# Taylor polynomial question

1. Aug 21, 2007

### smk037

1. The problem statement, all variables and given/known data

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

2. Relevant equations

f(x) = ln(1-x)

3. 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.

2. Aug 21, 2007

### Dick

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

3. Aug 21, 2007

### smk037

I think I got it

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

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

4. Aug 21, 2007

### smk037

Thanks a lot, Dick, I appreciate it.

5. Aug 21, 2007

### smk037

Thanks a lot.

6. Aug 22, 2007

### 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

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