This Theorem says

statement: $\int_{-1}^{1} \frac{T_n(x)T_m(x)}{ \sqrt{1+x^2} } dx = 0 ;$ when $n\ne m $

proof: "substitute $x= cos \theta$ " and that's it.

So I am wondering should I start with this

$\int_{-1}^{1} \frac{cos(\theta n)cos(\theta m)}{\sqrt{1+x^2}} dx $ or with this

$\int_{-1}^{1} \frac{cos(\theta n)cos(\theta m)}{\sqrt{1+cos^2 \theta}} dx $ I order to verify the proof.

note: T(x) is Chebyshev polynomial.


When you do this, you apply the substitution rule for integrals, fully. Everything with $x$ in it, including the denominator, the $dx$, and the limits, transform.

On the other hand, the statement that you're trying to prove is incorrect. The correct form has $\sqrt{1-x^2}$ in the denominator instead.

  • $\begingroup$ $\int_{-1}^{1} \frac{cos(\theta n)cos(\theta m)}{\sqrt{1+cos^2 \theta}} dx = \int_{0}^{\pi} cos(\theta m). cos(\theta n) d\theta$ I wonder how this come about when we change $x$ to $cos(\theta)$ where does the denominator go. $\endgroup$ – tt z Mar 10 '19 at 4:42
  • $\begingroup$ It goes into the $d\theta$. Well, it does if you use the correct version with a minus sign in that square root. $\endgroup$ – jmerry Mar 10 '19 at 4:44
  • $\begingroup$ I see, trig substitutions. $\endgroup$ – tt z Mar 10 '19 at 4:57

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