# 6.4 Working with taylor series  (Page 4/11)

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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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why our wants are limited
nooo want is unlimited but resources are limited
Ruchi
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our wants are not limited but rather the resources
Moses
as we know that there are two principle of microeconomics scarcity of resources and they have alternative uses...
Ruchi
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Mathias
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demand is something wt we called in economic theory of demand it simply means if price of product is increase then demand of product will decrease
Ruchi
inverse relationship between demand and price
Ruchi
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Ruchi
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Shikhar
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Ruchi
because demand is increase
Patience
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Aziz
Because of the Marketing and purchasing power of people.
AmarbirSingh
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yaqoob
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Ruchi
see that some product which increases day by day is comes under normal good which is used by consumer
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Shamamet
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ATTAH
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Ruchi
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Shikhar
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Ruchi
Importance of economics
the nature and significance of economics studies
Deborah
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deman is amount of goods and services a consumer is willing and able to buy or purchase at a given price.
Sainabou
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Gul
Demand refers to as quantities of a goods and services in which consumers are willing and able to purchase at a given period of time and demand can also be defined as the desire or willingness and backed by the ability to pay.
Yeah
Mathias
What is Choice
Kofi
Choice refers to the ability of a consumer or producer to decide which good, service or resource to purchase or provide from a range of possible options. Being free to chose is regarded as a fundamental indicator of economic well being and development.
Shonal
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Nathan
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rules
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Nathan
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Paradigm shift it is the reconcilliation of fedural goods in production
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