Riemann Zeta Function of 4/Proof 5

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Proof

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:

$1$) The number of terms added to calculate the coefficient of the $x^3$ term is $\dbinom k 1 = k$
$2$) The number of terms added to calculate the coefficient of the $x^5$ term is $\dbinom k 2$
$3$) For $k \ge 1$, the coefficient of $x^3$ in $\ds P_k = - \dfrac 1 {\pi^2} \sum_{i \mathop = 1}^k \dfrac 1 {i^2}$
$4$) 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
\(\ds \) \(=\) \(\ds \dfrac {\pi^4} {36} - \dfrac {\pi^4} {60}\)
\(\ds \) \(=\) \(\ds \dfrac {5\pi^4} {180} - \dfrac {3\pi^4} {180}\)
\(\ds \) \(=\) \(\ds \frac {2\pi^4} {180}\)
\(\ds \) \(=\) \(\ds \frac {\pi^4} {90}\)

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Riemann_Zeta_Function_of_4/Proof_5&oldid=647895"