Riemann Zeta Function of 4

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Theorem

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

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

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 $x \in \closedint {-\pi} \pi$:

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

Setting $x = \pi$:

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

$\blacksquare$


Proof 2

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

$\ds x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$


Hence:

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

$\blacksquare$


Proof 3


This theorem requires a proof.
In particular: $\ds \int_0^1\int_0^1\int_0^1\int_0^1 \frac 1 {1 - x_1x_2x_3x_4} \rd x_1 \rd x_2 \rd x_3 \rd x_4$. Use Beukers-Calabi-Kolk sub.
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Proof 4

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

$\blacksquare$


Proof 5

Create a multiplication table where the column down the left hand side and the row across the top each contains the terms of zeta function of $2$:

$\begin {array} {c|cccccccccc} \paren {\map \zeta 2}^2 & \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {4^2} } & \cdots \\ \hline \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {1^4} } & \paren {\dfrac 1 {1^2} } \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {1^2} } \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {1^2} } \paren {\dfrac 1 {4^2} } & \cdots \\ \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {2^2} } \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {2^4} } & \paren {\dfrac 1 {2^2} } \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {2^2} } \paren {\dfrac 1 {4^2} } & \cdots \\ \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {3^2} } \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {3^2} } \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {3^4} } & \paren {\dfrac 1 {3^2} } \paren {\dfrac 1 {4^2} } & \cdots \\ \paren {\dfrac 1 {4^2} } & \paren {\dfrac 1 {4^2} } \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {4^2} } \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {4^2} } \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {4^4} } & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end {array}$


The sum of all of the entries in this table is equal to $\paren {\map \zeta 2}^2$.

$\map \zeta 4$ is the sum of the entries along the main diagonal.


We have:

\(\ds \paren {\map \zeta 2}^2\) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^\infty {\frac 1 {i^2} } } \paren {\sum_{j \mathop = 1}^\infty {\frac 1 {j^2} } }\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^\infty \sum_{j \mathop = 1}^\infty {\frac 1 {i^2} } {\frac 1 {j^2} }\) Product of Absolutely Convergent Series
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^{\infty} {\frac 1 {i^4} } + \sum_{i \mathop = 1}^\infty \sum_{j \mathop = {i + 1} }^\infty {\frac 1 {i^2} } {\frac 1 {j^2} } + \sum_{j \mathop = 1}^\infty \sum_{i \mathop = {j + 1} }^\infty {\frac 1 {i^2} } {\frac 1 {j^2} }\) $\paren {i = j} + \paren {j > i} + \paren {j < i}$
\(\ds \) \(=\) \(\ds \map \zeta 4 + 2 \sum_{i \mathop = 1}^\infty \sum_{j \mathop = {i + 1} }^\infty {\frac 1 {i^2} } {\frac 1 {j^2} }\)


Let:

\(\ds P_k\) \(=\) \(\ds x \prod_{n \mathop = 1}^k \paren {1 - \frac {x^2} {n^2 \pi^2} }\)
\(\ds \) \(=\) \(\ds x \paren {1 - \dfrac {x^2} {1 \pi^2} } \paren {1 - \dfrac {x^2} {2^2 \pi^2} } \paren {1 - \dfrac {x^2} {3^2 \pi^2} } \cdots \paren {1 - \dfrac {x^2} {k^2 \pi^2} }\)

Therefore:

\(\ds P_1\) \(=\) \(\ds x - \frac {x^3} {\pi^2} \paren {\dfrac 1 {1^2} }\)
\(\ds P_2\) \(=\) \(\ds x - \frac {x^3} {\pi^2} \paren {\dfrac 1 {1^2} + \dfrac 1 {2^2} } + \frac {x^5} {\pi^4} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} }\)
\(\ds P_3\) \(=\) \(\ds x - \frac {x^3} {\pi^2} \paren {\dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} } + \frac {x^5} {\pi^4} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} + \dfrac 1 {1^2} \dfrac 1 {3^2} + \dfrac 1 {2^2} \dfrac 1 {3^2} } - \frac {x^7} {\pi^6} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} \dfrac 1 {3^2} }\)
\(\ds P_4\) \(=\) \(\ds x - \frac {x^3} {\pi^2} \paren {\dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} } + \frac {x^5} {\pi^4} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} + \dfrac 1 {1^2} \dfrac 1 {3^2} + \dfrac 1 {1^2} \dfrac 1 {4^2} + \dfrac 1 {2^2} \dfrac 1 {3^2} + \dfrac 1 {2^2} \dfrac 1 {4^2} + \dfrac 1 {3^2} \dfrac 1 {4^2} } - \frac {x^7} {\pi^6} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} \dfrac 1 {3^2} + \dfrac 1 {1^2} \dfrac 1 {2^2} \dfrac 1 {4^2} + \dfrac 1 {1^2} \dfrac 1 {3^2} \dfrac 1 {4^2} + \dfrac 1 {2^2} \dfrac 1 {3^2} \dfrac 1 {4^2} } + \frac {x^9} {\pi^8} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} \dfrac 1 {3^2} \dfrac 1 {4^2} }\)

We make the following observations:

$(1): \quad$ The number of terms added to calculate the coefficient of the $x^3$ term is $\dbinom k 1 = k$
$(2): \quad$ The number of terms added to calculate the coefficient of the $x^5$ term is $\dbinom k 2$
$(3): \quad$ For $k \ge 1$, the coefficient of $x^3$ in $\ds P_k = - \dfrac 1 {\pi^2} \sum_{i \mathop = 1}^k \dfrac 1 {i^2}$
$(4): \quad$ For $k \ge 2$, the coefficient of $x^5$ in $\ds P_k = \dfrac 1 {\pi^4} \sum_{i \mathop = 1}^{k - 1} \sum_{j \mathop = {i + 1} }^k \paren {\frac 1 {i^2} } \paren {\frac 1 {j^2} } $


Expanding the product out to k, we get:

\(\ds P_k\) \(=\) \(\ds x - \dfrac {x^3} {\pi^2} \sum_{i \mathop = 1}^k \dfrac 1 {i^2} + \frac {x^5} {\pi^4} \sum_{i \mathop = 1}^{k - 1} \sum_{j \mathop = {i + 1} }^k \paren {\frac 1 {i^2} } \paren {\frac 1 {j^2} } - \cdots\)

Now recall the following two representations of the Sine of x:

\(\ds \sin x\) \(=\) \(\ds x \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }\) Euler Formula for Sine Function
\(\ds \sin x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots\) Power Series Expansion for Sine Function

Notice that by taking the limit of $P_k$ as $k \to \infty$, we obtain precisely the Euler Formula for Sine Function.

Equating the coefficient of $x^5$ in the Euler Formula for Sine Function with the Power Series Expansion for Sine Function, we have:

$\ds \lim_{k \mathop \to \infty} \dfrac 1 {\pi^4} \sum_{i \mathop = 1}^{k - 1} \sum_{j \mathop = {i + 1} }^k \paren {\frac 1 {i^2} } \paren {\frac 1 {j^2} } = \frac 1 {5!}$

Therefore:

$\ds \sum_{i \mathop = 1}^\infty \sum_{j \mathop = {i + 1} }^\infty \paren {\frac 1 {i^2} } \paren {\frac 1 {j^2} } = \frac {\pi^4} {5!}$


Therefore:

\(\ds \paren {\map \zeta 2}^2\) \(=\) \(\ds \map \zeta 4 + 2 \sum_{i \mathop = 1}^\infty \sum_{j \mathop = {i + 1} }^\infty {\frac 1 {i^2} } {\frac 1 {j^2} }\)
\(\ds \paren {\map \zeta 2}^2\) \(=\) \(\ds \map \zeta 4 + 2 \dfrac {\pi^4} {5!}\)
\(\ds \map \zeta 4\) \(=\) \(\ds \paren {\map \zeta 2}^2 - 2 \dfrac {\pi^4} {5!}\) Rearranging
\(\ds \) \(=\) \(\ds \paren{\dfrac {\pi^2} 6 }^2 - \dfrac {\pi^4} {60}\) Basel Problem
\(\ds \) \(=\) \(\ds \dfrac {\pi^4} {36} - \dfrac {\pi^4} {60}\)
\(\ds \) \(=\) \(\ds \dfrac {5\pi^4} {180} - \dfrac {3\pi^4} {180}\)
\(\ds \) \(=\) \(\ds \frac {2\pi^4} {180}\)
\(\ds \) \(=\) \(\ds \frac {\pi^4} {90}\)

$\blacksquare$


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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