Calculate roots from $\frac{x \cosh(x) - \sinh(x)}{x^2}$

Question

I want to solve the following equation

$$f(x) = \frac{x \cosh(x) - \sinh(x)}{x^2} = 0$$

Because the term above is undefined for $x = 0$ I calcuted

$$\lim_{x \rightarrow 0}\frac{x \cosh(x) - \sinh(x)}{x^2} =0$$

So the function has one root at "$x =0$", but how can I show that for any $x \neq 0$ it does not have any root?

I transformed it to

$$x = \frac{\sinh(x)}{\cosh(x)} =\tanh(x)$$

But I do not get a idea how to show it formally. I would appreciate any hint.

Answer 1

If you're looking for a solution without complex analysis (assuming from your real analysis tag), we can apply calculus.

Let $f(x) = x - \tanh(x)$. Then $f'(x) = 1 - \text{sech}^2(x) = \tanh^2(x) \geq 0,$ with equality if and only if $x = 0$. Since $f$ is smooth, it is increasing on the interval $(-\infty,\infty)$, so the only solution is $x = 0$. But the original function is not defined at $x = 0$, so the equation has no real solutions.

Answer 2

$$\frac{x \cosh(x) - \sinh(x)}{x^2} = 0 \iff x\cosh x = \sinh x$$

This reduces to $$\tanh x = x$$ which has roots $x = i\pi n$ for $n \in \mathbb{Z}$. The only real root is $x=0$ where your expression is undefined. The rest of the roots are all imaginary.

Can you think of a calculus-based way (hint: think intersection between curve and tangents) to prove that $\tanh x =x$ has no non-origin solution?

Answer 3

The function $\tanh x$ is concave for $x\ge0$ and convex for $x\le 0$ (second derivative test) and its tangent at origin is $y=x$, hence its graph is above its tangents for $x>0$, below its tangents for $x<0$, which proves it can't intersect the tangent at origin at another point.