Can we use FFT to compute the first n values of a self-convolution form in O(nlogn)?

Правка en2, от duckladydinh, 2025-01-24 21:50:02

Hi

I am learning FFT. ChatGPT told me that FFT could assist in solving self-convolution form like Catalan number where the (n+1)-th element is some convolution of the first n elements. For example:

c[n + 1] = sum(c[i][n — i]) for i in 0..n

Unfortunately, no matter where I look (and how hard I press ChatGPT), I couldn't find a single website/book/paper detailing this approach.

My Question: Is it actually possible to compute the first n elements of a self-convolution form like Catalan using FFT in, for example, O(nlogn) or less than O(n^2)?

Thanks a lot

История

 
 
 
 
Правки
 
 
  Rev. Язык Кто Когда Δ Комментарий
en4 Английский duckladydinh 2025-01-26 17:11:26 0 (published)
en3 Английский duckladydinh 2025-01-26 17:10:58 838 Tiny change: 'i\n\n---\n##### Updat' -> 'i\n\n---\n\n#### Updat' (saved to drafts)
en2 Английский duckladydinh 2025-01-24 21:50:02 5 Tiny change: 'n solving convolutio' -> 'n solving self-convolutio'
en1 Английский duckladydinh 2025-01-24 21:49:44 661 Initial revision (published)