#### thrownaway

##### New member

- Joined
- Feb 1, 2021

- Messages
- 3

- Thread starter thrownaway
- Start date

- Joined
- Feb 1, 2021

- Messages
- 3

- Joined
- Jun 18, 2007

- Messages
- 24,181

Have you studied "Taylor series expansion" (My grandson calls it Taylor Swift thingy)?Hello,

I've been having trouble with an expansion of a function to a power series. I can't seem to get it to a clean geometric series (1/1-r) and I'm stuck where to go.

View attachment 25283 is f(x), to be expanded with center c=0

Please show us what you have tried and

Please follow the rules of posting in this forum, as enunciated at:

Please share your work/thoughts about this problem.

- Joined
- Feb 1, 2021

- Messages
- 3

Well, I first went about trying to find a form 1/1-r and turn to a geometric series. I obtained 6x/(16+2x^3) * 6x/(16+2x^3) . However, this is not of form 1/1-r. So I tried to factor out 6x, and obtained 1/(16/6x + x^2/3). This is not sufficient either. I am not sure if this is the right line of analysis or if I need to start over. Let me know anything else I can helpHave you studied "Taylor series expansion" (My grandson calls it Taylor Swift thingy)?

Please show us what you have tried andexactlywhere you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

Please share your work/thoughts about this problem.

- Joined
- Jun 18, 2007

- Messages
- 24,181

Use Taylor series expansion.Well, I first went about trying to find a form 1/1-r and turn to a geometric series. I obtained 6x/(16+2x^3) * 6x/(16+2x^3) . However, this is not of form 1/1-r. So I tried to factor out 6x, and obtained 1/(16/6x + x^2/3). This is not sufficient either. I am not sure if this is the right line of analysis or if I need to start over. Let me know anything else I can help

or

you could use synthetic division.

- Joined
- Oct 29, 2019

- Messages
- 852

\(\displaystyle \frac{1}{\left(1-y\right)^2} = \sum_{n=1}^{\infty}{n\cdot y^{\left(n-1\right)}}\) for \(\displaystyle |y|<1\)

If you want to use this identity, then start by turning the red value below into a "1" by multiplying the numerator and denominator by the same amount

\(\displaystyle \frac{6x^2}{\left(\color{red}16\color{black}+2x^3\right)^2}\)

...and continue