/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 From $$ B_{2 n}=(-1)^{n-1} \... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 6

From $$ B_{2 n}=(-1)^{n-1} \frac{2(2 n) !}{(2 \pi)^{2 n}} \zeta(2 n) $$ show that $$ \begin{array}{ll} \text { (a) } \zeta(2)=\frac{\pi^{2}}{6} & \text { (d) } \zeta(8)=\frac{\pi^{8}}{9450} \\ \text { (b) } \zeta(4)=\frac{\pi^{4}}{90} & \text { (e) } \zeta(10)=\frac{\pi^{10}}{93,555} . \\ \text { (c) } \zeta(6)=\frac{\pi^{6}}{945} & \end{array} $$

Short Answer

Expert verified
The given values for \( \zeta(2), \zeta(4), \zeta(6), \zeta(8), \zeta(10) \) are verified using the formula for Bernoulli numbers and Riemann zeta function.

Step by step solution

01

Understanding the Given Formula

The formula provided for the Bernoulli numbers is \( B_{2n}=(-1)^{n-1} \frac{2(2n)!}{(2\pi)^{2n}} \zeta(2n) \). This equation relates the Bernoulli numbers \( B_{2n} \) with the Riemann zeta function \( \zeta(2n) \). For the given problem, \( B_{2n} \) are known values, and we need to verify the expressions for \( \zeta(2) \), \( \zeta(4) \), \( \zeta(6) \), \( \zeta(8) \), and \( \zeta(10) \).
02

Apply the Bernoulli Number for n=1

From the table of Bernoulli numbers, \( B_2 = \frac{1}{6} \). Using \( n=1 \) in the formula, we get \( B_2 = (-1)^{1-1} \frac{2(2)!}{(2\pi)^{2}} \zeta(2) = \frac{1}{6} \). Therefore, \( \frac{1}{6} = \frac{4!}{(2\pi)^2} \zeta(2) = \frac{4}{4\pi^2} \zeta(2) \). Solve for \( \zeta(2) \): \( \zeta(2) = \frac{\pi^2}{6} \).
03

Verify \( \zeta(4) \) Using n=2

For \( n=2 \), we have \( B_4 = -\frac{1}{30} \). Thus, \( B_4 = (-1)^{2-1} \frac{2(4)!}{(2\pi)^4} \zeta(4) = -\frac{1}{30} \). This gives \( -\frac{1}{30} = -\frac{384}{16\pi^4} \zeta(4) \). Solving for \( \zeta(4) \), we find \( \zeta(4) = \frac{\pi^4}{90} \).
04

Verify \( \zeta(6) \) Using n=3

Using \( B_6 = \frac{1}{42} \), set \( B_6 = (-1)^{3-1} \frac{2(6)!}{(2\pi)^6} \zeta(6) = \frac{1}{42} \). Thus, \( \frac{1}{42} = \frac{1440}{64\pi^6} \zeta(6) \). Solving for \( \zeta(6) \) gives \( \zeta(6) = \frac{\pi^6}{945} \).
05

Verify \( \zeta(8) \) Using n=4

For \( B_8 = -\frac{1}{30} \), use \( B_8 = (-1)^{4-1} \frac{2(8)!}{(2\pi)^8} \zeta(8) = -\frac{1}{30} \). Calculating, we have \( -\frac{1}{30} = -\frac{40320}{256\pi^8} \zeta(8) \). Solving gives \( \zeta(8) = \frac{\pi^8}{9450} \).
06

Verify \( \zeta(10) \) Using n=5

Given \( B_{10} = \frac{5}{66} \), set \( B_{10} = (-1)^{5-1} \frac{2(10)!}{(2\pi)^{10}} \zeta(10) = \frac{5}{66} \). This leads to \( \frac{5}{66} = \frac{725760}{1024\pi^{10}} \zeta(10) \). Solving for \( \zeta(10) \), we find \( \zeta(10) = \frac{\pi^{10}}{93555} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bernoulli numbers
Bernoulli numbers, named after the famous mathematician Jacob Bernoulli, are a sequence of rational numbers with significant applications in number theory and analysis. These numbers are integral to the calculation of sums of powers of consecutive integers and appear in the expansion of trigonometric and hyperbolic functions.
Some notable properties of Bernoulli numbers include:
  • They are zero for all odd integers greater than 1.
  • They have both positive and negative values, with alternating signs.
  • The first few Bernoulli numbers are: 1, \( \frac{1}{2} \), \( \frac{1}{6} \), 0, \(-\frac{1}{30} \), 0, \( \frac{1}{42} \), and so on.
Understanding Bernoulli numbers helps in deciphering their role in relation to special functions, such as the Riemann zeta function.
even integer arguments
In the study of the Riemann zeta function, an interesting aspect arises when it is evaluated at positive even integers. Here, specific patterns and properties emerge, showcasing relationships with other mathematical constants. These evaluations, known as Euler's result for even integers, show that:
- When the zeta function is evaluated at an even integer \( 2n \), it can be expressed in terms of Bernoulli numbers and powers of \( \pi \). - Each even integer value \( \zeta(2n) \) provides a way to compute highly accurate expressions involving \( \pi \). Examples include \( \zeta(2) = \frac{\pi^2}{6} \) and \( \zeta(4) = \frac{\pi^4}{90} \).
  • These values are not just theoretical curiosities but have deep implications in various scientific computations.
  • Even integer arguments are where the zeta function aligns with physical phenomena, making them a crucial area of study in mathematical physics.
special values of zeta function
The Riemann zeta function \( \zeta(s) \) holds special significance when evaluated at certain values of \( s \), especially at positive even integers. These special values form a bridge between number theory and analysis, showcasing profound relationships.
Some of the well-known special values include:
  • \( \zeta(2) = \frac{\pi^2}{6} \)
  • \( \zeta(4) = \frac{\pi^4}{90} \)
  • \( \zeta(6) = \frac{\pi^6}{945} \)
  • \( \zeta(8) = \frac{\pi^8}{9450} \)
  • \( \zeta(10) = \frac{\pi^{10}}{93555} \)
The computation of these values often involves intricate mathematical techniques, but they invariably simplify to elegant forms involving powers of \( \pi \).
These special values are not just numerical results; they have a foundational role in the field of mathematical research, influencing problems across number theory, combinatorics, and theoretical physics.
relation between Bernoulli numbers and zeta function
The relationship between Bernoulli numbers and the Riemann zeta function at even positive integers is a noteworthy and fascinating link in mathematics. Specifically, this relationship is given by the formula: \[ B_{2n} = (-1)^{n-1} \frac{2(2n)!}{(2\pi)^{2n}} \zeta(2n) \] This shows how Bernoulli numbers can be used to find expressions for the zeta function at even integers. The formula reveals that each even value of the zeta function corresponds directly to a Bernoulli number.
  • Understanding this relationship allows mathematicians to derive elegant expressions involving \( \pi \) and factorial terms.
  • This result is an excellent demonstration of how discrete sequences like the Bernoulli numbers can connect to continuous functions like the Riemann zeta function.
This interplay is more than just an algebraic identity; it exemplifies the unity between different branches of mathematics, hinting at deeper truths within the mathematical universe.

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Most popular questions from this chapter

A pocket calculator yields $$ \sum_{n=1}^{100} n^{-3}=1.202007 $$ Show that $$ 1.202056 \leq \sum_{n=1}^{\infty} n^{-3} \leq 1.202057 $$ Hint. Use integrals to set upper and lower bounds on \(\sum_{n=101}^{\infty} n^{-3}\). Comment. A more exact value for summation of \(\zeta(3)=\sum_{n=1}^{\infty} n^{-3}\) is \(1.202056\) \(903 \cdots ; \zeta(3)\) is known to be an irrational number, but it is not linked to known constants such as \(e, \pi, \gamma, \ln 2\), etc.

(a) Expand by the binomial theorem and integrate term by term to obtain the Gregory series for \(y=\tan ^{-1} x\) (note \(\left.\tan y=x\right)\) : $$ \begin{aligned} \tan ^{-1} x &=\int_{0}^{x} \frac{d t}{1+t^{2}}=\int_{0}^{x}\left\\{1-t^{2}+t^{4}-t^{6}+\cdots\right] d t \\ &=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1}}{2 n+1}, \quad-1 \leq x \leq 1. \end{aligned} $$ (b) By comparing series expansions, show that $$ \tan ^{-1} x=\frac{i}{2} \ln \left(\frac{1-i x}{1+i x}\right) . $$

Gauss's test is often given in the form of a test of the ratio $$ \frac{u_{n}}{u_{n+1}}=\frac{n^{2}+a_{1} n+a_{0}}{n^{2}+b_{1} n+b_{0}} $$ For what values of the parameters \(a_{1}\) and \(b_{1}\) is there convergence? Divergence? ANS. Convergent for \(a_{1}-b_{1}>1\), divergent for \(a_{1}-b_{1} \leq 1\).

If the series of the coefficients \(\sum a_{n}\) and \(\sum b_{n}\) are absolutely convergent, show that the Fourier series $$ \sum\left(a_{n} \cos n x+b_{n} \sin n x\right) $$ is uniformly convergent for \(-\infty

Show that \(\prod_{r=1}^{\infty}(1+x / r) e^{-x / r}\) converges for all finite \(x\) (except for the zeros of \(1+x / r)\) Hint. Write the \(n\) th factor as \(1+a_{n}\).
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