- #1

songoku

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- Homework Statement
- Find the Taylor series of ##f(x) = \frac{1}{(2 - x)^2}## centered at 0

##a. f(x) = \sum_{n=0}^\infty \frac{n+1}{2^{n+2}} x^n##

##b. f(x) = \sum_{n=0}^\infty (-1)^n \frac{n+1}{2^{n+2}} x^n##

##c. f(x) = \sum_{n=0}^\infty \frac{n}{2^{n}} x^n##

##d. f(x) = \sum_{n=0}^\infty \frac{n+1}{2^{n+1}} x^n##

##e. f(x) = \sum_{n=0}^\infty \frac{n+2}{2^{n}} x^n##

- Relevant Equations
- Taylor series

Maclaurin series

Standard power series

Because the Taylor series centered at 0, it is same as Maclaurin series. My attempts:

1st attempt

[itex]

\begin{align}

\frac{1}{1-x} = \sum_{n=0}^\infty x^n\\

\\

\frac{1}{x} = \frac{1}{1-(1-x)} = \sum_{n=0}^\infty (1-x)^n\\

\\

\frac{1}{x^2} = \sum_{n=0}^\infty (1-x^2)^n\\

\\

\frac{1}{(2-x)^2} = \sum_{n=0}^\infty (1-(2-x)^2)^n\\

\\

= \sum_{n=0}^\infty (-x^2 + 4x -3)^n\\

\\

= \sum_{n=0}^\infty (-1)^n (x-3)^n (x-1)^n

\end{align}

[/itex]

Then how to continue?2nd attempt: I do the derivative manually and put it into Maclaurin series formula and I end up with:

## f(x) = \sum_{n=0}^\infty \frac{(n+1)!}{2^{(n+2)}} x^n##

Is my answer on 2nd attempt wrong? Or the choices of the question are wrong?

Thanks

1st attempt

[itex]

\begin{align}

\frac{1}{1-x} = \sum_{n=0}^\infty x^n\\

\\

\frac{1}{x} = \frac{1}{1-(1-x)} = \sum_{n=0}^\infty (1-x)^n\\

\\

\frac{1}{x^2} = \sum_{n=0}^\infty (1-x^2)^n\\

\\

\frac{1}{(2-x)^2} = \sum_{n=0}^\infty (1-(2-x)^2)^n\\

\\

= \sum_{n=0}^\infty (-x^2 + 4x -3)^n\\

\\

= \sum_{n=0}^\infty (-1)^n (x-3)^n (x-1)^n

\end{align}

[/itex]

Then how to continue?2nd attempt: I do the derivative manually and put it into Maclaurin series formula and I end up with:

## f(x) = \sum_{n=0}^\infty \frac{(n+1)!}{2^{(n+2)}} x^n##

Is my answer on 2nd attempt wrong? Or the choices of the question are wrong?

Thanks