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Calculus 2 : Calculus II

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Example Questions

Example Question #3081 : Calculus Ii

Give the Maclaurin series for the function

$f(x) = \frac{1}{6^{x}}$

up to the third term.

Possible Answers:

$f (x) = 1 -(\ln 6 ) x +\left [ \frac{( \ln 6)^{2}}{2} \right ]x^{2} -...$

$f (x) = 1 - \left [\frac{(\ln 6 ) }{6} x \right ]+ \left [ \frac{( \ln 6)^{2}}{72} \right ] x^{2} -...$

$f (x) = 1 + \left [ \frac{( \ln 6)^{2}}{2} x^{2} \right ]+ \left [ \frac{( \ln 6)^{4}}{12} \right ] x^{4} +...$

$f (x) = 1 + \frac{( \ln 6)^{2}}{72} x^{2} + \frac{( \ln 6)^{4}}{15,552} x^{4} +...$

$f (x) = (\ln 6 ) x -\left [ \frac{( \ln 6)^{2}}{2} \right ]x^{2}+\left [ \frac{( \ln 6)^{3}}{6 \right ]} x^{3} -...$

Correct answer:

$f (x) = 1 -(\ln 6 ) x +\left [ \frac{( \ln 6)^{2}}{2} \right ]x^{2} -...$

Explanation:

Rewrite this function as $f(x) = \frac{1}{6^{x}} =\left ( \frac{1}{6 } \right )^{x} =\left ( e^{\ln\frac{1}{6} } \right \) ^{x}=e ^{-x\ln 6}$ .

The Maclaurin series for $e^{x}$ , taken to the third term, is $e ^{x} = 1 + x + \frac{x^{2}}{2}+...$ .

Substitute $-x \ln 6$ for :

$f(x) =e ^{-x\ln 6}$

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

$= 1 -(\ln 6 ) x +\left [ \frac{( \ln 6)^{2}}{2} \right ]x^{2} +...$

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Example Question #3082 : Calculus Ii

Give the polar form of the equation of a circle with center at (5, -4) and radius .

Possible Answers:

$r ^{2} - 8 r \sin \theta + 10 r \cos \theta = 215$

$r ^{2} - 8 r \cos \theta + 10 r \sin \theta = 2 5$

$r ^{2} - 10 r \sin \theta + 8 r \cos \theta = 2 5$

$r ^{2} - 10 r \cos \theta + 8 r \sin \theta = 215$

$r ^{2} - 10 r \sin \theta + 8 r \cos \theta = 215$

Correct answer:

$r ^{2} - 10 r \cos \theta + 8 r \sin \theta = 215$

Explanation:

This circle will have equation

$(x-5)^{2} + (y+4) ^{2} = 16^{2}$ .

Rewrite this as follows:

$x ^{2} - 10x + 25 + y ^{2} + 8y + 16 = 256$

$x ^{2} - 10x + 25 + y ^{2} + 8y + 16 - 41 = 256- 41$

$x ^{2}+ y ^{2} - 10x + 8y = 215$

$r ^{2} - 10 r \cos \theta + 8 r \sin \theta = 215$

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Example Question #1 : Maclaurin Series

Suppose that $\small f(x)=\cos x^3$ . Calculate $\small f^{(48)}(0)$ .

Possible Answers:

$\small \frac{48!}{12!}$

$\small 1$

$\small \frac{48!}{16!}$

Correct answer:

$\small \frac{48!}{16!}$

Explanation:

Let's find the power series of $\small \cos x^3$ centered at $\small x=0$ to find $\small f^{(48)}(0)$ . We have

$\small \small \small \small \cos x^3= \sum_{n=0}^\infty \frac{(-1)^n(x^3)^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{(-1)^nx^{6n}}{(2n)!}=1-\frac{x^6}{2!}+\frac{x^{12}}{4!}-...$

This series is much easier to differentiate than the expression $\small \cos x^3$ . We must look at term $\small x^{48}$ , which is the only constant term left after differentiating 48 times. This is the only important term, because when we plug in $\small x=0$ , all of the non-constant terms are zero. So we must have

$\small \small (\cos x^3)^{(48)}|_{x=0}= \left( \frac{x^{48}}{16!}\right )^{(48)}|_{x=0}=\frac{48!}{16!}$

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Example Question #1 : Maclaurin Series

What is the value of the following infinite series?

$\small \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{3^{2n+1}(2n+1)!}$

Possible Answers:

$\small \frac{\sqrt{3}}{2}$

$\small \arctan{\pi/3}$

$\small \frac{1}{2}$

$\small \pi/4$

Correct answer:

$\small \frac{\sqrt{3}}{2}$

Explanation:

We can recognize this series as $\small \sin(x)$ since the power series is

$\small \sin x=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}$

with the value $\small \frac{\pi}{3}$ plugged into $\small x$ since

$\small \small \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{3^{2n+1}(2n+1)!}=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(\frac{\pi}{3} \right )^{2n+1}$ .

So then we have

$\small \small \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{3^{2n+1}(2n+1)!}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$ .

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Example Question #3 : Maclaurin Series

What is the value of the following infinite series?

$\small \small \small \sum_{n=0}^\infty \frac{1+(-1)^n}{n!}$

Possible Answers:

$\small \frac{1}{2}(e+e^{-1})$

$\small e+e^{-1}$

The infinite series diverges.

$\small e^2+e^{-2}$

$\small e+e^{1/2}$

Correct answer:

$\small e+e^{-1}$

Explanation:

The infinite series can be computed easily by splitting up the two components of the numerator:

$\small \small \small \small \sum_{n=0}^\infty \frac{1+(-1)^n}{n!}=\sum_{n=0}^\infty \frac{1}{n!}+\sum_{n=0}^\infty \frac{(-1)^n}{n!}$

Now we recall the MacLaurin series for the exponential function $\small e^x$ , which is

$\small e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$

which converges for all $\small x$ . We can see that the two infinite series are $\small e^x$ with $\small x=1,-1$ , respectively. So we have

$\small \small \small \small \small \sum_{n=0}^\infty \frac{1+(-1)^n}{n!}=\sum_{n=0}^\infty \frac{1}{n!}+\sum_{n=0}^\infty \frac{(-1)^n}{n!}=e^1+e^{-1}$

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Example Question #4 : Maclaurin Series

Find the value of the infinite series.

$\small \sum_{n=0}^{\infty} \frac{\pi^{4n+3}(-1)^n}{(2n+1)}$

Possible Answers:

$\small \pi \sin \pi^2$

$\small \pi \cos \pi^4$

The series does not converge.

$\small \cos \pi^3$

$\small -\pi\cos \pi^2$

Correct answer:

$\small \pi \sin \pi^2$

Explanation:

We can evaluate the series

$\small \sum_{n=0}^{\infty} \frac{\pi^{4n+3}(-1)^n}{(2n+1)}$

by recognizing it as a power series of a known function with a value plugged in for $\small x$ . In particular, it looks similar to $\small \sin x$ :

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

After manipulating the series, we get

$\small \small \sum_{n=0}^{\infty} \frac{\pi^{4n+3}(-1)^n}{(2n+1)}=\pi\sum_{n=0}^{\infty} \frac{\pi^{4n+2}(-1)^n}{(2n+1)}=\pi \sum_{n=0}^{\infty} \frac{(\pi^2)^{2n+1}(-1)^n}{(2n+1)}$ .

Now it suffices to evalute $\small \sum_{n=0}^{\infty} \frac{(\pi^2)^{2n+1}(-1)^n}{(2n+1)}$ , which is $\small \sin \pi^2$ .

So the infinite series has value

$\small \small \small \sum_{n=0}^{\infty} \frac{\pi^{4n+3}(-1)^n}{(2n+1)}=\pi \sum_{n=0}^{\infty} \frac{(\pi^2)^{2n+1}(-1)^n}{(2n+1)}=\pi \sin \pi^2$ .

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Example Question #5 : Maclaurin Series

Find the value of the following infinite series:

$\small \sum_{n=0}^\infty \frac{\pi^{2n+1}(-1)^n}{6^{2n+1}(2n+1)!}$

Possible Answers:

$\small -\frac{1}{2}$

$\small \frac{\sqrt{3}}{2}$

$\small -\frac{\sqrt{3}}{2}$

$\small \frac{1}{2}$

$\small \frac{\sqrt{2}}{2}$

Correct answer:

$\small \frac{1}{2}$

Explanation:

After doing the following manipulation:

$\small \small \sum_{n=0}^\infty \frac{\pi^{2n+1}(-1)^n}{6^{2n+1}(2n+1)!}=\sum_{n=0}^\infty \left(\frac{\pi}{6} \right )^{2n+1}\frac{(-1)^n}{(2n+1)!}$

We can see that this is the power series

$\small \sin x=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$ with $\small \small x=\frac{\pi}{6}$ plugged in.

So we have

$\small \small \small \sum_{n=0}^\infty \frac{\pi^{2n+1}(-1)^n}{6^{2n+1}(2n+1)!}=\sin \frac{\pi}{6}=\frac{1}{2}$

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Example Question #6 : Maclaurin Series

Find the value of the following series.

$\small \sum_{k=0}^\infty\frac{2^k+(-1)^k}{k!}$

Possible Answers:

$\small e^{-2}+e^{-1}$

Divergent.

$\small e^2+e^{-1}$

$\small e^{1/2}-e^{-1}$

$\small e^2-e^{-1}$

Correct answer:

$\small e^2+e^{-1}$

Explanation:

We can split up the sum to get

$\small \small \sum_{k=0}^\infty\frac{2^k+(-1)^k}{k!}=\sum_{k=0}^\infty\frac{2^k}{k!}+\sum_{k=0}^\infty\frac{(-1)^k}{k!}$ .

We know that the power series for $\small e^x$ is

$\small \sum_{n=0}^\infty \frac{x^n}{n!}$

and that each sum,

$\small \sum_{k=0}^\infty\frac{2^k}{k!}$

and

$\small \sum_{k=0}^\infty\frac{(-1)^k}{k!}$

are simply $\small e^{x}$ with $\small x=2,-1$ plugged in, respectively.

Thus,

$\small \small \small \sum_{k=0}^\infty\frac{2^k+(-1)^k}{k!}=\sum_{k=0}^\infty\frac{2^k}{k!}+\sum_{k=0}^\infty\frac{(-1)^k}{k!}=e^2+e^{-1}$ .

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Example Question #7 : Maclaurin Series

Find the value of the infinite series.

$\small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}$

Possible Answers:

Infinite series does not converge.

$\small \sqrt{5}$

$\small 4$

$\small 5$

$\small \ln 5$

Correct answer:

$\small 4$

Explanation:

The series

$\small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}$ looks similar to the series for $\small e^x$ , which is $\small e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$

but the series we want to simplify starts at $\small n=1$ , so we can fix this by adding a $\small 1$ and subtracting a $\small 1$ , to leave the value unchanged, i.e.,

$\small \small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}=-1+1+\sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}=-1+\sum_{n=0}^\infty \frac{(\ln 5)^n}{n!}$ .

So now we have $\small e^x$ with $\small x=\ln 5$ , which gives us $\small e^{\ln 5}=5$ .

So then we have:

$\small \small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}=-1+\sum_{n=0}^\infty \frac{(\ln 5)^n}{n!}=-1+5=4$

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Example Question #1 : Maclaurin Series

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

f(x)=x^3+2x^2+1

Possible Answers:

0-1

1+0

0+0

0+1

Correct answer:

1+0

Explanation:

The Maclaurin series of a function is simply the Taylor series of a function, but about x=0 (so a=0 in the formula):

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

To write out the first two terms (n=0 and n=1), we must find the first derivative of the function (because the zeroth derivative is the function itself):

f'(x)=3x^2+4x

The derivative was found using the following rule:

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

Next, use the general form, plugging in n=0 for the first term and n=1 for the second term:

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

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #3081 : Calculus Ii

Example Question #3082 : Calculus Ii

Example Question #1 : Maclaurin Series

Example Question #1 : Maclaurin Series

Example Question #3 : Maclaurin Series

Example Question #4 : Maclaurin Series

Example Question #5 : Maclaurin Series

Example Question #6 : Maclaurin Series

Example Question #7 : Maclaurin Series

Example Question #1 : Maclaurin Series

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