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:
$$f(x) = \frac{1}{1- (x)}$$ 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?

 

[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

$$\frac{1}{1+x} = 1-x+x^2-x^3+\cdots$$

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

 

[jegues]

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?

 

[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.

 

[jegues]

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?

 

[fzero]

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

$$\frac{1}{1-x} = \frac{1}{1 + (-x)} = \frac{1}{3 + (-x -2)} = \frac{1}{3} \left( \frac{1}{1 +(\text{this is a plus sign}) (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{-x-2}{3})^{n} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n}} = \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n+1}}$$

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

 

[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]

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?

 

[fzero]

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.

 

[jegues]

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!

 

[fzero]

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!

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

$$\frac{1}{1\pm a(x-2)}$$.

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