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**visualbenjie****Guest**

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

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,630

The Fibonacci sequence follows the simple rule of adding the previous two numbers to get the next.

0,1,1,2,3,5,8,13,21,34, etc

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

0 49

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.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Visualbenjie****Guest**

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.

**visualbenjie****Guest**

well....

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,630

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!)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,630

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)

1+4w+4w^2+4wn_1

- 1+4w+4w^2+4wn_2

= 0+ 0+ 0+4w(n_2-n_1)

= 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)

simplifying:

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

Simplifying:

n_3-n_2 = n_1

or

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?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**VisualBenjie****Guest**

Thank you very much! That is a great help. I really appriciate it. Thanks :)