Solutions to Equations Involving Exponential and Trig Functions

[Bashyboy]

Homework Statement

Show that $e^x = x$ does not have any solutions, and show that $\sec x = e^{-x^2}$ has only one solution.

Homework Equations

The Attempt at a Solution

Here is my proof of the first proposition: Since $e^x$ is concave up on $\Bbb{R}$, it must lie above all of its tangent lines. In particular $e^x \ge x+1 > x$ for every $x \in \Bbb{R}, which means there cannot be a solution to $e^x=x$.

Before I move on to the second part, I would like to point out that I have been using this connection between concavity and tangent lines but I have never read a precise statement of it. Would anyone know to how precisely state this equivalence, as I am interested in proving it.

Here is my work for the second part. Clearly $x=0$ is a solution, and in fact this is where the maximum value of $e^{-x^2}$ occurs. Since it is strictly in increasing on $(-\infty, 0]$, it attains this value only once; and since it is strictly decreasing on $(0,\infty)$), it will never attain this maximum value. Now, if $x$ is such that $e^{-x^2} = \sec x$, then $|e^{-x^2}| = | \sec x |$ or $e^{-x^2} \ge 1$, which happens if and only if $x =0$. So there are no other solutions besides $x=0$.

How does this sound? Any improvements?

 

[Ray Vickson]

Homework Statement

Show that $e^x = x$ does not have any solutions, and show that $\sec x = e^{-x^2}$ has only one solution.

Homework Equations

The Attempt at a Solution

Here is my proof of the first proposition: Since $e^x$ is concave up on $\Bbb{R}$, it must lie above all of its tangent lines. In particular $e^x \ge x+1 > x$ for every $x \in \Bbb{R}, which means there cannot be a solution to $e^x=x$.

Before I move on to the second part, I would like to point out that I have been using this connection between concavity and tangent lines but I have never read a precise statement of it. Would anyone know to how precisely state this equivalence, as I am interested in proving it.

Here is my work for the second part. Clearly $x=0$ is a solution, and in fact this is where the maximum value of $e^{-x^2}$ occurs. Since it is strictly in increasing on $(-\infty, 0]$, it attains this value only once; and since it is strictly decreasing on $(0,\infty)$), it will never attain this maximum value. Now, if $x$ is such that $e^{-x^2} = \sec x$, then $|e^{-x^2}| = | \sec x |$ or $e^{-x^2} \ge 1$, which happens if and only if $x =0$. So there are no other solutions besides $x=0$.

How does this sound? Any improvements?

The function $\sec(x)$ is a bit more complicated than you have described it. Look at $x \geq 0$. The function $\sec(x)$ is strictly convex (and $>0$) on $(0,\pi/2)$, strictly concave (and $< 0$) on $(\pi/2, 3 \pi/2)$, strictly convex ($>0$) on $(3 \pi/2, 5 \pi/2)$, etc. The function $e^{-x^2}$ is strictly concave, $>0$, strictly decreasing on $(0, 1/\sqrt{2})$ and strictly convex, $> 0$, strictly decreasing on $(1/\sqrt{2}, \infty)$. Both $e^{-x^2}$ and $\sec(x)$ are even functions of $x$, so the behavior for $x < 0$ is a mirror-image of that for $x > 0$.

Note: I use the more modern terminology "convex" instead of "concave up" and "concave" instead of "concave down". This is entirely in line with what is done in the optimization literature and most application areas these days.

As to your question about tangent lines: just Google "convex functions" to see a host of relevant articles.

Actually, I see that most of them just state the result without proof, so here is a little proof that has only a few, simple, missing steps.
Suppose $f(x)$ is a continuously differentiable convex function on an interval $I$ and let $x_0 \in I$. Consider the function $g(x) = f(x) - f'(x_0)(x-x_0)$. It is a convex function minus a linear function, so is still convex and continuously differentiable; also we have $g(x_0) = 0$ and $g'(x_0) = 0.$

We want to show that $g(x) \geq 0$ on $I$. Suppose not, so there is a $z \in I$ with $z \neq x_0$ and $g(z) < 0$. Suppose $z > x_0$; the other case is similar. By convexity, the graph of $g(x)$ lies below (or on) the line segment joining $(x_0,0)$ to $(z,g(z)$, and that means that the slope of the graph $y = g(x)$ must be less than (or equal to) the slope of the line segment, which is non-zero. This is a contradiction, so no such point $z > x_0$ can exist.

The result is true as well for a non-differentiable $f(x)$ where the right-hand and left-hand derivatives may be different at a few points, but a bit more work may be involved in that case.