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 #262 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

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

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

Report an Error

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

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

Report an Error

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

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

Report an Error

Example Question #271 : Series In Calculus

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

$f(x)=e^{x-2}+x^3+5$

Possible Answers:

$0+13(x-2)+\frac{13(x-2)^2}{2}$

$14+13(x-2)+\frac{13(x-2)^2}2{}$

0+0+0

14+13(x-2)+13(x-2)^2

Correct answer:

$14+13(x-2)+\frac{13(x-2)^2}2{}$

Explanation:

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

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

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

$f'(x)=e^{x-2}+3x^2$

$f''(x)=e^{x-2}+6x$

The derivatives were found using the rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{du}{dx}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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

$\frac{14(x-2)^0}{0!}+\frac{13(x-2)^1}{1!}+\frac{13(x-2)^2}{2!}$

which simplified becomes

$14+13(x-2)+\frac{13(x-2)^2}2{}$

Report an Error

Example Question #272 : Series In Calculus

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

$f(x)=(x^2+3)^2+9\cos(x)$

Possible Answers:

$[9+9\cos(1)]+[24-9\sin(1)](x-1)+\frac{[66-9\cos(1)](x-1)^2}{2}$

0+0+0

$[9+9\cos(1)]+[24+9\sin(1)](x-1)+\frac{[66-9\cos(1)](x-1)^2}{2}$

$0+[24+9\sin(1)](x-1)+\frac{[66-9\cos(1)](x-1)^2}{2}$

Correct answer:

$[9+9\cos(1)]+[24-9\sin(1)](x-1)+\frac{[66-9\cos(1)](x-1)^2}{2}$

Explanation:

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

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

First, we must find the zeroth, first, and second derivative of the function, where the zeroth derivative is the function itself:

$f'(x)=6x^5+18x^2-9\sin(x)$

$f''(x)=30x^4+36x-9\cos(x)$

The derivatives were found using the following rules:

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

Next, use the above formula to write out the first three terms (n=0, 1, 2) :

$\frac{[9+9\cos(1)](x-1)^0}{0!}+\frac{[24-9\sin(1)](x-1)^1}{1!}+\frac{[66-9\cos(1)](x-1)^2}{2!}$

which simplified becomes

$[9+9\cos(1)]+[24-9\sin(1)](x-1)+\frac{[66-9\cos(1)](x-1)^2}{2}$

Report an Error

Example Question #273 : Series In Calculus

Write the first two terms of the Taylor series about a=5 for the following function:

f(x)=x^2+2x

Possible Answers:

0+12(x-5)

35+12(x-5)

0+0

35+0

Correct answer:

35+12(x-5)

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, 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)=2x+2

The derivative was 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 two terms:

$\frac{35(x-5)^0}{0!}+\frac{12(x-5)^1}{1!}$

which simplified becomes

35+12(x-5)

Report an Error

Example Question #3056 : Calculus Ii

Find the Taylor series expansion of cos(x) at $x= \pi/2$ .

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the Taylor series expansion of cos(x) at $x= \pi/2$ , 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)=cos(x) and $x_{0}= \pi/2$ .

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

$f(\pi/2)=cos(\pi/2)=0$

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

$f'(\pi/2)=-sin(\pi/2)=-1$

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

$f''(\pi/2)=-cos(\pi/2)=0$

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

$f'''(\pi/2)=sin(\pi/2)=1$

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

$f^{4}(\pi/2)=cos(\pi/2)=0$

Substituting these values into the taylor polynomial we get

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

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

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

The Taylor polynomial simplifies to

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

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

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

Report an Error

Example Question #274 : Series In Calculus

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the Taylor series expansion of $\frac{1}{x}$ at x=1 , 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, $f(x)=\frac{1}{x}$ and $x_{0}=1$ .

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

$f(1)=\frac{1}{1}=1$

$f'(x)=\frac{d}{dx}\left [ \frac{1}{x} \right ]=\frac{d}{dx}\left [ \left ( x\right )^{-1} ]=-1( x\right )^{-2}$

$f'(1)=-1( 1 )^{-2}=-1=-1(1!)$

$f''(x)=\frac{d}{dx}[-1( x)^{-2}]=(-1)(-2)( x)^{-3}=2( x)^{-3}$

$f''(1)=2( 1)^{-3}=2=2!$

$f'''(x)=\frac{d}{dx}[2( x)^{-3}]=-6( x)^{-4}$

$f'''(1)=-6( 1)^{-4}=-6=-1(3!)$

Substituting these values into the taylor polynomial we get

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

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

The sign of each term oscillates, starting with a positive term. We represent this mathematically as $(-1)^{k}$

The Taylor polynomial simplifies to

$p_{n}x=1-(x-1)+(x-1)^{2}-(x-1)^{3}+....+(-1)^{n}(x-1)^{n}$

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

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

Report an Error

Example Question #3058 : Calculus Ii

Find the Taylor series expansion of ln(x) at x=2 .

Possible Answers:

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

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

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

ln(x)=ln(2)

Correct answer:

Explanation:

To find the Taylor series expansion of ln(x) at x=2 , 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)=ln(x) and $x_{0}=2$ .

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

f(2)=ln(2)

$f'(x)=\frac{d}{dx}\left [ ln(x) \right ]=\frac{1}{x} \right$

$f'(2)=\frac{1}{2} \right$

$f''(x)=\frac{d}{dx}\left [ \frac{1}{x} \right ]=\frac{d}{dx}\left [ \left ( x\right )^{-1} ]=-1( x\right )^{-2}$

$f''(2)=-1 \left( 2\right )^{-2}=(-1)\frac{1!}{2^2}$

$f'''(x)=\frac{d}{dx}[-1( x)^{-2}]=(-1)(-2)( x)^{-3}=2( x)^{-3}$

$f'''(2)=2( 2)^{-3}=\frac{2!}{2^3}$

$f^{4}(x)=\frac{d}{dx}[2( x)^{-3}]=-6( x)^{-4}$

$f^{4}(2)=-6( 2)^{-4}=(-1)\frac{3!}{2^4}$

Substituting these values into the taylor polynomial we get

$p_{n}x=ln(2)+\frac{1}{2}(x-2)-\frac{1!}{2^{2}*2!}(x-2)^{2}+\frac{2!}{2^{3}*3!}(x-2)^{3}-\frac{3!}{2^{4}*4!}(x-2)^{4}+....+\frac{f^{n}(2)}{n!}(x-2)^{n}$

$p_{n}x=ln(2)+\frac{1}{2}(x-2)-\frac{1}{2^{2}*2}(x-2)^{2}+\frac{1}{2^{3}*3}(x-2)^{3}-\frac{1}{2^{4}*4}(x-2)^{4}+....+\frac{f^{n}(2)}{n!}(x-2)^{n}$

The first term, ln(2) , does not follow the pattern of the other terms, so we will bring this term outside the summation and start the summation at instead of .

The sign of each term oscillates, starting with a positive term. We represent this mathematically as $(-1)^{k}$

The Taylor polynomial simplifies to

$p_{n}x=ln(2)+\frac{1}{2}(x-2)-\frac{1}{2^{2}*2}(x-2)^{2}+\frac{1}{2^{3}*3}(x-2)^{3}-\frac{1}{2^{4}*4}(x-2)^{4}+....+(-1)^{n-1}\frac{1}{2^{n}*n}(x-2)^{n}$

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

Report an Error

Example Question #271 : Series In Calculus

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

Possible Answers:

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

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

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

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

Correct answer:

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

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.

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

$f'(x)=e^{-x}(-1)=-e^{-x}$

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

$f''(x)=-e^{-x}(-1)=e^{-x}$

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

We can see that all the derivatives will be or , with the first term being position. This is represented mathematically as $(-1)^{k}$

The Taylor polynomial simplifies to

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

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

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

Report an Error

View Calculus 2 Tutors

Kurosh
Certified Tutor

University of tehran, Bachelor of Engineering, Electrical Engineering.

View Calculus 2 Tutors

James
Certified Tutor

Purdue University-Main Campus, Doctor of Education, Mathematics Teacher Education.

View Calculus 2 Tutors

Jeremy
Certified Tutor

University of Illinois at Chicago, Bachelor of Science, Physics.

All Calculus 2 Resources

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

Popular Subjects

Calculus Tutors in Seattle, Biology Tutors in Atlanta, GRE Tutors in San Francisco-Bay Area, French Tutors in New York City, Math Tutors in Los Angeles, Computer Science Tutors in Washington DC, MCAT Tutors in Boston, Algebra Tutors in Chicago, GRE Tutors in Philadelphia, LSAT Tutors in Seattle

Popular Courses & Classes

ISEE Courses & Classes in Denver, SSAT Courses & Classes in Boston, LSAT Courses & Classes in Philadelphia, GMAT Courses & Classes in Phoenix, LSAT Courses & Classes in Denver, Spanish Courses & Classes in Boston, ISEE Courses & Classes in Phoenix, LSAT Courses & Classes in New York City, ISEE Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Los Angeles

Popular Test Prep

MCAT Test Prep in Seattle, ACT Test Prep in Dallas Fort Worth, MCAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Los Angeles, LSAT Test Prep in Philadelphia, GRE Test Prep in Boston, LSAT Test Prep in New York City, ISEE Test Prep in San Francisco-Bay Area, SSAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Phoenix

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 #262 : Series In Calculus

Example Question #263 : Series In Calculus

Example Question #264 : Series In Calculus

Example Question #271 : Series In Calculus

Example Question #272 : Series In Calculus

Example Question #273 : Series In Calculus

Example Question #3056 : Calculus Ii

Example Question #274 : Series In Calculus

Example Question #3058 : Calculus Ii

Example Question #271 : Series In Calculus

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: