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I was reading a Wikipedia article on Chebyshev polynomials and got stuck in around the end of the article where the author takes advantage of orthogonality to compute the coefficients of the Chebyshev expansion of $ \log (x+1) $. I will be happy if someone explains the process to get $$a_0 = - \log(2)\quad\mbox{and}\quad a_n = \frac{- \pi (-1)^n}{n} \quad\mbox{ for $ n \in \mathbb{N} $.}$$ Any help is appreciated.

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  • $\begingroup$ @stevengregory , my function is log(1+x) not 1+x. $\endgroup$ – Aleph-null Sep 27 '17 at 5:39
  • $\begingroup$ So was mine. I made a typo. $a_0 = \displaystyle \frac{1}{\pi}\int_{-1}^{1} \dfrac{\ln(1+x)}{\sqrt{1-x^2}} dx = \frac{1}{\pi}(-\pi \ln 2)=-\ln 2$ $\endgroup$ – steven gregory Sep 27 '17 at 6:55
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Note that for $t\in (0,\pi)$, $$ \begin{align} \log(1+\cos(t)) &=\log\left(\frac{e^{it}+2+e^{-it}}{2}\right)\\ &=\log\left(1+e^{it}\right)+\log\left(1+e^{-it}\right)-\log(2)\\ &=\sum_{n=1}^\infty(-1)^{n+1}\frac{e^{int}}{n}+\sum_{n=1}^\infty(-1)^{n+1}\frac{e^{-int}}{n}-\log(2)\\ &=2\sum_{n=1}^\infty(-1)^{n+1}\frac{\cos(nt)}{n}-\log(2). \end{align} $$ Hence, by letting $x=\cos(t)\in (-1,1)$, we obtain $$\log(1+x)=\sum_{n=1}^\infty\frac{2(-1)^{n+1}}{n}T_n(x)-\log(2)T_0(x)$$ which implies that $$a_0=-\log(2)\quad\mbox{and}\quad a_n=\frac{2(-1)^{n+1}}{n}\mbox{ for $n>0$}$$ where $\ln(1+x)=\sum_{n\geq 0}a_n T_n(x)$, which is a bit different from what we read on the wikipage.

P.S. Going back to the wikipage: \begin{align*} \int_{-1}^1\frac{\log(1+x)T_m(x)}{\sqrt{1-x^2}}\,dx&=\int_{\pi}^0\frac{\log(1+\cos(t))\cos(mt)}{|\sin(t)|}\,(-\sin(t)dt)\\ &=\int_0^{\pi}\left(2\sum_{n=1}^\infty(-1)^{n+1}\frac{\cos(nt)}{n}-\log(2)\right)\cos(mt) dt\\ &=\begin{cases} -\pi\log(2) &\mbox{if $m=0$,}\\ \dfrac{(-1)^{m+1}\pi}{m}&\mbox{if $m>0$.} \end{cases} \end{align*} On the other hand if $\ln(1+x)=\sum_{n\geq 0}a_n T_n(x)$, \begin{align*} \int_{-1}^1\frac{\log(1+x)T_m(x)}{\sqrt{1-x^2}}\,dx&=a_m\int_{-1}^1\frac{T_m(x)^2}{\sqrt{1-x^2}}\,dx =\begin{cases} \pi a_0 &\mbox{if $m=0$,}\\ \dfrac{\pi}{2}a_m&\mbox{if $m>0$.} \end{cases} \end{align*}

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  • $\begingroup$ I knew that property of orthogonality, my question was about the computation of $ a_n $ in case of $ f(x) = log(1+x) $. Thanks though! $\endgroup$ – Aleph-null Sep 27 '17 at 5:36
  • $\begingroup$ Since $\cos nt=T_n(\cos t)$, you could have made the substitution $x=\cos t$ immediately. Integration of an infinite series term by term would need some justification; the series doesn't converge uniformly. $\endgroup$ – user436658 Sep 27 '17 at 7:07
  • $\begingroup$ @ProfessorVector Good point! Thanks. $\endgroup$ – Robert Z Sep 27 '17 at 7:22
  • $\begingroup$ @AlPinky I edited my answer. Is it better now? Any further doubt? $\endgroup$ – Robert Z Sep 27 '17 at 15:40
  • $\begingroup$ @RobertZ, Yes, much better. Thanks man, I appreciate your help. $\endgroup$ – Aleph-null Sep 27 '17 at 17:04

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