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I calculated the fourier series of the given function, but all of the fourier coefficient was 0, it sounds strange!

user577215664
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Hanieh
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  • Could you show how the first two or three coefficients are zero? – Eric Towers Apr 05 '18 at 05:38
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    No. $\sin 0 + \cos 0 = 1$. $\sin \pi + \cos \pi = -1$. – Eric Towers Apr 05 '18 at 05:49
  • $\sin x + \cos x = \sqrt{2}\cos(x - \frac\pi 4)$ . This function's period is $2\pi$ . – mr_e_man Apr 05 '18 at 05:57
  • sure, the period of the function is 2pi, for the first coefficient, we have the integral of the function on [0.2pi], also we know that integral of sinx and cosx on its period is 0, so the firs coefficient come to 0, for the second & third one, we just need the use of trigonometric identities to calculate the integral: an = (1/pi)integral(sinx cosnx + cosx cosnx) from 0 to 2pi = 1/2pi {-(cos(n+1)x/n+1)-(cos(1-n)x/1-n)+(sin(1+n)x/1+n)+(sin(1-n)x/1-n)} from 0 to 2pi = 0 (n is a natural number) we have the similar calculation for bn and it comes to 0 too – Hanieh Apr 05 '18 at 06:46

1 Answers1

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$f(x)$ is already in Fourier Series representation.

But if you want to do the integral to get the coefficients:

$$ a_1=\frac{1}{\pi}\int_0^{2\pi}\left(\left(\sin x+\cos x\right)\cdot\cos\left(x\right)\right)dx = 1\\ b_1=\frac{1}{\pi}\int_0^{2\pi}\left(\left(\sin x+\cos x\right)\cdot\sin\left(x\right)\right)dx = 1 $$

Danny
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