Finding roots of a nonlinear equation

Question

I plotted the functions $y=2x$ and $y=\ln(1+x)$ and they have two intersecting points. So the equation $2x=\ln(1+x)$ is supposed to have two roots. One of them is obviously $x=0$. How can I get the other one?

Answer 1

There is no analytic solution to this equation in terms of elementary functions. Yet, good approximations can still be obtained.

For example, let the root be $r=-1+a$ and expand the function

$$f(x) = e^{2x} - x -1$$

around $x=-1$ to get,

$$f(r) = 0 \approx f(-1) +f'(-1)a + \frac 12 f''(-1) a^2\tag{1}$$

Evaluate,

$$f(-1) = e^{-2}, \>\>\> f'(-1) = 2e^{-2}-1, \>\>\> f''(-1) = 4e^{-2}$$

Plug above values into (1) to obtain the approximate root,

$$r \approx -1 + \frac 14\left( e^2 -2 - \sqrt{e^4-4e^2-4}\right) = -0.799$$

Answer 2

If $x = -t/2-1$, the equation says $\ln(-t/2) = -t - 2$. Exponentiating, $-t/2 = \exp(-t-2)$, i.e. $t \exp(t) = -2 \exp(-2)$. Thus $t$ is one of the branches of $W(-2 \exp(-2))$, where $W$ is the Lambert W function. Since $-1/e < -2 \exp(-2) < 0$, there are two real branches, the principal branch and the $-1$ branch. In this case the $-1$ branch is $-2$, leading to $x=0$, the principal branch is approximately $-0.4063757400$, leading to $x = -t/2-1 \approx -0.7968121301$.