Exploiting Geometric Series with Power Series for Taylors Series

• jegues
In summary: If you want to confirm this, you should take the limit of |(1\pm a(x-2))| as x approaches 2. If the limit is less than 1, you're good to go.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. If you want to confirm this, you should take the limit of |(1\pm a(x-2))| as x approaches 2. If the limit is less than 1, you're good to go.So when doing the expansion, I should
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?

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

EDIT: The post above this one was removed.

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].

vela said:
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?

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.

fzero said:
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?

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

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

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

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.

vela said:
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?

jegues said:
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.

fzero said:
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!

jegues said:
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.

1. What is the purpose of using power series in Taylor's series?

The purpose of using power series in Taylor's series is to approximate a function with a polynomial. This allows for easier calculations and analysis of the function's behavior. Additionally, power series can be used to represent functions that are not easily expressed in polynomial form.

2. How are geometric series used in Taylor's series?

Geometric series are used in Taylor's series by taking advantage of their properties, such as their ability to be easily differentiated and integrated. By using geometric series, we can find the coefficients of the power series and use them to construct the Taylor polynomial for a given function.

3. Can Taylor's series be used to find exact values of a function?

No, Taylor's series can only provide an approximation of a function near a specific point. As the number of terms in the series increases, the accuracy of the approximation also increases, but it will never be an exact representation of the function.

4. What is the relationship between Taylor's series and calculus?

Taylor's series is closely related to calculus as it involves finding the derivatives and integrals of a function. The coefficients of the power series are determined by these derivatives, making it a useful tool for solving calculus problems and analyzing functions.

5. Are there any limitations or drawbacks to using Taylor's series?

One limitation of using Taylor's series is that it can only approximate a function near a specific point. If the function has a singularity or behaves differently in other regions, the Taylor series may not accurately represent the function. Additionally, finding the coefficients of the power series can be a tedious and time-consuming process for more complex functions.

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