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For the computation of the lower (attracting) fixpoint $L_0$ for $b^{L_0} = L_0 $ (where the base $b$ is in the range for convergence $1 \lt b \lt e^{1/e}$) there is the simple formula $$L_0 = \exp(-W(-\log(b))) \qquad \qquad \text{where $W$ is the Lambert-W-function}$$ or if we directly use the logs for $b$ and $L$ writing $\beta := \log(b)$ and $\lambda := \log(L)$ $$\lambda_0 = -W(-\beta) \phantom{\qquad \qquad \text{where $W$ is the Lambert-W-function}}$$ (This has also the nice property that $\beta = \lambda_0 / L_0 $ or $b=L_0^{1/L_0}$.)

I think there is a related formula for the upper (repelling) fixpoint $L_1$ too, but I can't remember it from the top of my head (some simple variations didn't work so far). Before beginning a long search in older texts and program-sources (the searchkey were much un-sharp) - could someone refresh my memory? A reference would suffice...

Example: let $b=\sqrt 2$ then $L_0 = 2$ because $b^2 = 2$ and $L_1=4$ because $b^4 = 4$

Remark: I know how to approximate $L_1$ using the Newton-algorithm - I specifically think to remember that there is some variant with the Lambert W (which is what I want here)

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Yes. If you use the other real-valued branch of $W$, denoted $W_{-1}$, you will get this fixed point:

$$\lambda_U = -W_{-1}(-\beta)$$

and

$$L_U = \exp(-W_{-1}(-\log(b)))$$.

But I do not know if or believe that $W_{-1}$ can be expressed through $W = W_0$, the principal branch (unlike simpler multivalued relations like $\sqrt{...}$ and $\log$ in the complex plane.).

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  • $\begingroup$ Ahh, thank you. I had tried $W_1()$ instead... I think now, there was even a routine to compute this easily in the "Mount Meru" of information in tetration-forum. I'll see... $\endgroup$ – Gottfried Helms Oct 7 '17 at 7:31
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Another solution which does not involve the LambertW function involves generating the following series inverse, which I will explain in steps. $$f=\sqrt{2(\exp(x)-x-1)};\;\;\;\;\text{xfixed}=f^{-1}(x)$$ $$\text{xfixed}= x - \frac{x^2}{6}+ \frac{x^3}{36} - \frac{x^4}{270} + \frac{x^5}{4320} + \frac{x^6}{17010} - \frac{139x^7}{5443200} ...$$ In pari-gp, this series would be "serreverse(sqrt((exp(x)-x-1)*2))". I don't have a closed form for the series. By plugging both square roots into this series, you can get either fixed point L for the base=b.

$$L_b = \frac{\text{xfixed}\left(\pm\sqrt{-2(\ln(\ln(b))+1)}\right)-\ln(\ln(b))}{\ln(b)}$$

One interesting thing about the square root radical is that if $b=\exp(1/e)$, then the radical is zero, which makes sense since $b=\exp(1/e)$ has only one fixed point. Using a 16 term series for xfixed, I got about 19 decimal digits of precision for the two fixed points of $b=\sqrt{2}$ So, why does this work? We must first show that iterating $$z \mapsto b^z$$ is exactly congruent with iterating $$y \mapsto \exp(y)+k;\;\;k=\ln(\ln(b));\;\;\;y=z\ln(b)+\ln(\ln(b))$$

So then our fixed point problem for $z \mapsto b^z$ becomes equivalent to finding the fixed point for $y \mapsto \exp(y)+k$. If we have the fixed point for y, then we can get the fixed point for $z \mapsto b^z$ by using this equation based on the congruency. $$z = \frac{y-\ln(ln(b))}{\ln(b)}$$

Next, we do some algebra to get the fixed point for $$y = \exp(y)+k;\;\;\;k=\ln\ln(b))$$ $$y = k + 1 + y + \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{24} + \frac{y^5}{5!} ...$$ $$ -(k+1) = \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{4!} ...$$ $$ -2(k+1) = y^2 + \frac{2y^3}{6} + \frac{2y^4}{24} + \frac{2y^5}{5!} ...$$ Now we take the square root of both sides. $$ \sqrt{-2(k+1)} = f(y) = y\cdot \sqrt{1 + \frac{2y}{6} + \frac{2y^2}{24} + \frac{2y^3}{5!} ...}$$ $$ \sqrt{-2(k+1)} = f(y) = y + \frac{y^2}{6} + \frac{y^3}{36} + \frac{y^4}{270} + \frac{y^5}{2592} + \frac{17y^6}{544320} + \frac{11y^7}{5443200}+...$$

The square root term on the right has a nice defined formal Taylor series in y; which leads exactly to the $f$ in the 2nd line of my post. So then we can get k from y. We need the reverse, so we take the inverse of Taylor series for f. $$ y = f^{-1}\left(\pm\sqrt{-2(k+1)}\right)=f^{-1}\left(\pm\sqrt{-2(\ln(ln(b))+1)}\right)$$

Now we have the fixed point $y=\exp(y)+k$, and the final step is to generate the fixed point $z=b^z$ from y using $z = \frac{y-\ln(ln(b))}{\ln(b)}$

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  • $\begingroup$ A miraculous method, indeed... $\endgroup$ – Gottfried Helms Oct 8 '17 at 21:51
  • $\begingroup$ Thanks Gottfried :) This equation also works for both real and complex bases for b. $\endgroup$ – Sheldon L Oct 8 '17 at 21:53

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