- #1

simmonj7

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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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