Differentiation of power series

In summary, the mistake in the attempt at a solution was made when plugging in -2 instead of using the power series for e^-x. By plugging in the correct value, -1, the equation becomes 0=1-2+3/2!-4/3!+... which is equivalent to the desired sum, thus proving that 4 = \sum from n = 1 to \infty (-2)^{n+1} (n+2)/n! by considering d/dx(x^{2}e^{-x}).
  • #1
simmonj7
66
0

Homework Statement



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

Homework Equations



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

The Attempt at a Solution



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

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

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

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

Now the right hand side would equal the sumation which was stated in the problem i.e. [tex]\sum[/tex] (-2)[tex]^{n+1}[/tex] (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:
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  • #2
Is there a pair of parentheses missing or do you really mean
[tex](-2)^{n+1}n+2/n![/tex]
as opposed to
[tex](-2)^{n+1}(n+2)/n![/tex]
?
 
  • #3
I mean (-2)^n+1 (n+2)/n!
 
  • #4
Your mistake is in your first step:
[tex]e^{-x} \neq\sum_{n=0}^{\infty}\frac{x^{n}}{n!}[/tex]
 
  • #5
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.
 
  • #6
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).
 
  • #7
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
[tex]-\sum_{n=0}^{n=\infty}(n+2)\frac{{(-x)}^{n+1}}{n!}[/tex]
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.
 
  • #8
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...
 
  • #9
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.
 
  • #10
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?
 
  • #11
Ok, but do you see what number you can stick into
[tex]-\sum_{n=0}^{n=\infty}(n+2)\frac{{(-x)}^{n+1}}{n!}[/tex]
to get the sum you want to appear?
 
  • #12
Ah! That's perfect kanato! Thanks!
 
  • #13
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.
 

1. What is the definition of a power series?

A power series is an infinite series of the form f(x) = ∑n=0 an(x-c)n, where an are constants, c is a fixed point, and x is a variable. This type of series is used to represent functions as a sum of terms with increasing powers of x.

2. How is the radius of convergence determined for a power series?

The radius of convergence of a power series is determined by the ratio test. The ratio test states that the series will converge if the limit of the absolute value of the ratio of consecutive terms is less than 1. The radius of convergence is then equal to the distance between the center c and the nearest point where the series diverges.

3. What is the process for differentiating a power series term by term?

To differentiate a power series term by term, first determine the radius of convergence. Then, differentiate each term of the series using standard differentiation rules for polynomials. Finally, plug in the original value of x to find the value of the derivative at that point.

4. Can a power series be integrated term by term?

Yes, a power series can be integrated term by term as long as the series is convergent. To integrate a power series, first determine the radius of convergence. Then, integrate each term of the series using standard integration rules for polynomials. Finally, plug in the original value of x to find the value of the integral at that point.

5. How is the Taylor series related to the differentiation of power series?

The Taylor series is a special type of power series that represents a function as a sum of terms with increasing powers of x centered at c. The coefficients of the Taylor series can be found by differentiating the function and evaluating at c. Therefore, the Taylor series is closely related to the differentiation of power series, as it allows us to find the derivatives of a function using a power series representation.

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