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Chebyshev Polynomials Proof Problem

Prove (or disprove): Define \(T_n(x) \) as the Chebyshev polynomial of the first kind with degree \(n\). If \(p\) is an odd prime, then \(\sqrt{\frac{T_p(x) - 1}{x-1}} \) is an irreducible polynomials over the rational numbers.


By definition \( T_n(x) = \cos(n \cos^{-1}(x)) \). Listing out the first few odd primes of \(n\) shows that (for simplicity sake, we take the positive root only)

\( T_3(x) - 1 = (4x^3-3x) - 1 = (x-1)(2x+1)^2 \\ \Rightarrow \sqrt{\frac{T_3(x) - 1}{x-1}} = 2x + 1 \)

\( T_5(x) - 1 = (16x^5-20x^3+5x) - 1 = (x-1)(4x^2+2x-1)^2 \\ \Rightarrow \sqrt{\frac{T_5(x) - 1}{x-1}} = 4x^2+2x-1 \)

\( T_7(x) - 1 = (64x^7-112x^5+56x^3-7x) - 1 = (x-1)(8x^3+4x^2-4x-1)^2 \\ \Rightarrow \sqrt{\frac{T_7(x) - 1}{x-1}} = 8x^3+4x^2-4x-1 \)

\(\begin{eqnarray} T_{11}(x) - 1 &=& (1024 x^{11}-2816 x^9+2816 x^7-1232 x^5+220 x^3-11x) - 1 \\ &=& (x-1) (32 x^5+16 x^4-32 x^3-12 x^2+6 x+1)^2 \\ \Rightarrow \sqrt{\frac{T_{11}(x) - 1}{x-1}} &=& 32 x^5+16 x^4-32 x^3-12 x^2+6 x+1 \end{eqnarray}\)

All these polynomials are irreducible polynomials over the rationals. Is it true for all odd prime \(p\)?

Note by Pi Han Goh
2 years, 1 month ago

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The solution is as I expected click here.I hope you don't mind that I copied the problem to math overflow.It uses only the multiplicative property of the degrees of number fields and that \(x^{p-1}+x^{p-2}+...+1\), the pth cyclotomic polynomial, is irreducible (can be proven by shifting x to x+1 and using eisenstein). Bogdan Simeonov · 2 years, 1 month ago

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@Bogdan Simeonov Haha, it's a bit of an overkill to use OverFlow! I was about to post this in StackExchange if there's no answer. Thanks! Pi Han Goh · 2 years, 1 month ago

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@Pi Han Goh I posted it on stackexchange, but no one provided a good solution :D Bogdan Simeonov · 2 years, 1 month ago

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This post needs to be seen by more people so we can actually solve it lol.Also I still cannot understand why Eisenstein works here (what is special about the coeffs) Bogdan Simeonov · 2 years, 1 month ago

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I think that the identity \(T_n(x)=cos(n.arccosx)\) can give us the roots of \(T_n(x)-1\) Bogdan Simeonov · 2 years, 1 month ago

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@Bogdan Simeonov I'm listening. Please continue on your elaboration. Pi Han Goh · 2 years, 1 month ago

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@Pi Han Goh Well cosx is 1 only when x is 2.pi.something.So that means that \(n.arrcosx\) must be 2.pi.something.So \(x=cos(2.\pi.k/n)\) are the roots of T_n(x)-1.Problem is, I don't know when they double roots and how to prove that T is their minimal polynomial.It is interesting that these roots are the real parts of the roots of unity. Bogdan Simeonov · 2 years, 1 month ago

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@Bogdan Simeonov Also the identity \((T_n(x)-1)/(x-1)(T_n(x)+1)/(x+1)=U_{n-1}\) might help, since we know the factorisation of U. Bogdan Simeonov · 2 years, 1 month ago

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If you shift x to x+1 the irreducibility of these can be proven with Eisenstein's criterion. Bogdan Simeonov · 2 years, 1 month ago

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@Bogdan Simeonov By the way do you have a solution or just spotted that they are irreducible for small values?Vecause I can't figure out what identity to use to make that square root in a better form.Eisenstein still works though Bogdan Simeonov · 2 years, 1 month ago

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@Bogdan Simeonov I don't have a solution, hence the "or disprove" statement. I got inspired by solving via the Challenge Master notes comment on my solution here. Pi Han Goh · 2 years, 1 month ago

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@Pi Han Goh So it is perfectly possible that a counterexample exists.Well I'm going to read up on Tschebyscheff Polynomials, any reccomendations?Also, I have found a paper that states that (T_p(x)/x/) is irreducible for all primes p.But I don't know how to deal with that -1. Bogdan Simeonov · 2 years, 1 month ago

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@Bogdan Simeonov Recommendation? Haha, no, I read up a lot before posting this, and I'm still pretty stumped. Link please? Pi Han Goh · 2 years, 1 month ago

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@Bogdan Simeonov Thanks! But how do I prove that it's true for ALL odd primes \(p\)? Pi Han Goh · 2 years, 1 month ago

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