Riemann Zeta Function of 4/Proof 5

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Theorem

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

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


Proof

\(\displaystyle \displaystyle \paren {\map \zeta 2}^2\) \(=\) \(\displaystyle \map \zeta 4 + 2 \dfrac {\pi^4} {5!}\) Illustrated below.

Zeta of 4 can be seen along the main diagonal in the illustration below.

The upper right and lower left quadrants are equal and are each equal to the sum associated with the 4th power term in the sin(x)/x expansion 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}} & \paren {\dfrac {1} {25}} & \paren {\dfrac {1} {36}} & \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}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {25}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {36}} & \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}} & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {25}} & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {36}} & \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}} & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {25}} & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {36}} & \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 & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {25}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {36}} & \cdots \\ \paren {\dfrac {1} {25}} & \paren {\dfrac {1} {25}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {25}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {25}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {25}} \paren {\dfrac {1} {16}} & \paren {\dfrac {1} {25}}^2 & \paren {\dfrac {1} {25}} \paren {\dfrac {1} {36}} & \cdots \\ \paren {\dfrac {1} {36}} & \paren {\dfrac {1} {36}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {36}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {36}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {36}} \paren {\dfrac {1} {16}} & \paren {\dfrac {1} {36}} \paren {\dfrac {1} {25}} & \paren {\dfrac {1} {36}}^2 & \cdots \\ \cdots & \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

$\blacksquare$