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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #5 : Series And Functions

A worm crawls up a wall during the day and slides down slowly during the night. The first day the worm crawls one meter up the wall. The first night the worm slides down a third of a meter. The second day the worm regains one third of the lost progress and slides down one third of that distance regained on the second night. This pattern of motion continues...

Which of the following is a correct expression for the distance the worm has travelled after the second night?

Possible Answers:

$1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}$

$1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}$

$1-\frac{2}{3}+\frac{1}{3}-\frac{1}{9}$

$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}$

Correct answer:

$1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}$

Explanation:

This is just a finite sum of the distance the worm travelled and fell each day.

The first day the worm traveled a postive one meter.

Over the night he lost one third of a meter. This can be translated into math terms as $-\frac{1}{3}$ .

He continued this ratio progression each day.

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Example Question #2 : Series In Calculus

How high off the ground will the worm eventually end up if he keeps at it forever?

Possible Answers:

$\frac{4}{3}m$

$\frac{3}{4}m$

Correct answer:

$\frac{3}{4}m$

Explanation:

This is a geometric series with an a=1 and $r=-\frac{1}{3}$ so will converge to:

$\frac{a}{1-r}=\frac{1}{1--\frac{1}{3}}=\frac{1}{1+\frac{1}{3}}=\frac{1}{\frac{4}{3}}=\frac{3}{4}$

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Example Question #2 : Series In Calculus

Which is the correct Maclaurin series representation for $f(x)=e^{2x}$ ?

Possible Answers:

$f(x)=1-\frac{4x^2}{2!}+\frac{16x^4}{4!}+...$

$f(x)=2x-\frac{8x^3}{3!}+\frac{32x^5}{5!}+...$

$f(x)=1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+...$

f(x)=1+2x+4x^2+8x^3+...

Correct answer:

$f(x)=1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+...$

Explanation:

The general form for the Maclaurin series for e^x is

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

To find the series representation for $f(x)=e^{2x}$ , simply substitute in place of in the series for e^x .

$e^{2x}=1+2x+\frac{(2x)^2}{2!}+\frac{(2x)^3}{3!}+...$

$=1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+...$

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Example Question #11 : Series In Calculus

What do we mean when we say an infinite series $\sum_{n=0}^{\infty} a_n$ converges?

Possible Answers:

None of the other choices

The sequence partial sums of the sequence s_n = a_n converges as $n \to \infty$ .

The sequence partial sums of the sequence a_n , also denoted a_n converges as $n \to \infty$ .

The sequence partial sums of the sequence a_n , denoted s_n = a_0 +a_1 +a_2 + ... + a_n converges as $n \to \infty$ .

The sequence a_n converges as $n \to \infty$ .

Correct answer:

The sequence partial sums of the sequence a_n , denoted s_n = a_0 +a_1 +a_2 + ... + a_n converges as $n \to \infty$ .

Explanation:

This is the definition of a convergent infinite series.

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Example Question #2791 : Calculus Ii

What is the sum of the following geometric series:

$2, 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}+...?$

Possible Answers:

Cannot be determined.

Correct answer:

Explanation:

Since this is a geometric series with a rate between and , we can use the following equation to find the sum:

$S_{\infty}=\frac{a_{1}}{1-r}$ , where $a_{1}$ is the starting number in the sequence, and is the common divisor between successive terms in the sequence. In this sequence, to go from one number to the next, we multiply by $\frac{1}{2}.$ Now, we plug everything into the equation:

$S_{\infty}=\frac{a_{1}}{1-r}=\frac{2}{1-0.5}=\frac{2}{0.5}=4.$

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Example Question #13 : Introduction To Series In Calculus

Find the infinite sum of the following geometric series:

-240, 60, 15, +...

Possible Answers:

Cannot be determined

192

Correct answer:

Explanation:

Since this is a geometric series with a rate between and , we can use the following equation to find the sum:

$S_{\infty}=\frac{a_{1}}{1-r}=\frac{-240}{1--0.25}=\frac{-240}{1.25}=-192.$

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Example Question #2791 : Calculus Ii

Find the infinite sum of the following series:

80, 40, 20, 10, 5+...

Possible Answers:

100

125

160

120

Correct answer:

160

Explanation:

For the sum of an infinite series, we have the following formula:

$S_{\infty}=\frac{a_{1}}{1-r}$ , where $a_{1}$ is the first term in the series and is the rate at which our series is changing between consecutive numbers in the series. Plugging all of the relevant information for this series, we get:

$S_{\infty}=\frac{80}{1-0.5}=160.$

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Example Question #13 : Series And Functions

Find the infinite sum of the following series:

-500,100, -20, 4+...

Possible Answers:

$\frac{-1250}{3}$

500

Correct answer:

$\frac{-1250}{3}$

Explanation:

For the sum of an infinite series, we have the following formula:

$S_{\infty}=\frac{-500}{1--0.2}=\frac{-1250}{3}=-416.666.$

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Example Question #16 : Series And Functions

A) Find a power series representation of the function,

$f(x)=\frac{4x}{x+2}$

B) Determine the power series radius of convergence.

Possible Answers:

A) Power Series for

$f(x)=4x\sum_{n=1}^{\infty}\left(\frac{x^{n+1}}{2^{n+1}} \right )$

B) Radius of Convergence

A) Power Series for ,

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

B) Radius of Convergence

A) Power Series for ,

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

B) Radius of Convergence

A) Power Series for ,

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

B) Radius of Convergence

A) Power Series for

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

B) Radius of Convergence

Correct answer:

A) Power Series for

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

B) Radius of Convergence

Explanation:

$f(x)=\frac{4x}{x+2}$

This function can be easily written as a power series using the formula for a convergent geometric series.

____________________________________________________________

$\sum_{n=1}^{\infty}ar^n =\frac{a}{1-r}$

For any $\left | r\right |<1$

____________________________________________________________

First let's make some modifications to the function so we can compare it to the form of a convergent geometric series:

$f(x)=\frac{4x}{x+2}=x\left(\frac{4}{2-(-x)} \right )=x\left(\frac{2}{1-\left(-\frac{x}{2} \right )} \right )$

Notice if we take $r =- \frac{x}{2}$ and a =2 we can write f(x) in the form,

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

We can find the radius of convergence by applying the condition $\left | r\right |<1$ .

_____________________________________________________________

Finding Radius of Convergence.

$\left | \: -\frac{x}{2}\right |<1$

Case 1

$- \frac{x}{2}<1\: \: \: \: \: \: \Rightarrow \: \: \: \: \: \: \frac{x}{2}>-1\: \: \: \: \: \: \:\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, x>-2$

Case 2

$-\frac{x}{2}>-1\: \: \, \: \: \Rightarrow\, \, \, \, \, \, \, \, \frac{x}{2}<1\, \, \, \, \, \, \Rightarrow\, \, \, \, \, \, \, \, \, x<2$

Combing both cases gives the interval of convergence,

-2<x<2

Therefore the radius of convergence is .

____________________________________________________________

We can continue simplifying our most recent expression of f(x) .

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

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

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Example Question #2791 : Calculus Ii

Does the following series converge or diverge:

$\sum_{n=0}^{\infty }\frac{2x^2-x}{3x^2-7}$

Possible Answers:

Converge

Diverge

Cannot be determined with the given information.

Correct answer:

Diverge

Explanation:

To test if this series diverges, before using a higher test, we may use the test for divergence.

The test for divergence informs that if the sequence does not approach 0 as n approaches infinity then the series diverges (NOTE: This only shows divergence, the converse is not true, that is, the test for divergence cannot be used to show convergence.).

We note that as

$\lim_{x\rightarrow \infty }\frac{2x^2-x}{3x^2-7}=\frac{2}{3}$ ,

this is derived from the fact that to find the limit as x approaches infinity of a function, one must first find the horizontal asymptote. Since this function is a rational expression with the highest power in both the numerator and denominator, the horizontal asymptote is equal to the quotient of the leading coefficients of both the numerator and denominator, which in this case is 2/3.

Since the limit as x tends to infinity of this series is a nonzero value, we may conclude that the series diverges by the Test for Divergence.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #5 : Series And Functions

Example Question #2 : Series In Calculus

Example Question #2 : Series In Calculus

Example Question #11 : Series In Calculus

Example Question #2791 : Calculus Ii

Example Question #13 : Introduction To Series In Calculus

Example Question #2791 : Calculus Ii

Example Question #13 : Series And Functions

Example Question #16 : Series And Functions

Example Question #2791 : Calculus Ii

All Calculus 2 Resources

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