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Q- 31 E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 1

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Question: In Problems 29–34, determine the Taylor series about the point X₀ for the given functions and values of X₀.
31. $f (x) = \frac{1 + x}{1 - x} .$ x₀ = 0 ,

The required expression is $1 + \sum_{n - 1}^{\infty} 2 {(x)}^{n} .$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Taylor series

For a function ^{$f (x)$} the Taylor series expansion about a point ^$x_{0}$is given by, $f (x - x_{0}) = f (x_{0}) + f' (x_{0)} . (x - x_{0}) + f'' (x_{0}) . \frac{{(x - x_{0})}^{2}}{2!} + f''' (x_{0}) \frac{{(x - x_{0})}^{3}}{3!} + . . . .$

Step 2: Derivatives of function at x0

We have to calculate the Taylor series expansion for, f(x) = $\frac{1 + x}{1 - x}$ at x₀=0.

The function f(x) can be further simplified for easier calculations,

$\frac{1 + x}{1 - x} = \frac{- (1 + x)}{x - 1} = \frac{- (2 + x - 1)}{x - 1} = \frac{- 2}{x - 1} - \frac{x - 1}{x - 1} = \frac{2}{1 - x} - 1$

Calculating the derivatives of function at x₀.

$f (x) = \frac{2}{1 - x} - 1$ then ^{$f (x_{0}) = 1$}

$f' (x) = \frac{2}{{(1 - x)}^{2}}$ then $f' (x_{0}) = 2$

$f'' (x_{}) = \frac{4}{{(1 - x)}^{2}}$ then $f'' (x_{0}) = 4$

$f''' (x) = \frac{12}{{(1 - x)}^{4}}$ then $f''' (x_{0}) = 12$

$f'''' (x_{0}) = \frac{48}{{(1 - x)}^{5}}$ then $f'''' (x_{0}) = 48$

Step 3: Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at x₀=0, then,

$\frac{1 + x}{1 - x} = 1 - 2 . (x - 0) + 4 \frac{{(x - 0)}^{2}}{2!} - 12 . \frac{{(x - 0)}^{3}}{3!} + 48 . \frac{{(x - 0)}^{4}}{4!} + . . . .$

= $1 + 2 x + 2 x^{2} + 2 x^{3} + 2 x^{4} + . . . .$

= $1 + \sum_{n - 1}^{\infty} 2 {(x)}^{n}$

Hence, the required expression is

Most popular questions for Math Textbooks

Competing Species. Let pi(t) denote, respectively, the populations of three competing species $S_{i}, i = 1, 2, 3 .$ Suppose these species have the same growth rates, and the maximum population that the habitat can support is the same for each species. (We assume it to be one unit.) Also, suppose the competitive advantage that $S_{1}$ has over $S_{2}$ is the same as that of $S_{2}$ over $S_{3}$ and over. This situation is modeled by the system

$p'_{1} = p_{1} (1 - p_{1} - {ap}_{2} - {bp}_{3}) p'_{2} = p_{2} (1 - b p_{1} - p_{2} - {ap}_{3}) p'_{3} = p_{3} (1 - a p_{1} - {bp}_{2} - p_{3})$

where a and b are positive constants. To demonstrate the population dynamics of this system when a = b = 0.5, use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the populations over the time interval [0, 10] under each of the following initial conditions:

$(a) p_{1} (0) = 1 . 0, p_{2} = 0.1, p_{3} = 0.1 (b) p_{1} (0) = 0 . 1, p_{2} = 1.0, p_{3} = 0.1 (c) p_{1} (0) = 0 . 1, p_{2} = 0.1, p_{3} = 1.0$

In problems 1-4 Use Euler’s method to approximate the solution to the given initial value problem at the points $x = 0.1, 0.2, 0.3, 0.4$ , and $0.5$ , using steps of size $0.1 (h = 0.1)$ .

$\frac{d y}{d x} = \frac{x}{y}, y (0) = - 1$

Verify that $x^{2} + {cy}^{2} = 1,$ where c is an arbitrary non-zero constant, is a one-parameter family of implicit solutions to $\frac{dy}{dx} = \frac{xy}{x^{2} - 1}$ and graph several of the solution curves using the same coordinate axes.

Show that $ϕ (x) = c_{1} \sin x + c_{2} \cos x,$ is a solution to $\frac{d^{2} y}{{dx}^{2}} + y = 0$ for any choice of the constants $c_{1}$ and $c_{2}$ . Thus, $c_{1} \sin x + c_{2} \cos x,$ is a two-parameter family of solutions to the differential equation.

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

$(\cos x \cos y + 2 x) dx - (\sin x \sin y + 2 y) dy = 0$

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