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homework 3 solutions

# homework 3 solutions - Homework 3 Solutions and remarks on...

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Homework 3: Solutions and remarks on selected problems Greenberg, 4.3. In the solutions to this problem I will sometimes cite the following fact, which you may take as given: a rational function (that is, the ratio of two polynomial functions), is analytic at all points at which it is defined. Also note that sin x and cos x are analytic at all points and that the Taylor series of sin x or cos x about any point x 0 have infinite radii of convergence. The Taylor series for these functions about x 0 = 0 and their radii of convergence are given on the inside back cover of the text. (a). In the equation y 00 - x 3 y 0 + xy the coefficients p ( x ) = - x 3 and q ( x ) = x are analytic at values of x . All points of this equation are ordinary points. (b). Rewrite the equation as y 00 - [(cos x ) /x ] y 0 + [5 /x ] y = 0. Then p ( x ) and q ( x ) are analytic at all points except x 0 = 0. Since the functions xp ( x ) = cos x and xq ( x ) = 5 are analytic, x 0 = 0 is a regular singular point. Now the power series for cos( x ) about x 0 = 0 has an infinite radius of convergence; the function which is equal to the constant 5 is given as a power series all of whose terms except the first are zero, so its power series representation also has an infinite radius of convergence. Therefore, when the Frobenius method is applied it will produce series solutions, centered at x 0 = 0, with infinite radii of convergence . (c). Write the equation in the form y 00 - [1 / ( x 2 - 3)] y . Here, p ( x ) 0 and q ( x ) = 1 / ( x 2 - 3). The function q is analytic at all points except ± 3. Consider x 0 = 3, ( x - 3) q ( x ) = 1 / ( x + 3). This is a rational function defined for x 0 = 3, hence ( x - 3) q ( x ) is analytic at x 0 = 3, and hence x 0 = 3 is a regular singular point. Using the geometric series, 1 x + 3 = 1 2 3 1 1 + (( x - 3) / 2 3) = 1 3 X n =0 ( - 1) n x - 3 2 3 !

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