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| − | | + | [https://polybox.ethz.ch/public.php?service=files&t=3cd27159ac0fca5b53c53fe4cff98dcc] |
| − | =Problem=
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| − | Show, by using the Fourier series, that
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| − | $$\sum_{k=1}^\infty \frac{1}{k^n} = - \frac{(2 \pi i)^n}{2(n!)} B_n,\ n \in 2 \mathbb{Z}_{>0}$$
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| − | where \(B_n\) are the Bernoulli numbers defined as
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| − | $$\frac{z}{e^z - 1} = \sum_{n=0}^\infty B_n \frac{z^n}{n!}$$
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| − | ''Hint:'' consider \( B_n(x) := \sum_{k \in \mathbb{Z} \setminus \{0\}} k^{-n} e^{ikx},\ n \in \mathbb{Z}_{\geq 2},\ x \in \mathbb{R} \)
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| − | =Solution Sketch=
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| − | The solution below is extremely comprehensive, if not expletive, because I felt that the [http://www.math.ethz.ch/~gruppe5/group5/group5.php?lectures/mmp/hs11/ml03.pdf Musterlösung] of the MMP Series that treated the problem was lacking some fairly important steps/guidelines. Therefore find a '''TL;DR''' here:
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| − | # See that \(B_n(0) = 2 \sum_{k=1}^\infty k^{-n}\). We now want to find the \(B_n(0)\) for \(n \in 2\mathbb{Z}_{>0}\)
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| − | # We find a polynomial with Fourier coefficients \(c_0=0\) and \(c_k = \frac{1}{k^2}\). <br/> On \([0,2\pi]\), this polynomial will thus be equal to \(B_2(x)\)
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| − | # \(B'_n(x) = iB_{n-1}(x)\), therefore all \(B_n(x)\) will be polynomials, too. We find a recursion relation for \(B_n(x)\)
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| − | # Using the \(2\pi\)-periodicity of \(B_n(x)\) (for all \(n \geq 2\)), we find a recursion relation for the \(B_n(0)\)
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| − | # We prove the postulated identity by relating the definition for the \(B_n\) given in the problem to the recursion for the \(B_n(0)\) found above.
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| − | =Solution=
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| − | As suggested in the problem, we will have a look at the convergent fourier series
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| − | $$B_n(x) := \sum_{k \in \mathbb{Z} \setminus \{0\}} k^{-n} e^{ikx},\ n \in \mathbb{Z}_{\geq 2},\ x \in \mathbb{R}$$
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| − | with Fourier coefficients
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| − | $$c_k = \begin{cases}
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| − | 0 & \text{for } k = 0\\
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| − | \frac{1}{k^n} & \text{for } k \neq 0
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| − | \end{cases}$$
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| − | It is obvious that \( B_n(x) \) is \(2\pi\)-periodic and, by differentiating every term separately, that
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| − | $$B'_n(x) = iB_{n-1}(x)$$
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| − | Also, for \(n \in 2\mathbb{Z}_{>0}\), we see that
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| − | $$B_n(0) = \sum_{k \in \mathbb{Z} \setminus \{0\}} k^{-n} e^{ik\,0}
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| − | = \sum_{k=-\infty}^{-1} k^{-n} + \sum_{k=1}^\infty k^{-n}
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| − | = 2 \sum_{k=1}^\infty k^{-n}
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| − | \tag{$\ddagger$}$$
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| − | where the last step follows from the fact that \(n\) is even.
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| − | <hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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| − | We will first show that \( B_2(x) = f(x) := \pi^2/3 - \pi x + x^2/2 \) for \( x \in [0, 2\pi]\).
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| − | Recall the definition of the Fourier coefficients for \(2\pi\)-periodic functions in \(C^\infty(\mathbb{R})\):
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| − | $$c_k = \frac{1}{2\pi} \int_0^{2\pi} f(x) \: e^{-ikx} \: dx$$
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| − | We can easily verify the following identities:
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| − | $$\begin{align}
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| − | \frac{1}{2\pi} \int_0^{2\pi} e^{-ikx} \: dx &\ =\
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| − | \begin{cases}
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| − | 1 & \text{for } k = 0\\
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| − | 0 & \text{for } k \neq 0
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| − | \end{cases} \\
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| − | \frac{1}{2\pi} \int_0^{2\pi} x \: e^{-ikx} \: dx &\ =\
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| − | \begin{cases}
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| − | \pi & \text{for } k = 0\\
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| − | \frac{i}{k} & \text{for } k \neq 0
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| − | \end{cases} \\
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| − | \frac{1}{2\pi} \int_0^{2\pi} x^2 \: e^{-ikx} \: dx &\ =\
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| − | \begin{cases}
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| − | \frac{4}{3} \pi^2 & \text{for } k = 0\\
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| − | \frac{2(1+ik\pi)}{k^2} & \text{for } k \neq 0
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| − | \end{cases}
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| − | \end{align}$$
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| − | Calculating the Fourier coefficients for \(f(x)\) then yields
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| − | $$\begin{align}
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| − | c_k &= \frac{1}{2\pi} \int_0^{2\pi} f(x) \: e^{-ikx} \: dx\\
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| − | &= \frac{1}{2\pi} \left(
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| − | \frac{\pi^2}{3} \int_0^{2\pi} e^{-ikx} \: dx
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| − | - \pi \int_0^{2\pi} x \: e^{-ikx} \: dx
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| − | + \frac{1}{2} \int_0^{2\pi} x^2 \: e^{-ikx}
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| − | \right) \\
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| − | &= \begin{cases}
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| − | 0 & \text{for } k = 0\\
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| − | \frac{1}{k^2} & \text{for } k \neq 0
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| − | \end{cases}
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| − | \end{align}$$
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| − | which proves that \(B_2(x) = f(x)\) for \(x \in [0, 2\pi]\).
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| − | Thus, we have shown that the \(B_n|_{[0, 2\pi]}\) are polynomials, which are recursively defined by \(B'_n = iB_{n-1}\) and \(\int_0^{2\pi} B_n(x) dx = -i(B_{n+1}(2\pi) - B_{n+1}(0)) = 0\). We also add
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| − | $$B_1(x) = -iB'_2(x) = i\pi - ix \qquad B_0(x) = -iB'_1(x) = -1$$
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| − | <hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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| − | By induction over \(n\), we will now prove that
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| − | $$B_n(x) = \sum_{k=0}^n B_{n-k}(0) \frac{(ix)^k}{k!}$$
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| − | Base case for \(n=0\):
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| − | $$B_0(0) \frac{(ix)^0}{0!} = B_0(0) = -1 = B_0(x)$$
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| − | Inductive step (\(n\!-\!1 \rightarrow n\)):
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| − | $$\begin{align}
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| − | B_n(x) &= \int_0^x iB_{n-1}(\tilde x)\ d\tilde x + B_n(0) \\
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| − | &= \int_0^x i\left(\sum_{k=0}^{n-1} B_{(n-1)-k}(0) \frac{(i\tilde x)^k}{k!} \right) d\tilde x + B_n(0) \\
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| − | &= \left( \sum_{k=0}^{n-1} B_{(n-1)-k}(0) \frac{i^{k+1}}{k!} \int_0^x \tilde x^k\ d\tilde x \right) + B_n(0) \\
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| − | &= \left( \sum_{k=0}^{n-1} B_{(n-1)-k}(0) \frac{i^{k+1}}{k!} \frac{x^{k+1}}{k+1} \right) + B_n(0) \\
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| − | &= \left( \sum_{k=0}^{n-1} B_{n-(k+1)}(0) \frac{(ix)^{k+1}}{(k+1)!} \right) + B_n(0) \\
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| − | &= \left( \sum_{k=1}^n B_{n-k}(0) \frac{(ix)^k}{k!} \right) + B_n(0) \\
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| − | &= \sum_{k=0}^n B_{n-k}(0) \frac{(ix)^k}{k!}
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| − | \end{align}$$
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| − | <hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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| − | As \(B_n\) is \(2\pi\)-periodic for all \(n \geq 2\), the above identity leads us to the relation
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| − | $$0 = B_n(2\pi) - B_n(0) = \sum_{k=1}^n B_{n-k}(0) \frac{(2\pi i)^k}{k!} \tag{*}$$
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| − | for all \(n \geq 2\), which determines all \(B_n(0)\) from \(B_0(0) = -1\)
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| − | <hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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| − | The definition of the \(B_n\) in the problem (the Bernoulli numbers, not the functions!) can be transformed to:
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| − | $$\begin{align}
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| − | & \frac{z}{e^z - 1} = \sum_{n=0}^\infty B_n \frac{z^n}{n!} \\
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| − | \Leftrightarrow\ & z = \left( \sum_{n=0}^\infty B_n \frac{z^n}{n!} \right) \left( \sum_{k=1}^\infty \frac{z^k}{k!} \right) \\
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| − | \Leftrightarrow\ & z = \left( B_0 + B_1z + \frac{B_2}{2}z^2 + \frac{B_3}{3!}z^3 + ... \right) \left( z + \frac{z^2}{2} + \frac{z^3}{3!} + \frac{z^4}{4!} + ... \right) \\
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| − | \Leftrightarrow\ & z = B_0z + \left( \frac{B_0}{2!} + B_1 \right)z^2 + \left( \frac{B_0}{3!} + \frac{B_1}{2!} + \frac{B_2}{2!} \right)z^3 + \left( \frac{B_0}{4!} + \frac{B_1}{3!} + \frac{B_2}{2!2!} + \frac{B_3}{3!} \right)z^4 \: +\ ...
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| − | \end{align}$$
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| − | which, through equating coefficients yields \(B_0 = 1\) and for \(n \geq 2\)
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| − | $$\sum_{k=1}^n \frac{B_{n-k}}{(n-k)!k!} = 0 \tag{$\dagger$}$$
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| − | <hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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| − | <span style="color:#339;font-size:80%;">The following proof by induction doesn't appear in the aforementioned Musterlösung, where the final step simply falls out of thin air. It is not unlikely, however, that I simply don't understand what they do. If this is the case, feel free to simplify the following mess of equations. --Nik</span>
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| − | We now now prove by induction that for \(n \geq 0\)
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| − | $$B_n(0) = - \frac{(2\pi i)^n B_n}{n!}$$
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| − | Base case for \(n=0\):
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| − | $$B_0(0) = -1 = -B_0$$
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| − | Inductive step (\(n\!-\!1 \rightarrow n\)):
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| − | $$\begin{align}
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| − | \sum_{k=1}^{n+1} B_{n+1-k}(0) \frac{(2\pi i)^k}{k!} \ &\overset{(*)}{=} \ 0 \\
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| − | \Leftrightarrow \ B_n(0)\: 2\pi i \ &= \: - \sum_{k=2}^{n+1} B_{n+1-k}(0) \frac{(2\pi i)^k}{k!} \\
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| − | \overset{\text{Hyp.}}{\Leftrightarrow} \ B_n(0)\: 2\pi i \ &= \: - \sum_{k=2}^{n+1} \left( - \frac{(2\pi i)^{n+1-k} B_{n+1-k}}{(n+1-k)!} \right) \frac{(2\pi i)^k}{k!} \\
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| − | \Leftrightarrow \ B_n(0)\: 2\pi i \ &= \ (2\pi i)^{n+1} \sum_{k=2}^{n+1} \frac{B_{n+1-k}}{(n+1-k)!\:k!} \\
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| − | \overset{(\dagger)}{\Leftrightarrow} \ B_n(0)\: 2\pi i \ &= \ (2\pi i)^{n+1} \left(- \frac{B_n}{n!} \right) \\
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| − | B_n(0) \ &= \: - \frac{(2\pi i)^n B_n}{n!}
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| − | \end{align}$$
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| − | This, together with equation \((\ddagger)\), proves the identity in the problem.
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| − | <p style="text-align:right;">\(\square\)</p>
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| − | =Alternate Solution=
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| − | <span style="color:#339;font-size:80%;">This solution has not yet been proof-read, but was proposed by Alessio (aka [[User:Milfhunter|Milfhunter]], veeery classy) as their TA solved the same problem in their tutor class.</span>
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| − | Let $$\zeta(s) = \sum\limits_{n=1}^\infty \frac{1}{n^s} $$ for $$s \in \mathbb{C}, \Re (s) > 1$$
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| − | Consider
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| − | $$ E(z) := \frac{z}{e^z -1} := \sum\limits_{n=0}^\infty B_n \frac{z^n}{n!} \quad\quad\quad z \in \mathbb{R} $$
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| − | $$\lim\limits_{z \to 0}{E(z)} = \lim\limits_{z \to 0}{\frac{1}{e^z}} = 1 = \lim\limits_{z \to 0}{\sum\limits_{n=0}^\infty B_n \frac{z^n}{n!}} $$
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| − | Where in the first equality we used l'Hospital's rule and with
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| − | $$ \lim\limits_{z \to 0}{E(z)} = E(\lim\limits_{z \to 0}{z}) = \sum\limits_{n=0}^\infty \lim\limits_{z \to 0}{B_n \frac{z^n}{n!}} =B_0 $$
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| − | Where we can put the limit inside because E is continuous and bounded in a neighborhood of z_0 = 0.
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| − | $$ \rightarrow B_0 = 1 $$
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| − | Now we repeat the same thing for E'
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| − | \begin{align*}
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| − | \lim\limits_{z \to 0}{E'(z)} & = \lim\limits_{z \to 0}{\frac{e^z -1 -z e^z}{e^{2z} -2e^z+1}} \\
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| − | & = \lim\limits_{z \to 0}{\frac{e^z -z e^z -e^z}{2e^{2z} -2e^z}} \\
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| − | & = \lim\limits_{z \to 0}{\frac{-z e^z - e^z}{4e^{2z}-2e^z}} \\
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| − | & = -\frac{1}{2}
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| − | \end{align*}
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| − | Here we again used l'Hospital's rule for the second and third equality. For the same reasons as above we get
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| − | $$ \lim\limits_{z \to 0}{E'(z)} = B_1 $$
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| − | and therefore
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| − | $$\rightarrow b_1 = - \frac{1}{2}$$
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| − | $$\Rightarrow E(z) = 1 - \frac{1}{2} z + \mathcal{O}(z^2)$$
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| − | Now define \(f(z) = \frac{z}{2} \coth (\frac{z}{2})\) on \(]-\frac{1}{2}, \frac{1}{2}]\)
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| − | \begin{align*}
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| − | f(z) & = \frac{z}{2} \coth (\frac{z}{2}) \\
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| − | & = \frac{z e^{\frac{z}{2}} + z e^{-\frac{z}{2}}}{2e^{\frac{z}{2}} - 2 e^{-\frac{z}{2}}} \\
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| − | & = \frac{z e^{\frac{z}{2}} + z e^{-\frac{z}{2}}}{2e^{\frac{z}{2}}(e^z-1)} \\
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| − | & = \frac{z}{2} \frac{e^z+1}{e^z-1} \\
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| − | & = \frac{z}{2} \left( 1 + \frac{2}{e^z-1} \right) \\
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| − | & = \frac{z}{2} + \frac{z}{e^z-1} \\
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| − | & = \frac{z}{2} + E(z)
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| − | \end{align*}
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| − | Since z and coth are odd functions, f(z) is an even function. From the part above we know that $$E(z) = 1 - \frac{1}{2} z + \mathcal{O}(z^2)$$ Since the addition of z/2 kills the uneven term and f(z) has to be even, the only remaining coefficients are even:
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| − | $$ f(z) = \frac{z}{2} + E(z) = \sum\limits_{n=0}^\infty B_{2n} \frac{z^{2n}}{(2n)!} $$
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| − | Now define \(g(x) = \cosh (z x)\) on \(]-\frac{1}{2}, \frac{1}{2}]\)
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| − | with \(z \in \mathbb{R} \setminus \{0\}\), continued periodically on \(\mathbb{R}\)
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| − | Calculating the Fourier coefficients of g:
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| − | $$\alpha_0 = \int\limits_{-\frac{1}{2}} ^{\frac{1}{2}} \cosh(zx)dx = \frac{2}{z}\sinh (\frac{z}{2})$$
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| − | \begin{align*}
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| − | \alpha_n & = \int\limits_{-\frac{1}{2}} ^{\frac{1}{2}} \cosh(zx)\cos(2\pi n x)dx \\
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| − | & = \left[\cosh(zx)\frac{1}{2\pi n}\sin(2\pi nx) \right]_{-\frac{1}{2}}^{\frac{1}{2}} - \int\limits_{-\frac{1}{2}} ^{\frac{1}{2}} z \sinh (zx) \frac{1}{2\pi n} \sin (2\pi nx) dx \\
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| − | & = -z \int\limits_{-\frac{1}{2}} ^{\frac{1}{2}}\sinh (zx) \frac{1}{2\pi n} \sin (2\pi nx) dx \\
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| − | & = -z \left[ \left[-\sinh (zx) \left(\frac{1}{2\pi n}\right)^2 \cos (2\pi nx) \right]_{-\frac{1}{2}}^{\frac{1}{2}} + z \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \cosh (zx)\left(\frac{1}{2\pi n}\right)^2 \cos (2\pi nx) dx \right]
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| − | \end{align*}
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| − | $$ \Rightarrow \left(1 + \left(\frac{z}{2\pi n}\right)^2\right)\int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \cosh (zx) \cos (2\pi nx) dx = 2z\left(\frac{z}{2\pi n}\right)^2 (-1)^n \sinh (\frac{z}{2}) $$
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| − | $$ \rightarrow \alpha_n = \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \cosh (zx) \cos (2\pi nx) dx = \frac{2z}{(2\pi n)^2 + z^2} (-1)^n \sinh (\frac{z}{2}) $$
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| − | Where we use $$\cos(2\pi n x)$$ because the integrand is even. Therefore we get for the Fourieseries of g:
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| − | $$ \cosh (zx) = \frac{2}{z} \sinh (\frac{z}{2}) \left[ 1 + \sum\limits_{n \in \mathbb{Z} \\ n \neq 0} (-1)^n \frac{2z}{4{\pi}^2 n^2 + z^2} \cos (2\pi n x) \right] $$
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| − | $$ = \frac{2}{z} \sinh(\frac{z}{2}) \left[ 1 + 2\sum\limits_{n=1}^\infty (-1)^n \frac{2z}{4{\pi}^2 n^2 + z^2} \cos (2\pi n x) \right]$$
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| − | Now we chose $$x = \frac{1}{2}$$:
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| − | $$ \cosh (\frac{z}{2}) = \frac{2}{z} \sinh(\frac{z}{2}) \left[ 1 + 2\sum\limits_{n=1}^\infty (-1)^n \frac{z^2}{4\pi^2 n^2} \frac{1}{1+\frac{z^2}{4\pi^2 n^2}} \right] $$
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| − | $$ \underbrace{\frac{z}{2} \frac{\cosh (\frac{z}{2})}{\sinh(\frac{z}{2})}}_{f(z)}
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| − | = 1 + 2\sum\limits_{n=1}^\infty (-1)^n \frac{z^2}{4{\pi}^2 n^2} \frac{1}{1+\frac{z^2}{4{\pi}^2 n^2}} $$
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| − | $$ = 1 + \sum\limits_{n=1}^\infty \sum\limits_{k=0}^\infty \frac{2z^2}{4{\pi}^2 n^2} \left(-\frac{z^2}{4\pi^2 n^2}\right)^k$$
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| − | Because both series are absolute convergent and uniform convergent we get after some calculation:
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| − | $$ f(z) = 1 + \sum\limits_{k = 1}^\infty 2 (-1)^{k-1} \zeta(2k) \frac{z^{2k}}{(2\pi)^{2k}} $$
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| − | but above we also calculated
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| − | $$ f(z) =1 + \sum\limits_{n=1}^\infty B_{2n} \frac{z^{2n}}{(2n)!} $$
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| − | With a coefficient comparison we get:
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| − | $$ \zeta(2n) = - \frac{B_{2n}(-1)^{2n}(2\pi)^{2n}}{2(2n)!} = -\frac{(2\pi i)^{2n}}{2(2n)!}B_{2n} \quad\quad\quad \forall n \in \mathbb{N} $$
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| − | which is what we had to show.
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