6
$\begingroup$

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 $n$?

$\endgroup$
  • $\begingroup$ You mean how fast does $x_n$, the intersection point for $n$ grow with respect to $n$. $\endgroup$ – Michael Burr Mar 29 '18 at 8:50
5
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $\varepsilon\to0$. $\endgroup$ – John Bentin Mar 29 '18 at 10:50
  • $\begingroup$ @JohnBentin: $\varepsilon$ itself is a function of $n$ so that is no real contradiction. $\endgroup$ – Jack D'Aurizio Mar 29 '18 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 ! $\endgroup$ – Claude Leibovici Mar 29 '18 at 16:20
2
$\begingroup$

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 $\ddot \smile$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ On the other hand, you do not need an explicit formula for computing the asymptotic behaviour of the solution. $\endgroup$ – Jack D'Aurizio Mar 29 '18 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. $\endgroup$ – Jaideep Khare Mar 29 '18 at 9:06
1
$\begingroup$

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 $n\to\infty$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.