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{\bf Question}


Chebyshev's equation may be written $$ {d \over
{dx}}\left[(1-x^2)^{1 \over 2}{{dy} \over
{dx}}\right]+n^2(1-x^2)^{-{1 \over 2}}y=0 $$ where $n$ is a
positive integer. The solutions which satisfy the boundary
conditions $T_n(-1)=(-1)^n$ and $T_n(1)=1$ are polynomials of
degree $n$. Find the first three Chebyshev polynomials. Prove that
if $m \neq n$ then $$ \int_{-1}^1 (1-x^2)^{-{1 \over
2}}T_m(x)T_n(x) dx =0 $$


\vspace{.5in}

{\bf Answer}

To find the Chebyshev polynomials $T_n(x)$ one substitutes a
general polynomial into the differential equation and uses the
boundary conditions. We illustrate this with $T_2(x)$.

The general form of $T_2(x)$ is a quadratic so that
$T_2(x)=a_0+a_1x+a_2x^2$. Using $T_2(1)=1$ gives $a_0+a_1+a_2=1$. Using
$T_2(-1)=1$ gives  $a_0-a_1+a_2=1$. Subtracting the equations gives
$a_1=0$ and hence $a_0=1-a_2$. Writing $a_2$ as $a$ we see that
$T_2(x)=ax^2+(1-a)$.

We now substitute this into Chebyshev's equation (with $n=2$) and
obtain

$\{2a(1-2x^2)+4ax^2+4(1-a)\}(1-x^2)^{-1/2}=0, \quad \Rightarrow
\quad 4-2a=0$. So that $a=2$ and $T_2(x)=2x^2-1$.

Now to show orthogonality. since $T_n$ and $T_m$ are solutions of
Chebeyshev's equation we have:

\begin{eqnarray} \frac{d}{dt}\left[(1-x^2)\frac{dT_n}{{dx}}\right]
+n^2(1-x^2)^{-1/2}T_n & = & 0 \\ \frac{d}{dt}\left[
(1-x^2)\frac{dT_m}{dx}\right]+m^2(1-x^2)^{-1/2}T_m & = & 0
\end{eqnarray}

Multiplying equation (1) by $T_m$ and equation (2) by $T_n$,
subtracting and integrating by parts between $-1$ and $1$ gives:

$\ds(n^2-m^2)\int_{-1}^1(1-x^2)^{-1/2}T_nT_mdx \\ =
\int_{-1}^1\left\{T_n\left[{d \over {dx}}(1-x^2){{dT_m} \over
{dx}}\right] -T_m\left[{d \over {dx}}(1-x^2){{dT_n} \over
{dx}}\right]\right\}dx \\ =  \left[(1-x^2)\left(T_n{{dT_m} \over
{dx}}-T_m{{dT_n} \over {dx}}\right)\right]_{-1}^1 \\
-\int_{-1}^1\left\{(1-x^2){{dT_n} \over {dx}}{{dT_m} \over {dx}}
-(1-x^2){{dT_m} \over {dx}}{{dT_n} \over {dx}}\right\}dx \\  = 0
$

Since the first term vanishes at $\pm 1$, and the terms and the
integral cancel.



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