Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Taylor Series

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

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

Example Questions

← Previous 1 2 3 4 5 Next →

Example Question #5 : Maclaurin 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+21(x-1)+10(x-1)^2

11+21x+10x^2

0+0+0

11+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

Report an Error

Example Question #1 : Maclaurin Series For Exponential And Trigonometric Functions

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:

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

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

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

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

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

Report an Error

Example Question #11 : 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:

0+0+0

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

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

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

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

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

Report an Error

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+1

0+0

$0-(x-\pi)$

Correct answer:

$0+(x-\pi)$

Explanation:

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

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

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

$0+(x-\pi)$

Report an Error

Example Question #13 : Taylor 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:

0+0+0+0

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

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 .

Report an Error

Example Question #11 : Taylor And Maclaurin Series

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:

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

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

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

0+0+0

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

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

Report an Error

Example Question #15 : Taylor 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

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

Report an Error

Example Question #16 : Taylor Series

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:

$0+(x-\pi)$

0+0

$0+2(x-\pi)$

$0+(x-\pi)$

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

$0+(x-\pi)$

Report an Error

Example Question #17 : Taylor Series

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:

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

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

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

0+0+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

Report an Error

Example Question #18 : Taylor Series

What is the power series of the function below?

$f(x)=\sin 5x$

Possible Answers:

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

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

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

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

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

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

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

Report an Error

← Previous 1 2 3 4 5 Next →

View Calculus 2 Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

View Calculus 2 Tutors

Kurosh
Certified Tutor

University of tehran, Bachelor of Engineering, Electrical Engineering.

View Calculus 2 Tutors

Filiberto
Certified Tutor

University of Arizona, Bachelor of Science, Electrical Engineering.

All Calculus 2 Resources

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

Popular Subjects

Calculus Tutors in San Diego, Calculus Tutors in Seattle, Algebra Tutors in Miami, Computer Science Tutors in Chicago, GMAT Tutors in Denver, Spanish Tutors in Philadelphia, Spanish Tutors in Seattle, SAT Tutors in Atlanta, Math Tutors in Washington DC, LSAT Tutors in San Francisco-Bay Area

Popular Courses & Classes

ISEE Courses & Classes in Houston, Spanish Courses & Classes in San Diego, ACT Courses & Classes in Philadelphia, ISEE Courses & Classes in San Diego, SSAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Houston, SSAT Courses & Classes in Washington DC, ISEE Courses & Classes in Washington DC, SSAT Courses & Classes in Miami, ACT Courses & Classes in Miami

Popular Test Prep

GMAT Test Prep in Atlanta, GMAT Test Prep in Los Angeles, GRE Test Prep in Boston, MCAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Boston, GMAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Washington DC, ISEE Test Prep in Los Angeles, ISEE Test Prep in Houston, LSAT Test Prep in Houston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 2 : Taylor Series

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #5 : Maclaurin Series

Example Question #1 : Maclaurin Series For Exponential And Trigonometric Functions

Example Question #11 : Taylor Series

Example Question #12 : Taylor Series

Example Question #13 : Taylor Series

Example Question #11 : Taylor And Maclaurin Series

Example Question #15 : Taylor Series

Example Question #16 : Taylor Series

Example Question #17 : Taylor Series

Example Question #18 : Taylor Series

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: