Riemann Zeta Function of 4

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 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 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