Wikipedia:Reference desk/Archives/Mathematics/2019 January 4
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January 4
editFactoring trigonometric polynomials
editLet
be a real trigonometric polynomial of degree N and let T have 2N roots {z1, ... ,z2N} in the interval [0, 2π). Does it follow that
for some k?
Also, if a0 = 0, what can you say about the roots {z1, ... ,z2N}? (For example if N=1, a0 = 0 implies z1=z2±π.) --RDBury (talk) 11:15, 4 January 2019 (UTC)
- Yes. If we write , we see that both expressions on the RH-sides (the trigonometric polynomial and the sine product) have the form for some -degree complex polynomial , and in both cases, the polynomial has distinct complex roots . So the corresponding polynomials coincide up to a multiplicative factor , as you stated. pma 14:45, 5 January 2019 (UTC)
- Thanks, I had a feeling converting to complex variables was the key but I was still missing some steps. By similar reasoning, I think for part 2 the answer is that the Nth symmetric polynomial in {eiz1, ... ,eiz2N} is 0. For N=1 this is eiz1+eiz2 = 0 which reduces easily to z1=z2±π, butfor N≥2 the relationship isn't that simple. --RDBury (talk) 21:50, 5 January 2019 (UTC)
- I agree... We may write (using a more standard instead of for the constant term of the trigonometric polynomial)
- where and for , and with roots . Then the -degree coefficient vanishes iff
- , the sum being extended over all subsets of of cardinality but I can't see a more geometric equivalent condition on the inscribed -gon with vertices , even for ... pma 00:02, 6 January 2019 (UTC)
- I think that you meant ? I tried to obtain the expression for and in the end obtained the following expression:
- .
- Therefore
- As for I can only observe that it can not be arbitrary large because otherwise there will be no roots. Ruslik_Zero 17:43, 6 January 2019 (UTC)
- Yes, thank you, in fact I forgot a factor 1/2 in front of the sum (fixed now).pma 22:48, 6 January 2019 (UTC)
- I think that you meant ? I tried to obtain the expression for and in the end obtained the following expression:
- Thanks, I had a feeling converting to complex variables was the key but I was still missing some steps. By similar reasoning, I think for part 2 the answer is that the Nth symmetric polynomial in {eiz1, ... ,eiz2N} is 0. For N=1 this is eiz1+eiz2 = 0 which reduces easily to z1=z2±π, butfor N≥2 the relationship isn't that simple. --RDBury (talk) 21:50, 5 January 2019 (UTC)
- Yes. If we write , we see that both expressions on the RH-sides (the trigonometric polynomial and the sine product) have the form for some -degree complex polynomial , and in both cases, the polynomial has distinct complex roots . So the corresponding polynomials coincide up to a multiplicative factor , as you stated. pma 14:45, 5 January 2019 (UTC)