2
$\begingroup$

I have come across the following identity $$ -\log|x-y|=\log 2+\sum_{n\geq 1}\frac{2}{n}T_n(x)T_n(y)\qquad -1\leq x,y\leq 1 $$ where $T_n(x)$ are Chebyshev polynomials. The identity seems to hold numerically, but I am at a loss on how to prove it, or whether there is a standard reference where I could look it up. Any help is appreciated!

$\endgroup$
  • $\begingroup$ Did you read it somewhere or did you stumble upon it after some random computation ? (Do you already have a reference, even without a proof ?) $\endgroup$ – charmd Jun 5 '18 at 9:00
  • $\begingroup$ @charMD I have read it somewhere - and got rather stunned by it! - but I cannot retrieve the reference...from what I remember, the paper I found it in did not contain a proof anyway... $\endgroup$ – Pierpaolo Vivo Jun 5 '18 at 9:22
2
$\begingroup$

The identity follows easily from $T_n(x)=\cos(n\,\arccos(x)).$ Letting $X=\arccos(x)$ and $Y=\arccos(y)$, use the logarithmic power series with a complex argument, and subsequent simplification will yield

$$\sum_{n=1}^\infty \frac{\cos(nX)\cos(nY)}{n} = \frac{-1}{4} \Big( \log(2-2\cos(X+Y)) +\log(2-2\cos(X-Y)) \Big) ,$$

which is equivalent to

$$\sum_{n=1}^\infty \frac{T_n(x) \, T_n(y)}{n} + \frac{\log2}{2}= \frac{-1}{4} \Big( \log(1-\cos(\arccos(x)+\arccos(y))) + \\\log(1-\cos(\arccos(x)-\arccos(y))) \Big) .$$

Use $\cos(\arccos(x) \pm \arccos(y))= xy\, \mp \sqrt{1-x^2}\,\sqrt{1-y^2}.$ With the domain for your $x$ and $y$, the two logarithms can be combined and algebra will complete the proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.