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Question

Suppose that you want to find the Fourier series of f(x)=x+x2f(x)=x+x^{2}. Explain why to find bkb_k you would need only to integrate xsin(kπxl)x \sin \left(\frac{k \pi x}{l}\right) and to find aka_k you would need only to integrate x2cos(kπxl)x^{2} \cos \left(\frac{k \pi x}{l}\right).

Solution

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

f(x)=a02+n=0ancos(2πnxT)+bnsin(2πnxT)(1)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 ana_{n} and bnb_{n} are given by:

an=2TT/2T/2f(x)cos(2πnxT)dx,n=0,1,2,(2)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:

bn=2TT/2T/2f(x)sin(2πnxT)dx,n=1,2,3,(3)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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