Inverse function and fourier series

Click For Summary
The discussion centers on whether the inverse function of a Fourier series can be obtained by reflecting the series through the line y=x. It highlights that while the Fourier expansion can include complex terms, the concept of reflection is more straightforward in the context of real functions. The complexity arises when considering the nature of the function, as the inverse function is not simply f(-1)(x) but rather x=f(-1)(z) when dealing with complex variables. Participants note that expressing the reciprocal of a Fourier series remains a challenging and unresolved problem in the field. The conversation emphasizes the need for clarity regarding what is being reflected in the context of complex versus real functions.
zetafunction
Messages
371
Reaction score
0
let be the Fourier expansion of the function

f(x) = \sum_{m=-\infty}^{m=\infty}c_{m} exp(imx)

valid on the interval (-1,1) , from this can we obtain the inverse function

f^{-1} (x) by reflection of the Fourier series through the line y=x ??
 
Physics news on Phys.org
can we obtain the inverse function f(-1)(x) by reflection of the Fourier series through the line y=x ??
Since exp(imx)=cos(mx)+i sin(mx), the function f(x) is complexe z=f(x)
Formally, the inverse function would not be f(-1)(x) but rather x=f(-1)(z)
What do you mean in writing "by reflection through the line y=x " ?
This would have a sens in case of reals y=f(x) and x=f(-1)(y). Presently it is not a real y, but a complex z instead of y.
What do you reflect through the line y=x ? Is it the curve representing y(x)= real part of z as a function of x ?, or is it the curve representing y(x)= imaginary part of z as a function of x ? or is it the curve representing y(x)= module of z as a function of x ? or is it the curve representing y(x)= argument of z as a function of x ?
 
assume function f(x) is real, then the Fourier series will include sines and cosines only .. but no exponential quantities.
 
assume function f(x) is real, then the Fourier series will include sines and cosines only .. but no exponential quantities
Sure ! if f(x) is assumed to be real the general terms of the Fourier series will be :
amcos(mx)+bmsin(mx) which is real.
But in the first question this was not the case : The general term was cmexp(i m x) = cm( cos(mx) + i sin(mx) ) which is not real.
So the question was raised in the general case of f(x) complex, even if the coefficients cm are complex (and/or real).
 
zetafunction said:
let be the Fourier expansion of the function

f(x) = \sum_{m=-\infty}^{m=\infty}c_{m} exp(imx)

valid on the interval (-1,1) , from this can we obtain the inverse function

f^{-1} (x) by reflection of the Fourier series through the line y=x ??

I'm interested in this as well if anyone has any ideas.
 
Well, the problem cannot be raised in terms of reflexion through the line y=x on a real space.
Seriously, expressing the reciprocal of a Fourier series is a very difficult problem, still open :
http://www.jstor.org/pss/2031811
http://www.jstor.org/pss/2034097
 
Last edited by a moderator:

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 139 ·
5
Replies
139
Views
10K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K