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I need help solving the following function for $x$

$$g(x) = x - x \cdot \cosh\left(\frac{1}{2x}\right)$$

As I have never used hyperbolic functions, all my attempts at solving this have failed miserably. The closest that I got was (using hightschool math)

$$\frac{1}{2x} = \cosh^{-1}\left(1 - \frac{g}{x}\right)$$

but that is going nowhere.

Note: a similar question has been asked here but it did not help me.

Wendelin
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1 Answers1

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Hint

You cannot solve for $x$ the equation $$g(x) = x - x \, \cosh\left(\frac{1}{2x}\right)$$ because it is a transcendental equation.

But what you could do is to expand the rhs as a series for large values of $x$ (better will be to let $x=\frac 1t$) and you will get $$g(x)=\sum_{n=0}^\infty \frac {a_n}{x^n}$$ Take very few terms and use series reversion to have an approximation of $x$ as a function of $g(x)$.

You will (may be) be surprised to see how good it is.

Claude Leibovici
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  • If I knew more math I'ld do what you suggested but it's still helpful because now I am not wasting my time trying solve the equation. – Wendelin Jan 12 '22 at 13:32
  • @Wendelin. First make $x=\frac 1t$ and expand as a series around $t=0$ (two or three terms, no more). Look how good is the approximation of the rhs. Now, you have a small polynomial equation. Do it; if you report in the question, I shall try to help you. – Claude Leibovici Jan 12 '22 at 13:38