W11 - Examples

Maclaurin series of e to the x

Maclaurin series of e to the x

What is the Maclaurin series of $f(x)=e^x$?

Solution

Using $\frac{d}{dx}{e}^x=e^x$ repeatedly, we see that $f^{(n)}(x)=e^x$ for all $n$.

So $f^{(n)}(0)=e^0=1$ for all $n$. Therefore $a_n=\frac{1}{n!}$ for all $n$ by the Derivative-Coefficient Identity:

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots =\sum _{n=0}^{\infty }\frac{x^n}{n!}$$ Link to original

Maclaurin series of cos x

Maclaurin series of cos x

Find the Maclaurin series representation of $\cos x$.

Solution

Use the Derivative-Coefficient Identity to solve for the coefficients:

$$a_n=\frac{f^{(n)}(0)}{n!}$$

$n$	$f^{(n)}(x)$	$f^{(n)}(0)$	$a_n$
0	$\cos x$	$1$	$1$
1	$-\sin x$	$0$	$0$
2	$-\cos x$	$-1$	$-1/2$
3	$\sin x$	$0$	$0$
4	$\cos x$	$1$	$1/24$
5	$-\sin x$	$0$	$0$
$⋮$	$⋮$	$⋮$	$⋮$

By studying this pattern, we find the series:

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots =\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n}}{(2n)!}$$

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Maclaurin series from other Maclaurin series

Maclaurin series from other Maclaurin series

(a) Find the Maclaurin series of $\sin x$ using the Maclaurin series of $\cos x$.

(b) Find the Maclaurin series of $f(x)=x^2{e}^{-5x}$ using the Maclaurin series of $e^x$.

(c) Using (b), find the value of $f^{(22)}(0)$.

Solution

(a)

Remember that $\frac{d}{dx}\cos x=-\sin x$. Let us differentiate the cosine series by terms:

$$\begin{array}{c}1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots \stackrel{\frac{d}{dx}}{\gg \gg }0-2\frac{x^1}{2!}+4\frac{x^3}{4!}-6\frac{x^5}{6!}+\cdots \\ \\ \gg \gg -\frac{x^1}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}-\cdots \end{array}$$

Take negative to get:

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$

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(b)

$$e^u=1+\frac{u^1}{1!}+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots$$

Set $u=-5x$:

$$e^{-5x}=1+\frac{(-5x)}{1!}+\frac{(-5x{)}^2}{2!}+\frac{(-5x{)}^3}{3!}+\cdots =\sum _{n=0}^{\infty }(-1{)}^n\frac{5^n}{n!}{x}^n$$

Multiply all terms by $x^2$:

$$\begin{array}{rl}{x}^2{e}^{-5x}\gg \gg & x^2\left(1+\frac{(-5x)}{1!}+\frac{(-5x{)}^2}{2!}+\frac{(-5x{)}^3}{3!}+\cdots \right)\\ & \\ \gg \gg & x^2-5x^3+\frac{25}{2}{x}^4-\frac{125}{3!}{x}^5+\cdots \\ & \\ \gg \gg & \sum _{n=0}^{\infty }(-1{)}^n\frac{5^n}{n!}{x}^{n+2}\end{array}$$

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(c)

For any series:

$$f(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots$$

we have:

$$f^{(n)}(0)=n!\cdot a_n$$

We can use this to compute $a_{22}$. From the series formula:

$$\sum _{n=0}^{\infty }(-1{)}^n\frac{5^n}{n!}{x}^{n+2}$$

we see that:

$$a_{n+2}=(-1{)}^n\frac{5^n}{n!}$$

Power, NOT term number

The coefficient with $a_{n+2}$ corresponds to the term having $x^{n+2}$, not necessarily the $(n+2{)}^{\text{th}}$ term of the series.

Therefore:

$$\begin{array}{c}{a}_{22}=(-1{)}^{20}\frac{5^{20}}{20!}\gg \gg 5^{20}\frac{1}{20!}\\ \\ f^{(22)}(0)=22!\cdot a_{22}\gg \gg 5^{20}\cdot \frac{22!}{20!}\gg \gg 5^{20}\cdot 22\cdot 21\end{array}$$ Link to original

Computing a Taylor series

Computing a Taylor series

Find the first five terms of the Taylor series of $f(x)=\sqrt{x+1}$ centered at $c=3$.

Solution

A Taylor series is just a Maclaurin series centered at a nonzero number.

General format of a Taylor series:

$$f(x)=a_0+a_1(x-c)+a_2(x-c{)}^2+a_3(x-c{)}^3+\cdots$$

The coefficients satisfy $f^{(n)}(c)=n!a_n$.

Find the coefficients by computing the derivatives and evaluating at $x=3$:

$$\begin{array}{rlrl}f(x)& =(x+1{)}^{1/2},& f(3)& =2\\ f^{\prime }(x)& =\frac{1}{2}(x+1{)}^{-1/2},& f^{\prime }(3)& =\frac{1}{4}\\ f^{\prime \prime }(x)& =-\frac{1}{4}(x+1{)}^{-3/2},& f^{\prime \prime }(3)& =-\frac{1}{32}\\ f^{\prime \prime \prime }(x)& =\frac{3}{8}(x+1{)}^{-5/2},& f^{\prime \prime \prime }(3)& =\frac{3}{256}\\ f^{(4)}(x)& =-\frac{15}{16}(x+1{)}^{-7/2},& f^{(4)}(3)& =-\frac{15}{2048}\end{array}$$

The first terms of the series:

$$\begin{array}{c}f(x)=\sqrt{x+1}\\ \\ =2+\frac{1}{4}(x-3)-\frac{1}{64}(x-3{)}^2+\frac{1}{512}(x-3{)}^3-\frac{5}{16,384}(x-3{)}^4+\cdots \end{array}$$Link to original

Taylor polynomial approximations

Taylor polynomial approximations

Let $f(x)=\sin x$ and let $T_n(x)$ be the Taylor polynomials expanded around $c=0$.

By considering the alternating series error bound, find the first $n$ for which $T_n(0.02)$ must have error less than $10^{-6}$.

Solution

Write the Maclaurin series of $\sin x$ because we are expanding around $c=0$:

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots =\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}$$

This series is alternating when we insert $x=0.02$, so the AST error bound formula applies (“Next Term Bound”):

$$|E_n|\le a_{n+1}$$

Find smallest $n$ such that $a_{n+1}\le 10^{-6}$, and then we know:

$$|E_n|\le a_{n+1}\le 10^{-6}\gg \gg |E_n|\le 10^{-6}$$

Plug $x=0.02$ in the series for $\sin x$:

$$a_{2n+1}=\frac{(0.02{)}^{2n+1}}{(2n+1)!}$$

Solve for the first time $a_{2n+1}\le 10^{-6}$ by listing the values:

$$\begin{array}{c}\frac{0.02^1}{1!}=0.02,\frac{0.02^3}{3!}\approx 1.33\times 10^{-6},\\ \\ \frac{0.02^5}{5!}\approx 2.67\times 10^{-11},\dots \end{array}$$

The first time $a_{2n+1}$ is below $10^{-6}$ happens when $2n+1=5$.

This is NOT the same $n$ as in $T_n$. That $n$ is the highest power of $x$ allowed.

The sum of prior terms is $T_4(0.02)$.

Since $T_4(x)=T_3(x)$ because there is no $x^4$ term, the final answer is $n=3$.

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Taylor polynomials to approximate a definite integral

Taylor polynomials to approximate a definite integral

Approximate ${\int }_0^{0.3}{e}^{-x^2}dx$ using a Taylor polynomial with an error no greater than $10^{-5}$.

Solution

Plug $u=-x^2$ into the series of $e^u$:

$$\begin{array}{c}{e}^u=\sum _{n=0}^{\infty }\frac{u^n}{n!}\\ \\ \gg \gg e^{-x^2}=\sum _{n=0}^{\infty }\frac{(-x^2{)}^n}{n!}\gg \gg \sum _{n=0}^{\infty }(-1{)}^n\frac{1}{n!}{x}^{2n}\end{array}$$

Find an antiderivative by terms:

$$\begin{array}{c}\int \sum _{n=0}^{\infty }(-1{)}^n\frac{1}{n!}{x}^{2n}dx\\ \\ \sum _{n=0}^{\infty }(-1{)}^n\frac{1}{n!}\frac{x^{2n+1}}{2n+1}\\ \\ \gg \gg x-\frac{1}{1!(3)}{x}^3+\frac{1}{2!(5)}{x}^5-\frac{1}{3!(7)}{x}^7+\cdots \end{array}$$

Plug in bounds for definite integral:

$$\begin{array}{c}{\int }_0^{0.3}\sum _{n=0}^{\infty }(-1{)}^n\frac{1}{n!}{x}^{2n}={\sum _{n=0}^{\infty }(-1{)}^n\frac{1}{n!}\frac{x^{2n+1}}{2n+1}|}_0^{0.3}\\ \\ \gg \gg \sum _{n=0}^{\infty }(-1{)}^n\frac{1}{n!}\frac{(0.3{)}^{2n+1}}{2n+1}\\ \\ \gg \gg 0.3-\frac{(0.3{)}^3}{3}+\frac{(0.3{)}^5}{10}-\frac{(0.3{)}^7}{42}+\cdots \end{array}$$

Notice alternating series pattern. Apply error bound formula, “Next Term Bound”:

$$\frac{(0.3{)}^5}{10}\approx 2.43\times 10^{-4},\frac{(0.3{)}^7}{42}\approx 5.21\times 10^{-6}$$

So we can guarantee an error less than $5.21\times 10^{-6}$ by summing the first terms through $\frac{(0.3{)}^5}{10}$:

$$0.3-\frac{(0.3{)}^3}{3}+\frac{(0.3{)}^5}{10}=0.291243$$Link to original