Find the Taylor series of a function

[songoku]

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

$\begin{align} \frac{1}{1-x} = \sum_{n=0}^\infty x^n\\ \\ \frac{1}{x} = \frac{1}{1-(1-x)} = \sum_{n=0}^\infty (1-x)^n\\ \\ \frac{1}{x^2} = \sum_{n=0}^\infty (1-x^2)^n\\ \\ \frac{1}{(2-x)^2} = \sum_{n=0}^\infty (1-(2-x)^2)^n\\ \\ = \sum_{n=0}^\infty (-x^2 + 4x -3)^n\\ \\ = \sum_{n=0}^\infty (-1)^n (x-3)^n (x-1)^n \end{align}$

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

 

[PeroK]

Problem 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

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

You're quite close. You probably just made a mistake somewhere.

PS what about using the binomial series?

 

[songoku]

You're quite close. You probably just made a mistake somewhere.

PS what about using the binomial series?

I found my mistake. The answer is (a)

How about my first method? Can it be continued to get the answer or it is just wrong from the beginning?

Thanks

 

[PeroK]

I found my mistake. The answer is (a)

How about my first method? Can it be continued to get the answer or it is just wrong from the beginning?

Thanks

I can't see immediately where it goes wrong, but if you try $x = 0$ you have a divergent series. I suspect there are issues with convergence somewhere along the way.

 

[songoku]

I can't see immediately where it goes wrong, but if you try $x = 0$ you have a divergent series. I suspect there are issues with convergence somewhere along the way.

Is it correct if I say like this: since I derive the series from $\frac{1}{1-x}$ then x ≠ 1. I used the previous series to find the series of $\frac{1}{x}$ so x ≠ 0. Finally, I used it again to get series for $\frac{1}{(2-x)^2}$ so x ≠ 2.
In total, x can not be 0, 1, and 2?

Thanks

 

[PeroK]

Is it correct if I say like this: since I derive the series from $\frac{1}{1-x}$ then x ≠ 1. I used the previous series to find the series of $\frac{1}{x}$ so x ≠ 0. Finally, I used it again to get series for $\frac{1}{(2-x)^2}$ so x ≠ 2.
In total, x can not be 0, 1, and 2?

Thanks

I'm not so sure. The series must converge for $x = 0$. You've got problems for $x > 1$ as well.

 

[songoku]

I'm not so sure. The series must converge for $x = 0$. You've got problems for $x > 1$ as well.

Ok, I'll think about it again. Thank you for the help Perok

 

[Mark44]

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

If you notice that $\frac d{dx}\left(\frac{-1}{x - 2}\right) = \frac 1 {(x - 2)^2} = \frac 1 {(2 - x)^2}$, you can find the Maclaurin series for $g(x) = \frac {-1}{x - 2}$, and then differentiate w.r.t. x, giving you the series for the function in your problem. I suspect that this is what the author of the problem had in mind.

I believe @PeroK's hint in post #2 is related to this idea.

 

[songoku]

If you notice that $\frac d{dx}\left(\frac{-1}{x - 2}\right) = \frac 1 {(x - 2)^2} = \frac 1 {(2 - x)^2}$, you can find the Maclaurin series for $g(x) = \frac {-1}{x - 2}$, and then differentiate w.r.t. x, giving you the series for the function in your problem. I suspect that this is what the author of the problem had in mind.

I believe @PeroK's hint in post #2 is related to this idea.

Thanks Mark44

 

[PeroK]

If you notice that $\frac d{dx}\left(\frac{-1}{x - 2}\right) = \frac 1 {(x - 2)^2} = \frac 1 {(2 - x)^2}$, you can find the Maclaurin series for $g(x) = \frac {-1}{x - 2}$, and then differentiate w.r.t. x, giving you the series for the function in your problem. I suspect that this is what the author of the problem had in mind.

I believe @PeroK's hint in post #2 is related to this idea.

I rewrote the function as

$f(x) = \frac14 (1 - \frac{x}{2})^{-2}$

and used a binomial expansion.

 

[Mark44]

"More than one way to skin a cat ..."