Approximate solution of $\sinh ^{2 r}(z)+\cosh ^{2 r}(z)=a$

Question

Making this question more general : find the zero of function $$\color{blue}{f(x) = 1 + x^r - a(x-1)^r }\qquad \text{with}\quad a >2\quad \text{and}\quad r >2 $$ Its maximum value is attained at $$x_{\text{max}}=\frac{1}{1-a^{\frac{1}{1-r}}}$$ and the solution looked for is larger than $x_{\text{max}}$. Expanded as series for large values of $r$ $$x_{\text{max}}=\frac{r}{\log (a)}+\left(\frac{1}{2}-\frac{1}{\log (a)}\right)+\frac{\log (a)}{12r}+O\left(\frac{1}{r^2}\right)\quad \implies \quad x_0=\frac{r}{\log (a)}+\frac{1}{2}$$ Making the story short, let $x=\coth ^2(z)$ and search for the zero of $$\color{blue}{g(z)=\sinh ^{2 r}(z)+\cosh ^{2 r}(z)-a}\qquad \text{with}\qquad z_0=\tanh ^{-1}\left(\sqrt{\frac{2 \log (a)}{2r+\log (a)}}\right)$$

$$g(z_0)=\left(\frac{2 \log (a)}{2 r-\log (a)}\right)^r+\left(\frac{2r+\log (a)}{2 r-\log (a)}\right)^r-a$$ seems to be always positive; the same for $g'(z_0)$ and $g''(z_0)$ (not proved).

So, we know that $z_0$ is an overestimate of the solution and, by Darboux theorem, Newton method will converge without any overshoot.

For illustration, using $a=r$, some results $$\left( \begin{array}{cccc} r & z_0 & z_1 & \text{solution} \\ 2 & 0.609103 & 0.575619 & 0.573108 \\ 3 & 0.627536 & 0.616024 & 0.615680 \\ 4 & 0.609103 & 0.604249 & 0.604179 \\ 5 & 0.585371 & 0.582415 & 0.582386 \\ 6 & 0.562373 & 0.560133 & 0.560114 \\ 7 & 0.541375 & 0.539547 & 0.539532 \\ 8 & 0.522489 & 0.520955 & 0.520944 \\ 9 & 0.505525 & 0.504221 & 0.504212 \\ 10 & 0.490240 & 0.489118 & 0.489111 \\ 20 & 0.392248 & 0.391872 & 0.391871 \\ 30 & 0.340085 & 0.339900 & 0.339899 \\ 40 & 0.306130 & 0.306020 & 0.306020 \\ 50 & 0.281615 & 0.281542 & 0.281542 \\ 60 & 0.262766 & 0.262714 & 0.262714 \\ 70 & 0.247645 & 0.247607 & 0.247607 \\ 80 & 0.235141 & 0.235111 & 0.235111 \\ 90 & 0.224558 & 0.224535 & 0.224535 \\ 100 & 0.215440 & 0.215421 & 0.215421 \end{array} \right)$$

For sure, one single iteration of Halley or Householder methods leads to much better approximations of the solution.

The problem is that all the formulae except $z_0$ are quite ugly.

My questions

• how to prove that $g(z_0)$, $g'(z_0)$, $g''(z_0)$ are always positive $\forall a,r >2$ ?
• could it be possible to generate a better estimate $z_0$ ? May be some perturbation method I am not aware of

Answer 1

Partly answering my own question.

In the initial problem, I mentioned that, in the vicinity of the solution, $\log[g(z)]$ is closer to linearity than $g(z)$ is. So, looking for the zero of $$\color{blue}{h(z)=\log\big[\sinh ^{2 r}(z)+\cosh ^{2 r}(z) \big]-k}\qquad\text{with} \quad k=\log(a)$$

Forgetting the problem in $x$, expanding $[h(z)+k]$ as series around $z=0$ and continuing with series reversion, for $n \geq 3$, $$z = \sqrt{\frac{k}{r}}\,\,\Bigg[1+ \frac k {12r}+O(k^2) \Bigg]\implies \color{blue}{z_0=\sqrt{\frac{k}{r}}\,\,\Bigg[1+ \frac k {12r} \Bigg]}$$ which is quite simple.

What it seems is that $h(z_0)<0$, $h'(z_0)>0$ and $h''(z_0)>0$. So, using Newton method, the first iterate $z_1$ is here again an overestimate of the solution.

Repeating the same calculations as in my post (using $k=\log(r)$), it seems that the apparently small difference in the initial estimates

$$ \sqrt{t}\left(1+\frac{t}{12}\right)-\tanh ^{-1}\left(\sqrt{\frac{2t}{t+2}}\right)=-\frac{7 }{160}t^{5/2}+O\left(t^{7/2}\right)\quad \quad t=\frac k r $$ has a significant impact (quite better approximations).

$$\left( \begin{array}{cccc} r & z_0 & z_1 & \text{solution} \\ 2 & 0.605707 & 0.573910 & 0.573108 \\ 3 & 0.623615 & 0.615726 & 0.615680 \\ 4 & 0.605707 & 0.604181 & 0.604179 \\ 5 & 0.582570 & 0.582386 & 0.582386 \\ 6 & 0.560067 & 0.560114 & 0.560114 \\ 7 & 0.539459 & 0.539532 & 0.539532 \\ 8 & 0.520877 & 0.520944 & 0.520944 \\ 9 & 0.504154 & 0.504212 & 0.504212 \\ 10 & 0.489060 & 0.489111 & 0.489111 \\ 20 & 0.391854 & 0.391871 & 0.391871 \\ 30 & 0.339891 & 0.339899 & 0.339899 \\ 40 & 0.306015 & 0.306020 & 0.306020 \\ 50 & 0.281539 & 0.281542 & 0.281542 \\ 60 & 0.262712 & 0.262714 & 0.262714 \\ 70 & 0.247605 & 0.247607 & 0.247607 \\ 80 & 0.235110 & 0.235111 & 0.235111 \\ 90 & 0.224534 & 0.224535 & 0.224535 \\ 100 & 0.215420 & 0.215421 & 0.215421 \end{array} \right)$$