# Equation $x^{\frac{n+1}{n}}=x+1$

Let $n$ be a positive integer. What is the positive value of $x$ such that $x^{\frac{n+1}{n}}=x+1$?

This equation has a unique solution because the function $x^{\frac{n+1}{n}}-x$ is increasing. However, I'm not sure if we can get a closed form for $x$. If not, how fast does $x_n$, the solution for $n$, grow asymptotically in terms of $n$?

• You mean how fast does $x_n$, the intersection point for $n$ grow with respect to $n$. – Michael Burr Mar 29 '18 at 8:50

We may assume $n\geq 2$.
Of course $x^{\frac{n+1}{n}}=x+1$ has a unique positive solution, since $x^{\frac{n+1}{n}}$ is increasing and convex.
Let $x=z^n$. The problem boils down to solving $$z^{n+1} = z^n+1$$ and we may notice that $z^{n+1}-z^n-1$ is negative at $z=1$ and positive at $z=1+\frac{\log n}{n}$.
By applying Newton's method with starting point $z_0=1+\frac{\log n}{n}$ we may derive accurate approximations for the positive real root of $z^{n+1}+z^n-1$ and the asymptotic behaviour for the positive real root of $x^{\frac{n+1}{n}}=x+1$, which has to be close to $\frac{n}{\log n}$.

• For large $n$, the term $(\ln n)/n^2$ in the asymptotic formula on the last line becomes small—call it $\varepsilon$. Then, if we set $x=1+\varepsilon$, the LHS of the original equation is $$1+\left(1+\frac1n\right)\varepsilon+O(\varepsilon^2),$$whereas the RHS is $2+\varepsilon$, which is asymptotically different as $\varepsilon\to0$. – John Bentin Mar 29 '18 at 10:50
• @JohnBentin: $\varepsilon$ itself is a function of $n$ so that is no real contradiction. – Jack D'Aurizio Mar 29 '18 at 11:00
• More than likely of no interest (excep, may be, for a numerical solution). If $f(z)=z^{n+1}-z^n-1$, $f(z)<0$ for $z=1+k\frac{\log n}{n}$ where $k\approx 0.7447735012040$. Nice bounds ! – Claude Leibovici Mar 29 '18 at 16:20

Raise to power $n$ both the sides.

You get $$x^{n+1}=(1+x)^n$$

Which, on further simplification, is a polynomial in $x$, with degree $n+1$.

Sadly, we don't have any general formula for solution of this polynomial, unless it's a quadratic or a cubic. (For quartic, the formula is too tedious)

Though, numerical approximations always have your back $\ddot \smile$

• On the other hand, you do not need an explicit formula for computing the asymptotic behaviour of the solution. – Jack D'Aurizio Mar 29 '18 at 8:56
• @JackD'Aurizio Yes, I get it. BTW, I was saying "sadly" because, we don't have a closed form for this. – Jaideep Khare Mar 29 '18 at 9:06

Asymptotically, The solution $x$ grows as $n/\ln n$. To see this, write the original equation as$$\frac1n\ln x=\ln\left(1+\frac1x\right).$$Since $x>1$, the RHS may be expanded as a power series:$$\frac1n\ln x=\frac1x-\frac1{2x^2}+\frac1{3x^3}-\cdots.$$ As $n$ increases, so must $x$, and asymptotically we get $x\ln x\sim n$, which implies $$x\sim\frac n{\ln n}\quad(n\to\infty),$$since $(\ln\ln n)/\ln n\to0$ as $n\to\infty$.

If $n=1$ then we have $$x^2= x+1$$ so $$x^2-x-1=0 \Longrightarrow x_{1,2}= {1\pm \sqrt{5}\over 2}$$

Actually, the solutions of this type of trinomial polynomials had been solved quite long time ago, around (1990), and basically from more general solution in standard modern mathematical notations, where a general formula had been discovered and announced publically for this general trinomial polynomial of n’th degree $$ax^n + bx^m + c = 0$$, where ($n > m$) are positive integers, ($abc \neq 0$) are rational numbers, so it is so easy task now to apply it for our particular question by noting that ($a = - b = -c = 1$), and applying the reduced or simplified form for $$z^n + z^m = 1$$, provided by Jack above

\begin{align} z &= \sum_{k=0}^\infty \frac{(-1)^k \prod_{i=1}^{k-1}(km-in+1)}{k!n^k} \\ &= 1-\frac{1}{n}+\frac{2m-n+1}{2!n^2}- \frac{(3m-2n+1)(3m-n+1)}{3!n^3} +\frac{(4m-3n+1)(4m-2n+1)(4m-n+1)}{4!n^4} -\frac{(5m-n+1)(5m-2n+1)(5m-3n+1)(5m-4n+1)}{5!n^5} +\dotsb \end{align}

Substituting for ($m = n – 1$), we get:

\begin{align} z &= \sum_{k=0}^\infty \frac{(-1)^k \prod_{i=1}^{k-1}(k(n – 1)-in+1)}{k!n^k} \\ &= 1-\frac{1}{n}+\frac{n - 1}{2!n^2}- \frac{(n-2)(2n-2)}{3!n^3} +\frac{(n-3)(2n-3)(3n-3)}{4!n^4} -\frac{(n-4)(2n-4)(3n-4)(4n-4)}{5!n^5} +\dotsb \end{align}

However, this series form can be further reduced more, where too many interesting cases