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I hereby demand that mathematicians define base 9 as the base we use, since zero is included in the count. This indicates that 9 is the largest integer which can fill out any individual digit of a number.

Amazing job dissecting the problem. Interesting fact about your findings: 7+32x-1 = -25 27+32x-1 = -5 31+16x-2 = -1 These numbers are all part of the loops in the negative number range.

Beautiful and great video.

When I look at the logistic map, I can't help but feel like it may be useful to describe quantum superposition sampling, or provide a way to model virtual particle creation/annihilation in the vacuum energy density.

Awesome. Thank you.

Good video! Just missing the definition of omega^omega to get the complete picture

N= positive odd number. N changes to (3N+1)/2 (3N+1)/2 could be: 1- (3N+1)/2 = positive odd integer 2- (3N+1)/2 = positive even integer 1- assume (3N+1)/2 = positive odd integer, Since N = positive odd integer, it could be a value for (3N+1)/2 So, (3N+1)/2 = N 3N + 1 = 2N 3N - 2N = -1 N = -1, which contradicts with N = positive odd integer. So, the assumption (3N+1)/2 = positive odd integer is false. 2- assume (3N+1)/2 = positive even integer, Since N + 1 = positive even integer, it could be a value for (3N+1)/2 So, (3N+1)/2 = N + 1 3N + 1 = 2N + 2 3N - 2N = 2 - 1 N = 1, which does not contradict with N = positive odd integer. So, the assumption (3N+1)/2 = positive even integer is true, and (3N+1)/2 will change to smaller value (3N+1)/4 < N, getting toward the destination 1. If N = 1, (3N+1)/4 will equal 1, which is a term within the destination loop 1 → 2 → 1. So, N = positive odd integer, just changes to (3N+1)/2 = positive even integer, which changes to (3N+1)/4 < N, getting toward the destination 1. So, Collatz conjecture is true. Eng. Mahmoud Attalla. WhatsApp: +20 1112669096.

N= positive odd number. N changes to (3N+1)/2 (3N+1)/2 could be: 1- (3N+1)/2 = positive odd integer 2- (3N+1)/2 = positive even integer 1- assume (3N+1)/2 = positive odd integer. Since N = positive odd integer, it could be a value for (3N+1)/2 So, (3N+1)/2 = N 3N + 1 = 2N 3N - 2N = -1 N = -1, which contradicts with N = positive odd integer. So, the assumption (3N+1)/2 = positive odd integer is false. 2- assume (3N+1)/2 = positive even integer. Since N + 1 = positive even integer, it could be a value for (3N+1)/2 So, (3N+1)/2 = N + 1 3N + 1 = 2N + 2 3N - 2N = 2 - 1 N = 1, which does not contradict with N = positive odd integer. So, the assumption (3N+1)/2 = positive even integer is true, and (3N+1)/2 will change to smaller value (3N+1)/4 < N, getting toward the destination 1. If N = 1, (3N+1)/4 will equal 1, which is a term within the destination loop 1 → 2 → 1. So, N = positive odd integer, just changes to (3N+1)/2 = positive even integer, which changes to (3N+1)/4 < N, getting toward the destination 1. So, Collatz conjecture is true. Eng. Mahmoud Attalla. WhatsApp: +20 1112669096.

Thanks!

this really cleared up some things I was confused about with the logistic map! but arent the logistic map and the quadratic map describing different things after all? I just cant see how their recurrence formula are the same (-rx^2+rx is not x^2+c). and the fractal of the logistic map (which is equivalent to the mandelbort fractal) have nothing to do with each other! i mean, there isnt even any symetry or anything... anyways, maybe i didnt understand correctly, so sorry if i cause any headaches

I do not believe it: at step n we have an integer g_n, write it in hereditary notation of base n and replace n with n+1, get an integer f_n > g_n, subtract 1 to get g_{n+1}=f_n-1 >= g_n. The sequence never decreases.

Thank you for the nice and simplified presentation. Adrien Douady and John H. Hubbard showed the connection of these two diagrams many years ago.

Nice vid, which helps me a lot.

😲😲👍👍

I thought it was somewhat intuitive that the -1s would eventually, after killing off the (n-1)^0 term, moves the last term n^1 (in the polynomial n^n^n^... + .... + n^1) to n^0, which would stop the last term from growing and then slowly kill it off. Then eventually after k ~ n^0 turns it would move the (n+k-1)^2 to (n+k)^1, and then eventually from (n+k')^1 to (n+k')^0, etc, slowly killing all the powers and reaches 0. I am surprised that we need the omega notation at all to formulate a rigorous proof.

The goodstein sequences will go to zero because of the -1. If the -1 didn't exist, then the number would keep getting larger and larger.

Now do: Relationship between Mandelbrot set & Riemann Zeta! /jk

Love the Ghibli-style music at the end. btw, Desdenova sent me 😉

A new pattern I found with triangles-> Something I found to think about: there are triangles of twins( numbers one apart in the Collatz)a represents the top number of the triangle and below it is a+1 and the restriction is a+2 is prime.Twins are vertical in these triangles and reach a connection in the same number of moves as the rest of the twin ex: 20,21 both reach 16 after 3 iterations. I’m not sure why 144, 145, and 146 have this behavior, reaching 94 after several moves. The triangles “start” with a(2^n -1)+(a-2), where the start are the values on the left. PS. a=-1 is an exception but is a triangle . What happens in the middle and on are that when you apply the Collatz sequence to it the twins and connections form. I think I am the first to find these “Collatz Triangles” and that they are helpful to the proof. There are patterns with a+1= 0 and 2 mod4 but I’ll leave those for someone to figure out. -Isaac D.

5:03 1) their indexes are divisible by two (2x) So you end up at indexes of type 3x. (2steps for 2+8x'th odd, ending up at 3+12x'th odd,4+8x'th odd, would therefore end up at 9+18x and so on but after 4steps) 2)indexes of type 3x(not divisible by two) always point to ones of type 4x+1(where x is now a different offset) therefore you can apply 4x+1->x(3 steps but even, so check other branch a step back) (or 4(x)+3->4(3x-1/2)+6(x-1/2)+3(non- odd math makes it hard) but then you are at an even odd x and not at an odd odd x and you need to check step 1) 3)leaving you at number smaller than the initial by ~3/4 (you can derive the exact value) What's your point, the conjecture is already true because you can't construct a number with infinite power as infinity is not a number. So you stop increasing after n steps and go into a spiral down. The same 1241 loop but with the end being smaller than the initial and longer in length.(to be peer reviewed) It's hard to keep two numbering systems parallelly in your head while describing complex topics. So that's the excuse for the potential mistakes.

7:11 Technically Linear map is Ax, where A is a matrix and x is a vector whicle Ax +b, where be is another vector is an Affine Map.

谢谢！

Problem 3n+1 solved! ruclips.net/video/ewZnfVO3wX8/видео.html

awesome video! I just made a video on the same topic, though I talked about a different relationship between the two. I thought about mentioning the fact that they're linear transformations of each other, but I'm glad I didn't since your video covers it so well. I must say, you definitely covered a lot of the introductory topics much more concisely than I 😅 great job

Hi everyone! For full transparency, I mentioned in the video that I pick and chose the 10k and 100k simulations out of a larger sample that had decent approximations of pi. However even for the 10mill simulation at the end, I manually chose 100k samples that had at least 1-2 correct decimal places to get the approximation at the end (I forgot to state this in the video). You can try simulating it yourself on this website mste.illinois.edu/activity/buffon/. As mentioned in the video, the issue with this approach is that it relies on probability. I was able to get pi to 5 decimal places in only a sample of 3mill but other times only 3 decimal places with samples 3-4mill so I think luck and how the random numbers are generated plays a big part in estimating pi using this method. Please let me know if you know how to increase the accuracy using this method..

Very nice!

very nice video, easy to follow for the layman, and those with some math knowledge can enjoy this too as you kept is brief and reasonably assume we know what, for example, an integral is. perhaps you can speak a bit stronger, but that is just something all small/new RUclipsrs face, and a better mic/more videos will improve that sole issue. Good job.

good explanation man , and extremely good visualisation, Thank You!!!

Wow, cool visualisation! Keep up the good work 😊 👏👏💯

i don't think you can subtract 1 from infinity and spect it to be less then infinity 🤔. Just like the sum of all integers is not -1/12

This is so cool. This is like the ultimate illustration of the proverb "slow and steady wins the race" :)

As a 9th grader who is interested in both fractals and the logistic map, I’d say your interpretation of both was honestly beautiful and your explanations for both the map and the Mandelbrot set are amazing, thank you

2500: YAY I REACHED (2.5342x10^10^7). 1 minute later ... 2500: NO! 45th term: ?^?^?+?^?-1=ZERO NO! I REACHED ZERO!

What I like to visualize here is a timer ticking its way to zero. The base doesn't really matter until you need to need to tick down from zero in the ones place.

My favorite part were the animations starting at around 5:15 onward, I don't think I've ever seen the cycles in the Mandelbrot set visualized this nicely. Also, the music fits extremely well!

14:25 How does subtracting from an infinite ordinal make it finite? Is that just how it works?

nice

Of course, the "construction" of the set of natural number requires the Axiom of infinity :)

What a journey!!

Every appearence of the mandelbrot in any math field it's worth my attention. I recently discovered your channel and let me tell you I love it. Keep it up with great videos!

Do these Goodstein sequence just increase monotonically to a maximum, and then the very next term is 0? Or, do they hit a maximum and then monotonically decrease to 0? Or, is the eventual decrease even monotonic at all?

jewish investment

Underrated

collatz conjecture 2.

Why is the second term in the new sequence not (w+1)^(w+1)^(w+1)?

no background music please

Some tasks require an infinite number of steps to finish. Do we know, or is there anything to suggest the number of steps before this terminates? The visuals of the video were very well done. If I could say one thing, it is that the pauses in speech when something complex is being shown made it more difficult for me to understand. Not complaining, because you did an excellent job explaining.

Very good presentation, but It would be very helpful to first show the G(3) sequence getting to zero as a starting point before going for the full proof. Until I did some further reading on the topic I was thinking that any Goldstein sequence only could reach zero after ω number of steps, which would be by definition infinite, and thus never actually happening.

Love the video! Just a note, I think the videos might "feel" better if you used a different kind of music. Just my two cents!

The problem seems to be intimately related to binary and also to powers of 3 which tend to give long sequences. Eg 9 to 28 to 14 to 7 to 22 to 11 to 34 to 17 to 52 to 26 to 13 to 40 to 20 to 10 to 5 to 16... like turning a power of 3 into a power of 2 at some point.