# Differentiation of power series

## Homework Statement

Show that 4 = $$\sum$$ from n = 1 to $$\infty$$ (-2)$$^{n+1}$$ (n+2)/n! by considering d/dx(x$$^{2}$$e$$^{-x}$$).

## Homework Equations

Power series for e$$^{x}$$ = $$\sum$$ x$$^{n}$$/n! from 0 to $$\infty$$.

## The Attempt at a Solution

So I started with the power series for e$$^{-x}$$ = $$\sum$$ -x$$^{n}$$/n! from 0 to $$\infty$$ = 1 - x + x$$^{2}$$/2! - x$$^{3}$$/3! +...

I multiplied both sides of the equation by x$$^{2}$$ so I got x$$^{2}$$e$$^{-x}$$ = x$$^{2}$$ - x$$^{3}$$ + x$$^{4}$$/2 - x$$^{5}$$/6 +...

Now I took the derivative of both sides: 2xe$$^{-x}$$ - x$$^{2}$$e$$^{-x}$$ = 2x - 3x$$^{2}$$ + 4$$^{3}$$/2 - 5x$$^{4}$$/6 +...

Now I figured that if I just plugged in x= -2, I would get (-2)(-2)e$$^{2}$$ - (-2)$$^{2}$$e$$^{2}$$ = 2(-2) -3(-2)$$^{2}$$ + 4(-2)$$^{3}$$/2 -...

Now the right hand side would equal the sumation which was stated in the problem i.e. $$\sum$$ (-2)$$^{n+1}$$ (n+2)/n! however the summation I get starts at n = 0 rather than n = 1.

Also, as you can clearly see, the left hand side of my equation does not equal 4 like it is supposed to.

If anyone could tell me where I went wrong that would be greatly appreciated. Thanks!

Last edited:

LeonhardEuler
Gold Member
Is there a pair of parentheses missing or do you really mean
$$(-2)^{n+1}n+2/n!$$
as opposed to
$$(-2)^{n+1}(n+2)/n!$$
?

I mean (-2)^n+1 (n+2)/n!

LeonhardEuler
Gold Member
$$e^{-x} \neq\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$$

That's another typo. I know e^-x equals the sum of (-x)^n/n! so that is not my mistake. If you looked at the work I did you would see that I used that sumation to get my answer rather than the one in the typo.

LeonhardEuler
Gold Member
I see. Then the mistake happens when you plug in -2 and say that it's the sum you were looking for. The alternating - signs in your expression cancel the - signs in the (-2)^(n+1).

LeonhardEuler
Gold Member
By the way, I think it's a lot easier in this case and most cases to deal with the expressions kept in terms of 'n' instead of using a "...".
e.g., writing the derivative as
$$-\sum_{n=0}^{n=\infty}(n+2)\frac{{(-x)}^{n+1}}{n!}$$
2x - 3x + 4/2 - 5x/6 +...

I think you will tend to make fewer mistakes this way and they will be easier to correct.

Ok...I see what you're saying but I'm not quite sure what you are trying to say here as far as where I was supposed to go instead. And your explanation still doesn't cover what went wrong with the left hand side either and that it does not equal 4...

I have been taught to do it the way I am doing it so I am going to stick with what I've been taught for power series.

I think you're on the right track. Notice that the left hand side of your equation is zero. Also, you mention that your series goes from 0 to infinity rather than 1 to infinity. Look at the value of the first term in that series. What would your equation be if you moved that value to the other side?

LeonhardEuler
Gold Member
Ok, but do you see what number you can stick into
$$-\sum_{n=0}^{n=\infty}(n+2)\frac{{(-x)}^{n+1}}{n!}$$
to get the sum you want to appear?

Ah! That's perfect kanato! Thanks!

LeonhardEuler
Gold Member
Except the left hand side isn't 0 when you plug in -2. That was another mistake on your part. But it's the right method once you do plug in the right number.