# orthogonality of Chebyshev polynomials from recursion

In this entry, we shall demonstrate the orthogonality relation of the Chebyshev polynomials  from their recursion relation  . Recall that this relation reads as

 $T_{n+1}(x)-2xT_{n}(x)+T_{n-1}=0$

with initial conditions  $T_{0}(x)=1$ and $T_{1}(x)=x$. The relation we seek to demonstrate is

 $\int_{-1}^{+1}dx\,{T_{m}(x)T_{n}(x)\over\sqrt{1-x^{2}}}=0$

when $m\neq n$.

We start with the observation that $T_{n}$ is an even function when $n$ is even and an odd function when $n$ is odd. That this is true for $T_{0}$ and $T_{1}$ follows immediately from their definitions. When $n>1$, we may induce this from the recursion. Suppose that $T_{m}(-x)=(-1)^{m}T_{m}(x)$ when $m. Then we have

 $\displaystyle T_{n+1}(-x)$ $\displaystyle=2(-x)T_{n}(-x)-T_{n-1}(-x)$ $\displaystyle=-(-1)^{n}2xT_{n}(x)-(-1)^{n-1}T_{n-1}(x)$ $\displaystyle=(-1)^{n+1}(2xT_{n}(x)-T_{n-1}(x))$ $\displaystyle=(-1)^{n+1}T_{n+1}(x).$

From this observation, we may immediately conclude half of orthogonality. Suppose that $m$ and $n$ are nonnegative integers whose difference is odd. Then $T_{m}(-x)T_{n}(-x)=-T_{m}(x)T_{n}(x)$, so we have

 $\int_{-1}^{+1}dx\,{T_{m}(x)T_{n}(x)\over\sqrt{1-x^{2}}}=0$

because the integrand is an odd function of $x$.

To cover the remaining cases, we shall proceed by induction  . Assume that $T_{k}$ is orthogonal  to $T_{m}$ whenever $m\leq n$ and $k\leq n$ and $m\neq k$. By the conclusions  of last paragraph, we know that $T_{n+1}$ is orthogonal to $T_{n}$. Assume then that $m\leq n-1$. Using the recursion, we have

 $\displaystyle\int_{-1}^{+1}dx\,{T_{m}(x)T_{n+1}(x)\over\sqrt{1-x^{2}}}$ $\displaystyle=2\int_{-1}^{+1}dx\,{xT_{m}(x)T_{n}(x)\over\sqrt{1-x^{2}}}-\int_{% -1}^{+1}dx\,{T_{m}(x)T_{n-1}(x)\over\sqrt{1-x^{2}}}$ $\displaystyle=\int_{-1}^{+1}dx\,{T_{m+1}(x)T_{n}(x)\over\sqrt{1-x^{2}}}+\int_{% -1}^{+1}dx\,{T_{m-1}(x)T_{n}(x)\over\sqrt{1-x^{2}}}-\int_{-1}^{+1}dx\,{T_{m}(x% )T_{n-1}(x)\over\sqrt{1-x^{2}}}$

By our assumption  , each of the three integrals is zero, hence $T_{n+1}$ is orthogonal to $T_{m}$, so we conclude that $T_{k}$ is orthogonal to $T_{m}$ when $m\leq n+1$ and $k\leq n+1$ and $m\neq k$.

Title orthogonality of Chebyshev polynomials from recursion OrthogonalityOfChebyshevPolynomialsFromRecursion 2013-03-22 18:54:46 2013-03-22 18:54:46 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Proof msc 33C45 msc 33D45 msc 42C05