Home / AP Calculus BC: 10.9 Determining Absolute or Conditional  Convergence – Exam Style questions with Answer- FRQ

AP Calculus BC: 10.9 Determining Absolute or Conditional  Convergence – Exam Style questions with Answer- FRQ

Question

(a) Topic-10.14-Finding Taylor or Maclaurin Series for a Function

(b) Topic-10.9-Determining Absolute or Conditional Convergence

(c) Topic-10.11-Finding Taylor Polynomial Approximations of Functions

(d) Topic-10.12-Lagrange Error Bound

Let $f$ be the function given by $f(x) = e^{-x^2}$.
(a) Write the first four nonzero terms and the general term of the Taylor series for $f$ about $x = 0$.
(b) Use your answer to part (a) to find
\(
\lim_{x \to 0} \frac{1 – x^2 – f(x)}{x^4}.
\)
(c) Write the first four nonzero terms of the Taylor series for $\int_0^x e^{-t^2} \, dt$ about $x = 0$. Use the first two terms of your answer to estimate
\(
\int_0^{1/2} e^{-t^2} \, dt.
\)
(d) Explain why the estimate found in part (c) differs from the actual value of $\int_0^{1/2} e^{-t^2} \, dt$ by less than $\frac{1}{200}$.

▶️Answer/Explanation

\(\textbf{6(a)}\)
\(
e^{-x^2} = 1 + \frac{(-x^2)}{1!} + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \dots + \frac{(-x^2)^n}{n!} + \dots
\)
\(
= 1 – x^2 + \frac{x^4}{2} – \frac{x^6}{6} + \dots + \frac{(-1)^n x^{2n}}{n!}.
\)

\(\textbf{6(b)}\)
\(
\frac{1 – x^2 – f(x)}{x^4} = \frac{-\frac{1}{2}x^2 + \frac{x^4}{6} + \sum_{n=4}^\infty \frac{(-1)^{n+1} x^{2n-4}}{n!}}{x^4}.
\)
Thus,
\(
\lim_{x \to 0} \frac{1 – x^2 – f(x)}{x^4} = -\frac{1}{2}.
\)

\(\textbf{6(c)}\)
\(
\int_0^x e^{-t^2} \, dt = \int_0^x \left( 1 – t^2 + \frac{t^4}{2} – \frac{t^6}{6} + \dots + \frac{(-1)^n t^{2n}}{n!} \right) \, dt
\)
\(
= x – \frac{x^3}{3} + \frac{x^5}{10} – \frac{x^7}{42} + \dots.
\)
Using the first two terms of this series, we estimate that
\(
\int_0^{1/2} e^{-t^2} \, dt \approx \frac{1}{2} – \frac{\left(\frac{1}{2}\right)^3}{3} = \frac{11}{24}.
\)

\(\textbf{6(d)}\)
\(
\int_0^{1/2} e^{-t^2} \, dt – \frac{11}{24} < \left(\frac{1}{2}\right)^5 \frac{1}{10} = \frac{1}{320} < \frac{1}{200},
\)
since
\(
\int_0^{1/2} e^{-t^2} \, dt = \sum_{n=0}^\infty \frac{(-1)^n \left(\frac{1}{2}\right)^{2n+1}}{n!(2n+1)},
\)
which is an alternating series with individual terms that decrease in absolute value to $0$.

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