TITLE: Why is $\log(1+e^x) - \frac{x}{2}$ even?
QUESTION [5 upvotes]: I'm dealing with Fourier series and I'm trying to figure out $\log(1+e^x) - \frac{x}{2}$ is even??? I've tried the $f(-x) = f(x)$ method but it doesn't give me the equality. But I've plotted it, and it is even? :S

REPLY [3 votes]: Why do you say that checking $f(-x)=f(x)$ doesn't work? Of course it does. $\ddot\smile$ $$\ln(1+e^{-x})-\frac{-x}2=\ln\frac{e^x+1}{e^x}+\frac x2=\ln(e^x+1)-\ln e^x+\frac x2=\\=\ln(1+e^x)-x+\frac x2=\ln(1+e^x)-\frac x2$$