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Calculus and Beyond Homework Help
Finding the sum of an infinite series using Fourier
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[QUOTE="John Jacke, post: 5796140, member: 587181"] [h2]Homework Statement [/h2] Trying to find the sum of (-1)[SUP]3n+1[/SUP]/(2n-1)[SUP]3[/SUP]. by using term-by-term integration on the cosine Fourier series x= L/2-4L/π[SUP]2[/SUP]∑cos(((2n-1)πx)/L)/(2n-1)[SUP]2[/SUP]. [h2]Homework Equations[/h2] Shown below [h2]The Attempt at a Solution[/h2] When integrating and substituting Lx/2 for x's sine Fourier series I get ∑L[SUP]2[/SUP]/π(2n-1)[(-1)[SUP]n+1[/SUP]-4/π[SUP]2[/SUP](2n-1)[SUP]2[/SUP]]sin(((2n-1)πx)/L) for the final series.To find the sum of the series in class we would find B[SUB]n[/SUB] explicitly using the formula and then set it equal to this coefficient found by term by term integration. However when I do that they are the same except for that the only difference is that one term alternates and one does not in sign. So explicitly solving for Bn gives me what you see above except the first term doesn't alternate. So setting them equal cancels everything except those terms. Is this supposed to mean anything? I've gotten the correct answer but was off by a factor of 2. It should (I think) be π[SUP]3[/SUP]/32. [/QUOTE]
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Calculus and Beyond Homework Help
Finding the sum of an infinite series using Fourier
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