Solving this exponential

Question

I'm trying to solve

$$ e^\lambda (1-\lambda^2) = 1$$

I know it has a solution at $\lambda = 0$. How do I get the second solution? Wolfram Alpha gives something around 0.71, but doesn't show an expression. Can I only get that numerically?

Answer 1

Here's one way to attack the problem:

Rearrange $e^{\lambda} \left (1- \lambda ^2 \right )=1$ as $1- \lambda^2 =e^{-\lambda}$.

Then use the series expression for $e^{-\lambda}$:

$1- \lambda^2 =1-\lambda+\frac {\lambda ^2}2 -\frac {\lambda ^3}6 + \frac {\lambda ^4}{4!}+...$

You can then decide on how many terms you want to take it to and solve the resulting polynomial to gain an approximation to $\lambda$

Example 1 (three terms):

$1- \lambda^2 \approx 1-\lambda+\frac {\lambda ^2}2$ yields $\frac {3\lambda ^2}2-\lambda=0 \Rightarrow \lambda = \frac 23$

Example 2 (four terms):

$1- \lambda^2 \approx 1-\lambda+\frac {\lambda ^2}2-\frac {\lambda ^3}6$ yields $\frac {\lambda ^3}6 - \frac {3\lambda ^2}2+\lambda=0$

$\Rightarrow \lambda ^3 - 9 \lambda ^2+6 \lambda=0$

$\Rightarrow \lambda ^2 - 9 \lambda+6 =0$ or $\lambda=0$

$\Rightarrow \lambda = \frac {9 + \sqrt {57}}2 $ or $\lambda = \frac {9 - \sqrt {57}}2 $ or $\lambda=0$

$\lambda = \frac {9 + \sqrt {57}}2 $ would not converge sufficiently in just four terms.

Using $\lambda = \frac {9 - \sqrt {57}}2 $ gives $\lambda \approx 0.725$ - you can see that we are approaching the estimate that Wolfram gave.