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danielakkerma
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Homework Statement
Calculate the following integral:
[tex]
\int_{0}^{2\pi}(\sum_{k=0}^{\infty} \frac{\cos(kx)}{3^k})^2 dx
[/tex]
Homework Equations
Parseval's identity: [tex] \frac{1}{2 \pi} \int_{-\pi}^{\pi} {|f(x)|^2 dx} = \sum_{n=0}^{\infty} {|a_n|^2+|b_n|^2} [/tex]
Where a_n, and b_n are the trigonometric Fourier coefficients; factors for cos(nx) and sin(nx) respectively.
The Attempt at a Solution
Since the question posited the integrand as a sum of a trigonometric polynomial, I let f(x) = Ʃ(cos(kx)/(3^k)).
Then, squaring it, according to Parseval's identity, and taking the integral as shown, should yield the sum of the absolute value of the coefficients(times two Pi -- a constant in the definition of the orthonormal inner product, in the Hilbert space).
In this case:
[tex] \int |f(x)|^2 dx = 2 \pi \sum |a_n|^2 = \sum_{k=0}^{\infty} (\frac{1}{9})^k = 2 \pi \frac{9}{8} = \frac{18 \pi}{8} [/tex]
However, Mathematica claims it's 17π/8!
http://www.wolframalpha.com/input/?...)]/(3^k),+{k,+0,+Infinity}]^2),+{x,+0,+2+Pi}]
I'm quite at a loss as to where I have erred,
And am, as ever, reliant on your help,
Thanks for your time and attention,
Daniel
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