Hello,

I have this problem, that requires us to derive the taylor series based at x = -1 for f(x) from scratch, then find an interval of converge for the series, including endpoint behavior, and finally approximating:

ln((-1 + 2*Pi)/(Pi)) accurate to within 10^-25.

f(x) = ln(x + 2)

Okay, so this is what I have done.

I first found the first 6 derivatives to find a pattern. So they are as follows:

f'(x) = 1/(x+2)

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

f^3(x) = 2/(x+2)^3

f^4(x) = -6(x+2)^4

f^5(x) = 24/(x+2)^5

f^6(x) = -120/(x+2)^6

And so forth.

So then I expanded it to:

1*(x+2)^(-1)

1*(-1)*(x+2)^(-2)

1*(-1)*(-2)*(x+2)^(-3)

1*(-1)*(-2)*(-3)*(x+3)^(-4)

And so forth.

Okay, so the coefficients are factorials. The coefficient for the nth derivative is just:

(+ or -)(n-1)!

Now, the nth derivative can be written as follows:

(+ or -)(n-1)!(x+2)^(-n)

The signs alternate, so I need a factor of (-1)^n or (-1)^(n+1), whichever matches up correctly

For n = 1,2,3,..., the signs are +,-,+,..., therefore I need to use the (-1)^(n+1)

So I have (-1)^(n+1)(n-1)!(x+2)^(-n)

What I really wanted was the nth derivative evaluated at x=-1, so substituted -1 in for x to get:

(-1)^(n+1)(n-1)!(1)^(-n)= (-1)^(n+1)(n-1)!

Now I have to use the Taylor series expansion formula:

S(x) = sum(a_n(x-x_0)^n, where a_n = [S^n(x_0)]/n!

So now I have sum((A_n/n!)(x+1)^n, x, 0, infinity).

A_n is the nth deriv. evaluated at x = -1.

Now what. How do I evaluate the other parts and further expand this if its not done so already. How do I get an approximation accurate to within such a small error bound.

Book has the following result: Sum from n=1 to infinity of [(-1)^(n+1) / n] * (x+1)^n ... I was close just don't really see it. How would I do the remaining parts for accuracy and approximating the function above?

Thanks for your help.