So I came across this equation:
xe^x=1
how would you solve that? there seems to be no analytical way of solving it. my teacher said something about iteration but any help would be appreciated. thanks!
I agree that x = 0 cannot be a solution.
Use the Lambert W function or an approximation method, like the Newton Raphson Method, to solve for x.
The Lambert W function has a simple solution, or if you plan to use the Newton Raphson method: $\displaystyle f(x) = xe^x -1$ then $\displaystyle f'(x) = xe^x + e^x$
Newton's method - Wikipedia, the free encyclopedia
$\displaystyle f(x) = xe^x - 1$
$\displaystyle f'(x) = e^x(x+1)$
$\displaystyle f(0) = -1$ , $\displaystyle f(1) = e-1 $
since f(x) is continuous, there exists a zero between x = 0 and x = 1
Newton's method ...
$\displaystyle x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$
let $\displaystyle x_0 = 0.5$
$\displaystyle x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0.571020439808...$
$\displaystyle x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.567155568744...$
$\displaystyle x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 0.567143290533...$
continued iterations will be extremely close to $\displaystyle x_3$