Special Functions – Orthogonal polynomials – Jacobi polynomials

The Jacobi polynomials are orthogonal on the interval \((-1,1)\) with respect to the beta distribution \(w(x)=(1-x)^{\alpha}(1+x)^{\beta}\). They can be defined by means of their Rodrigues formula:

\[P_n^{(\alpha,\beta)}(x)=\frac{(-1)^n}{2^n\,n!}\,\frac{1}{w(x)}\,D^n\left[w(x)\,(1-x^2)^n\right] =\frac{(-1)^n}{2^n\,n!}\,(1-x)^{-\alpha}(1+x)^{-\beta}\,D^n\left[(1-x)^{n+\alpha}(1+x)^{n+\beta}\right],\quad n=0,1,2,\ldots.\tag1\]

Using Leibniz' rule we have

\begin{align*} D^n\left[(1-x)^{n+\alpha}(1+x)^{n+\beta}\right]&=\sum_{k=0}^n\binom{n}{k}D^k(1-x)^{n+\alpha}D^{n-k}(1+x)^{n+\beta}\\[2.5mm] &=\sum_{k=0}^n\binom{n}{k}(-1)^k(n+\alpha)(n+\alpha-1)\cdots(n+\alpha-k+1)(1-x)^{n+\alpha-k}\\[2.5mm] &{}\hspace{50mm}{}\times(n+\beta)(n+\beta-1)\cdots(\beta+k+1)(1+x)^{\beta+k}\\[2.5mm] &=n!\,\sum_{k=0}^n(-1)^k\binom{n+\alpha}{k}\binom{n+\beta}{n-k}(1-x)^{n+\alpha-k}(1+x)^{\beta+k},\quad n=0,1,2,\ldots. \end{align*}

This implies that

\[P_n^{(\alpha,\beta)}(x)=\frac{(-1)^n}{2^n}\,\sum_{k=0}^n(-1)^k\binom{n+\alpha}{k}\binom{n+\beta}{n-k}(1-x)^{n-k}(1+x)^k,\quad n=0,1,2,\ldots.\]

This shows that \(P_n^{(\alpha,\beta)}(x)\) is a polynomial of degree \(n\). Note that we have the symmetry

\[P_n^{(\alpha,\beta)}(-x)=(-1)^nP_n^{(\beta,\alpha)}(x),\quad n=0,1,2,\ldots\tag2\]

and

\[P_n^{(\alpha,\beta)}(1)=\binom{n+\alpha}{n}\quad\text{and}\quad P_n^{(\alpha,\beta)}(-1)=(-1)^n\binom{n+\beta}{n},\quad n=0,1,2,\ldots.\]

In order to find a hypergeometric representation we write for \(x\neq 1\)

\[P_n^{(\alpha,\beta)}(x)=\left(\frac{x-1}{2}\right)^n\sum_{k=0}^n\binom{n+\alpha}{n} \binom{n+\beta}{n-k}\left(\frac{x+1}{x-1}\right)^k,\quad n=0,1,2,\ldots.\]

Now we have for \(x\neq 1\) \[\left(\frac{x+1}{x-1}\right)^k=\left(1+\frac{2}{x-1}\right)^k =\sum_{i=0}^k\binom{k}{i}\left(\frac{2}{x-1}\right)^i,\quad k=0,1,2,\ldots.\]

Now we obtain by changing the order of summations for \(x\neq 1\)

\begin{align*} P_n^{(\alpha,\beta)}(x)&=\left(\frac{x-1}{2}\right)^n\sum_{i=0}^n\sum_{k=i}^n \binom{n+\alpha}{k}\binom{n+\beta}{n-k}\binom{k}{i}\left(\frac{2}{x-1}\right)^i\\[2.5mm] &=\left(\frac{x-1}{2}\right)^n\sum_{i=0}^n\sum_{k=0}^{n-i} \binom{n+\alpha}{i+k}\binom{n+\beta}{n-i-k}\binom{i+k}{i}\left(\frac{2}{x-1}\right)^i,\quad n=0,1,2,\ldots. \end{align*}

Now we reverse the order in the first sum to find for \(x\neq 1\) and \(n=0,1,2,\ldots\)

\begin{align*} P_n^{(\alpha,\beta)}(x)&=\left(\frac{x-1}{2}\right)^n\sum_{i=0}^n\sum_{k=0}^n \binom{n+\alpha}{n-i+k}\binom{n+\beta}{i-k}\binom{n-i+k}{n-i}\left(\frac{2}{x-1}\right)^{n-i}\\ &=\sum_{i=0}^n\sum_{k=0}^n\binom{n+\alpha}{n-i+k}\binom{n+\beta}{i-k} \binom{n-i+k}{n-i}\left(\frac{x-1}{2}\right)^i\\ &=\sum_{i=0}^n\sum_{k=0}^n\frac{\Gamma(n+\alpha+1)}{(n-i+k)!\,\Gamma(i-k+\alpha+1)}\frac{\Gamma(n+\beta+1)}{(i-k)!\,\Gamma(n-i+k+\beta+1)}\, \frac{(n-i+k)!}{(n-i)!\,k!}\left(\frac{x-1}{2}\right)^i\\ &=\frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{n!}\sum_{i=0}^n \frac{(-n)_i}{\Gamma(i+\alpha+1)\,i!\,\Gamma(n-i+\beta+1)}\left(\frac{1-x}{2}\right)^i\sum_{k=0}^n\frac{(-i)_k(-i-\alpha-1)_k}{(n-i+\beta+1)_k\,k!}. \end{align*}

Since \(i\in\{0,1,2,\ldots,n\}\) we have by using the Chu-Vandermonde summation formula

\[\sum_{k=0}^n\frac{(-i)_k(-i-\alpha-1)_k}{(n-i+\beta+1)_k\,k!} ={}_2F_1\left(\genfrac{}{}{0pt}{}{-i,-i-\alpha-1}{n-i+\beta+1}\,;\,1\right)=\frac{(n+\alpha+\beta+1)_i}{(n-i+\beta+1)_i}.\]

Hence we have by using \(\Gamma(n-i+\beta+1)\,(n-i+\beta+1)_i=\Gamma(n+\beta+1)\)

\begin{align*} P_n^{(\alpha,\beta)}(x)&=\frac{\Gamma(n+\alpha+1)}{n!}\sum_{i=0}^n \frac{(-n)_i(n+\alpha+\beta+1)_i}{\Gamma(i+\alpha+1)\,i!}\left(\frac{1-x}{2}\right)^i\\ &=\frac{\Gamma(n+\alpha+1)}{n!\,\Gamma(\alpha+1)}\sum_{i=0}^n \frac{(-n)_i(n+\alpha+\beta+1)_i}{(\alpha+1)_i\,i!}\left(\frac{1-x}{2}\right)^i,\quad n=0,1,2,\ldots. \end{align*}

This proves the hypergeometric representation

\[P_n^{(\alpha,\beta)}(x)=\binom{n+\alpha}{n}\,{}_2F_1\left(\genfrac{}{}{0pt}{}{-n,n+\alpha+\beta+1}{\alpha+1}\,;\,\frac{1-x}{2}\right),\quad n=0,1,2,\ldots.\]

Note that this result also holds for \(x=1\). In view of the symmetry (2) we also have

\[P_n^{(\alpha,\beta)}(x)=(-1)^n\binom{n+\beta}{n}\,{}_2F_1\left(\genfrac{}{}{0pt}{}{-n,n+\alpha+\beta+1}{\beta+1}\,;\,\frac{1+x}{2}\right),\quad n=0,1,2,\ldots.\]

Note that the hypergeometric representation implies that

\begin{align*} \frac{d}{dx}P_n^{(\alpha,\beta)}(x)&=\binom{n+\alpha}{n}\,\frac{(-n)(n+\alpha+\beta+1)}{\alpha+1}\cdot\left(-\frac{1}{2}\right) {}_2F_1\left(\genfrac{}{}{0pt}{}{-n+1,n+\alpha+\beta+2}{\alpha+2}\,;\,\frac{1-x}{2}\right)\\ &=\frac{n+\alpha+\beta+1}{2}\,\binom{n+\alpha}{n-1}\,{}_2F_1\left(\genfrac{}{}{0pt}{}{-n+1,n+\alpha+\beta+2}{\alpha+2}\,;\,\frac{1-x}{2}\right)\\ &=\frac{n+\alpha+\beta+1}{2}\,P_{n-1}^{(\alpha+1,\beta+1)}(x),\quad n=1,2,3,\ldots. \end{align*}

Another consequence of the hypergeometric representation is that the leading coefficient of the polynomial \(P_n^{(\alpha,\beta)}(x)\) equals

\[k_n=\binom{n+\alpha}{n}\,\frac{(-n)_n(n+\alpha+\beta+1)_n}{(\alpha+1)_n\,n!}\,\frac{(-1)^n}{2^n} =\frac{(n+\alpha+\beta+1)_n}{2^n\,n!},\quad n=0,1,2,\ldots.\]

Now it can be shown that the Jacobi polynomials satisfy the orthogonality relation

\[\int_{-1}^1(1-x)^{\alpha}(1+x)^{\beta}P_m^{(\alpha,\beta)}(x)P_n^{(\alpha,\beta)}(x)\,dx =\frac{2^{\alpha+\beta+1}\,\Gamma(n+\alpha+1)\,\Gamma(n+\beta+1)} {(2n+\alpha+\beta+1)\,\Gamma(n+\alpha+\beta+1)\,n!}\,\delta_{mn}\]

for \(\alpha>-1\), \(\beta>-1\) and \(m,n\in\{0,1,2,\ldots\}\). This can be shown by using the definition (1) and integration by parts. The value of the integral in the case \(m=n\) can be computed by using the leading coefficient and then writing the integral in terms of a beta integral:

\begin{align*} \int_{-1}^1(1-x)^{\alpha}(1+x)^{\beta}\left\{P_n^{(\alpha,\beta)}(x)\right\}^2\,dx &=\frac{(-1)^n}{2^n\,n!}\int_{-1}^1P_n^{(\alpha,\beta)}(x)D^n\left[(1-x)^{n+\alpha}(1+x)^{n+\beta}\right]\,dx\\ &=\frac{1}{2^n\,n!}\int_{-1}^1D^nP_n^{(\alpha,\beta)}(x)(1-x)^{n+\alpha}(1+x)^{n+\beta}\,dx\\ &=\frac{(n+\alpha+\beta+1)_n}{2^{2n}\,n!}\int_{-1}^1(1-x)^{n+\alpha}(1+x)^{n+\beta}\,dx\\ &=\frac{\Gamma(2n+\alpha+\beta+1)}{\Gamma(n+\alpha+\beta+1)\,2^{2n}\,n!}\int_{-1}^1(1-x)^{n+\alpha}(1+x)^{n+\beta}\,dx,\quad n=0,1,2,\ldots \end{align*}

and using the substitution \(1-x=2t\)

\begin{align*} \int_{-1}^1(1-x)^{n+\alpha}(1+x)^{n+\beta}\,dx &=\int_0^1(2t)^{n+\alpha}(2-2t)^{n+\beta}\,2\,dt=2^{2n+\alpha+\beta+1}\int_0^1t^{n+\alpha}(1-t)^{n+\beta}\,dt\\[2.5mm] &=2^{2n+\alpha+\beta+1}\,B(n+\alpha+1,n+\beta+1) =2^{2n+\alpha+\beta+1}\,\frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{\Gamma(2n+\alpha+\beta+2)}\\ &=2^{2n+\alpha+\beta+1}\,\frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{(2n+\alpha+\beta+1)\,\Gamma(2n+\alpha+\beta+1)},\quad n=0,1,2,\ldots. \end{align*}

The Jacobi polynomials \(P_n^{(\alpha,\beta)}(x)\) satisfy the second-order linear differential equation

\[(1-x^2)y''(x)+\left[\beta-\alpha-(\alpha+\beta+2)x\right]y'(x)+n(n+\alpha+\beta+1)y(x)=0,\quad n=0,1,2,\ldots.\]

A generating function for the Jacobi polynomials is given by

\[\frac{2^{\alpha+\beta}}{R(1+R-t)^{\alpha}(1+R+t)^{\beta}} =\sum_{n=0}^{\infty}P_n^{(\alpha,\beta)}(x)t^n,\quad R=\sqrt{1-2xt+t^2}.\]

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Last modified on May 22, 2021