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Given the Fourier series:

$$F(x)=\sin{x}+\sum_{n=1}^\infty \frac{1}{5^n} \cos{nx}$$

How do I find $a_0, a_1, a_2$ when $$a_0=\frac{1}{\pi} \int_{-\pi}^\pi f(x) dx$$ and $$a_n=\frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos{n x} dx$$

? Also, what is f(x) in this case?

I'm aware that this is a basic question, but I simply can't make sense of this for some reason.

HorizonsMaths
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Steve
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    Your expression is already in the form of a Fourier series, so you just need to read off the coefficients from what you have. There is no constant term, so $a_0=0$, and for example $a_1=\frac{1}{5 \pi}$, etc. – Paul May 23 '16 at 16:35
  • So $\frac {1}{5^n}$ is my $f(x)$ in this case? My professor's answers are $a_0=0, a_1=\frac {1}{5}$ and $a_2=\frac {1}{5^2}$ – Steve May 23 '16 at 16:38
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    Note that $\int_{-\pi}^\pi \cos(nx) \cos(mx) = 0$ for $n \ne m \in \mathbb{Z}$. Now you should be able to evaluate the integrals using this orthogonality. I think $f(x)$ is intended to be $F(x)$ if those are your professor's answers. – Eric Thoma May 23 '16 at 16:45

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