The Remarkable Number 1/89 (2004) • Hacker News Board

Re: The Remarkable Number 1/89 (2004)

#111

Post by zeroimpl » Fri, Oct 30, 2020, 4:12 AM UTC

chrismorgan wrote: post #34

On the decimal expansion part, 1⁄7 has always fascinated me, having something very similar going on. Doubling from 7, you get 14, 28, 56; and 1⁄7 is 0.1̅4̅2̅8̅5̅7̅, 2⁄7 is 0.2̅8̅5̅7̅1̅4̅, 3⁄7 is 0.4̅2̅8̅5̅7̅1̅, &c. (just changing which digit you start the recurring sequence with). https://en.wikipedia.org/wiki/142,857 talks about it a bit more; the doubling sequence thing is covered in the section 1⁄7 as an infinite…

I discovered this in my late teens and thought it was super cool and made me want to understand more about repeating decimals. After playing around for a bit I realized you could form arbitrary repeating decimals by dividing by 9, 99, 999 etc. So for example, 1/7 = 142857/999999. Or written another way, 999999/7=142857.

Re: The Remarkable Number 1/89 (2004)

#112

Post by cscheid » Fri, Oct 30, 2020, 4:14 AM UTC

redvenom wrote: post #102

cscheid wrote: post #33

Earlier quoted context omitted.

If you read this and you’re curious, there’s a proof that is fairly easy to follow if you understand eigenvectors. Write the operation that takes the two last elements of the sequence and produces the following two, notice it’s linear, then analyze the eigenvalues of the associated matrix and relate the original operation to the power method.

You actually don't need eigenvectors for this proof. Sketch: Write x = lim a_n/a_{n-1} where a_n is the nth term of Fibonacci. Replace a_n with a_{n-1} + a_{n-2}, which will give you a quadratic equation in x. Solving this quadratic equation gives you the golden ratio.

I believe that gives you the result starting from Fibonacci, but not that any starting pair of numbers eventually lands at the golden ratio.

Re: The Remarkable Number 1/89 (2004)

#113

Post by mjcohen » Fri, Oct 30, 2020, 4:35 AM UTC

Another surprising fraction:

987654312/123456789 = 8

And this works in any base.

Re: The Remarkable Number 1/89 (2004)

#114

Post by ehsankia » Fri, Oct 30, 2020, 9:38 AM UTC

Jtsummers wrote: post #100

ehsankia wrote: post #78

Earlier quoted context omitted.

It tickles me that there was a "Fibonacci Quarterly" where people shared their favorite new Fibonacci facts on a quarterly basis.

Not was, is. I was curious about that as well. https://www.fq.math.ca/list-of-issues.html

Wow, almost 60 years of Fibonacci content!

Re: The Remarkable Number 1/89 (2004)

#115

Post by redvenom » Fri, Oct 30, 2020, 6:56 PM UTC

cscheid wrote: post #112

redvenom wrote: post #102

Earlier quoted context omitted.

You actually don't need eigenvectors for this proof. Sketch: Write x = lim a_n/a_{n-1} where a_n is the nth term of Fibonacci. Replace a_n with a_{n-1} + a_{n-2}, which will give you a quadratic equation in x. Solving this quadratic equation gives you the golden ratio.

I believe that gives you the result starting from Fibonacci, but not that any starting pair of numbers eventually lands at the golden ratio.

No, it works for any pair of numbers, as long as they are not both zero. I am only using the recursive relation of the Fibonacci sequence, not the starting terms.

Re: The Remarkable Number 1/89 (2004)

#116

Post by cleansingfire » Sat, Oct 31, 2020, 7:30 PM UTC

dan-robertson wrote: post #87

cleansingfire wrote: post #41

Earlier quoted context omitted.

This seems like a pretty basic calculus problem. Constants become irrelevant at the limit, so that the relationship (ratio) is all that's left.

Well it isn’t really calculus as there’s no differentiation (I guess you could consider the last step where you take a limit to be calculus), but it’s a bit like differential equations. You can write down the recurrence relation: a_(n+2) = a_(n+1) + a_n Observe that there is a linear solution space (I.e. if you add solutions point wise or multiply each value by the same scalar, you get solutions), and the values a_0…

You're right, of course. Thanks for the explanation. I was thinking particularly of the comment about the behavior with arbitrary initial numbers. Start with 1, 5000, and it still converges quickly to the golden ratio, as the parent comment mentioned. I enjoy watching the initial constants disappear.