Riemann Zeta Function of 4

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Theorem

The Riemann zeta function of $4$ is given by:

\(\displaystyle \map \zeta 4\) \(=\) \(\displaystyle \dfrac 1 {1^4} + \dfrac 1 {2^4} + \dfrac 1 {3^4} + \dfrac 1 {4^4} + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi^4} {90}\)
\(\displaystyle \) \(\approx\) \(\displaystyle 1 \cdotp 08232 \, 3 \ldots\)

This sequence is A013662 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof 1

By Fourier Series of Fourth Power of x, for $x \in \closedint {-\pi} \pi$:

$\displaystyle x^4 = \frac {\pi^4} 5 + \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \, \map \cos {n \pi} \, \map \cos {n x}$

Setting $x = \pi$:

\(\displaystyle \pi^4\) \(=\) \(\displaystyle \frac {\pi^4} 5 + \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \, \map {\cos^2} {n \pi}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {4 \pi^4} 5\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2} {n^4} - \sum_{n \mathop = 1}^\infty \frac {48} {n^4}\) Cosine of Multiple of Pi
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^4} 5\) \(=\) \(\displaystyle 2 \pi^2 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} - 12 \sum_{n \mathop = 1}^\infty \frac 1 {n^4}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^4} 3 - 12 \sum_{n \mathop = 1}^\infty \frac 1 {n^4}\) Basel Problem
\(\displaystyle \leadsto \ \ \) \(\displaystyle 12 \sum_{n \mathop = 1}^\infty \frac 1 {n^4}\) \(=\) \(\displaystyle \frac {\pi^4} 3 - \frac {\pi^4} 5\) rearranging
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \pi^4} {15}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^4}\) \(=\) \(\displaystyle \frac {\pi^4} {90}\)

$\blacksquare$


Proof 2

By Fourier Series of x squared, for $x \in \left[{- \pi \,.\,.\, \pi}\right]$:

$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\left({-1}\right)^n \frac 4 {n^2} \cos n x}\right)$


Hence:

\(\displaystyle \frac 1 \pi \int_{-\pi}^\pi x^4 \, \mathrm d x\) \(=\) \(\displaystyle \frac 1 2 \left({\frac {2 \pi^2} 3}\right)^2 + \sum_{n \mathop = 1}^\infty \left({\frac {4 \left({-1}\right)^n} {n^2} }\right)^2\) Parseval's Theorem
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 2 \pi \int_0^\pi x^4 \, \mathrm d x\) \(=\) \(\displaystyle \frac {2 \pi^4} 9 + \sum_{n \mathop = 1}^\infty \frac {16} {n^4}\) Definite Integral of Even Function
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {2 \pi^4} 5\) \(=\) \(\displaystyle \frac {2 \pi^4} 9 + \sum_{n \mathop = 1}^\infty \frac {16} {n^4}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {8 \pi^4} {45}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {16} {n^4}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^4}\) \(=\) \(\displaystyle \frac {\pi^4} {90}\)

$\blacksquare$


Proof 3


Proof 4

\(\displaystyle \map \zeta 4\) \(=\) \(\displaystyle \paren {-1}^3 \dfrac {B_4 2^3 \pi^4} {4!}\) Riemann Zeta Function at Even Integers
\(\displaystyle \) \(=\) \(\displaystyle \paren {-1}^3 \paren {-\dfrac 1 {30} } \dfrac {2^3 \pi^4} {4!}\) Definition of Sequence of Bernoulli Numbers
\(\displaystyle \) \(=\) \(\displaystyle \paren {\dfrac 1 {30} } \paren {\dfrac 8 {24} } \pi^4\) Definition of Factorial
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi^4} {90}\) simplifying

$\blacksquare$


Proof 5

From the Basel Problem we have:

$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$

Therefore:

\(\displaystyle \map \zeta 2\) \(=\) \(\displaystyle \paren {\frac 1 {1^2 } + \frac 1 {2^2 } + \frac 1 {3^2 } + \frac 1 {4^2 } + \cdots }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }\)


Taking the square of $\map \zeta 2$, we have:

\(\displaystyle \displaystyle \paren {\map \zeta 2}^2\) \(=\) \(\displaystyle \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots } \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\sum_{i \mathop = 1}^{\infty} {\frac 1 {i^2 } } } \paren {\sum_{j \mathop = 1}^{\infty} {\frac 1 {j^2 } } }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{\infty} \sum_{j \mathop = 1}^{\infty} {\frac 1 {i^2 } } {\frac 1 {j^2 } }\) Product of Absolutely Convergent Series
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{\infty} {\frac 1 {i^4 } } + \sum_{i \mathop = 2}^{\infty} \sum_{j \mathop = 1}^{i-1} {\frac 1 {i^2 } } {\frac 1 {j^2 } } + \sum_{j \mathop = 2}^{\infty} \sum_{i \mathop = 1}^{j - 1} {\frac 1 {i^2 } } {\frac 1 {j^2 } }\) $\paren {i = j } + \paren {j \lt i } + \paren {j \gt i }$
\(\displaystyle \) \(=\) \(\displaystyle \map \zeta 4 + 2 \dfrac {\pi^4} {5!}\) The sums $\paren {j < i }$ and $\paren {j > i }$ are symmetric and each equal to the coefficient of the 5th power term in the sin(x) expansion See Basel_Problem/Proof_2


$\begin{array}{r|cccccccccc} \displaystyle \paren {\map \zeta 2}^2 & \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {16}} & \cdots \\ \hline \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {1}}^2 & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {16}} & \cdots \\ \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {4}}^2 & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {16}} & \cdots \\ \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {9}}^2 & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {16}} & \cdots \\ \paren {\dfrac {1} {16}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {16}}^2 & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \end{array}$


\(\displaystyle \map \zeta 4\) \(=\) \(\displaystyle \paren {\map \zeta 2}^2 - 2 \dfrac {\pi^4} {5!}\) Rearranging
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi^4} {36} - \dfrac {\pi^4} {60}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi^4} {90}\) simplifying

$\blacksquare$


The decimal expansion can be found by an application of arithmetic.


Historical Note

The Riemann Zeta Function of 4 was solved by Leonhard Euler, using the same technique as for the Basel Problem.


If only my brother were alive now.
-- Johann Bernoulli


Sources

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