What is the root of $a(x) = 1 + x^r - 2(x-1)^r$?

Question

I have been working on my thesis and I have stumble upon the following equation that is giving me a tough time: $$a(x) = 1 + x^r - 2(x-1)^r$$

What is the root of this equation? As in, for what values of $x$ will $a(x) = 0$? Is there a unique root? There are a couple of constraints:

• $x \geq 1$
• $r \geq 2$

The constraint on $r$ is not so serious however, though I doubt "pretty" solutions will arise if $p < 1$.

I plotted this function into Mathematica and came out with a curve which is not monotonic with $x$, thus making it difficult to prove the existence of a unique root.

I have tried to take the derivative of the function in order to evaluate for critical points. I seem to be getting $x = \dfrac{2^{1/(r-1)}}{2^{1/(r-1)}-1}$ but this is not it. When I plug-in this critical point into the original equation, it does not equal to zero.

To my knowledge and skills, I feel stumped.

Answer 1

Another approach

Let $y=\cosh^2(z)$ (that is to say $x=\coth ^2(z)$) and consider now that we need to find the zero of function $$g(z)=\log\Big[\sinh ^{2 r}(z)+\cosh ^{2 r}(z)\Big]-\log(2)$$ which is very close to linearity.

Using one single iteration of halley method with $$z_0=\tanh ^{-1}\left(\sqrt{\frac{2 \log (2)}{2 r+\log (2)}}\right)\implies z_1=z_0- \frac {2 \,g(z_0)\, g'(z_0)} {2 \,{[g'(z_0)]}^2 - g(z_0)\, g''(z_0)}$$

Some results $$\left( \begin{array}{ccc} r & z_1 & \text{solution} \\ 10 & \color{red}{0.264800234264}95279706 & 0.26480023426440294083 \\ 20 & \color{red}{0.186702998748}20460378 & 0.18670299874819865061 \\ 30 & \color{red}{0.15229581653017}421982 & 0.15229581653017379595 \\ 40 & \color{red}{0.131828617848412}35565 & 0.13182861784841229053 \\ 50 & \color{red}{0.1178770690152299}6479 & 0.11787706901522994955 \\ 60 & \color{red}{0.10758584167182339}878 & 0.10758584167182339413 \\ 70 & \color{red}{0.09959144191558136}447 & 0.09959144191558136276 \\ 80 & \color{red}{0.09314965796552352}962 & 0.09314965796552352891 \\ 90 & \color{red}{0.087815296380881823}72 & 0.08781529638088182339 \\ 100 & \color{red}{0.083303559678972018}36 & 0.08330355967897201819 \end{array} \right)$$

You may suppose that this is my prefered solution.

Answer 2

Looking at the equation, my intuition was that the solution $x_*$ will be linear with $r$. This has been checked by a contour plot of $a(x)=0$ for $1\leq x \leq 100$ and $2\leq r \leq 100$.

A few calculations $$\left( \begin{array}{cc} r & x_* \\ 10 & 14.9327 \\ 20 & 29.3568 \\ 30 & 43.7828 \\ 40 & 58.2092 \\ 50 & 72.6359 \\ 60 & 87.0627 \\ 70 & 101.489 \\ 80 & 115.916 \\ 90 & 130.343 \\ 100 & 144.770 \end{array} \right)$$ A quick and dirty linear regression gives with $R^2>0.999999$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 0.50404 & 0.000730 & \{0.50232,0.50577\} \\ \beta & 1.44265 & 0.000012 & \{1.44262,1.44268\} \\ \end{array}$$

Extrapolated for $r=1000$, this gives $x_*=1443.16$ while the solution is $1443.20$.

There is another (very small) root.

In any manner, ^$$a'(x)=0 \implies \left(\frac{x}{x-1}\right)^{r-1}=2 \implies x_{\text{max}}=\frac{2^{\frac{1}{r-1}}}{2^{\frac{1}{r-1}}-1}$$ and $x_* >x_{\text{max}}$.

For large $r$ $$x_{\text{max}}=\frac{r}{\log (2)}+\left(\frac{1}{2}-\frac{1}{\log (2)}\right)+\frac{\log (2)}{12r}+\frac{\log (2)}{12 r^2}+O\left(\frac{1}{r^3}\right)$$ For $r=1000$, this gives $x_{\text{max}}=1441.75$ (quite close to the solution).

If you want to numerically solve, use Newton method with $x_0$ given by the empirical correlation. For example, for $r=10000$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 14427.0000000000 \\ 1 & 14427.6548458823 \\ 2 & 14427.4757848783 \\ 3 & 14427.4508530546 \\ 4 & 14427.4504147990 \\ 5 & 14427.4504146659 \end{array} \right)$$

Notice that $\beta \sim \frac 1{\log(2)}$ given in the expansion. So, the simplest approximation is $$x_*=\frac 12 +\frac r{\log(2)} $$

Edit

The last proposed approximation can be improved. Let $$x_*=\frac 12 +\frac r{\log(2)}+\epsilon(r) $$ Expand $a(x)$ as a series up to $O\left(\epsilon ^2\right)$ and use series reversion to obtain $$\epsilon(r)=\frac{1}{2 r \log (2)} \, \frac{\log ^r(4)-2 (2 r-\log (2))^r+(2 r+\log (2))^r}{2 (2 r-\log (2))^{r-1}-(2 r+\log (2))^{r-1}}$$ For the case where $r=10000$ looked at just before end, this would give as an estimate $x=14427.4504146659$ which is the solution.

Trying with $r=2^k$, some results $$\left( \begin{array}{ccc} k & \text{estimate} & \text{solution} \\ 1 & \color{red}{3.7}754225843653752403 & 3.7320508075688772935 \\ 2 & \color{red}{6.290}7262317521153027 & 6.2905203826719496086 \\ 3 & \color{red}{12.048}811554320985251 & 12.048779744481548435 \\ 4 & \color{red}{23.58673}9175499452336 & 23.586730682870478481 \\ 5 & \color{red}{46.66804}8555883121630 & 46.668046365114270141 \\ 6 & \color{red}{92.833385}706672057396 & 92.833385150520588998 \\ 7 &\color{red}{ 185.165416}64135828893 & 185.16541650126243983 \\ 8 & \color{red}{369.8301561}3655115031 & 369.83015610139490248 \\ 9 & \color{red}{739.1599737}6087533816 & 739.15997375206973774 \\ 10 & \color{red}{1477.8197782}809635256 & 1477.8197782787600580 \\ 11 & \color{red}{2955.13947194}53791096 & 2955.1394719448279842 \\ 12 & \color{red}{5909.778901583}4474134 & 5909.7789015833095997 \\ 13 & \color{red}{11819.0577820134}80435 & 11819.057782013445977 \\ 14 & \color{red}{23637.615553450313953} & 23637.615553450313953 \end{array} \right)$$

This could be still much improved expand to $O\left(\epsilon ^n\right)$ $(n \geq 3)$ but $\epsilon(r)$ would become quite nasty. For example, for $k=2$ and $n=3$ this would give $6.290516793$ which is much better.

Answer 3

I prefer to add a second answer since the approach is quite different.

Let $x=\frac{y}{y-1}$ and consider that we look for the zero of function $$f(y)=y^r+(y-1)^r-2$$ which is more pleasant since $f(1)=-1$ and $f(2)=2^r-1$ which show that the solution is close to $1^+$.

The initial guess, taken from the first answer, is $$y_0=\frac{2 r+\log (2)}{2 r-\log (2)}$$ Trying to improve as before $$y_1=\frac{2 r+\log (2)}{2 r-\log (2)}-\frac 1r\, \frac{\left(\frac{2\log (2)}{2 r-\log (2)}\right)^r+\left(\frac{2 r+\log (2)}{2 r-\log (2)}\right)^r-2}{\left(\frac{2 \log (2)}{2 r-\log (2)}\right)^{r-1}+\left(\frac{2 r+\log (2)}{2 r-\log (2)}\right)^{r-1}}$$

$$\left( \begin{array}{ccc} r & y_0 & y_1 &\text{solution} \\ 10 & 1.07180322831922 & \color{red}{1.07177346}625573 & 1.07177346253610 \\ 20 & 1.03526851583585 & \color{red}{1.035264923}95977 & 1.03526492384138 \\ 30 & 1.02337494396036 & \color{red}{1.02337389}201245 & 1.02337389199677 \\ 40 & 1.01748013332815 & \color{red}{1.01747969210}642 & 1.01747969210269 \\ 50 & 1.01395970491221 & \color{red}{1.01395947979}125 & 1.01395947979003 \\ 60 & 1.01161957027915 & \color{red}{1.01161944030}242 & 1.01161944030192 \\ 70 & 1.00995137233426 & \color{red}{1.009951290618}34 & 1.00995129061812 \\ 80 & 1.00870203846592 & \color{red}{1.00870198379052} & 1.00870198379052 \\ 90 & 1.00773140758048 & \color{red}{1.00773136921722} & 1.00773136921722 \\ 100 & 1.00695557800201 & \color{red}{1.00695555005676} & 1.00695555005676 \end{array} \right)$$ and, if required, better could be still done. For example, adding a second term is the expansion, for $r=10$, we have $y=\color{red}{1.071773462536}75$ to be compared to the exact solution $y=\color{red}{1.07177346253610}$.