Problem 14 from the MIT 18.01 course final exam:
(a) Find the Taylor series centered at a = 0 for ln(1 + x).
(b) Determine the radius of convergence for this Taylor series.
(c) Use the first two non-zero terms of the power series found in part (a) to approximate ln(3/2).
(d) Using Taylor’s inequality, give an upper bound on the error in the approximation from part (c).
the answer of part (d) is:
The upper bound of the error in (c)’s approximation is found using
Taylor’s inequality for an approximation of n terms:
$$ |R_n(x)|\le M_n \frac{|x^{n+1}|}{(n+1)!} $$
where x = 1/2 and n = 2. In addition,
$$ M_n \ge f^{(n+1)}(x) \implies M_2 \ge \frac{2}{(1+x)^3} $$
for all $|x|\le 1/2$; the maximum of $M_2$ in this range is for x = -1/2, which gives $M_2 = 16$. Putting these numbers into the above formula,
$$ R_n(.5) \le 16\frac{(.5)^3}{3!} =\frac{1}{3} $$ Regarding part (d), the answer given states that the range of x is |x| ≤ 1/2. However, i am confused because the radius of convergence is −1 < x < 1 and x₀ = 1/2. i am wondering why the range of x is |x| ≤ 1/2 instead of −3/2 < x < 1/2.