Question
Find the Fourier series of on
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Answer :
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Consider the (2\pi)-periodic function (f(t)=t^2) defined on ([-\pi,\pi]). Since (t^2) is an even function, its Fourier series contains only cosine terms. Thus the Fourier series has the form [ f(t)\sim \frac{a_0}{2}+\sum_{n=1}^{\infty} a_n\cos(nt), ] where [ a_0=\frac{1}{\pi}\int_{-\pi}^{\pi} t^2,dt,\qquad a_n=\frac{1}{\pi}\int_{-\pi}^{\pi} t^2\cos(nt),dt. ] First, [ a_0=\frac{1}{\pi}\int_{-\pi}^{\pi} t^2,dt=\frac{2}{\pi}\int_{0}^{\pi} t^2,dt =\frac{2}{\pi}\left[\frac{t^3}{3}\right]_{0}^{\pi} =\frac{2\pi^2}{3}. ] Hence (a_0/2=\pi^2/3). Next, for (n\ge1), [ a_n=\frac{2}{\pi}\int_{0}^{\pi} t^2\cos(nt),dt.
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