# Inverting $\frac{\xi}{2}(1+\tanh(\xi))=\lambda$ using the Lagrange-Burmann Theorem

For my quantum mechanics homework, I developed the transcendental equation $$\frac{\xi}{2}(1+\tanh(\xi))$$ for the well-posedness of symmetric potential formed from two delta functions. The professor encourages us to use a numerical tool to solve the equation $$\frac{\xi}{2}(1+\tanh(\xi))=\lambda$$ for $$\xi(\lambda)$$; however, I was curious if the Lagrange inversion theorem could be employed instead.

Taking $$B_n$$ to represent a Bernoulli number, we have

$$\begin{eqnarray} \tanh x &=& x - \frac {x^3} {3} + \frac {2x^5} {15} - \frac {17x^7} {315} + \cdots = \sum_{n=1}^\infty \frac{2^{2n}(2^{2n}-1)B_{2n} x^{2n-1}}{(2n)!}, \left |x \right | < \frac {\pi} {2} \\ \end{eqnarray}$$

could I pose the transcendental equation as

$$\frac{\xi}{2} + \sum_{n=1}^\infty \frac{2^{2n-1}(2^{2n}-1)B_{2n} \xi^{2n}}{(2n)!} = \lambda$$

Excerpting from Wikipedia, if $$g$$ shall be the inverse of $$f$$, where $$f$$ is given by a formal power series as $$f(w) = \sum_{k=0}^{\infty}f_k\frac{w^k}{k!}$$ $$g(z) = \sum_{k=0}^{\infty}g_k\frac{z^k}{k!}$$ then $$g_1=\frac{1}{f_1}$$ and $$g_n = \frac{1}{f_1^n}\sum_{k=1}^{n-1}(-1)^k n^{(k)} \mathcal B_{n-1,k}\left(\frac{f_2}{2f_1},\frac{f_3}{3 f_1},\dots,\frac{f_{n-k+1}}{(n-k+1)f_1}\right)$$ where $$n^{(k)}$$ is the rising factorial and $$\mathcal B$$ is a Bell polynomial.

Question: Are there any further simplifications that I can use? Currently, the presence of Bell polynomials seems discouraging as the performance of this algorithm. Also, can I work around the stipulation $$|\xi|<\frac{\pi}{2}$$? There are some solutions that exist outside of that regime for sufficiently high $$\lambda$$. Could I partition the function $$\frac{\xi}{2}(1+\tanh(\xi))$$ into intervals of length $$\frac{\pi}{2}$$ and apply the result to each of them?

• @Tyma Gaidash Good point, that would remove the need for the $2^{n-1}$ factor in the series solution. Do you know of any combinatorial identities for Bell polynomials? Jan 3, 2023 at 2:26

Too long for a comment.

You can write the equation as you did and use the standard series reversion to have $$\xi=\sum_{n=1}^\infty {a_n}\,\lambda^n$$ The problem is that the coefficients are almost exploding $$\left\{2,-4,16,-\frac{224}{3},384,-\frac{10496}{5},\frac{538624}{45 },-\frac{22171648}{315},\frac{26697728}{63},\cdots\right\}$$

Much better would be to use some $$[n+1,n]$$ Padé approximant $$P_n$$. The simplest would be $$P_2=\frac {\xi \left(\xi ^2+3 \xi +3\right) } {2 \left(\xi ^2+3\right) }$$ Just to give an idea $$\Phi_2=\int_0^{\frac \pi 2} \Big[\frac{1}{2} \xi (1+\tanh (\xi ))-P_2\Big]^2\, d\xi\sim \pi \times 10^{-4}$$

So, a good approximation will be obtained solbing the cubic equation $$\xi ^3+(3-2 \lambda ) \xi ^2+3 \xi -6 \lambda=0$$ Then, the first estimate $$\xi_0=\frac{2 \lambda -3}{3}+\frac{4}{3} \sqrt{\lambda(3-\lambda ) } \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{16 \lambda ^3-72 \lambda ^2+216 \lambda +27}{16 (\lambda(3-\lambda ) )^{3/2}}\right)\right)$$ To polish the root, perform one single iteration of Newton method $$\left( \begin{array}{cccc} \lambda & \xi_0 & \xi_1 & \text{solution} \\ 0.1 & 0.17103 & 0.17103 & 0.17103 \\ 0.2 & 0.30803 & 0.30802 & 0.30802 \\ 0.3 & 0.42764 & 0.42757 & 0.42757\\ 0.4 & 0.53696 & 0.53673 & 0.53673 \\ 0.5 & 0.63980 & 0.63923 & 0.63923 \\ 0.6 & 0.73851 & 0.73732 & 0.73732 \\ 0.7 & 0.83465 & 0.83245 & 0.83245 \\ 0.8 & 0.92936 & 0.92563 & 0.92563 \\ 0.9 & 1.02353 & 1.01759 & 1.01759 \\ 1.0 & 1.11785 & 1.10886 & 1.10886 \\ 1.1 & 1.21291 & 1.19982 & 1.19982 \\ 1.2 & 1.30920 & 1.29078 & 1.29079 \\ 1.3 & 1.40718 & 1.38194 & 1.38196 \\ 1.4 & 1.50722 & 1.47346 & 1.47350 \\ 1.5 & 1.60970 & 1.56545 & 1.56551 \\ \end{array} \right)$$

You have made me sixty years younger since this was part of my thesis work.

Edit

I do not know how she did but my wife found a copy of my thesis work. In fact, I also proposed a better approximation, namely $$\frac{\xi}{2}(1+\tanh(\xi))\sim \xi\,\,\frac{\frac{185}{352}+\frac{124 }{339}\xi+\frac{48 }{89}\xi ^2 } {1+\frac{154 }{247}\xi^2 }$$ which gives a norm equal to $$8.68\times 10^{-7}$$ ($$360$$ times smaller than the previous one).

For $$\lambda=1.5$$, this gives $$\xi_0=1.56682$$.

• A charming backstory (+1) Jan 3, 2023 at 14:32
• @FShrike. Thanks ! Being blind and messy and so old, I don't know how I could have been able to find it. Moreover, I did not remember my last approximation. Cheers ;-) Jan 3, 2023 at 14:37

We can re-write the equation as $$\lambda = \frac{\xi }{{1 + {\rm e}^{ - 2\xi } }},$$ i.e., $$- 2\lambda = \frac{{ - 2\xi }}{{1 + {\rm e}^{ - 2\xi } }}.$$ Consequently, $$2\lambda {\rm e}^{ - 2\lambda } = \frac{{2\xi }}{{1 + {\rm e}^{ - 2\xi } }}\exp \left( {\frac{{ - 2\xi }}{{1 + {\rm e}^{ - 2\xi } }}} \right)$$ or $$2\lambda {\rm e}^{ - 2\lambda } = \frac{{2\xi }}{{{\rm e}^{2\xi } + 1}}\exp \left( {\frac{{2\xi }}{{{\rm e}^{2\xi } + 1}}} \right).$$ Hence, in terms of the Lambert $$W$$-function, $$\frac{{2\xi }}{{{\rm e}^{2\xi } + 1}}=W(2\lambda {\rm e}^{ - 2\lambda } ).$$ But from the original equation, $$\frac{{2\xi }}{{{\rm e}^{2\xi } + 1}} = 2(\xi - \lambda ).$$ Thus, $$\boxed{ \xi = \lambda + \frac{1}{2}W(2\lambda {\rm e}^{ - 2\lambda } ).}$$ For example, if $$|2\lambda {\rm e}^{ - 2\lambda }|<\frac{1}{\mathrm{e}}$$ (which includes $$\lambda \ge 0$$), then $$\xi = \lambda + \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{( - n)^{n - 1} }}{{n!}}(2\lambda {\rm e}^{ - 2\lambda } )^n } .$$ Indeed, this follows by taking the principal brach of $$W$$ and using its standard Maclaurin series. Alternatively, by $$\mathrm{A}038049$$, $$\xi = - \frac{1}{2}\sum\limits_{n = 1}^\infty {\left( {\sum\limits_{k = 0}^n {\binom{n}{k}k^{n - 1} } } \right)\frac{{( - 2\lambda )^n }}{{n!}}}$$ provided $$|\lambda| < \frac{1}{2}W\!\left( {\frac{1}{{\rm e}}} \right) = 0.13923227 \ldots$$ ($$W$$ being the principal branch).

• The first step for the other answers should have been to check if Lambert W solved the equation. You were the first to notice it. Well done (+1) Jan 3, 2023 at 15:19
• This is more than elegant. Why didn't you post it 60 years ago ? Thanks Jan 3, 2023 at 15:22
• @ClaudeLeibovici Thanks. My father was 1 year old 60 years ago. :D
– Gary
Jan 3, 2023 at 15:26
• I do not see the problem here. Aren't you able to travel in time ? Jan 3, 2023 at 15:43
• Let's the closed-form functions include Lambert W. Because representations for the closed-form functions are known, we get hints i.a. for its series representations.
– IV_
Jan 3, 2023 at 15:59

Here is a later answer using Lagrange reversion.You can simplify the equation into:

$$y(\tanh(y)+1)=z\implies y=\sum_{n=1}^\infty \frac{z^n}{n!}\left.\frac{d^{n-1}}{dt^{n-1}}(\tanh(t)+1)^{-n}\right|_{t=0},z=2\lambda$$

The result requires finding $$n$$th derivatives. Rearranging and applying the binomial theorem:

$$\left.\frac{d^{n-1}}{dt^{n-1}}(\tanh(t)+1)^{-n}\right|_{t=0}=2^{-n}\left.\frac{d^{n-1}}{dt^{n-1}}(e^{-2t}+1)^n\right|_{t=0}=2^{-n}\sum_{k=0}^n\binom nk\left.\frac{d^{n-1}}{dt^{n-1}}e^{-2kt}\right|_{t=0}$$

After removing the $$n=1$$ term for no convergence problems. A simplification and radius of convergence is due to @Gary:

$$\boxed{y(\tanh(y)+1)=z\implies y=z-\frac12 \sum_{n=2}^\infty\sum_{k=0}^n\frac{k^{n-1}(-z)^n}{k!(n-k)!}}$$

Shown here without Bell polynomials. Additionally, $$k$$’s upper bound can go up to $$\infty$$ with interchangeable sums in that case.. The radius of convergence uses Lambert W$$(z)$$: $$|z|<\text W\left(\frac 1e\right),z\in\Bbb C$$.

Switching the sums and limiting the index uses the lower regularized gamma function P$$(a,z)$$:

$$y=z-\frac12\lim_{c\to 0}\sum_{n=c,1+c,2+c,\dots}\frac{(-n z)^n\text P(2-n,-nz)}{e^{nz}nn!}$$

Shown here

• Note that the final result may be written $$y = - \frac{1}{2}\sum\limits_{n =1}^\infty {a_n \frac{{( - z)^n }}{{n!}}}$$ with $$a_n = \sum\limits_{k = 0}^n {\binom{n}{k}k^{n - 1} } .$$ This is A038049 in the OEIS.
– Gary
Jan 3, 2023 at 14:13
• From the asymptotics of the $a_n$, the region of convergence is $\left| z \right| < W(1/{\rm e}) = 0.27846454 \ldots$.
– Gary
Jan 3, 2023 at 14:20
• @Gary Thanks. Your information was added to the post. Jan 3, 2023 at 14:20