Consider the positive real roots of the function

$f(x,t)={x}^{n}-x+t$

where the exponent *n* takes on arbitrary positive real values. For *t* = 0 the two positive real roots are located at zero and unity. Values of these roots for nonzero *t* can be evaluated with Lagrange inversion in terms of a convergent power series.

Lagrange inversion can be stated in three different forms, each useful in appropriate circumstances. Firstly, for a linear functional relationship
*c* is

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

Secondly, an arbitrary function $F\left(x\right)$ of this 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. This form is useful for treating the inversion of functions whose first derivative vanishes.

Thirdly, if the original functional relationship is rearranged to the form
$x=a+t\phantom{\rule{.2em}{0ex}}\phi \left(x\right)$ ,
an arbitrary function of its inverse can be expanded in terms of the parameter *t* in the form

$F\left(x\right)=F\left(a\right)+\sum _{m=1}^{\infty}\frac{{t}^{m}}{m!}\frac{{d}^{m-1}}{d{a}^{m-1}}\left[{F}^{\prime}\left(a\right)\left[\phi \right(a){]}^{m}\right]$

as given on page 133 of Whittaker and Watson. This form also goes by the name Lagrange reversion. It is useful for giving a power series fully expanded in its variable.

For the root in the vicinity of zero, the exponent *n* will be treated as an integer when taking limits but will ultimately appear as part of an analytically extendable expression. The justification for this method is that is leads to an expansion that matches numerical solutions for that root.

The multiple derivatives in the first form of Lagrange inversion can be evaluated straightforwardly with a negative binomial series,

$\begin{array}{l}\frac{{d}^{m-1}}{d{x}^{m-1}}(\frac{x}{x-{x}^{n}}{)}^{m}=\frac{{d}^{m-1}}{d{x}^{m-1}}(1-{x}^{n-1}{)}^{-m}\\ \phantom{\frac{{d}^{m-1}}{d{x}^{m-1}}(\frac{x}{x-{x}^{n}}{)}^{m}}=\frac{{d}^{m-1}}{d{x}^{m-1}}\sum _{l=0}^{\infty}\left(\genfrac{}{}{0ex}{}{m+l-1}{l}\right)\phantom{\rule{.3em}{0ex}}{x}^{(n-1)l}\\ \frac{{d}^{m-1}}{d{x}^{m-1}}(\frac{x}{x-{x}^{n}}{)}^{m}=\sum _{(n-1)l\ge m-1}^{\infty}\left(\genfrac{}{}{0ex}{}{m+l-1}{l}\right)\frac{\Gamma \left[\right(n-1)l+1]}{\Gamma \left[\right(n-1)l-m+2]}\phantom{\rule{.3em}{0ex}}{x}^{(n-1)l-m+1}\end{array}$

where the change in the lower limit of summation comes from treating *n* temporarily as an integer. The only terms that remain in the limit
$x\to 0$ are those for which

$(n-1)l=m-1$

and since all values are at this point integral, this becomes a condition on the allowed values of *m*. The root in the vicinity of zero can be expanded as

$x=\sum _{l=0}^{\infty}\left(\genfrac{}{}{0ex}{}{nl}{l}\right)\frac{\Gamma \left[\right(n-1)l+1]}{\Gamma \left[\right(n-1)l+2]}{t}^{(n-1)l+1}=\sum _{l=0}^{\infty}\frac{\Gamma [nl+1]}{\Gamma \left[\right(n-1)l+2]}\frac{{t}^{(n-1)l+1}}{l!}$

Since the exponent *n* now appears in analytic functions, the expansion can be taken as a representation of the root for arbitrary real values of the exponent.

In order to apply the ratio test to find the radius of convergence of this series, first evaluate a ratio of gamma functions with offset arguments using 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 ratio test for the given expansion leads to the limit

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

Since the series has a power of
$n-1$
in addition to integral steps, it will converge as long as the parameter *t* is less than the corresponding root of the inverse of the ratio test limit:

${x}_{\mathrm{zero}}=\sum _{l=0}^{\infty}\frac{\Gamma [nl+1]}{\Gamma \left[\right(n-1)l+2]}\frac{{t}^{(n-1)l+1}}{l!}\phantom{\rule{2em}{0ex}},\phantom{\rule{2em}{0ex}}t<\frac{n-1}{{n}^{n/(n-1)}}$

For the root in the vicinity of unity, the multiple derivatives in the first form of Lagrange inversion are not so easily evaluated. One approach is to make a change of variable $u=x-1$ , so that derivatives with respect to the two variables are identical. The expression subject to differentiation becomes

$\begin{array}{l}(\frac{x-1}{x-{x}^{n}}{)}^{m}=(\frac{u}{1+u-(1+u{)}^{n}}{)}^{m}\\ \phantom{(\frac{x-1}{x-{x}^{n}}{)}^{m}}=[(1-n)-\sum _{l=2}^{\infty}\frac{\Gamma (n+1)}{\Gamma (n-l+1)}\frac{{u}^{l-1}}{l!}{]}^{-m}\\ (\frac{x-1}{x-{x}^{n}}{)}^{m}=\frac{1}{(1-n{)}^{m}}\sum _{r=0}^{\infty}\left(\genfrac{}{}{0ex}{}{m+r-1}{r}\right)(\frac{1}{1-n}\sum _{l=2}^{\infty}\frac{\Gamma (n+1)}{\Gamma (n-l)}\frac{{u}^{l-1}}{l!}{)}^{r}\end{array}$

using an extension of the binomial theorem to continuous exponents followed by another negative binomial series.

Since the limit
$x\to 1$
is equivalent to
$u\to 0$
the only terms that will survive are those for which the total exponent is equal to *m* − 1 . This condition will be written as

$\sum _{i=1}^{r}{l}_{i}=m-1$

Each term in a power of the infinite series will have a coefficient after differentiation of the form

$\frac{(m-1)!}{(1-n{)}^{r}}\sum _{{l}_{i}}\prod _{i=1}^{r}\frac{\Gamma (n+1)}{\Gamma (n-{l}_{i})}\frac{1}{({l}_{i}+1)!}$

where the sum over indices is subject to the condition directly above. Combining all terms with equal exponents coming from all powers of the infinite series is akin to evaluating the coefficients of a general multinomial expansion: not a simple task.

It is simpler by far to use one of the alternate forms of Lagrange inversion, in this case the third. Rearrange the equation for the roots,

${x}^{n-1}+\frac{t}{x}=1\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}y=1+t\phantom{\rule{.2em}{0ex}}{y}^{n/(n-1)}$

with the relation
${x}^{n-1}y=1$
between the two variables. With the choice of arbitrary function

$\begin{array}{l}x=1-\frac{1}{n-1}\sum _{m=1}^{\infty}\frac{{t}^{m}}{m!}\frac{{d}^{m-1}}{d{a}^{m-1}}[{a}^{n(m-1)/(n-1)}{]}_{a=1}\\ \phantom{x}=1-\frac{1}{n-1}\sum _{m=1}^{\infty}\frac{\Gamma (\frac{n(m-1)}{n-1}+1)}{\Gamma (\frac{n(m-1)}{n-1}-m+2)}\frac{{t}^{m}}{m!}\\ x=1-\frac{1}{n-1}\sum _{m=1}^{\infty}\frac{\Gamma \left(\frac{nm-1}{n-1}\right)}{\Gamma \left(\frac{n+m-2}{n-1}\right)}\frac{{t}^{m}}{m!}\end{array}$

Note that there is no need to temporarily restrict the exponent *n* to integral values for this expansion. One can also verify by brute force evaluation in Mathematica that the coefficients arising from differentiating the expression above in the variable *u* are identical, at least for those that can be evaluated in a reasonable amount of time.

Applying the ratio test to this expansion, along with the expression above for the asymptotic ratio of two gamma functions with offset arguments, leads to the limit

$\begin{array}{l}L=\underset{m\to \infty}{lim}\frac{\Gamma (\frac{nm}{n-1}+1)}{\Gamma (\frac{nm}{n-1}-\frac{1}{n-1})}\frac{\Gamma (\frac{m}{n-1}+\frac{n-2}{n-1})}{\Gamma (\frac{m}{n-1}+1)}\frac{m!}{(m+1)!}\\ \phantom{L}=\underset{m\to \infty}{lim}(\frac{nm}{n-1}{)}^{\frac{n}{n-1}}(\frac{m}{n-1}{)}^{-\frac{1}{n-1}}\frac{1}{m+1}\\ L=\frac{{n}^{n/(n-1)}}{n-1}\end{array}$

Since the series has only integral steps, it converges as long as the parameter *t* is less than the inverse of the ratio test limit:

${x}_{\mathrm{unity}}=1-\frac{1}{n-1}\sum _{m=1}^{\infty}\frac{\Gamma \left(\frac{nm-1}{n-1}\right)}{\Gamma \left(\frac{n+m-2}{n-1}\right)}\frac{{t}^{m}}{m!}\phantom{\rule{2em}{0ex}},\phantom{\rule{2em}{0ex}}t<\frac{n-1}{{n}^{n/(n-1)}}$

The parameter *t* in this expansion parametrizes deviation from unity as opposed to deviation from zero in the previous result. The route to this expansion is the approach taken in this fun paper.

As the parameter increases in both expansions, the positive real roots move in from zero and unity until they coalesce at the maximum of the original function:

$\begin{array}{c}\frac{d}{dx}({x}^{n}-x+t)=0\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}{x}_{\mathrm{max}}=\frac{1}{{n}^{1/(n-1)}}\\ \\ {t}_{\mathrm{max}}={x}_{\mathrm{max}}-{x}_{\mathrm{max}}^{n}=\frac{n-1}{{n}^{n/(n-1)}}\end{array}$

Since this is the common radius of convergence for the two expansions, each expansion converges for the entire domain between its initial value and the point of coalescence.

Setting *n* = 5 in the original function, the expansions

$\sum _{l=0}^{\infty}\frac{\Gamma (5l+1)}{\Gamma (4l+2)}\frac{{t}^{4l+1}}{l!}\phantom{\rule{5em}{0ex}}1-\frac{1}{4}\sum _{m=1}^{\infty}\frac{\Gamma \left(\frac{5m-1}{4}\right)}{\Gamma \left(\frac{m+3}{4}\right)}\frac{{t}^{m}}{m!}$

are positive real solutions to the quintic equation

${x}^{5}-x+t=0$

as can be easily checked by numerical evaluation. This wonderful result is apparently not very well known even among professional mathematicians.

The function determining the two real turning points of classical mechanical bound states in the well of a power potential in two or more dimensions is of the form

$g(x,c)={x}^{k+2}-{x}^{2}+c$

where the exponent *k* takes on arbitrary real values. The real roots of this function are easily soluble for two related systems, the simple harmonic oscillator and the Kepler potential, but cannot be determined exactly in general as an analytic function of the exponent.

For *c* = 0 , a system with angular momentum of zero, the turning in this notation are located at zero and unity. One can apply previous techniques to evaluate the turning points for nonzero angular momentum in terms of a convergent power series. This will be done for positive values of the exponent *k* but the method is easily extendable to negative values greater than minus two.

For the turning point near zero, the function to be inverted has a first derivative of zero and the first form of Lagrange inversion cannot be used. To use the second form of Lagrange inversion, make a change of variable $y={x}^{2}$ ,

$g(y,c)={y}^{1+k/2}-y+c$

Choose the arbitrary function $F(\mathrm{y)}=\sqrt{y}$ to invert the change of variable. Multiple derivatives can again be evaluated straightforwardly with a negative binomial series,

$\begin{array}{l}\frac{{d}^{m-1}}{d{y}^{m-1}}\left[\frac{d\sqrt{y}}{dy}(\frac{y}{y-{y}^{1+k/2}}{)}^{m}\right]=\frac{1}{2}\frac{{d}^{m-1}}{d{y}^{m-1}}{y}^{-1/2}(1-{y}^{k/2}{)}^{-m}\\ \phantom{\rule{6em}{0ex}}=\frac{1}{2}\frac{{d}^{m-1}}{d{y}^{m-1}}\sum _{l=0}^{\infty}\left(\genfrac{}{}{0ex}{}{m+l-1}{l}\right)\phantom{\rule{.3em}{0ex}}{y}^{(kl-1)/2}\\ \phantom{\rule{6em}{0ex}}=\frac{1}{2}\sum _{\frac{kl-1}{2}\ge m-1}^{\infty}\left(\genfrac{}{}{0ex}{}{m+l-1}{l}\right)\frac{\Gamma (\frac{kl-1}{2}+1)}{\Gamma (\frac{kl-1}{2}-m+2)}\phantom{\rule{.3em}{0ex}}{y}^{(kl-2m+1)/2}\end{array}$

where the change in the lower limit of summation removes terms that become singular in the limit
*y* using the previous result for the root near zero:

$y=\sum _{l=0}^{\infty}{a}_{l}{c}^{(n-1)l+1}=c\phantom{\rule{.3em}{0ex}}\sum _{l=0}^{\infty}{a}_{l}{c}^{kl/2}$

The desired expansion is the square root of this series, which will have terms of the form

$x=\sqrt{y}=\sqrt{c}\phantom{\rule{.3em}{0ex}}[1+\sum _{l=1}^{\infty}{a}_{l}{c}^{kl/2}{]}^{\frac{1}{2}}=\sqrt{c}\phantom{\rule{.3em}{0ex}}\sum _{l=0}^{\infty}{b}_{l}{c}^{kl/2}$

using a binomial series. This indicates that in the limit

$m=\frac{kl+1}{2}$

which corresponds to the lower limit of the summation. While this condition is not strictly true for integral indices, it does produce the correct form of the expansion. The combinatoric part the each coefficient is

$\frac{1}{m!}\left(\genfrac{}{}{0ex}{}{m+l-1}{l}\right)\frac{\Gamma (\frac{kl-1}{2}+1)}{\Gamma (\frac{kl-1}{2}-m+2)}=\frac{1}{l!}\frac{\Gamma (\frac{kl+1}{2}+l)}{\Gamma (\frac{kl+1}{2}+1)}$

With the inclusion of the factor of one half, the resulting expansion for the original inverse function is

$x=\frac{\sqrt{c}}{2}\phantom{\rule{.3em}{0ex}}\sum _{l=0}^{\infty}\frac{\Gamma (\frac{kl+1}{2}+l)}{\Gamma (\frac{kl+1}{2}+1)}\frac{{c}^{kl/2}}{l!}=\frac{\sqrt{c}}{2}\phantom{\rule{.3em}{0ex}}\sum _{l=0}^{\infty}\frac{\Gamma (\frac{k+2}{2}l+\frac{1}{2})}{\Gamma (\frac{k}{2}l+\frac{3}{2})}\frac{{c}^{kl/2}}{l!}$

Further justification for this result is that it matches numerical evaluations of the root, as can be easily verified in Mathematica. Applying the ratio test to this series leads to the limit

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

Since the series has a power of
$k/2$
in addition to integral steps, it will converge as long as the parameter *c* is less than the corresponding root of the inverse of the ratio test limit:

${x}_{\mathrm{zero}}=\frac{\sqrt{c}}{2}\phantom{\rule{.3em}{0ex}}\sum _{l=0}^{\infty}\frac{\Gamma \left(\frac{(k+2)l+1}{2}\right)}{\Gamma \left(\frac{kl+3}{2}\right)}\frac{{c}^{kl/2}}{l!}\phantom{\rule{2em}{0ex}},\phantom{\rule{2em}{0ex}}c<\frac{k}{k+2}(\frac{2}{k+2}{)}^{2/k}$

For the turning point near unity, rearrange the equation determining the roots

${x}^{k}+\frac{c}{{x}^{2}}=1\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}y=1+c\phantom{\rule{.2em}{0ex}}{y}^{(k+2)/k}$

with the relation
${x}^{k}y=1$
between the two variables. With the choice of arbitrary function

$\begin{array}{l}x=1-\frac{1}{k}\phantom{\rule{.2em}{0ex}}\sum _{m=1}^{\infty}\frac{{c}^{m}}{m!}\frac{{d}^{m-1}}{d{a}^{m-1}}[{a}^{\left[k\right(m-1)+2m-1]/k}{]}_{a=1}\\ \phantom{x}=1-\frac{1}{k}\phantom{\rule{.2em}{0ex}}\sum _{m=1}^{\infty}\frac{\Gamma (\frac{k(m-1)+2m-1}{k}+1)}{\Gamma (\frac{k(m-1)+2m-1}{k}-m+2)}\frac{{c}^{m}}{m!}\\ x=1-\frac{1}{k}\phantom{\rule{.2em}{0ex}}\sum _{m=1}^{\infty}\frac{\Gamma \left(\frac{km+2m-1}{k}\right)}{\Gamma \left(\frac{k+2m-1}{k}\right)}\frac{{c}^{m}}{m!}\end{array}$

Applying the ratio test to this expansion, along with the expression above for the asymptotic ratio of two gamma functions with offset arguments, leads to the limit

$\begin{array}{l}L=\underset{m\to \infty}{lim}\frac{\Gamma (\frac{k+2}{k}m+\frac{k+1}{k})}{\Gamma (\frac{k+2}{k}m-\frac{1}{k})}\frac{\Gamma (\frac{2}{k}m+\frac{k-1}{k})}{\Gamma (\frac{2}{k}m+\frac{k+1}{k})}\frac{m!}{(m+1)!}\\ \phantom{L}=\underset{m\to \infty}{lim}(\frac{k+2}{k}m{)}^{\frac{k+2}{k}}(\frac{2}{k}m{)}^{-\frac{2}{k}}\frac{1}{m+1}\\ L=\frac{k+2}{k}(\frac{k+2}{2}{)}^{2/k}\end{array}$

Since the series has only integral steps, it converges as long as the parameter *c* is less than the inverse of the ratio test limit:

${x}_{\mathrm{unity}}=1-\frac{1}{k}\phantom{\rule{.2em}{0ex}}\sum _{m=1}^{\infty}\frac{\Gamma \left(\frac{km+2m-1}{k}\right)}{\Gamma \left(\frac{k+2m-1}{k}\right)}\frac{{c}^{m}}{m!}\phantom{\rule{2em}{0ex}},\phantom{\rule{2em}{0ex}}c<\frac{k}{k+2}(\frac{2}{k+2}{)}^{2/k}$

The parameter *c* in this expansion parametrizes deviation from unity as opposed to deviation from zero in the previous result. As the parameter increases in both expansions, the turning points move in from zero and unity until they coalesce at the maximum of the original function:

$\begin{array}{c}\frac{d}{dx}({x}^{k+2}-{x}^{2}+c)=0\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}{x}_{\mathrm{max}}=(\frac{2}{k+2}{)}^{1/k}\\ \\ {c}_{\mathrm{max}}={x}_{\mathrm{max}}^{2}-{x}_{\mathrm{max}}^{k+2}=\frac{k}{k+2}(\frac{2}{k+2}{)}^{2/k}\end{array}$

Since this is the common radius of convergence for the two expansions, each expansion converges for the entire domain between its initial value and the point of coalescence.

The common radius of convergence for the two expansions has physical significance: it corresponds to the maximum allowed value of angular momentum for bound orbits, which are circular for this maximum.

*Uploaded 2015.07.07 — Updated 2015.07.21*
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