This page will document orthogonal polynomials that can be expressed in terms of the Gauss hypergeometric function. Some of this information is not readily accessible on the web, and this page helps to correct that deficiency.

Gauss Hypergeometric Function

Mathematical Tools

Legendre Polynomials

Associated Legendre Polynomials

Gegenbauer Polynomials

Associated Gegenbauer Polynomials

Jacobi Polynomials

The function is defined by

$\begin{array}{l}F(a,b;c;z)=\sum _{k=0}^{\infty}\frac{\Gamma (a+k)}{\Gamma \left(a\right)}\frac{\Gamma (b+k)}{\Gamma \left(b\right)}\frac{\Gamma \left(c\right)}{\Gamma (c+k)}\frac{{z}^{k}}{k!}\\ \phantom{F(a,b;c;z)}=1+\frac{ab}{c}\phantom{\rule{.2em}{0ex}}z+\frac{a(a+1)b(b+1)}{c(c+1)}\frac{{z}^{2}}{2!}+\cdots \end{array}$

The function satisfies the linear second-order differential equation

$z(1-z)\frac{{d}^{2}F}{d{z}^{2}}+[c-(a+b+1\left)z\right]\frac{dF}{dz}-abF=0$

which can be verified rather easily by substitution of the first form of the series. The change of variable

$z=\frac{1-x}{2}\phantom{\rule{4em}{0ex}}\frac{d}{dz}=-2\frac{d}{dx}\phantom{\rule{4em}{0ex}}\frac{{d}^{2}}{d{z}^{2}}=4\frac{{d}^{2}}{d{x}^{2}}$

leads to an alternate form of the equation useful for identifying orthogonal polynomials:

$(1-{x}^{2})\frac{{d}^{2}F}{d{x}^{2}}+[a+b+1-2c-(a+b+1\left)x\right]\frac{dF}{dx}-abF=0$

The derivative of the function with respect to its independent variable follows trivially from the second form of the series,

$\frac{d}{dz}F(a,b;c;z)=\frac{ab}{c}F(a+1,b+1;c+1;z)$

and generalizes as

$\frac{{d}^{n}}{d{z}^{n}}F(a,b;c;z)=\frac{\Gamma (a+n)\Gamma (b+n)\Gamma \left(c\right)}{\Gamma \left(a\right)\Gamma \left(b\right)\Gamma (c+n)}F(a+n,b+n;c+n;z)$

In the following sections the usual notation ${}_{2}F_{1}(a,b;c;z)$ will appear for consistency with other sources.

The binomial formula is

$(a+b{)}^{n}=\sum _{k=0}^{n}\left(\genfrac{}{}{0ex}{}{n}{k}\right){a}^{n-k}{b}^{k}=\sum _{k=0}^{n}\left(\genfrac{}{}{0ex}{}{n}{k}\right){a}^{k}{b}^{n-k}$

with binomial coefficients

$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{(n-k)!\phantom{\rule{.2em}{0ex}}k!}=\left(\genfrac{}{}{0ex}{}{n}{n-k}\right)$

Applying the binomial formula to each term of the product identity

$(1+x{)}^{n}(1+x{)}^{m}=(1+x{)}^{n+m}$

and equating powers of *x* on both sides of the equation, it is straightforward to see that

$\sum _{k=0}^{p}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left(\genfrac{}{}{0ex}{}{m}{p-k}\right)=\left(\genfrac{}{}{0ex}{}{n+m}{p}\right)$

where $p<n,m$ . This is known as the Vandermonde identity.

A generalization of the binomial formula for arbitrary exponents is

$\begin{array}{l}(1-x{)}^{r}=1-rx+\frac{r(r-1)}{2!}{x}^{2}-\frac{r(r-1)(r-2)}{3!}{x}^{3}+\cdots \\ \phantom{(1-x{)}^{r}}=\sum _{k=0}^{\infty}\frac{\Gamma (-r+k)}{\Gamma (-r)}\frac{{x}^{k}}{k!}\end{array}$

The series can alternately be written with a falling factorial, leading to the relationship

$\frac{\Gamma (-r+k)}{\Gamma (-r)}=(-1{)}^{k}\frac{\Gamma (r+1)}{\Gamma (r-k+1)}$

If the binomials in the product identity leading to the Vandermonde identity are allowed to have nonintegral exponents, then one can represent them as

$(1+x{)}^{n}=\sum _{k=0}^{\infty}\frac{\Gamma (n+1)}{\Gamma (n-k+1)}\frac{{x}^{k}}{k!}$

Even though both sides of the product identity are now infinite, there are still a finite number of terms in powers of *x* to equate on both sides. This means the Vandermonde identity continues to hold for arbitrary values in the numerators of the binomial coefficients.

The duplication formula for the gamma function is

$\Gamma \left(2x\right)={2}^{2x-1}\frac{\Gamma \left(x\right)\Gamma (x+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}$

Multiple derivatives of a single power with integral exponent can be shown by induction to equal

$\frac{{d}^{n}}{d{x}^{n}}{x}^{p}=\{\phantom{\rule{1em}{0ex}}\begin{array}{cc}\frac{p!}{(p-n)!}{x}^{p-n}& ,\phantom{\rule{1em}{0ex}}p\ge n\\ 0& ,\phantom{\rule{1em}{0ex}}p<n\end{array}$

which generalizes for arbitrary exponent to

$\frac{{d}^{n}}{d{x}^{n}}{x}^{p}=\{\phantom{\rule{1em}{0ex}}\begin{array}{cc}\frac{\Gamma (p+1)}{\Gamma (p-n+1)}{x}^{p-n}& ,\phantom{\rule{1em}{0ex}}p\ge n\\ 0& ,\phantom{\rule{1em}{0ex}}p<n\end{array}$

For conciseness of notation, one can work with factorials until the gamma functions become critical.

Consider consecutive derivatives of a product, arranged in a triangle:

$\begin{array}{c}AB\\ {A}^{\prime}B+A{B}^{\prime}\\ {A}^{\u2033}B+2{A}^{\prime}{B}^{\prime}+A{B}^{\u2033}\\ {A}^{\u2034}B+3{A}^{\u2033}{B}^{\prime}+3{A}^{\prime}{B}^{\u2033}+A{B}^{\u2034}\end{array}$

The coefficients here are clearly binomial coefficients, and the general rule is

$\frac{{d}^{n}}{d{x}^{n}}\left[AB\right]=\sum _{k=0}^{n}\left(\genfrac{}{}{0ex}{}{n}{k}\right){A}^{(n-k)}{B}^{\left(k\right)}=\sum _{k=0}^{n}\left(\genfrac{}{}{0ex}{}{n}{k}\right){A}^{\left(k\right)}{B}^{(n-k)}$

This result is called the general Leibniz rule or the *n*th product rule.

Under the trigonometric change of variable

$x=cos\theta \phantom{\rule{3em}{0ex}}\frac{d}{dx}=-\frac{1}{sin\theta}\frac{d}{d\theta}\phantom{\rule{3em}{0ex}}\frac{{d}^{2}}{d{x}^{2}}=\frac{1}{{sin}^{2}\theta}\frac{{d}^{2}}{d{\theta}^{2}}-\frac{cos\theta}{{sin}^{3}\theta}\frac{d}{d\theta}$

The general second-order differential equation

$a\left(x\right)\frac{{d}^{2}f}{d{x}^{2}}+b\left(x\right)\frac{df}{dx}+c\left(x\right)f=0$

has the angular form

$\frac{a(cos\theta )}{{sin}^{2}\theta}\frac{{d}^{2}f}{d{\theta}^{2}}-[\frac{a(cos\theta )cos\theta}{{sin}^{3}\theta}+\frac{b(cos\theta )}{sin\theta}]\frac{df}{d\theta}+c(cos\theta )f=0$

These polynomials are defined by the generating function

$\frac{1}{\sqrt{1-2xt+{t}^{2}}}=\sum _{n=0}^{\infty}{P}_{n}\left(x\right){t}^{n}$

The left-hand side can be interpreted as an inverse potential in three dimensions. Applying the generalized binomial formula to it gives

$\frac{1}{\sqrt{1-2xt+{t}^{2}}}=\sum _{n=0}^{\infty}\frac{\Gamma (\frac{1}{2}+n)}{\Gamma \left(\frac{1}{2}\right)}\frac{(2xt-{t}^{2}{)}^{n}}{n!}=\sum _{n=0}^{\infty}{c}_{n}{t}^{n}(2x-t{)}^{n}$

The coefficient here can be simplified using the duplication formula for the gamma function:

${c}_{n}=\frac{1}{n!}\frac{\Gamma (n+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}=\frac{1}{n!}\frac{\Gamma \left(2n\right)}{{2}^{2n-1}\Gamma \left(n\right)}=\frac{1}{n!}\frac{(2n-1)!}{{2}^{2n-1}(n-1)!}=\frac{\left(2n\right)!}{{2}^{2n}(n!{)}^{2}}$

The simplification can also be done in terms of double factorials, with the same result. It should be kept in mind that the coefficient need not be simplified in this way, especially in comparison with Gegenbauer polynomials below.

Returning to the generalized binomial expansion, a first instinct is to expand each term with the binomial formula and attempt to reorder the nested summation. This unfortunately is more confusing than productive. Instead look at a few terms in the form

${c}_{n}{t}^{n}(2x-t{)}^{n}+{c}_{n-1}{t}^{n-1}(2x-t{)}^{n-1}+{c}_{n-2}{t}^{n-2}(2x-t{)}^{n-2}+\cdots $

Starting from the leading term in ${t}^{n}$ one can pick out matching terms in the subsequent partial sum. Since the overall exponent decreases by one in each large term, this must be compensated by a product from inside the binomial that is one higher at each step. And since the exponent of the binomial also decreases by one at each step, the power of the coefficient of this compensating product necessarily must decrease by two at each step. One thus selects the terms

${c}_{n}(2x{)}^{n}-{c}_{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{n-2}\right)(2x{)}^{n-2}+{c}_{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{n-4}\right)(2x{)}^{n-4}-\cdots $

from which the pattern becomes clear:

${P}_{n}\left(x\right)=\sum _{k=0}^{int(n/2)}(-1{)}^{k}{c}_{n-k}\left(\genfrac{}{}{0ex}{}{n-k}{n-2k}\right)(2x{)}^{n-2k}$

The upper limit on the summation is set to keep the power of the independent variable positive, as the original sum has no negative powers. The change in the power of the argument by two means that the polynomials are either odd or even.

With the simplified coefficient the sum can be written

${P}_{n}\left(x\right)=\sum _{k=0}^{int(n/2)}(-1{)}^{k}\frac{(2n-2k)!}{{2}^{n}k!(n-k)!(n-2k)!}{x}^{n-2k}$

or by including a factorial of the index

${P}_{n}\left(x\right)=\frac{1}{{2}^{n}n!}\sum _{k=0}^{int(n/2)}(-1{)}^{k}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\frac{(2n-2k)!}{(n-2k)!}{x}^{n-2k}$

The significance of this last form is that, using the multiple derivative of a single power with integral exponent, one has

${P}_{n}\left(x\right)=\frac{1}{{2}^{n}n!}\frac{{d}^{n}}{d{x}^{n}}\sum _{k=0}^{int(n/2)}\left(\genfrac{}{}{0ex}{}{n}{k}\right)(-1{)}^{k}({x}^{2}{)}^{n-k}=\frac{1}{{2}^{n}n!}\frac{{d}^{n}}{d{x}^{n}}({x}^{2}-1{)}^{n}$

where the added terms in the binomial summation are zero under the derivative. This last result is the Rodrigues formula of the Legendre polynomials, an alternate definition and a means to determine the differential equation satisfied by them.

Begin with an identity involving the binomial in the Rodrigues formula:

$({x}^{2}-1)\frac{d}{dx}({x}^{2}-1{)}^{n}=2nx({x}^{2}-1{)}^{n}$

The leading factor on the left-hand side has nonzero derivatives to the second order, while that on the right-hand side has them to first order. Taking account of the existing derivative, if one differentiates both sides $n+1$ times with respect to the independent variable,

$\frac{{d}^{n+1}}{d{x}^{n+1}}[({x}^{2}-1)\frac{d}{dx}({x}^{2}-1{)}^{n}=2nx({x}^{2}-1{)}^{n}]$

then the resulting derivatives of the binomial will all be greater than or equal to the polynomial index. With the abbreviation ${P}^{\left(k\right)}=\frac{{d}^{k}}{d{x}^{k}}({x}^{2}-1{)}^{n}$ one has

$\begin{array}{l}({x}^{2}-1){P}^{(n+2)}+2(n+1)x{P}^{(n+1)}+n(n+1){P}^{\left(n\right)}\\ \phantom{\rule{4em}{0ex}}=2nx{P}^{(n+1)}+2n(n+1){P}^{\left(n\right)}\end{array}$

Moving all terms to the right-hand side and multiplying by the normalization factor leads to the second-order differential equation

$(1-{x}^{2})\frac{{d}^{2}{P}_{n}}{d{x}^{2}}-2x\frac{d{P}_{n}}{dx}+n(n+1){P}_{n}=0$

Under the trigonometric change of variable above, the differential equation becomes

$\frac{{d}^{2}{P}_{n}}{d{\theta}^{2}}+\frac{cos\theta}{sin\theta}\frac{d{P}_{n}}{d\theta}+n(n+1){P}_{n}=0$

Comparing the unchanged equation with the alternate form of the Gauss hypergeometric differential equation above gives the unambiguous representation

${P}_{n}\left(x\right)={}_{2}F_{1}(-n\phantom{\rule{.2em}{0ex}},n+1\phantom{\rule{.2em}{0ex}};1\phantom{\rule{.2em}{0ex}};\frac{1-x}{2}\phantom{\rule{.2em}{0ex}})$

due to the symmetry of the Gauss hypergeometric function under interchange of its first two parameters.

To verify that the normalization is correct as given, write the general term of the hypergeometric series using the relation among gamma function products above arising from rewriting the generalized binomial formula:

$\begin{array}{l}\frac{\Gamma (-n+k)}{\Gamma (-n)}\frac{\Gamma (n+k+1)}{\Gamma (n+1)}\frac{\Gamma \left(1\right)}{\Gamma (k+1)}\frac{1}{k!}(\frac{1-x}{2}{)}^{k}\\ \phantom{\rule{4em}{0ex}}=(-1{)}^{k}\frac{\Gamma (n+1)}{\Gamma (n-k+1)}\frac{\Gamma (n+k+1)}{\Gamma (n+1)}\frac{1}{(k!{)}^{2}}(\frac{1-x}{2}{)}^{k}\\ \phantom{\rule{4em}{0ex}}=(-1{)}^{k}\frac{\Gamma (n+k+1)}{\Gamma (n-k+1)}\frac{1}{(k!{)}^{2}}(\frac{1-x}{2}{)}^{k}\end{array}$

The series terminates at $k=n$ , for which value the general term here clearly matches the leading term of the series above.

These polynomials are defined by mathematicians as

${P}_{n}^{m}\left(x\right)=(1-{x}^{2}{)}^{m/2}\frac{{d}^{m}}{d{x}^{m}}{P}_{n}\left(x\right)$

Physicists often include an additional factor of $(-1{)}^{m}$ for consistency with angular momentum ladder operators.

For the differential equation satisfied by these polynomials, take *m* derivatives of the equation for the regular polynomials in terms of the abbreviation introduced in the previous section:

$\begin{array}{l}\frac{{d}^{m}}{d{x}^{m}}[(1-{x}^{2}){P}^{(n+2)}-2x{P}^{(n+1)}+n(n+1){P}^{\left(n\right)}=0]\\ (1-{x}^{2}){P}^{(n+m+2)}-2(m+1)x{P}^{(n+m+1)}\\ \phantom{\rule{7em}{0ex}}+\left[n\right(n+1)-m(m+1\left)\right]{P}^{(n+m)}=0\end{array}$

The multiple derivatives of the binomial from the Rodrigues formula can be replaced with

${P}^{(n+m)}=(1-{x}^{2}{)}^{-m/2}{P}_{n}^{m}$

After clearing a common factor, the first and second derivatives of the combination on the right-hand side are

$\begin{array}{l}(1-{x}^{2}{)}^{m/2}\frac{d}{dx}\left[(1-{x}^{2}{)}^{-m/2}{P}_{n}^{m}\right]=\frac{d{P}_{n}^{m}}{dx}+\frac{mx}{1-{x}^{2}}{P}_{n}^{m}\\ \\ (1-{x}^{2}{)}^{m/2}\frac{{d}^{2}}{d{x}^{2}}\left[(1-{x}^{2}{)}^{-m/2}{P}_{n}^{m}\right]\\ \phantom{\rule{4em}{0ex}}=\frac{{d}^{2}{P}_{n}^{m}}{d{x}^{2}}+\frac{2mx}{1-{x}^{2}}\frac{d{P}_{n}^{m}}{dx}+[\frac{m}{1-{x}^{2}}+\frac{m(m+2){x}^{2}}{(1-{x}^{2}{)}^{2}}]{P}_{n}^{m}\end{array}$

Combining all terms leads to the second-order differential equation

$(1-{x}^{2})\frac{{d}^{2}{P}_{n}^{m}}{d{x}^{2}}-2x\frac{d{P}_{n}^{m}}{dx}+[n(n+1)-\frac{{m}^{2}}{1-{x}^{2}}]{P}_{n}^{m}=0$

This equation differs from that of the regular polynomials only in the coefficient of the nondifferentiated function. Under the trigonometric change of variable above it becomes

$\frac{{d}^{2}{P}_{n}^{m}}{d{\theta}^{2}}+\frac{cos\theta}{sin\theta}\frac{d{P}_{n}^{m}}{d\theta}+[n(n+1)-\frac{{m}^{2}}{{sin}^{2}\theta}]{P}_{n}^{m}=0$

For a representation of these polynomials in terms of the Gauss hypergeometric function, carry out the multiple derivatives on the representation of the regular polynomials,

$\begin{array}{l}{P}_{n}^{m}\left(x\right)=\frac{(-1{)}^{m}}{{2}^{m}}\frac{\Gamma (-n+m)(n+m)!}{\Gamma (-n)\phantom{\rule{.2em}{0ex}}n!\phantom{\rule{.2em}{0ex}}m!}(1-{x}^{2}{)}^{m/2}\\ \phantom{\rule{5em}{0ex}}\times {}_{2}F_{1}(-n+m\phantom{\rule{.2em}{0ex}},n+m+1\phantom{\rule{.2em}{0ex}};m+1\phantom{\rule{.2em}{0ex}};\frac{1-x}{2}\phantom{\rule{.2em}{0ex}})\end{array}$

where the leading factor comes from differentiating the independent variable and gamma functions have been converted to factorials where appropriate. Using the relation among gamma function products above arising from rewriting the generalized binomial formula, this representation becomes

$\begin{array}{l}{P}_{n}^{m}\left(x\right)=\frac{1}{{2}^{m}}\frac{(n+m)!}{(n-m)!\phantom{\rule{.2em}{0ex}}m!}(1-{x}^{2}{)}^{m/2}\\ \phantom{\rule{5em}{0ex}}\times {}_{2}F_{1}(-n+m\phantom{\rule{.2em}{0ex}},n+m+1\phantom{\rule{.2em}{0ex}};m+1\phantom{\rule{.2em}{0ex}};\frac{1-x}{2}\phantom{\rule{.2em}{0ex}})\end{array}$

These polynomials are defined by the generating function

$\frac{1}{(1-2xt+{t}^{2}{)}^{\alpha}}=\sum _{n=0}^{\infty}{C}_{n}^{\left(\alpha \right)}\left(x\right){t}^{n}$

The left-hand side can be interpreted as an inverse potential in $2\alpha +2$ dimensions, generalizing Legendre polynomials to an arbitrary number of dimensions. Applying the generalized binomial formula as before gives

$\frac{1}{(1-2xt+{t}^{2}{)}^{\alpha}}=\sum _{n=0}^{\infty}\frac{\Gamma (\alpha +n)}{\Gamma \left(\alpha \right)}\frac{(2xt-{t}^{2}{)}^{n}}{n!}=\sum _{n=0}^{\infty}{c}_{n}{t}^{n}(2x-t{)}^{n}$

The process of selecting matching terms in the binomial expansions proceeds precisely as for Legendre polynomials. With a result above one can immediately write

$\begin{array}{l}{C}_{n}^{\left(\alpha \right)}\left(x\right)=\sum _{k=0}^{int(n/2)}(-1{)}^{k}{c}_{n-k}\left(\genfrac{}{}{0ex}{}{n-k}{n-2k}\right)(2x{)}^{n-2k}\\ \phantom{{C}_{n}^{\left(\alpha \right)}\left(x\right)}=\sum _{k=0}^{int(n/2)}(-1{)}^{k}\frac{\Gamma (\alpha +n-k)}{\Gamma \left(\alpha \right)}\frac{1}{(n-2k)!\phantom{\rule{.2em}{0ex}}k!}(2x{)}^{n-2k}\end{array}$

The coefficient in this series cannot be simplified immediately as is the case for Legendre polynomials.

In the context of determining a Rodrigues formula for Gegenbauer polynomials, slightly rewrite the multiple derivative of the Rodrigues formula for the Legendre polynomials and apply the *n*th product rule:

$\begin{array}{l}\frac{{d}^{n}}{d{x}^{n}}({x}^{2}-1{)}^{n}=\frac{{d}^{n}}{d{x}^{n}}(x+1{)}^{n}(x-1{)}^{n}\\ \phantom{\frac{{d}^{n}}{d{x}^{n}}({x}^{2}-1{)}^{n}}=\sum _{l=0}^{n}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\frac{n!}{(n-l)!}(x+1{)}^{n-l}\frac{n!}{l!}(x-1{)}^{l}\\ \frac{{d}^{n}}{d{x}^{n}}({x}^{2}-1{)}^{n}=n!\sum _{l=0}^{n}(\genfrac{}{}{0ex}{}{n}{l}{)}^{2}(x+1{)}^{n-l}(x-1{)}^{l}\end{array}$

Due to interchange symmetry of the indices $n-l$ and $l$ in the binomial coefficients, it is straightforward to see that the final sum has even or odd parity,

$\begin{array}{l}\sum _{l=0}^{n}(\genfrac{}{}{0ex}{}{n}{l}{)}^{2}(-x+1{)}^{n-l}(-x-1{)}^{l}\\ \phantom{\rule{4em}{0ex}}=(-1{)}^{n}\sum _{l=0}^{n}(\genfrac{}{}{0ex}{}{n}{l}{)}^{2}(x-1{)}^{n-l}(x+1{)}^{l}\\ \phantom{\rule{4em}{0ex}}=(-1{)}^{n}\sum _{l=0}^{n}(\genfrac{}{}{0ex}{}{n}{l}{)}^{2}(x+1{)}^{n-l}(x-1{)}^{l}\end{array}$

and this corresponds to the fact that exponents in the series for Legendre polynomials move in steps of two. This sum over products of binomials does not appear to be easily expressible in single powers of the independent variable, but at least one can use the Vandermonde identity to determine the coefficient of the leading power,

$n!\sum _{l=0}^{n}(\genfrac{}{}{0ex}{}{n}{l}{)}^{2}=n!\sum _{l=0}^{n}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{n}{n-l}\right)=n!\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=\frac{\left(2n\right)!}{n!}$

which is indeed the coefficient of the leading term in the series above for Legendre polynomials, apart from overall factors outside the summation.

For Gegenbauer polynomials, modify the multiple derivative for Legendre polynomials by including an arbitrary power. For conciseness of notation, work initially in factorials:

$\begin{array}{l}({x}^{2}-1{)}^{-p}\frac{{d}^{n}}{d{x}^{n}}({x}^{2}-1{)}^{n+p}=({x}^{2}-1{)}^{-p}\frac{{d}^{n}}{d{x}^{n}}(x+1{)}^{n+p}(x-1{)}^{n+p}\\ \phantom{\rule{4em}{0ex}}=\sum _{l=0}^{n}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\frac{(n+p)!}{(n+p-l)!}(x+1{)}^{n-l}\frac{(n+p)!}{(p+l)!}(x-1{)}^{l}\\ \phantom{\rule{4em}{0ex}}=n!\sum _{l=0}^{n}\left(\genfrac{}{}{0ex}{}{n+p}{l}\right)\left(\genfrac{}{}{0ex}{}{n+p}{n-l}\right)(x+1{)}^{n-l}(x-1{)}^{l}\end{array}$

Due to the patent interchange symmetry in $n-l$ and $l$ , this sum has even or odd parity just as for the Legendre polynomials. This corresponds to the fact that exponents in the series for Gegenbauer polynomials again move in steps of two.

The sum again does not appear to be easily expressible in single powers of the independent variable, but one can use the Vandermonde identity to determine the coefficient of the leading power:

$n!\sum _{l=0}^{n}\left(\genfrac{}{}{0ex}{}{n+p}{l}\right)\left(\genfrac{}{}{0ex}{}{n+p}{n-l}\right)=n!\left(\genfrac{}{}{0ex}{}{2n+2p}{n}\right)=\frac{(2n+2p)!}{(n+2p)!}=\frac{\Gamma (2n+2p+1)}{\Gamma (n+2p+1)}$

This quantity, multiplied by some normalizing constant and for a suitable choice of the introduced power, needs to equal the coefficient of the leading term in the series for the polynomials:

$c(n,p)\frac{\Gamma (2n+2p+1)}{\Gamma (n+2p+1)}=\frac{{2}^{n}}{n!}\frac{\Gamma (\alpha +n)}{\Gamma \left(\alpha \right)}$

One way to motivate the choice of power is to notice that the coefficients of *n* and *p* in the denominator are different, making it the odd duck out. This can be remedied with a normalizing constant dependent on these indices. Let the left-hand side include two rising factorials,

$c\left(n\right)\frac{\Gamma (p+1)}{\Gamma (n+p+1)}\frac{\Gamma (n+2p+1)}{\Gamma (2p+1)}\frac{\Gamma (2n+2p+1)}{\Gamma (n+2p+1)}=c\left(n\right)\frac{\Gamma \left(p\right)}{\Gamma \left(2p\right)}\frac{\Gamma (2n+2p)}{\Gamma (n+p)}$

and notice the ratios of single and doubled arguments. With the duplication formula this becomes

$c\left(n\right)\frac{\Gamma \left(p\right)}{\Gamma \left(2p\right)}\frac{\Gamma (2n+2p)}{\Gamma (n+p)}={2}^{2n}c\left(n\right)\frac{\Gamma (n+p+\frac{1}{2})}{\Gamma (p+\frac{1}{2})}$

It is now clear that with the choices

$p=\alpha -\frac{1}{2}\phantom{\rule{5em}{0ex}}c\left(n\right)=\frac{1}{{2}^{n}n!}$

this left-hand side will be equivalent to the coefficient of the leading term of the series as required. With all the pieces in place, the Rodrigues formula for Gegenbauer polynomials is

${C}_{n}^{\left(\alpha \right)}\left(x\right)=\frac{1}{{2}^{n}n!}\frac{\Gamma (\alpha +\frac{1}{2})}{\Gamma (n+\alpha +\frac{1}{2})}\frac{\Gamma (n+2\alpha )}{\Gamma \left(2\alpha \right)}({x}^{2}-1{)}^{-\alpha +1/2}\frac{{d}^{n}}{d{x}^{n}}({x}^{2}-1{)}^{n+\alpha -1/2}$

For $\alpha =\frac{1}{2}$ this reduces to the Rodrigues formula for Legendre polynomials. This is expected from a comparison of the generating functions for the two types of polynomials.

For the differential equation satisfied by Gegenbauer polynomials, begin with an identity involving the binomial in their Rodrigues formula,

$({x}^{2}-1)\frac{d}{dx}({x}^{2}-1{)}^{n+p}=2(n+p)x({x}^{2}-1{)}^{n+p}$

and again differentiate $n+1$ times with respect to the independent variable, for the same reason as for Legendre polynomials. With the abbreviation ${C}^{\left(k\right)}=\frac{{d}^{k}}{d{x}^{k}}({x}^{2}-1{)}^{n+p}$ one has

$\begin{array}{l}({x}^{2}-1){C}^{(n+2)}+2(n+1)x{C}^{(n+1)}+n(n+1){C}^{\left(n\right)}\\ \phantom{\rule{4em}{0ex}}=2(n+p)x{C}^{(n+1)}+2(n+p)(n+1){C}^{\left(n\right)}\end{array}$

Moving all terms to the right-hand side gives the second-order differential equation

$(1-{x}^{2})\frac{{d}^{2}{C}^{\left(n\right)}}{d{x}^{2}}+2(p-1)x\frac{d{C}^{\left(n\right)}}{dx}+(n+1)(n+2p){C}^{\left(n\right)}=0$

The multiple derivatives of the binomial from the Rodrigues formula can be replaced with

${C}^{\left(n\right)}=({x}^{2}-1{)}^{p}{C}_{n}^{\left(\alpha \right)}$

where the normalization is omitted for simplicity. After clearing a common factor, the first and second derivatives of the combination on the right-hand side are

$\begin{array}{l}({x}^{2}-1{)}^{-p}\frac{d}{dx}\left[({x}^{2}-1{)}^{p}{C}_{n}^{\left(\alpha \right)}\right]=\frac{d{C}_{n}^{\left(\alpha \right)}}{dx}+\frac{2px}{{x}^{2}-1}{C}_{n}^{\left(\alpha \right)}\\ \\ ({x}^{2}-1{)}^{-p}\frac{{d}^{2}}{d{x}^{2}}\left[({x}^{2}-1{)}^{p}{C}_{n}^{\left(\alpha \right)}\right]\\ \phantom{\rule{4em}{0ex}}=\frac{{d}^{2}{C}_{n}^{\left(\alpha \right)}}{d{x}^{2}}+\frac{4px}{{x}^{2}-1}\frac{d{C}_{n}^{\left(\alpha \right)}}{dx}+[\frac{2p}{{x}^{2}-1}+\frac{4p(p-1){x}^{2}}{({x}^{2}-1{)}^{2}}]{C}_{n}^{\left(\alpha \right)}\end{array}$

Combining all terms leads to the second-order differential equation

$(1-{x}^{2})\frac{{d}^{2}{C}_{n}^{\left(\alpha \right)}}{d{x}^{2}}-2(p+1)x\frac{d{C}_{n}^{\left(\alpha \right)}}{dx}+n(n+2p+1){C}_{n}^{\left(\alpha \right)}=0$

which after substitution for the arbitrary power becomes

$(1-{x}^{2})\frac{{d}^{2}{C}_{n}^{\left(\alpha \right)}}{d{x}^{2}}-(2\alpha +1)x\frac{d{C}_{n}^{\left(\alpha \right)}}{dx}+n(n+2\alpha ){C}_{n}^{\left(\alpha \right)}=0$

Under the trigonometric change of variable above, the differential equation becomes

$\frac{{d}^{2}{C}_{n}^{\left(\alpha \right)}}{d{\theta}^{2}}+2\alpha \frac{cos\theta}{sin\theta}\frac{d{C}_{n}^{\left(\alpha \right)}}{d\theta}+n(n+2\alpha ){C}_{n}^{\left(\alpha \right)}=0$

For $\alpha =\frac{1}{2}$ both equations reduce to those for Legendre polynomials as expected.

Comparing the unchanged equation with the alternate form of the Gauss hypergeometric differential equation above gives the preliminary representation

${C}_{n}^{\left(\alpha \right)}\left(x\right)\approx {}_{2}F_{1}(-n\phantom{\rule{.2em}{0ex}},n+2\alpha \phantom{\rule{.2em}{0ex}};\alpha +\frac{1}{2}\phantom{\rule{.2em}{0ex}};\frac{1-x}{2}\phantom{\rule{.2em}{0ex}})$

apart from a normalization factor. To determine this write the general term of the hypergeometric series using the relation among gamma function products above arising from rewriting the generalized binomial formula:

$\begin{array}{l}\frac{\Gamma (-n+k)}{\Gamma (-n)}\frac{\Gamma (n+2\alpha +k)}{\Gamma (n+2\alpha )}\frac{\Gamma (\alpha +\frac{1}{2})}{\Gamma (\alpha +\frac{1}{2}+k)}\frac{1}{k!}(\frac{1-x}{2}{)}^{k}\\ \phantom{\rule{3em}{0ex}}=(-1{)}^{k}\frac{\Gamma (n+1)}{\Gamma (n-k+1)}\frac{\Gamma (n+2\alpha +k)}{\Gamma (n+2\alpha )}\frac{\Gamma (\alpha +\frac{1}{2})}{\Gamma (\alpha +\frac{1}{2}+k)}\frac{1}{k!}(\frac{1-x}{2}{)}^{k}\end{array}$

The series terminates at $k=n$ , for which value the general term here must be proportional to the leading term of the series above:

$c(n,\alpha )\frac{1}{{2}^{n}}\frac{\Gamma (2n+2\alpha )}{\Gamma (n+2\alpha )}\frac{\Gamma (\alpha +\frac{1}{2})}{\Gamma (\alpha +\frac{1}{2}+n)}=\frac{{2}^{n}}{n!}\frac{\Gamma (\alpha +n)}{\Gamma \left(\alpha \right)}$

This easily solved using the duplication formula, giving the simple result

$c(n,\alpha )=\frac{1}{n!}\frac{\Gamma (n+2\alpha )}{\Gamma \left(2\alpha \right)}$

and the final normalized representation

${C}_{n}^{\left(\alpha \right)}\left(x\right)=\frac{1}{n!}\frac{\Gamma (n+2\alpha )}{\Gamma \left(2\alpha \right)}{}_{2}F_{1}(-n\phantom{\rule{.2em}{0ex}},n+2\alpha \phantom{\rule{.2em}{0ex}};\alpha +\frac{1}{2}\phantom{\rule{.2em}{0ex}};\frac{1-x}{2}\phantom{\rule{.2em}{0ex}})$

which as expected reduces to that of Legendre polynomials for $\alpha =\frac{1}{2}$ .

These polynomials are defined relative to the Gegenbauer polynomials just as in the case of associated Legendre polynomials:

${C}_{n,m}^{\left(\alpha \right)}\left(x\right)=(1-{x}^{2}{)}^{m/2}\frac{{d}^{m}}{d{x}^{m}}{C}_{n}^{\left(\alpha \right)}\left(x\right)$

There does not appear to be consistent notation for these polynomials. The third index is given here in the lower position for compactness.

For the differential equation satisfied by these polynomials, take *m* derivatives of the equation for the regular polynomials in terms of the abbreviation introduced in the previous section:

$\begin{array}{l}\frac{{d}^{m}}{d{x}^{m}}[(1-{x}^{2}){C}^{(n+2)}-(2\alpha +1)x{C}^{(n+1)}+n(n+2\alpha ){C}^{\left(n\right)}=0]\\ (1-{x}^{2}){C}^{(n+m+2)}-(2m+2\alpha +1)x{C}^{(n+m+1)}\\ \phantom{\rule{7em}{0ex}}+\left[n\right(n+2\alpha )-m(m+2\alpha \left)\right]{C}^{(n+m)}=0\end{array}$

The multiple derivatives of the binomial from the Rodrigues formula can be replaced with

${C}^{(n+m)}=(1-{x}^{2}{)}^{-m/2}{C}_{n,m}^{\left(\alpha \right)}$

After clearing a common factor, the first and second derivatives of the combination on the right-hand side are

$\begin{array}{l}(1-{x}^{2}{)}^{m/2}\frac{d}{dx}\left[(1-{x}^{2}{)}^{-m/2}{C}_{n,m}^{\left(\alpha \right)}\right]=\frac{d{C}_{n,m}^{\left(\alpha \right)}}{dx}+\frac{mx}{1-{x}^{2}}{C}_{n,m}^{\left(\alpha \right)}\\ \\ (1-{x}^{2}{)}^{m/2}\frac{{d}^{2}}{d{x}^{2}}\left[(1-{x}^{2}{)}^{-m/2}{C}_{n,m}^{\left(\alpha \right)}\right]\\ \phantom{\rule{4em}{0ex}}=\frac{{d}^{2}{C}_{n,m}^{\left(\alpha \right)}}{d{x}^{2}}+\frac{2mx}{1-{x}^{2}}\frac{d{C}_{n,m}^{\left(\alpha \right)}}{dx}+[\frac{m}{1-{x}^{2}}+\frac{m(m+2){x}^{2}}{(1-{x}^{2}{)}^{2}}]{C}_{n,m}^{\left(\alpha \right)}\end{array}$

Combining all terms leads to the second-order differential equation

$(1-{x}^{2})\frac{{d}^{2}{C}_{n,m}^{\left(\alpha \right)}}{d{x}^{2}}-(2\alpha +1)x\frac{d{C}_{n,m}^{\left(\alpha \right)}}{dx}+[n(n+2\alpha )-\frac{m(m+2\alpha -1)}{1-{x}^{2}}]{C}_{n,m}^{\left(\alpha \right)}=0$

This equation again differs from that of the regular polynomials only in the coefficient of the nondifferentiated function. Under the trigonometric change of variable above it becomes

$\frac{{d}^{2}{C}_{n,m}^{\left(\alpha \right)}}{d{\theta}^{2}}+2\alpha \frac{cos\theta}{sin\theta}\frac{d{C}_{n,m}^{\left(\alpha \right)}}{d\theta}+[n(n+2\alpha )-\frac{m(m+2\alpha -1)}{{sin}^{2}\theta}]{C}_{n,m}^{\left(\alpha \right)}=0$

For a representation of these polynomials in terms of the Gauss hypergeometric function, carry out the multiple derivatives on the representation of the regular polynomials,

$\begin{array}{l}{C}_{n,m}^{\left(\alpha \right)}\left(x\right)=\frac{(-1{)}^{m}}{{2}^{m}}\frac{\Gamma (-n+m)}{\Gamma (-n)}\frac{\Gamma (n+2\alpha +m)}{\Gamma (n+2\alpha )}\frac{\Gamma (\alpha +\frac{1}{2})}{\Gamma (\alpha +\frac{1}{2}+m)}(1-{x}^{2}{)}^{m/2}\\ \phantom{\rule{5em}{0ex}}\times \frac{1}{n!}\frac{\Gamma (n+2\alpha )}{\Gamma \left(2\alpha \right)}{}_{2}F_{1}(-n+m\phantom{\rule{.2em}{0ex}},n+2\alpha +m\phantom{\rule{.2em}{0ex}};\alpha +\frac{1}{2}+m\phantom{\rule{.2em}{0ex}};\frac{1-x}{2}\phantom{\rule{.2em}{0ex}})\end{array}$

where the leading factor comes from differentiating the independent variable. Using the relation among gamma function products above arising from rewriting the generalized binomial formula, this representation becomes

$\begin{array}{l}{C}_{n,m}^{\left(\alpha \right)}\left(x\right)=\frac{1}{{2}^{m}(n-m)!}\frac{\Gamma (n+2\alpha +m)}{\Gamma \left(2\alpha \right)}\frac{\Gamma (\alpha +\frac{1}{2})}{\Gamma (\alpha +\frac{1}{2}+m)}(1-{x}^{2}{)}^{m/2}\\ \phantom{\rule{5em}{0ex}}\times {}_{2}F_{1}(-n+m\phantom{\rule{.2em}{0ex}},n+2\alpha +m\phantom{\rule{.2em}{0ex}};\alpha +\frac{1}{2}+m\phantom{\rule{.2em}{0ex}};\frac{1-x}{2}\phantom{\rule{.2em}{0ex}})\end{array}$

which as expected reduces to that of associated Legendre polynomials for $\alpha =\frac{1}{2}$ .

These polynomials can be defined by the Rodrigues formula

${P}_{n}^{(\alpha ,\beta )}\left(x\right)=\frac{1}{{2}^{n}n!}(x-1{)}^{-\alpha}(x+1{)}^{-\beta}\frac{{d}^{n}}{d{x}^{n}}(x-1{)}^{n+\alpha}(x+1{)}^{n+\beta}$

For equal parameters these polynomials are proportional to Gegenbauer polynomials, as can be readily seen from the Rodrigues formula above for the latter. The formula here separates the two factors of the binomials occurring till now.

Working in factorials for conciseness of notation, the explicit derivatives are

${P}_{n}^{(\alpha ,\beta )}\left(x\right)=\frac{1}{{2}^{n}n!}\sum _{k=0}^{n}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\frac{(n+\alpha )!}{(n+\alpha -k)!}\frac{(n+\beta )!}{(\beta +k)!}(x-1{)}^{n-k}(x+1{)}^{k}$

which due to the interchange symmetry of the indices $n-k$ and $k$ can be written in the two equivalent forms

$\begin{array}{l}{P}_{n}^{(\alpha ,\beta )}\left(x\right)=\frac{1}{{2}^{n}}\sum _{k=0}^{n}\left(\genfrac{}{}{0ex}{}{n+\alpha}{k}\right)\left(\genfrac{}{}{0ex}{}{n+\beta}{n-k}\right)(x-1{)}^{n-k}(x+1{)}^{k}\\ \phantom{{P}_{n}^{(\alpha ,\beta )}\left(x\right)}=\frac{1}{{2}^{n}}\sum _{k=0}^{n}\left(\genfrac{}{}{0ex}{}{n+\alpha}{n-k}\right)\left(\genfrac{}{}{0ex}{}{n+\beta}{k}\right)(x-1{)}^{k}(x+1{)}^{n-k}\end{array}$

where it is understood that factorials will be replaced by gamma functions for nonintegral arguments.

For the differential equation satisfied by these polynomials, begin with an n identity involving the binomials in their Rodrigues formula,

$\begin{array}{l}({x}^{2}-1)\frac{d}{dx}(x-1{)}^{n+\alpha}(x+1{)}^{n+\beta}\\ \phantom{\rule{4em}{0ex}}=(n+\alpha )(x+1)(x-1{)}^{n+\alpha}(x+1{)}^{n+\beta}\\ \phantom{\rule{8em}{0ex}}+(n+\beta )(x-1)(x-1{)}^{n+\alpha}(x+1{)}^{n+\beta}\end{array}$

and again differentiate $n+1$ times with respect to the independent variable, for the same reason as for Legendre polynomials. With the abbreviation ${P}^{\left(k\right)}=\frac{{d}^{k}}{d{x}^{k}}(x-1{)}^{n+\alpha}(x+1{)}^{n+\beta}$ one has

$\begin{array}{l}({x}^{2}-1){P}^{(n+2)}+2(n+1)x{P}^{(n+1)}+n(n+1){P}^{\left(n\right)}\\ \phantom{\rule{3em}{0ex}}=(n+\alpha )(x+1){P}^{(n+1)}+(n+\alpha )(n+1){P}^{\left(n\right)}\\ \phantom{\rule{6em}{0ex}}+(n+\beta )(x-1){P}^{(n+1)}+(n+\beta )(n+1){P}^{\left(n\right)}\end{array}$

Moving all terms to the right-hand side gives the second-order differential equation

$\begin{array}{l}(1-{x}^{2})\frac{{d}^{2}{P}^{\left(n\right)}}{d{x}^{2}}+[\alpha -\beta +(\alpha +\beta -2\left)x\right]\frac{d{P}^{\left(n\right)}}{dx}\\ \hfill +(n+1)(n+\alpha +\beta ){P}^{\left(n\right)}=0\end{array}$

The multiple derivatives of the binomials from the Rodrigues formula can be replaced with

${P}^{\left(n\right)}=(x-1{)}^{\alpha}(x+1{)}^{\beta}{P}_{n}^{(\alpha ,\beta )}$

where the normalization is omitted for simplicity. After clearing common factors, the first and second derivatives of the combination on the right-hand side are

$\begin{array}{l}(x-1{)}^{-\alpha}(x+1{)}^{-\beta}\frac{d}{dx}\left[(x-1{)}^{\alpha}(x+1{)}^{\beta}{P}_{n}^{(\alpha ,\beta )}\right]\\ \phantom{\rule{4em}{0ex}}=\frac{d{P}_{n}^{(\alpha ,\beta )}}{dx}+[\frac{\alpha}{x-1}+\frac{\beta}{x+1}]{P}_{n}^{(\alpha ,\beta )}\\ \\ (x-1{)}^{-\alpha}(x+1{)}^{-\beta}\frac{{d}^{2}}{d{x}^{2}}\left[(x-1{)}^{\alpha}(x+1{)}^{\beta}{P}_{n}^{(\alpha ,\beta )}\right]\\ \phantom{\rule{4em}{0ex}}=\frac{{d}^{2}{P}_{n}^{(\alpha ,\beta )}}{d{x}^{2}}+2[\frac{\alpha}{x-1}+\frac{\beta}{x+1}]\frac{d{P}_{n}^{(\alpha ,\beta )}}{dx}\\ \phantom{\rule{8em}{0ex}}+[\frac{\alpha (\alpha -1)}{(x-1{)}^{2}}+\frac{2\alpha \beta}{{x}^{2}-1}+\frac{\beta (\beta -1)}{(x+1{)}^{2}}]{P}_{n}^{(\alpha ,\beta )}\end{array}$

Combining all terms leads to the second-order differential equation

$\begin{array}{l}(1-{x}^{2})\frac{{d}^{2}{P}_{n}^{(\alpha ,\beta )}}{d{x}^{2}}+[\beta -\alpha -(\alpha +\beta +2\left)x\right]\frac{d{P}_{n}^{(\alpha ,\beta )}}{dx}\\ \hfill +n(n+\alpha +\beta +1){P}_{n}^{(\alpha ,\beta )}=0\end{array}$

The trigonometric form of this equation is not physically useful. Comparing this equation with the alternate form of the Gauss hypergeometric differential equation above gives the preliminary representation

${P}_{n}^{(\alpha ,\beta )}\left(x\right)\approx {}_{2}F_{1}(-n\phantom{\rule{.2em}{0ex}},n+\alpha +\beta +1\phantom{\rule{.2em}{0ex}};\alpha +1\phantom{\rule{.2em}{0ex}};\frac{1-x}{2}\phantom{\rule{.2em}{0ex}})$

apart from a normalization factor. As for Gegenbauer polynomials, one can use the Vandermonde identity to determine the coefficient of the leading power of the summation form:

$\frac{1}{{2}^{n}}\sum _{k=0}^{n}\left(\genfrac{}{}{0ex}{}{n+\alpha}{k}\right)\left(\genfrac{}{}{0ex}{}{n+\beta}{n-k}\right)=\frac{1}{{2}^{n}}\left(\genfrac{}{}{0ex}{}{2n+\alpha +\beta}{n}\right)$

The general term of the hypergeometric series can again be written using the relation among gamma function products above arising from rewriting the generalized binomial formula:

$\begin{array}{l}\frac{\Gamma (-n+k)}{\Gamma (-n)}\frac{\Gamma (n+\alpha +\beta +k+1)}{\Gamma (n+\alpha +\beta +1)}\frac{\Gamma (\alpha +1)}{\Gamma (\alpha +k+1)}\frac{1}{k!}(\frac{1-x}{2}{)}^{k}\\ \phantom{\rule{2em}{0ex}}=(-1{)}^{k}\frac{\Gamma (n+1)}{\Gamma (n-k+1)}\frac{\Gamma (n+\alpha +\beta +k+1)}{\Gamma (n+\alpha +\beta +1)}\frac{\Gamma (\alpha +1)}{\Gamma (\alpha +k+1)}\frac{1}{k!}(\frac{1-x}{2}{)}^{k}\end{array}$

The series terminates at $k=n$ , for which value the general term here must be proportional to the quantity just given:

$\begin{array}{c}c(n,\alpha ,\beta )\frac{\Gamma (n+\alpha +\beta +k+1)}{\Gamma (n+\alpha +\beta +1)}\frac{\Gamma (\alpha +1)}{\Gamma (\alpha +k+1)}=\frac{\Gamma (2n+\alpha +\beta +1)}{n!\Gamma (n+\alpha +\beta +1)}\\ c(n,\alpha ,\beta )=\frac{\Gamma (n+\alpha +1)}{n!\Gamma (\alpha +1)}=\left(\genfrac{}{}{0ex}{}{n+a}{n}\right)\end{array}$

which is written as a binomial coefficient for conciseness. The final normalized representation for Jacobi polynomials is

${P}_{n}^{(\alpha ,\beta )}\left(x\right)=\left(\genfrac{}{}{0ex}{}{n+a}{n}\right){}_{2}F_{1}(-n\phantom{\rule{.2em}{0ex}},n+\alpha +\beta +1\phantom{\rule{.2em}{0ex}};\alpha +1\phantom{\rule{.2em}{0ex}};\frac{1-x}{2}\phantom{\rule{.2em}{0ex}})$

On can presumably define associated Jacobi polynomials in the same way as for Legendre and Gegenbauer polynomials, but since there is no apparent physical application of the former this will be left for possible future addition.