## Related questions with answers

Suppose that you want to find the Fourier series of $f(x)=x+x^{2}$. Explain why to find $b_k$ you would need only to integrate $x \sin \left(\frac{k \pi x}{l}\right)$ and to find $a_k$ you would need only to integrate $x^{2} \cos \left(\frac{k \pi x}{l}\right)$.

Solution

VerifiedAny continuous periodic function $f(x)$ with period $T$, with $f^{\prime}(x)$ continous at $[-T / 2, T / 2]$ except at a finite number of jump discontinuities, can be expanded in a convergent Fourier series given by:

$f(x)=\frac{a_{0}}{2}+\sum_{n=0}^{\infty} a_{n} \cos \left(\frac{2 \pi n x}{T}\right)+b_{n} \sin \left(\frac{2 \pi n x}{T}\right)\tag{1}$

where the Fourier coefficients $a_{n}$ and $b_{n}$ are given by:

$a_{n}=\frac{2}{T} \int_{-T / 2}^{T / 2} f(x) \cos \left(\frac{2 \pi n x}{T}\right) d x, \quad n=0,1,2, \ldots \tag{2}$

and:

$b_{n}=\frac{2}{T} \int_{-T / 2}^{T / 2} f(x) \sin\left(\frac{2 \pi n x}{T}\right) d x, \quad n=1,2,3, \ldots\tag{3}$

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