Problem 60 - 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. $$f(x)=\cos x, a=\pi / 2$$

Step-by-Step Solution

Answer

Question: Prove that the limit of the remainder term \(R_n(x)\) in the Taylor series centered at \(a = \pi/2\) for the function \(f(x) = \cos x\) is zero as \(n\) approaches infinity for all \(x\) in the interval of convergence. Answer: To prove this, we first determined the Taylor series for \(f(x)\) centered at \(a = \pi/2\), which is given by: $$T_{n}(x) = \sum_{k=0, k\, odd}^{n} \frac{(-1)^{(k-1)/2}(x-\pi/2)^k}{k!}$$ Next, we calculated the remainder term, \(R_n(x)\), using Taylor's theorem, and found that its absolute value can be bounded by: $$\left|R_n(x)\right|\leq \frac{(x-\pi/2)^{n+1}}{(n+1)!}$$ Finally, we showed that the limit of the absolute value of the remainder as \(n\) approaches infinity is zero since the factorial grows much faster than the power function, causing the fraction to tend to zero: $$\lim_{n \to \infty} \left|R_n(x)\right| \leq \lim_{n \to \infty} \frac{(x-\pi/2)^{n+1}}{(n+1)!} = 0$$ Thus, the limit of the remainder term, \(R_n(x)\), as \(n\) approaches infinity is indeed zero for all \(x\) in the interval of convergence.

Step 1: Determine the Taylor series for \(f(x)\) centered at \(a\).

Firstly, we need to determine the Taylor series centered at the point \(a=\pi/2\) for the function \(f(x)=\cos x\). The n-th degree Taylor polynomial for \(f\) centered at \(a\) is given by: $$T_{n}(x) = \sum_{k=0}^{n}\frac{f^{(k)}(a)(x-a)^k}{k!}$$ The derivatives of \(\cos x\) are periodic (alternates between \(\pm \cos x\) and \(\pm \sin x\)), so we can derive a general formula: $$f^{(k)}(x) = \begin{cases} (-1)^{m}\cos x & \text{if}\ k = 4m, \\ (-1)^{m}\sin x & \text{if}\ k = 4m + 1, \\ (-1)^{m+1}\cos x & \text{if}\ k = 4m + 2, \\ (-1)^{m+1}\sin x & \text{if}\ k = 4m + 3, \\ \end{cases}$$ where \(m\) is a non-negative integer. Now, let's compute \(f^{(k)}(\pi/2)\): $$f^{(k)}(\pi/2) = \begin{cases} 0 & \text{if}\ k = 4m, \\ (-1)^{m} & \text{if}\ k = 4m + 1, \\ 0 & \text{if}\ k = 4m + 2, \\ (-1)^{m+1} & \text{if}\ k = 4m + 3. \\ \end{cases}$$ Now, we can write down the Taylor series \(T_n(x)\) for \(f(x)\) centered at \(\pi/2\): $$T_{n}(x) = \sum_{k=0}^{n} \frac{f^{(k)}(\pi/2)(x-\pi/2)^k}{k!} = \sum_{k=0, k\, odd}^{n} \frac{(-1)^{(k-1)/2}(x-\pi/2)^k}{k!}$$

Step 2: Calculate the remainder \(R_{n}(x)\)

According to Taylor's theorem, the remainder is given by: $$R_n(x) = \frac{f^{(n+1)}(c)(x-\pi/2)^{n+1}}{(n+1)!}$$ for some \(c\) between \(x\) and \(\pi/2\). Since both \(\cos c\) and \(\sin c\) are bounded between -1 and 1, the remainder can be estimated using the bounds: $$\left|R_n(x)\right|\leq \frac{(x-\pi/2)^{n+1}}{(n+1)!}$$

Step 3: Show the limit of the remainder as \(n\) approaches infinity is zero

In order to show that \(\lim_{n \to \infty} R_{n}(x) = 0\), we must prove that for any given \(x\) in the interval of convergence, the absolute value of the remainder goes to zero as \(n\) approaches infinity. From the estimate obtained in Step 2, we have: $$\lim_{n \to \infty} \left|R_n(x)\right| \leq \lim_{n \to \infty} \frac{(x-\pi/2)^{n+1}}{(n+1)!} = 0$$ since the factorial grows much faster than the power function, causing the fraction to tend to zero. Therefore, the limit of the remainder as \(n\) approaches infinity is indeed zero for all \(x\) in the interval of convergence. $$\lim_{n \to \infty} R_n(x) = 0$$