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seen Dec 16 at 4:00

Jun
19
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.
Jun
19
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)?
Jun
18
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...
Jun
18
revised Does a closed form sum for this fourier series exist?
Added missing a_0 to the fourier series
Jun
18
asked Does a closed form sum for this fourier series exist?
Jun
18
accepted Fourier-Series of a part-wise defined function?
Jun
18
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.
Jun
18
revised Fourier-Series of a part-wise defined function?
edited body
Jun
18
asked Fourier-Series of a part-wise defined function?
Jun
17
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).
Jun
17
accepted Partial Integration - Where did I go wrong?
Jun
17
asked Partial Integration - Where did I go wrong?
May
30
comment Lagrange multiplier - find minima of a function satisfying a condition
Thanks. Will do.
May
30
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.
May
30
asked Lagrange multiplier - find minima of a function satisfying a condition
May
6
accepted WolframAlpha blows simple substitution?
May
6
awarded  Supporter
May
6
awarded  Editor
May
6
revised WolframAlpha blows simple substitution?
added 215 characters in body
May
6
awarded  Commentator