Help expanding a function to a power series

thrownaway

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Feb 1, 2021
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3
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.
1614033069557.png is f(x), to be expanded with center c=0
 

Subhotosh Khan

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Jun 18, 2007
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24,181
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:


Please share your work/thoughts about this problem.
 

thrownaway

New member
Joined
Feb 1, 2021
Messages
3
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:


Please share your work/thoughts about this problem.
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
Joined
Jun 18, 2007
Messages
24,181
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
Joined
Oct 29, 2019
Messages
852
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
 
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