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| − | =Problem=
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| − | Compute
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| − | $$\sum_{j=1}^\infty \frac{1}{j^4}$$
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| − | using Parseval identity.
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| − | ''Hint:'' consider \(f(x) = \frac{1}{2} x^2\)
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| − |
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| − | =Solution Sketch=
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| − | # Calculate the Fourier coefficients \(\hat f(x)\)
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| − | # Apply Parseval identity
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| − | # ???
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| − | # Profit
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| − |
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| − | =Solution=
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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 consider the continuous continuation of \(f(x)\) on \([-\pi, \pi]\) and calculate the Fourier coefficients:
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| − |
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| − | For \(k=0\):
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| − | $$\hat f(0) = \frac{1}{2\pi}\int_{-\pi}^\pi \frac{x^2}{2} dx = \frac{\pi^2}{6}$$
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| − |
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| − | and for \(k \neq 0\):
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| − | $$\begin{align}
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| − | \hat f(k) &= \frac{1}{2\pi} \int_{-\pi}^\pi \frac{x^2}{2} e^{-ikx}\: dx \\
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| − | &= \frac{1}{2\pi} \int_0^\pi x^2 \cos(kx)\: dx \\
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| − | &= - \frac{1}{k\pi} \int_0^\pi x \sin(kx)\: dx \\
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| − | &= \left[ \frac{1}{k^2\pi}x \cos(kx) \right]_0^\pi - \frac{1}{k^2\pi} \int_0^\pi \cos(kx)\: dx \\
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| − | &= \frac{\cos(k\pi)}{k^2} \\
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| − | &= \frac{(-1)^k}{k^2}
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| − | \end{align}$$
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| − | Now, remember Parseval's identity:
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| − | $$\sum_{k=-\infty}^\infty \left| \hat f(k) \right|^2 = \frac{1}{2\pi} \int_{-\pi}^\pi \left| f(x) \right|^2 \: dx \tag{*}$$
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| − | Applied to our \(f(x)\):
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| − | $$\frac{\pi^4}{20} = \frac{1}{2\pi} \int_{-\pi}^\pi \left| \frac{1}{2} x^2 \right|^2 \: dx \overset{(*)}{=} \sum_{k=-\infty}^\infty \left| \hat f(k) \right|^2 = \frac{\pi^4}{36} + 2 \sum_{k=1}^\infty \frac{1}{k^4}$$
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| − | And thus, it follows that
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| − | $$\sum_{k=1}^\infty \frac{1}{k^4} = \frac{\pi^4 (\frac{1}{20} - \frac{1}{36})}{2} = \frac{\pi^4}{90}$$
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