Finding order of an infinite series

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SUMMARY

The discussion centers on approximating a truncated Fourier series for cos(αx) at x=π, specifically the expression F_{N}(π) = cos(απ) + (2α/π) sin(απ) Σ (1/(k² - α²)) from k=N+1 to ∞. The goal is to show that this series approximates to F_{N}(π) = cos(απ) + (2α/Nπ) sin(απ), indicating that the infinite sum is of order 1/N. Participants suggest using integration techniques, particularly partial fraction decomposition, to simplify the evaluation of the infinite sum.

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  • Understanding of Fourier series, specifically truncated Fourier series.
  • Knowledge of calculus, particularly integration techniques.
  • Familiarity with series convergence and asymptotic analysis.
  • Ability to manipulate mathematical expressions involving limits and sums.
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  • Study the method of partial fraction decomposition for integrals.
  • Learn about the convergence properties of infinite series.
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Students and researchers in mathematics, particularly those focusing on Fourier analysis, calculus, and series convergence. This discussion is beneficial for anyone looking to deepen their understanding of truncated series and integration techniques.

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.
 

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