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AP Calculus BC : Taylor Series

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Example Questions

Example Question #21 : Taylor Polynomial Approximation

$\begin{align*}&\text{Approximate }f(1) \\&\text{Where }f(x)=- cos(5x) - sin(2x)\\&\text{With a Taylor series centered at }\frac{(5\cdot \pi )}{2}\text{, using }3\text{ terms}\end{align*}$

Possible Answers:

47.98

-335.85

-47.98

7089.24

Correct answer:

-47.98

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }- cos(5x) - sin(2x)\text{ at }x=1\\&\text{This expansion becomes:}\\&f(x)\approx (- cos(5a) - sin(2a))+(5sin(5a) - 2cos(2a))(x-a)^{1}+\frac{25cos(5a) + 4sin(2a)}{2}(x-a)^{2}\\&x=1\\&a=\frac{(5\cdot \pi )}{2}\\&f(1)\approx-47.98\end{align*}$

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Example Question #22 : Taylor Polynomial Approximation

$\begin{align*}&\text{For the function: }f(x)=-cos(4x)^{2}\\&\text{Approximate }f(3)\\&\text{Using a Taylor expansion centered at }6\cdot \pi\text{ to }2\text{ terms}\end{align*}$

Possible Answers:

-9.00

4019.33

-1.00

-0.17

Correct answer:

-1.00

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }-cos(4x)^{2}\text{ at }x=3\\&\text{This expansion becomes:}\\&f(x)\approx (-cos(4a)^{2})+(8cos(4a)sin(4a))(x-a)^{1}\\&x=3\\&a=6\cdot \pi\\&f(3)\approx-1.00\end{align*}$

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Example Question #23 : Taylor Polynomial Approximation

$\begin{align*}&\text{Approximate }f(5) \\&\text{Where }f(x)=sin(6x) - cos(2x)\\&\text{By using a Taylor expansion to }3\text{ terms, centered at }\pi\end{align*}$

Possible Answers:

23.97

-5.24

17.06

-213.00

Correct answer:

17.06

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }sin(6x) - cos(2x)\text{ at }x=5\\&\text{This expansion becomes:}\\&f(x)\approx (sin(6a) - cos(2a))+(6cos(6a) + 2sin(2a))(x-a)^{1}+\frac{4cos(2a) - 36sin(6a)}{2}(x-a)^{2}\\&x=5\\&a=\pi\\&f(5)\approx17.06\end{align*}$

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Example Question #24 : Taylor Polynomial Approximation

$\begin{align*}&\text{Approximate }f(6) \\&\text{Where }f(x)=- cos(9x) - cos(x)\\&\text{With a Taylor series centered at }-\frac{(5\cdot \pi )}{2}\text{, using }3\text{ terms}\end{align*}$

Possible Answers:

-138.54

323377.21

692.70

138.54

Correct answer:

-138.54

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }- cos(9x) - cos(x)\text{ at }x=6\\&\text{This expansion becomes:}\\&f(x)\approx (- cos(9a) - cos(a))+(9sin(9a) + sin(a))(x-a)^{1}+\frac{81cos(9a) + cos(a)}{2}(x-a)^{2}\\&x=6\\&a=-\frac{(5\cdot \pi )}{2}\\&f(6)\approx-138.54\end{align*}$

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Example Question #1 : Taylor And Maclaurin Series

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

Possible Answers:

$f(x)= \frac{x^{5}}{3}+2$

$f(x)=\ln|x+3|$

$f(x)=x+3\sin(x)$

$f(x)= x^{2}+\sqrt{x}+1$

Correct answer:

$f(x)= \frac{x^{5}}{3}+2$

Explanation:

Recall the Maclaurin series formula:

$f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$

$=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

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Example Question #1 : Taylor And Maclaurin Series

Let $f(x)=(1+\frac{x}{2})^{1/3}$

Find the the first three terms of the Taylor Series for centered at .

Possible Answers:

$T_3=1+\frac{x}{3}+4x^2$

$T_{3}=0+0+0$

$T_{2}=0+\frac{x}{6} -\frac{x^2}{18}$

$T_{2}(x)=1+\frac{x}{6}+\frac{x^2}{36}$

$T_{2}(x)=1+\frac{x}{6}-\frac{x^2}{36}$

Correct answer:

$T_{2}(x)=1+\frac{x}{6}-\frac{x^2}{36}$

Explanation:

Using the formula of a binomial series centered at 0:

$(1+x)^k =\sum_{n=0}^{\infty } \binom{k}{n}x^n$ ,

where we replace with $\frac{x}{2}$ and $k=\frac{1}{3}$ , we get:

$\sum_{n=0}^{2}\binom{\frac{1}{3}}{n}\left(\frac{x}{2}\right)^n$ for the first 3 terms.

Then, we find the terms where,

$T_2(x)=\binom{\frac{1}{3}}{0}(\frac{x}{2})^0 + \binom{\frac{1}{3}}{1}(\frac{x}{2})^1+\binom{\frac{1}{3}}{2}(\frac{x}{2})^2=1+\frac{x}{6}- \frac{x^2}{36}$

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Example Question #1 : Taylor And Maclaurin Series

Write the first three terms of the Taylor series for the following function about x=1 :

f(x)=x^2+3x+2

Possible Answers:

5(x-1)+(x-1)^2

2-5(x-1)+(x-1)^2

2+5(x-1)+(x-1)^2

6+5(x-1)+(x-1)^2

Correct answer:

6+5(x-1)+(x-1)^2

Explanation:

The general form for the Taylor series (of a function f(x)) about x=a is the following:

$\sum_{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(x-a)^n$ .

Because we only want the first three terms, we simply plug in a=1, and then n=0, 1, and 2 for the first three terms (starting at n=0).

The hardest part, then, is finding the zeroth, first, and second derivative of the function given:

$f^{0}(1)=f(1)=6$

f'(1)=2(1)+3=5

f''(1)=2

The derivative was found using the following rules:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Then, simply plug in the remaining information and write out the terms:

$\frac{6(x-1)^0}{0!}+\frac{5(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}=6+5(x-1)+(x-1)^2$

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Example Question #1 : Taylor Series

Write out the first four terms of the Taylor series about a=1 for the following function:

f(x)=x^2+2x+1

Possible Answers:

4+4(x-1)+(x-1)^2+0

0+4+4(x-1)+(x-1)^2

4+4(x-1)^2+(x-1)^3+0

0+0+0+0

Correct answer:

4+4(x-1)+(x-1)^2+0

Explanation:

The Taylor series about x=a of any function is given by the following:

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

So, we must find the zeroth, first, second, and third derivatives of the function (for n=0, 1, 2, and 3 which makes the first four terms):

$f^{0}(x)=f(x)$

f'(x)=2x+2

f''(x)=2

f'''(x)=0

The derivatives were found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, evaluated at x=a=1, and plugging in the correct n where appropriate, we get the following:

$\frac{4(x-1)^0}{0!}+\frac{4(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}+\frac{0(x-1)^3}{3!}$

which when simplified is equal to

4+4(x-1)+(x-1)^2+0 .

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Example Question #251 : Series In Calculus

Write out the first three terms of the Taylor series about a=1 for the following function:

f(x)=10x^2+x

Possible Answers:

11+21(x-1)+10(x-1)^2

0+0+0

11+21x+10x^2

0+21(x-1)+10(x-1)^2

Correct answer:

11+21(x-1)+10(x-1)^2

Explanation:

The general formula for the Taylor series about x=a for a given function is

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

We must find the zeroth, first, and second derivative of the function (for n=0, 1, and 2). The zeroth derivative is just the function itself.

f'(x)=20x+1

f''(x)=20

The derivatives were found using the following rule:
$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now, follow the above formula to write out the first three terms:

$\frac{11(x-1)^0}{0!}+\frac{21(x-1)^1}{1!}+\frac{20(x-1)^2}{2!}$

which simplified becomes

11+21(x-1)+10(x-1)^2

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Example Question #21 : Taylor Series

Find the Taylor series expansion of $\frac{1}{1-x^{2}}$ at x=0 .

Possible Answers:

$\frac{1}{1-x^{2}}=\sum_{k=0}^{\infty }(x)^{2k}=1+(x)^{2}+(x)^{4}+...+(x)^{2n}+...\textup{}$

$\frac{1}{1-x^{2}}=\sum_{k=0}^{\infty }(2x)^{k}=1+(x)^{2}+(x)^{4}+...+(x)^{2n}+...\textup{}$

$\frac{1}{1-x^{2}}=\sum_{k=0}^{\infty }(x)^{k}=1+(x)^{2}+(x)^{4}+...+(x)^{2n}+...\textup{}$

$\frac{1}{1-x^{2}}=\sum_{k=0}^{\infty }(x)^{k+2}=1+(x)^{2}+(x)^{4}+...+(x)^{2n}+...\textup{}$

Correct answer:

$\frac{1}{1-x^{2}}=\sum_{k=0}^{\infty }(x)^{2k}=1+(x)^{2}+(x)^{4}+...+(x)^{2n}+...\textup{}$

Explanation:

The Taylor series is defined as

$f(x)=\sum_{k=0}^{\infty }\frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}$ where the expression within the summation is the nth term of the Taylor polynomial.

To find the Taylor series expansion of $\frac{1}{1-x}$ at x=0 , we first need to find an expression for the nth term of the Taylor polynomial.

The nth term of the Taylor polynomial is defined as

$p_{n}x=f(x_{0})+f'(x_{0})(x-x_{0})+\frac{f''(x_{0})}{2!}(x-x_{0})^{2}+\frac{f'''(x_{0})}{3!}(x-x_{0})^{3}+....+\frac{f^{n}(x_{0})}{n!}(x-x_{0})^{n}$

For, $f(x)=\frac{1}{1-x}$ and $x_{0}=0$ .

We need to find terms in the taylor polynomial until we can determine the pattern for the nth term.

$f(0)=\frac{1}{1-0}=1$

$f'(x)=\frac{d}{dx}\left [ \frac{1}{1-x} \right ]=\frac{d}{dx}\left [ \left ( 1-x\right )^{-1} ]=-1( 1-x\right )^{-2}(-1)=( 1-x)^{-2}$

$f'(0)=-1( 1-(0) )^{-2}=1=1!$

$f''(x)=\frac{d}{dx}[( 1-x)^{-2}]=-2( 1-x)^{-3}(-1)=2( 1-x)^{-3}$

$f''(0)=2( 1-0)^{-3}=2=2!$

$f'''(x)=\frac{d}{dx}[2( 1-x)^{-3}]=-6( 1-x)^{-4}(-1)=6( 1-x)^{-4}$

$f'''(0)=6( 1-0)^{-4}=6=3!$

Substituting these values into the taylor polynomial we get

$p_{n}x=1+1(x)+\frac{2!}{2!}(x)^{2}+\frac{3!}{3!}(x)^{3}+....+\frac{f^{n}(0)}{n!}(x)^{n}$

$p_{n}x=1+(x)+(x)^{2}+(x)^{3}+....+\frac{f^{n}(0)}{n!}(x)^{n}$

The Taylor polynomial simplifies to

$p_{n}x=1+(x)+(x)^{2}+(x)^{3}+....+(x)^{n}$

Now that we know the nth term of the Taylor polynomial, we can find the Taylor series.

The Taylor series is defined as

$f(x)=\sum_{k=0}^{\infty }\frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}$ where the expression within the summation is the nth term of the Taylor polynomial.

For this problem, the taylor series is

$\frac{1}{1-x}=\sum_{k=0}^{\infty }(x)^{k}=1+x+(x)^{2}+(x)^{3}+...+(x)^{n}+...$

Use the above to calculate the taylor series expansion of $\frac{1}{1-x^{2}}$ at x=0 ,

$\frac{1}{1-x^{2}}=\sum_{k=0}^{\infty }(x^{2})^{k}=\sum_{k=0}^{\infty }(x)^{2k}=1+(x)^{2}+(x)^{4}+...+(x)^{2n}+...\textup{}$

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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 #21 : Taylor Polynomial Approximation

Example Question #22 : Taylor Polynomial Approximation

Example Question #23 : Taylor Polynomial Approximation

Example Question #24 : Taylor Polynomial Approximation

Example Question #1 : Taylor And Maclaurin Series

Example Question #1 : Taylor And Maclaurin Series

Example Question #1 : Taylor And Maclaurin Series

Example Question #1 : Taylor Series

Example Question #251 : Series In Calculus

Example Question #21 : Taylor Series

All AP Calculus BC Resources

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