1
$\begingroup$

This may seem backwards since a fourier series isn't typically used this way but I'm trying to prove whether or not the sum of sin and cos waves could produce a sin wave with a wave length that is not in any of the summed waves.

I don't intend the use of the fourier series as a restriction. It just seemed an obvious place to start thinking about this problem.

The restriction is, make a sin of finite wavelength L by summing any sin's and cos's so long as they do not have that same wavelength L.

$\endgroup$
  • $\begingroup$ Only a constant wave... $\endgroup$ – Thomas Andrews Mar 16 '15 at 4:23
1
$\begingroup$

If you are working with a finite interval, you have available to you the sine and cosine waves with frequencies that are multiples of $\frac {2\pi}L$ These waves are all orthogonal, so you cannot approximate any one as a sum of the others.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If I'm understanding this correctly it's saying that many shorter and shorter wavelength sins could NOT be added up to make a sin with a wavelength that is bigger than any of the others? Even if they were allowed to be phase shifted? $\endgroup$ – candied_orange Mar 16 '15 at 4:53
  • $\begingroup$ That is correct. The phase shift doesn't help because you already have the sines and cosines of that wavelength. The combination can represent any phase shift. $\endgroup$ – Ross Millikan Mar 16 '15 at 4:56
  • $\begingroup$ I'm not 100% sure I understand the finite interval concern. I intend that the desired sin wave not have an infinite wavelength. I do not intend to restrict the fourier transformation to a particular method. $\endgroup$ – candied_orange Mar 16 '15 at 5:13
  • $\begingroup$ The finite interval is needed to make sure the function is periodic, so that it can be expressed as a Fourier series. The point then is that you need all the terms of the Fourier series available because you cannot express one in terms of the others. Using an infinite interval and the Fourier integral raises convergence concerns, but I think comes to the same conclusion as long as you demand that the sine wave continue for all time. I do't understand what you mean by "not restrict the Fourier transformation to a particular method". What new freedom do you want to give? $\endgroup$ – Ross Millikan Mar 16 '15 at 5:19
  • $\begingroup$ So if I allowed the wave length to decay there might still be a way? $\endgroup$ – candied_orange Mar 16 '15 at 5:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.