I've spent some time trying to prove that the function:

$$f(x)=\frac{\exp x}{(\exp x+1)^2}$$

is even.

I tried expanding the different $\exp x$ as power series, but I had a very difficult time trying to track the different indices. Is that the correct way to proceed or is there some other property that I am no taking into account ?


  • 5
    $\begingroup$ You are on the wrong track, power series don't help here. Directly establish $f(x)=f(-x)$. Hint: $e^x/(e^x+1)^2=1/(e^x+2+e^{-x})$. $\endgroup$
    – user65203
    Mar 21 '16 at 7:10


$$\frac{e^x}{(e^x+1)^2} = \frac{e^x}{e^{2x}+2e^x+1}.$$

Now try multiplying the numerator and denominator by a certain term to get something more symmetric looking. An expanded hint below in the spoiler text if you want to try to figure it out for yourself from here without getting the full answer.

Multiply and divide by $e^{-x}$.

  • 2
    $\begingroup$ LOL, as Gandalf said: "Absurdly simple, like most riddles when you see the answer." Thanks for your quick help Cameron! $\endgroup$ Mar 21 '16 at 3:33
  • 18
    $\begingroup$ You can also multiply both numerator and denominator by $e^{-2x}$ (without expanding). That will give you $$\frac{e^{-x}}{(1+e^{-x})^2}$$which is $f(-x)$. $\endgroup$
    – Clement C.
    Mar 21 '16 at 3:35
  • 1
    $\begingroup$ @ClementC. Neat :) $\endgroup$ Mar 21 '16 at 3:39
  • $\begingroup$ @davidsaezsan you are very welcome $\endgroup$ Mar 21 '16 at 3:39

Start with $$ f(-x) = \frac{\exp(-x)}{(\exp(-x)+1)^2}. $$ Multiplying the numerator by $\exp(2x)$ yields $\exp x$.

Multiplying the denominator by $\exp(2x) = (\exp x)^2$ yields $$ \Big( (\exp x) \, (\exp(-x)+1)) \Big)^2 = \Big( 1+ \exp x\Big)^2. $$


Why are you trying expanding into series? Just use the definition – you need to show $f(-x) = f(x)$, so:

$$ \begin{align} f(-x) & = \frac{e^{-x}}{(e^{-x}+1)^2} \cdot \frac{e^{2x}}{e^{2x}} \\ & = \frac{e^{-x}\cdot e^{2x}}{(e^{-x}+1)^2\cdot (e^x)^2} \\ & = \frac{e^x}{((e^{-x}+1)\cdot e^x)^2} \\ & = \frac{e^x}{(1 + e^x)^2} \\ & = f(x) \end{align} $$


You could also reduce the expression to a symmetric form $$ f(x) = \frac{e^x}{(e^x+1)^2} \cdot \frac{e^{-x}}{e^{-x}} = \frac 1{(e^x+1)^2\cdot (e^{-x/2})^2} = \frac 1{(e^{x/2}+e^{-x/2})^2} $$ and conclude both terms in the denominator replace each other with the $x$ sign switch, thus preserving the value of a sum, hence $f$ is even.


Something that may be more generally useful: instead of proving that $f(x) = f(-x)$ directly, it's sometimes easier to show that $f(x) - f(-x) = 0$. Writing out $f(x)$ and $f(-x)$ and trying to combine terms may suggest a simplification that you wouldn't think of otherwise. In this case, you'd start with $$\frac{e^x}{(e^x + 1)^2} - \frac{e^{-x}}{(e^{-x} + 1)^2}$$ (I don't think there's any point in hiding the specific steps, since they appear in several other answers.)
You might recognize immediately that multiplying one of these terms by $e^{\pm 2x}$ on the top and bottom converts it into the other term. If not, you can try the classic technique for adding fractions with different denominators: $$\frac{e^x(e^{-x} + 1)^2}{(e^x + 1)^2(e^{-x} + 1)^2} - \frac{e^{-x}(e^{x} + 1)^2}{(e^{-x} + 1)^2(e^x + 1)^2}$$ and then it should be pretty clear that you should expand the numerators, after which the result is fairly obvious.

The same idea goes for proving that a function is odd, just show that $f(x) + f(-x) = 0$ instead of $f(x) - f(-x)$.


Just for fun:

The power series of the given function is awful to process. But that of its inverse is straightforward:


By cancellation of the odd powers, the series is $$4+x^2+2\frac{x^4}{4!}+2\frac{x^6}{6!}\cdots$$

Now, by an act of courage, let us compute the terms up to sixth degree:

$$f(x)=\frac14\frac1{1+(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots)}\\ =\frac14\left(1-\left(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots\right)+\left(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots\right)^2-\left(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots\right)^3\cdots\right)\\ =\frac14\left(1-\left(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots\right)+\left(\frac{x^4}{4^2}+2\frac{x^6}{4\cdot2\cdot4!}\cdots\right)-\left(\frac{x^6}{4^3}\cdots\right)\right)\\ =\frac14-\frac{x^2}4+\frac{13x^4}{48}-\frac{299x^6}{1440}\cdots$$

Obviously there can't be odd terms.


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