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Calculus 2 : Taylor and Maclaurin Series

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

Example Questions

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:

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

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

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

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

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

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

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

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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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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

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

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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

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

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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)

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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

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

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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

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

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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

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Calculus 2 : Taylor and Maclaurin Series

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