5
$\begingroup$

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

$\endgroup$
4
  • 5
    $\begingroup$ Formatting tips here. $\endgroup$
    – Em.
    May 24, 2016 at 3:39
  • $\begingroup$ This question seemed ripe for a Taylor series expansion, but I don't immediately see that it helps. $\endgroup$ May 24, 2016 at 4:22
  • $\begingroup$ It would work fine (although I don't know if OP has worked with those): the only thing keeping $ \ \log(1+e^x) \ $ from being even turns out to be the $ \ \frac{x}{2} \ $ term in the series. $\endgroup$ May 24, 2016 at 4:25
  • $\begingroup$ The Taylor series of $f(x) := \log(1+e^x)$ around $x_0=0$ will give you a null coefficients $f^{(n)}(0)=0$ for every odd terms $$ f^{(n)}(x_0) \frac{(x-x_0)^n}{n!} = f^{(n)}(0) \frac{x^n}{n!},$$ except for the first, which will be $f'(0)=1/2$, giving you the $f'(0) \frac{x}{1!}=x/2$ that will cancel out with the $-x/2$. Unfortunately, it will be tiresome to do all the math. I was doing it, but then took an arrow to the knee. =D $\endgroup$ May 24, 2016 at 14:22

3 Answers 3

27
$\begingroup$

$$\begin{align} \ln(1+e^x)-\frac x2 &= \ln(1+e^x)-\ln(e^{\frac x2}) \\ &= \ln\left((1+e^x)e^{\frac {-x}2}\right) \\ &=\ln(e^{\frac {-x}2} +e^{\frac x2}) \\ \end{align}$$ From this it should be obvious that the function is indeed even.

$\endgroup$
3
  • $\begingroup$ +1 It might be mentioned that you nicely "unmask" this function as being $ \ \ln ( \ \cosh \left[ \frac{x}{2} \right] \ ) \ $ , the logarithm of an even and always positive function, and thus expected to be even. $\endgroup$ May 24, 2016 at 4:36
  • $\begingroup$ @RecklessReckoner A little nitpicking, but shouldn't it be $ \ln \left( 2\cosh ( \frac{x}{2} ) \right )$? :) $\endgroup$
    – BigbearZzz
    May 24, 2016 at 4:39
  • $\begingroup$ Yes, I slipped a factor of $ \ 2 \ $ : I went back to check it using the "double-angle formula" for cosh and see that I goofed using the first formula I applied. $\endgroup$ May 24, 2016 at 4:52
6
$\begingroup$

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) \ $ .]

$\endgroup$
2
  • 1
    $\begingroup$ +1. Note that this apply to all (real-valued) functions, not only the continuous ones. $\endgroup$
    – Taladris
    May 24, 2016 at 4:18
  • 1
    $\begingroup$ I was "playing it safe": this function is continuous everywhere, so the technique would definitely work here. $\endgroup$ May 24, 2016 at 4:21
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.