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Calculus 2 : Series in Calculus

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Example Questions

Example Question #11 : Taylor And Maclaurin Series

Write out the first four terms of the Taylor series about a=1 for the following function:

f(x)=x^3+3x^2+5x+1

Possible Answers:

10+20(x-1)+(x-1)^3

10+14(x-1)+6(x-1)^2+(x-1)^3

0+14(x-1)+6(x-1)^2+(x-1)^3

0+0+0+0

Correct answer:

10+14(x-1)+6(x-1)^2+(x-1)^3

Explanation:

The Taylor series about x=a for a function is

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

For the first four terms (n=0, 1, 2, 3), we must find the zeroth, first, second, and third derivative of the function. The zeroth derivative is the function itself.

f'(x)=3x^2+6x+5

f''(x)=6x+6

f'''(x)=6

The derivatives were found using the following rule:

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

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

$\frac{10(x-1)^0}{0!}+\frac{14(x-1)^1}{1!}+\frac{12(x-1)^2}{2!}+\frac{6(x-1)^3}{3!}$

which simplified becomes

10+14(x-1)+6(x-1)^2+(x-1)^3 .

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Example Question #262 : Series In Calculus

Write out the first three terms of the Taylor series about $a=4\pi$ for the following function:

$f(x)=\sin^2(x)$

Possible Answers:

$1+0+(x-4\pi)^2$

$0+0+(x-4\pi)^2$

0+0+0

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

Correct answer:

$0+0+(x-4\pi)^2$

Explanation:

The Taylor series about x=a for a given function is

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

So, for the first three terms (n=0, 1, 2), we must find the zeroth, first, and second derivatives of the function, where the zeroth derivative is just the function itself:

$f'(x)=2\sin(x)\cos(x)$

$f''(x)=-2\sin^2(x)+2\cos^2(x)$

The derivatives were found using the following rules:

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

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ ,

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

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

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'\(x)$

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

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

which simplified becomes

$0+0+(x-4\pi)^2$ .

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Example Question #261 : Series In Calculus

Write out the first four terms for the Taylor series about $a=\frac{\pi}{2}$ for the following function:

$f(x)=2\cos(x)+3x^3$

Possible Answers:

$\frac{3\pi^3}{8}+(2+\frac{9\pi^2}{4})(x-\frac{\pi}{2})-\frac{(9\pi)(x-\frac{\pi}{2})^2}{2}+\frac{10(x-\frac{\pi}{2})^3}{3}$

$\frac{3\pi^3}{8}+(-2+\frac{9\pi^2}{4})(x-\frac{\pi}{2})+\frac{(9\pi)(x-\frac{\pi}{2})^2}{2}+\frac{20(x-\frac{\pi}{2})^3}{3}$

0+0+0+0

$\frac{3\pi^3}{8}+(-2+\frac{9\pi^2}{4})(x-\frac{\pi}{2})+\frac{(9\pi)(x-\frac{\pi}{2})^2}{2}+\frac{10(x-\frac{\pi}{2})^3}{3}$

Correct answer:

$\frac{3\pi^3}{8}+(-2+\frac{9\pi^2}{4})(x-\frac{\pi}{2})+\frac{(9\pi)(x-\frac{\pi}{2})^2}{2}+\frac{10(x-\frac{\pi}{2})^3}{3}$

Explanation:

The Taylor series about x=a for a funtion is given by

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

For the first four terms (n=0, 1, 2, 3), we must find the zeroth, first, second, and third derivative of the function, where the zeroth derivative is just the function itself:

$f'(x)=-2\sin(x)+9x^2$

$f''(x)-2\cos(x)+18x$

$f'''(x)=2\sin(x)+18$

The derivatives were found using the following rules:

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

Now, use the above formula to write out the first four terms (after some distribution we get the final answer):

$\frac{3(\frac{\pi}{2})^3(x-\frac{\pi}{2})^0}{0!}+\frac{(-2+9(\frac{\pi}{2})^2)(x-\frac{\pi}{2})^1}{1!}+\frac{18(\frac{\pi}{2})(x-\frac{\pi}{2})^2}{2!}+\frac{20(x-\frac{\pi}{2})^3}{3!}$

which simplified becomes

$\frac{3\pi^3}{8}+(-2+\frac{9\pi^2}{4})(x-\frac{\pi}{2})+\frac{(9\pi)(x-\frac{\pi}{2})^2}{2}+\frac{10(x-\frac{\pi}{2})^3}{3}$

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Example Question #264 : Series In Calculus

Write out the first two terms of the Taylor series about $a=\pi$ for the following function:

$f(x)=e^{x-\pi}+\cos^3(x)$

Possible Answers:

$0+2(x-\pi)$

0+0

$0+(x-\pi)$

Correct answer:

$0+(x-\pi)$

Explanation:

The Taylor series about x=a for a function is given by

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

First, we must find the zeroth and first derivative of the function (n=0, 1), where the zeroth derivative is just the function itself:

$f'(x)=e^{x-\pi}-3\cos^2(x)\sin(x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ ,

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

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x)) \cdot g'(x)$ ,

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

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

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

which simplified becomes

$0+(x-\pi)$

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Example Question #265 : Series In Calculus

Write out the first three terms of the Taylor series about $a=\pi$ for the following function:

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

Possible Answers:

$0+2(x-\pi)+0$

0+0+0

$1+2(x-\pi)+0$

$0+(x-\pi)+0$

Correct answer:

0+0+0

Explanation:

The Taylor series about x=a for a function is given by

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

For the first three terms (n=0, 1, 2) we must find the zeroth, first, and second derivative of the function, where the zeroth derivative is just the function itself:

$f'(x)=\cos(x)+\sec^2(x)$

$f''(x)-\sin(x)+2\sec^2(x)\tan(x)$

The derivatives were found using the following rules:

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

$\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ ,

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

$\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x)) \cdot g'(x)$

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

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

which simplified becomes

0+0+0

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Example Question #13 : Taylor And Maclaurin Series

What is the power series of the function below?

$f(x)=\sin 5x$

Possible Answers:

$\sin 5x = \sum_{n=1}^{\infty} \frac{(-1)^n5^{2n+1}}{(2n+1)!}x^{2n+1}$

$\sin 5x = 5\sum_{n=1}^{\infty} \frac{(-1)^n5^{2n}}{(2n+1)!}x^{2n+1}$

$\sin 5x = \sum_{n=0}^{\infty} \frac{(-1)^n5^{2n+2}}{(2n+1)!}x^{2n+1}$

$\sin 5x = 5\sum_{n=0}^{\infty} \frac{(-1)^n5^{2n}}{(2n+1)!}x^{2n+1}$

Correct answer:

$\sin 5x = 5\sum_{n=0}^{\infty} \frac{(-1)^n5^{2n}}{(2n+1)!}x^{2n+1}$

Explanation:

It is known that the power series for $\sin x$ at a=0 is

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

So we just need to plug this in with instead of to get

$\sin 5x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}(5x)^{2n+1}=\sum_{n=0}^{\infty} \frac{(-1)^n5^{2n+1}}{(2n+1)!}x^{2n+1}$

$=5\sum_{n=0}^{\infty} \frac{(-1)^n5^{2n}}{(2n+1)!}x^{2n+1}$

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Example Question #261 : Series In Calculus

Write out the first three terms of the Taylor series about x=1 for the following function:

f(x)=x^3+9x+20

Possible Answers:

$30+12(x-1)+\frac{15(x-1)^2}{2}$

0+0+0

30+12(x-1)+3(x-1)^2

30(x-1)+12(x-1)+3(x-1)^2

Correct answer:

30+12(x-1)+3(x-1)^2

Explanation:

The Taylor series about x=a for a given function is:

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

For the first three terms (n=0, 1, 2), we must find the zeroth, first, and second derivatives, the zeroth derivative being the function itself:

f'(x)=3x^2+9

f''(x)=6x

The derivatives were found using the following rule:

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

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

$\frac{30(x-1)^0}{0!}+\frac{12(x-1)^1}{1!}+\frac{6(x-1)^2}{2!}$

which simplified becomes

30+12(x-1)+3(x-1)^2

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Example Question #262 : Series In Calculus

Find the taylor series expansion of $e^{x}$ at x=0 .

Possible Answers:

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

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

$\sum_{k=0}^{\infty }\frac{1}{k!}(x)^{k}=e^{x}$

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

Correct answer:

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

Explanation:

To find the Taylor series expansion of $e^{x}$ 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

$p_{n}x=f(x_{0})+f'(x_{0})(x-x_{0})+\frac{f''(x_{0})}{2!}(x-x_{0})^{2}+\frac{f'''(x_{0})}{3!}(x-x_{0})^{3}+....+\frac{f^{n}(x_{0})}{n!}(x-x_{0})^{n}$

In this problem, $f(x)=e^{x}$ and $x_{0}=0$ .

We need to continue finding terms in the taylor polynomial until we can determine the pattern for the nth term.

$f(0)=e^{0}=1$

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

$f'(0)=e^{0}=1$

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

$f''(0)=e^{0}=1$

We can see that all the derivatives will be 1 or $f^{n}(x_{0})=1$

The Taylor polynomial simplifies to

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

Now that we know the nth term of the Taylor polynomial, we can find the Taylor series.

The Taylor series is defined as

$e^{x}=\sum_{k=0}^{\infty }\frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}$ where the expression within the summation is the nth term of the Taylor polynomial.

For this problem, the taylor series is

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

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

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Example Question #263 : Series In Calculus

Find the taylor series expansion of sin(x) at x=0 .

Possible Answers:

$\sum_{k=0}^{\infty }\frac{(-1)^{k}(x^{2k+1})}{(2k+1)!}=sin(x)-\frac{cos(x)}{3!}$

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

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

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

Correct answer:

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

Explanation:

To find the Taylor series expansion of sin(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

$p_{n}x=f(x_{0})+f'(x_{0})(x-x_{0})+\frac{f''(x_{0})}{2!}(x-x_{0})^{2}+\frac{f'''(x_{0})}{3!}(x-x_{0})^{3}+....+\frac{f^{n}(x_{0})}{n!}(x-x_{0})^{n}$

In this problem, f(x)=sin(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)=sin(0)=0

$f'(x)=\frac{d}{dx}sin(x)=cos(x)$

f'(0)=cos(0)=1

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

f''(0)=-sin(0)=0

$f'''(x)=\frac{d}{dx}[-sin(x)]=-cos(x)$

f'''(0)=-cos(0)=-1

$f^{4}(x)=\frac{d}{dx}[-cos(x)]=sin(x)$

$f^{4}(0)=sin(0)=0$

Substituting these values into the taylor polynomial we get

$p_{n}x=sin(0)+cos(0)(x)+\frac{-sin(0)}{2!}(x)^{2}+\frac{-cos(0)}{3!}(x)^{3}+\frac{sin(0)}{4!}(x)^{4}+....+\frac{f^{n}(0)}{n!}(x)^{n}$

$p_{n}x=0+x+0+\frac{-1}{3!}(x)^{3}+0+....+\frac{f^{n}(0)}{n!}(x)^{n}$

All the even derivatives disappear and all the odd derivatives oscillate in sign, starting with a positive term. Mathematically, we represent odd numbers as (2k+1) or . We represent oscillating signs as $(-1)^{k}$

The Taylor polynomial simplifies to

$p_{n}x=0+x+0+\frac{-1}{3!}(x)^{3}+0+....+\frac{(-1)^{k}(x^{2k+1})}{(2k+1)!}$

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

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

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Example Question #264 : Series In Calculus

Find the Taylor series expansion of $\frac{1}{1-x}$ at x=0

Possible Answers:

$\frac{1}{1-x}=\sum_{k=0}^{\infty }(x)^{k}=1+2+3+...$

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

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

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

Correct answer:

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

Explanation:

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

$p_{n}x=f(x_{0})+f'(x_{0})(x-x_{0})+\frac{f''(x_{0})}{2!}(x-x_{0})^{2}+\frac{f'''(x_{0})}{3!}(x-x_{0})^{3}+....+\frac{f^{n}(x_{0})}{n!}(x-x_{0})^{n}$

In this problem, $f(x)=\frac{1}{1-x}$ 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

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

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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 #11 : Taylor And Maclaurin Series

Example Question #262 : Series In Calculus

Example Question #261 : Series In Calculus

Example Question #264 : Series In Calculus

Example Question #265 : Series In Calculus

Example Question #13 : Taylor And Maclaurin Series

Example Question #261 : Series In Calculus

Example Question #262 : Series In Calculus

Example Question #263 : Series In Calculus

Example Question #264 : Series In Calculus

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