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.

This theorem requires a proof. In particular: Analysis needed of the Kepler Equation You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Also see

• Definition:Kepler's Equation

Sources

• 1925: Édouard Goursat: Cours d'Analyse Mathématique: Volume $\text { 2 }$ (4th ed.)
• 1983: François Le Lionnais and Jean Brette: Les Nombres Remarquables ... (previous) ... (next): $1,19966 78640 25773 4 \ldots$