6.4 Working with taylor series  (Page 4/11)

 Page 4 / 11

Use power series to solve ${y}^{\prime }=2y,\phantom{\rule{0.5em}{0ex}}y\left(0\right)=5.$

$y=5{e}^{2x}$

We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation

${y}^{\prime }-xy=0$

is known as Airy’s equation . It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.

Power series solution of airy’s equation

Use power series to solve

$y\text{″}-xy=0$

with the initial conditions $y\left(0\right)=a$ and $y\prime \left(0\right)=b.$

We look for a solution of the form

$y=\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\text{⋯}.$

Differentiating this function term by term, we obtain

$\begin{array}{ccc}\hfill {y}^{\prime }& =\hfill & {c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\text{⋯},\hfill \\ \hfill y\text{″}& =\hfill & 2·1{c}_{2}+3·2{c}_{3}x+4·3{c}_{4}{x}^{2}+\text{⋯}.\hfill \end{array}$

If y satisfies the equation $y\text{″}=xy,$ then

$2·1{c}_{2}+3·2{c}_{3}x+4·3{c}_{4}{x}^{2}+\text{⋯}=x\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\text{⋯}\right).$

Using [link] on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,

$\begin{array}{c}2·1{c}_{2}=0,\hfill \\ 3·2{c}_{3}={c}_{0},\hfill \\ 4·3{c}_{4}={c}_{1},\hfill \\ 5·4{c}_{5}={c}_{2},\hfill \\ \hfill \text{⋮}.\hfill \end{array}$

More generally, for $n\ge 3,$ we have $n·\left(n-1\right){c}_{n}={c}_{n-3}.$ In fact, all coefficients can be written in terms of ${c}_{0}$ and ${c}_{1}.$ To see this, first note that ${c}_{2}=0.$ Then

$\begin{array}{}\\ \\ {c}_{3}=\frac{{c}_{0}}{3·2},\hfill \\ {c}_{4}=\frac{{c}_{1}}{4·3}.\hfill \end{array}$

For ${c}_{5},{c}_{6},{c}_{7},$ we see that

$\begin{array}{}\\ \\ {c}_{5}=\frac{{c}_{2}}{5·4}=0,\hfill \\ {c}_{6}=\frac{{c}_{3}}{6·5}=\frac{{c}_{0}}{6·5·3·2},\hfill \\ {c}_{7}=\frac{{c}_{4}}{7·6}=\frac{{c}_{1}}{7·6·4·3}.\hfill \end{array}$

Therefore, the series solution of the differential equation is given by

$y={c}_{0}+{c}_{1}x+0·{x}^{2}+\frac{{c}_{0}}{3·2}{x}^{3}+\frac{{c}_{1}}{4·3}{x}^{4}+0·{x}^{5}+\frac{{c}_{0}}{6·5·3·2}{x}^{6}+\frac{{c}_{1}}{7·6·4·3}{x}^{7}+\text{⋯}.$

The initial condition $y\left(0\right)=a$ implies ${c}_{0}=a.$ Differentiating this series term by term and using the fact that ${y}^{\prime }\left(0\right)=b,$ we conclude that ${c}_{1}=b.$ Therefore, the solution of this initial-value problem is

$y=a\left(1+\frac{{x}^{3}}{3·2}+\frac{{x}^{6}}{6·5·3·2}+\text{⋯}\right)+b\left(x+\frac{{x}^{4}}{4·3}+\frac{{x}^{7}}{7·6·4·3}+\text{⋯}\right).$

Use power series to solve $y\text{″}+{x}^{2}y=0$ with the initial condition $y\left(0\right)=a$ and ${y}^{\prime }\left(0\right)=b.$

$y=a\left(1-\frac{{x}^{4}}{3·4}+\frac{{x}^{8}}{3·4·7·8}-\text{⋯}\right)+b\left(x-\frac{{x}^{5}}{4·5}+\frac{{x}^{9}}{4·5·8·9}-\text{⋯}\right)$

Evaluating nonelementary integrals

Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.

One integral that arises often in applications in probability theory is $\int {e}^{\text{−}{x}^{2}}dx.$ Unfortunately, the antiderivative of the integrand ${e}^{\text{−}{x}^{2}}$ is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term “elementary function” is not synonymous with noncomplicated function. For example, the function $f\left(x\right)=\sqrt{{x}^{2}-3x}+{e}^{{x}^{3}}-\text{sin}\left(5x+4\right)$ is an elementary function, although not a particularly simple-looking function. Any integral of the form $\int f\left(x\right)\phantom{\rule{0.1em}{0ex}}dx$ where the antiderivative of $f$ cannot be written as an elementary function is considered a nonelementary integral    .

Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering $\int {e}^{\text{−}{x}^{2}}dx.$

Using taylor series to evaluate a definite integral

1. Express $\int {e}^{\text{−}{x}^{2}}dx$ as an infinite series.
2. Evaluate ${\int }_{0}^{1}{e}^{\text{−}{x}^{2}}dx$ to within an error of $0.01.$
1. The Maclaurin series for ${e}^{\text{−}{x}^{2}}$ is given by
$\begin{array}{cc}\hfill {e}^{\text{−}{x}^{2}}& =\sum _{n=0}^{\infty }\frac{{\left(\text{−}{x}^{2}\right)}^{n}}{n\text{!}}\hfill \\ & =1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\text{⋯}+{\left(-1\right)}^{n}\frac{{x}^{2n}}{n\text{!}}+\text{⋯}\hfill \\ & =\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n}}{n\text{!}}.\hfill \end{array}$

Therefore,
$\begin{array}{cc}\hfill \int {e}^{\text{−}{x}^{2}}dx& =\int \left(1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\text{⋯}+{\left(-1\right)}^{n}\frac{{x}^{2n}}{n\text{!}}+\text{⋯}\right)\phantom{\rule{0.1em}{0ex}}dx\hfill \\ & =C+x-\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5.2\text{!}}-\frac{{x}^{7}}{7.3\text{!}}+\text{⋯}+{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)n\text{!}}+\text{⋯}.\hfill \end{array}$
2. Using the result from part a. we have
${\int }_{0}^{1}{e}^{\text{−}{x}^{2}}dx=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}-\text{⋯}.$

The sum of the first four terms is approximately $0.74.$ By the alternating series test, this estimate is accurate to within an error of less than $\frac{1}{216}\approx 0.0046296<0.01.$

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