MATH 286 Sections D1 & X1: Assignment 9Solutions for the graded problems.Section 9.1, #24 [3 points] The graph of f (t) = | sin t|, −π < t < π, is just the periodicextension of the graph of the sine curve on the interval [0, π ]. To find the Fourier seriesf (t) ∼a02+∞∑n=1ancos nt + bnsin nt , (L = π here!)observe first that f (t) is an even function, sobn=1πZπ−πf (t) sin ntdt = 0.Now we calculate the an’s:a0=1πZπ−πf (t)dt =2πZπ0sin tdt =4π,and for n ≥ 1an=1πZπ−πf (t) cos ntdt =1πZπ−π| sin t| cos ntdt =2πZπ0sin t cos ntdt=1πZπ0sin((1 − n)t) + sin((1 + n)t)dt (using sin x cos y =sin(x − y) + sin(x + y)2)=− cos((1 − n)π) + 1(1 − n)π+− cos(1 + n)π + 1(1 + n)π=n0 n odd2(1−n)πn even+n0 n odd2(1+n)πn even=n0 n odd4(1−n2)πn even.Section 9.2, #16 [3 points] Let f be a period two function such that f (t) = 0 if −1 <t < 0 and f (t) = t if 0 < t < 1.(a). To show thatf (t) =14+∑n odd−2n2πcos nπt +∞∑n=1(−1)n+1nπsin nπt,we first observe that f is a piecewise smooth function. Therefore by Theorem 1 of Section9.2, f equals its Fourier series at each point where f is continuous. I.e.,f (t) =a02+∞∑n=1ancos nπt + bnsin nπt ( f continuous at t).1We therefore just need to check that the coefficients of these two series match, whichthey indeed do:a0=Z1−1f (t)dt =Z10tdt =12an=Z1−1f (t) cos nπtdt =Z10t cos nπ tdt =n0 n even−2n2πn oddbn=Z1−1f (t) sin nπtdt =Z10t sin nπ tdt =(−1)n+1nπ.(b). Since t = 0 is a point of continuity for f , we conclude from (a) that0 = f (0) =14+∑n odd−2n2πcos nπ0 +∞∑n=1(−1)n+1nπsin nπ0 =14+∑n odd−2n2π.Rearranging, we get the required formula:∑n oddn−2= π2/8.Section 9.3, #2 [4 points] For f (t) = 1 − t, 0 < t < 1, the Fourier sine series isf (t) =∞∑n=1bnsin nπt (0 < t < 1),wherebn=21Z10f (t) sin nπtdt = 2Z10(1 − t) sin n π tdt =2πn.The Fourier cosine series isf (t) =a02+∞∑n=1ancos nπt (0 < t < 1),wherean=21Z10f (t) cos nπtdt = 2Z10(1 − t) cos n π tdt =n1 n = 04n2π2n odd0 n
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