Math Problem Statement

ln(0, 2π)((π - x)/2)^2 = π^2/12 + Σ(cos(nx)/n^2), then 1/1^4 + 1/2^4 + 1/3^4 + ... = ?

Solution

The image presents a mathematical problem involving a series and an equation, which appears to be related to Fourier series or trigonometric sums.

The given expression is: ln(0,2π)(πx2)2=π212+n=1cos(nx)n2\ln(0, 2\pi)\left(\frac{\pi - x}{2}\right)^2 = \frac{\pi^2}{12} + \sum_{n=1}^{\infty} \frac{\cos(nx)}{n^2} It asks to evaluate the following sum: n=11n4=114+124+134+144+\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots

The given options are:

  • (a) π490\frac{\pi^4}{90}
  • (b) π496\frac{\pi^4}{96}
  • (c) π492\frac{\pi^4}{92}
  • (d) π494\frac{\pi^4}{94}

Solution:

The series in the problem is the Riemann zeta function evaluated at s=4s = 4: ζ(4)=n=11n4\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} It is known that: ζ(4)=π490\zeta(4) = \frac{\pi^4}{90}

Thus, the value of the sum is π490\frac{\pi^4}{90}, which matches option (a).

Final Answer:

The correct answer is (a) π490\frac{\pi^4}{90}.

Would you like further details or clarification on any part of the solution?

Here are 5 relative questions that expand the previous explanation:

  1. What is the Riemann zeta function, and how is it generally defined?
  2. Can you explain how the value of ζ(4)\zeta(4) is derived?
  3. What other important values of the Riemann zeta function are commonly used in mathematical problems?
  4. How are Fourier series related to trigonometric sums like the one shown in the equation?
  5. Can this approach be extended to other sums involving higher powers of nn?

Tip: The Riemann zeta function has many important applications in both pure and applied mathematics, particularly in number theory and physics.

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Math Problem Analysis

Mathematical Concepts

Riemann zeta function
Infinite series
Trigonometric sums

Formulas

ln(0, 2π)((π - x)/2)^2 = π^2/12 + Σ(cos(nx)/n^2)
Σ(1/n^4) = ζ(4)
ζ(4) = π^4/90

Theorems

Riemann zeta function

Suitable Grade Level

Undergraduate level or advanced high school (Grades 11-12)