Root of exponential equation

Question

I am trying to find the roots of the equation

$$ e^{x} -\cos x = 0. $$

Used the Lambert W function to arrive at $$ x = W(x\cos x), $$ but I don't know how to proceed from there to get the explicit roots. Any help is much welcome

Answer 1

Since $-1\le \cos x\le 1$ and $e^x>0$ with $e^x>1, \; x>0$, by IVT we have infinitely many solutions for $x\le 0$, one is the trivial $x=0$ the others can by found by numerical methods.

Notably we have $2$ not trivial roots for each pair of intervals with $x<0$ such that $0<\cos x<1$ that is for $n=0,1,2,\dots$ that is

$$-\frac{\pi}2 -2\pi n \pi <x<-2\pi n$$

$$-2\pi-2\pi n<x<-\frac32 \pi -2\pi n$$

Note also that for $|x|$ very large, roots are well approximated by the roots for $\cos x=0$ with $x<0$ that is $x\approx -\frac{\pi}2-n\pi$.

Answer 2

You could use numerical methods

You could use maclaurin's expansion

$e^x-\cos x = x+x^2+\frac{x^3}{6}$

Upto 3 degree would be good

Now this becomes a cubic equation $x=0$

Now this becomes a quadratic equation

$1+x+\frac{x^2}{6}$ Roots are -4.725,-1.281

For large negative x $e^{-x} \approx 0$ so $\cos x = 0$ which gives $x \approx -(2n+1)\frac{\pi}{2}$ for $n=1,2,3....$$

Answer 3

A supposed hint is that by using the Mclaurin-series for both the $e^x$ and $cos(x)$, we can arrive at this;

$$x^2\sum_{k=0}^{\infty}\frac{x^{4k}}{(4k+2)!}+\frac{1}{2}\sinh\left(x\right)=0$$

Now take,

$a=\sum_{k=0}^{\infty}\frac{x^{4k}}{(4k+2)!}$

Now we can get different expressions for the same thing; $$\ln\left(-2x^{2}a+\sqrt{4x^{4}a^{2}+1}\right)-x=0$$ $$e^{x}4x^{2}a+e^{2x}-1=0$$

Answer 4

The previous answers already told where are more or less located the roots of the equation.

We can obtaing quite good approximations of them if, instead of Taylor expansions, we use the simplest $[1,1]$ Padé approximant. For a function $f(x)$ developed around $x=a$, it would write $$f(x) \sim \frac{f(a)+ \left(f'(a)-\frac{f(a) f''(a)}{2 f'(a)}\right)(x-a) } {1-\frac{ f''(a)}{2 f'(a)}(x-a) }$$ This is equivalent to a Taylor series up to $O\left((x-a)^3\right)$ and then the approximation $$x=a+\frac{2 f(a) f'(a)}{f(a) f''(a)-2 f'(a)^2}$$ Using $a=-(2n-1)\frac \pi 2$, in the table below are given the estimates and the solutions for the first roots. $$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 1 & -1.299002559 & -1.292695719 \\ 2 & -4.721292643 & -4.721292759 \\ 3 & -7.853593280 & -7.853593280 \\ 4 & -10.99559106 & -10.99559106 \\ 5 & -14.13716622 & -14.13716622 \\ 6 & -17.27875963 & -17.27875963 \\ 7 & -20.42035225 & -20.42035225 \\ 8 & -23.56194490 & -23.56194490 \\ 9 & -26.70353756 & -26.70353756 \\ 10 & -29.84513021 & -29.84513021 \end{array} \right)$$