Power series into function

[Anthrax]

Hi there!

Yah, the problem is I can't find a way to convert the power series 1+ x + x² + x³+...
into its equivalent function f(x) = 1/1-x.

I tried to search it on youtube but they only gave me the reverse of the process. So, I hope someone help me

Thank you !

 

[Deleted member 4993]

Hi there!

Yah, the problem is I can't find a way to convert the power series 1+ x + x² + x³+...
into its equivalent function f(x) = 1/1-x.

I tried to search it on youtube but they only gave me the reverse of the process. So, I hope someone help me

Thank you !

Have you tried to expand 1/(1-x) through Taylor's series expansion?

Or long division?

 

[Dr.Peterson]

Yah, the problem is I can't find a way to convert the power series 1+ x + x² + x³+...
into its equivalent function f(x) = 1/1-x.

I tried to search it on youtube but they only gave me the reverse of the process. So, I hope someone help me

I don't know that there is a general way to reverse the series expansion process, in order to find a functional form for any series.

But in this case, do you recognize that it is a geometric series? And do you know the formula for that (or a way to derive it)?

 

[Anthrax]

Have you tried to expand 1/(1-x) through Taylor's series expansion?

Or long division?

I did long division and get 1+x+x²+x³... as its quotient.

 

[Anthrax]

I don't know that there is a general way to reverse the series expansion process, in order to find a functional form for any series.

But in this case, do you recognize that it is a geometric series? And do you know the formula for that (or a way to derive it)?

Sorry, I know only its summation representation but that's not what I need.

 

[Dr.Peterson]

Sorry, I know only its summation representation but that's not what I need.

Then look it up! It's a valuable thing to know, and it's the answer to your question, so you do want to learn it.

 

[Anthrax]

Then look it up! It's a valuable thing to know, and it's the answer to your question, so you do want to learn it.

Wow, thank you sir

 

[Steven G]

For the record, since 1/1=1, 1/1-x = 1-x. I suspect that you wanted to write 1/(1-x)

 

[HallsofIvy]

Here is what I would do: let $\displaystyle S= 1+ x+ x^2+ x^3+ \cdot\cdot\cdot+ x^n+ \cdot\cdot\cdot$.

Then $\displaystyle S- 1= x+ x^2+ x^3+ \cdot\cdot\cdot+ x^n+ \cdot\cdot\cdot$.
Factor out an x: $\displaystyle S- 1= x(1+ x+ x^2\cdot\cdot\cdot+ x^{n-1}\cdot\cdot\cdot)$.

Since that is an infinite sum it is still equal to S so
$\displaystyle S- 1= xS$
$\displaystyle S- xS= 1$
$\displaystyle S(1- x)= 1$
$\displaystyle S= \frac{1}{1-x}$.

This is, of course, assuming that the sum converges which requires that |x|<1.