3
$\begingroup$

Is there any simplification or other interesting transformation of:

$\log_e{(1+e^x)}$

(where $x \in \mathbb{R}$) ?

  • 0
    @robjohn, I know how to prove the expression is not equal to a fixed candidate function, such as $\frac{1}{2} \left( x + \sqrt{1 + x^2}\right),$ but I do not know how to answer Why Not at a precalculus level.2012-10-20

2 Answers 2

3

You can use the Taylor series of $\log$: $ \log(1+e^x) = x+\log(1+e^{-x}) = x + e^{-x} - \frac{e^{-2x}}{2} + \frac{e^{-3x}}{3} - \frac{e^{-4x}}{4} + \cdots. $ This assumes $x \geq 0$. If $x \leq 0$ you should use $ \log(1+e^x) = e^x - \frac{e^{2x}}{2} + \frac{e^{3x}}{3} - \frac{e^{4x}}{4} + \cdots. $ Both series converge quickly only if $|x|$ is large.

  • 1
    Write $1 + e^x = e^x(1+e^{-x})$ and uses the laws of logarithms.2017-06-22
2

Manipulations of this and similar quantities like $ \frac{e^x}{1+e^x}$, or $e^{-x} (1 + e^x)$, and their logarithms, are common in logistic regression in statistics.

There is no closed form simplification but you can expand $\log(1 \pm t) = \pm t + \frac{t^2}{2} \pm \frac{t^2}{3} + \cdots$ in a power series for specially chosen $t$ (smaller than 1 in absolute value) to illustrate how $f(x) = \log (1 + e^x)$ is approximately $x$. Possibilities include

  • to write $f(x) = x + \log (1 + e^{-x})$ and use $t = e^{-x}$ to get a series valid for positive $x$,

  • or $t = \frac{1}{1+e^x}$, for which $\log (1 + e^x) = x - \log(1-t) = x + t + \frac{t^2}{2} + \frac{t^3}{3} + \cdots$, is valid for all $x$.