103 and so a n 1 a n x n 1 for any given x the ratio

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103
and so a n + 1 a n = | x | n + 1 For any given x the ratio goes to zero as n gets large. Thus the series converges for all x . What about the Taylor series of e x about 3? f ( x ) = e x f (3) = e 3 f ( x ) = e x f (3) = e 3 . . . . . . f ( n ) ( x ) = e x f ( n ) (3) = e 3 e x = n =0 f ( n ) (3) n ! ( x - 3) n = n =0 e 3 n ! ( x - 3) n Note that most (i.e. differentiable) functions have a power series, but for a given power series it is unlikely we could find a corresponding function. Example. Find the Taylor series and interval of convergence of f ( x ) = 4 3 x - 2 around x = 0. 5.1.9 Method 1 f ( x ) = 4 3 x - 2 f ( x ) = - 4 · 3 (3 x - 2) 2 f ( x ) = 4 · 3 2 · 2 (3 x - 2) 3 f ( x ) = - 4 · 3 3 · 2 · 3 (3 x - 2) 4 f iv ( x ) = 4 · 3 4 · 2 · 3 · 4 (3 x - 2) 5 f n ( x ) = ( - 1) n 4 · 3 n · n ! (3 x - 2) n +1 f (0) = 4 - 2 f (0) = - 4 · 3 2 2 f (0) = 4 · 3 2 · 2 - 2 3 f (0) = - 4 · 3 3 · 2 · 3 2 4 f iv (0) = 4 · 3 4 · 2 · 3 · 4 - 2 5 f n (0) = - 4 · 3 n · n ! 2 n +1 f ( x ) = n =0 4 · 3 n · n ! - 2 n +1 x n n ! = n =0 4 · 3 n - 2 n +1 x n lim n →∞ a n +1 a n = lim n →∞ 43 n +1 - 2 n +2 x n +1 · - 2 n +1 43 n x n = 3 2 | x | < 1 | x | < 2 3 Thus, the radius of convergence is 2 3 , so - 2 3 < x < 2 3 . If x = 2 3 : n =0 4 · 3 n - 2 n +1 2 n 3 n = n =0 - 2 = -∞ If x = - 2 3 : n =0 4 · 3 n - 2 n +1 ( - 1) n 2 n 3 n = n =0 2( - 1) n +1 This sum diverges. Therefore, - 2 3 < x < 2 3 104
5.1.10 Method 2 1 1 - y = 1 + y + y 2 + y 3 + ... = n =0 y n | y | < 1 4 3 x - 2 = - 4 2 - 3 x = - 4 2 1 1 - 3 x 2 = - 4 2 n =0 3 x 2 n = - 4 2 n =0 3 n x n 2 n = n =0 - 4 · 3 n · x n 2 n +1 We know that 3 x 2 < 1 so | x | < 2 3 and we test endpoints individually as before. Exercise. Find the Taylor series about 0 for sin x and cos x . 5.2 Series solutions 5.2.1 Problem We try to find solutions to P ( x ) y + Q ( x ) y + R ( x ) y = 0 as infinite series. We do this because thus far we have dealt almost exclusively with the case of constant coefficients. In general it is hard or impossible to find “nice” solutions to non-constant coefficient problems. 5.2.2 Ordinary point x = x 0 is an ordinary point if P ( x 0 ) = 0. i.e. we can divide through by P ( x 0 ) so y + Q ( x ) P ( x ) y + R ( x ) P ( x ) y = 0 is well defined at (and around) x = x 0 . 5.2.3 Singular point x = x 0 is a singular point if P ( x 0 ) = 0. A singular point means a degenerate second order problem at x = x 0 . 5.2.4 Regular singular point x = x 0 is a regular singular point if it satisfies the following conditions i ) P ( x 0 ) = 0 ii ) ( x - x 0 ) Q ( x ) P ( x ) and ( x - x 0 ) 2 R ( x ) P ( x ) have convergent Taylor series about x 0 . If P, Q and R are polynomials then the conditions for a regular singular point are i ) lim x x 0 ( x - x 0 ) Q ( x ) P ( x ) is finite. ii ) lim x x 0 ( x - x 0 ) 2 R ( x ) P ( x ) is finite . Specifically, ( x - x 0 ) Q ( x ) P ( x ) and ( x - x 0 ) 2 R ( x ) P ( x ) are defined at x = x 0 (and infinitely differentiable). 105
5.3 Euler-Cauchy equation 5.3.1 Equation x 2 y + αxy + βy = 0 Example. Show that x = 0 is a regular singular point of x 2 y + αxy + βy = 0. We have P ( x ) = x 2 Q ( x ) = αx R ( x ) = β P (0) = 0 is a singular point Q ( x ) P ( x ) = α x x Q ( x ) P ( x ) = α = α + 0 x + 0 x 2 + ... is the Taylor series R ( x ) P ( x ) = β x 2 x 2 R ( x ) P ( x ) = β = β + 0 x + 0 x 2 + ... is the Taylor series 5.4 Worked Example 5.4.1 Problem Try to find a series solution of y - 2 ty - 2 y = 0 5.4.2 Proposed solution Try y = n =0 a n t n If you prefer, use the longhand notation y = a 0 + a 1 t + a 2 t 2 + ...

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