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For $x>0$, I want to prove that $(x+1)^x-x^x(x-1)$ only has one root

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  • $\begingroup$ Perhaps not super relevant, but note that $x^x>0$, assuming $x>0$. $\endgroup$
    – The Count
    Commented Jun 6, 2019 at 0:58
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    $\begingroup$ Also, I'd be fascinated to know what context this emerged from. $\endgroup$
    – The Count
    Commented Jun 6, 2019 at 1:05
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    $\begingroup$ Why the vote too close? I am willing to edit to prevent closure. How did it emerge? I was just playing around with numbers and came across this and thought it was a fascinating graph.... nothing more. $\endgroup$
    – user351797
    Commented Jun 6, 2019 at 1:13
  • $\begingroup$ The reason I asked for context was purely out of my own curiosity. I could not imagine a situation where this came up. Playing around is a good thing, though, and I support it. As for the close vote... it may be that no effort appears on your part to solve the problem or get started. Just a thought. $\endgroup$
    – The Count
    Commented Jun 6, 2019 at 1:15
  • $\begingroup$ $1$ is not a root. The root is roughly $x=3.40298$ (using a plotting function). This, however, is not a rigorous proof. $\endgroup$
    – user351797
    Commented Jun 6, 2019 at 2:26

2 Answers 2

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My answer is definitely going to pull from some ideas of Robert Israel.

So first we re-write the equation as $$(1+1/x)^x = x-1$$ as previously suggested. Then we consider these as two separate functions and ask when they intersect.

Let $f(x) = (1+1/x)^x$ and let $g(x)=x-1$.

Motivated from the following answer How to prove $(1+1/x)^x$ is increasing when $x>0$? we see that $\log(f(x))' = \log(1+\frac{1}{x})-\frac{1}{x+1}$ is increasing and hence $f(x)$ is too. Likewise one can easily show that $\log(f(x))'' = -\frac{1}{x(1+x)^2}<0$ this shows that $f$ is log-convex and hence convex (this follows from the following The composition of two convex functions is convex since $Exp$ is convex and $\log(f)$ is convex hence $f=Exp[\log(f)]$ is too).

$g(1)=0$ and $f(1)=2$ so $g<f$ at $x=1$. Now that we have established $f$ is convex it follows that once $g$ (a straight line) is greater than $f$ it is always greater. It follows there is one and only one solution for the equation equivalently $g$ intersects $f$ exactly once.

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Write your equation as

$$ (1+1/x)^x = x-1 $$

Note that $x > 1$ is required for both sides to have the same sign, and then

$$ x \log(1+1/x) - \log(x-1) = 0$$

Call the left side $g(x)$. Now show that

  1. $g(x)$ is convex.
  2. $\lim_{x \to 1+} g(x) = +\infty$
  3. $\lim_{x \to \infty} g(x) = -\infty$.

EDIT: As Clclstdnt comments, $g'' = \frac{x^3+x^2+3x-1}{x(x^2-1)^2}$. For $x > 1$, both numerator and denominator are positive, so $g$ is convex on $(1,\infty)$. Since $\lim_{x \to \infty} g(x) = -\infty$, this implies $g' < 0$ (i.e. if $g'(b) \ge 0$ for some $b$, we'd have $g'(x) \ge 0$ for $x \ge b$, and then $\lim_{x\to\infty} g(x)$ couldn't be $-\infty$). That tells you $g$ has at most one zero. On the other hand, (2) and (3) and the Intermediate Value Theorem say there is at least one.

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    $\begingroup$ Can you please explain why showing these three conditions is sufficient to prove the result? Also I don't think that g(x) is convex (everywhere) because the second derivative of it is $\frac{x^3+x^2+3x-1}{x(x^2-1)^2}$ $\endgroup$
    – user351797
    Commented Jun 6, 2019 at 1:21
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    $\begingroup$ I really feel this answer needs more explanation. All this shows is that $g(x)$ is convex and then that $h(x)=(1+1/x)^x - x+1$ is log-convex and hence convex. But $h(x)$ is not convex... in fact, if anything, it is concave for $x>2.5$ (that's not a precise bound). So not only do I not see why you did this... I also don't see that your argument follows. $\endgroup$
    – user351797
    Commented Jun 6, 2019 at 2:23
  • $\begingroup$ You don't need $h$ to be convex. The zeros of $h$ are the zeros of $g$. $\endgroup$ Commented Jun 6, 2019 at 12:31

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