I tried and got zero for $a_0, a_n$ and $ b_n$. I do not know if that is correct, but that just looks weird. However, the function looks weird too. Help?
Fourier coefficients for $f(x)=\sin(14x)+30\cos(25x)$?
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analysis
fourier-series
1 Answers
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All coefficients are zero except $a_{25}$ and $b_{14}$.
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0At one point I get in this: $\int_0^\pi 30 \cos (nx) \cos(25x)dx$ If I apply orthogonality relations, I get your result, but let's say I didn't notice that and continued. Through addition theorems I would get: $\int_0^\pi \cos(nx+25x) + \cos (nx-25x) dx$ After integrating I would get somewhere $1_(n-25)$, so that means I can't even put 25 in the equation. What is wrong with my thinking? – 2017-01-26
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0And when $n=25$ that integral is? – 2017-01-26
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0Sorry, meant to write 1/(n-25). – 2017-01-26
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1That would mean that for $n=25$ the computation cannot be done like this. – 2017-01-26