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 Jun19 accepted Does a closed form sum for this fourier series exist? Jun19 comment Does a closed form sum for this fourier series exist? Yep, as it turns out my fourier series was wrong (again). It should have been $\frac{(a+b)(-1)^{n+1}}{n}$ where it says b-a in the series (I missed a '-' in the scalar product). Yeah, but I don't think I'm gonna include a closed form for the sum in my homework after all - what you just said is a little over my head, we just briefly touched fourier series in 2 or 3 classes and then went on to the next topic. Jun19 comment Does a closed form sum for this fourier series exist? I guess that makes sense, except I have no idea how to prove these sums either. Do you happen to know what causes my problem with (a=b)? Jun18 comment Does a closed form sum for this fourier series exist? This isn't part of the question title - but I now noticed that this doesn't work with a=b - the sum of the series would become 0. But f(x) is not constant 0 for a=b... But I don't recall having made assumptions about a and b not being equal in my calculations... Jun18 revised Does a closed form sum for this fourier series exist? Added missing a_0 to the fourier series Jun18 asked Does a closed form sum for this fourier series exist? Jun18 accepted Fourier-Series of a part-wise defined function? Jun18 comment Fourier-Series of a part-wise defined function? I see. Unfortunately I don't have the time to fix this before I hand it in, but it will certainly be useful in the future. Thanks. Jun18 revised Fourier-Series of a part-wise defined function? edited body Jun18 asked Fourier-Series of a part-wise defined function? Jun17 comment Partial Integration - Where did I go wrong? Thanks. I "fixed" the sign when I wrote this by accident (I had it right in my notes, then for some reason when I wrote it here I thought it was wrong). Jun17 accepted Partial Integration - Where did I go wrong? Jun17 asked Partial Integration - Where did I go wrong? May30 comment Lagrange multiplier - find minima of a function satisfying a condition Thanks. Will do. May30 comment Lagrange multiplier - find minima of a function satisfying a condition I don't know, I am using the notation I am used to. But unless the distance happens to be 1, why would I be able to use the square of the distance rather than the distance? EDIT: Actually, that makes sense because I can just take the square root when I have the result... EDIT2: Your signs seem off though. May30 asked Lagrange multiplier - find minima of a function satisfying a condition May6 accepted WolframAlpha blows simple substitution? May6 awarded Supporter May6 awarded Editor May6 revised WolframAlpha blows simple substitution? added 215 characters in body