AP Calculus BC 10.2 Working with Geometric Series - FRQs - Exam Style Questions
No-Calc Question
The function \(f\) is defined by the power series \[ f(x)=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots+\frac{(-1)^n x^{2n+1}}{2n+1}+\cdots \] for all real numbers \(x\) for which the series converges.
Most-appropriate topic codes (CED):
• TOPIC 10.10: Alternating Series Error Bound — part (b)
• TOPIC 10.15: Representing Functions as Power Series (differentiate power series) — part (c)
• TOPIC 10.2: Working with Geometric Series (sum of a geometric series) — part (d)
▶️ Answer/Explanation
(a) Interval of convergence (ratio test)
General term: \(a_n=\dfrac{(-1)^n x^{2n+1}}{2n+1}\).
\(\displaystyle \left|\frac{a_{n+1}}{a_n}\right| = \left|x^2\right|\cdot \frac{2n+1}{2n+3}\xrightarrow[n\to\infty]{} |x|^2.\)
Require \( |x|^2<1 \Rightarrow |x|<1\). Test endpoints:
\(x=1:\ 1-\frac13+\frac15-\cdots\) converges (alternating).
\(x=-1:\ -1+\frac13-\frac15+\cdots\) converges (alternating).
\(\boxed{\text{Interval of convergence }[-1,\,1]}\).
(b) Error bound at \(x=\tfrac12\)
The series is alternating with decreasing term magnitudes; first term \(=\tfrac12\).
Alternating series error \(\le\) magnitude of next term:
\(\displaystyle \left|f\!\left(\tfrac12\right)-\tfrac12\right| \le \left|\frac{\left(\tfrac12\right)^3}{3}\right| = \frac{1}{24} < \frac{1}{10}.\)
(c) Series for \(f'(x)\)
Differentiate term-by-term: \[ f'(x)=1-x^{2}+x^{4}-x^{6}+\cdots+(-1)^n x^{2n}+\cdots \] First four nonzero terms: \(1 – x^{2} + x^{4} – x^{6}\).
General term: \(\displaystyle \sum_{n=0}^{\infty}(-1)^n x^{2n}\).
(d) Evaluate \(f’\!\left(\tfrac{1}{6}\right)\)
From (c), \(f'(x)=\sum_{n=0}^{\infty}(-1)^n x^{2n}\) is geometric with ratio \(r=-x^{2}\).
At \(x=\tfrac16\): \(a=1,\ r=-\tfrac{1}{36}\).
Sum \(=\dfrac{a}{1-r}=\dfrac{1}{1-(-1/36)}=\dfrac{1}{1+1/36}=\dfrac{36}{37}.\)
\(\boxed{f’\!\left(\tfrac{1}{6}\right)=\dfrac{36}{37}}\).