Assignment: Pick any exercise in 8.8 of the calculus textbook. Solve and post your solution. If you have issues doing so, discuss them. ## Solution
Show that the Taylor series for f(x), as given in key idea 8.8.1, show that lim Rn(x) = 0 as n approaches infinity (page 496)
The Taylor series for sin(x) is given in Key Idea 8.8.1 and is as follows:
\[\Sigma ^\infty _{n = 0} (-1)^n \frac{x^{2n+1}}{(2n + 1)}\]
The general formula for the magnitude of R(x) is provided in 8.7.1:
\[ \vert R_n(x)\vert = \frac{max \vert f^{n+1}(z_x) \vert}{(n + 1)!} \vert x^{(n+1)}\vert\]
where z is a value that exists on the interval I. The absolute value of f(x) = sin(x) is never greater than 1. Taking serial derivatives of sin(x) also only generates the function +/- cos(x) which also has a max value of 1. This means that R_n(x) is bounded in the folloiwng manner:
\[-\frac {\vert x^{n+1}\vert}{(n+1)!} \leq R_n(x) \leq \frac {\vert x^{n+1}\vert}{(n+1)!}\]
The limit of the denominator (n + 1)! approaches infinity much faster than the exponential term x^n in the numerator. The limit of the left and right-handed sides both approach zero, and by the Squeeze theorem the center term R_n(x) also approaches zero for all values of x.