Root of Equation e^x (x - 1) = e^-x (x + 1)

Theorem
The equation:
 * $e^x \paren {x - 1} = e^{-x} \paren {x + 1}$

has a root:
 * $x = 1 \cdotp 19966 \, 78640 \, 25773 \, 4 \ldots$

Proof
Let $\map f x = e^x \paren {x - 1} - e^{-x} \paren {x + 1}$.

Then if $\map f c = 0$, $c$ is a root of $e^x \paren {x - 1} = e^{-x} \paren {x + 1}$.

Notice that:


 * $\map f 1 = e^1 \paren {1 - 1} - e^{-1} \paren {1 + 1} = -\dfrac 2 e < 0$


 * $\map f 2 = e^2 \paren {2 - 1} - e^{-2} \paren {2 + 1} = e^2 - \dfrac 3 {e^2} > 0$

By Intermediate Value Theorem:


 * $\exists c \in \openint 1 2: \map f c = 0$.

This shows that our equation has a root between $1$ and $2$.

The exact value of this root can be found using any numerical method, e.g. Newton's Method.

Also see

 * Definition:Kepler's Equation