Assume that you have a circle with radius $r_0$, then you keep adding cosine modes as below:


if you plot this as below by matlab:


You will see that you can get different shapes by changing the number of modes (i.e. here n=4) or coefficients (i.e. $a_i$) and even omitting some modes (i.e. $a_k=0$) you will get different shapes.

My question is that how can you decompose for instance an oval or a square in cosine modes, and find their coefficients (somehow in the way of fourier decomposition), is it ever possible?

But I think making any shape would be possible by my method mentioned above, but I am interested to know how I can come from an arbitrary shape to its corresponding cosine modes.

  • $\begingroup$ I don't have an answer for you, but this is a really interesting question. I'm experimenting and doing research with it right now. I can tell you that some shapes are impossible. (I won't explain why right now). I'll get back to you. $\endgroup$ – Polygon Jun 4 '16 at 2:45
  • $\begingroup$ @Polygon yes it is a good question, but could you give a simple example at least or a hint that help me in my thinking on it. $\endgroup$ – Soyol Jun 4 '16 at 3:01
  • $\begingroup$ @soeil I don't have any examples or hints. It's pretty complicated. I don't even know if it's possible to make squares or ellipses, but I'm working on it. $\endgroup$ – Polygon Jun 4 '16 at 3:22
  • $\begingroup$ @soeil I've been working on this, and I hate to say it, but I'm pretty sure a square is impossible. I'll keep trying different things and keep you updated. $\endgroup$ – Polygon Jun 5 '16 at 2:18
  • 1
    $\begingroup$ The title also mentions sin modes, the text does not. Do you want to include these? Essentially you have a periodic function $r(\theta)$. I'd say that writing this as a linear combination $\sum_ka_k\sin(k\theta)+b_k\cos(k\theta)$ is not just “somehow in the way of fourier decomposition” but in fact is Fourier decomposition. Did I miss some relevant difference here? If $r(\theta)$ is not a function but a relation (i.e. if you have more than two points on the shape for every line through the origin), then you can't get the shape via this method. $\endgroup$ – MvG Jun 6 '16 at 11:29

Your Answer

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

Browse other questions tagged or ask your own question.