As we know, the Taylor Expansion we usually see is

\begin{equation} e^W = \sum_{n=0}^{\infty}\frac{W^n}{n!} \end{equation}

but today I see another equation:

\begin{equation} e^{-W}+We^{-W} = \sum_{n=0}^{\infty}(n+1)\frac{W^n}{n!} \end{equation}

I can not figure out how did they derive out the second equation.

I saw these when trying to solve a exercise on a probability exercise manual.

Please support and thanks in advance.

  • $\begingroup$ Are you sure of the negative signs? $\endgroup$ – Hasan Saad Oct 4 '15 at 20:06
  • $\begingroup$ I also feel strange about the negative sign on the exponential, but since these are from the solution manual, so I kind of can not confirm my doubt... $\endgroup$ – Catherine Chen Oct 4 '15 at 20:08
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    $\begingroup$ I think the solution manual is wrong, as in there's a typo. You might want to read what's there right after it to see what result it uses. $\endgroup$ – Hasan Saad Oct 4 '15 at 20:09
  • $\begingroup$ Ah I see, in the next step they multiple the equation by a factor of (e^(-W))*W, then they get the final result W+W^2, yes now I confirm the intermediate step of this solution is wrong... $\endgroup$ – Catherine Chen Oct 4 '15 at 20:18

$We^W=\sum_{n=0}^\infty \frac{W^{n+1}}{n!}$

Now, differentiate both sides,

$We^W+e^W=\sum_{n=1}^\infty (n+1)\frac{W^{n}}{n!}$

Thus, we have the desired result.

  • $\begingroup$ OP wanted $We^{-W}+e^{-W}$ $\endgroup$ – Aleksandar Oct 4 '15 at 20:15
  • $\begingroup$ The exponential should be positive. $\endgroup$ – Catherine Chen Oct 4 '15 at 20:20
  • $\begingroup$ @CatherineChen If the exponential is positive how come you marked his answer as correct? $\endgroup$ – Aleksandar Oct 4 '15 at 20:24
  • $\begingroup$ @Aleksandar She actually wanted $We^W+e^W$. Read her comment on the original post. $\endgroup$ – Hasan Saad Oct 4 '15 at 20:33
  • $\begingroup$ @Aleksandar I actually want to mark both correct but I can only mark one...Hasan derive it from one direction and you derive it from another direction, so both of your answers are correct. So the solution manual is wrong. $\endgroup$ – Catherine Chen Oct 4 '15 at 20:42

$$e^{-W}+We^{-W}=(W+1)e^{-W}=(W+1)\sum_{n=0}^{\infty} \frac{(-W)^n}{n!}=\sum_{n=0}^{\infty} \frac{(W+1)(-W)^n}{n!}$$

I believe you are wrong.

  • $\begingroup$ Yes your derivation is correct. $\endgroup$ – Catherine Chen Oct 4 '15 at 20:19

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