Why Are All Coefficients Zero in My Fourier Series Calculation of Sin(2x)?

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SUMMARY

The Fourier series of the function f(x) = sin(2x) should not yield all coefficients (a0, an, bn) as zero when calculated correctly. The basis functions used, specifically {1, cos(x), cos(2x), sin(x), sin(2x), ...}, along with the interval of length 2π, dictate that the odd nature of the sine function results in some coefficients being zero, but not all. Specifically, the coefficient b2 should be non-zero, indicating a miscalculation in the user's approach.

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Zeitblom
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I have to find the Fourier series of f(x)=sin(2x)
but i always get all the coefficients (a0, an and bn) equal zero.
is it right?
Thank you
 
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It depends on what basis functions and interval you use. If the often used basis {1,cos(x),cos(2x),...,sin(x),sin(2x),...} with interval length 2pi the answer should be obvious.
 
sin is an odd function, so half your coefficients should be zero, not all of them
 
Zeitblom said:
I have to find the Fourier series of f(x)=sin(2x)
but i always get all the coefficients (a0, an and bn) equal zero.
is it right?
Thank you

Did you check b_2 carefully? It shouldn't be 0.
 

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