# Chebyshev polynomials of first kind

I know the chebyshev polynomials of the first kind can be approximated using the cosine function, where $T_n(\cos \theta)=\cos(n \theta)$ and I know that chebyshev polynomials are a family of orthogonal polynomials. How would I prove that $$\int_{-1}^1 \frac{T_h(x)T_i(x)}{\sqrt{1-x^2}}dx=0, h \neq i$$

I can use $x=\cos(\theta)$, but I don't understand how to prove, and what this really means. Isn't it trivial because they're orthogonal?

Can anyone clarify?

Thank you.

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It is indeed "trivial because they're orthogonal", but there's also the question of how to prove that they're orthogonal. And that means you want to find that integral without relying on that fact, which is not yet proved at that point. – Michael Hardy Nov 16 '12 at 1:19
I suppose the easiest way would be to use the differential equation. – Pragabhava Nov 16 '12 at 1:19
Is there a way to do it without the differential equation? I'm supposed to use change of vairables... – A A Nov 16 '12 at 1:21

It is indeed "trivial because they're orthogonal", but there's also the question of how to prove that they're orthogonal. And that means you want to find that integral without relying on that fact, which is not yet proved at that point.

If you know that trigonometric identity, then you might write \begin{align} x & = \cos\theta \\[6pt] dx & = -\sin\theta\,d\theta \\[6pt] \sqrt{1-x^2} & = \sin\theta \\ \end{align} $$\int_{-1}^1 \frac{T_h(x) T_i(x)}{\sqrt{1-x^2}} \, dx = \int_\pi^0 \frac{\cos(h\theta)\cos(i\theta)}{\sin\theta} (-\sin\theta\,d\theta) = \int_0^\pi\cos(h\theta)\cos(i\theta)\,d\theta.$$ Then one way to procede might use the identity $$\cos\alpha\cos\beta = \frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}.$$ A symmetry argument shows quickly, without using antiderivatives, that the integral of cosine of a nonzero integer multiple of $\theta$ from $0$ to $\pi$ is $0$.

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Got it, thanks! – A A Nov 16 '12 at 1:36

Yes if you already know that they are orthogonal then this is trivial but how do you show that they are orthogonal in the first place. Evaluating this integral proves orthogonality. For

$$\int_{-1}^1 \frac{T_h(x) T_i(x)}{\sqrt{1-x^2}} \, dx$$

start with the substitution $x=\cos(\theta), dx=-\sin(\theta) \, d\theta$ and you get

$$\int_0^\pi \cos(h\theta) \cos(i\theta) \, d\theta$$

where the bounds become 0 and pi but then they flip because of the minus sign and the denominator simplifies to sine which cancels with the sine in dx. Then we do different cases with $h=i$ and $h\neq i$. The second case will always be zero which can be done using trig addition/subtraction formulas.

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Using Tschebyscheff differential equation in it's autoadjoint form we have, for $T_n$ and $T_m$ \begin{align} \frac{d}{dx}\left\{\sqrt{1-x^2} T_n'\right\} + \frac{n^2}{\sqrt{1-x^2}} T_n &= 0 \\ \frac{d}{dx}\left\{\sqrt{1-x^2} T_m'\right\} + \frac{m^2}{\sqrt{1-x^2}} T_m &= 0 \end{align}

Multiplying the first equation by $T_m$ and the second by $T_n$ and substracting them \begin{align} \frac{d}{dx}\left\{\sqrt{1-x^2} T_n'\right \}T_m - \frac{d}{dx}\left\{\sqrt{1-x^2} T_m'\right\}T_n + \frac{n^2 - m^2}{\sqrt{1-x^2}} T_m T_n &= 0 \\ \frac{d}{dx}\left\{\sqrt{1-x^2} (T_n'T_m - T_m'T_n)\right\} + \frac{n^2 - m^2}{\sqrt{1-x^2}} T_m T_n &= 0 \end{align}

Now, the wronskian $W(T_n,T_m)$ is $$W(T_n,T_m) = T_n T_m' - T_n' T_m = \frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{\sqrt{1-x^2}}$$ and then $$\int_{-1}^1 \frac{T_n(x) T_m(x)}{\sqrt{1-x^2}} dx = \int_{-1}^1 \frac{d}{d x} \left\{\frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{n^2 - m^2}\right\}dx$$ If $n \neq m$, it's easy to see that the integral is zero. If $n \to m$, we have that $$\lim_{n \to m} \frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{n^2 - m^2} = \begin{cases}-\frac{\theta}{2}, & m \neq 0, \\ \\ - \theta, & m = 0 \end{cases}$$ and then $$\int_{-1}^1 \frac{T_n(x) T_m(x)}{\sqrt{1-x^2}} dx = \begin{cases} 0, & m \neq n, \\ \\ \frac{\pi}{2} & m = n \neq 0, \\ \\ \pi & m = n = 0. \end{cases}$$

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