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AP Calculus BC : Maclaurin Series

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Example Question #31 : Taylor Series

$\begin{align*}&\text{Approximate }f(4) \\&\text{Where }f(x)=-2^{(3x)}cos(9x)\\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align*}$

Possible Answers:

151.02

-4228.63

604.09

-4832.72

Correct answer:

604.09

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Like the Taylor series, it is used to approximate an unknown}\\&\text{value of a function at a given point, if values of the function}\\&\text{and its derivatives are known at another point. In the case}\\&\text{of the Maclaurin series, this latter point has the value of}\\&\text{zero (note that for many functions, such as the trigonometric}\\&\text{functions, f(0) is readily known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-2^{(3x)}cos(9x)\text{ at }x=4\\&\text{This expansion becomes:}\\&f(4)\approx \frac{-2^{(3(0))}cos(9(0))}{1}(4)^{0}+\frac{9\cdot 2^{(3(0))}sin(9(0)) - 3\cdot 2^{(3(0))}cos(9(0))ln(2)}{1}(4)^{1}+\frac{81\cdot 2^{(3(0))}cos(9(0)) + 54\cdot 2^{(3(0))}sin(9(0))ln(2) - 9\cdot 2^{(3(0))}cos(9(0))ln(2)^{2}}{2}(4)^{2}\\&f(4)\approx604.09\end{align*}$

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Example Question #31 : Polynomial Approximations And Series

$\begin{align*}&\text{Approximate }f(2.5) \\&\text{Where }f(x)=-cos(6x)e^{(3x)}\\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align*}$

Possible Answers:

10.84

-607.00

-151.75

75.88

Correct answer:

75.88

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Like the Taylor series, it is used to approximate an unknown}\\&\text{value of a function at a given point, if values of the function}\\&\text{and its derivatives are known at another point. In the case}\\&\text{of the Maclaurin series, this latter point has the value of}\\&\text{zero (note that for many functions, such as the trigonometric}\\&\text{functions, f(0) is readily known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-cos(6x)e^{(3x)}\text{ at }x=2.5\\&\text{This expansion becomes:}\\&f(2.5)\approx \frac{-cos(6(0))e^{(3(0))}}{1}(2.5)^{0}+\frac{6sin(6(0))e^{(3(0))} - 3cos(6(0))e^{(3(0))}}{1}(2.5)^{1}+\frac{27cos(6(0))e^{(3(0))} + 36sin(6(0))e^{(3(0))}}{2}(2.5)^{2}\\&f(2.5)\approx75.88\end{align*}$

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Example Question #31 : Polynomial Approximations And Series

$\begin{align*}&\text{Approximate }f(2.9) \\&\text{Where }f(x)=-3^{(4x)}cos(9x)\\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align*}$

Possible Answers:

1228.29

245.66

736.97

2210.92

Correct answer:

245.66

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-3^{(4x)}cos(9x)\text{ at }x=2.9\\&\text{This expansion becomes:}\\&f(2.9)\approx (-3^{(4(0))}cos(9(0)))(2.9)^{0}+(9\cdot 3^{(4(0))}sin(9(0)) - 4\cdot 3^{(4(0))}cos(9(0))ln(3))(2.9)^{1}+\frac{81\cdot 3^{(4(0))}cos(9(0)) + 72\cdot 3^{(4(0))}sin(9(0))ln(3) - 16\cdot 3^{(4(0))}cos(9(0))ln(3)^{2}}{2}(2.9)^{2}\\&f(2.9)\approx245.66\end{align*}$

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Example Question #31 : Taylor Series

$\begin{align*}&\text{For the function: }f(x)=cos(4x)sin(x)\\&\text{Approximate }f(3.1)\\&\text{Using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Possible Answers:

0.52

-9.30

24.80

3.10

Correct answer:

3.10

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }cos(4x)sin(x)\text{ at }x=3.1\\&\text{This expansion becomes:}\\&f(3.1)\approx (cos(4(0))sin((0)))(3.1)^{0}+(cos(4(0))cos((0)) - 4sin(4(0))sin((0)))(3.1)^{1}+\frac{- 17cos(4(0))sin((0)) - 8sin(4(0))cos((0))}{2}(3.1)^{2}\\&f(3.1)\approx3.10\end{align*}$

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Example Question #35 : Taylor Series

$\begin{align*}&\text{Approximate }f(1.8) \\&\text{Where }f(x)=-cos(5x)e^{(-3x)}\\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Possible Answers:

30.32

-15.16

-151.60

3.37

Correct answer:

30.32

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-cos(5x)e^{(-3x)}\text{ at }x=1.8\\&\text{This expansion becomes:}\\&f(1.8)\approx (-cos(5(0))e^{(-3(0))})(1.8)^{0}+(3cos(5(0))e^{(-3(0))} + 5sin(5(0))e^{(-3(0))})(1.8)^{1}+\frac{16cos(5(0))e^{(-3(0))} - 30sin(5(0))e^{(-3(0))}}{2}(1.8)^{2}\\&f(1.8)\approx30.32\end{align*}$

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Example Question #31 : Taylor Series

$\begin{align*}&\text{Approximate }f(3.4) \\&\text{Where }f(x)=-sin(5x)sin(6x)\\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Possible Answers:

173.40

2774.40

-346.80

-1734.00

Correct answer:

-346.80

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-sin(5x)sin(6x)\text{ at }x=3.4\\&\text{This expansion becomes:}\\&f(3.4)\approx (-sin(5(0))sin(6(0)))(3.4)^{0}+(- 5cos(5(0))sin(6(0)) - 6cos(6(0))sin(5(0)))(3.4)^{1}+\frac{61sin(5(0))sin(6(0)) - 60cos(5(0))cos(6(0))}{2}(3.4)^{2}\\&f(3.4)\approx-346.80\end{align*}$

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Example Question #41 : Ap Calculus Bc

$\begin{align*}&\text{Approximate }f(3) \\&\text{Where }f(x)=-cos(5x)e^{(-3x)}\\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align*}$

Possible Answers:

-8.89

80.00

480.00

160.00

Correct answer:

80.00

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-cos(5x)e^{(-3x)}\text{ at }x=3\\&\text{This expansion becomes:}\\&f(3)\approx (-cos(5(0))e^{(-3(0))})(3)^{0}+(3cos(5(0))e^{(-3(0))} + 5sin(5(0))e^{(-3(0))})(3)^{1}+\\&\frac{16cos(5(0))e^{(-3(0))} - 30sin(5(0))e^{(-3(0))}}{2}(3)^{2}\\&f(3)\approx80.00\end{align*}$

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Example Question #41 : Ap Calculus Bc

$\begin{align*}&\text{Approximate }f(2.1) \\&\text{Where }f(x)=-sin(3x)sin(7x)\\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Possible Answers:

-10.29

-92.61

15.43

-370.44

Correct answer:

-92.61

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-sin(3x)sin(7x)\text{ at }x=2.1\\&\text{This expansion becomes:}\\&f(2.1)\approx (-sin(3(0))sin(7(0)))(2.1)^{0}+(- 3cos(3(0))sin(7(0)) - 7cos(7(0))sin(3(0)))(2.1)^{1}+\\&\frac{58sin(3(0))sin(7(0)) - 42cos(3(0))cos(7(0))}{2}(2.1)^{2}\\&f(2.1)\approx-92.61\end{align*}$

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Example Question #41 : Polynomial Approximations And Series

$\begin{align*}&\text{Approximate }f(1.6) \\&\text{Where }f(x)=-cos(3x)cos(7x)\\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align*}$

Possible Answers:

73.24

8.14

-24.41

-366.20

Correct answer:

73.24

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-cos(3x)cos(7x)\text{ at }x=1.6\\&\text{This expansion becomes:}\\&f(1.6)\approx (-cos(3(0))cos(7(0)))(1.6)^{0}+(7cos(3(0))sin(7(0)) + 3cos(7(0))sin(3(0)))(1.6)^{1}+\\&\frac{58cos(3(0))cos(7(0)) - 42sin(3(0))sin(7(0))}{2}(1.6)^{2}\\&f(1.6)\approx73.24\end{align*}$

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Example Question #11 : Maclaurin Series

$\begin{align*}&\text{For the function: }f(x)=-sin(2x)^{2}\\&\text{Approximate }f(1.9)\\&\text{Using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Possible Answers:

2.06

1.60

-3.61

-14.44

Correct answer:

-14.44

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-sin(2x)^{2}\text{ at }x=1.9\\&\text{This expansion becomes:}\\&f(1.9)\approx (-sin(2(0))^{2})(1.9)^{0}+(-4cos(2(0))sin(2(0)))(1.9)^{1}+\\&\frac{8sin(2(0))^{2} - 8cos(2(0))^{2}}{2}(1.9)^{2}\\&f(1.9)\approx-14.44\end{align*}$

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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 : Taylor Series

Example Question #31 : Polynomial Approximations And Series

Example Question #31 : Polynomial Approximations And Series

Example Question #31 : Taylor Series

Example Question #35 : Taylor Series

Example Question #31 : Taylor Series

Example Question #41 : Ap Calculus Bc

Example Question #41 : Ap Calculus Bc

Example Question #41 : Polynomial Approximations And Series

Example Question #11 : Maclaurin Series

All AP Calculus BC Resources

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