Why is $\log(1+e^x) - \frac{x}{2}$ even? 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
 A: Another way of treating this is that a continuous function $ \ f(x) \ $ can be "separated" into "even" and "odd" components,
$$ f_e(x) \ = \ \frac{f(x) \ +  \ f(-x)}{2} \ \ \ \text{and} \ \ \  f_o(x) \ = \ \frac{f(x) \ -  \ f(-x)}{2} \ \ . $$
Here, we have
$$   f_o(x) \ = \ \frac{[ \ \log(1+e^x) \ - \ \frac{x}{2} \ ] \ -  \ [ \ \log(1+e^{-x}) \ - \ \frac{(-x)}{2} \ ]}{2}  $$
$$ = \ \frac{ \ \log(1+e^x) \  -  \  \ \log(1+e^{-x}) \ - \ x }{2}  $$
$$ = \ \frac{1}{2} \ \left[ \ \log \left(\frac{1+e^x}{1+e^{-x}} \right)   \  - \ x \ \right] \  = \ \frac{1}{2} \ \left[ \ \log \left(\frac{e^x \ [1+e^x]}{e^x+1} \right)   \  - \ x \ \right] \ \ $$
$$ = \ \frac{1}{2} \ [ \ \log (e^x  )   \  - \ x \ ] \ \ = \ \ \frac{1}{2} \ [ \ x  \  - \ x \ ] \ = \ 0 \ \ . $$
Our function has zero "odd component", so it is purely even.  [We could also have shown that $ \ f_e(x) \ = \ f(x) \ $ .]
A: Why do you say that checking $f(-x)=f(x)$ doesn't work? Of course it does. $\ddot\smile$ $$\ln(1+e^{-x})-\frac{-x}2=\ln\frac{e^x+1}{e^x}+\frac x2=\ln(e^x+1)-\ln e^x+\frac x2=\\=\ln(1+e^x)-x+\frac x2=\ln(1+e^x)-\frac x2$$
A: $$\begin{align}
\ln(1+e^x)-\frac x2  &=  \ln(1+e^x)-\ln(e^{\frac x2}) \\
&= \ln\left((1+e^x)e^{\frac {-x}2}\right) \\
&=\ln(e^{\frac {-x}2} +e^{\frac x2}) \\
\end{align}$$
From this it should be obvious that the function is indeed even.
