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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

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    $\begingroup$ Formatting tips here. $\endgroup$ – Em. May 24 '16 at 3:39
  • $\begingroup$ This question seemed ripe for a Taylor series expansion, but I don't immediately see that it helps. $\endgroup$ – Austin Mohr May 24 '16 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. $\endgroup$ – colormegone May 24 '16 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 $\endgroup$ – Marcelo Ventura May 24 '16 at 14:22
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$$\begin{align} \ln(1+e^x)-\frac x2 &= \ln(1+e^x)-\ln(e^{\frac x2}) \\ &= \ln\left((1+e^x)e^{\frac {-x}2}\right) \\ &=\ln(e^{\frac {-x}2} +e^{\frac x2}) \\ \end{align}$$ From this it should be obvious that the function is indeed even.

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  • $\begingroup$ +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. $\endgroup$ – colormegone May 24 '16 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 '16 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. $\endgroup$ – colormegone May 24 '16 at 4:52
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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) \ $ .]

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    $\begingroup$ +1. Note that this apply to all (real-valued) functions, not only the continuous ones. $\endgroup$ – Taladris May 24 '16 at 4:18
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    $\begingroup$ I was "playing it safe": this function is continuous everywhere, so the technique would definitely work here. $\endgroup$ – colormegone May 24 '16 at 4:21
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Why do you say that checking $f(-x)=f(x)$ doesn't work? Of course it does. $\ddot\smile$ $$\ln(1+e^{-x})-\frac{-x}2=\ln\frac{e^x+1}{e^x}+\frac x2=\ln(e^x+1)-\ln e^x+\frac x2=\\=\ln(1+e^x)-x+\frac x2=\ln(1+e^x)-\frac x2$$

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