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Calculus 2 : Taylor Series

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Example Questions

Example Question #31 : Taylor Series

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

Possible Answers:

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

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

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

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

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

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

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Example Question #271 : Series In Calculus

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 ,

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

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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.

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

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

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

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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

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

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

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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

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

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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,

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

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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

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

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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

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

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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

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

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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.

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Calculus 2 : Taylor Series

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