# Prove that $(x+1)^x-x^x(x-1)$ only has one (real) root

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

• Perhaps not super relevant, but note that $x^x>0$, assuming $x>0$. – The Count Jun 6 '19 at 0:58
• Also, I'd be fascinated to know what context this emerged from. – The Count Jun 6 '19 at 1:05
• 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. – Clclstdnt Jun 6 '19 at 1:13
• 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. – The Count Jun 6 '19 at 1:15
• $1$ is not a root. The root is roughly $x=3.40298$ (using a plotting function). This, however, is not a rigorous proof. – Clclstdnt Jun 6 '19 at 2:26

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 $$g at $$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.

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

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.

• 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 $\frac{x^3+x^2+3x-1}{x(x^2-1)^2}$ – Clclstdnt Jun 6 '19 at 1:21
• 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. – Clclstdnt Jun 6 '19 at 2:23
• You don't need $h$ to be convex. The zeros of $h$ are the zeros of $g$. – Robert Israel Jun 6 '19 at 12:31