we will find the solution to the following lhcc recurrence: a_1 = 1a_n-1 + 6a_n-2 or n>= 2 with initial conditions a_0=4,a_1 = 6. The first step in any problem like this is to find the characteristics equation by trying a solution of the geometric format a_n =r^n
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The recurrence relation is: \[ a_n = a_{n-1} + 6a_{n-2} \quad \text{for} \quad n \geq 2 \] with initial conditions: \[ a_0 = 4, \quad a_1 = 6 \] Show more…
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We will find the solution to the following lhcc recurrence: an = -2an-1 + 3an-2 for n ≥ 2 with initial conditions a0 = 1, a1 = 6. The first step in any problem like this is to find the characteristic equation by trying a solution of the "geometric" format an = r^n. (We assume also r ≠ 0). In this case we get: r^n = -2r^{n-1} + 3r^{n-2}. Since we are assuming r ≠ 0 we can divide by the smallest power of r, i.e., r^{n-2} to get the characteristic equation: r^2 = -2r + 3. (Notice since our lhcc recurrence was degree 2, the characteristic equation is degree 2.) Find the two roots of the characteristic equation r1 and r2. When entering your answers use r1 ≤ r2: r1 = , r2 = Since the roots are distinct, the general theory (Theorem 1 in section 5.2 of Rosen) tells us that the general solution to our lhcc recurrence looks like: an = α1(r1)^n + α2(r2)^n for suitable constants α1, α2. To find the values of these constants we have to use the initial conditions a0 = 1, a1 = 6. These yield by using n=0 and n=1 in the formula above: 1 = α1(r1)^0 + α2(r2)^0 and 6 = α1(r1)^1 + α2(r2)^1 By plugging in your previously found numerical values for r1 and r2 and doing some algebra, find α1, α2: [Be careful to note that (-x)^n ≠ -(x^n) when n is even, for example (-3)^2 ≠ -(3^2).] α1 = α2 = Note the final solution of the recurrence is: an = α1(r1)^n + α2(r2)^n where the numbers ri, αi have been found by your work. This gives an explicit numerical formula in terms of n for the an.
Adi S.
Find the general solution of the degree k LHCR (the signs are alternating on the right-hand side). [Hint: you *can* factor the characteristic equation! Your final answer will be simpler than you may expect.]
Madhur L.
The solution of the second order recurrence relation is given by the formula an = an-1 + 6an-2, where a0 = 1 and a1 = 5.
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