Taylor Polynomial of 6th Degree for ln(1-x^2) with c=0

[shan]

I just want to check my answer. The question asks for the Taylor polynomial of degree 6 for ln(1-x^2) for -1<x<1 with c=0.
I got tired after differentiating 6 times so I'm worried I made some mistakes along the way. The question also said: hint: evaluate the derivatives using the formula ln(1-x^2)=ln(1+x)+ln(1-x)

So I differentiated ln(1+x)+ln(1-x) to get 1/(1+x) - 1/(1-x) for the first derivative and so on until the 6th derivative and I got:

-2(x)^2/2! -12(x)^4/4! - 240(x^6)/6!

 

[HallsofIvy]

Yes, that's correct.

(Of course, it can be simplified to -x^2- (1/2)x^4- (1/3)x^6.)

 

[shan]

Yay, thank you very much :)

There's also something else I want to check. The last part of the question asks:
Write f(x) = ln(1-x^2) as a Taylor series.

I'm a little confused with the distinction between a Taylor polynomial and a Taylor series. With the previous question on the Taylor polynomial, there was a specific number of terms (Taylor polynomial to degree 6). Is the Taylor series different in that you need to write it as an ongoing series? In which case, do I write it as

ln(1-x^2) = -x^2- (1/2)x^4- (1/3)x^6 -...
centered at c=0 and -1<x<1

or do I need to write it using the sigma notation?

sigma (n=0 to infinity) f^(n)(0)(x^n)/n!

 

[Data]

Sigma notation is likely what they're looking for, though it probably doesn't matter as long as you include the general term (ie. in all likelihood they want you to find the general expression for $\left[ \frac{d^n}{dx^n} \ln (1-x^2) \right]_{x=0}$, [call it $a_n$], and then write your series as $\sum_{n=0}^\infty \frac{a_n}{n!} x^n$).

 

[shan]

I don't understand, why would I find the general expression for $\left[ \frac{d^n}{dx^n} \ln (1-x^2) \right]_{x=0}$ to find $\sum_{n=0}^\infty \frac{a_n}{n!} x^n$ ??