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AP Calculus BC : Taylor Series

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Example Questions

Example Question #1 : Maclaurin Series

Use the first 3 terms of the Maclaurin series to approximate $\sin (\frac{\pi}{8})$

Possible Answers:

0.3928

0.3827

Correct answer:

0.3827

Explanation:

The Maclaurin series for sine is $\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!}x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5! } - ...$

Plugging in $\frac{\pi }{ 8 }$ for x gives:

.3927 - .0101 + .0001 = .3827

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Example Question #1 : Maclaurin Series

Use the first 4 terms of the Maclaurin series to approximate $e^{0.78}$

Possible Answers:

1.78

2.18147

1.3042

2.16329

Correct answer:

2.16329

Explanation:

The Maclaurin series for e^x is $\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2 }{2! } + \frac{x^3}{3!} + ...$

Plugging in 0.78 for x gives:

1 + .78 + .3042 + .079092 = 2.163292

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Example Question #3 : Maclaurin Series

Use the first four terms of the Maclaurin series to approximate $e^{0.38}$ to 6 decimal places.

Possible Answers:

1.460782

1.469385

1.461345

1.462285

Correct answer:

1.461345

Explanation:

The Maclaurin series for e^x is $\sum_{n=0}^{\infty } \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

Plugging in x = 0.38 gives:

$1 + 0.38 + .0722 + .09145 \overline{3} = 1.461345$

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Example Question #2 : Maclaurin Series

$\begin{align*}&\text{Approximate }f(1.2) \\&\text{Where }f(x)=-cos(x)\\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Possible Answers:

2.52

-0.28

-0.07

1.40

Correct answer:

-0.28

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Like the Taylor series, it is used to approximate an unknown}\\&\text{value of a function at a given point, if values of the function}\\&\text{and its derivatives are known at another point. In the case}\\&\text{of the Maclaurin series, this latter point has the value of}\\&\text{zero (note that for many functions, such as the trigonometric}\\&\text{functions, f(0) is readily known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-cos(x)\text{ at }x=1.2\\&\text{This expansion becomes:}\\&f(1.2)\approx \frac{-cos((0))}{1}(1.2)^{0}+\frac{sin((0))}{1}(1.2)^{1}+\frac{cos((0))}{2}(1.2)^{2}\\&f(1.2)\approx-0.28\end{align*}$

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Example Question #11 : Maclaurin 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:

-4228.63

-4832.72

151.02

604.09

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 #12 : Maclaurin 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:

75.88

10.84

-151.75

-607.00

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 #13 : Maclaurin 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:

736.97

245.66

2210.92

1228.29

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 #14 : Maclaurin 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

24.80

3.10

-9.30

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 #15 : Maclaurin 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

3.37

-151.60

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 #16 : Maclaurin 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:

2774.40

-346.80

173.40

-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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AP Calculus BC : Taylor Series

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Maclaurin Series

Example Question #1 : Maclaurin Series

Example Question #3 : Maclaurin Series

Example Question #2 : Maclaurin Series

Example Question #11 : Maclaurin Series

Example Question #12 : Maclaurin Series

Example Question #13 : Maclaurin Series

Example Question #14 : Maclaurin Series

Example Question #15 : Maclaurin Series

Example Question #16 : Maclaurin Series

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