1 (1 x) 4 by taking derivatives and rescaling appropriately. Conjecture what the general formula for the series for 1/(1 x) n.

Transcription

1 Math 365 Wednesday 3/20/9 8.2 and 8.4 Exercise 35. Consider the recurrence relation a n =8a n 2 6a n 4 + F (n). (a) Write the associated homogeneous recursion relation and solve for its general solution {h n }. (b) For each of the following F (n), write the general form for the particular solution (don t solve for the unknowns). (i) F (n) =n 3 (ii) F (n) =n2 n (iii) F (n) =(n 2 2)( 2) n (iv) F (n) =2 (v) F (n) =( 2) n (vi) F (n) =n 2 4 n (vii) F (n) =n 4 2 n (c) Find the general solution to (d) Find the general solution to a n =8a n 2 6a n 4 +( 2) n. a n =8a n 2 6a n 4 + n 3. (e) Pick an example of appropriate initial conditions for the sequence in part (c), and solve for the corresponding specific solution. Check your answer by computing the first 6 terms of the sequence both recursively and using your closed formula. Exercise 36. (a) Explicitly compute the series for ( x) 3 and ( x) 4 by taking derivatives and rescaling appropriately. Conjecture what the general formula for the series for /( x) n. (b) Substitute y =3x into the series for y to get the series for. What is the series for 3x /( 4x)? What is the series for /( + x)? (c) Substitute y = x 3 into the series for y to get the series for. What is the series for x3 /( x 4 )? What is the series for /( 2x 3 )? d (d) Use the fact that dx (ln( x)) = to compute the series for ln( x). (Integrate, x and pick your +C to make it so that evaluating your series at x = 0 matches correctly with evaluating ln( x).)

2 Today: Section 8.2, continued. Last time: To solve recurrence relations, the best we can do is make educated guesses, according to what kind of relation we have. A linear homogeneous recurrence relation of degree k with constant coe cients is a recurrence relation of the form a n c a n ` c 2 a n 2 ` `c k a n k, where c,c 2,...,c k are real numbers, and c k 0. Solutions to this kind of relation come in the form r n,wherer is a root of the characteristic equation, which is obtained by plugging in r n, dividing through by r n k, and solving for 0: r k c r k c 2 r k 2 c k r c k 0. The roots of this equation are called the characteristic roots. Example from last time: plugging a n r n into the recursion relation gets which is true if and only if a n a n ` a n 2 ` a n 3 r n r n ` r n 2 ` r n 3, 0 r n ` r n r n 2 r n 3 r n 3 pr 3 ` r 2 r q which is true if and only if r 0 or 0 r 3 ` r 2 r looooooooooomooooooooooon characteristic equation pr ` q 2 pr q. The characteristic roots are r (with multiplicity ) and r 2 (with multiplicity 2).

3 Theorem: Solving linear homogeneous recurrences Let c,c 2,...,c k be real numbers. Suppose that the characteristic equation has roots r,r 2,...,r` with multiplicities m,m 2,...,m`. Then a sequence ta n u npn is a solution of the recurrence relation a n c a n ` c 2 a n 2 ` `c k a n k if and only if a n p pnqr n ` p 2 pnqr n 2 ` `p`pnqr ǹ, where p i pnq are polynomials in n of degree m i. Example: a n a n ` 5a n 2 a n 3 8a n 4 4a n 5. Characteristic equation: 0 r 5 r 4 5r 3 ` r 2 ` 8r ` 4 pr ` q 3 pr 2q 2. General solution: a n p 0 ` n ` 2 n 2 qp q n `p 0 ` nqp2q n. Non-homogeneous equations Suppose your recurrence is linear and constant coe cient in a i s, but is not homogeneous. In other words, it is in the form a n c a n ` c 2 a n 2 ` `c k a n k ` F pnq where F pnq is a function only in n (no a i s). The relation h n c h n ` c 2 h n 2 ` `c k h n k (so that a n h n ` F pnq) is called the associated homogeneous recurrence relation. Ex: a n 3a n ` 2n. F pnq 2n, Assoc. hom: h n 3h n. Ex: a n 5a n 6a n 2 ` 7 n. F pnq 7 n, Assoc. hom: h n 5h n 6h n 2. The following theorem says that if we can find one solution to a n, then the general solutions to h n will help us build all the rest of the solutions to a n.

4 Theorem: Solving non-homogeneous equations (a) If tâ n u npn is one solution of the non-homogeneous linear recurrence relation with constant coe cients a n c a n ` c 2 a n 2 ` `c k a n k ` F pnq, then every solution is of the form ta n â n ` h n u npn,where th n u npn is a solution of the associated homogeneous recurrence relation. (b) Finding â n : where Qpnq is a polynomial in n, and R is a constant, then there is a solution to a n of the form â n n m qpnqr n where degpqpnqq degpqpnqq, and m mult. of R in the characteristic equation (possibly 0). Ex: Find all solutions to a n 3a n ` 2n. What is the solution with a 3?. Break the sequence into two parts: homogeneous h n and a where m mult of R, andq n b 0 ` b n ` b 2 n 2 ` `b d n d Example: a n 4a n ` 3a n 2 8a n 3 ` 4 n Homog: h n 4h n ` 3h n 2 8h n 3 and F pnq 4 n. Char eq: 0 r 3 4r 2 3r ` 8 pr ` 2qpr 3q 2 Homog sol: h n p 2q n `p ` Particular solution guess: â n b4 n (gives b 32{3) General sol: a n p 2q n `p ` ` n.

5 . Break the sequence into two parts: homogeneous h n and a where m mult of R, andq n b 0 ` b n ` b 2 n 2 ` `b d n d Example: a n 4a n ` 3a n 2 8a n 3 ` n4 n Homog: h n 4h n ` 3h n 2 8h n 3 and F pnq n4 n. Char eq: 0 r 3 4r 2 3r ` 8 pr`2qpr 3q 2 Homog sol: h n p 2q n `p ` Particular solution guess: â n pb 0 ` b nq4 n (gives b 0 376{9 and b 63{3) General sol: a n p 2q n `p ` ` p nq4n.. Break the sequence into two parts: homogeneous h n and a where m mult of R, andq n b 0 ` b n ` b 2 n 2 ` `b d n d Example: a n 4a n ` 3a n 2 8a n 3 ` 3 n 2 Homog: h n 4h n ` 3h n 2 8h n 3 and F pnq 3 n. Char eq: 0 r 3 4r 2 3r ` 8 pr`2qpr 3q 2 Homog sol: h n p 2q n `p ` Particular solution guess: â n bn 3 n (gives b 3{2) General sol: a n p 2q n `p ` ` 3 2 n2 3 n.

6 . Break the sequence into two parts: homogeneous h n and a where m mult of R, andq n b 0 ` b n ` b 2 n 2 ` `b d n d Example: a n 4a n ` 3a n 2 8a n 3 ` n3 n Homog: h n 4h n ` 3h n 2 8h n 3 and F pnq 3 n. Char eq: 0 r 3 4r 2 3r ` 8 pr`2qpr 3q 2 Homog sol: h n p 2q n `p ` Particular solution guess: â n pb 0 ` b nqn 2 3 n (gives b 0 2{50 and b {0) General sol: a n p 2q n `p ` ` p 2 50 ` 0 nqn2 3 n. Section 8.4: Generating functions. Taylor series to know and love: nÿ ˆn ˆn p ` xq n x k ` k x ` x n n x ÿ x k ` x ` x 2 ` `x n x e x x k ` x ` x 2 ` x k {k! ` x ` x2 2 ` x3 3! Combining series: Let f pxq f pxq`gpxq ˆn x 2 ` `x n 2 a k x k and gpxq pa k`b k qx k and fpxqgpxq (finite) (finite) (infinite) ` (infinite) b k x k.then kÿ a i b k i x k.

7 Example: Compute the series for using p xq 2 x 8 xk. Approach : Use the multiplication rule, k kÿ b k x a i b k i x k, on a k x k p xq 2 x Here, a i b i for all i. So kÿ a i b k i k k x x x. kÿ k `. Thus p xq 2 pk ` qx k ` 2x ` 3x 2 ` 4x 3 `. Example: Compute the series for p xq 2 using x 8 xk. Approach 2: Use derivatives, noting that d dx Thus, x d dx p xq p qp xq 2 p q p xq 2. p xq 2 d dx p xq d x k dx d dx xk kx k (change summation: let j k ) pj ` qx j ` 2x ` 3x 2 ` 4x 3 `. j 0 You try: Exercise 36