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Note: throughout this question, I'll be using the following notation convention:

$$f^{(n)}(x)=\frac{d^nf}{dx^n}(x)$$

I was browsing through Wikipedia and even MSE's related questions searching for a proof for the Lagrange Inversion Theorem. I'm letting the function $g$ be $f$'s inverse, and I'm letting $f$ be analytic at $a$ while also maintaining the property that $f'(a) \neq 0$.

$g$ is also assumed to be analytic, therefore suggesting that

$$g(f(w))=\displaystyle\sum_{n=0}^{\infty}g^{(n)}(f(a))\frac{(f(w)-f(a))^n}{n!}=w.$$

Expanding past the zeroth term, we get

$$g(f(w))=a+\displaystyle\sum_{n=1}^{\infty}g^{(n)}(f(a))\frac{(f(w)-f(a))^n}{n!}.$$

Now, letting $f(w)=z$, we have

$$g(z)=a+\displaystyle\sum_{n=1}^{\infty}g^{(n)}(f(a))\frac{(z-f(a))^n}{n!}.$$

Wikipedia states that

$$g(z)=a+\displaystyle\sum_{n=1}^{\infty}\lim_{w \to a}\Bigg[\frac{d^{n-1}}{dw^{n-1}}\Bigg(\frac{w-a}{f(w)-f(a)}\Bigg)^n\Bigg]\frac{(z-f(a))^n}{n!},$$

which suggests the following:

$$g^{(n)}(f(a))=\lim_{w \to a}\Bigg[\frac{d^{n-1}}{dw^{n-1}}\Bigg(\frac{w-a}{f(w)-f(a)}\Bigg)^n\Bigg].$$

It looks like a residue of some weird function at $a$, but the nature of the function itself and the reason why it shows up from a derivative rather than an integral are really stumping me. I've read a few pdf's related to it, but one gave an answer that was too far from the equation itself for me to draw a connection, and the other was in French (my high school French classes have failed me, needless to say).

This is primarily just me wanting to find an inverse polylogarithm, a question that has already been asked here on MSE with no avail, but also for me to understand this theorem more precisely.

If someone could please point out where this residue-ish thing comes from and why it looks the way it does (i.e., no factorial or $2\pi i$ term), I'd be greatly appreciative. If instead you were able to give me a derivation or a proof of the polynomial approximation given by the commenter Simon here, that would be great too. A few more things that may or may not be useful specifically for the polylogarithms are the following equations:

$$z _{s+1}F_s(1,\dots,1;2,\dots,2;z)=\operatorname{Li}_s(z)$$

$$(z-1) _2F_1(1,1;2;1-z)=\log(z)$$

$$\operatorname{Li}_1(z)=-\log(1-z)$$

$$\operatorname{Li}_0(z)=\frac{z}{1-z}$$

$$\operatorname{Li}_1^{-1}(z)=1-e^{-z}$$

$$\operatorname{Li}_0^{-1}(z)=\frac{z}{1+z}$$

were $_pF_q(a_1,\dots,a_p;b_1,\dots,b_q;z)$ denotes the generalized hypergeometric function and $\log$ denotes the natural logarithm.

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    $\begingroup$ It looks like Simons's few terms come from the direct series reversion (mathworld.wolfram.com/SeriesReversion.html) of the known series for the Polylogarithm. This can be given in Mathematica by InverseSeries[Series[PolyLog[s,x],{x,0,10}]] in case that helps. The mathworld article quotes "A derivation of the explicit formula for the nth term is given by Morse and Feshbach (1953)", I've looked at the book in the library before and it is a very short and opaque derivation. $\endgroup$ – Benedict W. J. Irwin Jul 4 '18 at 15:50
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    $\begingroup$ If you take the coefficients of that reversion for the Polylog and multiply by $\Gamma(s)$, then take an inverse Mellin transform, they appear to map to sums of simple exponentials weighted by the coefficients in A111785. For the coefficient of $z^n$ there are $P(n)$ exponentials, where $P(n)$ are the partition numbers: A000041. This might be a way of encoding this inverse function. The powers of the exponentials appear to be given by A074139, which looks nice. $\endgroup$ – Benedict W. J. Irwin Jul 4 '18 at 16:06
  • $\begingroup$ Thanks for the comments and the answer! They're super helpful! $\endgroup$ – 46andpi Jul 9 '18 at 0:29
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I will write what I have found here because it seems interesting enough. It is more a demonstration of how to group the terms in the messy reversion into references to documented (but complicated) sequences.

For the Polylogarithm we have the series representation $$ \mathrm{Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s} $$ if we perform a series reversion on this (term by term) we end up with an expansion for the inverse function $$ \mathrm{Li}^{-1}_s(z) = \sum_{k=1}^\infty a_k z^k $$ the first few coefficients are $$ a_1 = 1 \\ a_2 = -2^{-s} \\ a_3 = 2^{1-2s} - 3^{-s} \\ a_4 = 5 6^{-s} - 8^{-s}(5+2^s) \\ \cdots $$ there may be a pattern in there somewhere, but the terms seem to grow quite large and complicated rather quickly. For some reason I considered looking at the inverse Mellin transform of these coefficients, multipled by a gamma function, we can denote these functions $e_k(x)$ $$ e_k(x) = \mathcal{M}^{-1}[\Gamma[s]a_k(s)](s) $$ these begin \begin{equation} e_1(x) = e^{-x} \\ e_2(x) = -e^{-2x} \\ e_3(x) = 2 e^{-4x} - e^{-3 x} \\ e_4(x) = -5 e^{-8 x} + 5 e^{-6x} - e^{-4 x} \\ \cdots \end{equation} in each term $e_k(x)$ there are $P(k-1)$ exponential functions, where $P(k)$ from $k=0$ goes like $1,1,2,3,5,7,\cdots$ and are the partition numbers A000041. The coefficients in this grid of exponentials goes like: $$ \alpha_1=[ +1]\\ \alpha_2=[ -1]\\ \alpha_3=[ -1, 2]\\ \alpha_4=[ -1, 5, -5]\\ \alpha_5=[ -1, 6, 3, -21, 14]\\ \alpha_6=[ -1, 7, 7, -28, -28, 84, -42]\\ \alpha_7=[ -1, 8, 8, 4, -36, -72, -12, 120, 180, -330, 132] $$ , and appear to be given by A111785. Interestingly, the powers of the exponentials also appear to have a sequence, the terms go like A074139, which is titled "Number of divisors of A036035(n).". A036035 is titled "Least integer of each prime signature, in graded (reflected or not) colexicographic order of exponents."

We can recreate a coefficient by performing the Mellin transform and dividing through by $\Gamma(s)$ $$ a_k(s) = \frac{1}{\Gamma(s)}\mathcal{M}[e_k(x)](s) = \frac{1}{\Gamma(s)}\int_0^\infty x^{s-1}e_k(x) \; dx $$ Then we can write $$ e_k(x) = \sum_{l=1}^{P(k-1)} \alpha_{kl}e^{-\beta_{kl} x} $$ where $\beta_k$ are the anaolgous rows of A074139 $$ \beta_1=[1]\\ \beta_2=[2]\\ \beta_3=[3, 4]\\ \beta_4=[4, 6, 8]\\ \beta_5=[5, 8, 9, 12, 16]\\ \beta_6=[6, 10, 12, 16, 18, 24, 32]\\ $$ we we know then the Mellin transform of $a e^{-bx}=a b^{-s}\Gamma(s)$, and it's as simple as, grouping the terms which are now understood $$ a_k(s) = \frac{1}{\Gamma(s)}\int_0^\infty x^{s-1}\sum_{l=1}^{P(k-1)} \alpha_{kl}e^{-\beta_{kl} x} \; dx $$ $$ a_k(s) = \sum_{l=1}^{P(k-1)}\frac{1}{\Gamma(s)}\int_0^\infty x^{s-1} \alpha_{kl}e^{-\beta_{kl} x} \; dx $$ $$ a_k(s) = \sum_{l=1}^{P(k-1)}\frac{1}{\Gamma(s)}\alpha_{kl}\beta_{kl}^{-s}\Gamma(s) $$ $$ a_k(s) = \sum_{l=1}^{P(k-1)}\alpha_{kl}\beta_{kl}^{-s} $$ $$ \mathrm{Li}^{-1}_s(z) = \sum_{k=1}^\infty a_k(s)z^{k} = \sum_{k=1}^\infty \sum_{l=1}^{P(k-1)}\alpha_{kl}\beta_{kl}^{-s}z^k $$ the limitation here is understanding where the terms in the sequences referenced come from, and finding whether any tractable forms exist for $\alpha$ and $\beta$. We already know that $\beta_{kl} = \sigma_0(\gamma_{kl})$ for the $\gamma$ in A036035, and divisor counting function $\sigma$. $$ \mathrm{Li}^{-1}_s(z) = \sum_{k=1}^\infty \sum_{l=1}^{P(k-1)}\frac{\alpha_{kl}z^k}{\sigma^s(\gamma_{kl})} $$

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    $\begingroup$ This answer was truly great. Despite the limitation mentioned in the end, this certainly gives much more than I've found anywhere else, and it's more in depth than any other source I've seen so far. Thank you so much, and sorry for the late response to this answer. $\endgroup$ – 46andpi Jul 9 '18 at 0:28
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    $\begingroup$ @46andpi I'm glad it was useful, I have looked a bit deeper and there seem to be some nice functions defined if $\sigma$ is replaced with other functions such as $\lambda$ and $\Omega$ the Liouville lambda and prime Omega functions respectively. Also if the $\gamma$ coefficients are raised to the $n^{th}$ power in the last equation, then that seems to describe the inverse of $\frac{z}{n^s} \Phi(z,s,\frac{1}{n})$, with $\Phi$ the (Hurwitz) generalisation of the Lerch transcendent. $\endgroup$ – Benedict W. J. Irwin Jul 9 '18 at 10:50

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