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  1.  @(p,q)ifft(fft([p q*0]).*fft([q p*0]))(1:end-1)
  2.  
  3.   ans([1 1], [1 4 6 4 1])
Success #stdin #stdout 0.52s 121728KB
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ans =

@(p, q) ifft (fft ([p, q * 0]) .* fft ([q, p * 0])) (1:end - 1)

ans =

    1.00000    5.00000   10.00000   10.00000    5.00000    1.00000
https://ideone.com/To0xFl
language:
Octave (octave 4.4.1)
created:
7 years ago
visibility:
public

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