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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 $n$?

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  • $\begingroup$ You mean how fast does $x_n$, the intersection point for $n$ grow with respect to $n$. $\endgroup$ Mar 29, 2018 at 8:50

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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 $\frac{n}{\log n}$.

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  • $\begingroup$ 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 $\varepsilon\to0$. $\endgroup$ Mar 29, 2018 at 10:50
  • $\begingroup$ @JohnBentin: $\varepsilon$ itself is a function of $n$ so that is no real contradiction. $\endgroup$ Mar 29, 2018 at 11:00
  • $\begingroup$ 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 ! $\endgroup$ Mar 29, 2018 at 16:20
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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 $\ddot \smile$

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    $\begingroup$ On the other hand, you do not need an explicit formula for computing the asymptotic behaviour of the solution. $\endgroup$ Mar 29, 2018 at 8:56
  • $\begingroup$ @JackD'Aurizio Yes, I get it. BTW, I was saying "sadly" because, we don't have a closed form for this. $\endgroup$ Mar 29, 2018 at 9:06
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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 $n\to\infty$.

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

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