Power Series in Differential Equations
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1 Power Series in Differential Equations Prof. Doug Hundley Whitman College April 18, 2014
2 Last time: Review of Power Series (5.1) The series a n (x x 0 ) n can converge either:
3 Last time: Review of Power Series (5.1) The series a n (x x 0 ) n can converge either: Only at x = x 0
4 Last time: Review of Power Series (5.1) The series a n (x x 0 ) n can converge either: Only at x = x 0 for all x.
5 Last time: Review of Power Series (5.1) The series a n (x x 0 ) n can converge either: Only at x = x 0 for all x. for x x 0 < ρ and diverges x x 0 > ρ.
6 Last time: Review of Power Series (5.1) The series a n (x x 0 ) n can converge either: Only at x = x 0 for all x. for x x 0 < ρ and diverges x x 0 > ρ. In this case, ρ is the RADIUS of CONVERGENCE.
7 Last time: Review of Power Series (5.1) The series a n (x x 0 ) n can converge either: Only at x = x 0 for all x. for x x 0 < ρ and diverges x x 0 > ρ. In this case, ρ is the RADIUS of CONVERGENCE. Check the endpoints, x x 0 = ρ, separately to find the INTERVAL of CONVERGENCE.
8 The Ratio Test (to determine the radius of conv) For a power series, the ratio test takes the following form: a n+1 x x 0 n [ ] a n+1 a n+1 lim n a n x x 0 n = lim x x 0 = lim x x 0 n a n n a n If the limit in the brackets is r, then overall the limit is: a n+1 x x 0 n lim n a n x x 0 n = r x x 0 Conclusion: If r x x 0 < 1, the power series converges absolutely. Equivalently, the series converges absolutely if: x x 0 < 1 r = ρ and this ρ is the RADIUS of CONVERGENCE.
9 Example 2 Find the radius and interval of convergence: n=1 ( 1) n (x + 1)n n3n
10 Example 2 Find the radius and interval of convergence: n=1 Algebra before the limit to simplify: Now the limit: ( lim n ( 1) n (x + 1)n n3n x + 1 n+1 (n + 1)3 n+1 n3 n x + 1 n = n x + 1 n n n + 1 ) x + 1 x + 1 = < 1 x + 1 < 3 3 3
11 Continuing the example Find the interval of convergence: x + 1 < 3 3 < x + 1 < 3 4 < x < 2
12 Continuing the example Find the interval of convergence: x + 1 < 3 3 < x + 1 < 3 4 < x < 2 Checking endpoints: Substitute x = 4 into the sum: n=1 ( 1) n ( 3) n n3 n = n=1 Substitute x = 2 into the sum: 1 n Divergent n=1 ( 1) n (3) n n3 n = ( 1) n n=1 Radius of convergence: 3. Interval of convergence: ( 4, 2] n
13 The Taylor Series Given a function f and a base point x 0, the Taylor series for f at x 0 is given by: f (x 0 ) + f (x 0 )(x x 0 ) + f (x 0 ) 2! or (x x 0 ) 2 + f (x 0 ) (x x 0 ) 3 + 3! f (n) (x 0 ) (x x 0 ) n n! If f (x) is equal to its Taylor series, then f is said to be analytic. Definition: The Maclaurin series for f is the Taylor series based at x 0 = 0.
14 Example: Find the Maclaurin series for f (x) = e x at x = 0. SOLUTION: Since f (n) (0) = 1 for all n, e x = 1 + x x x 3 + = The radius of convergence is. 1 n! x n
15 Template Series sin(x) = 1 1 x = x n e x = ( 1) n (2n + 1)! x 2n+1 cos(x) = 1 n! x n ( 1) n (2n)! x 2n
16 Algebra on the Index, Example Simplify to one sum that uses the term x n : m(m 1)a m x m 2 + x ka k x k 1 = m=2 k=1 SOLUTION: Try writing out the first few terms:? ( C n ) x n m(m 1)a m x m 2 = 2a a 3 x + 3 4a 4 x a 5 x 3 + m=2 n=? ka k x k = a 1 x + 2a 2 x 2 + 3a 3 x 3 + k=1 Powers of x don t line up: To write this as a single sum, we need to manipulate the sums so that the powers of x line up.
17 SOLUTION 1 Pad the second equation by starting at k = 0: m(m 1)a m x m 2 = 2a a 3 x + 4 3a 4 x a 5 x 3 + m=2 ka k x k = 0 + a 1 x + 2a 2 x 2 + 3a 3 x 3 + k=0 Substitute n = m 2 (or m = n + 2) into the first sum, and n = k into the second sum: m(m 1)a m x m 2 = m=2 ka k x k = k=0 (n + 2)(n + 1)a n+2 x n na n x n
18 Finishing the solution: m(m 1)a m x m 2 + x ka k x k 1 m=2 k=1 = (n + 2)(n + 1)a n+2 x n + na n x n = ((n + 2)(n + 1)a n+2 + na n ) x n
19 Alternate Solution: We could have started both indices using x 1 instead of x 0. Here are the sums again: m(m 1)a m x m 2 = 2a 2 + [ 3 2a 3 x + 4 3a 4 x a 5 x 3 + ] m=2 In this case, ka k x k = a 1 x + 2a 2 x 2 + 3a 3 x 3 + k=1 m(m 1)a m x m 2 = 2a 2 + m=2 and let n = m 2 to get n=1 m(m 1)a m x m 2 m=3 2a 2 + (n + 2)(n + 2)a n+2 x n + na n x n n=1
20 Using Series in DEs Given y + p(x)y + q(x)y = 0, y(x 0 ) = y 0 and y (x 0 ) = v 0, assume y, p, q are analytic at x 0. Ansatz: so that y (t) = y(t) = a n (x x 0 ) n na n (x x 0 ) n 1 and y (t) = n=1 n(n 1)a n (x x 0 ) n 2 n=2
21 The Big Picture Substituting the series into the DE will give something like: (...) + (...) + (...) = 0 We will want to write this in the form: ( Cn )x n = 0 Then we will set C n = 0 for each n.
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