power series representation of (2 + x)^(-2)

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nsievers

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Jun 9, 2012
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Hello all,

My class went over power series representation a few days ago and I am having some trouble. Here is the problem I'm working on, and how I have tried to solve it:


"use differentiation to find a power series representation of f(x) = (2+x)^(-2)"

(2+x)^(-2) = d/dx (-1)/(2+x) = d/dx (-1/2)/(1-(-x/2))

d/dx (summation from n = 0 to infinity) (-1/2)(-x/2)^n

(summation from n = 1 to infinity) (n)(-1/2)(-x/2)^(n-1)

ANSWER: (summation from n = 0 to infinity) (n + 1)(-1/2)(-x/2)^n


Can someone please explain what it is that I am doing wrong?
 
nsievers said:
Hello all,

My class went over power series representation a few days ago and I am having some trouble. Here is the problem I'm working on, and how I have tried to solve it:


"use differentiation to find a power series representation of f(x) = (2+x)^(-2)"

(2+x)^(-2) = d/dx (-1)/(2+x) = d/dx (-1/2)/(1-(-x/2))

d/dx (summation from n = 0 to infinity) (-1/2)(-x/2)^n

(summation from n = 1 to infinity) (n)(-1/2)(-x/2)^(n-1)

ANSWER: (summation from n = 0 to infinity) (n + 1)(-1/2)(-x/2)^n


Can someone please explain what it is that I am doing wrong?
Click to expand...


You didn't take the derivative of (-x/2)^n correctly.
 
if d/dx (x/2)^n != (n)(x/2)^(n-1), then I'm in more trouble than I thought

The course I am taking is a single-variable calculus course. How would I solve this using only knowlege you would find in a Calculus I textbook?
 
Last edited:
nsievers said:
if d/dx (x/2)^n != (n)(x/2)^(n-1), then I'm in more trouble than I thought

The course I am taking is a single-variable calculus course. How would I solve this using only knowlege you would find in a Calculus I textbook?
Click to expand...

Since your 'inner' function is (x + 2), whose d/dx = 1, stop worrying so much.

BUT:

If f(x) = (x + 2)^-2 [ = zeroth derivative D0]

D1 = -2(x + 2)^-3

D2 = +6(x + 2)^-4

D3 = -24(x + 2)^-5 = (-1)^3 * 4! (x + 2)^-5

Looks as if

Dn = (-1)^n * (n + 1)! (x + 2)^(-(n+2))

Now do your power series stuff, using x = 0. Just plug in:

Dn(0) = (-1)^n * (n + 1)! (x + 2)^(-(n+2))
 
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