Problem 58 - Calculus Early Transcendentals

Question

Find the remainder in the Taylor series centered at the point a for the following functions. Then show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) for all \(x\) in the interval of convergence. $$f(x)=\cos 2 x, a=0$$

Step-by-Step Solution

Answer

Based on the solution above, create a short answer to the following: - Given the function $$f(x) = \cos(2x)$$, we found its Taylor series centered at a=0. It turned out to be: $$T_n(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k}2^{2k}}{(2k)!}x^{2k}$$ - We determined the remainder term $$R_n(x)$$ for the Taylor series, and showed that as n approaches infinity, the remainder converges to zero for all x in the interval of convergence: $$\lim _{n \rightarrow \infty} R_{n}(x) = 0$$

Step 1: 1. Find the Taylor series for f(x) centered at a=0

First, we need to find the Taylor series for the given function f(x) centered at a=0. Since $$f(x) = \cos(2x)$$, we will find the successive derivatives of f(x) and evaluate them at $$x=0$$. The n-th derivative of a function is given by: $$f ^{(n)}(x) = \frac{d^n f(x)}{d x^n}$$ Find the derivatives of f(x) evaluated at x=0: $$f(x) = \cos(2x) \Rightarrow f(0) = \cos(0) = 1$$ $$f'(x) = -2 \sin(2x) \Rightarrow f'(0) = 0$$ $$f''(x) = -4 \cos(2x) \Rightarrow f''(0) = -4$$ $$f'''(x) = 8 \sin(2x) \Rightarrow f'''(0) = 0$$ $$\vdots$$ Notice that for even n, $$f^{(n)}(0)=(-1)^{n/2}2^n$$ and for odd n, $$f^{(n)}(0)=0$$. Hence, the nth-degree Taylor polynomial centered at a=0 can be written as: $$T_n(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k}2^{2k}}{(2k)!}x^{2k}$$

Step 2: 2. Find the remainder R_n(x)

The remainder term in the Taylor series is given by the (n+1)-st term in the series, which we can write as: $$R_{n}(x) = \frac{f^{(2n+2)}(0)}{(2n+2)!}x^{2n+2}$$ Now we need to find the expression for the remainder term when x is in the interval of convergence.

Step 3: 3. Finding the limit of R_n(x) as n goes to infinity

In order to prove that the remainder converges to zero for all x, we need to find the upper bound of the maximum value R_n(x) can achieve in the interval of convergence. Since the original function is a cosine function with a period of $$\pi$$, the Taylor series converges everywhere. Now we will show that the remainder goes to zero as n goes to infinity. Using the Lagrange form of the remainder: $$R_n(x) = \frac{f^{(2n+2)}(c)}{(2n+2)!}x^{2n+2}$$ for some $$c \in (0, x)$$. Note that since $$f^{(2n+2)}(c) = (-1)^{n+1}2^{2n+2}\sin(2c)$$, we have that: $$|f^{(2n+2)}(c)| \leq 2^{2n+2}$$ Therefore, the absolute value of the remainder is given by: $$|R_n(x)| \leq \frac{2^{2n+2}}{(2n+2)!}|x|^{2n+2}$$ As n goes to infinity, the term $$\lim _{n \rightarrow \infty}\frac{2^{2n+2}}{(2n+2)!} \to 0$$ , and since x is in the interval of convergence, it follows that: $$\lim _{n \rightarrow \infty} R_{n}(x) = 0$$ This shows that the remainder converges to zero for all x in the interval of convergence as n approaches infinity, and we complete the task.