Our main goal in this chapter is to study how some of

This preview shows page 56 - 59 out of 199 pages.

Our main goal in this chapter is to study how some of these functions can be constructed via the process of finding solutions to differential equations by means of power series. As a first example, we consider our old friend, the equation of simple harmonic motion d 2 y dx 2 + y = 0 , (2.11) which we have already learned how to solve by other methods. Suppose for the moment that we don’t know the general solution and want to find it by means of power series. We could start by assuming that y = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + · · · = X n =0 a n x n . (2.12) It can be shown that the standard technique for differentiating polynomials term by term also works for power series, so we expect that dy dx = a 1 + 2 a 2 x + 3 a 3 x 2 + · · · = X n =1 na n x n - 1 . (Note that the last summation only goes from 1 to , since the term with n = 0 drops out of the sum.) Differentiating again yields d 2 y dx 2 = 2 a 2 + 3 · 2 a 3 x + 4 · 3 a 4 x 2 + · · · = X n =2 n ( n - 1) a n x n - 2 . We can replace n by m + 2 in the last summation so that d 2 y dx 2 = X m +2=2 ( m + 2)( m + 2 - 1) a m +2 x m +2 - 2 = X m =0 ( m + 2)( m + 1) a m +2 x m . The index m is a “dummy variable” in the summation and can be replaced by any other letter. Thus we are free to replace m by n and obtain the formula d 2 y dx 2 = X n =0 ( n + 2)( n + 1) a n +2 x n . 50
Substitution into equation(2.11) yields X n =0 ( n + 2)( n + 1) a n +2 x n + X n =0 a n x n = 0 , or X n =0 [( n + 2)( n + 1) a n +2 + a n ] x n = 0 . Recall that a polynomial is zero only if all its coefficients are zero. Similarly, a power series can be zero only if all of its coefficients are zero. It follows that ( n + 2)( n + 1) a n +2 + a n = 0 , or a n +2 = - a n ( n + 2)( n + 1) . (2.13) This is called a recursion formula for the coefficients a n . The first two coefficients a 0 and a 1 in the power series can be determined from the initial conditions, y (0) = a 0 , dy dx (0) = a 1 . Then the recursion formula can be used to determine the remaining coefficients by the process of induction. Indeed it follows from (2.13) with n = 0 that a 2 = - a 0 2 · 1 = - 1 2 a 0 . Similarly, it follows from (2.13) with n = 1 that a 3 = - a 1 3 · 2 = - 1 3! a 1 , and with n = 2 that a 4 = - a 2 4 · 3 = 1 4 · 3 1 2 a 0 = 1 4! a 0 . Continuing in this manner, we find that a 2 n = ( - 1) n (2 n )! a 0 , a 2 n +1 = ( - 1) n (2 n + 1)! a 1 . Substitution into (2.12) yields y = a 0 + a 1 x - 1 2! a 0 x 2 - 1 3! a 1 x 3 + 1 4! a 0 x 4 + · · · = a 0 1 - 1 2! x 2 + 1 4! x 4 - · · · + a 1 x - 1 3! x 3 + 1 5! x 5 - · · · . 51
We recognize that the expressions within parentheses are power series expan- sions of the functions sin x and cos x , and hence we obtain the familiar expression for the solution to the equation of simple harmonic motion, y = a 0 cos x + a 1 sin x. The method we have described—assuming a solution to the differential equa- tion of the form y ( x ) = X n =0 a n x n and solve for the coefficients a n —is surprisingly effective, particularly for the class of equations called second-order linear differential equations.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture