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Investigation on iteration of sequence |A-B|, |B-C|, |C-D|, |D-A|

by

Ana Kuzle

Problem: Place four numbers in the first row as follows

A B C D

For each successive row replace the entries by the absolute value of the difference of the entry just above and the entry just to the right in the previous row. In the fourth position use the absolute value of the difference of the fourth and the first (i.e. cycle)

|A-B| |B-C| |C-D| |D-A|.

I. I started my task by exploring the problem by plugging in a lot of number combinations and set of numbers. But every time after even four iterations the values in the entire row were zeros. Depending on the input numbers the number of iterations (rows) varied. You can see my trials bellow.

I did not understand what was happening and what the connection was so I decided to give it a mathematical approach. Therefore, to have more structured approach I started by exploring this problem with the set of natural numbers.

II. Let A, B, C and D be natural numbers. They can either be even (E) or odd (O). Let us take a quick look on possible combinations that can occur. We can have |E-E|=E, |E-O|=|O-E|=O or |O-O|=E. Therefore, from here we can see that regarding less of that after a certain number of iterations we will get all even numbers. If we think of it in a mathematical way we will have four numbers divisible by 2. After several iterations we will have four numbers divisible by 4, then by 8, then by 16 and etc.

Therefore, I concluded that that the sequence I get by iteration converges to zero since they are bounded (less than maximum) and because they are divisible by 2^n.

Mathematical proof of that can be found here.

III. In the beginning of my exploration I played with rational numbers as well. Even then the sequence converged to zero. Why is that so? What we do is to find LCD of all the numbers. Then we can subtract them and use the same reasoning as in II. applying it to the numerators. Thus, sequence of rational numbers also converges to zero.

IV. Investigating irrational numbers was a natural thing to do. As shown here(.nb file because Excel works with estimated values) the sequence is getting close to zero with each iteration but it does not hit zero no matter what irrational numbers I decided to plug in. Try to change number of steps in .nb file as well.

For this example I have 30 steps and I did not get a sequence of all zeroes.

V. Let us give a more mathematical touch to this solution. Let A, B, C and D be real numbers such that A<B<C<D and 0<<1 (A, B, C, D, non 0).

I observed this system of equations:

A-B=A

B-C=B

C-D=C

D-A=D.

Next iteration would be equal to the previous one multiplied by . Therefore, in the next iteration I would have and after n steps i would have .

By multiplying with I am controlling the behavior of given four real numbers. Because of the construction of iteration it will never hit zero.

Let us now solve equations mentioned before. By going from the bottom to the top and plugging them in to get

and since I divided by D.

Therefore, we have

.

Let f be the function of . Thus, . Then f(0)=1, f(1)=0 and f(1/2)=3/16.

Thus, because of The Bolzano–Weierstrass theorem for continuous functions it has a zero point between 0 and 1.

We could have approach this by calculating in Mathematica as well (see .nb file mentioned above). But I have rather used the BW theorem because my goal was just to show that 0<<1.