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I was motivated by this question. As shown there, one can easily come with unique integer solution to equation $$2^{x+1} = x^3(x-1),$$ but let's say I'd like to know total number of all real solutions. If we denote $$\begin{align*} f(x) &= 2^{x+1} \\ g(x) &= x^3(x-1) \\ h(x) &= f(x) - g(x) \end{align*}$$ then we can form the problem in couple of equivalent ways:

1) What is the number of roots of $h$?

2) What is the number of intersections of graphs of $f$ and $g$?

3) Let $k(x) = x - h(x)$. What is the number of fixed points of $k$, i.e. number of points such that $k(x) = x$?

Let's try to work a bit on given example.

$f$ is strictly increasing, and $g$ is strictly decreasing on $\langle-\infty,\frac 3 4\rangle$, thus, $h$ is strictly increasing continuous function on $\langle-\infty,\frac 3 4\rangle$ and has at most one root. Since $h(-1) = -1$, $h(0) = 2$, $h$ has exactly one root on $\langle-\infty,\frac 3 4\rangle$. What remains is to check $I = [ \frac 3 4, +\infty\rangle$. It's easy to check that both $f$ and $g$ are convex on $I$.

(Q1) Is there an upper limit to the number of intersections of graphs of convex functions? What about monotone convex functions? What about convex and concave function?

Let's try to use a simple lemma to find one more root of $h$.

Lemma If $f(a)>g(a)$ and $f'(x)> g'(x)$ on $\langle a,b\rangle$, then $f(x) > g(x)$ on $\langle a,b\rangle$.

In our example, we have $$ \begin{align*} f^{(n)}(x) &= 2^{x+1}(\ln 2)^n, \\ g'(x) &= 4x^3 - 3x^2, \\ g''(x) &= 12x^2 - 6x, \\ g'''(x) &= 24 x - 6, \\ g^{(4)}(x) &= 24. \end{align*}$$

Now, let's apply the above lemma in several steps: $f^{(5)}(x) = 2^{x+1}(\ln 2)^5 > 0 = g^{(5)}(x)$, so

  • $f^{(4)}(6) > g^{(4)}(6) \implies f^{(4)}(x) > g^{(4)}(x)$ on $\langle 6,+\infty\rangle$

  • $f'''(9) > g'''(9) \implies f'''(x) > g'''(x)$ on $\langle 9,+\infty\rangle$

  • $f''(11) > g''(11) \implies f''(x) > g''(x)$ on $\langle 11,+\infty\rangle$

  • $f'(13) > g'(13) \implies f'(x) > g'(x)$ on $\langle 13,+\infty\rangle$

Thus, $h$ is strictly increasing continuous function on $\langle 13,+\infty\rangle$, and $h(14) = -2904$, $h(15)=18286$, so we have exactly one root of $h$ on $\langle 13,+\infty\rangle$.

This method, although works in this example, seems very inefficient and hard (impossible?) to apply when not dealing with polynomials. Not to mention dealing with $[\frac 3 4, 13]$ seems rather gruesome in this case (we could argue similarly to above that there is exactly one root on $\langle \frac 3 2, 3\rangle$, assuming, of course, we didn't already know that $x=2$ is a root).

To wrap it up a bit, let's say we counted $3$ roots.

(Q2) Are those all? How do we know?

Here we relied on the fact that strictly monotone continuous function that takes a positive and negative value has exactly one root (exploiting the fact that continuous image of compact and connected set is compact and connected).

(Q3) Are there other tools at our disposal that we can use to count the number of roots?

At the end, let's leave the domain of real functions of one variable and talk about more general manifolds. I think it's more convenient to talk about fixed points in this setting. I'm aware of some of the fixed point theorems, but I'm wondering

(Q4) Can we in general count all the fixed points of a (smooth) map?

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Q4: No, even for one-variable functions. See Richardson's theorem.

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  • $\begingroup$ Thank you, I was rather skeptical about that one, but wasn't aware of the result. $\endgroup$ – Ennar Jul 19 '15 at 18:38

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