Differentiation of power series

  • Thread starter simmonj7
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  • #1
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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!
 
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Answers and Replies

  • #2
LeonhardEuler
Gold Member
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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
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I mean (-2)^n+1 (n+2)/n!
 
  • #4
LeonhardEuler
Gold Member
859
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Your mistake is in your first step:
[tex]e^{-x} \neq\sum_{n=0}^{\infty}\frac{x^{n}}{n!}[/tex]
 
  • #5
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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
LeonhardEuler
Gold Member
859
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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
LeonhardEuler
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859
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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
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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
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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
415
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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
LeonhardEuler
Gold Member
859
1
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
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Ah! That's perfect kanato! Thanks!
 
  • #13
LeonhardEuler
Gold Member
859
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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.
 

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