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# 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 $m<n$. 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 $m\neq k$.

## Mathematics Subject Classification

33C45*no label found*33D45

*no label found*42C05

*no label found*

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