How Do You Determine the nth Term of a Taylor Polynomial for ln(1-x)?

[smk037]

Homework Statement

Write down the Taylor Polynomial of degree n of the function f(x) at x=0

Homework Equations

f(x) = ln(1-x)

The Attempt at a Solution

f(x) = ln(1-x)

f'(x) = (-1)((1-x)^(-1))

f``(x) = (-1)((1-x)^(-2))

f```(x) = (-2)((1-x)^(-3))

f````(x) = (-6)((1-x)^(-4))

f`````(x) = (-24)((1-x)^(-5))

Pn(x) = -1x - ((x^2)/(2!)) - ((2(x^3))/(3!)) - ((6(x^4))/(4!)) - ((24(x^5))/(5!))

now, I rechecked all my derivatives, but I still can't find a pattern to make an nth term with.

any help would be appreciated.

thanks.

 

[Dick]

1, 2, 6, 24 etc is 1!, 2!, 3!, 4! etc. So you have (n-1)!/n!. What's that?

 

[smk037]

I think I got it

I got the numbers in the factorials to cancel out, and came up with

(-(x^n))/(n)

 

[smk037]

Thanks a lot, Dick, I appreciate it.

 

[smk037]

Thanks a lot.

 

[nicktacik]

Another way to do it:
You know that

$$\sum^{\infty}_{n=0} x^n = \frac{1}{1-x}$$

So by integration you get that

$$ln(1-x)=-\int{\sum^{\infty}_{n=0}x^n dx}$$