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Using the theory of Pell equations and the fact that the discriminant $5$ has both the forms $k^2 + 4$ and $k^2 - 4$, I stumbled across a proof of the identities

$$ (-1)^n T_{2n} (\tfrac{i}{2}) = T_n (\tfrac{3}{2}) $$ $$ (-i)^{2n+1} U_{2n+1} (\tfrac{i}{2}) = U_n (\tfrac{3}{2})$$

for all positive integers $n$. Here, $i$ is a square root of $-1$ and $T_n$ and $U_n$ denote Chebyshev polynomials of the first and second kinds.

My question is, are there general identities involving Chebyshev polynomials that specialize to these when appropriate inputs are substituted?

Barry Smith
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1 Answers1

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Partial quick answer: The Wolfram function site http://functions.wolfram.com/05.04.16.0006.01 lists the relation $$T_{2n}(z) = (-1)^n T_n(1 - 2 z^2)$$

If you substitute $z=\frac{i}{2}$ you get $$T_{2n}\left(\frac{i}{2}\right) = (-1)^n T_n\left(1 - 2 (\frac{i}{2})^2\right)= (-1)^n T_n\left(\frac{3}{2}\right)$$ The corresponding entry for $U$ (http://functions.wolfram.com/05.05.16.0004.01) needs some more work than direct substitution (maybe with recurrence formula).

gammatester
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  • Thanks! Didn't know about the Wolfram function site! It's funny, using the identity they list for U and recurrence as you mention, it is easy to show that $U_{2n+1}(z) = (-1)^n 2z U_n(1-2z^2)$, which seems to me the more natural analogue of the identity for $T$. I wonder why they don't list it... – Barry Smith Oct 06 '16 at 14:05
  • The your $U$ relation and the Wolfram $T$ relation are both listed at https://en.wikipedia.org/wiki/Chebyshev_polynomial_of_the_first_kind#Other_properties – gammatester Oct 06 '16 at 14:15
  • Don't know how I missed those! – Barry Smith Oct 06 '16 at 14:45