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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #3071 : Calculus Ii

Find the expression for the Taylor Series of $x^2\sin(x^2)$ .

Possible Answers:

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

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

$\sum_{n=0}^\infty\frac{(-1)^n2^{n+1}x^{n+1}}{(n+1)!}$

$\sum_{n=0}^\infty\frac{(-1)^n2^{n+1}x^{n+3}}{n+1}$

$\sum_{n=0}^\infty\frac{(-1)^nx^{2n+4}}{(n+1)!}$

Correct answer:

$\sum_{n=0}^\infty\frac{(-1)^nx^{2n+4}}{(n+1)!}$

Explanation:

To obtain the Taylor Series for $x^2\sin(x^2)$ , we start with the Taylor Series for $\sin(x) = \sum_{n=0}^\infty \frac{(-1)^nx^{n+1}}{(n+1)!}$ .

$\sin(x^2) = \sum_{n=0}^\infty \frac{(-1)^n(x^2)^{n+1}}{(n+1)!}$ . (Substitute with x^2 on both sides)

$x^2\sin(x^2) = x^2\sum_{n=0}^\infty \frac{(-1)^n(x^2)^{n+1}}{(n+1)!}$ . (Multiply both sides by x^2 ,

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

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Example Question #42 : Taylor Series

Use Taylor Expansion of around x=0 and determine the value of y''

Possible Answers:

$y''=\sum_{2}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

$y''=\sum_{0}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

$y''=\sum_{2}^{\infty} (n)(n-2)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

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

$y''=\sum_{1}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-1}$

Correct answer:

$y''=\sum_{2}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

Explanation:

We can write as a Mclaurin series by saying that:

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

Since y is just a polynomial expression:

$y''=\sum_{2}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

The reason we must bring the lower bound to 2 is because Taylor Series can ONLY be written as a summation of polynomials and no rational components.

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Example Question #291 : Series In Calculus

Give the Maclaurin series for the function

$f(x) = e ^{2x ^{2}}$

up to the third term.

Possible Answers:

$f(x) = 1 + 2x^{2} + 4x^{4}+...$

$f(x) = 2x^{2} - \frac{4x^{6}}{3} + \frac{4x^{10}}{15}-...$

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

$f(x) = 1 + 2x^{2} + 2x^{4}+...$

$f(x) = 2x^{2} - 4x^{4}+ 8x^{6}+...$

Correct answer:

$f(x) = 1 + 2x^{2} + 2x^{4}+...$

Explanation:

The Maclaurin series for $e^{x}$ , taken to the third term, is:

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

Substitute $2x^{2}$ for :

$f(x) = e ^{2x ^{2}}$

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

$e ^{x} = 1 + 2x^{2} + \frac{4x^{4}}{2}+...$

$e ^{x} = 1 + 2x^{2} + 2x^{4} +...$

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Example Question #3072 : Calculus Ii

Give the Maclaurin series for the function

$f(x) = \sqrt{3^x}$

up to the third term.

Possible Answers:

$f(x)= \left ( \frac{\ln 3}{2} \right ) x +\left [\frac{ ( \ln 3 )^{2} }{8} \; \right ] x^{2}+\left [ \frac{ ( \ln 3 )^{3} }{20 \right ]} \; x^{3}...$

$f(x)= 1 + ( \ln 3 ) x +\left [ \frac{ ( \ln 3 )^{2} }{2} \right ] \; x^{2}+...$

$f(x)= 1 + ( \ln 3 ) x +( \ln 3 )^{2} x^{2}+...$

$f(x)= 1 + \left ( \frac{\ln 3}{2} \right ) x +\left [ \frac{ ( \ln 3 )^{2} }{8} \right ] \; x^{2}+...$

$f(x)= ( \ln 3 ) x +( \ln 3 )^{2} x^{2}+( \ln 3 )^{3} x^{3} +...$

Correct answer:

$f(x)= 1 + \left ( \frac{\ln 3}{2} \right ) x +\left [ \frac{ ( \ln 3 )^{2} }{8} \right ] \; x^{2}+...$

Explanation:

Rewrite this function as

$f(x) = \sqrt{3^x} =\left ( \sqrt{3} \right )^x$

$= \left ( 3 ^{\frac{1}{2}} \right )^x = \left ( e^{\ln 3 ^{\frac{1}{2}}} \right )^x = \left ( e^{\frac{1}{2} \ln 3} \right )^x= \left ( e^{\frac{1}{2} \ln 3 \; \cdot x } \right )$

The Maclaurin series for $e^{x}$ , taken to the third term, is:

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

Substitute $\frac{1}{2} \ln 3 \; \cdot x$ for :

$f(x) = \sqrt{3^x} = 1 + \frac{1}{2} \ln 3 \; \cdot x + \frac{(\frac{1}{2} \ln 3 \; \cdot x)^{2}}{2}+...$ $= 1 + \frac{1}{2} \ln 3 \; \cdot x + \frac{\left $\frac{1}{4} \right$( \ln 3 )^{2} x^{2}}{2}+...$

$= 1 + \left ( \frac{\ln 3}{2} \right ) x +\left [ \frac{ ( \ln 3 )^{2} }{8} \right ] \; x^{2}+...$

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Example Question #292 : Series In Calculus

Give the Maclaurin series for the function

$f(x) = \cos \sqrt {2x}$

up to the third term.

Possible Answers:

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

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

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

$x^{2} - \frac{1}{6}x^{4} + \frac{1}{90} x^{6}...$

$x - \frac{1}{6}x^{2} + \frac{1}{90} x^{3}...$

Correct answer:

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

Explanation:

The Maclaurin series for $\cos x$ is

$\cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...$

Substitute $\sqrt{2x}$ for . The series becomes

$\cos \sqrt{2x} = 1 - \frac{\left ( \sqrt{2x} \right )^{2}}{2} + \frac{\left ( \sqrt{2x} \right)^{4}}{24} -...$

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

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

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Example Question #42 : Taylor And Maclaurin Series

Give the Maclaurin series for the function

$f(x) = \cos \left (2x^{2} \right )$

up to the third term.

Possible Answers:

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

$2x^{2} +\frac{4}{3} x^{6} + \frac{4}{15}x^{10} - ...$

$2x^{4} - \frac{ 2 }{3}x^{8} + \frac{4}{45}x^{12} ...$

$2x^{2} -\frac{4}{3} x^{6} + \frac{4}{15}x^{10} - ...$

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

Correct answer:

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

Explanation:

The Maclaurin series for $\cos x$ is

$\cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...$

Substitute $2x^{2}$ for . The series becomes

$\cos \left (2x^{2} \right ) = 1 - \frac{\left (2x^{2} \right )^{2}}{2} + \frac{\left (2x^{2} \right )^{4}}{24} -...$

$= 1 - \frac{ 4x^{4} }{2} + \frac{ 16x^{8} }{24} -...$

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

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Example Question #3072 : Calculus Ii

Give the Maclaurin series for the function

$f(x) = \cos \frac{4x}{3}$

up to the third term.

Possible Answers:

$\frac{4}{3}x -\frac{32}{81}x^{3} + \frac{138}{3,645}x^{5} -...$

$1 +\frac{8}{9} x^{2} + \frac{32}{243} x^{4} +...$

$1 -\frac{8}{9} x^{2} + \frac{32}{243} x^{4} -...$

$\frac{8}{9} x^{2} + \frac{32}{243} x^{4} + \frac{256}{32,805} x^{6}+...$

$\frac{4}{3}x +\frac{32}{81}x^{3} + \frac{138}{3,645}x^{5}+...$

Correct answer:

$1 -\frac{8}{9} x^{2} + \frac{32}{243} x^{4} -...$

Explanation:

The Maclaurin series for $\cos x$ is

$\cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...$

Substitute $\frac{4x}{3}$ for . The series becomes

$\cos \frac{4x}{3} = 1 - \frac{\left (\frac{4x}{3} \right )^{2}}{2} + \frac{\left (\frac{4x}{3} \right )^{4}}{24} -...$

$= 1 - \frac{\left (\frac{16x^{2}}{9} \right )}{2} + \frac{\left (\frac{256x^{4}}{81} \right )}{24} -...$

$= 1 -\frac{8}{9} x^{2} + \frac{32}{243} x^{4} -...$

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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:

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

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

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

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

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

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 #292 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

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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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #3071 : Calculus Ii

Example Question #42 : Taylor Series

Example Question #291 : Series In Calculus

Example Question #3072 : Calculus Ii

Example Question #292 : Series In Calculus

Example Question #42 : Taylor And Maclaurin Series

Example Question #3072 : Calculus Ii

Example Question #3081 : Calculus Ii

Example Question #292 : Series In Calculus

Example Question #1 : Maclaurin Series

All Calculus 2 Resources

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