# How to find the MacLaurin series of $\frac{1}{1+e^x}$

Mathcad software gives me the answer as: $$\frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$ I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1}$ as in the binomial MacLaurin expansion. I want until the $x^4$ term. So what i found is the following:

$$(1+e^x)^{-1} = 1 - e^x + e^{2x} + e^{3x} + e^{4x}+\cdots\tag1$$ Then by expanding each $e^{nx}$ term individually: $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots\tag2$$ $$e^{2x} = 1 + 2x + 2x^2 + \frac{3x^3}{2} + \frac{2x^4}{3}+\cdots\tag3$$ $$e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \frac{27x^4}{8}+\cdots\tag4$$ $$e^{4x} = 1 + 4x + 8x^2 + \frac{32x^3}{3} + \frac{32x^4}{3}+\cdots\tag5$$ So substituting $(2),(3),(4),(5)$ into $(1)$ i get: $$(1+e^x)^{-1} = 1 +2x+5x^2+ \frac{22x^3}{3} + \frac{95x^4}{12} +\cdots$$ Which isn't the correct result. Am i not allowed to expand this series binomially? I've seen on this site that this is an asymptotic expansion. However, i don't know about these and i haven't been able to find much information on this matter to solve this. If someone could help me understand how to solve it and why my approach isn't correct, i would be VERY grateful. Thanks in advance.

• We have $(1+e^x)^{-1}=1-e^x+e^{2x}-\cdots$ only if $x<0$. If $x>0$, you need to factor $e^x$ out of the denominator. Aug 21, 2015 at 12:55
• You forgot the series for $(1+u)^{-1}$ converges if and only if $\lvert u\rvert <1$, which is not the case of $\mathrm e^x$. Aug 21, 2015 at 13:05
• Bernard is correct, @RestlessC0bra: If $e^x<1$, we have $x<0$. We need it for the geometric series to converge. Otherwise, you can do what I (and mwomath) suggested, which is to factor $e^x$ out of the denominator. Aug 21, 2015 at 13:06
• Expanding $1/(1+a)$, you get $1-a+a^2-a^3+a^4-\cdots$, with alternating signs, not $1-a+a^2+a^4+ \cdots$, with the same sign in each term after the $1$st-degree term. It's not clear why you stopped after $(5)$. ${}\qquad{}$ Aug 21, 2015 at 13:07
• @RestlessC0bra : If $x$ is real, the $e^x>0$. But one needs $e^x<1$, which means $x<0$. ${}\qquad{}$ Aug 21, 2015 at 13:08

Let $\dfrac1{1+e^x}=a_0+a_1x+a_2x^2+\cdots$

$$\implies(1+e^x)\left(a_0+a_1x+a_2x^2+\cdots\right)=1$$

As $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$

$$\implies\left(2+\dfrac x1+\dfrac{x^2}2+\dfrac{x^3}{3!}+\cdots\right)\left(a_0+a_1x+a_2x^2+\cdots\right)=1$$

Comparing the coefficients of different powers of $x,$

$\implies1=2a_0\iff a_0=?$

$0=2a_1+a_0\iff a_1=?$

$0=2a_2+a_1+\dfrac{a_0}2$

and so on

• This is a nice, simple approach. Aug 21, 2015 at 17:33
• And what is the radius of convergence? Dec 27, 2019 at 21:59
• @BeesaFangirlDOTO, The Series should converge when the series of $e^x$ converges. Dec 28, 2019 at 10:46

Just a sketch of a proof to be expanded later (sorry, I am in a hurry): by taking the logarithmic derivative of the Weierstrass product of the $\cosh$ function we have the Taylor series of the $\tanh$ function, with the coefficients depending on values of the $\zeta$ function at even integers. With a little maquillage it is not difficult to turn that into the Taylor series of $\frac{1}{1+e^x}$. Or we may just play a bit with the generating function of Bernoulli numbers.

So, I was saying: $$\cosh z = \prod_{n\geq 0}\left(1+\frac{4z^2}{(2n+1)^2 \pi^2}\right)\tag{1}$$ hence: $$\tanh z = \sum_{n\geq 0}\frac{8z}{(2n+1)^2 \pi^2 + 4z^2}\tag{2}$$ leads to: $$\tanh z = \sum_{m\geq 0}(-1)^m z^{2m+1}\sum_{n\geq 0}\frac{2^{2m+3}}{\pi^{2m+2}(2n+1)^{2m+2}}\tag{3}$$ by expanding the general term of the RHS of $(2)$ as a geometric series. Computing the innermost sum in terms of the $\zeta$ function it follows that: $$\tanh z = \sum_{m\geq 0}\frac{2(-1)^m\left(2^{2m+2}-1\right) }{\pi^{2m+2}}\zeta(2m+2)\, z^{2m+1}\tag{4}$$ but since $\frac{1}{1+e^{z}}=\frac{1-\tanh(z/2)}{2}$ it follows that: $$\frac{1}{1+e^{z}}=\frac{1}{2}-\sum_{m\geq 0}\frac{(-1)^m\left(2^{2m+2}-1\right) }{2^{2m+1}\pi^{2m+2}}\zeta(2m+2)\, z^{2m+1}\tag{5}$$ so to solve the original problem it is enough to recall that $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$.

Moreover, through $(5)$ it is straightforward to compute the magnitude of the coefficients by simply approximating $\zeta(2m+2)$ with $1$. On the other hand, that is trivial also by considering that the radius of convergence of the Taylor series of $f(z)=\frac{1}{1+e^z}$ at $z=0$ is $\pi$, since $f(z)$ is a meromorphic function with two simple poles at $z=\pm \pi i$.

• I haven't delved that deep into mathematics yet, but i'd certainly want to see the proof. Aug 21, 2015 at 13:39
• @RestlessC0bra: please just wait a little, I'll type it later. Aug 21, 2015 at 13:40
• @RestlessC0bra: done ;) Aug 21, 2015 at 15:46

Since you're just looking for the first few terms: $$f(x)=\frac{1}{1+e^x} \implies f(0)=\frac 12$$ $$f'(x)=\frac{-e^x}{(1+e^x)^2} \implies f'(0)=-\frac 14$$ $$f''(x)=\frac{-e^x}{(1+e^x)^2}+2\frac{e^{2x}}{(1+e^x)^3}\implies f''(0)=0$$ $$f'''(x)=f''(x)+4\frac{e^{2x}}{(1+e^x)^3}-3\times 2\frac{e^x}{(1+e^x)^4}\implies f'''(0)=\frac 18$$

You get: $$\frac{1}{1+e^x}=\frac 1{2\times 0!} -\frac 1{4\times 1!} x+\frac 1{8\times 3!}x^3+\mathcal O (x^5)$$ $$\frac{1}{1+e^x}=\frac 1{2} -\frac 1{4} x+\frac 1{48}x^3+ \mathcal O (x^5)$$

• @RestlessC0bra Since this is a truncated Taylor series we need this "Big-O" notation to describe how closely the series approximates the given function. Aug 21, 2015 at 15:15
• I mark this as the accepted answer because it was the most simple for me. But i'll study all the other answers as well. Thank you all. Aug 21, 2015 at 20:23
• @RestlessC0bra - how come your username is spelt with a lower case "o" but when typing it with an "@" it appears with a zero or "0"? maybe now you can also consider using $\Theta$ instead... :) Aug 21, 2015 at 21:12
• @GeorgSaliba - what's the difference between $O(.)$ and $\Theta (.)$ here? Aug 21, 2015 at 21:13
• @hypergeometric Sorry, it should be $\mathcal O (.)$ because it's an upper bound not a tight bound (where $\Theta (.)$ would've been correct). Aug 21, 2015 at 21:21

Here's another way to derive it term by term: $$y^{-1}=1+e^x\Rightarrow -y^{-2}y'=e^x=y^{-1}-1\Rightarrow y'=y^2-y$$

Now you can continue to differentiate with the minimum of fuss:

$$y''=2yy'-y'$$ $$y'''=2yy''+2y'y'-y''$$ $$y^{(4)}=2yy'''+6y''y'-y'''$$

And so on, evaluating at each step.

$\begin{array}\\ f(x) &=\frac{1}{1+e^x}\\ &=\frac{1}{2+(e^x-1)}\\ &=\frac12\frac{1}{1+(e^x-1)/2}\\ &=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(e^x-1)^n}{2^n}\\ &=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(x+x^2/2+x^3/6+...)^n}{2^n}\\ &=\frac12\sum_{n=0}^{\infty}(-1)^n(x/2)^n(1+x/2+x^2/6+...)^n\\ &=\frac12\sum_{n=0}^{\infty}(-1)^n(x/2)^n(1+nx/2+...)\\ &=\frac12\left(1-\frac{x}{2}(1+\frac{x}{2}+...)+\frac{x^2}{4}(1+...)+...\right)\\ &=\frac12\left(1-\frac{x}{2}-\frac{x^2}{4} +\frac{x^2}{4}(1+...)+...\right)\\ &=\frac12\left(1-\frac{x}{2}+O(x^3)\right)\\ &=\frac12-\frac{x}{4}+O(x^3)\\ \end{array}$