Thread: Taylor series

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.

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)]

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

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

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

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

It seems the pattern is,

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

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

Originally Posted by @_@

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

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

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

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,*)
$\displaystyle \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:
$\displaystyle \lim_{k\to \infty} \left| \frac{(-1)^{k+1}}{k+1}\cdot \frac{k}{(-1)^k} \right|=1$
Thus the reciprocal of that is also 1.
Meaning the radius of convergence is 1.
Thus,
$\displaystyle |x|<1$ you can also write $\displaystyle -1<x<1$
The series converges absolutely.


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

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

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 $\displaystyle \ln |1-x|$ for $\displaystyle |x|<1$.

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.

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

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

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

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

Originally Posted by putnam120

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

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

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

To add, "If a function is infinitely differenciable at $\displaystyle x=a$ 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.

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?

Originally Posted by ThePerfectHacker

To add, "If a function is infinitely differenciable at $\displaystyle x=a$ 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.

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?

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

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