Algebra on the Index

Example: Algebra on the Index Rewrite the sum so that the generic term involves x n : ∞ X m=2 m(m − 1)am x m−2 Example: Algebra on the Index Rewrite the sum so that the generic term involves x n : ∞ X m(m − 1)am x m−2 m=2 SOLUTION: Let n = m − 2 (so that m = n + 2). Example: Algebra on the Index Rewrite the sum so that the generic term involves x n : ∞ X m(m − 1)am x m−2 m=2 SOLUTION: Let n = m − 2 (so that m = n + 2). Make the substitutions: ∞ X (n + 2)(n + 1)an+2 x n n=0 Example: Algebra on the Index Simplify to one sum that uses the term x n : ∞ X m=2 m(m − 1)am x m−2 +x ∞ X k=1 kak x k−1 = ? X n=? ( Cn ) xn Example: Algebra on the Index Simplify to one sum that uses the term x n : ∞ X m=2 m(m − 1)am x m−2 +x ∞ X k=1 kak x k−1 = ? X ( Cn ) xn n=? SOLUTION: Try writing out the first few terms: ∞ X m=2 m(m − 1)am x m−2 = 2a2 + 2 · 3a3 x + 3 · 4a4 x 2 + 4 · 5a5 x 3 + · · · Example: Algebra on the Index Simplify to one sum that uses the term x n : ∞ X m(m − 1)am x m=2 m−2 +x ∞ X k=1 kak x k−1 = ? X ( Cn ) xn n=? SOLUTION: Try writing out the first few terms: ∞ X m(m − 1)am x m−2 = 2a2 + 2 · 3a3 x + 3 · 4a4 x 2 + 4 · 5a5 x 3 + · · · m=2 ∞ X k=1 kak x k = a1 x + 2a2 x 2 + 3a3 x 3 + · · · Example: Algebra on the Index Simplify to one sum that uses the term x n : ∞ X m(m − 1)am x m=2 m−2 +x ∞ X k=1 kak x k−1 = ? X ( Cn ) xn n=? SOLUTION: Try writing out the first few terms: ∞ X m(m − 1)am x m−2 = 2a2 + 2 · 3a3 x + 3 · 4a4 x 2 + 4 · 5a5 x 3 + · · · m=2 ∞ X kak x k = a1 x + 2a2 x 2 + 3a3 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. SOLUTION 1 Pad the second equation by starting at k = 0: SOLUTION 1 Pad the second equation by starting at k = 0: ∞ X m=2 m(m − 1)am x m−2 = 2a2 + 3 · 2a3 x + 4 · 3a4 x 2 + 5 · 4a5 x 3 + · · · SOLUTION 1 Pad the second equation by starting at k = 0: ∞ X m(m − 1)am x m−2 = 2a2 + 3 · 2a3 x + 4 · 3a4 x 2 + 5 · 4a5 x 3 + · · · m=2 ∞ X k=0 kak x k = 0 + a1 x + 2a2 x 2 + 3a3 x 3 + · · · SOLUTION 1 Pad the second equation by starting at k = 0: ∞ X m(m − 1)am x m−2 = 2a2 + 3 · 2a3 x + 4 · 3a4 x 2 + 5 · 4a5 x 3 + · · · m=2 ∞ X kak x k = 0 + a1 x + 2a2 x 2 + 3a3 x 3 + · · · k=0 Substitute n = m − 2 (or m = n + 2) into the first sum, and n = k into the second sum: ∞ X m(m − 1)am x m−2 ∞ X = (n + 2)(n + 1)an+2 x n m=2 n=0 ∞ X k=0 k kak x = ∞ X n=0 nan x n Finishing the solution: ∞ X m(m − 1)am x m−2 + x m=2 = ∞ X n=0 ∞ X kak x k−1 k=1 (n + 2)(n + 1)an+2 x n + ∞ X n=0 nan x n Finishing the solution: ∞ X m(m − 1)am x m−2 + x m=2 = ∞ X kak x k−1 k=1 (n + 2)(n + 1)an+2 x n + n=0 = ∞ X ∞ X n=0 ∞ X nan x n n=0 ((n + 2)(n + 1)an+2 + nan ) x n Alternate Solution: We could have started both indices using x 1 instead of x 0 . Here are the sums again: ∞ X   m(m−1)am x m−2 = 2a2 + 3 · 2a3 x + 4 · 3a4 x 2 + 5 · 4a5 x 3 + · · · m=2 ∞ X k=1 kak x k = a1 x + 2a2 x 2 + 3a3 x 3 + · · · Alternate Solution: We could have started both indices using x 1 instead of x 0 . Here are the sums again: ∞ X   m(m−1)am x m−2 = 2a2 + 3 · 2a3 x + 4 · 3a4 x 2 + 5 · 4a5 x 3 + · · · m=2 ∞ X kak x k = a1 x + 2a2 x 2 + 3a3 x 3 + · · · k=1 In this case, ∞ X m(m − 1)am x m−2 = 2a2 + m=2 ∞ X m(m − 1)am x m−2 m=3 and let n = m − 2 to get 2a2 + ∞ X n=1 [(n + 2)(n + 1)an+2 + nan ] x n Using Series in DEs Given y 00 + p(x)y 0 + q(x)y = 0, y (x0 ) = y0 and y 0 (x0 ) = v0 , assume y , p, q are analytic at x0 . Ansatz: y (t) = ∞ X an (x − x0 )n n=0 so that 0 y (t) = ∞ X n=1 nan (x − x0 ) n−1 00 and y (t) = ∞ X n=2 n(n − 1)an (x − x0 )n−2 The Big Picture Substituting the series into the DE will give something like: X X X (...) + (...) + (...) = 0 We will want to write this in the form: X ( Cn )x n = 0 Then we will set Cn = 0 for each n. Example: Power Series into a DE Find the recurrence relation and the first four terms of a fundamental set of solutions to: y 00 − xy 0 − y = 0 x0 = 1 Example: Power Series into a DE Find the recurrence relation and the first four terms of a fundamental set of solutions to: y 00 − xy 0 − y = 0 Substitute y = ∞ X x0 = 1 an (x − 1)n : n=0 ∞ X n=2 n(n − 1)an (x − 1)n−2 − x ∞ X n=1 nan (x − 1)n−1 − ∞ X n=0 an (x − 1)n = 0 Look at/Simplify the second term: −x ∞ X nan (x − 1)n−1 = n=1 −x ∞ X nan (x − 1)n−1 + 1 n=1 ∞ X nan (x − 1)n−1 − 1 n=1 ∞ X nan (x − 1)n−1 n=1 to get something useful: − ∞ X n=1 n nan (x − 1) − ∞ X n=1 nan (x − 1)n−1 We now have 4 sums: ∞ X n(n − 1)an (x − 1)n−2 − n=2 − ∞ X n=1 ∞ X nan (x − 1)n n=1 nan (x − 1)n−1 − ∞ X n=0 an (x − 1)n = 0 We now have 4 sums: ∞ X n(n − 1)an (x − 1)n−2 − n=2 − ∞ X ∞ X nan (x − 1)n n=1 nan (x − 1)n−1 − n=1 Ready to simplify? (Check powers) ∞ X n=0 an (x − 1)n = 0 Reset the indices: ∞ X n−2 n(n − 1)an (x − 1) − n=2 − ∞ X n=1 ∞ X nan (x − 1)n n=0 n−1 nan (x − 1) − ∞ X n=0 an (x − 1)n = 0 Reset the indices: ∞ X n−2 n(n − 1)an (x − 1) − n=2 − ∞ X ∞ X nan (x − 1)n n=0 n−1 nan (x − 1) n=1 − ∞ X an (x − 1)n = 0 n=0 Make the substitutions: k =n−2 n =k +2 k=n n=k k =n−1 n =k +1 k =n n=k Reset the indices: ∞ X n−2 n(n − 1)an (x − 1) − n=2 − ∞ X ∞ X nan (x − 1)n n=0 n−1 nan (x − 1) n=1 − ∞ X an (x − 1)n = 0 n=0 Make the substitutions: k =n−2 n =k +2 k=n n=k k =n−1 n =k +1 k =n n=k To get: ∞ X k=0 ((k + 2)(k + 1)ak+2 − kak − (k + 1)ak+1 − ak ) (x − 1)k = 0 Principle: If P(x) = 0 for all x, then the coefficients of the polynomial are all zero. Principle: If P(x) = 0 for all x, then the coefficients of the polynomial are all zero. Therefore, ∞ X ((k + 2)(k + 1)ak+2 − kak − (k + 1)ak+1 − ak ) (x − 1)k = 0 k=0 means: ((k + 2)(k + 1)ak+2 − kak − (k + 1)ak+1 − ak = 0 Principle: If P(x) = 0 for all x, then the coefficients of the polynomial are all zero. Therefore, ∞ X ((k + 2)(k + 1)ak+2 − kak − (k + 1)ak+1 − ak ) (x − 1)k = 0 k=0 means: ((k + 2)(k + 1)ak+2 − kak − (k + 1)ak+1 − ak = 0 Solve for ak+2 : 1 (ak + ak+1 ) k +2 This is the recurrence relation. ak+2 = Build up a solution from the recurrence Find y1 , y2 that solve: y (1) = a0 = 1 y 0 (1) = a1 = 0 y (1) = a0 = 0 y 0 (1) = a1 = 1 (Therefore, W (y1 , y2 )(1) = 1) We do this for y1 in the first column, y2 in the second: k=0 a2 = 21 (a0 + a1 ) = 1 2 k=1 a3 = 13 (a1 + a2 ) = 1 6 k=2 a4 = 14 (a2 + a3 ) = 1 6 k = 3 a5 = 15 (a3 + a4 ) = 1 15 and k=0 a2 = 12 (a0 + a1 ) = 1 2 k=1 a3 = 13 (a1 + a2 ) = 1 2 k=2 a4 = 14 (a2 + a3 ) = 1 4 k = 3 a5 = 15 (a3 + a4 ) = 3 20 We now have two linearly independent solutions to the DE: 1 1 1 y1 (x) = 1 + (x − 1)2 + (x − 1)3 + (x − 1)4 + . . . 2 6 6 and 1 1 1 y2 (x) = (x − 1) + (x − 1)2 + (x − 1)3 + (x − 1)4 + . . . 2 2 4