Maclaurin Series

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AP Calculus BC › Maclaurin Series

Questions 1 - 10

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 $2+\frac{1}{3}x^5$ .

Find the value of the infinite series.

$\small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}$

$\small 4$

$\small 5$

Infinite series does not converge.

$\small \sqrt{5}$

$\small \ln 5$

Explanation

The series

$\small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}$ looks similar to the series for $\small e^x$ , which is $\small e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$

but the series we want to simplify starts at $\small n=1$ , so we can fix this by adding a $\small 1$ and subtracting a $\small 1$ , to leave the value unchanged, i.e.,

$\small \small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}=-1+1+\sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}=-1+\sum_{n=0}^\infty \frac{(\ln 5)^n}{n!}$ .

So now we have $\small e^x$ with $\small x=\ln 5$ , which gives us $\small e^{\ln 5}=5$ .

So then we have:

$\small \small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}=-1+\sum_{n=0}^\infty \frac{(\ln 5)^n}{n!}=-1+5=4$

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

$\frac{212}{3}$

$\frac{152}{3}$

$-\frac{352}{3}$

Explanation

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

Find the Maclaurin Series of the function

up to the fifth degree.

$x-\frac{x^3}{3!}+\frac{x^5}{5!}$

$x+\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!}+\frac{x^5}{5!}$

$1-\frac{x^2}{2!}+\frac{x^4}{4!}$

Explanation

The formula for an i-th degree Maclaurin Polynomial is

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

For the fifth degree polynomial, we must evaluate the function up to its fifth derivative.

$\begin{matrix} f(0)=sin(0)=0\ f'(0)=cos(0)=1\ f''(0)=-sin(0)=0\ f'''(0)=-cos(0)=-1\ f^{(4)}=sin(0)=0 \ \end{matrix}$

$f^{(5)}(0)=cos(0)=1$

The summation becomes

$\sum_{n=0}^{5}\frac{f^{(n)}(0)}{n!}x^n=\frac{f(0)}{0!}x^0+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4+\frac{f^{(5)}(0)}{5!}x^5$

And substituting for the values of the function and the first five derivative values, we get the Maclaurin Polynomial

$f(x)=sin(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}$

Write the first two terms of the Maclaurin series for the following function:

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

Explanation

The Maclaurin series for any function is simply the Taylor series for the function about a=0:

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

For the first two terms (n=0, 1) we must find the zeroth and first derivative of the function. The zeroth derivative is just the function itself.

$f'(x)=4x+\sin(x)$

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

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

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

Explanation

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

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

$\frac{50}{3}$

Explanation

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

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

Explanation

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

What is the value of the following infinite series?

$\small \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{3^{2n+1}(2n+1)!}$

$\small \frac{\sqrt{3}}{2}$

$\small \frac{1}{2}$

$\small \pi/4$

$\small \arctan{\pi/3}$

Explanation

We can recognize this series as $\small \sin(x)$ since the power series is

$\small \sin x=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}$

with the value $\small \frac{\pi}{3}$ plugged into $\small x$ since

$\small \small \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{3^{2n+1}(2n+1)!}=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(\frac{\pi}{3} \right )^{2n+1}$ .

So then we have

$\small \small \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{3^{2n+1}(2n+1)!}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$ .

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

$f(x)=10\cos(x)$

Explanation

The Maclaurin series for any function is simply the Taylor series of the function about a=0:

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

We first must find the zeroth, first, and second derivative of the function (for n=0, 1, and 2). The zeroth derivative is the function itself:

$f'(x)=-10\sin(x)$

$f''(x)=-10\cos(x)$

The derivatives were found using the following rules:

$\frac{\mathrm{d}}{\mathrm{d} x}\sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-\sin(x)$

Now, we just use the formula, with , to write out the first three terms of the series (n=0, 1, and 2):

$\frac{10(1)(x)^0}{0!}-\frac{10(0)(x)^1}{1!}-\frac{10(1)(x)^2}{2!}=10+0+5x^2$

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