# Find the Taylor series of a function

• songoku
In summary: Taylor series for a function."In summary, The Taylor series for a function that is centered at 0 is the same as the Maclaurin series. My attempt was (b), but it may not be the correct answer.
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

songoku said:
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
PeroK said:
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

songoku said:
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
PeroK said:
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

songoku said:
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
PeroK said:
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

songoku said:
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
Mark44 said:
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

Mark44 said:
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.

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

## 1. What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, where each term is calculated using the function's derivatives at a specific point. It is used to approximate a function and can be used to find the value of the function at any point.

## 2. How do you find the Taylor series of a function?

To find the Taylor series of a function, you need to calculate the function's derivatives at a specific point. Then, use these derivatives to construct the terms of the series. The general formula for a Taylor series is f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ..., where a is the point at which the derivatives are calculated.

## 3. What is the purpose of finding the Taylor series of a function?

The purpose of finding the Taylor series of a function is to approximate the value of the function at a specific point. It can also be used to study the behavior of a function and make predictions about its values at other points.

## 4. What are some common applications of Taylor series?

Taylor series are commonly used in calculus, physics, and engineering to approximate the behavior of functions. They are also used in numerical analysis to solve differential equations and in signal processing to analyze signals.

## 5. Can any function have a Taylor series?

Not all functions have a Taylor series. A function must be infinitely differentiable at a point in order for its Taylor series to exist. Additionally, the Taylor series may only converge within a certain interval, so it may not accurately represent the function outside of that interval.

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