Problem 57 - 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. $$f(x)=\sin x, a=0$$

Step-by-Step Solution

Answer

Question: Show that the remainder of the Taylor series for the function \(f(x) = \sin{x}\) around the point \(a = 0\) converges to 0 for all x in its interval of convergence. Answer: We can show that the limit of the remainder converges to 0 for all x by using the Lagrange form of the remainder: $$\lim_{n \rightarrow \infty} R_n(x) = \lim_{n \rightarrow \infty} \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}$$ Since the maximum value of \(|f^{(n+1)}(c)|\) is 1, we can apply the limit and get: $$\lim_{n \rightarrow \infty} \left|R_n(x) \right| \le \lim_{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+1)!} = 0$$ As the limit is less than or equal to 0 for all \(x\) in the interval of convergence, we can conclude that the limit is indeed 0: $$\lim_{n \rightarrow \infty} R_n(x) = 0$$

Step 1: Computing derivatives of sin(x)

We will need to compute some derivatives of \(f(x) = \sin{x}\) to form the Taylor series and the remainder. \begin{align*} f^{(1)}(x) &= \cos x \\ f^{(2)}(x) &= -\sin x \\ f^{(3)}(x) &= -\cos x \\ f^{(4)}(x) &= \sin x \end{align*} Notice that the pattern repeats every 4 derivatives. Additionally, the derivatives of \(\sin{x}\) and \(\cos{x}\) are bounded by \(-1\) and \(1\).

Step 2: Constructing the Taylor series

With these derivatives, we can construct the Taylor series for \(f(x) =\sin{x}\) around \(a = 0\): $$\sin x = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x - 0)^n = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$ Since \(f^{(n)}(0)\) are either \(0\) or \(\pm 1\), the Taylor series will have non-zero terms for odd values of \(n\). $$\sin{x} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k + 1)!} x^{2k + 1}$$

Step 3: Finding the remainder

Now we need to find the \(R_n(x)\), which is given by the Lagrange form of the remainder: $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - 0)^{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}$$ Since the \(n\)-th derivative cycles every 4 terms, we have: $$f^{(n+1)}(c) = \sin c, \cos c, -\sin c, -\cos c$$ Depending on the value of \(n\) modulo 4. Regardless, the maximum value for \(|f^{(n+1)}(c)|\) is 1.

Step 4: Taking the limit

Now we need to show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) for all \(x\) in the interval of convergence. Since the maximum value of \(|f^{(n+1)}(c)|\) is 1, we have: $$\left|R_n(x)\right| = \left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right| \le \frac{|x|^{n+1}}{(n+1)!}$$ Taking the limit, we get: $$\lim_{n \rightarrow \infty} \left|R_n(x) \right| \le \lim_{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+1)!} = 0$$ As the limit is less than or equal to 0 for all \(x\) in the interval of convergence, we can conclude that the limit is indeed 0: $$\lim _{n \rightarrow \infty} R_{n}(x)=0$$