MHB Fourier Cosine Series: Equivalence for {x}

Joystar77
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2. Fourier cosine series correspondence for f(x)= x, o < x < pi given by x ~ pi / 2 - 4/n, E infinity on top and n=1 on bottom. cos (an-1)/x / (2n-1)squared, (0 < x < pi).

Explain why this correspondence is actually an equality for 0 is less than or equal to x and x is less than or equal to pi. Then explain how we can write

{x} = pi /2 - 4 / n E infinity on top and n=1 on bottom

cos (2n - 1)x/ (2n-1) squared, (-n is less than or equal to x and x is less than or equal to pi)
 
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Sorry, I forgot to mention in the original thread that I am totally lost and don't know where to start at with this problem.
 
Here is the above problem and I hope that its easier to read. I used some Mathematical symbols, so hope that you can understand it. Any questions, then please ask me if your unable to understand or read the problem.

Fourier cosine series correspondence for f (x) = x, o < x < Pie given by x ~ Pie / 2 – 4 / n

∑_(n=1)^∞▒cos⁡〖(2n-1)x/(2n-1)〗 〖^2〗,(o<x<Pie)

Explain why this correspondence is actually an equality for 0 ≤ x ≤ Pie. Then explain how we can write

[x] = Pie / 2 – 4 / n ∑_(n=1)^∞▒cos⁡〖(2n-1)x/(2n-1)〗 〖^2〗 ( -n ≤ x ≤ Pie)
 
Joystar1977 said:
Here is the above problem and I hope that its easier to read. I used some Mathematical symbols, so hope that you can understand it. Any questions, then please ask me if your unable to understand or read the problem.

Fourier cosine series correspondence for f (x) = x, o < x < Pie given by x ~ Pie / 2 – 4 / n

∑_(n=1)^∞▒cos⁡〖(2n-1)x/(2n-1)〗 〖^2〗,(o<x<Pie)

Explain why this correspondence is actually an equality for 0 ≤ x ≤ Pie. Then explain how we can write

[x] = Pie / 2 – 4 / n ∑_(n=1)^∞▒cos⁡〖(2n-1)x/(2n-1)〗 〖^2〗 ( -n ≤ x ≤ Pie)
http://mathhelpboards.com/math-notes-49/fourier-series-integral-transform-notes-2860.html
 
Thanks! I will take a look at these Fourier Series Integral Transform Notes.

dwsmith said:
http://mathhelpboards.com/math-notes-49/fourier-series-integral-transform-notes-2860.html
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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