# 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

$${T}_{n+1}(x)-2x{T}_{n}(x)+{T}_{n-1}=0$$ |

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

$${\int}_{-1}^{+1}\mathit{d}x\frac{{T}_{m}(x){T}_{n}(x)}{\sqrt{1-{x}^{2}}}=0$$ |

when $m\ne n$.

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 $$. Then we have

${T}_{n+1}(-x)$ | $=2(-x){T}_{n}(-x)-{T}_{n-1}(-x)$ | ||

$=-{(-1)}^{n}2x{T}_{n}(x)-{(-1)}^{n-1}{T}_{n-1}(x)$ | |||

$={(-1)}^{n+1}(2x{T}_{n}(x)-{T}_{n-1}(x))$ | |||

$={(-1)}^{n+1}{T}_{n+1}(x).$ |

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

$${\int}_{-1}^{+1}\mathit{d}x\frac{{T}_{m}(x){T}_{n}(x)}{\sqrt{1-{x}^{2}}}=0$$ |

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 $m\le n$
and $k\le n$ and $m\ne k$. By the conclusions^{} of last
paragraph, we know that ${T}_{n+1}$ is orthogonal to ${T}_{n}$.
Assume then that $m\le n-1$. Using the recursion, we have

$\int}_{-1}^{+1}}\mathit{d}x{\displaystyle \frac{{T}_{m}(x){T}_{n+1}(x)}{\sqrt{1-{x}^{2}}$ | $=2{\displaystyle {\int}_{-1}^{+1}}\mathit{d}x{\displaystyle \frac{x{T}_{m}(x){T}_{n}(x)}{\sqrt{1-{x}^{2}}}}-{\displaystyle {\int}_{-1}^{+1}}\mathit{d}x{\displaystyle \frac{{T}_{m}(x){T}_{n-1}(x)}{\sqrt{1-{x}^{2}}}}$ | ||

$={\displaystyle {\int}_{-1}^{+1}}\mathit{d}x{\displaystyle \frac{{T}_{m+1}(x){T}_{n}(x)}{\sqrt{1-{x}^{2}}}}+{\displaystyle {\int}_{-1}^{+1}}\mathit{d}x{\displaystyle \frac{{T}_{m-1}(x){T}_{n}(x)}{\sqrt{1-{x}^{2}}}}-{\displaystyle {\int}_{-1}^{+1}}\mathit{d}x{\displaystyle \frac{{T}_{m}(x){T}_{n-1}(x)}{\sqrt{1-{x}^{2}}}}$ |

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 $m\le n+1$ and
$k\le n+1$ and $m\ne k$.

Title | orthogonality of Chebyshev polynomials from recursion |
---|---|

Canonical name | OrthogonalityOfChebyshevPolynomialsFromRecursion |

Date of creation | 2013-03-22 18:54:46 |

Last modified on | 2013-03-22 18:54:46 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 6 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 33C45 |

Classification | msc 33D45 |

Classification | msc 42C05 |