# Exploiting geometric series with power series to find Taylors series

• Sep 28th 2010, 05:34 PM
jegues
Exploiting geometric series with power series to find Taylors series
I'm confused between some formulae so I'm going to give some examples and you can let me know if what I'm writing is correct.

Find the Taylor series for....

EXAMPLE 1:
$f(x) = \frac{1}{1- (x)}$ around $x = 2$

Then,

$\frac{1}{1-(x)} = \frac{1}{3-(x+2)} = \frac{1}{3} \left( \frac{1}{1 - \frac{(x+2)}{3}} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{x+2}{3})^{n} = \sum_{n=0}^{\infty} \frac{(x+2)^{n}}{3^{n+1}}$ provided that $| \frac{x+2}{3} | < 1$ or, $-5 < x < 1$

Does this look correct?

EDIT: Now if I do the same question in another fashion...

$\frac{1}{1-x} = \frac{1}{1 + (-x)} = \frac{1}{3 + (-x -2)} = \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{-x-2}{3})^{n} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n}} = \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n+1}}$ provided that $| \frac{(-x - 2)}{3} | < 1$ or, $-5 < x < 1$

They aren't equal, in this one I have a $(-1)^{n}$ kicking around in my sum.

What am I doing wrong here?
• Sep 28th 2010, 06:35 PM
TheEmptySet
Quote:

Originally Posted by jegues
I'm confused between some formulae so I'm going to give some examples and you can let me know if what I'm writing is correct.

Find the Taylor series for....

EXAMPLE 1:
$f(x) = \frac{1}{1- (x)}$ around $x = 2$

Then,

$\frac{1}{1-(x)} = \frac{1}{3-(x+2)} = \frac{1}{3} \left( \frac{1}{1 - \frac{(x+2)}{3}} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{x+2}{3})^{n} = \sum_{n=0}^{\infty} \frac{(x+2)^{n}}{3^{n+1}}$ provided that $| \frac{x+2}{3} | < 1$ or, $-5 < x < 1$

Does this look correct?

EDIT: Now if I do the same question in another fashion...

$\frac{1}{1-x} = \frac{1}{1 + (-x)} = \frac{1}{3 + (-x -2)} = \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{-x-2}{3})^{n} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n}} = \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n+1}}$ provided that $| \frac{(-x - 2)}{3} | < 1$ or, $-5 < x < 1$

They aren't equal, in this one I have a $(-1)^{n}$ kicking around in my sum.

What am I doing wrong here?

The 2nd series is not centered at x=2 it is centered at x=-2

Notice that

$\displaystyle \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right)=\frac{1}{3} \left( \frac{1}{1 - (\frac{x+2}{3})} \right)$
• Sep 28th 2010, 07:24 PM
jegues
Quote:

Originally Posted by TheEmptySet
The 2nd series is not centered at x=2 it is centered at x=-2

Notice that

$\displaystyle \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right)=\frac{1}{3} \left( \frac{1}{1 - (\frac{x+2}{3})} \right)$

But that exact term is also in my 1st derivation of the question.

Are they both centered at x = -2?
• Sep 29th 2010, 06:11 AM
HallsofIvy
They are both centered at x= -2 but there is a missing sign in your second calculation.

For a geometric series, $\displaytype\sum_{n=0}^r ar^n= \frac{a}{1- r}$.

In your first calculation, you have $\frac{\frac{1}{3}}{1- \frac{x+2}{3}}$ so $a= \frac{1}{3}$ and $r= \frac{x+2}{3}$. Your calculations are correct.

In your second calculation, you have $\frac{\frac{1}{3}}{1+ \frac{-x-2}{3}}$ and that is NOT of the correct form for the geometric series. You have $\frac{a}{1+ r}$, not $\frac{a}{1- r}$

You could then say $\frac{a}{1+r}= \frac{a}{1-(-r)}= \sum_{n=0}^\infty a(-r)^n$ and replacing "r" with "-r" cancels that incorrect $(-1)^n$