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cc72501
cc72501
18.11.2020 • 
Mathematics

Find the solution to the following lhcc recurrence: an=8anâ1â16anâ2 for nâ¥2 with initial conditions a0=4,a1=7. The first step as usual is to find the characteristic equation by trying a solution of the "geometric" format an=rnan=rn. (We assume also râ 0). In this case we get: rn=8r^nâ1â16r^nâ2. Since we are assuming râ 0râ 0 we can divide by the smallest power of r, i.e., rnâ2 to get the characteristic equation:r^2=8râ16. (Notice since our lhcc recurrence was degree 2, the characteristic equation is degree 2.)This characteristic equation has a single root rr. (We say the root has multiplicity 2). Find r.r=?Since the root is repeated, the general theory (Theorem 2 in Section 8.2 of Rosen) tells us that the general solution to our lhcc recurrence looks like:an=α1(r)n+α2n(r)n for suitable constants α1, α2.To find the values of these constants we have to use the initial conditions a0=4,a1=7. These yield by using n=0 and n=1 in the formula above:4=α1(r)^0+α20(r)^0and7=α1(r)^1+α21(r)^1By plugging in your previously found numerical value for rr and doing some algebra, find α1,,α2:α1=α2=Note the final solution of the recurrence is:an=α1(r)^n+α2n(r)^nwhere the numbers r,αi have been found by your work. This gives an explicit numerical formula in terms of n for the an.

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