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Calculus 2 : Taylor and Maclaurin Series

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Example Questions

Example Question #11 : Taylor Series

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

f(x)=10x^2+x

Possible Answers:

0+0+0

11+21x+10x^2

11+21(x-1)+10(x-1)^2

0+21(x-1)+10(x-1)^2

Correct answer:

11+21(x-1)+10(x-1)^2

Explanation:

The general formula for the Taylor series about x=a for a given function is

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

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

f'(x)=20x+1

f''(x)=20

The derivatives were found using the following rule:
$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now, follow the above formula to write out the first three terms:

$\frac{11(x-1)^0}{0!}+\frac{21(x-1)^1}{1!}+\frac{20(x-1)^2}{2!}$

which simplified becomes

11+21(x-1)+10(x-1)^2

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Example Question #51 : Ap Calculus Bc

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

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

Possible Answers:

0+(1+e^1)+[(2+e^1)(x-1)]

$(1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}$

(1+e^1)+[(2+e^1)(x-1)]+(x-1)^2

0+0+(x-1)^2

Correct answer:

$(1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}$

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. The zeroth derivative is just the function itself.

f'(x)=2x+e^x

f''(x)=2+e^x

The derivatives were found using the following rules:

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

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

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

$\frac{(1+e^1)(x-1)^0}{0!}+\frac{[(2+e^1)(x-1)^1]}{1!}+\frac{(2+e^1)(x-1)^2}{2!}$

which simplified becomes

$(1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}$

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Example Question #12 : Taylor Series

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

$f(x)=3\cos(x)$

Possible Answers:

$3+0-3\frac{(x-\pi)^2}{2}$

$0+0+3\frac{(x-\pi)^2}{2}$

0+0+0

$-3+0-3\frac{(x-\pi)^2}{2}$

Correct answer:

$3+0-3\frac{(x-\pi)^2}{2}$

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

f^0=3cos(x)

$f'(x)=-3\sin(x)$

$f''(x)=-3\cos(x)$

The derivatives were found using the following rules:

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

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

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

$\frac{3(x-\pi)^0}{0!}-3\cdot \frac{0(x-\pi)^1}{1!}-\frac{3(x-\pi)^2}{2!}$

which simplified becomes

$3+0-3\frac{(x-\pi)^2}{2}$ .

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Example Question #12 : Taylor Series

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

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

Possible Answers:

$0-(x-\pi)$

0+0

$0+(x-\pi)$

0+1

Correct answer:

$0+(x-\pi)$

Explanation:

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

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

Now, for the first two terms (n=0, 1) we must find the zeroth and first derivative of the function, the zeroth derivative being the function itself:

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

The derivative was found using the following rules:

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

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

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

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

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

which simplified becomes

$0+(x-\pi)$

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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 #12 : Taylor And Maclaurin Series

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:

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

0+0+0+0

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

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

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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Calculus 2 : Taylor and Maclaurin Series

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Taylor Series

Example Question #51 : Ap Calculus Bc

Example Question #12 : Taylor Series

Example Question #12 : Taylor Series

Example Question #11 : Taylor And Maclaurin Series

Example Question #262 : Series In Calculus

Example Question #12 : Taylor And Maclaurin Series

Example Question #264 : Series In Calculus

Example Question #265 : Series In Calculus

Example Question #13 : Taylor And Maclaurin Series

All Calculus 2 Resources

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