$e^x = \cos(x) + 1$ with elementary functions [closed]

This question just got asked by a user and then closed, even though unanswered. I have doubts that it is even possible to express the root in terms of elementary functions.

By attempting to solve $$ e^x = \cos(x) + 1 $$ I quickly recognised that this probably has no simple solutions. Now I'm asking myself the question whether there is some sequence that converges to the root of $$ e^x-\cos(x) -1 . $$ So far I couldn't find one. Does somebody have any ideas? (I should find the solution in terms of elementary functions / constants, the latter consisting of $e$ and $\pi$, i.e. in particular no numerical solution but an "exact" in this sense)

Solution 1:

Consider that we look for the negative zero's of function $$f(x)=e^x-\cos(x)-1$$ they are located close to $a=-(2n+1)\pi$.

Using a series expansion around $x=a$, we have as an approximation $$f(x)=e^{a}+e^{a}(x-a)+\frac 12(e^{a}-1)(x-a)^2+O\left((x-a)^3\right)$$ and then the roots $$x_\pm=a+\frac{e^a \pm\sqrt{e^a \left(2-e^a\right)}}{1-e^a}\quad \text{with}\qquad a=-(2n+1)\pi$$ Some results $$\left( \begin{array}{ccccc} n & \text{estimate} & \text{exact} & \text{estimate} & \text{exact} \\ 0 & -3.400353505 & -3.400592904 & -2.792500392 & -2.789129646 \\ 1 & -9.437402316 & -9.437402397 & -9.411992194 & -9.411992104 \\ 2 & -15.70851212 & -15.70851212 & -15.70741411 & -15.70741411 \\ 3 & -21.99117230 & -21.99117230 & -21.99112485 & -21.99112485 \\ 4 & -28.27433491 & -28.27433491 & -28.27433286 & -28.27433286 \\ 5 & -34.55751923 & -34.55751923 & -34.55751915 & -34.55751915 \end{array} \right)$$

This is the simplest approximation.

It can be improved using the $[2,m]$ Padé approximant of $f(x)$ and the solutions are still obtained at the price of a quadratic equation in $(x-a)$. The formulae will not be given here but the results for the first negative root are

$$\left( \begin{array}{cc} m & x_- & x_+ \\ 0 & -3.40035350459848 & -2.79250039185374 \\ 1 & -3.39992453955299 & -2.79144452980351 \\ 2 & -3.40059832774563 & -2.78918524129760 \\ 3 & -3.40059015587550 & -2.78914785893247 \\ 4 & -3.40059296364765 & -2.78913037876924 \\ 5 & -3.40059289328230 & -2.78912978260528 \\ \cdots & \cdots & \cdots \\ \infty & -3.40059290398227 & -2.78912964643395 \end{array} \right)$$

Edit

This could be improved using a better choice of $a$. Expanding the first derivative as a series around $x=-(2n+1)\pi$ and then series reversion, we have the maximum value at $$a'=a+\frac{e^a}{1-e^a}+\frac{e^{3 a}}{2 \left(1-e^a\right)^3}+\frac{e^{3 a} \left(1+2 e^{2 a}\right)}{6 \left(1-e^a\right)^5}+\cdots$$ Using this truncated series, for $n=0$, $a'=\color{red}{-3.09636}404$ while the "exact" value is $ \color{red}{-3.09636393}$

The process used in the first part of this answer is now repeatable with $a'$ instead of $a$.