Differentiation of power series

[simmonj7]

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!

 

[LeonhardEuler]

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!
?

 

[simmonj7]

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

 

[LeonhardEuler]

Your mistake is in your first step:
e^{-x} \neq\sum_{n=0}^{\infty}\frac{x^{n}}{n!}

 

[simmonj7]

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]

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]

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!}
instead of
2x - 3x + 4/2 - 5x/6 +...

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

 

[simmonj7]

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...

 

[simmonj7]

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.

 

[kanato]

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]

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?

 

[simmonj7]

Ah! That's perfect kanato! Thanks!

 

[LeonhardEuler]

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.