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.