Recursie Algorithms. Is my solution ok?

  • Thread starter ptex
  • Start date
  • #1
42
0
Code:
This is the question I must solve;
Solve the given recurrence relation for the given inital conditions.
(This means give a formula in terms of n, not in terms of previous entries)
a[sub]n[/sub] = 7a[sub]n-1[/sub] - 12a[sub]n-2[/sub]
a[sub]0[/sub] = 3 a[sub]1[/sub] = 10
Now I am not sure what that means but I think this will solve the question 

let me know if I am not even close;
Input = n
Output = X(n)

procdure find(n)
 if n = 3 or n = 10 then
  return (n)
 return(find(n-1)+find(n-2))
end find(n)
 

Answers and Replies

  • #2
jamesrc
Science Advisor
Gold Member
476
1
I'm fairly certain that what they are looking for is a single equation for a as a function of n. Simple example:

Given: an = an-1 + 1
and a0 = 11

Then:

an = 11+n
 
  • #3
42
0
Code:
Solve the given recurrence relation for the given inital conditions.
(This means give a formula in terms of n, not in terms of previous entries)
a[sub]n[/sub] = 7a[sub]n-1[/sub] - 12a[sub]n-2[/sub]
a[sub]0[/sub] = 3 a[sub]1[/sub] = 10

So;
a[sub]2[/sub] = 7a[sub]2-1[/sub] - 12a[sub]2-2[/sub]
a[sub]2[/sub] = 7a[sub]1[/sub] - 12a[sub]0[/sub]
a[sub]2[/sub] = 7(10) - 12(3)
a[sub]2[/sub] = 70 - 36
a[sub]2[/sub] = 34?
 
  • #5
42
0
Oh not that again I thought I was done with that. Ok I know what to do thanks again.
 
  • #6
42
0
I come up with all mixed fractions when I get a using the same method as I did on that problem? So that does not work.
 
  • #7
1,036
1
It works. You need to work on your algebra.
 

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