Finding All the Solutions of an Equation
The question is as follows:
Find all solutions of $2^x = \frac{1}{x}.$
I graphed the equations $y = 2^x$ and $y = \frac{1}{x}$ to find the intersection point of $(0.641, 1.56)$. The two functions did not intersect anywhere else at all. Is there an algebraic way to solve the equation? Any help will be greatly appreciated.
$\endgroup$ 45 Answers
$\begingroup$If you do not want to use Lambert function, you may have a rather good approximation of $(x\, 2^x-1)$ using the simplest $[1,1]$ Padé approximant which , built for a function $f(x)$ at $x=a$, is given by $$f(x)=\frac{f(a)+ \frac{2 f'(a)^2-f(a) f''(a)}{2 f'(a)}(x-a)} {1-\frac{f''(a)}{2 f'(a)}(x-a) }$$
Applied to your case, using $a=1$. it would write as $$\frac{1+A(x-1)}{1+B(x-1)}$$ with $$A=\frac{4+3 \log ^2(2)+6 \log (2)}{2 (1+\log (2))}\qquad \text{and}\qquad B=-\frac{\log ^2(2)+2 \log (2)}{2 (1+\log (2))}$$ Solving for $0$, $x=1-\frac{1}{A} \approx 0.64727$ which is quite close to the solution.
$\endgroup$ $\begingroup$Use the following iteration to approximate the solution
$$x_{n+1}=\frac{1}{2^{x_n}}$$
$\endgroup$ 1 $\begingroup$Let $$f (x)=x\ln(2)+\ln (x) $$ for $x>0$.
$f $ is continuous at $(0,1] $.
$$f (1)>0$$ $$\lim_{x\to 0^+}f (x)=-\infty $$
$$f'(x)=\ln (2)+\frac {1}{x}>0$$
thus the equation $f (x)=0$ has only one root between $0$ and $1$.
$\endgroup$ $\begingroup$It is not amenable to an algebraic solution. Being transcendental only numerical iteration methods ( e.g., Newton-Raphson) will work. Apart from the correct root/solution you obtained by graphing there is one more root $x=-\infty$ where your sketched curves are asymptotically approaching the x-axis.
$\endgroup$ $\begingroup$\begin{align} 2^x&=\frac1x \tag{1}\label{1} . \end{align}
There is a long-time known Lambert W functionthat is used to make it easy to solve equations like this.
\eqref{1} can be solved in terms of $\operatorname{W}$ as follows.
\begin{align} \exp(\ln(2^x))&=\frac1x ,\\ \exp(x\ln2)&=\frac1x ,\\ x\exp(x\ln2)&=1 ,\\ x\ln2\exp(x\ln2)&=\ln2 ,\\ \operatorname{W}(x\ln2\exp(x\ln2)) &=\operatorname{W}(\ln2); ,\\ x\ln2&=\operatorname{W}(\ln2) ,\\ x&=\frac{\operatorname{W}(\ln2)}{\ln2} \tag{2}\label{2} . \end{align}
And since the argument of $\operatorname{W}$ is $\ln2>0$, we know that there is only one real root in \eqref{2} and hence, in \eqref{1},
\begin{align} x&=\frac{\operatorname{W}(\ln2)}{\ln2} \approx\frac{0.4444360910}{0.6931471806} =0.6411857444 . \end{align}
And here is just a one (python) way to get the numbers:
import numpy as np
from scipy.special import lambertw as W
a=np.log(2)
print a
# 0.69314718056
print W(a)
# (0.444436091019+0j)
print W(a)/a
# (0.641185744505+0j)
