary edentulous cast (Fig. 10). The whole assembly was placed on the adjusting table of the milling machine (Girrbach, Germany) and was made parallel to the platform of the milling machine. Holes were drilled on to the transparent transfer template and the thermoplastic base plate SP1 with the help of 0.5 mm round diamond point (Fig. 11). The distances between the reference points, i.e. A-B, A-C, A-D, D-F, B-E, C-G, E-F, F-G, B-D, D-G, C-D, D-E
Fig. 1: Thermoplastic sheet
Fig. 2: Transparent thermoplastic sheet
T Shankar et al
844 JAYPEE
Fig. 3: Metallic eden mold
Fig. 4: Vacuum pressed thermoplastic sheet
Fig. 5: Thermoplastic denture baseplate
Fig. 6: Transparent transfer plate
Fig. 7: RS tension plate assembly
Fig. 8: RS tension plate assembly with flask
of all three thermoplastic denture base plates were measured and recorded with the help of traveling microscope and were used for comparison later with the processed acrylic denture bases (Fig. 12).
Method of Fabrication of Acrylic Denture Bases
Group A: Ten standardized maxillary denture bases were fabricated by conventional clamping method cured by long-curing cycle (74°C for 8 hours). 8A Comparative Evaluation of the Dimensional Accuracy of Heat Polymerized Acrylic Resin Denture Base Clamped
The Journal of Contemporary Dental Practice, November-December 2012;13(6):842-849
845
JCDP
Fig. 9: Reference points on SP1
Fig. 10: Milling machine
Fig. 11: Markings on SP1 and transfer template
Fig. 12: Traveling microscope
Group A ’: Ten standardized maxillary record bases were fabricated by RS Tension 7 clamping method/New-Press Technique cured by long curing cycle. 7
Measuring the Denture Bases
The processed denture bases (A, A ’) were stored in distilled water for 7 days and then the distances between the following points (A-B, A-C, A-D, D-F, B-E, C-G, E-F, F-G, B-D, D-G, C-D, D-E) were measured and recorded using a traveling microscope.
Statistical Analysis of the Data
All the 20 denture bases were measured, readings were obtained, and data was tabulated. These readings were compared with the standard denture base plate readings, i.e. A and A ’ with SP1 was done and analyzed with the help of one-way ANOVA test and HSD multiple comparison test.
RESULTS
Table 1 shows one-way analysis of variance (ANOVA) was used to analyze the mean distances between all 7 points on SP1, A and A‘, p-value (<0.0005) was found very highly significant at all distances between the two groups. Table 2 shows that HSD multiple comparison test was used to analyse the mean distances between all 7 points between the groups SP1 and A, SP1 and A ’ and A and A ’.p-value (<0.0005) was found very highly significant at distances C-G, E-F, F-G, D-G.
DISCUSSION
Heat activated acrylic denture base resins have been used for over half a century for its excellent esthetics and physical properties. The ease with which methylmethacrylate resins can be processed has made them the material of choice in complete denture fabrication. In PMMA resins, a certain lack of dimensional accuracy has been accepted as one of its prime disadvantages. This T Shankar et al
846 JAYPEE
Table 1: Comparison of SP1, A, A’ at various distances by one-way ANOVA test
Mean Std. Min. Max. pdeviation
A-B Sp1 15.400 0.0067 15.4 15.4 <0.0005 A15.364 0.0084 15.4 15.4 A’15.380 0.0133 15.4 15.4 A-C Sp1 15.8940 0.00843 15.88 15.90 <0.0005 A15.8480 0.00789 15.84 15.86 A’15.8600 0.0 15.86 15.86 A-D Sp1 23.00 0.007 23 23.0 <0.0005 A22.97 0.017 23 23.0 A’22.99 0.017 23 23.0 D-F Sp1 15.700 0.0067 15.7 15.7 <0.0005 A15.628 0.0103 15.6 15.6 A’15.634 0.0084 15.6 15.6 B-E Sp1 29.082 0.0140 29.1 29.1 <0.0005 A29.000 0.0133 29.0 29.0 A’29.010 0.0094 29.0 29.0 C-G Sp1 29.694 0.0084 29.7 29.7 <0.0005 A29.598 0.0169 29.6 29.6 A’29.630 0.0133 29.6 29.6 E-F Sp1 23.5000 0.00667 23.49 23.51 <0.0005 A23.4380 0.00422 23.43 23.44 A’23.4380 0.02150 23.40 23.46 F-G Sp1 24.5960 0.00516 24.59 24.60 <0.0005 A24.4920 0.01687 24.48 24.52 A’24.5100 0.00943 24.50 24.52 B-D Sp1 23.1500 0.00667 23.14 23.16 <0.0005 A23.0720 0.03910 23.00 23.10 A’23.0960 0.00843 23.08 23.10 D-G Sp1 29.0000 0.00667 28.99 29.01 <0.0005 A28.9240 0.01578 28.90 28.94 A’28.9400 0.01333 28.92 28.96 C-D Sp1 21.8940 0.00843 21.88 21.90 <0.0005 A21.7340 0.00843 21.72 21.74 A’21.7600 0.01886 21.74 21.78 D-E Sp1 27.7500 0.00667 27.74 27.76 <0.0005 A27.5400 0.01333 27.52 27.56 A’27.6600 0.02667 27.62 27.68
Table 2: Comparison of SP1-A, SP1-A ’, and A-A ’at various distances by HSD multiple comparison test
Mean difference p-value
A-B Sp1 A0.0360 <0.0005 A’0.0200 <0.0005 AA’–0.0160 0.003 A-C Sp1 A0.04600 <0.0005 A’0.03400 <0.0005 AA’–0.01200 0.001 A-D Sp1 A0.028 <0.0005 A’0.008 0.434 AA’–0.020 0.011 D-F Sp1 A0.0720 <0.0005 A’0.0660 <0.0005 AA’–0.0060 0.280 B-E Sp1 A0.0820 <0.0005 A’0.0720 <0.0005 AA’–0.0100 0.188 C-G Sp1 A0.0960 <0.0005 A’0.0640 <0.0005 AA’–0.0180 0.005 E-F Sp1 A0.06200 <0.0005 A’0.06200 <0.0005 AA’0.0320 <0.0005 F-G Sp1 A0.10400 <0.0005 A’0.08600 <0.0005 AA’–0.0260 <0.0005 B-D Sp1 A0.07800 <0.0005 A’0.05400 <0.0005 AA’–0.02400 0.074 D-G Sp1 A0.07600 <0.0005 A’0.06000 <0.0005 AA’–0.01600 0.022 C-D Sp1 A0.16000 <0.0005 A’0.13400 <0.0005 AA’–0.01100 00.020 D-E Sp1 A0.21000 <0.0005 A’0.09000 <0.0005 AA’–0.12000 <0.0005
dimensional change in the denture base can be influenced by polymerization shrinkage, the resin flasking method and time–temperature correlation during the polymerization procedure. When PMMA resin is mixed at recommended polymer: monomer ratio; density of the mass changes from 0.94 to 1.19g/cm 3 . This change in density results in volumetric shrinkage of 7% and a corresponding linear shrinkage of 2%. The greater the linear shrinkage the greater the discrepancy observed in the initial fit of the denture. In a study Dixon et al 9 reviewed the properties of denture base polymers and concluded that linear shrinkage generally is observed to be less than 1%. When the acrylic resin flasking procedure is performed by conventional method, the flask is placed in traditional clamps after final pressing in a hydraulic press. This condition may lead to the release of residual internal stresses from the acrylic resin dough before polymerization. The release of residual internal stress along with polymerization shrinkage, thermal contraction during flask cooling and strain accompanying stress release during deflasking may cause diminished adaptation of the denture to the tissues. 6An alternative technique was recognized by Rafael and Saide 6 , and they introduced an improved processing technique (New Press Technique) with the help of RS tension plate clamping, which proved to decrease the magnitude of the gap between the denture base and the tissue surface. This system suggests that the RS Tension plates maintained the acrylic dough under constant flask pressure conditions because the flask halves remained firmly in contact when the flask was removed from the hydraulic press. This condition inhibits the premature release of the residual internal stresses from the acrylic resin dough before polymerization shrinkage. Glass transition temperature of a polymer is an important physical property which may have a major effect on the dimensional stability of denture base materials. Jerolimov V et al 7 investigated variations in glass transition temperature A Comparative Evaluation of the Dimensional Accuracy of Heat Polymerized Acrylic Resin Denture Base Clamped
The Journal of Contemporary Dental Practice, November-December 2012;13(6):842-849 847
JCDP
in different curing cycles using thermo mechanical analysis and they concluded that different curing cycles produced variations in glass transition temperature of up to 20°C, and these variations resulted in polymerization shrinkage. Therefore, polymerization shrinkage, flask clamping method and time temperature correlation play a critical role in decreasing the probability of unfavorable dimensional changes in acrylic based denture base resins. In this study, a total of 20 standardized maxillary record bases were fabricated with seven reference points. Twenty maxillary record bases were fabricated by conventional clamping method, out of which 10 Acrylic denture bases were cured by long curing cycle (Group A) in conventional clamping technique and 10 Acrylic denture bases were cured by long curing cycle (Group A’) by New Press or RS tension clamping method. The distances between the reference points, i.e. A-B, A-C, A-D, D-F, B-E, C-G, E-F, F-G, B-D, D-G, C-D, D-E of all three thermoplastic denture base plates were measured and recorded with the help of traveling microscope and were used for comparison with the measured and recorded readings of processed acrylic denture bases. The data obtained was analyzed by using the one-way analysis of variance and HSD multiple comparison test. The comparative results of thermoplastic denture base plate SP1(control) (Table 3) and denture bases cured by long curing cycle using conventional method and New press method, i.e. A and A ’ respectively (Table 4), using one-way ANOVA showed that the p value (<0.0005) was very highly significant at all 12 distances, i.e. A-B, A-C, A-D, D-F, B-E, C-G, E-F, F-G, B-D, D-G, C-D, D-E between the two groups (see Table 1). It is concluded that the denture bases fabricated by New Press method were dimensionally more stable when compared to those fabricated with conventional method. This is in accordance with the previous study done by Consani RLX, Domitti SS, Consani S 6 who studied the effect of New Press system on the dimensional stability of denture bases. They suggested that New Press system is able to produce a more accurate denture base, minimizing the inaccuracies associated with the conventional clamping technique. The multiple comparison test between thermoplastic denture base plate SP1(control) and denture bases cured by long curing cycle using conventional method (A) and New press method (A ’) showed that the p value (<0.0005) was very highly significant at all the 12 distances A-B, A-C, D-F, B-E, C-G, E-F, F-G, B-D, D-G, C-D, D-E, A-D. The multiple comparison between A and A ’, showed that p-value (<0.0005) was very highly significant only at three distances, i.e. E-F, F-G and D-E and other nine distances were not significant (p > 0.05). This shows that the denture bases fabricated by long curing cycle and New Press method (C ’)were dimensionally more stable than the denture bases fabricated by conventional method using long curing cycle (C). The clamping method (New Press/conventional method) had the least impact on the dimensions of the denture bases cured by long curing cycle with only 3 distances showing significant changes. The mean value of the total of all the distances between the points, of standard denture base plate SP1was 23.219 mm. When compared to this, the similar mean value of group A ’ was 23.159 mm which shows that there was an overall shrinkage of 0.060 mm. This was the least among
Table 4: Average readings of processed acrylic denture bases A and A ’
A-B A-C A-D D-F B-E C-G E-F F-G B-D D-G C-D D-E
A15.364 15.848 22.972 15.628 29.00 29.598 23.438 24.492 23.072 28.924 21.734 27.54 15.36 15.86 22.96 15.64 29.02 29.58 23.44 24.48 23.00 28.92 21.74 27.52 15.36 15.84 22.96 15.62 29.00 29.58 23.43 24.48 23.08 28.94 21.74 27.54 15.36 15.85 22.98 15.62 29.00 29.60 23.44 24.50 23.08 28.92 21.73 27.56 A’15.38 15.86 22.992 15.634 29.01 29.63 23.438 24.51 23.096 28.94 21.76 27.66 15.36 15.86 23.00 15.64 29.00 29.64 23.44 24.50 23.10 28.92 21.74 27.62 15.38 15.86 22.96 15.63 29.02 29.61 23.45 24.50 23.08 28.94 21.78 27.64 15.38 15.86 23.00 15.64 29.00 29.62 23.44 24.51 23.10 28.96 21.78 27.68
Table 3: Average readings of standard denture base plates SP1
A-B A-C A-D D-F B-E C-G E-F F-G B-D D-G C-D D-E
SP1 15.40 15.95 23.00 15.70 29.10 29.70 23.55 24.56 23.13 29.05 21.86 27.74 15.39 15.93 22.98 15.68 29.08 29.68 23.54 24.59 23.12 29.04 21.88 27.75 15.40 15.95 22.97 15.67 29.10 29.70 23.55 24.58 23.13 29.05 21.88 27.74 15.38 15.94 23.00 15.70 29.06 29.69 23.53 24.59 23.14 29.03 21.87 27.73 15.40 15.95 23.00 15.70 29.10 29.70 23.55 24.57 23.13 29.05 21.88 27.72 T Shankar et al
848 JAYPEE
the groups, proving that the denture bases fabricated by the New Press method and long curing cycle demonstrated least shrinkage (Graph 1). The group A showed a mean value of 23.134, which showed 0.085 mm shrinkage. This shows that the denture bases fabricated by long curing cycle were the most dimensionally stable followed by the denture bases fabricated by short curing cycle and the denture bases fabricated by fast curing cycle being the least stable. This is because the long curing cycle is a well-controlled process which avoids the effects of uncontrolled temperature rise, boiling of monomer, denture base porosity and associated polymerization shrinkage, when compared to other two cycles. This is in accordance with the study done by Honorez P et al 10 who studied the effects of three processing cycles on some physical and chemical properties of heat cure acrylic resins and concluded that long curing cycle had resulted in low residual monomer content thereby indicating low polymerization shrinkage when compared to other two curing cycles. On overall analysis of data it was concluded that distance A-D (anterior slope of the palate) was the most dimensionally stable distance and distances E-F and F-G (posterior palatal area) were least dimensionally stable distances in all the groups. The reason for A-D being most dimensionally stable could be the presence of rugae which would impede the linear shrinkage due to better adaptability of the acrylic denture base to the underlying cast surface and the reason for E-F and F-G being least dimensionally stable is the large unsupported posterior palatal area this is in accordance with the previous study done by Firtell DN et al 5 who studied the distortion in posterior peripheral seal area related to processing temperature and concluded that the greatest effect of linear shrinkage on a maxillary denture is usually on the posterior palatal area, resulting in space between the palatal portion of the cast and the processed denture. It can be concluded that the denture bases fabricated by the New Press method using the long curing cycle would produce the most dimensionally stable denture bases.
CONCLUSION
Within the limits of the present study and on the basis of results obtained, it may be concluded that: 1. The denture bases fabricated by New Press method were dimensionally more stable when compared to those fabricated by conventional method. 2. In all the fabricated denture bases, the distance A-D, i.e. the anterior slope of the palate was the most dimensionally stable region. 3. In all the fabricated denture bases, the distance E-F and F-G, i.e. the posterior palatal seal area was the least dimensionally stable region.
REFERENCES
Rueggeberg FA. From vulcanite to vinyl, a history of resins in restorative dentistry. J Prosthet Dent 2002;87:364-79. 2. Peyton FA. History of resins in dentistry. Dent Clin North Am 1975;19:211-22. 3. Craig RG, Powers JM. Restorative Dental Materials (11th edn). St. Louis, MO, Mosby 2002:636-89. 4. Lechner SK, Thomas GA. Changes caused by processing complete denture. J Prosthet Dent 1951;1:551-59. 5. Firtell DN. Posterior peripheral seal distortion related to processing temperature. J Prosthet Dent 1981:598-601. 6. Consani RL, Domitti SS, Consani S. Effect of a new tension system, used in acrylic resin flasking, on the dimensional stability of denture bases. J Prosthet Dent 2002;88(3):285-89. 7. Jerelimov V, Jagger RG, Milward. Effect of curing cycle on acrylic denture base glass transition temperature. J Dent 1991;19(4):245-48. 8. Mainieri ET, Boone ME, Potter RH. Tooth movement and dimensional change of denture base materials using two investment methods; J Prosthet Dent 1980:44(4):368-73. 9. Dixon DL. Linear dimensional variability of three denture base resins after processing and in water storage. J Prosthet Dent 1992:68(1):196-200. 10. Honorez P, Catalan A, Anganes U. The effect of three processing cycles on some physical and chemical properties of heat-cured acrylic resins. J Prosthet Dent 1989;61(4):510-17.
ABOUT THE AUTHORS T Shankar
Professor, Department of Prosthodontics, Kalinga Institute of Dental Sciences, Bhubaneswar, Odisha, India
Snigdha Gowd
Professor and Head, Department of Orthodontics, Kalinga Institute of Dental Sciences, Bhubaneswar, Odisha, India
Graph 1: Comparison of SP1, A and A ’
A Comparative Evaluation of the Dimensional Accuracy of Heat Polymerized Acrylic Resin Denture Base Clamped
The Journal of Contemporary Dental Practice, November-December 2012;13(6):842-849 849
JCDP
Syed Tauqheer Ahmed
Senior Lecturer, Department of Prosthodontics, The Oxford Dental College and Hospital, Bengaluru, Karnataka, India
V Vinod (Corresponding Author)
Senior Lecturer, Department of Prosthodontics, The Oxford Dental College, Bommanahalli, Bengaluru, Karnataka, India, e-mail: drvinod@mail.com, dvinnz@gmail.com
M Vikas Goud
Professor, Department of Prosthodontics, Private Practice, Hyderabad Andhra Pradesh, India
N Venugopal Rao
Reader, Department of Prosthodontics, Hi-Tech Dental College and Hospital, Bhubaneswar, Odisha, India |
9097 | https://lpsa.swarthmore.edu/LaplaceZTable/Common%20Laplace%20Transform%20Pairs.pdf | Common Laplace Transform Pairs Time Domain Function Laplace Domain Function Name Definition Unit Impulse ) (t δ 1 Unit Step (t) γ † s 1 Unit Ramp t 2 1 s Parabola t2 3 2 s Exponential at e− a s + 1 Asymptotic Exponential ) 1 ( 1 at e a − − ) ( 1 a s s + Dual Exponential ( ) bt at e e a b − − − − 1 ) )( ( 1 b s a s + + Asymptotic Dual Exponential ( ) at bt 1 1 1 be ae ab a b − − + − − ) )( ( 1 b s a s s + + Time multiplied Exponential at te− 2 ) ( 1 a s + Sine 0 sin( t) ω 0 2 2 0 s ω + ω Cosine 0 cos( t) ω 2 2 0 s s + ω Decaying Sine at d e sin( t) − ω d 2 2 d (s a) ω + + ω Decaying Cosine at d e cos( t) − ω 2 2 d s a (s a) + + + ω Generic Oscillatory Decay ( ) ( ) at d d d C aB e Bcos t sin t − − ω + ω ω ( ) 2 2 d Bs C s a + + + ω Prototype Second Order Lowpass, underdamped ( ) 0t 2 0 0 2 e sin 1 t 1 −ζω ω ω −ζ −ζ 2 0 2 2 0 0 s 2 s ω + ζω + ω Prototype Second Order Lowpass, underdamped - Step Response ( ) 0t 2 0 2 2 1 1 1 e sin 1 t 1 1 tan −ζω − − ω −ζ + φ −ζ −ζ φ = ζ 2 0 2 2 0 0 s(s 2 s ) ω + ζω + ω All time domain functions are implicitly=0 for t<0 (i.e. they are multiplied by unit step, γ(t)). †u(t) is more commonly used for the step, but is also used for other things. γ(t) is chosen to avoid confusion (and because in the Laplace domain it looks a little like a step function, Γ(s)). Common Laplace Transform Properties Name Illustration Definition of Transform L st 0 f (t) F(s) F(s) f (t)e dt − ∞ − ← → = ∫ Linearity ) ( ) ( ) ( ) ( 2 1 2 1 s BF s AF t Bf t Af L + → ← + First Derivative ) 0 ( ) ( ) ( − − → ← f s sF dt t df L Second Derivative ) 0 ( ) 0 ( ) ( ) ( 2 2 2 − −− − → ← f sf s F s dt t f d L nth Derivative ) 0 ( ) ( ) ( ) 1 ( 1 − − = − ∑ − → ← i n i i n n L n n f s s F s dt t f d Integral ) ( 1 ) ( 0 s F s d f L t → ← ∫ λ λ Time Multiplication ds s dF t tf L ) ( ) ( − → ← Time Delay L as f (t a) (t a) e F(s) − − γ − ← → γ(t) is unit step Complex Shift ) ( ) ( a s F e t f L at + → ← − Scaling ) (as aF a t f L → ← Convolution Property ) ( ) ( ) ( ) ( 2 1 2 1 s F s F t f t f L → ← Initial Value (Only if F(s) is strictly proper; order of numerator < order of denominator). ) ( lim ) ( lim 0 s sF t f s t ∞ → → = + Final Value (if final value exists; e.g., decaying exponentials or constants) ) ( lim ) ( lim 0 s sF t f s t → ∞ → = |
9098 | https://brainly.com/question/36749509 | [FREE] How does the Oswald efficiency factor (e_0) compare to the span efficiency factor, considering there is - brainly.com
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How does the Oswald efficiency factor (e 0) compare to the span efficiency factor, considering there is more drag on an entire aircraft than only on a wing?
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The Oswald efficiency factor and the span efficiency factor are two parameters used to assess the drag characteristics of an aircraft. The Oswald efficiency factor represents the efficiency of the entire aircraft in converting lift into drag, while the span efficiency factor focuses on the wing's efficiency in reducing induced drag. The two factors are not directly comparable but are significant in determining the overall drag of an aircraft.
Explanation
The Oswald efficiency factor, e0, and the span efficiency factor are two different parameters used to quantify the drag characteristics of an aircraft. The Oswald efficiency factor, e0, takes into account the entire aircraft's drag and represents the efficiency of the aircraft in converting lift into drag. On the other hand, the span efficiency factor specifically focuses on the wing's efficiency in reducing induced drag, which is drag caused by the creation of lift.
In general, a lower Oswald efficiency factor indicates higher drag for the entire aircraft, while a higher span efficiency factor indicates lower induced drag. Therefore, the two factors are not directly comparable, but they both play crucial roles in determining the overall drag characteristics of an aircraft.
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The Oswald efficiency factor measures the overall aerodynamic efficiency of an aircraft, while the span efficiency factor specifically evaluates the wing's ability to minimize induced drag. The two factors are important but assess different types of drag. Overall, the Oswald efficiency factor tends to consider the entirety of the aircraft, reflecting more drag than what is seen with only the wing's contribution.
Explanation
The Oswald efficiency factor (
e 0
) and the span efficiency factor are essential concepts in aerodynamics that help understand the drag characteristics of an aircraft.
Oswald Efficiency Factor (
e 0): This factor measures how effectively an aircraft converts lift into drag across the entire structure. It provides insight into the overall aerodynamic efficiency of the aircraft, encompassing contributions from the wings, fuselage, and other components. A higher
e 0
indicates better aerodynamic efficiency, meaning less drag for the amount of lift produced.
Span Efficiency Factor: This factor focuses specifically on the efficiency of the wing in reducing induced drag, which is the drag that occurs due to the generation of lift. The span efficiency factor assesses how well the wing's design minimizes the vortex induced by lift, which can create additional drag. A higher span efficiency factor suggests the wing is shaped to effectively reduce this drag.
While both factors are used to analyze drag, they are not directly comparable due to the different aspects of drag they assess. The Oswald efficiency factor includes all sources of drag on the aircraft, whereas the span efficiency factor is limited to the wing's contribution.
In conclusion, although the Oswald efficiency factor considers the entire aircraft and is generally higher due to the more significant overall drag forces, the span efficiency factor is focused and indicates how well the wing alone can operate with minimal induced drag. Both factors are crucial in improving aircraft design for better performance and efficiency.
Examples & Evidence
For example, a glider may have a high Oswald efficiency factor due to its streamlined design, resulting in less drag overall, while its wings may also have a high span efficiency factor, indicating they effectively reduce induced drag despite the entire aircraft's performance. Another example is a fighter jet, which may have a lower Oswald efficiency factor due to its design prioritizing speed and maneuverability but may still have a well-designed wing that maintains a favorable span efficiency factor to enhance its aerodynamic capability.
The concepts of the Oswald efficiency factor and span efficiency factor are well established in aerodynamics, varying by the aircraft's design and purpose, and are commonly discussed in aerodynamics texts and research articles.
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9099 | https://www.sciencedirect.com/science/article/abs/pii/S0022328X00002102 | Methods of enhancement of reactivity and selectivity of sodium borohydride for applications in organic synthesis - ScienceDirect
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Outline
Abstract
Keywords
1. Introduction
2. Hydroboration of alkenes and alkynes
3. Reduction of carboxylic acids
4. Reduction of amino acids and their derivatives
5. Reduction of carboxylic acid esters
6. Reduction of carboxylic acid amides
7. Reduction of nitriles
8. Reduction of acid chlorides
9. Reduction of nitro compounds
10. Reduction of aldehydes and ketones
11. Miscellaneous
12. Conclusions
Acknowledgements
References
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Journal of Organometallic Chemistry
Volume 609, Issues 1–2, 8 September 2000, Pages 137-151
Methods of enhancement of reactivity and selectivity of sodium borohydride for applications in organic synthesis
Author links open overlay panelMariappan Periasamy, Muniappan Thirumalaikumar
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Abstract
NaBH 4 does not reduce carboxylic acids, esters, amides and nitriles under ambient conditions. However, the reactivity of NaBH 4 can be enhanced by the addition of certain additives. For example, addition of iodine to NaBH 4 in THF provides H 3 B–THF that is useful for hydroborations and reductions of various functional groups. The aldehydes and ketones are reduced in a fast manner by the NaBH 4 reagent. Even so, the selectivities realised in such reductions can be enhanced using NaBH 4 along with another additive. In this article, various methods used for the enhancement of reactivity and selectivity of NaBH 4 in organic synthesis are described.
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Keywords
Sodium borohydride
Enhancement of reactivity
Additives
Reduction of organics
1. Introduction
Metal hydrides are valuable reagents in modern organic chemistry. The most frequently used hydride is the NaBH 4 reagent. It is a mild, inexpensive and invaluable reagent for applications in a wide range of reduction processes. It is the reagent of choice for the reduction of aldehydes and ketones to alcohols and imines or iminium salts to amines with protic solvents . The carboxylic acids, esters, amides and nitriles are more resistant towards NaBH 4. However, the reactivity of NaBH 4 can be enhanced by carrying out the reaction in the presence of certain additives. In this article, various methods of enhancement of reactivity and selectivity of NaBH 4 using additives for applications in organic synthesis are described.
2. Hydroboration of alkenes and alkynes
Hydroboration of carboncarbon multiple bonds provides a method for the synthesis of the valuable organoborane intermediates with high regio- and stereospecificities . Historically, Brown and Subba Rao discovered the hydroboration reaction during their investigation of the activation of NaBH 4 for the reduction of esters using AlCl 3. The use of BF 3 in the place of AlCl 3 led to more effective utilisation of the hydride for the generation of diborane, B 2 H 6 and borane Lewis base complexes (, , ) .(1)9 RCHCH 2+AlCl 3→diglyme 3 RCH 2 CH 2 3 B+AlH 3+3 NaCl(2)12 RCHCH 2+3 NaBH 4+4 BF 3→diglyme 4 RCH 2 CH 2 3 B+3 NaBF 4(3)RCH 2 CH 2 3 B+3 H 2 O 2+NaOH→3 RCH 2 CH 2 OH+NaB OH 4
Although several of these borane complexes are commercially available (e.g. H 3 B–THF, H 3 B–SMe 2 and H 3 B–NR 3), there have been sustained efforts towards the development of alternative, simple and convenient methods of generation of boranes in situ for hydroboration. In 1963, it was reported that a 1:1 mixture of NaBH 4 and CH 3 COOH hydroborates alkenes [8a]. Later, a modified procedure for hydroboranes using NaBH 4–CH 3 COOH was reported (Eq. (4)) [8b].(4)NaBH 4+CH 3 COOH→CH 3 COOBH 3 Na→2.H 2 O 2OH−1. n-C 4 H 9 CHCH 2 n-C 4 H 9 CH 2 CH 2 OH 75%
Selective hydroboration of olefinic moiety in the presence of carboxylic acid group was reported from this laboratory (Eq. (5)) .
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5.
Also, a new method of conversion of olefins to dialkyl ketone was developed (Eq. (6)) .(6)NaBH 4→2.n-C 8 H 17 CHCH 2 25°C,12 h 1.AcOH−THF→2.H 2 O 2−OH−1.CHCl 3−NaOMe n-C 8 H 17 CH 2 CH 2 2 80%CO
A method of conversion of terminal alkenes to carboxylic acids through hydroboration of olefins was also developed (Eq. (7)) . This method provides a simple, one-pot synthesis of carboxylic acids from terminal alkenes.
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(7).
Also, various combinations of metal salts and borohydrides, such as SnCl 4–NaBH 4, TiCl 4–NaBH 4, TiCl 4–PhCH 2 N+(Et)3 BH 4− and CoCl 2–NaBH 4 have been reported to effect hydroboration of olefins.
Whereas the CoCl 2–NaBH 4 combination behaves as a hydroborating agent when the reaction is carried out with THF for 2 h at room temperature (r.t.) before the addition of alkene, it works as a hydrogenating agent in methanol (Eq. (8)). This method has some advantages over the reported method using alcoholic medium .
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(8).
Chiral semicorrin 1–3 cobalt complexes can be prepared readily using CoCl 2 and the corresponding free ligands. These complexes are efficient enantioselective catalysts for the conjugate reduction of α,β-unsaturated carboxylates (Eq. (9)) and α,β-unsaturated carboxamides using NaBH 4.
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1. Download: Download full-size image
(9).
Chalcones undergo facile reduction on reaction with NiCl 2–NaBH 4 system to afford dihydrochalcones (Eq. (10)) . The use of copper or cobalt chloride in place of NiCl 2 is not as efficient for this application.
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(10).
The NaBH 4 reacts with I 2 to give diborane (Eq. (11)) [20a].(11)2 NaBH 4+I 2→B 2 H 6+H 2+2 NaI The reactive ‘H 3 B–THF’ species can be easily generated in situ by mixing NaBH 4 and I 2 in THF [20b]. Hydroboration of alkenes using this NaBH 4–I 2 system in THF followed by oxidation gives the corresponding anti-Markovnikov alcohols (Eqs. (12) and(13)) in good yields , .(12)n-C 8 H 17 CHCH 2→THF NaBH 4−I 2→NaOH H 2 O 2 n-C 8 H 17 CH 2 CH 2 OH 92%
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(13).
Later, it was reported that electrochemical oxidation of NaBH 4 using catalytic amounts of sodium iodide gives diborane that hydroborates olefins (Eq. (14)) .
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(14).
The Me 3 SiCl–PhCH 2 N+(Et)3 BH 4− reagent system has been reported to effect hydroboration of olefins to give anti-Markovnikov alcohols after oxidation (Eq. (15)) .(15)CH 3 CH 2 6 CHCH 2→Me 3 SiCl,CH 2 Cl 2 2.H 2 O 2OH−1.PhCH 2 N+Et 3 B−H 4 CH 3 CH 2 7 CH 2 OH 72%
Very recently, it has been reported that the tetrabutylammonium borohydride liberates diborane in solvents such as CH 2 Cl 2, CHCl 3 and CCl 4. A number of terminal, internal and cyclic alkenes were hydroborated using this borohydride (Eq. (16)) .
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(16).
The 1-alkynes undergo dihydroboration to yield the corresponding terminal alcohols after oxidation. Generally, the disubstituted alkynes give vinyl boranes that on oxidation offord ketones as the major product (, ) , . However, the diphenyl acetylene yields 1,2-diphenylethanol as the major product through dihydroboration under these conditions.(17)C 6 H 5 CCC 6 H 5→CHCl 3 NaBH 4−Bu 4 N+Cl−C 6 H 5 CH OH CH 2 C 6 H 5 83%(18)CH 3 CH 2 CCCH 2 CH 3→CHCl 3 NaBH 4Bu 4 N+Cl−CH 3 CH 2 CO CH 2 2 CH 3 90%
The PdCl 2–NaBH 4–polyethyleneglycol (PEG)–CH 2 Cl 2 system is effective for hydrogenation of carboncarbon triple bonds to the corresponding cis-alkenes (Eq. (19)) . This reagent has advantages of faster rates and higher selectivity.
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(19).
3. Reduction of carboxylic acids
The NaBH 4 gives acyloxyborohydride species on reaction with carboxylic acids in THF that hydroborate olefins. The acyloxy moieties in such acyloxyborohydrides remain unchanged under ambient conditions. However, one half of the acyloxy moiety undergoes reduction upon heating to give the corresponding alcohol (Eq. (20)) .
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(20).
A similar reaction was also observed using NaBH 4, RCOOH and catechol at 25°C (Eq. (21)) .
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(21).
Aliphatic carboxylic acids are reduced by NaBH 4 to the corresponding alcohols in good yields when RCOOH and CF 3 COOH are used in 1:1 ratio under ambient conditions (Eq. (22)). However, the aromatic acids give poor yields (e.g. benzoic acid 20%). Also, the NaBH 4–CF 3 COOH combination is good for the reduction of aliphatic carboxylic acids (65–95% yields). Again, aromatic acids give poor results under these conditions (30%).
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(22).
The ZnCl 2–NaBH 4 reagent system readily reduces both aliphatic and aromatic acids to the corresponding alcohols in refluxing THF (Eq. (23)) . The reaction requires only stoichiometric quantities of hydride for this conversion. Also, dicarboxylic acids are reduced to the corresponding diols under these conditions.(23)RCOOH→THF,Δ NaBH 4ZnCl 2 RCH 2 OH R=alkyl/aryl 7095%
The reagent prepared using ZrCl 4 and NaBH 4 reduces the carboxylic acids in excellent yields under mild conditions (Eq. (24)) .(24)PhCOOH→THF,rt,5 h NaBH 4ZrCl 4 PhCH 2 OH 85%
Carboxylic acids are reduced to the corresponding alcohols under ambient conditions by the NaBH 4–I 2 reagent system in very good yields with some selectivities (Eqs. (25) and(26)) .(25)CH 3 CH 2 8 COOH→THF 0−25°C NaBH 4−I 2 CH 3 CH 2 8 CH 2 OH 95%
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(26).
Further, selective reduction of the carboxylic acid group in an olefinic acid has also been achieved by forming the corresponding acyloxyborohydride before the addition of I 2 (Eq. (27)) .
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(27).
Cyanuric chloride–NaBH 4 reagent system has also been used to effect the reduction of carboxylic acids to alcohols under mild conditions (Eq. (28)) .
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(28).
Facile, chemoselective reduction of carboxylic acids to alcohols using a phosphonium hexafluorophosphate (BOP reagent)–NaBH 4 reagent system has been reported ((29), (30)) . Also, this method is convenient, rapid and chemoselective in several cases. For example, functional groups such as nitro, nitrile, azido and ester are unaffected under these conditions.
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(29).
1. Download: Download full-size image
(30).
4. Reduction of amino acids and their derivatives
Chiral amino alcohols are important class of compounds. They are useful in asymmetric transformations, synthesis of pharmaceuticals , resolution of racemic mixtures and in synthesis of insecticides . Obviously, several reagents are available (e.g. LiAlH 4, DIBAL , H 3 B–THF ) for the reduction of free as well as protected amino acids to the corresponding amino alcohols. However, these reagents suffer from disadvantages of cost, inflammability and tedious isolation procedures. Meyers and coworkers examined the reduction of amino acids using the NaBH 4–I 2 reagent system. The results indicate that it is an excellent reagent system for the conversion of amino acids to amino alcohols (Eq. (31)) .
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(31).
The N-acyl amino acids give the corresponding N-alkyl amino alcohols under these conditions (Eq. (32)) .
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(32).
However, the N-carbamate protecting group is unaffected under these conditions . Also, the reductions of pentachlorophenyl esters of the Boc protected amino acids and peptides to the corresponding alcohols have been reported (Eq. (33)) .
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(33).
The NaBH 4–I 2 reagent system is safe, simple and inexpensive. Hence, it is useful, especially in the large scale synthesis of chiral amino alcohols.
Amino acids are also reduced using the inexpensive NaBH 4–H 2 SO 4 reagent system in THF (Eq. (34)) . It is of interest to note that no racemization occurs in the reduction of amino acids using NaBH 4–I 2 or NaBH 4–H 2 SO 4.
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(34).
5. Reduction of carboxylic acid esters
The NaBH 4–ZnCl 2 reagent system exhibits powerful reducing properties in the presence of a tertiary amine. The carboxylic esters were smoothly reduced by this reagent to their corresponding alcohols (Eq. (35)) . Further, the reduction does not take place without the amine under these conditions.
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(35).
The NaBH 4–I 2 reagent system has also been used for this application. It readily reduces carboxylic acid esters under reflux conditions in good yields (Eq. (36)) .(36)PhCH 2 COOEt→70°C,0.5 h NaBH 4−I 2,THF→H 2 O PhCH 2 CH 2 OH 85%
6. Reduction of carboxylic acid amides
Numerous chemicals of importance in medicinal chemistry have been prepared through reduction of amides . It was found that the amides can be easily reduced to primary amines using NaBH 4–CoCl 2 system in good yields in hydroxylic as well as in non-hydroxylic solvents (Eq. (37)) .(37)n-C 3 H 7 CONH 2→NaBH 4−CoCl 2 n-C 3 H 7 CH 2 NH 2 70%
The NaBH 4–I 2 system is also useful for the reduction of amides (, , ) .(38)PhCONH 2→THF,Δ NaBH 4−I 2→NaOH PhCH 2 NH 2 70%(39)PhNHCOCH 3→PhNHCH 2 CH 3 75%(40)Ph CH 3 NCOCH 3→Ph CH 3 NCH 2 CH 3 74%
Further, reduction of amides containing sensitive functional groups can also be carried out using NaBH 4–I 2 reagent system under ambient conditions (Eq. (41)) .
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(41).
A new procedure for the highly selective reduction of tertiary amides to amines using NaBH 4–bis(2-bromoethyl)selenium dibromide 4 has been reported (Eq. 42) [48a,b].
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(42).
At a higher temperature, this reaction takes place smoothly in a shorter period. However, reduction of secondary and primary amides with this system in THF failed. Also, without the selenium compound, the reaction did not take place. It has been observed that the reaction of NaBH 4 with 4 produces borane (5) and bis(2-bromoethyl)selenide (6).(BrCH 2 CH 2)2 Se−BH 3(BrCH 2 CH 2)2 Se 5 6
7. Reduction of nitriles
The CoCl 2–NaBH 4 reagent system effectively reduces the nitriles in good yields in hydroxylic as well as non-hydroxylic solvents (Eq. (43)).(43)C 6 H 5 CH OH CN→NaBH 4−CoCl 2 C 6 H 5 CH OH CH 2 NH 2 80%
Mandelonitrile is smoothly reduced using the NaBH 4–CoCl 2 system . It is of interest to note that reduction by LiAlH 4 gives the α-hydroxy-β-phenylethylamine in low yields. Cupric salts–NaBH 4 combinations are also effective for reduction of nitriles. The halogenides, sulfates, carboxylates of cobalt, nickel, iridium, rhodium, osmium and platinum were also used along with NaBH 4 in several applications. Also, ZrCl 4–NaBH 4 reagent system has been used for the reduction of nitriles in excellent yields (Eq. (44)) .(44)PhCH 2 CN→THF,r.t.NaBH 4−ZrCl 4 PhCH 2 CH 2 NH 2 91%
Nitriles are also reduced by NaBH 4–I 2 system in THF under refluxing conditions (Eq. (45)) .(45)PhCN→THF NaBH 4−I 2→NaOH PhCH 2 NH 2 72%
Also, nitriles on reaction with a mixture of NaBH 4 and bis(2-bromoethyl)selenium dibromide (4), in boiling THF give the corresponding primary amines (Eq. (46)) .(46)RCN→2.HCl 1.NaBH 4−4, THF RCH 2 NH 2 HCl R=alkyl/aryl 35−73%
This reagent system does not affect many other substituents that are susceptible to H 3 B–THF [48b].
8. Reduction of acid chlorides
The acyl chlorides are efficiently reduced to the corresponding alcohols on reaction with Zn(BH 4)2 in presence of TMEDA (Eq. (47)) . Electron withdrawing group in the substrates enhances the rate of reduction. It is of interest to note that the selective reduction of acyl chlorides in the presence of other functional groups, such as chloro, nitro, ester and conjugated double bond can be achieved using this reagent system.(47)RCOCl→TMEDA NaBH 4−ZnCl 2 RCH 2 OH R=alkyl/aryl 86−98%
9. Reduction of nitro compounds
Nitro compounds are reduced to the corresponding amino compounds effectively using the NaBH 4–CuSO 4 system (Eq. (48)) . It was reported that this system also reduces ketones, aliphatic esters, olefins and nitriles.
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(48).
A novel reagent system, prepared using BiCl 3 and NaBH 4, reduces aromatic nitro compounds to the corresponding amines in good yields (Eq. (49)) . The functional groups such as Me, OH, NH 2, OMe, Cl in the aromatic ring do not have any marked effect on the rate of the reaction. Moreover, these functional groups survive during the reduction, making this process fairly general and selective.(49)RNO 2→THF NaBH 4−BiCl 3 RNH 2 R=alkyl/aryl 35−90%
The aromatic nitro compounds are also reduced in good yields to the corresponding N-aryl hydroxylamines using NaBH 4 in the presence of catalytic amounts of metallic selenium (Eq. (50)) .
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(50).
These selenium catalysed reductions were accelerated by electron withdrawing groups. The aliphatic nitro compounds gave the corresponding oximes under these conditions. The active species in this system is the hydrogen selenide anion . Hence, this method opens up a new way to the use of selenium as a redox catalyst in organic synthesis.
The N-aryl hydroxylamines have been also prepared through antimony catalysed NaBH 4 reduction of nitroarenes (Eq. (51)) .
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(51).
The novel NaBH 4–(NH 4)2 SO 4 reagent system has been used for the selective, rapid reduction of nitro compounds to the corresponding amino derivatives in good yields (Eq. (52)) .(52)RNO 2→EtOH,r.t.,30−120 min NaBH 4−NH 4 2 SO 4 RNH 2 R=aryl 70−90%
10. Reduction of aldehydes and ketones
The NaBH 4 is a reagent of choice for the reduction of aldehydes and ketones. The reactivity of NaBH 4 can be readily modified through its reaction with acetic acid (Eq. 53) , . With excess of acetic acid, triacyloxy borohydride is formed. The NaBH 4–CH 3 COOH reagent system has been used for reductions of enamines, imines, vinylogous carbamates, aromatic and aliphatic α,β-unsaturated tosylhydrazones, pyrylium salts . It has been also used for the reduction or reductive N-alkylation of amines, oximes and nitrogen containing heterocycles .
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(53).
The NaBH(OAc)3 reagent has been used for the chemoselective reduction of aldehydes in the presence of ketones . Also, selective reduction of ketones using NaBH 4 in glacial acetic acid solvent or NaBH 4 in THF using 3 equivalents of acetic acid were reported . Further, highly stereoselective reduction of ketones was also achieved using NaBH(OAc)3 in acetic acid (Eq. 54) .
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(54).
In the presence of ZrCl 4, NaBH 4 reduces the aldehydes selectively in high yields (Eq. (55)) .(55)PhCHO→THF,r.t.,5 h NaBH 4ZrCl 4 PhCH 2 OH 95%
Aromatic aldehydes are more selectively reduced than the related ketones by the reducing system consisting of NaBH 4 and SnCl 2 in THF (Eq. (56)) . It may of interest to note that selective reduction of aldehydes in the presence of ketones is ordinarily impracticable using the reducing agents such as alkali metal borohydrides, aluminohydrides and diborane .(56)PhCHO→NaBH 4−SnCl 2 PhCH 2 OH 96.8%+PhCOPh 92%
The NaBH 4 in combination with anhydrous AlCl 3 conveniently reduces diaryl and aryl alkyl ketones to methylenic hydrocarbons (Eq. (57)) .
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(57).
Although alkali metal borohydrides have received much attention in organic synthesis, the studies on the use of alkaline earth metals are limited. For example, α,β-unsaturated ketones more readily converted to allylic alcohols selectively using NaBH 4 in the presence of CaCl 2 (Eq. (58)) . Among the alkaline earth metal chlorides examined, CaCl 2 gives the best combination of good yields and selectivities in the NaBH 4 reduction of 2-cyclohexen-1-one. Further, this method provides a simple, inexpensive alternative procedure for the selective 1,2-reduction of α,β-unsaturated ketones.
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(58).
Metal salts mediated NaBH 4 reduction of α-alkyl-β-keto esters leads to different stereochemical control depending on the nature of the metal atom. For example, the strongly chelating TiCl 4 led to the syn isomer while non-chelating CeCl 3 gave anti isomer . The keto esters are also reduced using the ZnCl 2–NaBH 4 system to afford the corresponding hydroxy ester (Eq. (59)) .
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(59).
Stereoselective reduction of 3-keto-2-methyl esters and 3-keto-2-methyl amides with NaBH 4 and catalytic amount of MnCl 2 leads to the corresponding erythro-alcohols. Of the metal salts examined, MnCl 2 provides high selectivity ((60), (61)) .
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(60).
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(61).
The PdCl 2–NaBH 4 reagent system reduces aryl ketones, aryl chlorides and benzylic alcohols to the corresponding hydrocarbons (Eq. (62)). Also, certain hindered steroidal ketones are reduced to the corresponding alcohols in good yields.
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(62).
Moreover, this reagent is useful for the preparation of alkylbenzenes from aryl alkyl ketones. This reduction proceeds under mild conditions at r.t. and neutral pH in a short time with high selectivity. Also, the chlorine atoms attached to aromatic rings are removed reductively by this reagent.
The lanthanoid ion-promoted NaBH 4 reduction of enones to allylic alcohols is a process of immense interest, since the regio- and chemoselectivity observed are in sharp contrast to the reduction promoted using other additives . Among the lanthanoid chlorides examined, the CeCl 3 is the best reagent for selective 1,2-reduction of enones by NaBH 4 (Eq. (63)). Moreover, this reducing system does not affect carboxylic acids, esters, amides, halides, cyano and nitro groups.
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(63).
High chemo- and regioselective 1,2-reductions of unsaturated carbonyl compounds using NaBH 4 assisted by organolanthanide complexes, LnCpCl 2(THF)3 (Ln=Sm and Er) in methanol have been reported (Eq. (64)) .
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(64).
Finally, the CeCl 3–NaBH 4 system reduces the α′-diphenylphosphinoyl enones and ketones stereoselectively in good yields (Eq. (65)) .
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(65).
10.1. Reduction of ketones using NaBH 4 supported reagents
A novel reagent system consisting of NaBH 4 and amberlyst-15(H+) in THF is a powerful reducing agent for the reduction of unreactive ketones (Eq. (66)) . The reduction is fast, high yielding and the workup is very simple. Ketals, silyl ethers, acetates, allylic acetates, allylic γ-lactones, carboxylic esters, halides and isolated double bonds are not disturbed during the reduction.
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(66).
The NaBH 4 impregnated on neutral alumina reduces a wide variety of carbonyl compounds to the corresponding hydroxy derivatives in solution phase . The solid state reduction has also been achieved by mixing ketones with NaBH 4 and storing the mixture in a dry box for 5 days . The major disadvantage of these solid state methods is that they require long reaction time. Accordingly, a facile microwave assisted reduction of aldehydes and ketones by alumina supported NaBH 4 has been developed (Eq. (67)) .
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67.
Significant chemoselectivity is observed in the reduction of trans-cinnamaldehyde and the olefinic moiety remains intact under these conditions (Eq. (68)).
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(68).
A novel solid state reduction of organic functional groups using NaBH 4 impregnated on Merrifield's resin has been reported . Several aldehydes and ketones were also reduced using this reagent system.
10.2. Reduction of aldehydes and ketones under phase transfer catalysis
A simple, efficient and economical method for chemoselective reduction of aldehydes in the presence of ketones using stoichiometric amounts of NaBH 4 under phase-transfer catalysis was reported (Eq. (69)) . Aliphatic, aromatic and unsaturated aldehydes are reduced rapidly in high yields while dialkyl, aralkyl, diaryl and cyclic ketones were not affected under these conditions. Toluene and CH 2 Cl 2 have also been used as solvents while PhCH 2 NEt 3 Cl and Aliquat-336 worked well as catalysts.
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(69).
10.3. Asymmetric reduction of ketones using NaBH 4 and a chiral auxiliary
The use of NaBH 4–Me 3 SiCl reagent system in the β-hydroxysulfoximine catalysed asymmetric reduction of ketones affords the corresponding secondary alcohols in high yields with good enantioselectivities (Eq. (70)) .
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(70).
There are a few reports in the literature on asymmetric reduction in non-aqueous solutions using NaBH 4 modified by chiral ligands . For example, asymmetric reduction of acetophenone, propiophenone and 2-acetylnaphthalene using NaBH 4 and optically active (S)-lactic acid derivatives produced the corresponding optically active (R)-alcohols in up to 38.3% ee (Eq. (71)) .
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(71).
Also, successful asymmetric reduction of functionalized ketones using a NaBH 4–(L)-tartaric acid system was reported (Eq. (72)) . This system is effective for the reduction of keto esters which are expected to chelate to a reductant through both the carbonyl and alkoxycarbonyl groups.
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(72).
Unfortunately, lower reactivity was observed in the case of simple ketones and the products were almost racemic in these cases. A catalytic enantioselective borohydride reduction of aromatic ketones using an optically active cobalt(II) complex catalyst 7 was reported (Eq. (73)) .
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(73).
The lanthanoid complexes, tris{4-(l-menthyloxy)-1-(p-tolyl) butane-1,3-dionato}-lanthanoid(III), [Ln(l-moba-Me)3] facilitate enantioselective borohydride reduction of ketones (Eq. (74)) . The reaction is stoichiometric on the chiral lanthanoid complex. However, the host complexes are readily recovered in good yields for use again.
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(74).
The use of montmorillonite clay in the asymmetric reduction of ketones using (-)-N-dodecyl-N-methyl ephedrenium borohydride was investigated (Eq. (75)) . The rates of reductions are quite fast both in polar as well as in non-polar solvents. However, the asymmetric inductions realised were only poor.
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(75).
The chirally modified Cu 2+–montmorillonite has been prepared using (−)-2-amino-1-propanol and (S)-proline. The ketone pre-adsorbed on this clay was reduced by the addition of NaBH 4 in ethanol with low asymmetric inductions (Eq. (76)) . The advantage of this system is the reusability of the chiral moiety.
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(76).
The chiral reverse micelles formed from chiral surfactants 8 are useful for the asymmetric reduction of prochiral ketones with NaBH 4 (Eq. (77)) . It was noticed that the enantioselectivity of the reaction is affected by the structures of both surfactant and substrate. Also, the results depend on the composition of the micro environment. Further, it was reported that the presence of some sugars increases the enantiomeric excess of the products.
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(77).
10.4. Reductive amination of aldehydes and ketones
The reactions of aldehydes or ketones with NH 3, primary amines or secondary amines in the presence of reducing agents to obtain primary, secondary or tertiary amines, respectively are among the most useful and important processes for the synthesis of different amines. The choice of the reducing agent is very important for the success of the reaction since the reducing agent must reduce the imines (or iminium salts) intermediates selectively over aldehydes or ketones under the reaction conditions (Eq. (78)).
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(78).
The NaBH 4 suits well for such applications. For example, reductive alkylation of amines were be readily carried out using NaBH 4 in neat liquid carboxylic acids (Eq. (79)) . Also, reductive amination of aldehydes and ketones were reported using sodium triacetoxyborohydride .
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(79).
The scope of this transformation has been illustrated using aliphatic acyclic and cyclic ketones, aliphatic and aromatic aldehydes, primary and secondary amines including a variety of weakly basic and non basic amines.
Further, highly diastereoselective reductive amination of substituted cyclohexanones give axial amines using sodium triacyloxyborohydrides derived from various carboxylic acids (Eq. (80)) .
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(80).
Very recently, a novel one-pot reductive alkylation of amines, by S-ethylthioesters mediated by triethyl silane and sodium triacetoxyborohydride in the presence of Pd on carbon, was reported (Eq. (81)) .
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(81).
Also, the reaction of formaldehyde with primary and secondary aromatic amines using NaBH 4–H 2 SO 4 in THF lead to reductive amination (Eq. (82)) .
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(82).
Also, the reaction of the chiral O-protected-α-hydroxyketone with primary amines by NaBH 4 in the presence of magnesium perchlorate, Mg(ClO 4)2 led to the exclusive formation of erythro (1 R,2 S)-O-protected-N-substituted ethanolamines (Eq. (83)) .
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(83).
Evidently, the intermediacy of magnesium bidendate complex leads to a more rigid conformation resulting in a higher percentage of the erythro diastereomer .
A simple, mild and efficient procedure for reductive alkylations of dimethylamine , methylamine hydrochloride, triethylamine with carbonyl compounds using titanium(IV) isopropoxide and NaBH 4 has been reported (Eq. (84)).
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(84).
The titanium(IV) isopropoxide is a mild reagent compatible with a variety of potentially acid sensitive functional groups such as acetals, lactams, acetonide and t-butyldimethylsilyl ether.
The Ti(OPr i)4–NaBH 4 has been also used in the synthesis of unsymmetrically disubstituted ureas by the reductive amidation of aldehydes and monosubstituted ureas (Eq. (85)) .
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(85).
10.5. Reduction of epoxy ketones
The α,β-epoxy alcohols serve as versatile synthetic intermediates that can be efficiently elaborated into poly hydroxy derivatives with multiple chiral centers . A highly stereoselective reduction of α,β-epoxyketones using NaBH 4–ZnCl 2 reagent system has been reported . Also, a very simple alternative procedure for the stereoselective reduction of α,β-epoxyketones into erythro-α,β-epoxy alcohols with NaBH 4 in the presence of CaCl 2–LaCl 3 in methanol was reported (Eq. (86)) .
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(86).
In these cases, use of metal salt facilitates the formation of the erythro product. Also, the metal chlorides that have larger radii provided higher erythro-selectivity. For example, the lanthanum chloride gave the highest erythro-selectivity . Further, the reductions proceeded very slowly in ether or benzene because of insolubility of NaBH 4 and metal chlorides in these solvents.
Reduction of α,β-epoxy ketones under the Luche conditions using NaBH 4–CeCl 3 in methanol provides anti- (or erythro-) α,β-epoxy alcohols in high yields and with extremely high stereoselectivity (Eq. (87)) .
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(87).
11. Miscellaneous
11.1. Reduction of carbinols and benzylic alcohols
The NaBH 4 in combination with anhydrous AlCl 3 conveniently reduces some diaryl–arylalkylcarbinols to methylenic hydrocarbons (Eq. (88)) .
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(88).
The reduction of di- and triarylmethanols using NaBH 4 and TFA is ineffective in the case of non-benzylic and monobenzylic alcohols . These methods lead to rapid formation of a very weak reducing agent i.e. trifluoroacetoxy borohydrides which are slow in effecting hydride capture of the carbocation intermediate. It was found that dropwise addition of trifluoroacetic acid to a mixture of the substrate and NaBH 4 in THF provides a protonating medium rich in more reactive reducing species (Eq. (89)) . This method was found to be highly effective for the reduction of monobenzylic alcohols to give the corresponding hydrocarbons in moderate to high yields.
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(89).
11.2. Reduction of azides
Aroyl azides are reduced predominantly to the corresponding benzyl alcohols using NaBH 4 in methanol . However, they yield the corresponding amides using the NiCl 2–NaBH 4 system (Eq. (90)) .(90)p-X-C 6 H 4 CON 3→MeOH,0−10°C NaBH 4−NiCl 2·6 H 2 O p-X-C 6 H 4 CONH 2 80−100%
A facile reduction of alkyl, aryl and aroyl azides using NaBH 4–CuSO 4 system has also been achieved (Eq. (91)) .(91)PhCON 3→MeOH,0−5°C NaBH 4−CuSO 4 PhCONH 2 90%
A series of isoxazole azides were reduced selectively to isoxazole amines in good yields by NaBH 4 in the presence of 1,3-propanedithiol (Eq. (92)) . A solvent effect on this reaction was observed due to differences in the rate of decomposition of NaBH 4 in the alcoholic media. When the reduction was carried out in isopropanol under the same conditions, the azides were reduced to amines in quantitative yields.
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(92).
It is of interest to note that the reducing power of NaBH 4 increases in THF (or t-butyl alcohol) when methanol is added dropwise during the reaction .
11.3. Reduction of oximes and oxime ethers
A reagent system consisting of NaBH 4–LiCl–amberlyst–15(H+) in THF is powerful for the reduction of oximes (Eq. (93)) .
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(93).
Reducing agents prepared using ZrCl 4–NaBH 4 and chiral amino alcohols 9 have been successfully applied to the enantioselective reduction of oxime ethers . Optically active primary amines were obtained in high enantiomeric excess (≥95% ee) with good chemical yields (Eq. (94)).
The reduction was examined using different Lewis acids. High enantioselectivity was observed using ZnCl 2 and AlCl 3 (1:1) mixture, but the removal of the mixed salts posed a difficulty leading to lowering of the chemical yield. Reductions using ZrCl 4–NaBH 4 system gave high chemical and optical yields. Removal of the metal salts after the reaction and isolation of the product are easy in the case of ZrCl 4.
O-Acyl derivatives of aldoximes and ketoximes are reduced in good yields to the corresponding amines using the NaBH 4–I 2 system (Eq. (95)) .
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(94).
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(95).
11.4. Reduction of CN and NN functional groups
The BiCl 3–NaBH 4 system can be used effectively for the reduction of azomethines to the corresponding amines (Eq. (96)) .
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(96).
A catalytic enantioselective borohydride reduction of imines in the presence of optically active cobalt(II) catalyst 7 was achieved (Eq. (97)) .
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(97).
The NaBH 4–I 2 reducing system was successfully used for the reduction of azoarenes and azoxyarenes to the corresponding hydrazobenzenes in good yields (Eq. (98)) . The reaction did not take place in the absence of iodine.
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(98).
11.5. Reduction of alkyl and aryl halides
Dechlorination of 4-chlorobiphenyl with NaBH 4 was promoted by LiCl in high boiling inert solvents (Eq. (99)) .
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(99).
The 2,2,2-trichloroethoxy moiety is widely used as a protective group in organic synthesis. It can be easily deprotected by NaBH 4 under the catalysis of selenium in DMF (Eq. (100)) .
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(100).
The (PhSe)2–NaBH 4 system has been used as a reducing agent for the reduction of α-bromosulfones . This reduction did not proceed without (PhSe)2. Whereas iodomethyl sulfone also undergoes this reduction, the corresponding chloro derivative was not affected (, , ).(101)PhSO 2 CH 2 Br→EtOH,r.t.PhSe 2−NaBH 4 PhSO 2 CH 3 quantitative(102)PhSO 2 CH 2 I→PhSO 2 CH 3 95%(103)PhSO 2 CH 2 Cl→No reaction
This reagent system was also applied for the radical cyclization and intramolecular coupling reactions (Eq. (104)).
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(104).
There is no appreciable reaction between NaBH 4 and aryl halides under most conditions. It was discovered that in the presence of catalytic amount of tris(triphenylphosphine)nickel(0), [Ni(0)(Ph 3 P)3], NaBH 4 serves as a powerful nucleophilic reagent in the reduction of aryl bromides (Eq. (105)) . Use of DMF gives good results and ethanol, methanol and THF were proved to be ineffective. The aromatic chlorides also react, but much more slowly.
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(105).
The reduction of aryl halides by NaBH 4 catalysed by titanocene dichloride is highly effective . The adduct of DMF and NaBH 4 reduces simple aryl halides to give dechlorinated and dimethylamino-substituted products. In dimethyl–acetamide or in ether solvents only dechlorinated products are obtained (Eq. (106)).
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(106).
Nickel boride on borohydride exchange resin (BER) is a good alternative reagent to tributyltin hydride for the coupling of alkyl iodides with the electron deficient alkenes in methanol (Eq. (107)).
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(107).
11.6. Cleavage of ethers using NaBH 4 and additives
A facile cleavage of allylic ethers has been achieved by NaBH 4 using catalytic amount of Pd(PPh 3)4 (Eq. (108)) . The Pd(PPh 3)2 Cl 2 could also be used as a catalyst in this reaction.
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(108).
Also, a facile cleavage of allyl ethers has been achieved using the NaBH 4–I 2 reagent system (Eq. (109)) .
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(109).
11.7. Deoxygenation of phenols and enols using NaBH 4 and additives
Deoxygenation reaction of phenol and 1,3-dicarbonyl compounds was investigated . Phenols, enolizable 1,3-diketones and 3-keto esters are converted to the toluene–p-sulfonates that are reduced by the NiCl 2–NaBH 4 system to obtain the deoxygenated aromatic compounds, alcohols and esters, respectively ((110), (111)).
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(110).
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(111).
11.8. Reduction of sulfides using NaBH 4 and additives
It was found that the arylalkyl, diaryl and dialkyl sulfoxides were reduced to the corresponding sulfides by FeCl 3–NaBH 4 in excellent yield under mild conditions (Eq. (112)) . The sulfuroxygen bond is weakened by the coordination of the sulfoxide oxygen with the metal ion, rendering it more liable to borohydride reduction .
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112.
11.9. Reactions using Bu 3 SnCl–AIBN in combination with NaBH 4
The reaction of (Z)-3-iodoacrylic acid with electrophilic alkenes using tributyltin hydride, NaBH 4 and a catalytic amount of AIBN followed by NaF treatment produced the (E)-α,β-unsaturated carboxylic acids in a stereoselective manner (Eq. (113)) .
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(113).
Also, it has been reported that replacement of hydride in borohydride by an electron donating alkylamino group greatly enhances the reducing ability of the resulting reagents. Thus, sodium(dimethylamino)borohydride (NaDMAB) and (tert-butylamino)-borohydride (NaTBAB) are more effective for the conversion of aldehydes, ketones and esters to alcohols (, ) and primary amides to amines in good yields . Aryl halides are slowly converted to arenes, but alkyl halides and epoxides undergo unusual reactions with the amino portion of the reagents.(114)PhCHO→NaDMAB−THF PhCH 2 OH 90%(115)PhCOPh→NaDMAB−THF PhCH OH 99%Ph
Besides the use of Lewis acids and other additives for enhancing the reaction and selectivity of NaBH 4, different solvents can be also used to realise similar objectives. For example, the rate of reduction of a ketone is solvent dependent with the order of reactivity among alcohols being, MeOH>EtOH>i-PrOH>t-BuOH. Thus, the NaBH 4 reactivity may be attenuated by the nature of the hydroxylic solvent and its concentration .
12. Conclusions
There have been sustained efforts on the development of methods of activation of NaBH 4 using Lewis acids, carboxylic acids, metal salts and some other additives for synthetic applications. Metal halides, Lewis and protic acids and oxidising agents enhance the reactivity of NaBH 4 towards the reduction of various functional groups, including carboncarbon double bonds, carboncarbon triple bonds, carboxylic acids, acyl chlorides, carboxamides, oximes, sulfoxides, nitriles and nitro compounds. Also, certain additives enhance the selectivities in the reduction of aldehydes and ketones. Further, several recipes were reported for enantioselective reductions using NaBH 4. Since it is easy to handle NaBH 4 in large scale syntheses, it is anticipated that this area of research will continue to expand.
Acknowledgements
We thank the CSIR for financial support and for the award of Research Associate position to M.T. The research work in this laboratory on the development of new synthetic methods using NaBH 4 is also supported by the UGC under the Special Assistance Program in Organic Chemistry.
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References
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2014, Chemical Engineering Journal Citation Excerpt :
The excessive use of reducing agents could contaminate the resulting product and even be harmful to human health and the environment . Moreover, some oxygen functionalities in GO are selective and could not be removed completely with only one reductant treatment [30,31]. On the other hand, the thermal reduction route involves the use of high temperature to remove the oxygen functionalities, which would result in high production cost in addition to tedious control of experimental conditions. Show abstract A considerable amount of research has been devoted to the synthesis of graphene materials via graphene oxide (GO) precursor during recent years due to the fact that it is ease in processing, versatile, and scalable for mass production. Nevertheless, GO needs to be reduced in order to recover the unique properties of pristine graphene. Of the various reduction approaches, the electrochemical method provides a facile, fast, scalable, economic and environmentally benign pathway to the production of desirable quality graphene materials. The electrochemical approach can be undertaken via two different routes: the one-step route which involves direct electrochemical reduction of GO in suspension onto the substrate electrode whereas the two-step route requires pre-deposition of GO onto the substrate electrode prior to electrochemical reduction process. This paper first reviews the preparation methods and various properties of graphene oxide. This is followed by a discussion on the working parameters of the two different electrochemical routes and the associated electrochemical techniques used to produce graphene. This paper also provides reviews on the characteristic properties of the electrochemically reduced graphene through the analysis of various spectroscopic techniques, such as X-ray photoelectron spectroscopy, Raman spectroscopy, infrared spectroscopy, X-ray diffraction and electron microscopic.
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2012, Carbon Show abstract Graphene has attracted great interest for its excellent mechanical, electrical, thermal and optical properties. It can be produced by micro-mechanical exfoliation of highly ordered pyrolytic graphite, epitaxial growth, chemical vapor deposition, and the reduction of graphene oxide (GO). The first three methods can produce graphene with a relatively perfect structure and excellent properties, while in comparison, GO has two important characteristics: (1) it can be produced using inexpensive graphite as raw material by cost-effective chemical methods with a high yield, and (2) it is highly hydrophilic and can form stable aqueous colloids to facilitate the assembly of macroscopic structures by simple and cheap solution processes, both of which are important to the large-scale uses of graphene. A key topic in the research and applications of GO is the reduction, which partly restores the structure and properties of graphene. Different reduction processes result in different properties of reduced GO (rGO), which in turn affect the final performance of materials or devices composed of rGO. In this contribution, we review the state-of-art status of the reduction of GO on both techniques and mechanisms. The development in this field will speed the applications of graphene.
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2016, Chemical Reviews
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