Taylor development

**Ulysses** · April 29th 2011, 03:43 PM

Hi there. I have this exercise which I'm trying to solve now. It says:

Using that $\displaystyle\sum_{n=0}^{\infty}x^n=(1-x)^{-1}$ find one Taylor development for the function $f(x)=\ln(1-x)$

So, I've made some derivatives:
$f^1(x)=\displaystyle\frac{-1}{(1-x)},f^2(x)=\displaystyle\frac{-1}{(1-x)^2},f^3(x)=\displaystyle\frac{-2}{(1-x)^3},f^4(x)=\displaystyle\frac{-6}{(1-x)^4},f^5(x)=\displaystyle\frac{-24}{(1-x)^5}$

And then:

$\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac {f^b(x_0)(x-x_0)^n}{n!}=-\displaystyle\frac{(x-x_0)}{(1-x_0)}-\displaystyle\frac{(x-x_0)^2}{2(1-x_0)^2}-\displaystyle\frac{2(x-x_0)^3}{6(1-x_0)^3}-\displaystyle\frac{6(x-x_0)^4}{24(1-x_0)^4}-\displaystyle\frac{24(x-x_0)^5}{120(1-x_0)^5}+\ldots+-\displaystyle\frac{(x-x_0)^n}{n(1-x_0)^n}$

I have two problems with this. In the first place, the general expression that I've found (which is probably wrong) doesn't work for n=0, it does for the others values of n. I thought of starting the summation at 1, but I'm not sure if this is valid. In the second place I don't know how to use the relation the problem gives at the beginning. I can see that I have (1-x_0) for every term, but I couldn't make it fit inside the summation.

So this is what I got: $\displaystyle\sum_{n=1}^{\infty}-\displaystyle\frac{(x-x_0)^n}{n(1-x_0)^n}$

Bye there, thanks for your help and suggestions.

**TheEmptySet** · April 29th 2011, 03:49 PM

You didn't use the hint

Notice that

$f(x)=\ln(1-x) \implies f'(x)=\frac{-1}{1-x}$

But now by the hint we know this is a geometric series and gives

$f'(x)=\frac{-1}{1-x}=-\sum_{n=0}^{\infty}x^n$

But we want the power series for f not its derivative so lets integrate to get

$\int f'(x)dx =-\int \sum_{n=0}^{\infty}x^ndx=-\sum_{n=0}^{\infty}\int x^ndx$

This gives

$f(x)=C-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$

and since $f(0)=0 \implies C=0$

$f(x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$

**Ulysses** · April 29th 2011, 03:54 PM

Very nice. Thank you. I would never realized of it :P

BTW, the problem is implicitly telling that the function is taken on an interval where its uniformly convergent, right?

**TheCoffeeMachine** · April 29th 2011, 04:12 PM

Oops, I thought wrong; sorry. :)

**Ulysses** · April 29th 2011, 04:14 PM

I think not but now I don't need 'em anyway :P

**TheCoffeeMachine** · April 29th 2011, 04:18 PM

Originally Posted by Ulysses

I think not but now I don't need 'em anyway :P

Yeah, I missed that you accounted for the sign alternations in the denominators! :P

**TheEmptySet** · April 29th 2011, 04:52 PM

Originally Posted by Ulysses

Very nice. Thank you. I would never realized of it :P

BTW, the problem is implicitly telling that the function is taken on an interval where its uniformly convergent, right?

Yes the geometric series is uniformly convergent on |x|<1

**Ulysses** · April 30th 2011, 06:58 AM

Thank to all of you. Now I'm trying to solve the same exercise, but with the function $f(x)=(1+4x^2)^{-1}$ I've tried to make some algebraic work to get the expression I'm looking for, $(1-x)^{-1}$ but I didn't get it.

I've made the first derivative, and a few more, but I think its just necessary the first, as before. I get:
$f'(x)=\diplaystyle\frac{-8x}{(1+4x^2)^2}$

I thought of making $f(x)=\diplaystyle\frac{1}{(1+4x^2)}= \diplaystyle\frac{1-4x^2}{(1+4x^2)(1-4x^2)}=\diplaystyle\frac{1-4x^2}{(1-16x^2)}$
So, when I take the derivative in this last form I get:

$f'(x)=\diplaystyle\frac{-256x^3}{(1-16x^4)^2}$

I'm not sure how to proceed.

**TheEmptySet** · April 30th 2011, 07:11 AM

Originally Posted by Ulysses

Thank to all of you. Now I'm trying to solve the same exercise, but with the function $f(x)=(1+4x^2)^{-1}$ I've tried to make some algebraic work to get the expression I'm looking for, $(1-x)^{-1}$ but I didn't get it.

I've made the first derivative, and a few more, but I think its just necessary the first, as before. I get:
$f'(x)=\diplaystyle\frac{-8x}{(1+4x^2)^2}$

I thought of making $f(x)=\diplaystyle\frac{1}{(1+4x^2)}= \diplaystyle\frac{1-4x^2}{(1+4x^2)(1-4x^2)}=\diplaystyle\frac{1-4x^2}{(1-16x^2)}$
So, when I take the derivative in this last form I get:

$f'(x)=\diplaystyle\frac{-256x^3}{(1-16x^4)^2}$

I'm not sure how to proceed.

It is just function composition.

Let $y=-4x^2$

Then we know that

$f(x)=\frac{1}{1+4x^2}=\frac{1}{1-y}=f(y)$

But this is just a geometric series so we get

$f(y)=\frac{1}{1-y}=\sum_{n=0}^{\infty}y^n$

Now just plug back in $y=-4x^2$

$f(x)=\sum_{n=0}^{\infty}(-4x^2)^n=\sum_{n=0}^{\infty}(-1)^n4^nx^n$

Thread: Taylor development

LinkBack

Thread Tools

Display

Taylor development

Similar Math Help Forum Discussions

Mathematics for aplications development.

Taylor development

Complex analysis and Taylor development

Taylor's development question

shapes development

Search Tags