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:
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:
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.