5
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ @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? $\endgroup$
    – Talmsmen
    Jan 3, 2023 at 2:26

3 Answers 3

4
$\begingroup$

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 $\xi_0=1.56682$.

$\endgroup$
2
  • 3
    $\begingroup$ A charming backstory (+1) $\endgroup$
    – FShrike
    Jan 3, 2023 at 14:32
  • 1
    $\begingroup$ @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 ;-) $\endgroup$ Jan 3, 2023 at 14:37
4
$\begingroup$

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).

$\endgroup$
5
  • 2
    $\begingroup$ 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) $\endgroup$ Jan 3, 2023 at 15:19
  • $\begingroup$ This is more than elegant. Why didn't you post it 60 years ago ? Thanks $\endgroup$ Jan 3, 2023 at 15:22
  • $\begingroup$ @ClaudeLeibovici Thanks. My father was 1 year old 60 years ago. :D $\endgroup$
    – Gary
    Jan 3, 2023 at 15:26
  • 1
    $\begingroup$ I do not see the problem here. Aren't you able to travel in time ? $\endgroup$ Jan 3, 2023 at 15:43
  • $\begingroup$ 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. $\endgroup$
    – IV_
    Jan 3, 2023 at 15:59
3
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – Gary
    Jan 3, 2023 at 14:13
  • $\begingroup$ From the asymptotics of the $a_n$, the region of convergence is $ \left| z \right| < W(1/{\rm e}) = 0.27846454 \ldots$. $\endgroup$
    – Gary
    Jan 3, 2023 at 14:20
  • $\begingroup$ @Gary Thanks. Your information was added to the post. $\endgroup$ Jan 3, 2023 at 14:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .