T values are found thus: Take a grid of width W. T_n = 1 + 4w(1 + w + n)
When T_n is found for a Fibbanaci sequence i.e. T_0 T_1 T_1 T_2 T_3 T_5 T_8 T_13 etc... The differences between these values form a fibbanaci sequence, that is to say that they confom to the rule D (where d is the difference) D_n = D_(n - 1) + D_(n - 2). I want to prove that this is the case for any value of W. How do i do this
The Fibonacci sequence follows the simple rule of adding the previous two numbers to get the next.
That is stated in your problem as D_n = D_(n-1) + D_(n-2) (for example, 21=13+8)
Now, trying to decipher the problem, I believe it is saying that when calculating the formula 1 + 4w(1 + w + n), where n follows a fibonacci sequence, the difference between each successive answer also follows the fibonacci sequence.
Let's plug in some real numbers to get an idea of this.
Let us try w=3:
n T_n Difference
1 61 12
1 61 0
2 73 12
3 85 12
5 109 24
8 145 36
13 205 60
21 301 96
34 457 156
So, we have n following the Fibonacci sequence, we have calculated 1 + 4w(1 + w + n) for w=3 and different values of n, and we have calculated the difference. Interestingly, the difference also follows the rule that the previous two numbers add up to make the next. (12+0=12) (12+12=24) etc
If you believe I have stated the problem correctly, please let me know, and we can then work on the proof.
Yes That is the general idea. But i want to prove that is the case for any value of W. Any ideas? Thanks very much for taking the time to think about this.
Patience, visualbenjie, I have had many things to deal with today, and will need time to sit and think the proof (jeez, you could give me an easy one I could answer directly!)
OK, I believe we can look at the problem like this: we want to know the difference between two successive n-values for any w (this difference will be the "D" value mentioned in your questiom)
First let's expand out the function: 1+4w(1+w+n) = 1+4w+4w^2+4wn
Let us see what the difference between two n-values is (call them n_1 and n_2)
= 0+ 0+ 0+4w(n_2-n_1)
So, that would be the solution for one difference "D" (remember the D_n = D_(n-1) + D_(n-2) formula?)
Let us try to "plug this into" that the D_n = D_(n-1) + D_(n-2) formula. For ease of writing this on the screen I will use actual numbers, but you can substitute (n-1), etc:
4w(n_3-n_2) = 4w(n_2-n_1) + 4w(n_1-n_0)
n_3-n_2 = (n_2-n_1) + (n_1-n_0)
n_3-n_2 = n_2-n_0
And that last formula is indeed a Fibonacci sequence which we can prove by playing with it:
substitute n_2 for n_2=n_1+n_0
n_3-n_2 = (n_1+n_0)-n_0
n_3-n_2 = n_1
n_3 = n_2 + n_1
So, I have "shown" how this can be proved. You just need to neaten it up a little before calling it a proof, OK?
Thank you very much! That is a great help. I really appriciate it. Thanks :)