Fibonacci Multiplication

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Fibonacci Multiplication

01 02 13

This is a quite math-intensive post. It doesn’t involve mathematics beyond Algebra, but may be a bit confusing if you haven’t worked with the notation before.

The Fibonacci numbers are a series of numbers where the last two numbers in the sequence are added together to generate the next one.

$F_{n} = F_{n-2} + F_{n-1}$

To generate the famous Fibonacci Sequence, start with the “seed numbers” of $F_0 = 0$ and $F_1 = 1$ . This creates the sequence $0, 1, 1, 2, 3, 5, 8, 13, 21, 34...$ etc. Other sequences of numbers, such as the Lucas numbers, can be produced in this fashion, by starting with different seed numbers. (For example, the Lucas numbers start off $2, 1, 3, 4, 7, 11, 18...$ )

I haven’t covered Fibonacci numbers before, but everyone knows about them (at least, I hope so if you’re reading this post.) So I’m not going to go in depth about the Fibonacci numbers and how they relate to nature and all that. Instead, I want to ponder a similar idea.

* * *

What if we took the general idea behind the Fibonacci numbers and tweaked it slightly? See below. (We’ll call the series created by this rule G and the Fibbonacci series F.)

$G_n = G_{n-2} * G_{n-1}$

What if we multiplied the previous two numbers? If we use this rule and try to apply it to the familiar Fibonacci, we get a degenerate case: $0, 1, 0, 0, 0, 0...$ , or if we use $G_0 = 1$ & $G_1 = 1$ , we get a string of endless ones. So that is out of the question. From these observations, we can start making a list of properties of this series:

1. for a zero in the series, all further numbers generated will be a zero.
( $a,0,0,0,0,0,0...$ )

What about starting with $G_0 = 2, G_1 = 1$ , like the Lucas numbers? Quickly we see that this was a better choice. The sequence starts $2,1,2,2,4,8,32,256...$ and we see a pattern. $2,1,2^1,2^1,2^2,2^3,2^5,2^8...$ The exponents show the Fibbonacci Sequence. The pattern holds for $G_0 = 3 G_1 = 1$ , $3,1,3,3,9,27,243...$ which is $3,1,3^1,3^1,3^2,3^3,3^5...$ . We can generalize and say that the pattern, for $G_0 = a, G_1 = 1$ is $G_n = a^{F_{n-2}}$

Adding this to our list of properties:

2a. for a one (at m) in the series, the series progresses to $(m-1)^{F_{n-m}}$
( $a,1,a,a,a^2,a^3,a^5,a^8...$ )

Similarly, we can say that two equal numbers in sequence also generate a power series. (2a implies 2)

2. for two numbers equaling a in sequence starting at m, the series progresses to $a^{F_{n-m+1}}$
( $a,a,a^2,a^3,a^5,a^8...$ )

We can see why this is true by going back to multiplication of powers. $a^n * a^m = a^{n+m}$ . Since we are following essentially the same rule as for Fibonacci numbers, we get Fibonacci numbers in the exponents.

Moving on to seeds that aren’t one, let’s try $G_0 = 2, G_1 = 3$ . We get the series $2,3,6,18,108,1944...$ However, if we $G_0 = 3, G_1 = 2$ , we get $3,2,6,12,72,864$ The numbers are different, and increase slower.

3. Order matters.
( $G_0 = 2, G_1 = 3$ , $2,3,6,18,108,1944...$ ; $G_0 = 3, G_1 = 2$ , $3,2,6,12,72,864$ )

If we generalize (by letting $G_0 = a, G_1 = b$ ), we see the true nature of this sequence. $a,b,ab,ab^2,a^2b^3,a^3b^5,a^5b^8...$ The exponents are Fibonacci numbers, with b one step ahead of a.

4. For seeds $G_0 = a, G_1 = b; G_n = a^{F_{n-2}}b^{F_{n-1}}$
( $a,b,ab,ab^2,a^2b^3,a^3b^5,a^5b^8...$ )

That is the essence of this series. If we wanted the 100th term, we’d calculate $a^{135301852344706746049}b^{218922995834555169026}$ . Bulky, but it works. As you can imagine, for integer seeds, this will be a very large number. For fractional seeds, however, things start to get interesting.

* * *

If we use non-negative fractional seeds less than one for $G_1$ , there are two possibilities for what could happen, depending on the other seed. The series could converge to zero, or it could diverge to infinity. It all depends on the other seed, $G_0$ .

For example, take $G_0 = 3, G_1 = 0.5$ . This series converges to zero (to 10 decimal places) within 19 iterations, but for $G_0 = 4$ , it diverges to infinity, overflowing Excel within 21 iterations.

I would like to note a property of the sequence: how to tell if it will converge or diverge. The series will converge if two sequential values are both less than one. It will diverge if two sequential values are both greater than one. However, there is a period before one of these situations pops up, where the sequence is alternating between >1 and <1. I call it the uncertain period, and the iteration it breaks on the uncertain iteration. See picture.

We can numerically derive a “boundary” between convergence and divergence by trial and error. For 0.5, I derived a value of 3.06956450765297. (I could have kept going, but Microsoft Excel prevented me.) This value keeps up uncertainty for 35 iterations, before dropping off to zero. I don’t know if there’s an easy way to derive this (by a formula, say), so evaluating it numerically is the best I’ll be able to do.

Running the test again, but with varying $G_1$ seeds, it appears that the smaller the fractional seed, the larger the “boundary” value becomes.

And that is the extent of my studies on this sequence rule. Thanks for reading!

(WordPress messed up the title :/ )

Category:

Math

8 Comments

Comments on: "Fibonacci Multiplication" (8)

Mo said:

01/02/2013 at 9:06 pm

That’s actually really cool. I kinda wish I had any motivation to play with numbers like that in my spare time.

Since I don’t have the computer programs installed to try this out myself — at what fractional seed would the boundary value become, um, too small to show up in the pattern or nonexistent — or is there always a boundary when one seed is fractional? I’m sorry if that doesn’t make sense or my question looks stupid. I’m not too great at wording things. Anyway. :/
- Climbing Gecko said:
  
  01/06/2013 at 2:37 pm
  
  Great question, actually. As far as I can tell, there’s always a boundary for <1 fractional seeds. For 0.99 however, the boundary is between 1.01 and 1.02. So as you approach 1 with your fractional seed, the boundary approaches 1 as well.
  
  Also, all of your comments are going to spam. I think it might be your website link. Akismet is just overzealous, I guess.
  - Mo said:
    
    01/14/2013 at 8:00 pm
    
    Thanks, that makes more sense.
    
    About the spam thing — it’s probably the email address, actually? If I type in my real address, WordPress forces me to sign in because of that blog I tried making years ago before I decided blogger was better, lol. So. Mailinator. Which is supposed to be an anti-spam thing but is often associated with people who make alternate accounts for things. But I could be wrong.
    
    I was just linking to my Figment profile because promotionnn.
Kim said:

01/03/2013 at 6:16 am

Wow! I Impressive. You lost me in the 1st paragraph, but I guess that’s ok, because I probably don’t need to know it in order to go to the grocery store. (That’s about the most exciting place I go these days!) Keep up the good work!
geminigoddess said:

01/06/2013 at 11:44 am

You lost me after 2a. I’ll let you know if it makes more sense after this term of math classes ;) Awesome reasoning! Great to see how your brain works :)
"Goofy" Grandma said:

01/08/2013 at 12:29 pm

Since I was not able to attend college, your post JUST BLEW ME AWAY! It is hard to believe that one of our Grandsons is so intelligent, so I don’t mean to hurt any of my family members (those that have passed and are living at the present), but that “gift” probably came from your Dad’s side of the family. He and your Mom are intelligent also, but you passed them years ago in the math department (they told me so). Our prayer is that the day is coming when you will use your God-given talents fo help humanity in many ways, since you have so many interests.in the scientic fields. GREAT JOB!
Love you – Proud of you!
Grandpa and Grandma B
Math Teachers at Play 58 « Let's Play Math! said:

01/15/2013 at 5:45 am

[…] A former student of mine returns to blogging: Gecko plays around with number series in Fibonacci Multiplication. […]
David García said:

05/16/2014 at 4:27 pm

The thing is, 3.069564… is 2 to the phi-th power, where 2 is the multiplicative inverse of 0.5, the fractional seed you used. And phi, or the Golden Ratio, is (1+sqrt(5))/2, approximately 1.618.

The thing about phi is that, if you so on dividing consecutive terms of the Fibonacci sequence, you get numbers which get closer and closer to phi: 2/1=2, 3/2=1.5, 5/3=1.666…, 8/5=1.6, 13/8=1.6125, and so on. As you approach the infinity-th term of the sequence, this ratio will be phi.

The hundredth term of your sequence, (a^135301852344706746049)*(b^218922995834555169026), is thus very close to (a^135301852344706746049)*(b^(135301852344706746049*phi)), which is (a*(b^phi))^135301852344706746049. If this is going to be close to 1, or in the uncertainty zone, that means that a*(b^phi) is going to be even closer (in fact a lot closer) to 1.

So, in general, if you pick any value for b and you want the sequence to converge, you’ll need a=(1/b)^phi.