Website Title: Taylor Expansion for a Function on a Restricted Set

[sam_jones26]

Hi, could do with some help on this question if anyone can help.
Any help much appreciated, thanks

Q: Suppose a function f(.) defined on the set I = {x ∈ R∣x < 1} is as follows.
For each real number x∈ I , f(x) = 1/(1-x)
By using the Taylor expansion of this function, show that for any real number x such that
∣x∣ < 1,
f(x)= 1 + ∞
∑ x (to power j)
j =1

 

[Cyosis]

Hint: geometric series.

 

[sam_jones26]

how would i go about that then?
Thanks

 

[Cyosis]

What is the definition of a geometric series? Also what is the problem you're having with finding a Taylor series?

Show your work!

 

[sam_jones26]

for a geometric series:
S=a(k^n-1)/(k-1)

The questions is at the top, where i have to use taylor expansion to show that for any real number...

 

[Cyosis]

Yes I have read the question. It is now time for you to make an attempt and show your work in accordance with the rules of this forum.

You can either compare it to the geometric series or find the Taylor series by differentiation. Have you tried any of that?

 

[sam_jones26]

would the taylor series fror 1/1-x be:

1/1-a + x-a/(1-a)^2 +(x-a)^2/(1-a)^3 + etccc

or should i use the mMacLaurin series


1/1-x = 1 + x + x2 + ... + xn + ... = for |x| < 1


thanks

 

[Cyosis]

Yes find the series about the point x=0. Now how can you write that expression as an infinite series?

 

[sam_jones26]

how do i find the series about the point x=0?
Do i just sub in 0 to anywhere x is in the taylor series?

 

[Cyosis]

You have already done it in post #7 (also called the Maclaurin series). Now you just need to find an expression for your result using sum notation (and technically you would need to show that it converges for |x|<1).

 

[sam_jones26]

Okay thanks,
do i still use the taylor series though, like i showed in 7 or not?

 

[Cyosis]

That's what the question asks you to do.