Exploiting Geometric Series with Power Series for Taylors Series

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:
[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?

jegues

So is the first one correct, and the 2nd one incorrect?

EDIT: The post above this one was removed.

vela

In the second series, you're expanding

[tex]\frac{1}{1+x} = 1-x+x^2-x^3+\cdots[/tex]

You forgot the alternating sign, which cancels with the factor of [itex](-1)^n[/tex].

jegues

How do you see,

[tex]\frac{1}{1+x}[/tex], I can never see it in my work.

So is the 2nd one incorrect then, or correct and I'm just not understanding why?

fzero

A series expansion around [tex]x=2[/tex] is an expansion in [tex]x-2[/tex]. Also, for the 2nd derivation, you'll find that the expansion of [tex]1/(1+u)[/tex] has a factor of [tex](-1)^n[/tex] that cancels out the one that you introduced.

jegues

But the 2nd derivation is the same question as the first question, right?

I'm not doing,

[tex]1/(1+x)[/tex]

I'm doing [tex]1/(1-x)[/tex]

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 [tex](-1)^{n}[/tex] I introduced gets canceled, and that's it?

fzero

The 1/(1+u) appears in the 3rd part of the 2nd derivation:

Once you perform the proper expansion, everything falls into place.

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.

jegues

So would this be correct,

[tex]\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}[/tex]

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

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

[tex](1)^{n}[/tex] and my problems will be fixed, correct?

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.

jegues

It was suppose to be for x=2, there was no typo.

I guess if I'm doing,

[tex]\frac{1}{1 - x}[/tex] then I sub in, [tex] x + 2[/tex] or [tex] x - 2[/tex]?

Similarly,

if I'm doing,

[tex]\frac{1}{1 + (-x)}[/tex] then I sub in, [tex] x + 2[/tex] or [tex] x - 2[/tex]?

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

fzero

Around x=2 you will be expanding an expression of the form

[tex]\frac{1}{1\pm a(x-2)}[/tex].

The important thing is that x=2 is actually within the radius of convergence of your expansion.