1. ## Taylor Series Help

So tomorrow I need to explain a problem to the class, I know the correct answer is 3, i just do not know how to show the work necessary to explain it.

Here is the problem:

What is the coefficient of x^2 in the Taylor series for 1/((1+x)^2) about x=0.

(A) 1/6 (B) 1/3 (C) 1 (D)3 (E)6

Thanks for the help.

2. Originally Posted by TFD
So tomorrow I need to explain a problem to the class, I know the correct answer is 3, i just do not know how to show the work necessary to explain it.

Here is the problem:

What is the coefficient of x^2 in the Taylor series for 1/((1+x)^2) about x=0.

(A) 1/6 (B) 1/3 (C) 1 (D)3 (E)6

Thanks for the help.
The power series for this would be found by seeing that $\frac{1}{(1+x)^2}=\frac{-D[\frac{1}{1+x}]}{dx}$

So $\frac{1}{1+x}=\sum_{n=1}^{\infty}(-1)^{n}x^{n}$

so the oppositve of the derivative would be $-\sum_{n=0}^{\infty}n(-1)^{n}x^{n-1}$

so $a_n=(-1)^{n+1}nx^{n}$

so $a_3=(-1)^{4}{3}x^{3-1}=3x^2$

3. Originally Posted by TFD
So tomorrow I need to explain a problem to the class, I know the correct answer is 3, i just do not know how to show the work necessary to explain it.

Here is the problem:

What is the coefficient of x^2 in the Taylor series for 1/((1+x)^2) about x=0.

(A) 1/6 (B) 1/3 (C) 1 (D)3 (E)6

Thanks for the help.

Given $f(x)=\frac{1}{(1+x)^2}$

The coefficient of $x^2$ is going to be $\frac{f''(0)}{2!}$.

So first you have to compute $f''(0)$.

4. Originally Posted by Mathstud28
The power series for this would be found by seeing that $\frac{1}{(1+x)^2}=\frac{-D[\frac{1}{1+x}]}{dx}$

So $\frac{1}{1+x}=\sum_{n=1}^{\infty}(-1)^{n}x^{n}$

so the oppositve of the derivative would be $-\sum_{n=0}^{\infty}n(-1)^{n}x^{n-1}$

so $a_n=(-1)^{n+1}nx^{n}$

so $a_3=(-1)^{4}{3}x^{3-1}=3x^2$
Im a bit confused on what the oppositve derivative is?

5. Originally Posted by TFD
Im a bit confused on what the oppositve derivative is?
Since $\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n}$

and since the derivative of $\frac{1}{1+x}$ is $\frac{-1}{(1+x)^2}$ you can see that $\frac{1}{(1+x)^2}$ is the opposite(or negative) of the derivative of $\frac{1}{1+x}=\sum_{n=1}^{\infty}(-1)^{n}x^{n}$

so by differentiating we get $\frac{D[\sum_{n=0}^{\infty}(-1)^{n}x^{n}]}{dx}=\sum_{n=1}^{\infty}n(-1)^{x}x^{n-1}$ and since we said that $\frac{1}{(1+x)^2}$ was the opposite of the derivative the answer is $-1\sum_{n=1}^{\infty}n(-1)^{n}x^{n}=\sum_{n=1}^{\infty}n(-1)^{n+1}x^{n-1}$