Can 1/a Be Expanded Using Fourier Expansion?

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SUMMARY

The term 1/a can be represented as a constant function in Fourier expansion, yielding a Fourier series with coefficients for sine and cosine terms equal to zero. For the function f(z) = 1/z, a Fourier series can be derived using standard formulas, although it will not converge at z = 0. Additionally, the function f(x) = (1 - x/a) can also be expanded using Fourier series, provided it is periodic. Understanding these expansions is crucial for applying Fourier analysis effectively.

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kaizen.moto
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Dear all,

Iam just wondering whether the term 1/a can be expanded using Fourier expansion. If it does, please let me know how to to do this.

Thank for any kind help.
 
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What do you mean bu "the term 1/a"? Normally, we expand functions in Fourier series, not numbers. Of course, we can think of f(z)= 1/a as the constant function. In that case, the Fourier series is simply (1/a)+ 0 sin(z)+ 0 cos(z)+ ...

If you mean f(z)= 1/z, then, yes, you can find a Fourier series for 1/z using the standard formulas. It will not converge at z= 0, of course.
 


How about f(x) = (1 - x/a), what would be the solution after Fourier expansion?
 
Any constant is a finite Fourier series with all the sine and cosine terms having 0 coeifficients. There is nothing to calculate.
 


kaizen.moto said:
How about f(x) = (1 - x/a), what would be the solution after Fourier expansion?

Your f(x) needs to be periodic to have a FS.
 
Moderator's note: I copied several posts from the thread that was started in the Mathematics technical section.
 

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