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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.

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    $\begingroup$ It does not hold for $x = 0$ because the expression is undefined. $\endgroup$ – user217285 Jul 4 '15 at 23:14
  • $\begingroup$ I edited my post to ask my question more accurate, sorry for that. $\endgroup$ – DerJFK Jul 4 '15 at 23:21
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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.

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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.

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

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?

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    $\begingroup$ Those roots are where $\tanh x = 0$, not where $\tanh x = x$. $\endgroup$ – Antonio Vargas Jul 5 '15 at 0:28

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