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$\DeclareMathOperator\elim{lim}\DeclareMathOperator\Lim{Lim}\DeclareMathOperator\lmb{lmb}\DeclareMathOperator\Lmb{Lmb}\DeclareMathOperator\mts{mts}$There are two similar functions; they determine the dependence of the values ​​of similar equations

\begin{gather*} x = L e^{s x} \iff x = \elim_s(L) \\ x = e^{L x^s} \iff x = \lmb_s(L). \end{gather*}

For positive parameters, in the region of convergence of their power series, both have two positive roots. x = L e^x

x = e^{L x}

Through power series we can only obtain zero roots of these functions \begin{gather*} \Lim_{s, 0}(L) = L + 2s\frac{L^2}{2!} + (3s)^2\frac{L^3}{3!} + (4s)^3\frac{L^4}{4!} + \dotsb \\ \Lmb_{s, 0}(L) = 1 + L + (1 + 2s)\frac{L^2}{2!} + (1 + 3s)^2\frac{L^3}{3!} + (1 + 4s)^3\frac{L^4}{4!} + \dotsb. \end{gather*}

We need to find an analytical method, a formula for obtaining the first root of zero. It is known that the dependence of these roots is the same for both functions: $$ \frac{\Lim_{s, 1}(L)}{\Lim_{s, 0}(L)} = \left(\frac{\Lmb_{s, 1}(L)}{\Lmb_{s, 0}(L)}\right)^s. $$ lim is any root. Lim is a specific root, it is indicated by the number 0 or 1 after the parameter "s" I tried this: let's take special cases $s=1$, and $L=.01$ and $L=.1$.

\begin{gather*} a=\Lim_{1, 0}(.1) = 0.1118325591589629648 \\ b=\Lim_{1, 1}(.1) = 3.5771520639572972184 \\ b/a=31.9866780377671590 \\ A=\Lmb_{1, 0}(.01) = 0.01010152719853875327 \\ B=\Lmb_{1, 1}(.01) = 6.47277512439400469474 \\ B/A=640.7719344982143479. \end{gather*} It is very similar that in the formula that shows the dependence of the first root on the zero root, the natural logarithm of $1/L$ divided by $L$ is used \begin{gather*} 10\ln10=23.02585092994 \\ 100\ln100=460.5170185988091368 \end{gather*} There are still some factors and/or terms missing. Perhaps there is a method similar to different roots of the logarithm $$Ln_{1}(L)=Ln_{0}(L)+2pii=ln|L|+i(arg(L)+2pi)$$

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    $\begingroup$ root-systems is about something else entirely, so I have removed it. It's not quite clear to me what you're asking, starting with the fact that you seem to be defining a function of a real variable called "$\lim$" (?), and then introduce capitalised versions of your one-subscript functions with two subscripts, and an undefined 3-subscript function "mts", so I didn't venture further tagging beyond real-analysis, but please at least add a top-level tag. \\ Please use TeX, not images, for your equations. I have transcribed your images; please check if it represents what you intended. $\endgroup$
    – LSpice
    Commented May 19 at 14:59
  • $\begingroup$ lim is any root. Lim is a specific root, it is indicated by the number 0 or 1 after the parameter "s" $\endgroup$
    – Luke
    Commented May 19 at 16:45
  • $\begingroup$ In my first edit, I incorrectly transcribed the definitions of $b$ and $A$ according to your original version; my apologies. I notice that you fixed the subscripts, but the actual functions applied still did not match. I edited them again to match what you had in your original version. I hope that this was correct. $\endgroup$
    – LSpice
    Commented May 19 at 16:55
  • $\begingroup$ Thank you. Everything is correct now. But why doesn't anyone give advice? How can one find first root $$Lim_{s,1}(L)$$ via zero root $$Lim_{s,0}(L)$$ $\endgroup$
    – Luke
    Commented May 19 at 17:36
  • $\begingroup$ Re, for myself, it is because I don't know the answer, or even enough how to improve the question. However, the downvotes indicate that some users find this question "does not show any research effort; it is unclear or not useful." While no-one is obliged to leave a comment indicating the reason for their downvote, perhaps you could see if you can show your research effort, and make your question or its motivation clearer. $\endgroup$
    – LSpice
    Commented May 19 at 17:44

1 Answer 1

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You seek the two real solutions $x_0(L,s)$ and $x_1(L,s)$ of the transcendental equation $$x=Le^{sx},\;\;x,L,s>0.$$ These can be expressed in terms of the Lambert W function, $$x_0(L,s)=-\frac{W_0(-L s)}{s},\;\; x_1(L,s)=-\frac{W_{-1}(-L s)}{s}.$$ The expressions cannot be simplified in terms of elementary functions.

Similarly, the two real solutions $y_0(L,s)$ and $y_1(L,s)$ of the equation $x=e^{Lx^s}$ are given by $$y_0(L,s)=\left(-\frac{W_0(-L s)}{L s}\right)^{1/s},\; y_1(L,s)=\left(-\frac{W_{-1}(-L s)}{L s}\right)^{1/s},$$ hence the solutions $x_n$ and $y_n$ are related by $$y_n(L,s)=(x_n(L,s)/L)^{1/s}.$$

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  • $\begingroup$ This is right. Thank you. But I need to know the formula to find x1 from x0. Ln has a similar formula. $\endgroup$
    – Luke
    Commented May 19 at 18:46
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    $\begingroup$ the formula you are asking involves a ratio of Lambert W-functions: $$x_1=x_0 \frac{W_{-1}(-Ls)}{W_0(-Ls)}.$$ This cannot be simplified in terms of elementary functions. $\endgroup$
    – Carlo Beenakker
    Commented May 19 at 18:49
  • $\begingroup$ So no one knows how to find $$W_{-1}$$ through $$W_{0}$$? $\endgroup$
    – Luke
    Commented May 19 at 19:10
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    $\begingroup$ the two functions are related in a transcendental way, by the equation $W_0e^{W_0}=W_{-1}e^{W_{-1}}$; there is no relation in terms of elementary functions. $\endgroup$
    – Carlo Beenakker
    Commented May 19 at 19:17
  • $\begingroup$ Thank you, Carlo Beenakker! You understood everything correctly. It's a pity that I don't know English. $\endgroup$
    – Luke
    Commented May 19 at 19:30

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