Random Fibonacci Numbers - Numberphile
video description
I love everything related to the golden ratio and the fibonacci sequence, and especially when combined with randomness.
I created an excel file which generates the finbonacci, the golden ratio, a random fibonacci sequence R(n+2) = R(n+1) + R(n), the growth rates, and the Viswanath's number, which is the geometric average of all the growth rates.
I also computed a very interesting number that has never been done before, (maybe It will carry my name someday: ) which is the probability that the random fibonacci sequence breaks, that is when R(n) = 0. At this point the Viswanath's number cannot be computed as R(n+1)/R(n. And that has a probability higher than 50% that it occurs. Indeed of the first coin flip is a tails, the random fibonacci sequence breaks.
I can send you the file if you want.
Date: 2022-04-09
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Comments and reviews: 9
FLPhotoCatcher
OK Numberphile, question: I tried something based on this video. I picked a number at random that was between 1 and 100, and added that to -1- as my start to a modified -Fibonacci- number, and proceeded as usual in adding the previous two numbers. When I reached 1870, I tried dividing that by the previous sequence number, and got 1. 6154 - the approximate Golden Ratio.
A: Do you always get the Golden Ratio when starting with any two numbers, and summing as per the Fibonacci sequence?
B: When I divided by the number two spots back in the sequence, I got 2. 6154 - the Golden Ratio PLUS ONE! What's up with that?
reply
OK Numberphile, question: I tried something based on this video. I picked a number at random that was between 1 and 100, and added that to -1- as my start to a modified -Fibonacci- number, and proceeded as usual in adding the previous two numbers. When I reached 1870, I tried dividing that by the previous sequence number, and got 1. 6154 - the approximate Golden Ratio.
A: Do you always get the Golden Ratio when starting with any two numbers, and summing as per the Fibonacci sequence?
B: When I divided by the number two spots back in the sequence, I got 2. 6154 - the Golden Ratio PLUS ONE! What's up with that?
reply
MaLiN2223
If you guys are interested, I have managed to get the following.
Setup: 10, 000 experiments, 50% chance of +/-, iterations up to and including 200, 000
Mean: 1. 131952169616262201125718239
Std: 0. 0009075157148151106535621414414
Same for iterations up to 400, 000
Mean: 1. 131911381607980595294693675
Std: 0. 0006809854552570100368442605874
Same for iterations up to 1, 000, 000-
Mean: 1. 131925577620492136479641628
Std: 0. 0003989819273866980927847283544
We can nicely see how std goes down and mean closes to the number from the video
reply
If you guys are interested, I have managed to get the following.
Setup: 10, 000 experiments, 50% chance of +/-, iterations up to and including 200, 000
Mean: 1. 131952169616262201125718239
Std: 0. 0009075157148151106535621414414
Same for iterations up to 400, 000
Mean: 1. 131911381607980595294693675
Std: 0. 0006809854552570100368442605874
Same for iterations up to 1, 000, 000-
Mean: 1. 131925577620492136479641628
Std: 0. 0003989819273866980927847283544
We can nicely see how std goes down and mean closes to the number from the video
reply
vsm1
What if you have an unfair coin? Can you put probabilities in the equation?
At 100% -+-, 0% --- we get 1. 618. growth ratio.
At 50%/50% we get -1. 132 ratio.
At 33, 3%/66, 7% do we get -1 ratio which means it barely grows? Or does it only happen in plus-minus-munis pattern and every randomisation makes it grow, even if only slightly?
It looks like the ratio starts to grow again if we go further:
At 0%/100% (i. e. always minus) we get 1, 618. again, but with 3 positions delay (1, 1, 0 then -1, -1, -2, -3, -5, etc)
reply
What if you have an unfair coin? Can you put probabilities in the equation?
At 100% -+-, 0% --- we get 1. 618. growth ratio.
At 50%/50% we get -1. 132 ratio.
At 33, 3%/66, 7% do we get -1 ratio which means it barely grows? Or does it only happen in plus-minus-munis pattern and every randomisation makes it grow, even if only slightly?
It looks like the ratio starts to grow again if we go further:
At 0%/100% (i. e. always minus) we get 1, 618. again, but with 3 positions delay (1, 1, 0 then -1, -1, -2, -3, -5, etc)
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Klaas
One could improve on the formula given at 1: 58, by skipping the first part of the Fibonacci series where the ratios betweeen successive numbers are quite far from the Golden Ratio. As an example, the 20th Fibonacci number is 6764. So phi-1000000 could be approximated by 6765 - phi-999980. Skipping the first 1000 is most likely even better. Keep in mind that you need a lot of decimals in your phi value to get close to the correct answer!
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One could improve on the formula given at 1: 58, by skipping the first part of the Fibonacci series where the ratios betweeen successive numbers are quite far from the Golden Ratio. As an example, the 20th Fibonacci number is 6764. So phi-1000000 could be approximated by 6765 - phi-999980. Skipping the first 1000 is most likely even better. Keep in mind that you need a lot of decimals in your phi value to get close to the correct answer!
reply
MrVipitis
So any binary number can be made into one of these sequences. And each number has a growth ratio (if you take the rear most element or the moving average.
That means you can map integers to real numbers? But you can't as every integer has a limited number of binary digit so it's not an infinite sequence. But if you inverse numbers, you can map reals to reals using this and find some interesting bits of bijictives.
reply
So any binary number can be made into one of these sequences. And each number has a growth ratio (if you take the rear most element or the moving average.
That means you can map integers to real numbers? But you can't as every integer has a limited number of binary digit so it's not an infinite sequence. But if you inverse numbers, you can map reals to reals using this and find some interesting bits of bijictives.
reply
byteridr
What is if you do that with complex numbers. At first you take the length of one with a random angle and take this as your first complex number. Then you take the length of one with another random angle as your second complex number and add them to get the absolute value of your third complex number. You then cohoose another random angle to have your third complex number complete. Then add second and third and so on.
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What is if you do that with complex numbers. At first you take the length of one with a random angle and take this as your first complex number. Then you take the length of one with another random angle as your second complex number and add them to get the absolute value of your third complex number. You then cohoose another random angle to have your third complex number complete. Then add second and third and so on.
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Emma
Hi -Numberphile. I'm an undergraduate mathematics student and I'm taking my first Probability Theory course this year, so this blew my mind. I would like to know if that Viswanath's constant you mentioned is kind of the expected value of the quotient of the random variables R_(n+1) and R_n, i. e, Viswanath's constant = E( R_(n+1) / R_n, because it sounds reasonable to me.
Thank you so much for the great content; D
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Hi -Numberphile. I'm an undergraduate mathematics student and I'm taking my first Probability Theory course this year, so this blew my mind. I would like to know if that Viswanath's constant you mentioned is kind of the expected value of the quotient of the random variables R_(n+1) and R_n, i. e, Viswanath's constant = E( R_(n+1) / R_n, because it sounds reasonable to me.
Thank you so much for the great content; D
reply
Corey
How about a fun with Maths video about the cost of getting a Kg into space, and the cost of mining an asteroid with today's tech. I read somewhere it was about $20, 000 /Kg just to get something into low Earth orbit. What about all the way back from an astroid? The distances are so great. Propulsion, spacecraft, etc. I thought it may be a fun one for you if you haven't already done something like it. -Corey
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How about a fun with Maths video about the cost of getting a Kg into space, and the cost of mining an asteroid with today's tech. I read somewhere it was about $20, 000 /Kg just to get something into low Earth orbit. What about all the way back from an astroid? The distances are so great. Propulsion, spacecraft, etc. I thought it may be a fun one for you if you haven't already done something like it. -Corey
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gresach
The difference between phi-1, 000, 000 and F_1, 000, 000 was glossed as some kind of rounding error. But the ration is sqrt(5, because F_n = (phi-n - psi-n)/sqrt(5. Since psi = (1-sqrt(5)/2, it is transient, as you take it to bigger and bigger powers it comes increasingly close to zero. So psi-n/sqrt(5) will become closer and closer to being a whole number, namely F_n.
reply
The difference between phi-1, 000, 000 and F_1, 000, 000 was glossed as some kind of rounding error. But the ration is sqrt(5, because F_n = (phi-n - psi-n)/sqrt(5. Since psi = (1-sqrt(5)/2, it is transient, as you take it to bigger and bigger powers it comes increasingly close to zero. So psi-n/sqrt(5) will become closer and closer to being a whole number, namely F_n.
reply
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