Hi I am revising for my exams and I have the following inhomogeneous first order recurrence relation defined as follows:
f(0) = 2
f(n) = 6f(n-1) - 5, n > 0
I have tried for ages using the methods I have been taught to solve this but I cannot get a proper solution.
1. Integrate a new function g(n)
2. f(n) = 6^n.g(n)
3. => 6^n.g(n) = 6.6^(n-1) .g(n-1) -5
4. => g(n) = g(n-1)-5/6^n
5. => g(n) = sum(i=1, n)-5/6^i
6. => f(n) = 6^n.sum(i=1, n)-5/6^i
7. => *Evaluate the sum using geometric series forumla*
8. => sum(i = 1, n)-5/6^i = [sum(i = 1, n)a^i] -------> (a = -5/6)
9. => *sub a = -5/6 into geometric forumla [a(1-a^n)/(1-a)]*
10. => [(-5/6(1 - (-5/6)^n))/(1-(-5/6))]
11. => g(n) = [(-5/6(1 + (5/6)^n))/(1+5/6)]
12. => f(n) = 6^n . g(n) = 6^n[(-5/6(1 + (5/6)^n))/(1+5/6)]
13. => *sub in n = 0 to see if f(0) = 2*
I cannot get this working, however. f(0) [base case] doesn't equal 2...Where have I gone wrong??
Just to let you know here is the example I am following:
f(0) = 0
f(n) = 3f(n-1)+1, n>0
f(n) = 3^n.g(n)
3^n.g(n) = g(n-1)+(1/3)^n
g(n) = sum(i=1, n)(1/3)^i
f(n) = 3^n . sum(i=1, n)(1/3)^n
sum(i=1, n) = sum(i=1, n)(a^i) ----> a = 1/3
sub into geometric series formula gives:
1/2(1-(1/3^n)
Hence:
f(n) = 3^n/2(1-(1/3^n)) = 1/2(3^n - 1) = O(3^n)
Now my maths isn't great, I know enough to get about but I have followed the exact steps as my lecturer did in the example, but I cannot get a solution to fit f(0). I have to follow the methods used in above example and I am absolutely stumped as to where the issue is..