Which recurrence relation is greater?

swtlilsoni
Messages
16
Reaction score
0

Homework Statement


An+2=6An+1+6An
A1=A2=1
Bn+2=5Bn+1+9Bn
B1=B2=10100

a) is it true that Bn>An for every integer n > 0?
b) is it true that Bn>An for infinetly many integers n>0?


The Attempt at a Solution



It just seems like these are increasing functions since they both start with positive integers and are only addition. Thus the function with B has a greater initial value so it should be greater.
The only way this wouldn't be true is if the rate at which A is increasing is greater than the rate at which B is increasing.
However I do not know how to find the rate.
 
Physics news on Phys.org
Probably there is some nice trick for this, but personally I would try solving for the direct formula (for inspiration, check wikipedia)
 
Here is what my professor wrote (I don't completely understand what he did):
An= c1\alphan + c2\betan
c1,c2= 3 \pm \sqrt{}15
Bn= c3\deltan + c4\varpin
c3,c4= (5 \pm \sqrt{}61)/2

3+\sqrt{}15=\alpha=6.8
3-\sqrt{}15=.9n = approaching zero
(5 + \sqrt{}61)/2 = \delta = 6.5
(5 - \sqrt{}61)/2 = \varpi = -1.5

An > Bn for all n
since \alpha > \delta > \varpi

I have NO CLUE how he got this. It is possible I could have copied some of it wrong, but this is the general idea of what he did. I'm sure it could be helpful in figuring out how this problem should be done?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top