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The generating function for Chebyshev polynomials of the first kind is given as:

$$\sum_{n=0}^{\infty}T_n(x)t^n=\frac{1-xt}{1-2xt+t^2}$$

As I understand it, the sum of the coefficients of the generating function for Chebyshev polynomials is given by letting $t=1$. This seems to be clear from inspecting the LHS of the equation. However, the RHS of the equation simply reduces to $\frac{1}{2}$ and is no longer a function of $x$. Can anybody resolve this apparent anomaly? Thank you in advance!

PinkyWay
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Bernard Bollen
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  • According to https://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html, the generating function only works for $|x| \leq 1, |t| < 1$. – Gareth Ma Apr 08 '20 at 05:30

1 Answers1

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There is this $|t|<1$ restriction for this formula to be valid (the lhs series must be convergent)

The proof can be found page 116 of https://eprints.ucm.es/32785/1/T36277.pdf

The main steps are: $$ T_n(x) = \cos(n \arccos(x)) = \Re[\exp(i\,n\arccos(x))] $$ which is valid for any $x\in\mathbb{R}$

However, if $|t|<1$ $$ \sum_{n=0}^\infty t^n \exp(i\,n\arccos(x))=\sum_{n=0}^\infty (t \exp(i \arccos(x)))^n=\frac{1}{1-t\exp(i \arccos(x))} $$

Taking the real part of this gives you your generating function

Picaud Vincent
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    Thank you for both of those answers. That clears up that issue. Just wondering if it is at all possible to develop an expression for the infinite sum of Chebyshev polynomials as a function of x? – Bernard Bollen Apr 08 '20 at 05:54
  • @BernardBollen I do not think so as the series of terms $T_n(\cos\theta)=cos(n\theta)$ will not converge – Picaud Vincent Apr 08 '20 at 06:02