6 003 TA Office Hour Notes Spring 2010 Alison Laferriere alaferri mit edu Many of you had questions about the solution to the difference equation of the form y n y n 1 1 You can use recursion to solve for y n y n y n 1 2 2 y n 2 y n 2 2 3 3 2 y n 3 y n 3 4 n y 0 n 1 n 2 5 n y 0 n 1 X k 6 k 0 1 n n y 0 1 n y 0 1 1 7 8 9 Notice this is of the form y n A n B 10 In HW1 Problem 7 some of you did something similar to the calculation above using a table and noticing a pattern this was a bit simpler since 1 and y 0 1 in the HW or you might have assumed the answer would be of this form using boundary conditions to solve for A and B with 1 These produce equivalent results y y y 1 y 0 A B y B 1 A y 0 1 11 12 13 14 15 So again y n y 0 n 1 1 16 Yet another way to think about this solution parallels the homogeneous inhomogeneous solution of a differential equation with y n y 0 n the homogeneous solution i e the solution of the difference equation y n y n 1 17
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