# When does $(x+1)^s -x^s = x^{-s}$ have a solution?

Let $$0\lt s \lt 1$$. I want to find the range of $$s$$ for which the following equation has a solution for $$x\ge 0$$.

$$(x+1)^s -x^s = x^{-s}$$

After looking at the graphs of the LHS and RHS and Wolfram, I believe that if $$s$$ is greater than some unknown value, there is always one solution. For example, WolframAlpha gives no solutions for $$s=0.5$$ but one for $$s=0.6$$.

How can I obtain this transition point? Does its closed-form even exist?

I might add an idea. At that special value of $$s$$, the two curves must touch each other, giving the system $$(x+1)^s -x^s = x^{-s} \\ (x+1)^{s-1} -x^{s-1} = -x^{-s-1}$$ Maybe there’s a way to solve for $$s$$.

• For the critical $s$ do they "touch" at infinity? Apr 16, 2021 at 14:43
• You can do a bisection search in Alpha manually. I find a solution for $s=0.54$ around $x=2500$ in a spreadsheet. Now try $0.52$ and keep going. Someone more expert than I or having Mathematica could automate this. For $s=0.51$ and $x$ around $2 \cdot 10^{12}$ there is a solution Apr 16, 2021 at 14:50
• @Henry That’s a point, the solution could approach $\infty$ as we get closer to the critical point. Apr 16, 2021 at 14:52
• @RossMillikan A bisection search? I’m not very good at these computational things. It’s interesting that there is a solution for $s=0.51$. I’m guessing the transition point is $0.5$. Apr 16, 2021 at 14:53
• @RossMillikan. Yes, I guess that we need to look what happens when $x\to \infty$ Apr 16, 2021 at 14:56

Consider the two functions$$f=(x+1)^s -x^s - x^{-s}$$ $$g=(x+1)^{s-1} -x^{s-1} +x^{-s-1}$$ and the norm $$\Phi=f^2+g^2$$ My first work has been to look at the contour plot of the norm but, even if illustrative, the fact that $$x$$ increases more and more made the analysis not very concluding. I also tried to minimize $$\Phi$$ with respect to $$x$$ and $$s$$; it worked but, since $$x\to \infty$$, the result was depending on the working precision.

So, working with cross sections, I made $$x=10^k$$ and solves $$\Phi=0$$ for $$s$$. The results are $$\left( \begin{array}{cc} k & s_k \\ 0 & 1.000000 \\ 1 & 0.611256 \\ 2 & 0.562674 \\ 3 & 0.544074 \\ 4 & 0.534053 \\ 5 & 0.527757 \\ 6 & 0.523429 \\ 7 & 0.520269 \\ 8 & 0.517862 \\ 9 & 0.515966 \\ 10 & 0.514434 \\ 11 & 0.513170 \\ 12 & 0.512114 \\ 13 & 0.511181 \end{array} \right)$$

Based on these data, I performed a least square fit for the model $$s_k=\frac {1+a k} {1+b k}$$ which is almost perfect $$(R^2 > 0.999999)$$ (maximum error $$=0.0002$$) and the parameters are $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 1.75930 & 0.00353 & \{1.75143,1.76717\} \\ b & 3.51381 & 0.00668 & \{3.49893,3.52868\} \end{array}$$ predicting an asymptotic value of $$0.500681$$.

So, as other users concluded, there are solutions for $$0.5 < s \leq 1$$.

Taking this into account, a very good estimate of $$s_k$$ is given by $$\color{blue}{s_k=\frac{1+k\sqrt 3}{1+2k\sqrt 3}=\frac 12 \left(1+\frac 1 {1+2k\sqrt 3}\right)}$$

Using it as a starting point for Newton method, we have very fast convergence and obtain solutions for extremely large values of $$x$$ $$\left( \begin{array}{cc} k & s_k \\ 100 & 0.5014392210 \\ 200 & 0.5007206477 \\ 300 & 0.5004806627 \\ 400 & 0.5003605837 \\ 500 & 0.5002885086 \\ 600 & 0.5002404469 \\ 700 & 0.5002061115 \\ 800 & 0.5001803569 \\ 900 & 0.5001603237 \\ 1000 & 0.5001442959\\ 2000 & 0.5000721584 \\ 3000 & 0.5000481079 \\ 4000 & 0.5000360818 \\ 5000 & 0.5000288658 \\ 6000 & 0.5000240551 \\ 7000 & 0.5000206188 \\ 8000 & 0.5000180415 \\ 9000 & 0.5000160370 \\ 10000 & 0.5000144333 \end{array} \right)$$

Edit

If, around $$s=\frac 12$$, we expand $$f$$ and $$g$$ as series, we have $$f=\left(\sqrt{x+1}-\frac{x+1}{\sqrt{x}}\right)+\left(s-\frac{1}{2}\right) \left(\sqrt{x+1} \log (x+1)-\frac{(x-1) \log (x)}{\sqrt{x}}\right)+O\left(\left(s-\frac{1}{2}\right)^2\right)$$ and $$g=\left(\frac{1-x}{x^{3/2}}+\frac{1}{\sqrt{x+1}}\right)+\left(s-\frac{1}{2}\right) \left(\frac{\log (x+1)}{\sqrt{x+1}}-\frac{(x+1) \log (x)}{x^{3/2}}\right)+O\left(\left(s-\frac{1}{2}\right)^2\right)$$

Solving $$f=0$$ for $$s$$ and plugging the result in $$g$$, we need to solve for $$x$$ $$\left(3 x-4 \sqrt{x+1} \sqrt{x}+1\right) \log (x)+(x+1) \log (x+1)=0$$ which does not show solution.

For large $$x$$, the asymptotics of the above is $$1+\frac{\log (x)+1}{2 x}+O\left(\frac{1}{x^2}\right)$$

Update

Without any approximation, if $$f=g=0$$, writing $$f-x\,g =0 \implies s=\frac{\log (2(x+1))}{\log (x(x+1))}$$

Plugging $$s$$ in $$f$$ then reduces to problem to $$1+2x=x^{\frac{2 \log (2 (x+1))}{\log (x (x+1))}}$$ which does not show solutions.

Effectively, expanding the rhs as series $$x^{\frac{2 \log (2 (x+1))}{\log (x (x+1))}}=1+2x-\frac{\log (2)}{\log (x)}+\frac{\log (2)(\log(2)+2)-\log (x) (\log (x)+2)}{4 x \log ^2(x)}$$ and the first terms are the lhs (!!).

Then the solution is $$x\to \infty$$ and if this is the case $$s=\frac 12\left(1+\frac{\log (2)}{\log (x)} \right)$$

• Very useful! That approximation of $s_k$ is nice. Apr 17, 2021 at 11:51
• @Tavish. I am still working this very interesting ptoblem. Using the data for $10^{1000} \leq x \leq 10^{10000}$, the $2\sqrt 3$ is confirmed and the residues are $\leq 10^{11}$. What does $\sqrt 3$ is doing here ? Cheers :-) Apr 17, 2021 at 12:55
• Wow, I didn’t realize that $s$ could be written in terms of $x$ like that. Thank you, Claude. This completes the answer. Apr 17, 2021 at 14:08
• @Tavish. I thank you for the problem ! I had a lot of fun (at my age, it is good). Cheers :-) Apr 17, 2021 at 14:11
• @Tavish; Take care : this is valid if $f=g=0$ only Apr 18, 2021 at 5:18

I believe it fails at $$s=\frac 12$$ but will succeed at any $$s$$ above that. We have $$\sqrt{x+1}-\sqrt x \approx \frac 12 x^{-1/2}-\frac 18x^{-3/2}$$ as $$x \to \infty$$ The leading factor of $$\frac 12$$ says there will be no solution because the left side is always larger. For $$x=0.52$$ I found a solution around $$x=1.6\cdot 10^7$$ but closer than that I am running into precision errors in my spreadsheet. I tried to get Alpha to expand the $$s=0.51$$ case but it wouldn't.

• Is there any way to prove this? Apr 16, 2021 at 15:21
• That it has a solution for $s > 1/2$ ($f(0^+)=-\infty, f(x)>0$ for $x$ large) and no solution for $s\le 0$ ($f<0$) follows from $f(x)=(x+1)^s - x^s-x^{-s} = \int_0^1 (s (x+t)^{s-1}-x^{-s})dt$ Apr 16, 2021 at 15:36
• @reuns Can you elaborate on that? And it is assumed that $0\lt s \lt 1$. Apr 16, 2021 at 17:32

For a given value of $$s$$, instead of solving for $$x$$ $$f(x)=(x+1)^s -x^s - x^{-s}=0$$ it looks much better to solve instead $$\color{red}{h(x)=\log \left( (x(x+1))^s-x^{2 s}\right)=0}$$ Using what was proposed in the previous answer $$x_0=\exp\left(\frac{\log (10)}{\sqrt{3}} \,\frac{s-1}{1-2s} \right)$$ is a good starting point for Newton method (it could probably be significantly improved).
For $$s=0.52$$ used by @Ross Millikan, the iterates are
$$\left( \begin{array}{cc} n & x_n \\ 0 & 8.47624\times 10^6 \\ 1 & 1.18276\times 10^7 \\ 2 & 1.25635\times 10^7 \\ 3 & 1.25868\times 10^7 \end{array} \right)$$
Still with the idea of finding a better conditionning, it is definitrly better to let $$\color{red}{x=e^t}$$; (his makes $$f(x)$$ much more linear. For the same problem as above with the same initial conditions, Newton iterates are $$\left( \begin{array}{cc} n & t_n \\ 0 & 15.952777 \\ 1 & 16.348162 \end{array} \right)$$
For $$s=0.51$$ which was making problems $$\left( \begin{array}{cc} n & t_n \\ 0 & 32.570254 \\ 1 & 33.667228 \end{array} \right)$$
Now, a much better estimate $$t_0=\frac{764}{583} \theta+\frac{6}{1015}\theta^2-\frac{1}{7097}\theta^3 \qquad \text{with} \qquad \theta=\frac{s-1}{1-2 s}$$ For the two above examples, the estimates are $$16.3333$$ and $$33.5826$$.