# orthogonality of Chebyshev polynomials

By expanding the of de Moivre identity

 $\cos{n\varphi}\;=\;(\cos\varphi+i\sin\varphi)^{n}$

to sum, one obtains as real part certain terms containing power products of $\cos\varphi$ and $\sin\varphi$, the latter ones only with even exponents.  When these are expressed with cosines ($\sin^{2}\varphi=1-\cos^{2}\varphi$), the real part becomes a polynomial $T_{n}$ of degree $n$ in the argument (http://planetmath.org/Argument2) $\cos\varphi$:

 $\displaystyle\cos{n\varphi}\;=\;T_{n}(\cos\varphi)$ (1)

This can be written equivalently (http://planetmath.org/Equivalent3)

 $\displaystyle T_{n}(x)\;=\;\cos(n\arccos{x}).$ (2)

It’s a question of Chebyshev polynomial of first kind and of $n$ (cf. special cases of hypergeometric function).

For showing the orthogonality of $T_{m}$ and $T_{n}$ we start from the integral $\displaystyle\int_{0}^{\pi}\cos{m\varphi}\cos{n\varphi}\,d\varphi$, which via the substitution

 $\cos\varphi\,:=\,x,\quad dx\,=\,-\sin\varphi\,d\varphi\,=\,-\sqrt{1\!-\!x^{2}}% \,d\varphi$

changes to

 $\displaystyle\int_{0}^{\pi}\cos{m\varphi}\,\cos{n\varphi}\;d\varphi\;=\;-\!% \int_{1}^{-1}T_{m}(x)T_{n}(x)\frac{dx}{\sqrt{1\!-\!x^{2}}}.$ (3)

The left of this equation is evaluated by using the product formula in the entry trigonometric identities:

 $\displaystyle\int_{0}^{\pi}\!\cos{m\varphi}\,\cos{n\varphi}\;d\varphi\;=\;% \frac{1}{2}\int_{0}^{\pi}(\cos{(m\!-\!n)\varphi}+\cos{(m\!+\!n)\varphi})\,d% \varphi\;=\,\begin{cases}0\mbox{\, for\, }m\neq n,\\ \frac{\pi}{2}\mbox{\, for\, }m=n\neq 0.\end{cases}$

By (3), we thus have

 $\displaystyle\int_{-1}^{1}\!T_{m}(x)T_{n}(x)\frac{dx}{\sqrt{1\!-\!x^{2}}}\;=\;% \begin{cases}0\mbox{\, for\, }m\neq n,\\ \frac{\pi}{2}\mbox{\, for\, }m=n\neq 0,\end{cases}$

which means the orthogonality of the polynomials $T_{m}(x)$ and $T_{n}(x)$ weighted by $\frac{1}{\sqrt{1\!-\!x^{2}}}$.

Any Riemann integrable real function $f$, defined on  $[-1,\,1]$,  may be expanded to the series

 $f(x)\;=\;\frac{a_{0}}{2}T_{0}(x)+\sum_{j=1}^{\infty}a_{j}T_{j}(x),$

where

 $a_{j}\;=\;\frac{2}{\pi}\int_{-1}^{1}\!f(x)T_{j}(x)\frac{dx}{\sqrt{1\!-\!x^{2}}% }\qquad(j=0,\,1,\,2,\,\ldots)$

This concerns especially the polynomials  $f(x):=x^{n}$,  for which we obtain

 $\displaystyle x^{n}$ $\displaystyle\;=\;\cos^{n}\varphi\;=\;\cosh^{n}{i\varphi}\;=\;2^{-n}(e^{i% \varphi}+e^{-i\varphi})$ $\displaystyle\;=\;2^{-n}\left[{n\choose 0}(e^{ni\varphi}+e^{-ni\varphi})+{n% \choose 1}(e^{(n-2)i\varphi}+e^{-(n-2)i\varphi})+\ldots\right]$ $\displaystyle\;=\;2^{1-n}\left[{n\choose 0}T_{n}(x)+{n\choose 1}T_{n-2}(x)+{n% \choose 2}T_{n-4}(x)+\ldots\right]\!.$

(If $n$ is even, the last term contains $T_{0}(x)$ but its coefficient is only a half of the middle number of the Pascal’s triangle row in question.)  Explicitly:

$1\;=\;T_{0}$
$x\;=\;T_{1}$
$x^{2}\;=\;2^{-1}(T_{2}+T_{0})$
$x^{3}\;=\;2^{-2}(T_{3}+3T_{1})$
$x^{4}\;=\;2^{-3}(T_{4}+4T_{2}+3T_{0})$
$x^{5}\;=\;2^{-4}(T_{5}+5T_{3}+10T_{1})$
$x^{6}\;=\;2^{-5}(T_{6}+6T_{4}+15T_{2}+10T_{0})$
$x^{7}\;=\;2^{-6}(T_{7}+7T_{5}+21T_{3}+35T_{1})$
$x^{8}\;=\;2^{-7}(T_{8}+8T_{6}+28T_{4}+56T_{2}+36T_{0})$
$x^{9}\;=\;2^{-8}(T_{9}+9T_{7}+36T_{5}+84T_{3}+126T_{1})$
$\mbox{ }\;\cdots\qquad\cdots$

## References

• 1 Pentti Laasonen: Matemaattisia erikoisfunktioita.  Handout No. 261. Teknillisen Korkeakoulun Ylioppilaskunta; Otaniemi, Finland (1969).
 Title orthogonality of Chebyshev polynomials Canonical name OrthogonalityOfChebyshevPolynomials Date of creation 2013-03-22 18:54:42 Last modified on 2013-03-22 18:54:42 Owner pahio (2872) Last modified by pahio (2872) Numerical id 16 Author pahio (2872) Entry type Derivation Classification msc 33C45 Classification msc 33D45 Classification msc 42C05 Related topic OrthogonalPolynomials Related topic LaguerrePolynomial Related topic ChangeOfVariableInDefiniteIntegral Related topic DeterminationOfFourierCoefficients Related topic OrthogonalityOfLegendrePolynomials Related topic PropertiesOfOrthogonalPolynomials