Help regarding a question Parseval's Theorem

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SUMMARY

The discussion centers on applying Parseval's Theorem to the Fourier series of the function f(x) = x² over the interval (−1/2, 1/2). The Fourier series is expressed as f(x) = 1/12 - (1/π²)(cos 2πx - (1/4)cos 4πx + (1/9)cos 6πx). Participants confirm that using Parseval's Theorem, the series converges to the result ∑(n=1 to ∞) (1/n⁴) = π⁴/90. Key calculations include determining the coefficients a₀ and aₙ, with the sine coefficients bₙ being zero.

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TFM
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Homework Statement



The Fourier series for f(x) = x2 over the interval (−1/2, 1/2) is:

[tex]f(x) = \frac{1}{12}-\frac{1}{\pi^2} (cos 2\pi x - \frac{1}{2^2}cos4\pi x + \frac{1}{3^2}cos6\pi x) ...[/tex]

Using Parseval's Theorem, show that

[tex]\sum _{n = 1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}[/tex]

Homework Equations



Fourier's Series:

[tex]f(x) = \frac{1}{2}a_0 + \sum _{n = 1}^\infty a_n cos\frac{2\pi nx}{l} \sum _{n = 1}^\infty b_n sin \frac{2\pi nx}{l}[/tex]

[tex]a_0 = \frac{1}{l} \int^{\frac{l}{2}}_{\frac{-l}{2}}f(x) dx[/tex]

[tex]a_n = \frac{1}{l} \int^{\frac{l}{2}}_{\frac{-l}{2}}f(x) cos \frac{2\pi nx}{l} dx[/tex]


[tex]b_n = \frac{1}{l} \int^{\frac{l}{2}}_{\frac{-l}{2}}f(x) sin \frac{2\pi nx}{l} dx[/tex]

Parseval's Theorem:

[tex]\frac{1}{2\pi} \int^{\pi}_{-\pi}f(x)^2 dx = \frac{1}{4}a_0^2 + \frac{1}{2} \sum _{n = 1}^\infty a_n^2 + \frac{1}{2}\sum _{n = 1}^\infty b_n^2[/tex]

The Attempt at a Solution



See I'm not quite sure where to go from here. It says that it is a Fourier series, but it doesn't seem to fit with the definition of Fourier Series I have quoted below?

Any assitance would be greatly appreciated,

TFM
 
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You multiplied the cosine and sine portions of the Fourier series together, they're supposed to be added. At any rate

For your f(x), a0=1/6 and then you can distribute the [tex]\frac{-1}{\pi^2}[/tex] to calculate the other ai's

And then the sine component is just all zeroes (so all your bi's will be 0)
 
So firstly,

[tex]f(x) = \frac{1}{2}a_0 + \sum _{n = 1}^\infty a_n cos\frac{2\pi nx}{l} + \sum _{n = 1}^\infty b_n sin \frac{2\pi nx}{l}[/tex]

And

[tex]a_0 = \frac{1}{6}[/tex]

As there are no sins, this implies that b_n = 0 (since b_n is the sin component.)
Multiplying out the brackets:

[tex]f(x) = \frac{1}{12} - (\frac{1}{\pi^2}cos 2\pi x - \frac{1}{\pi^2}\frac{1}{2^2}cos4\pi x + \frac{1}{\pi^2}\frac{1}{3^2}cos6\pi x ...)[/tex]

[tex]f(x) = \frac{1}{12} - (\frac{1}{\pi^2}cos 2\pi x - \frac{1}{4\pi^2}cos4\pi x + \frac{1}{9\pi^2}cos6\pi x ...)[/tex]

So the a_n seems to be an increasing value:

[tex]a_n = frac{1}{\pi^2}, frac{1}{4\pi^2}, frac{1}{9\pi^2}...[/tex]

Does this seem okay so far?

?

TFM
 

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