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Calculus 2 : Series in Calculus

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Example Questions

Example Question #11 : Maclaurin Series

Write out the first three terms for the Maclaurin series of the following function:

$f(x)=10\cos(x)$

Possible Answers:

0+10x+0

10+0-5x^2

0+0+0

10+0+5x^2

Correct answer:

10+0-5x^2

Explanation:

The Maclaurin series for any function is simply the Taylor series of the function about a=0:

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

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

$f'(x)=-10\sin(x)$

$f''(x)=-10\cos(x)$

The derivatives were found using the following rules:

$\frac{\mathrm{d}}{\mathrm{d} x}\sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-\sin(x)$

Now, we just use the formula, with x=a=0 , to write out the first three terms of the series (n=0, 1, and 2):

$\frac{10(1)(x)^0}{0!}-\frac{10(0)(x)^1}{1!}-\frac{10(1)(x)^2}{2!}=10+0+5x^2$

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Example Question #312 : Series In Calculus

Write the first two terms Maclaurin series of the following function:

$f(x)=x\cos(x)$

Possible Answers:

0+0

0-x

x+1

0+x

Correct answer:

0+x

Explanation:

The Maclaurin series for a function is simply the Taylor series for the function about a=0:

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

We must find the first two terms of the series, corresponding to n=0 and n=1. We need the zeroth and first derivative of the function, the zeroth derivative being the the function itself:

$f'(x)=\cos(x)-x\sin(x)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

Now, use the formula above to write out the first two terms:

$\frac{0(x)^0}{0!}+\frac{1(x)^1}{1!}=0+x$

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Example Question #3091 : Calculus Ii

Write out the first five terms of the Maclaurin series for the following function:

$f(x)=\cos(x)$

Possible Answers:

0+0+0+0+0

$0+x+0+\frac{x^3}{6}+0$

$1+0-\frac{x^2}{2}+0+\frac{x^4}{24}$

$1+0+\frac{x^2}{2}+0+\frac{x^4}{4}$

Correct answer:

$1+0-\frac{x^2}{2}+0+\frac{x^4}{24}$

Explanation:

The Maclaurin series for any function is simply the Taylor series for the function about a=0:

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

First, we must find the zeroth through fourth derivative of the function. The zeroth derivative is simply the function itself.

$f'(x)=-\sin(x)$

$f''(x)=-\cos(x)$

$f'''(x)=\sin(x)$

$f^4(x)=\cos(x)$

The following rules were used for the derivatives:

$\frac{d}{dx}\cos(x)=-\sin(x)$ , $\frac{d}{dx}\sin(x)=-\cos(x)$

Next, we simply evaluate all of the derivatives at x=a=0 and then write out all of the terms:

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

which simplified is equal to

$1+0-\frac{x^2}{2}+0+\frac{x^4}{24}$

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Example Question #314 : Series In Calculus

Write out the first four terms of the Maclaurin series for the following function:

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

Possible Answers:

0+x+2x^2+0

1+x+2x^2+0

0+0+0+0

1+x+4x^2+0

Correct answer:

1+x+2x^2+0

Explanation:

The Maclaurin series is the Taylor series for a function about a=0:

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

We need to find the zeroth, first, second, and third derivative of the function (n=0, 1, 2, and 3). The zeroth derivative is simply the function itself.

f'(x)=4x+1

f''(x)=4

f'''(x)=0

The derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, use the above formula, with x=a=0 to write out the first four terms:

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

which simplified becomes

1+x+2x^2+0

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Example Question #315 : Series In Calculus

Write the first two terms of the Maclaurin series for the following function:

$f(x)=2x^2-\cos(x)$

Possible Answers:

1+0

-1+0

0+1

0+0

Correct answer:

-1+0

Explanation:

The Maclaurin series for any function is simply the Taylor series for the function about a=0:

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

For the first two terms (n=0, 1) we must find the zeroth and first derivative of the function. The zeroth derivative is just the function itself.

$f'(x)=4x+\sin(x)$

Now, using the above formula, write out the first two terms:

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

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Example Question #61 : Taylor And Maclaurin Series

Find the Maclaurin Series of the function

$f(x)=sin\,x$

up to the fifth degree.

Possible Answers:

$x+\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!}+\frac{x^5}{5!}$

DNE

$x-\frac{x^3}{3!}+\frac{x^5}{5!}$

$1-\frac{x^2}{2!}+\frac{x^4}{4!}$

Correct answer:

$x-\frac{x^3}{3!}+\frac{x^5}{5!}$

Explanation:

The formula for an i-th degree Maclaurin Polynomial is

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

For the fifth degree polynomial, we must evaluate the function up to its fifth derivative.

$\begin{matrix} f(0)=sin(0)=0\\ f'(0)=cos(0)=1\\ f''(0)=-sin(0)=0\\ f'''(0)=-cos(0)=-1\\ f^{(4)}=sin(0)=0 \\ \end{matrix}$

$f^{(5)}(0)=cos(0)=1$

The summation becomes

$\sum_{n=0}^{5}\frac{f^{(n)}(0)}{n!}x^n=\frac{f(0)}{0!}x^0+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4+\frac{f^{(5)}(0)}{5!}x^5$

And substituting for the values of the function and the first five derivative values, we get the Maclaurin Polynomial

$f(x)=sin(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}$

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Example Question #316 : Series In Calculus

Given the Maclaurin Series for the following function, write the summation of the series.

y(x)=e^x+sin(x)

Possible Answers:

$\sum_{n=0}^{\infty}x^n$

$\sum_{n=0}^{\infty}x^n\left(\frac{1}{n!}+\frac{(-1)^n*x^{n}}{(1+n)!}\right)$

$\sum_{n=0}^{\infty}x^n\left(\frac{1}{n!}+\frac{(-1)^n*x^{1+n}}{(1+2n)!}\right)$

$\sum_{n=0}^{\infty}x^n\left(\frac{1}{n!}+\frac{(-1)^n}{(1+2n)!}\right)$

Correct answer:

$\sum_{n=0}^{\infty}x^n\left(\frac{1}{n!}+\frac{(-1)^n*x^{1+n}}{(1+2n)!}\right)$

Explanation:

The Maclaurin series for e^x is given by

$e^x= \sum_{n=0}^{\infty }\frac{x^n}{n!}= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+...$

The Maclaurin Series for sin(x) is given by

$sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(x)^{2n+1}=x-\frac{x^3}{6}+\frac{x^5}{120}+...$

Since the question asked

e^x+sin(x)

To write this sum, we can take out x^n as a common factor for both series:

$\sum_{n=0}^{\infty}x^n\left(\frac{1}{n!}+\frac{(-1)^n*x^{1+n}}{(1+2n)!}\right)$

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Example Question #3101 : Calculus Ii

Find the radius of convergence for the power series $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ .

Possible Answers:

$\infty$

$\sqrt2$

Correct answer:

$\infty$

Explanation:

You might already recognize this as the power series representation for e^x . Since e^x is well defined for all $x \in \mathbb{R}$ , the radius of convergence is $\infty$ .

If we want to find the radius of covergence using convergence tests, we can use the Ratio Test here. We have-

$\lim_{k\to\infty} \left |\frac{a_{k+1}}{a_k} \right | = \lim_{k\to\infty} \left | \frac{x^{k+1}}{(k+1)!} \frac{k!}{x^k} \right | =\lim_{k\to\infty} \left | \frac{x}{k+1}\right | =$

$\lim_{k\to\infty} \frac{\left | x \right |}{k+1} = \left | x \right | \lim_{k\to\infty} \frac{1}{k+1} = \left | x \right | \cdot 0 = 0$

Since this limit equals regardless of the value of , and the Ratio Test indicates absolute convergence of a series when the above limit is less than , $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ converges for all $x \in \mathbb{R}$ . Hence the radius of convergence is $\infty$ .

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Example Question #3102 : Calculus Ii

One useful way of determining convergence for power series is via Asymptotic Comparison.

Which of the following functions grows fastest?

Possible Answers:

$9,999,999,999,999,999,999^{x}$

$x^{x}$

$x^{1,000,000,000,000}$

Correct answer:

$x^{x}$

Explanation:

The question does not ask which function is the largest at any given point, it asks which grows fastest. For this question we need to look at all of the terms and determine which function has the dominant term. In this case x^x is the dominate term therefore, it will grow the fastest.

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Example Question #3102 : Calculus Ii

Express cos(x^2) as a power series.

Possible Answers:

$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^{6}}{6!}+...$

$1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$

$1-\frac{x^4}{2!}+\frac{x^6}{4!}-\frac{x^8}{6!}+...$

$1-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+...$

$1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...$

Correct answer:

$1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...$

Explanation:

Write the correct definition of cosine as a power series.

$cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$

Replace with the term x^2 .

$cos(x^2)=1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...$

The correct answer is:

$1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...$

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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Maclaurin Series

Example Question #312 : Series In Calculus

Example Question #3091 : Calculus Ii

Example Question #314 : Series In Calculus

Example Question #315 : Series In Calculus

Example Question #61 : Taylor And Maclaurin Series

Example Question #316 : Series In Calculus

Example Question #3101 : Calculus Ii

Example Question #3102 : Calculus Ii

Example Question #3102 : Calculus Ii

All Calculus 2 Resources

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