# A particular Taylor Expansion

As we know, the Taylor Expansion we usually see is

$$e^W = \sum_{n=0}^{\infty}\frac{W^n}{n!}$$

but today I see another equation:

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

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.

• Are you sure of the negative signs? – Hasan Saad Oct 4 '15 at 20:06
• 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... – Catherine Chen Oct 4 '15 at 20:08
• 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. – Hasan Saad Oct 4 '15 at 20:09
• 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... – 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.

• OP wanted $We^{-W}+e^{-W}$ – Aleksandar Oct 4 '15 at 20:15
• The exponential should be positive. – Catherine Chen Oct 4 '15 at 20:20
• @CatherineChen If the exponential is positive how come you marked his answer as correct? – Aleksandar Oct 4 '15 at 20:24
• @Aleksandar She actually wanted $We^W+e^W$. Read her comment on the original post. – Hasan Saad Oct 4 '15 at 20:33
• @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. – 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.

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