## Exploiting Geometric Series with Power Series for 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?

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?
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
 So is the first one correct, and the 2nd one incorrect? EDIT: The post above this one was removed.
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus In the second series, you're expanding $$\frac{1}{1+x} = 1-x+x^2-x^3+\cdots$$ You forgot the alternating sign, which cancels with the factor of [itex](-1)^n[/tex].

## Exploiting Geometric Series with Power Series for Taylors Series

 Quote by vela In the second series, you're expanding $$\frac{1}{1+x} = 1-x+x^2-x^3+\cdots$$ You forgot the alternating sign, which cancels with the factor of [itex](-1)^n[/tex].
How do you see,

$$\frac{1}{1+x}$$, I can never see it in my work.

So is the 2nd one incorrect then, or correct and I'm just not understanding why?
 Recognitions: Gold Member Homework Help Science Advisor A series expansion around $$x=2$$ is an expansion in $$x-2$$. Also, for the 2nd derivation, you'll find that the expansion of $$1/(1+u)$$ has a factor of $$(-1)^n$$ that cancels out the one that you introduced.

 Quote by fzero A series expansion around $$x=2$$ is an expansion in $$x-2$$. Also, for the 2nd derivation, you'll find that the expansion of $$1/(1+u)$$ has a factor of $$(-1)^n$$ that cancels out the one that you introduced.
But the 2nd derivation is the same question as the first question, right?

I'm not doing,

$$1/(1+x)$$

I'm doing $$1/(1-x)$$

I'm sure you know this, and it is probably me who's misunderstanding something but I still can't see it.

Is it just that the $$(-1)^{n}$$ I introduced gets canceled, and that's it?

Recognitions:
Gold Member
Homework Help
The 1/(1+u) appears in the 3rd part of the 2nd derivation:

 Quote by jegues $$\frac{1}{1-x} = \frac{1}{1 + (-x)} = \frac{1}{3 + (-x -2)} = \frac{1}{3} \left( \frac{1}{1 +(\text{this is a plus sign}) (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{-x-2}{3})^{n} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n}} = \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n+1}}$$
Once you perform the proper expansion, everything falls into place.
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus When you do the expansion, you'd expanding a function of the form 1/(1+u), not 1/(1-u), where u=-(x+2)/3.

 Quote by vela When you do the expansion, you'd expanding a function of the form 1/(1+u), not 1/(1-u), where u=-(x+2)/3.
So would this be correct,

$$\frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} (\frac{-x-2}{3})^{n}$$

That would take care of the "+ sign", right?

But then when I pull the other, $$(-1)^{n}$$ out I'll get,

$$(1)^{n}$$ and my problems will be fixed, correct?

Recognitions:
Gold Member
Homework Help
 Quote by jegues So would this be correct, $$\frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} (\frac{-x-2}{3})^{n}$$ That would take care of the "+ sign", right? But then when I pull the other, $$(-1)^{n}$$ out I'll get, $$(1)^{n}$$ and my problems will be fixed, correct?

Yes, both expressions agree.

Are you meant to expand around x=2 or x=-2? Your OP says x=2, but your expansions are around x=-2. You have the proper radii of convergence, so I'm not sure if the first line of your example is a typo or not.

 Quote by fzero Yes, both expressions agree. Are you meant to expand around x=2 or x=-2? Your OP says x=2, but your expansions are around x=-2. You have the proper radii of convergence, so I'm not sure if the first line of your example is a typo or not.
It was suppose to be for x=2, there was no typo.

I guess if I'm doing,

$$\frac{1}{1 - x}$$ then I sub in, $$x + 2$$ or $$x - 2$$?

Similarly,

if I'm doing,

$$\frac{1}{1 + (-x)}$$ then I sub in, $$x + 2$$ or $$x - 2$$?

This should be the last bit of confusion I have to clear up!

Recognitions:
Gold Member
Homework Help
 Quote by jegues It was suppose to be for x=2, there was no typo. I guess if I'm doing, $$\frac{1}{1 - x}$$ then I sub in, $$x + 2$$ or $$x - 2$$? Similarly, if I'm doing, $$\frac{1}{1 + (-x)}$$ then I sub in, $$x + 2$$ or $$x - 2$$? This should be the last bit of confusion I have to clear up!
$$\frac{1}{1\pm a(x-2)}$$.