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There exists a generating function for the Chebyshev polynomials in the following form:

$$\sum\limits_{n=1}^{\infty}T_{n}(x) \frac{t^n}{n} = \ln\left( \frac{1}{\sqrt{ 1 - 2tx + t^2 }}\right)$$

Question: can one find a similar closed form for $\sum\limits_{n=1}^{\infty}T_{2n}(x) \frac{t^n}{n}$ ?

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3 Answers 3

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It suffices to observe that $$\sum\limits_{n=1}^{\infty}T_{n}(x) \frac{(\color{red}-t)^n}{n} = \ln\left( \frac{1}{\sqrt{ 1 \color{red}+ 2tx + t^2 }}\right)$$ Then $$\begin{align} \ln\left( \frac{1}{\sqrt{ 1 - 2tx + t^2 }}\right) + \ln\left( \frac{1}{\sqrt{ 1 + 2tx + t^2 }}\right) &= 2\sum\limits_{n\in \mathbb{N}, 2|n}T_{n}(x) \frac{t^n}{n} \\ &= 2\sum\limits_{m=1}^{\infty}T_{2m}(x)\frac{t^{2m}}{2m}\\ &= \sum\limits_{m=1}^{\infty}T_{2m}(x)\frac{(\color{red}{t^{2} })^m}{m} \end{align}$$ $$\implies \sum\limits_{n=1}^{\infty}T_{2n}(x) \frac{t^n}{n}=\ln\left( \frac{1}{\sqrt{ 1 - 2\sqrt{t}x + t }}\right) + \ln\left( \frac{1}{\sqrt{ 1 + 2\sqrt{t}x + t }}\right) $$

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  • $\begingroup$ Thank you ! Just a typo I guess, ie. the power 2 in the $t$ term should disappear: $$\implies \sum\limits_{n=1}^{\infty}T_{2n}(x) \frac{t^n}{n}=\ln\left( \frac{1}{\sqrt{ 1 - 2\sqrt{t}x + t }}\right) + \ln\left( \frac{1}{\sqrt{ 1 + 2\sqrt{t}x + t }}\right) $$ $\endgroup$
    – edrezen
    Mar 2 at 17:13
  • $\begingroup$ @edrezen Agreed, I corrected it. $\endgroup$
    – NN2
    Mar 2 at 17:25
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Just for the record, here is a little modification to NN2's answer in order to avoid the inner square root for $t$: $$\sum\limits_{n=1}^{\infty}T_{2n}(x) \frac{t^n}{n} = - \frac{1}{2} \, \ln \bigg ( 1 + 2\,(1-2x^2)\,t + t^2 \bigg) $$

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Following NN2's idea, I figured out a generalization for the formula. If we note:

$$\sum\limits_{n=1}^{\infty}T_{n}(x) \frac{t^n}{n} = - \frac{1}{2} \, \ln \bigg( 1 + P_0(x)\,t + t^2 \bigg) \ \ \ \ with \ \ \ P_0(x)=-2x $$ then we will have: $$\sum\limits_{n=1}^{\infty}T_{2n}(x) \frac{t^n}{n} = - \frac{1}{2} \, \ln \bigg( 1 + P_1(x)\,t + t^2 \bigg) $$ with $$ P_1(x)=2-P_0^2(x)$$

More generally, we will have : $$\sum\limits_{n=1}^{\infty}T_{n\,2^k}(x) \frac{t^n}{n} = - \frac{1}{2} \, \ln \bigg( 1 + P_k(x)\,t + t^2 \bigg) \ \ \ \ with \ \ \ P_{k}(x) = 2 - P_{k-1}^2(x) $$

The first $P_n$ polynomials are: \begin{equation} \begin{split} P_0(x) &= -2x \\ P_1(x) &= -4x^2 + 2 \\ P_2(x) &= -16x^4 + 16x^2 - 2 \\ P_3(x) &= -256x^8 + 512x^6 - 320x^4 + 64x^2 - 2 \\ \end{split} \end{equation}

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