# Any simplification of $\log_e{(1+e^x)}?$

Is there any simplification or other interesting transformation of:

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

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

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No. ${}{}{}{}{}$ – Will Jagy Oct 16 '12 at 2:15
@WillJagy Why not? – fhyve Oct 16 '12 at 2:41
@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. – Will Jagy Oct 20 '12 at 19:19

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 $x$.

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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.

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