# Approximating the root $x\in(0,1/2)$ of $\frac{(1-n)^{n-1}}{n^n}x^n+x-1=0$, where $0<n<1$

I have the following equation:

$$\frac{(1-n)^{n-1}}{n^n}x^n+x-1=0$$

where $$0.

I know a solution for $$x$$ always exists in the range $$(0, \frac{1}{2})$$, but I'm unsure if it can be solved for exactly. I was hoping there was a method to obtain an approximate analytical solution, or perhaps tighter analytical bounds on the solution, in terms of $$n$$.

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• This is not a polynomial equation. For that, $n$ should be a positive integer.
– Gary
Commented Aug 1 at 12:18
• Thanks, I've updated the question. Commented Aug 1 at 12:22
• This is trinomial equation for which there are series solutions Commented Aug 1 at 13:29

There is no closed-form expression for the solution unless $$n$$ is a simple fraction with denominator $$2$$, $$3$$, or $$4$$. Newton–Raphson gives rapidly converging results for this type of equation. You could simplify it a little by the scaling $$t:=\frac{1-n}nx\quad\text{with}\quad f(t):=\frac{t^n}{1-n}+\frac n{1-n}t-1.$$ Then, starting with some chosen $$t_0$$, we iterate with the function $$g(t):=t-\frac{f(t)}{f'(t)}=\frac{1-n}n\frac{1-t^n}{1+t^{n-1}}$$ to get $$t_1,t_2,$$ etc. For example, with $$n=\frac12$$, from $$t_0=\frac14$$, we get $$t_1=\frac16$$, and already $$t_2=0.1715476...$$, which compares with the exact solution $$3-2\surd2=0.1715725...$$ .

• If we use $t_0=\left(\frac{1-n}{n+1}\right)^{\frac{1}{ n}}$, by Darboux theorem, the solution will be reached without any overshoot. Commented Aug 2 at 10:11
• @Claude : I tried your suggestion, also for $n=\frac12$. That gives $t_0=\frac19$, which coincidentally yields $t_1=\frac16$ too. Perhaps your formula for $t_0$ results in faster convergence for values of $n$ away from $\frac12$; but anyway I don't know how you obtained it. Commented Aug 2 at 11:01

With the substitution $$u= \frac{{(1 - n)^{n - 1} }}{{n^n }}x^n ,$$ the equation becomes $$u + \frac{n}{(1 - n)^{1 - 1/n}}u^{1/n} - 1 = 0.$$ Employing the formula in this answer with $$\alpha=\beta=1/n$$ and $$\omega=-n/(1 - n)^{1 - 1/n}$$, we obtain $$u^{1/n} = \frac{1}{n}\sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{k!}}\left( {\frac{n}{{(1 - n)^{1 - 1/n} }}} \right)^k \frac{{\Gamma \big( {\frac{{k + 1}}{n}} \big)}}{{\Gamma \big( {\frac{{k + 1}}{n} - k + 1} \big)}}}.$$ Returning to the original variable, we find $$x = \frac{1}{n}\sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{k!}}\left( {\frac{n}{{(1 - n)^{1 - 1/n} }}} \right)^{k + 1} \frac{{\Gamma \big( {\frac{{k + 1}}{n}} \big)}}{{\Gamma \big( {\frac{{k + 1}}{n} - k + 1} \big)}}}.$$ The only thing that needs to be checked is the convergence of the series, as it is evaluated at the boundary of convergence. I anticipate that it will converge based on the alternating series test.

Addendum. Up to a constant factor that depends only on $$n$$, the $$k^{\text{th}}$$ term of the series is $$\frac{{( - 1)^k }}{{k^{3/2} }} + \mathcal{O}\!\left( {\frac{1}{{k^{5/2} }}} \right),$$ which proves the convergence. This asymptotics can be deduced from Stirling's formula.

Starting from @John Bentin's answer $$f(t)=t^n+nt+(n-1)$$ the function is upper bounded by $$g(t)=t^n+nt^n+(n-1)\quad \implies \quad t_0=\left(\frac{1-n}{1+n}\right)^{\frac{1}{ n}}$$

Since $$f(t_0)<0$$ and $$f''(t_0)<0$$, by Darboux theorem, Newton method will converge without any overshoot.

A little bit of analysis shows that a better initial estimate could be $$t_0==\left(\frac{1-n}{1+n}\right)^{\frac{2+n}{3n}}$$

The table reproduces the value of the estimate, the value of the first iterate of Newton method and the solution for a few values of $$n$$. $$\left( \begin{array}{cccc} n & t_0 & t_1 & \text{solution} \\ 0.05 & 0.254665 & 0.269110 & 0.269417 \\ 0.10 & 0.245442 & 0.259770 & 0.260065 \\ 0.15 & 0.235927 & 0.250108 & 0.250389 \\ 0.20 & 0.226116 & 0.240103 & 0.240368 \\ 0.25 & 0.216000 & 0.229732 & 0.229981 \\ 0.30 & 0.205566 & 0.218972 & 0.219202 \\ 0.35 & 0.194797 & 0.207796 & 0.208004 \\ 0.40 & 0.183673 & 0.196171 & 0.196357 \\ 0.45 & 0.172168 & 0.184064 & 0.184226 \\ 0.50 & 0.160250 & 0.171435 & 0.171573 \\ 0.55 & 0.147879 & 0.158239 & 0.158353 \\ 0.60 & 0.135007 & 0.144426 & 0.144515 \\ 0.65 & 0.121575 & 0.129933 & 0.130000 \\ 0.70 & 0.107507 & 0.114692 & 0.114739 \\ 0.75 & 0.092705 & 0.098614 & 0.098644 \\ 0.80 & 0.077040 & 0.081595 & 0.081612 \\ 0.85 & 0.060333 & 0.063495 & 0.063503 \\ 0.90 & 0.042318 & 0.044124 & 0.044126 \\ 0.95 & 0.022548 & 0.023179 & 0.023179 \\ \end{array} \right)$$