# I don't know if there's a mathod for solving this.

• Apr 28th 2011, 03:24 PM
scounged
I don't know if there's a mathod for solving this.
I'm just wondering if there's a way to solve this equation algebraically:

\$\displaystyle e^{x}\$+\$\displaystyle x=0\$

My calculator gives me that x is approximately -0.567.
I suspect that if there's a way to solve this it probably has something to do with complex numbers, but I don't know as I haven't started using them yet.

And btw, is there something wrong with Latex, or is it just my computer that is getting a little anxious over something?
• Apr 28th 2011, 03:27 PM
pickslides
There is no way to solve it algebraically. Your calculator has employed a numerical method to find an approximation
• Apr 28th 2011, 03:39 PM
scounged
Hmmm. That's too bad really, but ah well...
• Apr 28th 2011, 03:42 PM
pickslides
Never fear, you can solve it yourself using numerical methods. If you are interested research 'the bisection method' and if you know a little bit of calculus try 'newtons method'.
• Apr 28th 2011, 08:59 PM
mr fantastic
Quote:

Originally Posted by scounged
I'm just wondering if there's a way to solve this equation algebraically:

\$\displaystyle e^{x}\$+\$\displaystyle x=0\$

My calculator gives me that x is approximately -0.567.
I suspect that if there's a way to solve this it probably has something to do with complex numbers, but I don't know as I haven't started using them yet.

And btw, is there something wrong with Latex, or is it just my computer that is getting a little anxious over something?

\$\displaystyle e^x + x = 0 \Rightarrow e^x = -x \Rightarrow -x e^{-x} = 1 \Rightarrow -x = W(1) \Rightarrow x = -W(1)\$

where W(x) is the Lambert W-function:

solve Exp&#91;x&#93; &#43; x &#61; 0 - Wolfram|Alpha
• May 1st 2011, 09:38 AM
scounged
After doing a little bit of research I see now that x=-(Omega constant), or x=-W(1)