Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Calculus II

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

Example Question #21 : Taylor Series

Find the Taylor series expansion of $\frac{1}{1-x^{2}}$ at x=0 .

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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.

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

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

For, $f(x)=\frac{1}{1-x}$ 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

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

Use the above to calculate the taylor series expansion of $\frac{1}{1-x^{2}}$ at x=0 ,

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

Report an Error

Example Question #32 : Taylor Series

Find the Taylor series of $x^{2}e^{x}$ at x=0

Possible Answers:

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

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

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

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

Correct answer:

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

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

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

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

Since we know the taylor series of $e^{x}$ at x=0 is $\sum_{k=0}^{\infty }\frac{1}{k!}(x)^{k}=1+x+\frac{1}{2}(x)^{2}+...+\frac{1}{n!}(x)^{n}+...$

To find the the Taylor series expansion of $x^{2}e^{x}$ at x=0 , we multiply the expansion of $e^{x}$ by $x^{2}$ and simplify.

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

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

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

Report an Error

Example Question #33 : Taylor Series

Find the taylor series expansion of $x^{4}e^{x}$ at x=0 .

Possible Answers:

$\sum_{k=0}^{\infty }\frac{1}{k!}(x)^{k+4}=x^{4}+x^{5}+\frac{1}{2}x^{6}+\frac{1}{6}x^{7}+....+\frac{1}{n!}(x)^{n+4}+....$

$\sum_{k=0}^{\infty }\frac{1}{k!}(x)^{4k}=x^{4}+x^{5}+\frac{1}{2}x^{6}+\frac{1}{6}x^{7}+....+\frac{1}{n!}(x)^{n+4}+....$

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

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

Correct answer:

$\sum_{k=0}^{\infty }\frac{1}{k!}(x)^{k+4}=x^{4}+x^{5}+\frac{1}{2}x^{6}+\frac{1}{6}x^{7}+....+\frac{1}{n!}(x)^{n+4}+....$

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

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

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

The Taylor series expansion 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.

Since we found the taylor series expansion at x=0 is

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

The taylor series expansion of $x^{4}e^{x}$ at x=0 is

$x^{4}e^{x}=x^{4}\sum_{k=0}^{\infty }\frac{1}{k!}(x)^{k}=\sum_{k=0}^{\infty }\frac{1}{k!}x^{4}(x)^{k}=\sum_{k=0}^{\infty }\frac{1}{k!}(x)^{k+4}$

$\sum_{k=0}^{\infty }\frac{1}{k!}(x)^{k+4}=x^{4}+x^{5}+\frac{1}{2}x^{6}+\frac{1}{6}x^{7}+....+\frac{1}{n!}(x)^{n+4}+....$

Report an Error

Example Question #34 : Taylor Series

Find the taylor series expansion of sin(2x) at x=0 .

Possible Answers:

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

$sin(2x)=\sum_{k=0}^{\infty }\frac{(-1)^{k}((x)^{4k+2})}{(2k+1)!}=(2x)-\frac{(2x)^{3}}{3!}+...+\frac{(-1)^{k}((2x)^{2k+1})}{(2k+1)!}+...$

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

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

Correct answer:

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

Explanation:

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.

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.

We found the taylor series of sin(x) at x=0 is

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

For sin(2x) at x=0

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

Report an Error

Example Question #281 : Series In Calculus

Find the Taylor series expansion of $\frac{4}{1-x}$ at x=0 .

Possible Answers:

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

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

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

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

Correct answer:

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

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.

The taylor series is

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

The taylor series expansion of $\frac{4}{1-x}$ at x=0 is then,

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

Report an Error

Example Question #31 : Taylor And Maclaurin Series

Write out the first three terms of the Taylor series about x=3 for the following function:

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

Possible Answers:

$(e^3+36)+(e^3+12)(x-3)+\frac{(e^3+6)(x-3)^2}{2}$

(e^3+36)+(e^3+12)(x-3)+(e^3+6)(x-3)^2

$(e^3+36)+(e^3+12)(x+3)+\frac{(e^3+6)(x+3)^2}{2}$

$0+(e^3+12)(x-3)+\frac{(e^3+6)(x-3)^2}{2}$

Correct answer:

$(e^3+36)+(e^3+12)(x-3)+\frac{(e^3+6)(x-3)^2}{2}$

Explanation:

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

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

So, 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)=e^x+2(x+3)

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

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x}f(g(x)=f'(g(x))\cdot g'(x)$

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

$\frac{(e^3+36)(x-3)^0}{0!}+\frac{(e^3+12)(x-3)^1}{1!}+\frac{(e^3+6)(x-3)^2}{2!}$ which simplified becomes

$(e^3+36)+(e^3+12)(x-3)+\frac{(e^3+6)(x-3)^2}{2}$

Report an Error

Example Question #3062 : Calculus Ii

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

$f(x)=\cos(x-1)+x^2$

Possible Answers:

2+2(x-1)+(x-1)^2

0+0+0

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

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

Correct answer:

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

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, where the zeroth derivative is just the function itself:

$f'(x)=-\sin(x-1)+2x$

$f''(x)=-\cos(x-1)+2$

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

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

which simplified becomes

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

Report an Error

Example Question #32 : Taylor And Maclaurin Series

Write out the first two terms of the Taylor series about x=2 for the following function:

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

Possible Answers:

$0-4\sin(4)(x-2)$

$\cos(4)-4\sin(4)(x-2)$

$\cos(4)-\sin(4)(x-2)$

0+0

Correct answer:

$\cos(4)-4\sin(4)(x-2)$

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

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

The derivative was found using the following rules:

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

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

$\frac{\cos(4)(x-2)^0}{0!}-\frac{4\sin(4)(x-2)^1}{1!}$

which simplified becomes

$\cos(4)-4\sin(4)(x-2)$

Report an Error

Example Question #33 : Taylor And Maclaurin Series

What is the third non-zero term in the Maclaurin series of $\sin(x)$ ?

Possible Answers:

$\frac{x}{3}$

x^3

$\frac{x^5}{5!}$

$\frac{-x^5}{5}$

Correct answer:

$\frac{x^5}{5!}$

Explanation:

To answer this question we just have to keep taking derivatives and wait to see when we have three values that aren't 0.

first, we see that $\sin(0)=0$ . that's no good so we take the derivative.

$\cos(0) = 1$ , so our first non-zero term is .

Taking the derivative again we get

$-\sin(0)=0$

then,

$-\cos(0) = -1$ so our Maclaurin series looks like this: $x-\frac{x^3}{3!}$

We take the derivative again and get back around:

$\sin(0)=0$

$\cos(0)=1$ so our next term is $\frac{x^5}{5!}$ . And we can see that this is our third non-zero term.

Report an Error

Example Question #41 : Taylor And Maclaurin Series

Write the first three terms of the Taylor series about a= $2\pi$ for the following function:

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

Possible Answers:

$0+(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)$

$(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)+\frac{(9e^{6\pi}-1)(x-2\pi)^2}{2}$

$(e^{2\pi}+1)+3e^{2\pi}(x-2\pi)+\frac{(9e^{2\pi}-1)(x-2\pi)^2}{2}$

$(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)+(9e^{6\pi}-1)(x-2\pi)^2$

Correct answer:

$(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)+\frac{(9e^{6\pi}-1)(x-2\pi)^2}{2}$

Explanation:

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

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

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

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

$f''(x)=9e^{3x}-\cos(x)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^{ax}=ae^{ax}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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

$(e^{6\pi}+1)\frac{(x-2\pi)^0}{0!}+3e^{6\pi}\frac{(x-2\pi)^1}{1!}+(9e^{6\pi}-1)\frac{(x-2\pi)^2}{2!}$

which simplifies to

$(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)+\frac{(9e^{6\pi}-1)(x-2\pi)^2}{2}$

Report an Error

View Calculus 2 Tutors

Gary
Certified Tutor

SUNY at Albany, Bachelor of Science, Economics. Pace University-New York, Masters in Business Administration, Analysis and Fu...

View Calculus 2 Tutors

Satweka
Certified Tutor

The University of Texas at Dallas, Master of Science, Biomedical Engineering.

View Calculus 2 Tutors

Christopher
Certified Tutor

University of Michigan-Ann Arbor, Bachelor of Science, Computer Science.

All Calculus 2 Resources

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

Popular Subjects

ISEE Tutors in Philadelphia, Math Tutors in Philadelphia, Statistics Tutors in Phoenix, GMAT Tutors in New York City, Physics Tutors in Atlanta, Reading Tutors in Houston, LSAT Tutors in Miami, ACT Tutors in Houston, SAT Tutors in Washington DC, English Tutors in Dallas Fort Worth

Popular Courses & Classes

GRE Courses & Classes in Philadelphia, ACT Courses & Classes in Houston, SAT Courses & Classes in Houston, SAT Courses & Classes in Miami, LSAT Courses & Classes in Miami, SSAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in New York City, SAT Courses & Classes in Phoenix, ISEE Courses & Classes in Atlanta, LSAT Courses & Classes in Chicago

Popular Test Prep

SAT Test Prep in Phoenix, LSAT Test Prep in Seattle, LSAT Test Prep in Philadelphia, LSAT Test Prep in Houston, GRE Test Prep in Phoenix, MCAT Test Prep in Atlanta, MCAT Test Prep in San Francisco-Bay Area, GRE Test Prep in Atlanta, ISEE Test Prep in Boston, ISEE Test Prep in Atlanta

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 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Taylor Series

Example Question #32 : Taylor Series

Example Question #33 : Taylor Series

Example Question #34 : Taylor Series

Example Question #281 : Series In Calculus

Example Question #31 : Taylor And Maclaurin Series

Example Question #3062 : Calculus Ii

Example Question #32 : Taylor And Maclaurin Series

Example Question #33 : Taylor And Maclaurin Series

Example Question #41 : Taylor And Maclaurin Series

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: