Home / AP Calculus BC: 10.14 Finding Taylor or Maclaurin Series for  a Function – Exam Style questions with Answer- FRQ

AP Calculus BC: 10.14 Finding Taylor or Maclaurin Series for  a Function – Exam Style questions with Answer- FRQ

Question

Let \(f(x)=sin(x^{2})+cos x.\) The graph of \(y = \left | f^{(5)}(x) \right |\) is shown above.

(a) Write the first four nonzero terms of the Taylor series for sin x about x = 0, and write the first four nonzero terms of the Taylor series for \(sin (x^{2})\) about x = 0.
(b) Write the first four nonzero terms of the Taylor series for cos x about x = 0. Use this series and the series for \(sin (x^{2})\), found in part (a), to write the first four nonzero terms of the Taylor series for f about x = 0.
(c) Find the value of f(6) (0).
(d) Let P4 (x) be the fourth-degree Taylor polynomial for f about x = 0. Using information from the graph of \(y = \left | f^{(5)}(x) \right |\) shown above, show that \(\left | p_{4}\left ( \frac{1}{4} \right )-f\left ( \frac{1}{4} \right ) \right |<\frac{1}{3000}.\)

Answer/Explanation

Ans:

(a)

\(sin x\approx x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}\)

\(sin (x^{2})\approx x^{2}-\frac{x^{6}}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}\)

(b)

\(cos x\approx 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}\)

\(sin (x^{2})+cos (x)\approx1 -\frac{x^{2}}{2!}7x^{2}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}-\frac{x^{6}}{3!}\)

\(\approx1 -x^{2}\left ( \frac{1}{2}-1 \right )\frac{x^{4}}{4!}-x^{2}\left ( \frac{1}{6!}+\frac{1}{3!} \right )\)

(c)

By the definition of taylor series:

\(\frac{f^{t}(0)(x)^{6}}{6!}=x^{6}\left ( \frac{1}{6!}+\frac{1}{3!} \right )\)

\(\frac{f^{t}(0)}{6!}=-\left ( \frac{1+4.5.6}{6!} \right )\)

f(6) (0) = – (1+120)

= -121

(d)

\(\left | p_{4}\left ( \frac{1}{4} \right )-t\left ( \frac{1}{4} \right ) \right |\) is equivalent to the error of \(p_{4}\left ( \frac{1}{4} \right )\) Using the lagrangian error formulas we have 

\(E<\frac{f^{(n+1)}(x)x^{n+1}}{(n+1)!}=\frac{f^{(5)}\left ( \frac{1}{4} \right )\cdot \left ( \frac{1}{4} \right )^{5}}{5!}\)

\(=\frac{f^{(5)}\left ( \frac{1}{4} \right )}{120\cdot 102t}\)

Looking at the graph, we see that \(f^{(5)}\left ( \frac{1}{4} \right )\) is a number less that 40.

If it were 40, then the maximum error would be \(\frac{40}{120.102t}=\frac{1}{3.102t},\) which is less than 1/3000. Since \(f^{(5)}\left ( \frac{1}{4} \right )\) is less than 40 then the maximum error must be less than 1/3000 QE.D.

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