Difference between revisions of "MediaWiki:Sitenotice"

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(fabian could not sleep because of me using \hat f(k) and c_k interchangeably. hope this helps)
(Added reminder to sign messages and use)
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=Problem=
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''Please'' sign your messages when you write on Talk pages or add any other opinion based content! Just add <nowiki>~~~~</nowiki>, MediaWiki will automatically replace it with your username and the current date.<br/>
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Compute
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Also, please make thorough use of the ''Summary'' field when editing pages!<br/>
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$$\sum_{j=1}^\infty \frac{1}{j^4}$$
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(btw, there's a "dismiss" button over there \(\rightarrow\))<br/>
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using Parseval identity.
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[[User:Nik|Nik]] ([[User talk:Nik|talk]]) 18:41, 10 June 2015 (CEST)
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''Hint:'' consider \(f(x) = \frac{1}{2} x^2\)
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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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=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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$$\hat f(k) = \frac{1}{2\pi} \int_0^{2\pi} f(x) \: e^{-ikx} \: dx$$
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We consider the continuous continuation of \(f(x)\) on \([-\pi, \pi]\) and calculate the Fourier coefficients:
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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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and for \(k \neq 0\) (twice we integrate by parts):
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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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$$\frac{1}{2\pi} \int_{-\pi}^\pi \left| f(x) \right|^2 \: dx = \sum_{k=-\infty}^\infty \left| \hat f(k) \right|^2 \tag{*}$$
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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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Revision as of 16:41, 10 June 2015

Please sign your messages when you write on Talk pages or add any other opinion based content! Just add ~~~~, MediaWiki will automatically replace it with your username and the current date.
Also, please make thorough use of the Summary field when editing pages!
(btw, there's a "dismiss" button over there \(\rightarrow\))
Nik (talk) 18:41, 10 June 2015 (CEST)

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