3

The inspiration for the question is

Closed form of $$\frac{d}{dk}\text W_k(z)$$ Derivative of W-Lambert function with respect to its branch cuts experiment.

I also like making functions central to my work. So I though about “centralizing” the Generalized W-Lambert function so that the branch cut is the same as the argument:

$$\text W_k(z)=\text W_x(x)$$

I then considered what the value of $$\text W_{\pm \infty}(\pm \infty)$$ would be. Assume that the signs can be chosen in any order. I found a conjectured closed form. For simplicity, let’s consider the convergent real part of the expression. Let’s also further constrain our problem by choosing signs to be

$$\text{Re}(\text W_{ \infty}( \infty))$$

$$\lim_{x\to\pm \infty}\text {Re}(\text W_x(x))\mathop=^{\large ?} -\ln(2\pi)$$

The imaginary part turn out to be asymptotic to $$\pm 2\pi i \infty=\pm\infty i$$ so it is just an infinite complex number.

I found the possible result after using a discrete limit. For example, the value at $x=10^{10}$ is the following. I used the the real an imaginary part as well as different signs for completion.

$$\text W_x(x)= \text W_x(-x)= \text W_{x}(\pm ix) =-1.8378770663843454835603092354803385245074740939... + 6.2831853070225068442428720324366974719089270342... × 10^{10} i, \text W_{-x}(-x)= \text W_{-x}(x)=\text W_{-x}(\pm ix)=-1.8378770663343454835578092354801887222706941730... - 6.2831853067083475788838927085903664574425628653... × 10^{10} i $$

Note the $2\pi$ approximation in the scientific notation if the imaginary part. As said before, the real part is approximately:

$$-\ln(2\pi)= -1.8378770663843454835603092354803385245074740939... $$

Here are the Inverse Symbolic Calculator results.

How can you formally evaluate $$\text{Re}\lim_{x\to \infty} \text W_x(x)$$ using a discrete limit? Please correct me and give me feedback!

Just for fun, the Wright Omega function can also have the interesting identity that:

$$ω(z)\mathop=^\text{def} \text W_{\left \lceil\frac{\text{Im}(z)}{2\pi}-\frac12\right\rceil}\left(e^z\right)\implies -\ln(2\pi)=\lim_{x\to\infty} ω(\ln(x)+2i\pi x)$$

Where $$\left\lceil\frac{\text{Im}(z)}{2\pi}-\frac12\right\rceil =\frac{i\left(\ln\left(e^z\right)-z\right)}{2\pi}=\text{unwindK(z)}=\text K(z)$$

is called the Unwinding Number

Тyma Gaidash
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  • I have a question, does Tannery's Theorem work for complex function? – Rounak Sarkar Oct 07 '21 at 16:21
  • @RounakSarkar It is based on inequalities which are not well defined for complex numbers, but you can just try to switch the limit and summations to see what happens by each term. – Тyma Gaidash Oct 07 '21 at 17:16
  • See the edit in my answer. – Rounak Sarkar Oct 07 '21 at 17:43
  • @RounakSarkar There is also this useful integral representation, I tried splitting the limits and the integral itself seems to go to 0, but I cannot get the right answer. I saw the edit; we still need to know if it applies. – Тyma Gaidash Oct 07 '21 at 17:43
  • We don't need to define the Tannery's theorem for complex functions, we just need to define a real sequence which would be greater than or equal to the summand. Do you have any idea? And have you made any progress regarding that integral representation? – Rounak Sarkar Oct 08 '21 at 04:33
  • We don't actually need the Tannery's theorem if we can show that the real part of the derivative of the summand with respect to $x$ is positive for large $x$ using the theorem from this post, I have filled out all the necessary details, but If that's not possible then the only hope I see is to show that the Tannery's theorem applies, we don't need to worry about complex numbers because the real part of them is a real number. What do you say? – Rounak Sarkar Oct 08 '21 at 09:48
  • Another question. Notice that in the post I linked above, it is pointed out that the interchange between limit and sum works of the sequence is non-decreasing, since you want the discrete limit, can I just replace the derivative with $\Delta f(n,x)\geq 0$ and since for large $x$, $f(n,x+1)$ and $f(n,x)$ meaning $\lim_{x\rightarrow\infty}(f(n,x+1)-f(n,x))=0\geq 0$ and conclude that the theorem applies? If the answer is yes then my answer is complete. – Rounak Sarkar Oct 08 '21 at 13:10
  • I may have completed the answer. Please see it, and tell me if there is anything wrong. – Rounak Sarkar Oct 08 '21 at 15:07
  • @RounakSarkar Good job. I just wanted a discrete limit because the non-integer branches are not well defined, they are actually just not defined. – Тyma Gaidash Oct 08 '21 at 15:58
  • I think you asked a question involving the inverse beta and gamma regularized functions. I can't find it. Can you give me the link? – Rounak Sarkar Oct 17 '21 at 08:14
  • @RounakSarkar This is the main post. Again, the beta function on its own was closed, but the other two are still open and doing well, so I put it in the bolded link. There a couple others. Thanks again. – Тyma Gaidash Oct 17 '21 at 11:33
  • The bolded link is the main one, but these are all optional of course. – Тyma Gaidash Oct 17 '21 at 11:37

2 Answers2

3

$\underline{\text{A Reformulation}:-}$

I think we can use one of the asymptotic series expansion given in wolfram alpha:

$\textstyle\displaystyle{\Re W_x(x)=\Re\left(2x\pi i+\ln(x)-\ln(2x\pi i+\ln(x))-\sum_{n=1}^{\infty}(2x\pi i+\ln(x))^{-n}\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\ln^m(2x\pi i+\ln(x))\right)}$

For, $\ln(2x\pi i+\ln(x))$ we can write the expression inside the logarithm in the polar form:

$\ln(re^{i\theta})$ where $r=\sqrt{4x^2\pi^2+\ln^2(x)}$ and since $x>0$ we have $\theta=\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)$

So, $\ln(2x\pi i+\ln(x))=\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)$

And, $\textstyle\displaystyle{(2x\pi i+\ln(x))^{-n}=\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n}$

So, $\textstyle\displaystyle{\Re W_x(x)=\ln(x)-\frac{1}{2}\ln(4x^2\pi^2-\ln^2(x))-\Re\left(\sum_{n=1}^{\infty}\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\right)}$

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}\Re W_x(x)=\ln\left(\lim_{x\rightarrow\infty}\frac{x}{\sqrt{4x^2\pi^2-\ln^2(x)}}\right)-\lim_{x\rightarrow\infty}\left(\Re\left(\sum_{n=1}^{\infty}\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\right)\right)}$

Let's do the limit inside the natural log first-

Let, $\textstyle\displaystyle{L=\lim_{x\rightarrow\infty}\frac{x}{\sqrt{4x^2\pi^2+\ln^2(x)}}}$

$\textstyle\displaystyle{L^2=\lim_{x\rightarrow\infty}\frac{x^2}{4x^2\pi^2+\ln^2(x)}}$

$\textstyle\displaystyle{\overset{\text{L.H.}\frac{\infty}{\infty}}{=}\lim_{x\rightarrow\infty}\frac{2x}{8x\pi^2+\frac{2}{x}\ln(x)}}$

$\textstyle\displaystyle{=\lim_{x\rightarrow\infty}\frac{x^2}{4x^2\pi^2+\ln(x)}}$

$\textstyle\displaystyle{\overset{\text{L.H.}\frac{\infty}{\infty}}{=}\lim_{x\rightarrow\infty}\frac{2x}{8x\pi^2+\frac{1}{x}}}$

$\textstyle\displaystyle{=\lim_{x\rightarrow\infty}\frac{2x}{8x\pi^2}}$

$\textstyle\displaystyle{=\frac{1}{2\pi}}$

$\implies\ln(L)=-\ln(2\pi)$

Which leaves us with the conclusion that your conjecture will be true if:

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}\left(\sum_{n=1}^{\infty}\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right)=0}$

$\underline{\text{The Proof}:-}$

From this post we have that-

Assume $\sum_m f(m,1)>-\infty$. If for each $m$ we have $f(m,n)\le f(m,n')$ whenever $n\le n'$, then $$\lim_n\sum_{m=1}^\infty f(m,n) = \sum_{m=1}^\infty\lim_n f(m,n).\tag1$$

Define, $\textstyle\displaystyle{f(n,x):=\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]}$

To apply the theorem we see that the function is not defined at $1$ because of the $\ln(x)$ in the denominator inside the inverse tangent, but we can define it in terms of a limit-

$f(n,1):=\lim_{x\rightarrow 1}f(n,x)$

$\textstyle\displaystyle{=\Re\lim_{x\rightarrow 1}\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]}$

$\textstyle\displaystyle{=\Re\left[\lim_{x\rightarrow 1}\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\bigg]\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\lim_{x\rightarrow 1}\bigg[\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right]}$

$\textstyle\displaystyle{=\Re\left[0×\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\ln(2\pi)+i\frac{\pi}{2}\right)^m\right]}$

$\implies f(n,1)=0$

$\implies\sum_{n}f(n,1)=0>-\infty$

For the next condition I am not sure is the following reasoning correct or not, I would request the readers to correct me if I am wrong:

Since we are taking the discrete limit we have

$f(n,x)\leq f(n,x+1)$

$f(n,x+1)-f(n,x)\geq 0$

Since $x\rightarrow\infty$, we have-

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}(f(n,x+1)-f(n,x))\geq 0}$

Notice that $f(n,x+1)\sim f(n,x)$ as $x\rightarrow\infty$, so

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}(f(n,x+1)-f(n,x))\geq 0}$

$0\geq 0$

Which is trivially true.

So we can apply the theorem meaning that we can interchange the sum and the limit giving us-

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}\left(\sum_{n=1}^{\infty}\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right)}$

$\textstyle\displaystyle{=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\lim_{x\rightarrow\infty}\left(\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right)}$

Notice that $n\geq m$, so for $m=n$ we see that the limit is $0$ from Wolfram Alpha. And for $n>m$ we can write $n=m+k$ so we have-

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}\left(\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right)}$

$\textstyle\displaystyle{=\Re\bigg[\left(\lim_{x\rightarrow\infty}\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^k\right)\left(\lim_{x\rightarrow\infty}\left(\frac{\ln(x)+2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^m\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\right)\bigg]}$

Which is $0$ by this and this.

Which proves our statement.

$\underline{\text{Conclusion}:-}$

$\boxed{\textstyle\displaystyle{\lim_{x\rightarrow\infty}\Re\operatorname{W}_x(x)=-\ln(2\pi)}}$

$\underline{\text{Comment}:-}$

This was a really interesting equality. I never expected the branch cuts of the Lambert W Function to be this interesting(have a connection with $\pi$), that's why I never really looked into the branches of the productlog. And remember-

Every Time You're Not Running $\pi$ Is Getting closer.

Rounak Sarkar
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    "the whole thing is getting really complicated" is a true understatement. Your idea is very interesting. I hope and wish you be able to prove it. Cheers and $\to +1$. – Claude Leibovici Oct 07 '21 at 11:05
  • @Claude Leibovici. Repeated application of the binomial theorem led me to something really complicated. I stopped before I reached this expression. I can easily tell you how complicated it was, it was an expression which was a sum of 64 infinite sums whose sizes were almost the same as the size of the sum in the limit. The imaginary parts will vanish and give us a sum of 32 infinite sums. This answer was a part of a much more bigger answer. But after realizing how big the expression would be I just stopped and came here to just post this small result. – Rounak Sarkar Oct 07 '21 at 11:18
  • I believe $$ W_k (z) = \log z + 2\pi ik - \log (\log z + 2\pi ik) + o(1) $$ as $z\to \infty$ uniformly in $k$. – Gary Oct 07 '21 at 11:21
  • @Gary. If that is true then sum in the limit will trivially follow. Do you have any idea about how to prove it? – Rounak Sarkar Oct 07 '21 at 11:26
  • Well, Corless et al. shows that $$ W(z) = \operatorname{Log}z - \log \operatorname{Log}z + \mathcal{O}!\left( {\frac{{\log \operatorname{Log}z}}{{\operatorname{Log}z}}} \right) $$ for large $z$ and conclude that this approximates $W_k$ if $\operatorname{Log}z$ is chosen to be $\log z +2\pi i k$ where $\log$ is the principal branch. – Gary Oct 07 '21 at 11:31
  • (+1) I am not sure how helpful it is, but $$\text W_{-1}\left(-\frac{\ln(2\pi)}{2\pi}\right)=-\ln(2\pi)$$ – Тyma Gaidash Oct 07 '21 at 12:20
  • @Gary. Wouldn't that make the term in ,$o\left(\frac{\ln(\ln(x)+2\pi ik)}{\ln(x)}\right)$, not $o(1)$, right? Sorry if I am wrong, I am not that fluent with the o notation – Rounak Sarkar Oct 07 '21 at 12:35
  • @RounakSarkar You are right. – Gary Oct 07 '21 at 22:32
  • @Claude Leibovici. I have completed the answer but I am still not so sure if my reasoning is correct. Especially in the place where I tried to prove that $f(n,x)$ in non-decreasing for large discrete $x$. Please see it and tell me if there is anything wrong and if it is wrong then if there are any other ways to overcome it. – Rounak Sarkar Oct 08 '21 at 14:35
  • @Gary. Hey Gary I have completed my answer. However I am still not sure if my reasoning is correct, can you please see it? – Rounak Sarkar Oct 08 '21 at 15:22
1

This is not an answer since just based on numerical calculations.

Using $x=10^k$, define $$R_k=10^{k+1} \left(1+\frac{\Re\left(W_{10^k}\left(10^k\right)\right)}{\log (2 \pi )}\right)$$ $$I_k=10^{k+1}\left(10^{k+1} \left(1-\frac{\Im\left(W_{10^k}\left(10^k\right)\right)}{ 2\pi 10^k}\right)-\frac{5}{2}\right)$$ Some results for $0 \leq k \leq 9$ (I have not been able to compute for $k=10$)

$$\left( \begin{array}{ccc} k & R_k & I_k \\ 0 & 1.65388508388988 & 5.36676559479282 \\ 1 & 1.38001939449002 & 4.70977390795459 \\ 2 & 1.36217751498421 & 4.66071573895105 \\ 3 & 1.36045571658805 & 4.65592771371741 \\ 4 & 1.36028414014521 & 4.65545005440653 \\ 5 & 1.36026698851516 & 4.65540229987385 \\ 6 & 1.36026527341227 & 4.65539752453454 \\ 7 & 1.36026510190259 & 4.65539704700174 \\ 8 & 1.36026508475162 & 4.65539699924848 \\ 9 & 1.36026508303653 & 4.65539699447315 \end{array} \right)$$

These numbers have not been identified by inverse symbolic calculators.

Claude Leibovici
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