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AP Calculus BC : Maclaurin Series

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Example Questions

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:

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

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

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

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

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 #2 : 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:

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

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

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

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

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

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 #4 : 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:

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

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

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

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 #5 : Taylor And Maclaurin 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+0+0+0

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

4+4(x-1)^2+(x-1)^3+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 #1 : Maclaurin Series

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 #271 : Series In Calculus

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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Example Question #1 : Maclaurin Series

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

Possible Answers:

0.3827

0.3928

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 #2 : Maclaurin Series

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

Possible Answers:

2.18147

1.78

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:

1.461345

1.462285

1.460782

1.469385

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 #31 : Ap Calculus Bc

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

Possible Answers:

-0.28

-0.07

2.52

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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AP Calculus BC : Maclaurin Series

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Taylor And Maclaurin Series

Example Question #2 : Taylor And Maclaurin Series

Example Question #4 : Taylor And Maclaurin Series

Example Question #5 : Taylor And Maclaurin Series

Example Question #1 : Maclaurin Series

Example Question #271 : Series In Calculus

Example Question #1 : Maclaurin Series

Example Question #2 : Maclaurin Series

Example Question #3 : Maclaurin Series

Example Question #31 : Ap Calculus Bc

All AP Calculus BC Resources

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