AP Calculus BC - Maclaurin Series

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

$f(x)= \frac{x^{5}}{3}+2$

$f(x)= x^{2}+\sqrt{x}+1$

$f(x)=x+3\sin(x)$

$f(x)=\ln|x+3|$

Explanation

Recall the Maclaurin series formula:

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

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

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

$\begin{align*}&\text{Approximate }f(-4) \&\text{Where }f(0)=-12;f'(0)=-10;f''(0)=16;f'''(0)=8\&\text{By using a Maclaurin expansion.}\end{align*}$

$\frac{212}{3}$

$\frac{152}{3}$

$-\frac{352}{3}$

Explanation

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function values }f(0)=-12;f'(0)=-10;f''(0)=16;f'''(0)=8\&\text{ at }x=-4\&\text{This expansion becomes:}\&f(-4)\approx \frac{-12}{0!}(-4)^{0}+\frac{-10}{1!}(-4)^{1}+\frac{16}{2!}(-4)^{2}+\frac{8}{3!}(-4)^{3}\&f(-4)\approx\frac{212}{3}\end{align*}$

$\begin{align*}&\text{For the function values: }\&f(0)=-8;f'(0)=11;f''(0)=8\&\text{Approximate }f(2)\text{ using a Maclaurin expansion.}\end{align*}$

$\frac{50}{3}$

Explanation

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function values }f(0)=-8;f'(0)=11;f''(0)=8\&\text{ at }x=2\&\text{This expansion becomes:}\&f(2)\approx \frac{-8}{0!}(2)^{0}+\frac{11}{1!}(2)^{1}+\frac{8}{2!}(2)^{2}\&f(2)\approx30\end{align*}$

$\begin{align*}&\text{Approximate }f(3) \&\text{Where }f(x)=-cos(5x)e^{(-3x)}\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-cos(5x)e^{(-3x)}\text{ at }x=3\&\text{This expansion becomes:}\&f(3)\approx (-cos(5(0))e^{(-3(0))})(3)^{0}+(3cos(5(0))e^{(-3(0))} + 5sin(5(0))e^{(-3(0))})(3)^{1}+\&\frac{16cos(5(0))e^{(-3(0))} - 30sin(5(0))e^{(-3(0))}}{2}(3)^{2}\&f(3)\approx80.00\end{align*}$

$\begin{align*}&\text{Approximate }f(2.1) \&\text{Where }f(x)=-sin(3x)sin(7x)\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-sin(3x)sin(7x)\text{ at }x=2.1\&\text{This expansion becomes:}\&f(2.1)\approx (-sin(3(0))sin(7(0)))(2.1)^{0}+(- 3cos(3(0))sin(7(0)) - 7cos(7(0))sin(3(0)))(2.1)^{1}+\&\frac{58sin(3(0))sin(7(0)) - 42cos(3(0))cos(7(0))}{2}(2.1)^{2}\&f(2.1)\approx-92.61\end{align*}$

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

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.

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

$\begin{align*}&\text{Approximate }f(1.2) \&\text{Where }f(x)=-cos(x)\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Like the Taylor series, it is used to approximate an unknown}\&\text{value of a function at a given point, if values of the function}\&\text{and its derivatives are known at another point. In the case}\&\text{of the Maclaurin series, this latter point has the value of}\&\text{zero (note that for many functions, such as the trigonometric}\&\text{functions, f(0) is readily known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-cos(x)\text{ at }x=1.2\&\text{This expansion becomes:}\&f(1.2)\approx \frac{-cos((0))}{1}(1.2)^{0}+\frac{sin((0))}{1}(1.2)^{1}+\frac{cos((0))}{2}(1.2)^{2}\&f(1.2)\approx-0.28\end{align*}$

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

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

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

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

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

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 , 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 ,

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

$\begin{align*}&\text{Approximate }f(-8) \&\text{Where }f(0)=-20;f'(0)=-3;f''(0)=6\&\text{With a Maclaurin series expansion.}\end{align*}$

Explanation

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function values }f(0)=-20;f'(0)=-3;f''(0)=6\&\text{ at }x=-8\&\text{This expansion becomes:}\&f(-8)\approx \frac{-20}{0!}(-8)^{0}+\frac{-3}{1!}(-8)^{1}+\frac{6}{2!}(-8)^{2}\&f(-8)\approx196\end{align*}$

$\begin{align*}&\text{Approximate }f(3) \&\text{Where }f(0)=18;f'(0)=-13;f''(0)=-10\&\text{With a Maclaurin series expansion.}\end{align*}$

$-\frac{99}{2}$

Explanation

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function values }f(0)=18;f'(0)=-13;f''(0)=-10\&\text{ at }x=3\&\text{This expansion becomes:}\&f(3)\approx \frac{18}{0!}(3)^{0}+\frac{-13}{1!}(3)^{1}+\frac{-10}{2!}(3)^{2}\&f(3)\approx-66\end{align*}$

Maclaurin Series

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