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!

Printable View

- Nov 13th 2010, 09:45 AMYehiaSolving xe^x = 1
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! - Nov 13th 2010, 09:50 AMrtblue
Well, from intuition, we see that x=0 is a viable option.

$\displaystyle xe^x=1$

multiply both sides by xe.

$\displaystyle (xe^x)*(xe)=xe$

add the exponents on the left side, and we get:

$\displaystyle x+1=1$

$\displaystyle x=0$ - Nov 13th 2010, 09:57 AMYehia
- Nov 13th 2010, 10:11 AMFernandoRevilla
- Nov 13th 2010, 11:54 AMEducated
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 $\displaystyle f'(x) = xe^x + e^x$ - Nov 14th 2010, 06:58 AMrtblue
Look at the given function. If x = 0, then the whole left side becomes one. Anything to power zero is one.

- Nov 14th 2010, 07:26 AMUnknown008
Yes, but if x becomes 0, you have $\displaystyle 0e^0 = 0$

- Nov 14th 2010, 08:09 AMWilmer
x = ~.56714....

- Nov 14th 2010, 10:18 AMskeeter
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 $\displaystyle x_3$