# Math Help - Taylor development

1. ## Taylor development

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 $f(x)=\ln(1-x)$

$f^1(x)=\displaystyle\frac{-1}{(1-x)},f^2(x)=\displaystyle\frac{-1}{(1-x)^2},f^3(x)=\displaystyle\frac{-2}{(1-x)^3},f^4(x)=\displaystyle\frac{-6}{(1-x)^4},f^5(x)=\displaystyle\frac{-24}{(1-x)^5}$

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.

2. 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 $f(0)=0 \implies C=0$

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

3. 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?

4. Oops, I thought wrong; sorry. :)

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

6. 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

7. 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

8. 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.

9. 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 $y=-4x^2$

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