If $|x|<1$, then what is the coefficient of $x^3$ in expansion of $\dfrac{1}{e^x\cdot (1+x)}$?

My Try: So I rewrote this by taking $e^x$ to the numerator to get $\frac{e^{-x}}{(1+x)}$. My problem is, how can I deduce the coefficient of $x^3$ here, since I would need to divide the numerator by the denominator, which is a cumbersome task. Is there a simpler way? How can I solve problems like these? I'd love to know.

  • $\begingroup$ You just need that term. Do a few steps of the long division. $\endgroup$ – Hellen Sep 3 '17 at 14:08
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    $\begingroup$ Is really long division the only method? What if I was asked the coefficient of $x^200$? I would certainly not divide then. There has to be another, more logical way. $\endgroup$ – Tanuj Sep 3 '17 at 14:09
  • $\begingroup$ Not all the terms of all Taylor series have nice expressions. Some of them may take a lot of effort. I, too, hate it when I need to calculate the $x^5$ term of $\tan x$. Here you can also rewrite $1/(1+x)=1-x+x^2-x^3+\cdots$ and multiply that term-by-term with the series of $e^{-x}$. This is still easy. $\endgroup$ – Jyrki Lahtonen Sep 3 '17 at 14:12
  • $\begingroup$ @JyrkiLahtonen But in this case, both Taylor series are as nice as it gets... $\endgroup$ – Clement C. Sep 3 '17 at 14:17
  • $\begingroup$ @ClementC. Quite. But I still wouldn't want to calculate the $x^{200}$ term :-) $\endgroup$ – Jyrki Lahtonen Sep 3 '17 at 14:50

Hint. Note that $$\frac{1}{e^x\cdot(1+x)}=\frac{e^{-x}}{1-(-x)}=\left(\sum_{n=0}^{\infty} \frac{(-x)^n}{n!}\right)\cdot \left(\sum_{n=0}^{\infty} {(-x)^n}\right)\\ =\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}+o(x^3)\right)\cdot \left(1-x+x^2-x^3+o(x^3)\right).$$ Can you take it from here?

P.S. Along the same lines, it can be seen that, more generally, the coefficient of $x^n$ in the expansion of $\frac{1}{e^x\cdot(1+x)}$ is $$\sum_{k=0}^n\frac{(-1)^k}{k!}\cdot (-1)^{n-k}=(-1)^n\sum_{k=0}^n\frac{1}{k!}.$$ See the Cauchy product wiki page.

  • $\begingroup$ Okay, I see, but how does that help me? You know if there were two terms in the numerator and one in denominator I could easily solve. But not in this case. $\endgroup$ – Tanuj Sep 3 '17 at 14:11
  • $\begingroup$ @Tanuj Now you have to expand the product. $\endgroup$ – Robert Z Sep 3 '17 at 14:14

\begin{eqnarray*} \frac{e^{-x}}{1+x} &=&\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots \right) (1- x+x^2-x^3+\cdots ) \\ &=& \cdots +x^3 \underbrace{\left( -1-1-\frac{1}{2}-\frac{1}{6}\right)}_{\color{blue}{-\frac{8}{3}}}+ \cdots \end{eqnarray*}

  • $\begingroup$ Can you explain what exactly did you do? $\endgroup$ – Tanuj Sep 3 '17 at 14:12
  • $\begingroup$ Wow a eureka moment for me! You used that result of infinite Geometric Progression, right? $\endgroup$ – Tanuj Sep 3 '17 at 14:14
  • $\begingroup$ Just used the standard expansions for the functions ... $\endgroup$ – Donald Splutterwit Sep 3 '17 at 14:16

From the two Taylor series at $0$, to order $x^3$:$$ e^{-x} = 1-x+\frac{x^2}{2}-\frac{x^3}{6} + o(x^3) \tag{1} $$ and $$ \frac{1}{1+x} = 1-x+x^2-x^3+o(x^3)\tag{2} $$ we obtain $$\begin{align} \frac{e^{-x}}{1+x} &= \left(1-x+\frac{x^2}{2}-\frac{x^3}{6} + o(x^3)\right)\left(1-x+x^2-x^3+o(x^3)\right)\\ &= 1-x+x^2-x^3+o(x^3) - x+x^2-x^3+\frac{x^2}{2}-\frac{x^3}{2}-\frac{x^3}{6}\\ &= 1-2x+\frac{5}{2}x^2{\color{red}{-\frac{8}{3}}}x^3 +o(x^3) \end{align}$$ and you can just read off the coefficient.

Note that when expanding the product, we could have focused on only the terms $x^3$ (it would have been marginally faster, but would have required keeping track); and didn't expand any term $x^k$ with $k>3$, since anyway they are "swallowed" by the $o(x^3)$.


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