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DifferenceEquations

# DifferenceEquations - Thus we can simplify Y n = a n 3 b n...

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Determining a Formula for i k i = 0 n ! Let x 0 , x 1 , x 2 , be a sequence. Define X n = x i i = 0 n ! be the sum of the sequence up to the n th term. Define ! i ({ x j }) = x i " x i " 1 for i = 1,2, . Notice that ! i ({ X n }) = X i " X i " 1 = x j j = 0 i # " x j j = 0 i " 1 # = x i for i = 1,2, . Theorem. Suppose that X n = x i i = 0 n ! and that Y n is any sequence such that (1) X 0 = Y 0 and (2) ! i ({ X n }) = ! i ({ Y n }) for all i = 1,2, . Then X n ! Y n for all n = 0,1,2, . Proof. Clearly, X 0 = Y 0 by (1) of the assumptions. It is also clear that ! 1 ({ X n }) = x 1 = ! 1 ({ Y n }) = Y 1 " Y 0 = Y 1 " X 0 = Y 1 " x 0 and thus that Y 1 = x 0 + x 1 = X 1 . Similarly, ! 2 ({ X n }) = x 2 = ! 2 ({ Y n }) = Y 2 " Y 1 = Y 2 " X 1 = Y 2 " ( x 0 + x 1 ) and thus, Y 2 = x 0 + x 1 + x 2 = X 2 . It should be clear that we can continue this process inductively to show that Y n = X n for all n = 0,1,2, . Example. Find a formula for X n = i 2 i = 0 n ! . Solution. Let Y n = a ! n 3 + b ! n 2 + c ! n + d . We want to adjust the coefficients { a , b , c , d } so that X 0 = Y 0 and ! i ({ X n }) = ! i ({ Y n }) for all i = 1,2, . Then we can conclude from the above theorem that X n ! Y n for all n = 0,1,2, . Now X 0 = 0 = Y 0 = d gives us that d = 0 .
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Unformatted text preview: Thus, we can simplify Y n = a ! n 3 + b ! n 2 + c ! n . Now we compute: ! n ({ X i }) = n 2 ! n ({ Y i }) = Y n " Y n " 1 = a # n 3 + b # n 2 + c # n " ( a # ( n " 1) 3 + b # ( n " 1) 2 + c # ( n " 1)) This gives us the equation: ! n ({ X i }) = n 2 = 3 a " n 2 + (2 b # 3 a ) n + ( a # b + c ) = ! n ({ Y i }) For this equality to hold we must have that 3 a = 1 2 b ! 3 a = a ! b + c = These equations can be solved to get a = 1 3 , b = 1 2 , and c = 1 6 . Thus we can conclude that: i 2 i = n ! = 1 3 n 3 + 1 2 n 2 + 1 6 n for all n . Exercise. Find a formula for X n = i 3 i = n ! ....
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