Finding order of an infinite series

Moschops
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Homework Statement



I have been working on a truncated Fourier series. I have come up with a truncated series for cos (αx) and it matches my book, where in this case I'm letting x=π, and then I have shown as the book asks,

Truncated series = F_{N}(π) = cos (απ) + \frac{2α}{π} \sin{(απ)}\sum\limits_{k=N+1}^\infty\frac{1}{k^2 - α^2}

I am then asked to show that this approximates to almost the same thing:

F_{N}(π) = cos (απ) + \frac{2α}{Nπ} \sin{(απ)}

So I thought no worries, looks like I need to take that infinite sum there and simply show that it is of order 1/N.

Homework Equations


The Attempt at a Solution

The only way I know how to try that is to pretend the sum is actually an integral,

\int \limits_{N+1}^\inftydk \frac{1}{k^2 - α^2}, and I wound up with something like:

\frac{\frac{αk}{α^2 - k^2} + arctan \frac{k}{α} }{2α^3}

as the solution to the integral, with limits of k=infinity and k=N+1 in there for the definite integral (I can't figure out how to show that - I was lucky to manage what I have.

Anyway, having done that, it really doesn't look like it's of order 1/N to me. Am I heading the right way with this, or is there some other way to establish that it's of order 1/N?

This is the very last step in the question and I get the idea it's meant to be, if not trivial, at least quite an easy part.
 
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Moschops said:

Homework Statement



I have been working on a truncated Fourier series. I have come up with a truncated series for cos (αx) and it matches my book, where in this case I'm letting x=π, and then I have shown as the book asks,

Truncated series = F_{N}(π) = cos (απ) + \frac{2α}{π} \sin{(απ)}\sum\limits_{k=N+1}^\infty\frac{1}{k^2 - α^2}

I am then asked to show that this approximates to almost the same thing:

F_{N}(π) = cos (απ) + \frac{2α}{Nπ} \sin{(απ)}

So I thought no worries, looks like I need to take that infinite sum there and simply show that it is of order 1/N.

Homework Equations





The Attempt at a Solution




The only way I know how to try that is to pretend the sum is actually an integral,

\int \limits_{N+1}^\inftydk \frac{1}{k^2 - α^2}, and I wound up with something like:

\frac{\frac{αk}{α^2 - k^2} + arctan \frac{k}{α} }{2α^3}
I don't know if your result is correct. An easier way to do the integration is to decompose the integrand using partial fractions.

##\frac{1}{k^2 - α^2} = \frac{A}{k - α} + \frac{B}{k + α}##
Solve the equation above for A and B and then integrate.
Moschops said:
as the solution to the integral, with limits of k=infinity and k=N+1 in there for the definite integral (I can't figure out how to show that - I was lucky to manage what I have.

Anyway, having done that, it really doesn't look like it's of order 1/N to me. Am I heading the right way with this, or is there some other way to establish that it's of order 1/N?

This is the very last step in the question and I get the idea it's meant to be, if not trivial, at least quite an easy part.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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