Check a 1 n 1 y a 1 a 1 n y a 1 a 1 n y a 1 a 1 n y 0

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CHECK: (-a 1 ) n + 1 Y 0 + a 1 (-a 1 ) n Y 0 = -a 1 (-a 1 ) n Y 0 + a 1 (-a 1 ) n Y 0 = 0 (ii) Higher Order DFE Notice that if U 1 (n) and U 2 (n) are both solutions of the homo equation (H) Y n+k + a 1 Y n - 1 + k + ° + a k Y n = 0 then so is C 1 U 1 (n) + C 2 U 2 (n) where C 1 , C 2 0 oe . Two solutions of (H) are different iff ³ C such that U 1 (n) = CU 2 (n)
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S. Tanny MAT 344 Spring 1999 75 It can be shown that if U 1 (n), U 2 (n), ° , U k (n) are k different solutions of (H), then the general solution is S H (n) = where C i 0 oe . j k i 1 C i U i (n) Finding Different Solutions Y n + 2 + a 1 Y n + 1 + a 2 Y n = 0 Characteristic polynomial p( 8 ) = 8 2 + a 1 8 + a 2 Let p( 8 ) have roots 8 1 ´ 8 2 (Real 8 1 , 8 2 ). Then U 1 (n) = 8 1 n U 2 (n) = 8 2 n are different solutions, since 8 1 n + 2 + a 1 8 1 n + 1 + a 2 8 1 n = 8 1 n ( 8 1 2 + a 1 8 1 +a 2 ) = 0 and the same holds for 8 2 and clearly U 1 (n) ´ CU 2 (n) for any C 0 oe . Thus, the general solution is C 1 8 1 n + C 2 8 2 n Exercise : F n + 2 = F n + 1 + F n F 0 = F 1 = 1 F n + 2 - F n + 1 - F n = 0. P( 8 ) = 8 2 - 8 - 1 , roots 5 2 General Solution: S H (n) = C 1 + C 2 1 % 5 2 n 1 & 5 2 n F 0 = S H (0) = 1 Y C 1 + C 2 = 1 F 1 = S H (1) = 1 Y C 1 + C 2 = 1 1 % 5 2 1 & 5 2 ± C 1 = C 2 = - 1 5 1 % 5 2 1 5 1 & 5 2
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S. Tanny MAT 344 Spring 1999 76 ± F n = - 1 5 1 % 5 2 n % 1 1 5 1 & 5 2 n % 1 Since > 1 and < 1 , for n large 1 % 5 2 / 0 0 0 / 0 0 0 1 & 5 2 F n - 1 5 1 % 5 2 n % 1 In fact you can verify that for all n, < 0.5 / 0 0 0 0 / 0 0 0 0 1 5 1 & 5 2 n % 1 so that F n = “integer nearest” 1 5 1 % 5 2 n % 1 (Also notice that 1 - < 0 so that 5 is alternately above or below F n .) 1 5 1 % 5 2 n % 1 Also, P = J “Golden Mean F n % 1 F n 1 % 5 2 AB AC AC CB Suppose the two roots were the same p( 8 ) = ( 8 - 8 1 ) 2 Then U 1 (n) = 8 1 n is one solution, we need a second. Suppose the second looks like 8 1 n V(n) for some V(n), then we have V(n + 2) 8 1 n + 2 - 2 8 1 V(n + 1) 8 1 n + 1 + 8 1 2 V(n) 8 1 n = 0 Divide by 8 1 n + 2 to get V(n + 2) - 2 V(n + 1) + V(n) = 0
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S. Tanny MAT 344 Spring 1999 77 By inspection we notice that 2 possible solutions are V(n) = 1 (!!) and V(n) = n. This latter solution for V(n) gives a second (different) solution to the original equation. Thus, the general solution is S H (n) =C 1 n 8 1 C 2 n 8 1 n = 8 1 n (C 1 + C 2 n) Exercise : Y n + 2 - 4Y n + 1 + 4Y n = 0 P( 8 ) = 8 2 - 4 8 + 4 = ( 8 - 2) 2 Y n = (C 1 + C 2 n)2 n The final possibility is that the 2 roots are distinct but not real . Then they must be complex conjugates (since the coefficients are real ), say " ± i $ .
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  • Fall '06
  • miller
  • Trigraph, yn, Characteristic polynomial, Recurrence relation

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