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The explicit expression for the Chebyshev polynomials of the first kind is given as follows. $$ T_n(x) =n\sum _{k=0}^{n}(-2)^{k}{\frac {(n+k-1)!}{(n-k)!(2k)!}}(1-x)^{k}\qquad n>0 $$ However, no proof of this expression is available in any literature. Can anyone please provide the proof for this equation?

Thanks in advance.

Orkodip Mookherjee
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The usual recurrence for Chebyshev is $$ T_{n+1}(x) + T_{n-1}(x) = 2 x T_n(x), \quad T_0(x) = 1, \quad T_1(x) = x. \tag{1} $$ Consider the polynomial sequence $$ f_n(t) := \sum_{k=0}^n \frac{(n+k-1)!\, n}{(n-k)!\,(2k)!}t^k, \qquad f_0(t) := 1. \tag{2} $$ We can prove it satisfies the recursion $$ f_{n+1}(t) + f_{n-1}(t) = (2+t)f_n(x). \tag{3} $$ The method is to compare coefficients of $\,t^k\,$ using the identity $$ \frac{(n+k)(n+k-1)(n+1)}{(n+1-k)(n-k)} + (n-1) =\\ \frac{(n+k-1)(2n)}{(n-k)} + \frac{(2k)(2k-1)n}{(n+1-k)(n-k)}. \tag{4} $$ Finally, notice that if $\,2+t=2x\,$ then $\,1+t/2=x\,$ and therefore $\, T_n(x) = f_n(t).\,$

The idea is that the usual Chebyshev polynomials satisfy $$ |T_n(x)| \le 1 \qquad \text{ if } \quad -1\le x \le 1 \tag{5} $$ while the related polynomials satisfy $$ |f_n(t)| \le 1 \qquad \text{ if } \quad -4\le x \le 0.\tag{6} $$ and so their domains are shifted and scaled differently.

Somos
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  • Thanks for your help. Can you give any reference for the polynomial sequence $f_n(t)$ and its recursion relation ? – Orkodip Mookherjee Dec 22 '19 at 14:04
  • @OrkodipMookherjee I just used your explicit expression summation and simplified it only using powers of $t$ instead of $x$. – Somos Dec 22 '19 at 15:31