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#1 2022-11-13 04:38:26

MrDries
Guest

A very inefficient way of approximating the golden ratio

Consider the following:
We start with a positive integer: x
If x is even, do
x/2
If is odd, do
Floor function( x * y)
with y being some decimal number between 1 and 2

And repeat until a loop is reached. If 1 is reached, the next number will be 1 as well. So we reach a loop too.

An example:
x = 667
y = 1.003
x is odd, so we do: Floor Function( 667 * 1.003) = Floor Function( 669.001) = 669

And now we repeat with the new input of x being 669.
Thus we get a sequence, which goes as follows:
667 669 671 673 675 677 679 681 683 685 687 689 691 693 695 697 699 701 703 705 707 709 711 713 715 717 719 721 723 725 727 729 731 733 735 737 739 741 743 745 747 749 751 753 755 757 759 761 763 765 767 769 771 773 775 777 779 781 783 785 787 789 791 793 795 797 799 801 803 805 807 809 811 813 815 817 819 821 823 825 827 829 831 833 835 837 839 841 843 845 847 849 851 853 855 857 859 861 863 865 867 869 871 873 875 877 879 881 883 885 887 889 891 893 895 897 899 901 903 905 907 909 911 913 915 917 919 921 923 925 927 929 931 933 935 937 939 941 943 945 947 949 951 953 955 957 959 961 963 965 967 969 971 973 975 977 979 981 983 985 987 989 991 993 995 997 999 1001 1004 502 251 251

The sequence starts to repeat at 251, so it is stopped right there. The length of this sequence, as shown, is 172.

Can we find a longer one? Using values of x = 1 to 999 and values of y = 1.001 to 1.999, longer sequences can be found. A program running in Visual Basic on an ancient laptop, saved a sequence every time it found a new record, starting at y = 1.001.

Several new records were found along the way, for example: x = 449 and y = 1.618, gives a sequence length of 447. It ends in a loop of length 83.

A record length was found at x = 931 and y = 1.800. Length of the sequence was 20,092. Why so long and why at exactly 1.8? I have no idea. I hope Visual Basic is handling the huge numbers in this sequence correctly though.

Anyhow, the 1.618 caught my idea. It reminded me of the Golden Ratio. I thought it might be a fluke. But to be sure I ran the program again, now checking the value of y between 1.61800 and 1.61900, value of x from 1 to 999. Where would it find the longest sequence now?

Well, at y = 1.61803 and x = 637. The length of the sequence: 2,085. It forms a loop of length 1,235. Repeating the number 60,373.

Now, the Golden Ratio is approximately 1.6180339887498948482045868343656
So the program is further approximating the Golden Ratio in a most inefficient way. And for a totally unknown reason. I have no idea what's going on.

This process continues, at least for as far as I've tested it. Using values y = 1.61803390 to y = 1.61803400, the longest sequence was at y = 1.61803399 and x = 819. The length of the sequence? 26,560. It ends at 891,939. A number it had passed already, 13,415 steps before. Thereby creating a loop.

I have no idea why this happens. It also happens when doing the Ceiling Function. It didn't work while rounding numbers. At least, that's what I found.

#2 2022-11-14 00:41:46

Bob
Administrator
Registered: 2010-06-20
Posts: 9,584

Re: A very inefficient way of approximating the golden ratio

hi MrDries

Welcome to the forum.  I seem to remember that the golden ratio does have some interesting properties that may explain this but I'll have to do some research first.  Have you considered becoming a member of the forum?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2022-11-14 02:57:46

Dries
Member
Registered: 2021-01-05
Posts: 7

Re: A very inefficient way of approximating the golden ratio

It is my post. I lost my password and now I found it.
I don't really understand how these sequences can get so long. Some preference towards odd numbers in the sequences.

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#4 2022-11-14 07:29:35

Dries
Member
Registered: 2021-01-05
Posts: 7

Re: A very inefficient way of approximating the golden ratio

So in the sequence with x = 819 and y = 1.61803399, I get a sequence length of 26,560 with 15,683 odd numbers and 10,877 even numbers. That's peculiar. :-)

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#5 2022-11-14 22:31:56

Bob
Administrator
Registered: 2010-06-20
Posts: 9,584

Re: A very inefficient way of approximating the golden ratio

hi Dries

Welcome as a member!

The golden ratio has been identified as having the very worst rational approximations which means it is least well approximated by rationals.

It has long been known that plant seed development tends to follow in Fibonacci numbers and these directly lead to the golden ratio.  An expalnation for this and in the appendix a proof of the number theory statement above can be found here: https://onlinelibrary.wiley.com/doi/10. … 04.01185.x

I suspect that your discovery is related to this property although I don't think I could prove it.  There is also lots of interesting facts (enough to last you a year!) on the wiki page:
https://en.wikipedia.org/wiki/Golden_ra … quare_root

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#6 2022-11-15 02:31:44

Dries
Member
Registered: 2021-01-05
Posts: 7

Re: A very inefficient way of approximating the golden ratio

Thanks,

I don't know what it is.
It happens too when y approaches 4/3. Sequences get very long.
So for y= 1.3333333 and x= 1 to 999, it found a sequence of length 23,671 at x = 981.
There are 9,799 more odd numbers than even numbers in this sequence.
Some values of y do better than others.
Thanks for the reply though, I was just experimenting, hoping to discover something.
Not easy for me to figure it out. But I had some fun.

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