Taylor series

**@_@** · Nov 25th 2006, 07:56 PM

f(x) =ln (1-x)

a) Compute f'(x), f''(x), f'''(x). Spot the pattern and give an expression for f ^(n) (x) [the n-th derivative of f(x)]

b) Compute the MacLaurin series of f(x) (i.e. the Taylor series of f(x) around x=0)

c) Compute the radius of convergence and determine the interval of convergence of the series in b).

d) Determine the Taylor series of f'(x) around x=0. Can you do so without using b)?

e) How would you have computed part b) if you had first done part d)?

I did part a) (hope its right)

f'(x) = -1/(1-x)
f''(x) = -1/(1-x)^2
f'''(x) = -2/(1-x)^3

f^(n) (x) = -((n-1)!)/(1-x)^n
for n = 1,2,3,...

I am not too sure for the other ones, thanks for any help.

**ThePerfectHacker** · Nov 25th 2006, 08:03 PM

Originally Posted by @_@

f(x) =ln (1-x)

a) Compute f'(x), f''(x), f'''(x). Spot the pattern and give an expression for f ^(n) (x) [the n-th derivative of f(x)]

$f(x)=\ln (1-x)$
$f(0)=\ln (1-0)=0$
Thus,
$a_0 =\frac{f(0)}{0!}=0$

$f'(x)=-\frac{1}{1-x}$
$f'(0)=-\frac{1}{1-0}=-1$
Thus,
$a_1=\frac{f'(0)}{1!}=-1$

$f''(x)=\frac{1}{(1-x)^2}$
$f''(0)=\frac{1}{(1-0)^2}=1$
Thus,
$a_2=\frac{f''(0)}{2!}=\frac{1}{2}$

$f'''(x)=-\frac{2}{(1-x)^3}$
$f'''(0)=-\frac{2}{(1-0)^3}=-1$
Thus,
$a_3=\frac{f'''(0)}{3!}=-\frac{1}{3}$

It seems the pattern is,

$\ln (1-x)=-\frac{x}{1}+\frac{x^2}{2}-\frac{x^3}{3}+...$

**@_@** · Nov 26th 2006, 10:55 AM

but its asking for the n-th derivative of f(x), would my answer then be correct?

**ThePerfectHacker** · Nov 26th 2006, 11:10 AM

Originally Posted by @_@

but its asking for the n-th derivative of f(x), would my answer then be correct?

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

**@_@** · Nov 26th 2006, 11:28 AM

oh. i see, then how do you computer radius of convergence and its interval?
thanks.

**ThePerfectHacker** · Nov 26th 2006, 01:19 PM

Originally Posted by @_@

oh. i see, then how do you computer radius of convergence and its interval?
thanks.

The power series that we get is,*)
$\sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^k$

In order to find the radius of convergence we will use the generalized ratio test for power series:
$\lim_{k\to \infty} \left| \frac{(-1)^{k+1}}{k+1}\cdot \frac{k}{(-1)^k} \right|=1$
Thus the reciprocal of that is also 1.
Meaning the radius of convergence is 1.
Thus,
$|x|<1$ you can also write $-1<x<1$
The series converges absolutely.

*)Which is in fact not the power series for $\ln (1-x)$ .

**@_@** · Nov 26th 2006, 01:55 PM

why is it not the power series? Then where did you get this equation?
thanks.

**ThePerfectHacker** · Nov 26th 2006, 02:19 PM

Originally Posted by @_@

why is it not the power series? Then where did you get this equation?
thanks.

Because if a power series does exist. Then this has to be it. It does not mean that this is it. In fact this power series is for $\ln |1-x|$ for $|x|<1$ .

**@_@** · Nov 26th 2006, 02:55 PM

i am kind of lost right now, sorry about that.

so far, we have found part a) which is the n-th derivative of f(x),
so what about MacLaurin series of f(x), is it (-x/1) +(x^2/2) - (x^3/3) +...?

and also for f''(x), i don't know why i kept getting f''(x) = -1/(1-x)^2, which you don't have a negative.

thanks.

**ThePerfectHacker** · Nov 26th 2006, 03:00 PM

Originally Posted by @_@

i am kind of lost right now, sorry about that.

so far, we have found part a) which is the n-th derivative of f(x),
so what about MacLaurin series of f(x), is it (-x/1) +(x^2/2) - (x^3/3) +...?

and also for f''(x), i don't know why i kept getting f''(x) = -1/(1-x)^2, which you don't have a negative.

thanks.

Because you have,
$-\frac{1}{(1-x)^2}$
When you take the derivative you need to use the chain rule, the inner function is,
$1-x$ and you need to multiply by its derivative thus, $-1$ . But you already have a minus sign, so they die and you have a positive sign. Basic differenciation.

**putnam120** · Nov 26th 2006, 03:50 PM

The general formula for a Taylor Series of $f(x)$ around the point $x=a$ is

$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$ where $f^{(n)}(x)$ is the nth derivative.

The MacLaurin Series is just a Taylor Series centered around $a=0$ .

**ThePerfectHacker** · Nov 26th 2006, 03:57 PM

Originally Posted by putnam120

The general formula for a Taylor Series of $f(x)$ around the point $x=a$ is

$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x)}{n!}(x-a)^n$ where $f^{(n)}(x)$ is the nth derivative.

The MacLaurin Series is just a Taylor Series centered around $a=0$ .

To add, "If a function is infinitely differenciable at $x=a$ then the Taylor series is definied as a power series...."

But, the Taylor series of a function does not always represent the function. You need to show that the remainder term sequence is convergent to zero. Like here.

**@_@** · Nov 26th 2006, 04:11 PM

the taylor series would just be like this? f(x)= (-x/1) +(x^2/2) - (x^3/3) +...

how would you put it in a form like the one putnam120 provide?

**putnam120** · Nov 26th 2006, 05:01 PM

Originally Posted by ThePerfectHacker

To add, "If a function is infinitely differenciable at $x=a$ then the Taylor series is definied as a power series...."

But, the Taylor series of a function does not always represent the function. You need to show that the remainder term sequence is convergent to zero. Like here.

o yes. its just that most of the functions that i work w/ the remainder tends to zero. but its not too hard to prove.

**ThePerfectHacker** · Nov 26th 2006, 05:45 PM

Originally Posted by @_@

the taylor series would just be like this? f(x)= (-x/1) +(x^2/2) - (x^3/3) +...

how would you put it in a form like the one putnam120 provide?

$f(x)=\sum_{k=1}^{\infty} \frac{(-1)^k}{k}x^k$
That is the summation form.

(Again, $f(x)\not = \ln (1-x)$ )

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