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AP Calculus BC 10.10 Alternating Series Error Bound - FRQs - Exam Style Questions

AP Calculus BC-FRQs ad Answer - All Topics

No-Calc Question

A function \(f\) has derivatives of all orders for \(-1<x<1\). The derivatives of \(f\) satisfy: \[ f(0)=0,\qquad f'(0)=1,\qquad f^{(n+1)}(0) = -\,n\,f^{(n)}(0)\ \text{ for all } n\ge 1. \] The Maclaurin series for \(f\) converges to \(f(x)\) for \(|x|<1\).
(a) Show that the first four nonzero terms of the Maclaurin series for \(f\) are \(\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\), and write the general term of the Maclaurin series for \(f\).
(b) Determine whether the Maclaurin series in part (a) converges absolutely, converges conditionally, or diverges at \(x=1\). Explain your reasoning.
(c) Write the first four nonzero terms and the general term of the Maclaurin series for \(\displaystyle g(x)=\int_{0}^{x} f(t)\,dt\).
(d) Let \(P_{n}\!\left(\tfrac12\right)\) represent the \(n\)th-degree Taylor polynomial for \(g\) about \(x=0\) evaluated at \(x=\tfrac12\). Use the alternating series error bound to show that \(\displaystyle \big|P_{4}\!\left(\tfrac12\right)-g\!\left(\tfrac12\right)\big|<\frac{1}{500}\).

Most-appropriate topic codes (CED):

• TOPIC 10.14: Finding Taylor/Maclaurin Series (build via derivative data & recurrence) — parts (a), (c)
• TOPIC 10.13: Radius & Interval of Convergence / Endpoint behavior — part (b)
• TOPIC 10.10: Alternating Series Test & Error Bound — part (d)

▶️ Answer/Explanation

(a) Build the Maclaurin series from the recurrence
Use \(f^{(n+1)}(0)=-n\,f^{(n)}(0)\) with the base values \(f(0)=0,\ f'(0)=1\):
\(\displaystyle f”(0) = -1\cdot f'(0) = -1\) \(\displaystyle f^{(3)}(0) = -2\cdot f”(0) = 2\) \(\displaystyle f^{(4)}(0) = -3\cdot f^{(3)}(0) = -6\) Maclaurin: \(\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}\). Thus, \[ f(x)=0+\frac{1}{1!}x+\frac{-1}{2!}x^{2}+\frac{2}{3!}x^{3}+\frac{-6}{4!}x^{4}+\cdots =\boxed{x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots }. \] Pattern: \(\displaystyle \boxed{\text{general term }\, a_n(x)=(-1)^{\,n+1}\frac{x^{n}}{n}\quad (n\ge1)}.\)

(b) Behavior at \(x=1\)
At \(x=1\): \(\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}\) — the alternating harmonic series. • Not absolutely convergent because \(\sum \frac{1}{n}\) diverges. • Alternating with decreasing terms \(\to 0\) ⇒ converges by the Alternating Series Test. \(\Rightarrow\) \(\boxed{\text{Converges conditionally at } x=1.}\)

(c) Series for \(g(x)=\displaystyle\int_{0}^{x}f(t)\,dt\)
Integrate the series term-by-term (valid for \(|x|<1\)): \[ g(x)=\int_{0}^{x}\!\left(t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\frac{t^{4}}{4}+\cdots\right)dt =\boxed{\frac{x^{2}}{2}-\frac{x^{3}}{6}+\frac{x^{4}}{12}-\frac{x^{5}}{20}+\cdots }. \] General term (integrate \( (-1)^{n+1}\frac{t^{n}}{n} \)): \[ \boxed{\text{general term }\, b_n(x)=(-1)^{\,n+1}\frac{x^{\,n+1}}{n(n+1)}\quad (n\ge1)}. \]

(d) Alternating error at \(x=\tfrac12\) for \(P_4\)
The degree–4 Taylor polynomial for \(g\) keeps terms through \(x^{4}\): \(P_4(x)=\dfrac{x^{2}}{2}-\dfrac{x^{3}}{6}+\dfrac{x^{4}}{12}\). The next omitted (first unused) term is the \(x^{5}\) term: \(\displaystyle \frac{x^{5}}{20}\) in magnitude. Alternating Series Error Bound: \[ \big|P_{4}(\tfrac12)-g(\tfrac12)\big| \le \left|\frac{(\tfrac12)^{5}}{20}\right| = \frac{1}{32\cdot 20} = \frac{1}{640} < \frac{1}{500}. \] Hence \(\boxed{\big|P_{4}(\tfrac12)-g(\tfrac12)\big|<\tfrac{1}{500}}\).

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