Proof by Game (infinite Fibonacci sum)

Рет қаралды 6,598

11 ай бұрын

In this video, we describe a technique of using the probability of winning a simple board game to find a particular infinite sum. In this case, the infinite sum comes from multiplying the n-1st Fibonacci number by the nth power of 1/2 and adding over all positive integers n. The board game is played on a four spot board with a starting square, and ending square, a blank square, and one square that makes you go back two spaces.
If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoffee.com/VisualPr.... Thanks!
This animation is based on a proof by Kay P. Litchfield from the October 1994 issue of Mathematics Magazine (www.jstor.org/stable/2690848 p. 281).
The final problem posed is based on the Carousel problem from Playground column in the February 2022 issue of Math Horizons from then-editor Glen Whitney (doi.org/10.1080/10724117.2021... p. 31 and 33 )
For related videos, see:
• Adding powers of 1/2
• Geometric series: sum ...
• Beautiful Geometry beh...
• General Differentiated...
#math #infiniteseries #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #geometricseries #series #mathshorts #mathvideo #fibonacci #fibonaccisequence #fibonacciseries #infinitesum #boardgames
To learn more about animating with manim, check out:
manim.community
_______________
Music in this video:
Land on the Golden Gate by Chris Zabriskie is licensed under a Creative Commons Attribution 4.0 license. creativecommons.org/licenses/...
Source: chriszabriskie.com/stuntisland/
Artist: chriszabriskie.com/

Пікірлер: 55

@me0101001000 11 ай бұрын

I am so sorry to do this to you all.... Unfortunately, I lost the game.

@MathVisualProofs 11 ай бұрын

🤣

@trevorthieme5157 4 ай бұрын

@@MathVisualProofstrying to game the math system!

@jacksonstenger 11 ай бұрын

The solution to the last problem should be 1 as well, where you are allowed to move ahead by 1, 2, or 3 squares. If you land in the two rightmost squares, the game ends. The denominator (3^k) comes from the 3 possible choices each turn. The board fits the recursive formula since when you start at the leftmost square with k moves, a 3-jump gives you the same situation with k-1 moves, and a 2-jump followed by a 1-jump gives you the same situation with k-2 moves, and a 1-jump followed by a 2-jump also gives you the same situation with k-2 moves, and finally a 1-jump followed by a 1-jump followed by a 1-jump gives you the same situation with k-3 moves, which when added up gives a_k = a_{k-1} + 2*a_{k-2} + a_{k-3}.

@MathVisualProofs 11 ай бұрын

Nice!👍

@davidemasi__ 11 ай бұрын

This is such an uncommon solution, thank you for the amazing video

@MathVisualProofs 11 ай бұрын

Glad you enjoyed it!

@oelarnes 11 ай бұрын

the initial sum can be solved with a game as well: heads you win, tails flip again.

@MathVisualProofs 11 ай бұрын

Of course!

@Living_Murphys_Law 11 ай бұрын

Okay, that is actually incredibly cool.

@MathVisualProofs 11 ай бұрын

👍

@iamtraditi4075 11 ай бұрын

This was a great video! Thank you :)

@MathVisualProofs 11 ай бұрын

Glad you liked it! Thanks for watching!

@zalut_sky 11 ай бұрын

is there anything interesting about similar sequence, but with Fibonacci numbers swapped for 'not-so-winners' in the numerator? I think it can be interesting too! good video, as always

@MathVisualProofs 11 ай бұрын

Interesting for sure. I haven't thought about it. Let me know what you find! Thanks!

@Ninja20704 11 ай бұрын

One question, how do we know for sure you win eventually? I am thinking that while its statistically unlikely, you could just keep flipping H’s or keep flipping T’s and just keep landing back at the start. Could you maybe explain a bit more? Still, very creative solution.

@MathVisualProofs 11 ай бұрын

This is good to grapple with. The idea is that the probability of selecting the all T or the all H string (for instance) is zero. There are lots of strings that will keep you from winning but that entire set has probability zero.

@Ninja20704 11 ай бұрын

@@MathVisualProofs ok that makes more sense, thank you.

@olaf7441 11 ай бұрын

The technical term for this is that you will win "almost surely" - ie the probability of winning is 1 even though there is a possible outcome where you don't win.

@frimi8593 2 ай бұрын

@@MathVisualProofshow would you prove this? It seems intuitive that you are effectively guaranteed to win eventually, but I’m not sure if/how probabilities work with infinite figures. There is an uncountable amount of losing infinite sequencing while there is only a countable amount of winning terminating sequences (though an uncountable amount of infinite sequences that start with a winning sequence). The amount of losing sequences is the same as the amount of all possible infinite sequences, winning or losing, so how would one demonstrate that the probability is 0?

@Ihab.A 11 ай бұрын

Great video

@MathVisualProofs 11 ай бұрын

Thanks!

@avyakthaachar2.718 11 ай бұрын

Amazing solution !

@MathVisualProofs 11 ай бұрын

Pretty cool right!?

@avyakthaachar2.718 11 ай бұрын

@@MathVisualProofsYes indeed.

@jakobthomsen1595 11 ай бұрын

Interesting! Now I'm curious how this would be proved symbolically. Perhaps via Binet's formula? Partial Sums of Fibonacci sequence? Hm...

@MathVisualProofs 11 ай бұрын

Binet's formula is the way to go in my opinion. Then you end up with a sum of two geometric series and you can apply the geometric sum formula :)

@LittlePaimon 11 ай бұрын

good！

@MathVisualProofs 11 ай бұрын

Thanks!

@darkphantom_01 11 ай бұрын

😂 dame tu cosita 😂

@frimi8593 2 ай бұрын

An interesting idea for sure, but unfortunately there is a problem with it. Despite the fact that the number of winning moves in n flips DOES follow the Fibonacci sequence, there are not 2^n possible strings of flips for each n, as any string which already won for some flip move prior to n (such as any string beginning with “TH”) is impossible. In actuality the probability of winning this game would need to be represented as: Σ{n=1,inf}(F_{n-1}/(2^n-Σ{m=1,n}(F_m-2)*2^(n-m)) Where you subtract from the denominator all possible strings that start with a winning sequence, represented by each term in the Fibonacci sequence shifted over by 2 to match the original n-sum being shifted over by 1, then multiplied by the amount of possible sequences that can be made from the remaining flips

@dineshkumar007 11 ай бұрын

@MathVisualProofs How can I create These type math video please tell me sir.... These video edited by which application. Please reply me back.....

@MathVisualProofs 11 ай бұрын

I use manimgl to animate these. This is a python add-on developed by 3blue1brown.

@cheeseburgermonkey7104 11 ай бұрын

An 8-minute video? Sign me up

@MathVisualProofs 11 ай бұрын

What's better? This one (at 8) or my SoME3 video at 13? Neither one really has the visuals like my other videos. Would you prefer the others were longer too?

@cheeseburgermonkey7104 11 ай бұрын

@@MathVisualProofs Do whichever format you like most, I just like seeing cool proofs for cool math facts

@MathVisualProofs 11 ай бұрын

@@cheeseburgermonkey7104 😀

@davidemasi__ 11 ай бұрын

Is it possible to read the article containing the solution to the last question without paying for it?

@MathVisualProofs 11 ай бұрын

digitaleditions.sheridan.com/publication/?m=53548&i=741066&p=30&ver=html5

@davidemasi__ 11 ай бұрын

Thank you very much!

@JamesD2957 11 ай бұрын

the board game seemed like an example of setting up the series. How was the board game itself a proof?

@MathVisualProofs 11 ай бұрын

Using the board game to determine the probability of winning in two ways. One using the series but the sum principle from probability. The other using the fact that the game will end with probability one if you keep playing (this one requires some more thought)

@JamesD2957 11 ай бұрын

@@MathVisualProofshow do you know the game will end with probability one if you keep playing. That seems circular to me. You're saying we know the series adds up to one because we know the series must add up to one

@MathVisualProofs 11 ай бұрын

So the part about the game always ending requires some deeper thought I agree. But it is not circular. The idea is that the set of strings that would allow for an infinite game is a set of measure zero amongst all the binary strings. So with probability 1 the game ends.

@JamesD2957 11 ай бұрын

@@MathVisualProofsi think i can intuitively grasp what you're saying. But am I partly on the right track? You didn't formally prove that yet?

@JamesD2957 11 ай бұрын

@@MathVisualProofsthanks, btw

@azamshaikh6011 11 ай бұрын

Do you know about jee advanced mathematics

@MathVisualProofs 11 ай бұрын

Only superficially. So I know what it is, but I am not super familiar with the tests and what not.

@deep45789 11 ай бұрын

I didn't understand how you generalize the number of ways to win the games.

@MathVisualProofs 11 ай бұрын

Your first flip can either be a H or a T. If it is a H, then you are back at the start, so you now have to win in exactly n-1 moves. If your first flip is a T, then the second flip can be H or T. If it is H, you end, but then you win in 2 moves and not n (as long as n>2). If you flip T again, you are back at the start, so now you want to win in exactly n-2 moves. Therefore, the number of ways to win in exactly n moves is to flip H and win in n-1 moves OR to flip T and win in n-2 moves.

@deep45789 11 ай бұрын

@@MathVisualProofs yes! I got it. Thank you !