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I want to compute the sum of the fourier series for my 2pi periodic function $$y(x)=x^{2} $$ when $$x=6$$ its defined such that $$|x|\le \pi$$

I managed to compute the fourier series of my function and ended up with : $$y(x)=\frac{\pi^{2}}{3} + 2\sum_{n=-\infty}^\infty \frac{(-1)^{n}}{n^{2}}*e^{inx} $$

Therfor $$y(6)=\frac{\pi^{2}}{3} + 2\sum_{n=-\infty}^\infty \frac{(-1)^{n}}{n^{2}}*e^{in6} $$

But I have no clue how to evaluate this expression...any help?

The correct answer is $$(6-2\pi)^{2} $$

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    What is the range over which you are computing the Fourier series? – 2017-02-04
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    It made me a bit confused since it is $$-\pi\le x \le \pi $$ so 6 is not contained in that interval.. – 2017-02-04
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    In that case, the summation will give a $2\pi$-periodic function $\eta$ such that $\eta(x) = y(x)$ for $|x| \le \pi$ (I am using some convergence theorems here). Then $\eta(6) = \eta(6-2\pi) = y(6-2\pi) = (6-2\pi)^2$ (because $6-2\pi \in [-\pi,\pi]$). – 2017-02-04
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    I did not understand this answer....is there no way to use my fourier series representation of y(x) to compute the value when x=6? – 2017-02-04
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    Perhaps, but it is a lot more work. Why would you do it that way when you have an explicit answer? – 2017-02-04
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    Can i say that 6 is not contained in the interval, but 6-2pi is ? and then just plug in this value in y(x)=x^2 ? – 2017-02-04
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    That is what I did in my second comment... – 2017-02-04
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    Oh okay, the signs of new functions just made me confused.. – 2017-02-04
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    I have another question that I want to ask you, Im asked to find the fourier series of $$y(x)=max(cosx,0)$$ ..Im not sure about this question, because this function makes me confused – 2017-02-04
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    Im confused about how I should treat this function when integrating to find the fourier coefficients.. – 2017-02-04
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    You need to split the integral into a few parts, depending on the sign of $\cos$. – 2017-02-04
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    Why do I have to split it? Cosx>0 when x is between -pi/2 and pi/2 right? – 2017-02-04
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    Well, on the domain $[-\pi,\pi]$ you have $y(x) = \cos x$ for $|x| < {1 \over 2} \pi$ and zero outside. So the integral will be like $\int_{-\pi}^{\pi} y(x)... = \int_{-{1 \over 2} \pi}^{{1 \over 2} \pi} \cos x ...$. – 2017-02-04
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    But in this exercise they didnt specify any domain...is it the same domain? – 2017-02-04
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    Sorry, I have no way of knowing. It depends on what it is going to be used for, but I would imagine that it is the same domain. – 2017-02-04
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    It is going to be used to calculate the sum of 1/(4n^2 - 1) – 2017-02-04
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    I haven't done the computation, but I suspect $[-\pi,\pi]$. – 2017-02-04
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    Okay I will try with that interva then, thanks – 2017-02-04
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    Good luck! ${}{}$ – 2017-02-04
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    By the way, do you have any hint on which method should be used when integrating cosx*e^(inx) ... integration by parts doesnt look to work ;/ – 2017-02-04
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    I represent $\cos$ by a sum of exponentials. – 2017-02-04
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    Yeah , but this integral simplifies to just c1 and c(-1) terms? Because by orthogonality inspection all other terms will equal to 0? thats strange.. – 2017-02-04

1 Answers 1

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When dealing with even/odd functions, the real coefficients are particularly useful:

$$ f(x) \sim \frac{a_0}{2} + \sum_{n = 1}^{+\infty} \Big ( a_n \cos (nx) + b_n \sin (nx) \Big ) $$

where:

$$ b_n = 0 \qquad (\text{it's an even function}) $$ $$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \ \mathrm dx = \frac{2}{\pi} \left. \frac{x^3}{3} \right |_{0}^{\pi} = \frac{2}{3} \pi^2 $$ $$ \begin{aligned} a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos (nx) \ \mathrm dx \\ &= \frac{2}{\pi} \left [ \underbrace{\left. \frac{1}{n} x^2 \sin (nx) \right |_{0}^{\pi}}_{{} = 0, \ n \in \mathbb{N}} - \frac{2}{n} \int_{0}^{\pi} x \sin (nx) \ \mathrm dx \right ] \\ &= \frac{2}{\pi} \left [ \left. \frac{2}{n^2} x \cos (nx) \right |_{0}^{\pi} - \frac{2}{n^2} \underbrace{\int_{0}^{\pi} \cos (nx) \ \mathrm dx}_{{} = 0, \ n \in \mathbb{N}} \right ] \\ &= \frac{{4}}{n^2} (-1)^n \end{aligned} $$

Therefore:

$$ f(x) = \frac{\pi^2}{3} + 4 \sum_{n = 1}^{+\infty} \frac{(-1)^n}{n^2} \cos (nx) $$

In conclusion:

$$ f(6) = f(6 - 2 \pi) = (6 - 2\pi)^2 = \frac{\pi^2}{3} + 4 \sum_{n = 1}^{+\infty} \frac{(-1)^n}{n^2} \cos \big (n(6 - 2\pi) \big) $$ $$ \implies \sum_{n = 1}^{+\infty} \frac{(-1)^n}{n^2} \cos (6n) = \frac{1}{4} \left [ (6 - 2\pi)^2 - \frac{\pi^2}{3} \right ] $$