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AP Calculus BC : Maclaurin Series

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Example Questions

Example Question #31 : Maclaurin Series

$\begin{align*}&\text{Approximate }f(-2) \\&\text{Where }f(0)=8;f'(0)=-4;f''(0)=20\\&\text{By using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$-\frac{152}{3}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=8;f'(0)=-4;f''(0)=20\\&\text{ at }x=-2\\&\text{This expansion becomes:}\\&f(-2)\approx \frac{8}{0!}(-2)^{0}+\frac{-4}{1!}(-2)^{1}+\frac{20}{2!}(-2)^{2}\\&f(-2)\approx56\end{align*}$

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Example Question #52 : Polynomial Approximations And Series

$\begin{align*}&\text{Approximate }f(3) \\&\text{Where }f(0)=-5;f'(0)=-14;f''(0)=11\\&\text{With a Maclaurin series expansion.}\end{align*}$

Possible Answers:

$\frac{5}{2}$

$-\frac{57}{2}$

$-\frac{81}{2}$

Correct answer:

$\frac{5}{2}$

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-5;f'(0)=-14;f''(0)=11\\&\text{ at }x=3\\&\text{This expansion becomes:}\\&f(3)\approx \frac{-5}{0!}(3)^{0}+\frac{-14}{1!}(3)^{1}+\frac{11}{2!}(3)^{2}\\&f(3)\approx\frac{5}{2}\end{align*}$

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Example Question #53 : Polynomial Approximations And Series

$\begin{align*}&\text{For the function values: }\\&f(0)=-8;f'(0)=1;f''(0)=14;f'''(0)=4;f''''(0)=-7\\&\text{Approximate }f(-3)\text{ using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$-\frac{9}{8}$

$-\frac{273}{40}$

$\frac{83}{8}$

Correct answer:

$\frac{83}{8}$

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-8;f'(0)=1;f''(0)=14;f'''(0)=4;f''''(0)=-7\\&\text{ at }x=-3\\&\text{This expansion becomes:}\\&f(-3)\approx \frac{-8}{0!}(-3)^{0}+\frac{1}{1!}(-3)^{1}+\frac{14}{2!}(-3)^{2}+\frac{4}{3!}(-3)^{3}+\frac{-7}{4!}(-3)^{4}\\&f(-3)\approx\frac{83}{8}\end{align*}$

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AP Calculus BC : Maclaurin Series

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #31 : Maclaurin Series

Example Question #52 : Polynomial Approximations And Series

Example Question #53 : Polynomial Approximations And Series

All AP Calculus BC Resources

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