# Help expanding a function to a power series

#### thrownaway

##### New member
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.
is f(x), to be expanded with center c=0

#### Subhotosh Khan

##### Super Moderator
Staff member
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
Have you studied "Taylor series expansion" (My grandson calls it Taylor Swift thingy)?

Please show us what you have tried and exactly where you are stuck.

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

#### thrownaway

##### New member
Have you studied "Taylor series expansion" (My grandson calls it Taylor Swift thingy)?

Please show us what you have tried and exactly where you are stuck.

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

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

#### Subhotosh Khan

##### Super Moderator
Staff member
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
Use Taylor series expansion.

or

you could use synthetic division.

#### Cubist

##### Full Member
Or, have you studied the following yet (which can be obtained by differentiating the geometric series)...

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