- #1

jegues

- 1,097

- 3

**Find the Taylor series for...**

**EXAMPLE 1:**

[tex]f(x) = \frac{1}{1- (x)}[/tex] around [tex]x = 2[/tex]

Then,

[tex] \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}}[/tex] provided that [tex]| \frac{x+2}{3} | < 1[/tex] or, [tex] -5 < x < 1[/tex]

Does this look correct?

**Now if I do the same question in another fashion...**

[tex]\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}} [/tex] provided that [tex]| \frac{(-x - 2)}{3} | < 1[/tex] or, [tex] -5 < x < 1[/tex]

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

What am I doing wrong here?