A previous presentation considered inversion of the integral

$I(x,k)=\int \frac{dx}{\sqrt{1-x^k}}$

where the exponent k takes on arbitrary positive real values. The purpose of that presentation was to produce a practical power series approximation giving a decent representation of an inverse generalized hyperelliptic integral. The result was in the form of expansions about the two characteristic points, the origin and the turning point determined by the singularity in the denominator of the integrand. The expansion about the origin covered a surprisingly large portion of the domain of the independent variable, failing only close to the turning point.

The method used for determining the result was basically brute force expansion in Mathematica. This approach was taken because direct evaluation of the limits of multiple partial derivatives bogged down very quickly to the point of impracticality. With hindsight one realizes that with a more clever arrangement of the function to be differentiated, all multiple derivatives for an expansion about the origin can be evaluated trivially, thereby producing a complete expansion about the origin. This arrangement is achieved using partial Bell polynomials.

Partial Bell polynomials are defined by the generating function

$\frac{1}{p!}(\sum _{m=1}^{\infty }{x}_m\frac{t^m}{m!}{)}^p=\sum _{n=p}^{\infty }{B}_{n,p}(x_1,x_2,\dots ,x_{n-p+1})\frac{t^n}{n!}$

The coefficients appearing in the partial polynomials can be described as partitions among the independent variables of the polynomials, but this unfortunately does not lend itself to simple evaluation of the coefficients. There does not appear to be a particularly simple way to get these values other than expansion of the generating function and collection of terms. While the partial polynomials are not trivial to generate from scratch, they are known mathematical entities that can be used to express the complete expansion desired in a compact form.

Partial Bell polynomials are useful in this context because they describe function composition: given two functions expressed as power series,

$f\left(x\right)=\sum _{n=1}^{\infty }{a}_n\frac{x^n}{n!}g\left(x\right)=\sum _{n=1}^{\infty }{b}_n\frac{x^n}{n!}$

the composite function $g\left[f\right(x\left)\right]$ can be expressed as

$g\left[f\right(x\left)\right]=\sum _{n=1}^{\infty }\frac{x^n}{n!}\sum _{l=1}^n{b}_l{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1})$

Partial Bell polynomials are essentially a bookkeeping method for tracking the combinations appearing in an arbitrary power of an infinite series. The exact same coefficients appear when taking multiple derivatives of a composite function, so that an arbitrary derivative of the composite function $g\left[f\right(x\left)\right]$ can be expressed compactly with partial Bell polynomials in derivatives of $f\left(x\right)$ .

Turning to the inversion at hand, the generalized hyperelliptic integral can always be evaluated with a Gauss hypergeometric function

$I(x,k)=x{}_{2}F_1[\frac{1}{2},\frac{1}{k};\frac{1}{k}+1;x^k]=t$

where an independent variable for the inversion has been included in the final equality.

The singularity of the integrand at x = 1 determines the quarter period of the oscillatory motion described by the inverse of the integral. The value of the integral between zero and and that point is

$\frac{T}{4}=\frac{\Gamma (\frac{1}{k}+1)\Gamma \left(\frac{1}{2}\right)}{\Gamma (\frac{1}{k}+\frac{1}{2})\Gamma \left(1\right)}=\frac{\sqrt{\pi }\Gamma (\frac{1}{k}+1)}{\Gamma (\frac{1}{k}+\frac{1}{2})}$

using an expression for the value of the hypergeometric function at unity. The full period can be written equivalently

$T=4\sqrt{\pi }\frac{\Gamma (\frac{1}{k}+1)}{\Gamma (\frac{1}{k}+\frac{1}{2})}=\frac{2^{1+2/k}}{k}\frac{\Gamma (\frac{1}{k}{)}^2}{\Gamma \left(\frac{2}{k}\right)}$

using the Legendre duplication formula.

The inverse function will be evaluated with Lagrange inversion: for a function $t=f\left(x\right)$ , the inverse function expanded about a center c is

$x=f^{-1}\left(t\right)=c+\sum _{m=1}^{\infty }\frac{[t-f(c){]}^m}{m!}\underset{x\to c}{lim}\frac{d^{m-1}}{d{x}^{m-1}}[\frac{x-c}{f\left(x\right)-f\left(c\right)}{]}^m$

For the given integrated function and a center at the origin, this becomes

$x=\sum _{m=1}^{\infty }\frac{t^m}{m!}\underset{x\to 0}{lim}\frac{d^{m-1}}{d{x}^{m-1}}[\frac{1}{{}_{2}F_1[\frac{1}{2},\frac{1}{k};\frac{1}{k}+1;x^k]}{]}^m$

The expansion of the Gauss hypergeometric function about the origin is

$\begin{array}{l}{}_{2}F_1[\frac{1}{2},\frac{1}{k};\frac{1}{k}+1;x^k]=1+\sum _{n=1}^{\infty }\frac{\Gamma (\frac{1}{2}+n)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\frac{1}{k}+n)}{\Gamma \left(\frac{1}{k}\right)}\frac{\Gamma (\frac{1}{k}+1)}{\Gamma (\frac{1}{k}+n+1)}\frac{(x^k{)}^n}{n!}\\ \phantom{{}_{2}F_1[\frac{1}{2},\frac{1}{k};\frac{1}{k}+1;x^k]}=1+\sum _{n=1}^{\infty }\frac{1}{nk+1}\frac{\Gamma (n+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}\frac{(x^k{)}^n}{n!}\end{array}$

If one now takes

$f\left(x\right)=\sum _{n=1}^{\infty }{a}_n\frac{(x^k{)}^n}{n!}{a}_n=\frac{1}{nk+1}\frac{\Gamma (n+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}$

along with

$(\frac{1}{1+x}{)}^m=\sum _{n=0}^{\infty }(-1{)}^n\left(\genfrac{}{}{0ex}{}{m+n-1}{n}\right)x^n=1+g\left(x\right)$

$g\left(x\right)=\sum _{n=1}^{\infty }{b}_n\frac{x^n}{n!}{b}_n=(-1{)}^n\frac{(m+n-1)!}{(m-1)!}=(-1{)}^n\frac{\Gamma (m+n)}{\Gamma \left(m\right)}$

using a negative binomial series, the composite function can be expressed compactly with partial Bell polynomials:

$[\frac{1}{{}_{2}F_1[\frac{1}{2},\frac{1}{k};\frac{1}{k}+1;x^k]}{]}^m=1+\sum _{n=1}^{\infty }\frac{(x^k{)}^n}{n!}\sum _{l=1}^n{b}_l{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1})$

If the exponent k is temporarily restricted to integral values, the multiple derivative is simply

$\begin{array}{l}\frac{d^{m-1}}{d{x}^{m-1}}[\frac{1}{{}_{2}F_1[\frac{1}{2},\frac{1}{k};\frac{1}{k}+1;x^k]}{]}^m\\ =\sum _{nk\ge m-1}\frac{\left(nk\right)!}{(nk-m+1)!}\frac{x^{nk-m+1}}{n!}\sum _{l=1}^n{b}_l{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1}),m>1\end{array}$

The limit $x\to 0$ is now trivial: merely picking out terms for which

$nk-m+1=0\to m=nk+1$

Care is required in applying this restriction to the coefficients $b_n$ , since it holds for the summation that is one level higher than that containing the partial Bell polynomials. This is best captured by adding an additional index to these coefficients for an accurate result.

It is now straightforward to evaluate the first few terms of this series in Mathematica or the like to verify that it matches the previous practical result from brute force expansion.

Although this complete expansion was derived for integral values of the exponent k, it can clearly be extended to arbitrary positive values for the exponent. The only point of possible difficulty would be if the indices of the partial Bell polynomials depended on the exponent, but fortunately they do not.

An inverse function can now be defined piecewise by reflection about the quarter period of the motion and the x-axis as necessary, as long as the expansion converges up to the quarter period. To establish this using the ratio test, the limit will be evaluated in two parts. The contribution to the limit from the coefficients of the first summation is

$L_1=\underset{n\to \infty }{lim}\frac{nk+1}{nk+k+1}\frac{n!}{(n+1)!}=\frac{1}{n}$

Evaluating an accurate contribution to the limit from the second sum most strictly would need an explicit expression for all partial Bell polynomials as a function of the two indices. Lacking such an expression, the contribution will be considered for the indices l = 1 and l = n and the more stringent limit adopted. The partial Bell polynomials for these two cases are

$B_{n,1}(a_1,\dots ,a_n)=a_n{B}_{n,n}\left(a_1\right)=a_1^n$

These forms can be determined from the generating function given at the outset.

Evaluating the contributions from the second sum requires a ratio of gamma functions with offset arguments that can be found with Stirling’s approximation. For large x one has

$\begin{array}{l}\frac{\Gamma (x+a)}{\Gamma (x+b)}\approx \sqrt{\frac{x+a}{x+b}}\frac{e^{x+b}}{e^{x+a}}\frac{(x+a{)}^{x+a}}{(x+b{)}^{x+b}}\\ \phantom{\frac{\Gamma (x+a)}{\Gamma (x+b)}}\approx e^{b-a}\frac{(1+\frac{a}{x}{)}^x}{(1+\frac{b}{x}{)}^x}\frac{(x+a{)}^a}{(x+b{)}^b}\approx e^{b-a}{e}^{a-b}{x}^{a-b}=x^{a-b}\end{array}$

where a limit definition of the exponential function has been employed. The contribution for l = 1 is

$\begin{array}{l}{L}_2=\underset{n\to \infty }{lim}\frac{nk+1}{nk+k+1}\frac{\Gamma (n+\frac{3}{2})}{\Gamma (n+\frac{1}{2})}\frac{\Gamma (nk+k+2)}{\Gamma (nk+2)}\frac{\Gamma (nk+1)}{\Gamma (nk+k+1)}\\ \phantom{L_2}=\underset{n\to \infty }{lim}(n+\frac{1}{2})(nk{)}^k(nk{)}^{-k}=n\end{array}$

and the contribution for l = n is

$\begin{array}{l}{L}_2=\underset{n\to \infty }{lim}(\frac{1}{k+1}·\frac{1}{2})\frac{\Gamma \left[n\right(k+1)+k+2]}{\Gamma \left[n\right(k+1)+1]}\frac{\Gamma (nk+1)}{\Gamma (nk+k+1)}\\ \phantom{L_2}=\frac{1}{2(k+1)}\underset{n\to \infty }{lim}\left[n\right(k+1){]}^{k+1}(nk{)}^{-k}=\frac{n}{2}(1+\frac{1}{k}{)}^k\end{array}$

For k > 1 the contribution for l = n is the larger of the two. The total limit for the ratio test for k > 1 is

$L=\underset{n\to \infty }{lim}{L}_1{L}_2=\frac{1}{2}(1+\frac{1}{k}{)}^k$

With the inclusion of the independent variable, the series converges as long as

$\frac{1}{2}(1+\frac{1}{k}{)}^k{t}^k<1\to t<\frac{2^{1/k}}{1+\frac{1}{k}}$

This limit value is slightly under unity for k > 1 , so the radius of convergence is about unity. Since the quarter period of the oscillatory motion defined by the inverse generalized hyperelliptic integral is only unity when k → ∞ , the expansion unfortunately does not converge for the entire quarter period.

Numerical evaluation, however, indicates that the expansion is quite accurate over the entire quarter period when the series is truncated at around ten terms or so. Apparently the complete expansion is asymptotic to the inverse generalized hyperelliptic integral outside the region of convergence.

For a negative value of the exponent in the generalized hyperelliptic integral, first define a positive value p = −k. To avoid introducing an imaginary unit, the integral to be inverted will be taken to be

$\begin{array}{l}I(x,p)=\int \frac{dx}{\sqrt{\frac{1}{x^p}-1}}=\int dx\frac{x^{p/2}}{\sqrt{1-x^p}}\\ \phantom{I(x,p)}=\frac{2}{p+2}{x}^{(p+2)/2}{}_{2}F_1[\frac{1}{2},\frac{1}{p}+\frac{1}{2};\frac{1}{p}+\frac{3}{2};x^p]=t\end{array}$

where the integral is again evaluated with a Gauss hypergeometric function and an independent variable for the inversion has been included in the final equality.

The quarter period for this inversion is

$\frac{T}{4}=\frac{2}{p+2}\frac{\Gamma (\frac{1}{p}+\frac{3}{2})\Gamma \left(\frac{1}{2}\right)}{\Gamma (\frac{1}{p}+1)\Gamma \left(1\right)}=\frac{\sqrt{\pi }\Gamma (\frac{1}{p}+\frac{1}{2})}{\Gamma \left(\frac{1}{p}\right)}$

so that the full period can be written equivalently

$T=4\sqrt{\pi }\frac{\Gamma (\frac{1}{p}+\frac{1}{2})}{\Gamma \left(\frac{1}{p}\right)}=2^{3-2/p}\pi \frac{\Gamma \left(\frac{2}{p}\right)}{\Gamma (\frac{1}{p}{)}^2}$

Since the function to be inverted has a first derivative of zero, and alternate form of Lagrange inversion must be employed. Given the linear relation $t=f\left(x\right)$ , an arbitrary function $F\left(x\right)$ of the inverse function can be expanded in the form

$F\left(x\right)=F\left(c\right)+\sum _{m=1}^{\infty }\frac{[t-f(c){]}^m}{m!}\underset{x\to c}{lim}\frac{d^{m-1}}{d{x}^{m-1}}\left[F^{\prime }\left(x\right)[\frac{x-c}{f\left(x\right)-f\left(c\right)}{]}^m\right]$

This expansion appears on page 129 of Whittaker and Watson’s A Course of Modern Analysis with different notation. With the change of variable

$y=x^{(p+2)/2}s=\frac{p+2}{2}t$

the function to be inverted is

$y{}_{2}F_1[\frac{1}{2},\frac{1}{p}+\frac{1}{2};\frac{1}{p}+\frac{3}{2};y^{2p/(p+2)}]=s$

With the choice of arbitrary function $F\left(y\right)=y^{2/(p+2)}$ to undo the change of variable, the Lagrange inversion about the origin is

$x=\frac{2}{p+2}\sum _{m=1}^{\infty }\frac{s^m}{m!}\underset{y\to 0}{lim}\frac{d^{m-1}}{d{y}^{m-1}}\left[y^{-p/(p+2)}\right[\frac{1}{{}_{2}F_1[\frac{1}{2},\frac{1}{p}+\frac{1}{2};\frac{1}{p}+\frac{3}{2};y^{2p/(p+2)}]}{]}^m]$

The expansion of the Gauss hypergeometric function about the origin is

$\begin{array}{l}{}_{2}F_1[\frac{1}{2},\frac{1}{p}+\frac{1}{2};\frac{1}{p}+\frac{3}{2};y^{2p/(p+2)}]\\ =1+\sum _{n=1}^{\infty }\frac{\Gamma (\frac{1}{2}+n)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\frac{1}{p}+\frac{1}{2}+n)}{\Gamma (\frac{1}{p}+\frac{1}{2})}\frac{\Gamma (\frac{1}{p}+\frac{3}{2})}{\Gamma (\frac{1}{p}+\frac{3}{2}+n)}\frac{(y^{2p/(p+2)}{)}^n}{n!}\\ =1+\sum _{n=1}^{\infty }\frac{1}{\frac{2p}{p+2}n+1}\frac{\Gamma (n+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}\end{array}$

If one now takes

$f\left(y\right)=\sum _{n=1}^{\infty }{a}_n\frac{(y^{2p/(p+2)}{)}^n}{n!}{a}_n=\frac{1}{\frac{2p}{p+2}n+1}\frac{\Gamma (n+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}$

along with

$(\frac{1}{1+y}{)}^m=\sum _{n=0}^{\infty }(-1{)}^n\left(\genfrac{}{}{0ex}{}{m+n-1}{n}\right)y^n=1+g\left(y\right)$

$g\left(y\right)=\sum _{n=1}^{\infty }{b}_n\frac{y^n}{n!}{b}_n=(-1{)}^n\frac{(m+n-1)!}{(m-1)!}=(-1{)}^n\frac{\Gamma (m+n)}{\Gamma \left(m\right)}$

using a negative binomial series, the composite function can again be expressed compactly with partial Bell polynomials,

$\begin{array}{l}{y}^{-p/(p+2)}[\frac{1}{{}_{2}F_1[\frac{1}{2},\frac{1}{p}+\frac{1}{2};\frac{1}{p}+\frac{3}{2};y^{2p/(p+2)}]}{]}^m\\ =y^{-p/(p+2)}[1+\sum _{n=1}^{\infty }\frac{(y^{2p/(p+2)}{)}^n}{n!}\sum _{l=1}^n{b}_l{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1})]\end{array}$

The multiple derivative can be easily evaluated as

$\begin{array}{l}\frac{d^{m-1}}{d{y}^{m-1}}\left[y^{-p/(p+2)}[\frac{1}{{}_{2}F_1[\frac{1}{2},\frac{1}{p}+\frac{1}{2};\frac{1}{p}+\frac{3}{2};y^{2p/(p+2)}]}{]}^m\right]\\ =y^{-p/(p+2)-m+1}\\ +\sum _{(2n-1)p/(p+2)\ge m-1}\frac{\Gamma [\frac{(2n-1)p}{p+2}+1]}{\Gamma [\frac{(2n-1)p}{p+2}-m+2]}\frac{y^{(2n-1)p/(p+2)-m+1}}{n!}\\ \hfill \times \sum _{l=1}^n{b}_l{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1})\end{array}$

where the change to the lower limit of summation removes terms that become singular in the limit $y\to 0$ . To justify the utility of this, let $x\left(s\right)=s^{2/(p+2)}f\left(s\right)$ in the original equation to be inverted and cancel a single power of the scaled temporal variable:

$[f\left(s\right){]}^{(p+2)/2}{}_{2}F_1[\frac{1}{2},\frac{1}{p}+\frac{1}{2};\frac{1}{p}+\frac{3}{2};s^{2p/(p+2)}\left[f\right(s){]}^p]=1$

By inspection $f\left(s\right)$ can only be a function of powers of $s^{2p/(p+2)}$ , which means that the expansion for $x\left(s\right)$ contains terms of the form $s^{2(np+1)/(p+2)}$ . This can be achieved by picking out terms in the limit $y\to 0$ for which

$\frac{(2n-1)p}{p+2}-m+1=0\to m=\frac{2(np+1)}{p+2}$

While this condition is not strictly true for integral indices, it does produce the correct form of the expansion. Care is again required in applying this restriction to the coefficients $b_n$ , once more being best captured by adding an additional index to these coefficients.

Evaluating the first few terms again matches the previous practical result from brute force expansion, but less straightforwardly. Mathematica does not automatically simplify the ratio of gamma functions in $b_{n,l}$ which has to be rewritten as an explicit product to confirm the match.

The limit for the ratio test will again be evaluated in two parts. The contribution to the limit from the coefficients of the first summation is

$L_1=\underset{n\to \infty }{lim}\frac{np+1}{np+p+1}\frac{n!}{(n+1)!}=\frac{1}{n}$

The contribution from the second summation l = 1 is

$\begin{array}{l}{L}_2=\underset{n\to \infty }{lim}\frac{\frac{2p}{p+2}n+\frac{2p}{p+2}+1}{\frac{2p}{p+2}n+1}\frac{\Gamma (n+\frac{3}{2})}{\Gamma (n+\frac{1}{2})}\frac{\Gamma (\frac{2p}{p+2}n+\frac{2p+2}{p+2}+l)}{\Gamma (\frac{2p}{p+2}n+\frac{2}{p+2}+l)}\frac{\Gamma (\frac{2p}{p+2}n+\frac{2}{p+2})}{\Gamma (\frac{2p}{p+2}n+\frac{2p+2}{p+2})}\\ \phantom{L_2}=\underset{n\to \infty }{lim}(n+\frac{1}{2})(\frac{2p}{p+2}n{)}^{2p/(p+2)}(\frac{2p}{p+2}n{)}^{-2p/(p+2)}=n\end{array}$

and the contribution for l = n is

$\begin{array}{l}{L}_2=\underset{n\to \infty }{lim}(\frac{1}{\frac{2p}{p+2}+1}·\frac{1}{2})\frac{\Gamma (\frac{3p+2}{p+2}n+\frac{3p+4}{p+2})}{\Gamma (\frac{3p+2}{p+2}n+\frac{2}{p+2})}\frac{\Gamma (\frac{2p}{p+2}n+\frac{2}{p+2})}{\Gamma (\frac{2p}{p+2}n+\frac{2p+2}{p+2})}\\ \phantom{L_2}=\frac{p+2}{2(3p+2)}\underset{n\to \infty }{lim}(\frac{3p+2}{p+2}n{)}^{(3p+2)/(p+2)}(\frac{2p}{p+2}n{)}^{-2p/(p+2)}\\ L_2=\frac{n}{2}(\frac{3}{2}+\frac{1}{p}{)}^{2p/(p+2)}\end{array}$

With negative exponents the region of interest for physics is 0 < p < 2 , for which the contribution for l = 1 is larger. The total limit for the ratio test is

$L=\underset{n\to \infty }{lim}{L}_1{L}_2=1$

With the inclusion of the independent variable, the series converges as long as

$s^{2p/(p+2)}<1\to t<\frac{2}{p+2}$

which again does not cover the full quarter period of the motion. Numerical evaluation also again indicates that the complete expansion is asymptotic to the inverse generalized hyperelliptic integral outside the region of convergence, with far more terms allowed before the divergence manifests itself.

Uploaded 2015.07.30 — Updated 2017.01.19 analyticphysics.com