# Taylor Series, working backwards

1. Oct 30, 2012

### tinylights

1. The problem statement, all variables and given/known data
Okay, first there is an explanation of the Taylor Series equation. This I don't have a problem with. Then, we have this:

Consider the power series 2 - (2/3)x + (2/9)x^2 - (2/27)x^3. What rational
function does this power series represent?

2. Relevant equations / 3. The attempt at a solution

I basically am just stuck at the end. I do understand how this function relates to the Taylor Series equation and wrote a detailed explanation of what parts represent a, i, f(a), f'(a), etc. However, after solving and presenting this information:

f(a) = 2
f'(a) = -2/3
f''(a) = 4/9
f'''(a) = -12/27

I don't know how to sort of reverse it and find the relevant function. I know that the numerator is being multiplied by -1, then -2, then -3, and the denominator is being multiplied by 3, but how does one take that information and turn it into equations?

I don't want the answer - I just want a hint in the right direction, or a method by which I can find it. Thank you guys.

2. Oct 30, 2012

### Zondrina

3. Oct 30, 2012

### Dick

If you put u=(-1/3)*x then it looks like your series is 2*(u^0+u^1+u^2+u^3+...). Look familiar, enuf hints?

4. Oct 31, 2012

### tinylights

It actually doesn't. I am still flummoxed. :|

5. Oct 31, 2012

### Dick

Look up "geometric series".

6. Nov 1, 2012

### tinylights

Okay, so - a geometric sequence is represented by (1+r+r^2+r^3+r^4....) and can be expressed in the form 1/(1-r), so basically I have 2(1+u+u^2+u^3+u^4), which can equivalently be expressed in the form 2/(1-u). U = 1/3x, so we end up with 2/(1-1/3x)?

The answer seems to check out so I'll go with it. Thanks very much!

7. Nov 1, 2012

### Dick

Your series alternates signs. u=(1/3)x doesn't. How should you change it?

Last edited: Nov 1, 2012
8. Nov 1, 2012

### tinylights

Ahhh, you're right! Realized that as I was checking my work. It should be -1/3x.