How Infinity Explains the Finite | Infinite Series

[@TheGeneralThings]

Still watching this in 2019. We miss you Kelsey!

[@audreytazanou6965]

2021 here

[@pierfrancescopeperoni]

3ω+17488769 here.

[@alihesham8167]

i’m at 2021 2 years in the future from you

[@yashkrishnatery9082]

I started watching her in 2021. And wish she join it again

[@paullamar4111]

2022 now, and there still has not been a better series on pure mathematics, before or since. This is still the best.

[@appelelle]

Dear PBS. This and Space Time has one of the best sound work on KZbin. This could be a full audio book, and it would work out just as nice.

[@Mentat13]

Check Isaac Arthur

[@pbsvoices]

Made our day.

[@deldarel]

While you're right, for me it would be too difficult without the visual aid. This is not the 'baby's first mathematics' that series like 'crash course' have. I'm actually learning stuff here.

[@MeIsGurlNow]

he is not kidding really appritiate the work

[@jpphoton]

Now that you mention it, you're absolutely right!

[@bernardfinucane2061]

The difference between omega and any finite number is that you can't get to a high power of omega by multiplying it by a finite number. That's the trick that keeps the levels of the "hydra" tree distinct, even for arbitrarily large numbers of branches.

[@origamitraveler7425]

You're absolutely right! Thank you.

[@ianjehle]

Damn, I miss Infinite Series. I really really hope PBS considers bringing it back.

[@goodwillhart]

This is by far one of the best videos I've ever seen on KZbin. Congratulations on setting a new standard in communicating technical information in an interesting way!

[@vazixLT]

Please make a video about Gödel's incompleteness theorem.

[@iborn4music]

Dear PBS and Kelsey, I cant say that I overstand these things. In fact, I most certainly understand them, that is to say, I understand some basic tenets of math/space and thoroughly enjoy watching these videos. Both PBS Space Time and PBS Infinite series are making a marked impact on my life regardless. They seem to contain the wealth of information that I so crave, even if it does not seem to make a complete connection so as to achieve the "ah-ha" state of mind. As I softly focus on these sorts of things, I feel as though my intelligence is being washed by gentle water. Thank you for all you do. It is my hope that in the future I will be in a position where I may give a much more dignified and extravagant show of appreciation. Until then, may you be aware that there are those of us who feel deeply connected to the series in one way or another. Again, there is much appreciation for it. I'll be sticking around.

[@flymypg]

I really love where this channel is going. Finding accessible chunks of serious math(s) that can be meaningfully covered in a short KZbin video? Can't be easy! I particularly like how you used infinite ordinals BEFORE going into all the whys and hows of their applicability to this kind of problem. Just a few "whats" and we were on our way. Then mentioning Peano arithmetic to provide motivation for using infinite ordinals, again without going into the details. Can't wait for the explanation of that one! Truly excellent organization and presentation. How did you pull it off, and have it simultaneously be so engaging? Who are your beta testers? The last time I had homework like this was over 30 years ago. And I know I didn't enjoy it this much. I'm going to have to treat this as a class, and set aside enough time to do all the work. I believe the comparisons to Numberphile and Standupmaths aren't really relevant. Each is aimed at different but overlapping audiences. Infinite Series is now, by far, the most hard-core of the bunch. I almost never get homework from those other channels. There is no such thing as having too many good KZbin math(s) channels. I love them all. PBS Digital is on a roll. Space Time set the bar very high, and Infinite Series is now pushing it even higher. Well done!

[@meball2060]

PBS get this show back it’s one of the only shows on KZbin that are really good at explaining maths

[@HBS981]

I really like your music selection for this series. Would love to see what tracks you use.

[@JaapVersteegh]

I like how you came back to a closely related topic in a consecutive video. Please keep doing that. It might seem possibly boring, but it really helps with getting a grasp on the subject.

[@pbsinfiniteseries]

Thanks for the feedback! I've always found that I need to see a math concept two or three (or four or five or a hundred) times before it really sinks in.

[@deldarel]

I hope this channel stays true to its name, as in: I hope this video series keeps going forever.

[@9core]

rip lul

[@mitchkovacs1396]

Can you do a video on Peano Arithmetic? I hear it everywhere and I've always wondered about the details

[@RandomBurfness]

Mitch Kovacs There are five axioms. 1. There is a number that we call 0. 2. All numbers have a successor: call this successor succ(x). 3. For all x, succ(x) != 0. 4. For all x and y, if succ(x)=succ(y), we have that x=y. 5. There exists a set with the property that if x is in the set, then succ(x) is also contained within the set. This is how the counting numbers are axiomatised, and they are the foundations of Peano arithmetic.

[@Arycke]

The most important being of course s u c c .

[@monkeyvanya]

Great trick! It reminded me a bit of how my high school teacher introduced us to barycentric coordinates(en.wikipedia.org/wiki/Barycentric_coordinate_system) to solve some 2D geometric problems with intersecting circles. I thought "How can you go from 2 coordinates to 3 and make a problem any easier?" But they worked. It's funny how sometimes you need to step out of scope of your problem to find a solution.

[@FirstRisingSouI]

I'm a physics graduate student, and this stuff is still new to me. Wonderful experience.

[@wuyizhou]

LOVE THIS SERIES! It's so informative and entertaining at the same time! (great soundtrack too

[@alexandrag6364]

Just wanted to say thanks for making all these videos! I really like your channel and I'm exited to see what comes next after two weeks!

[@a52productions]

7:42 is that an aleph made from omegas?

[@HarkunwarSinghKochar]

I really loved it. I am just so addicted to your videos.

[@shubhamshinde3593]

God, she's awesome!!!

[@louisng114]

The part about leave the finite number structure being necessary reminds me of Gödel's incompleteness theorems.

[@fuckoffannoyingutube]

Kelsey, I have no issues following Matt and Space Time's videos, but you are killing me :D and yet I am waiting every week for a new episode to blow my head off. keep up the amazing work :)

[@fuckoffannoyingutube]

I guarantee you cannot make this statement without the proper data first :*

[@kronos_cm]

Hi, thank you so much for these videos. I find them easy to grasp and very entertaining. I have one question, isn't there a little typo in min 6:07 the third number in the sequence is written in hereditary base 4 notation and when is replaced with an infinite ordinal the exponent changes from 3 to 4, which would have made the ordinal omega to the omega instead of 4, right?

[@davidwright8432]

Charming, delightful, intriguing, amazing! Tickles the brain in Pleasing Places and Ways. Lotsa thanks! More, please!

[@EnriqueRegisPascalinRomo]

I watch PBS channels because those are among the BEST of anything you can watch over the crap of TV or Netflix. They are fun, smart, entertaining and enlightening. However, I must honestly admit that one huge reason I watch the Infinite Series channel is because of the intelligence, charisma and beauty of Kelsey.

[@darkracer86]

I paused the video when you said Deja Vú, and decided I may just worship your work forever. Perfectly placed.

[@vedantbhutra1118]

Aww man, I love these videos so much, I can't wait for two weeks for the next month!!! This channel is awesome! :D

This is maybe the best math teaching I have ever seen.

[@andrewharrison8436]

The series of infinite ordinals corresponding to a Goldstein series will lose 1 at a time till an omega is "exposed". The next reduction by 1 gives a major reduction in the size of the ordinal but spawns new terms and constants in proportion to the base that has been reached (that's a very hand wavey imprecise statement but that's the best my mind can manage). So reaching the next point where an omega is exposed takes more terms, so a higher base so a longer reduction to get to the next omega. OK, I accept that any Goldstein sequence terminates but the number of terms that takes is mind boggling and grows ridiculously fast as the starting number grows.

[@hagarbebado]

There is a mistake at 6:05: the exponent for the second term in the ordinal expansion for 1279 should be 3 instead of 4. The corrected line should read: 1279 w^(w+1) + w^3*3 + w^2*3 + w*3 + 3 It doesn't change te explanation, the terms are still decreasing either way, but just thought it might be worth to comment. Great video!

[@patrik9806]

In another sense, infinity is not necessary, because omega is just a label used to construct a tree-like structure (where Peano numbers are only list-like) which is where the real strength lies.

[@christosvoskresye]

Exactly. Still, it is something to show that this cannot be done using only Peano numbers.

[@t3st1221]

If you try to write a program to compute the goodstein sequence, you'll quickly run out of memory. Since 4 requires 3*2^(3*2^27+27)-3 steps, we can safely assume that we will encounter number at least as big as 2^(2^27) which would require 2^27 bits to be written out entirely which is more than the Zettabyte (2^24 bit) that exist in the world currently. Maybe some trick like factorial base or knut arrow notation can be used to keep storing the number on a much more compact memory but that would drastically slow down that process. Also that was just for the goodstein serie starting with 4.

[@triplebig]

It's been a long time since I've been this fascinated by a math video. Congrats! Excellent production! I do have a few questions... I'm a little confused by the proof in both these videos. You defined a mapping, and saw that this mapping always decreases, and the plugged that in a theorem that concerns that specific mapping. First of all, how do we know that at some point in the sequence the cardinals do not grow? this seemed like just an observation on the starting sequence, but it felt rather presumptuous. Going along, I thought the same thing you did, "why resort to this kind of stuff.. is there no other sort of mapping in graph theory or what not?". While I kind of got the gist of it from your explanation, it would be nice to understand a simple example of when peano fails and infinite cardinals succeed. Any references? Thanks!

[@IronFairy]

I miss Infinite Series so much

[@ViewtifulSam]

Great video! Really nice sequence and each video was also interesting by itself

[@relativemotion]

Man, I wish I had math teachers like you when I was in high school.

[@TheConnor12500]

Loving this channel

[@spudd86]

The main difficulty in writing a program to calculate Goodstien sequence values is that you'd pretty much have to use arbitrary width integers, and most languages don't have built in support for that, just fixed width integers of 8, 16, 32 and 64 bits, because that's what the CPU operates on.

[@MrMakae90]

Yeah, but *how* does the Kirby-Paris Theorem shows that we need infinities? You mentioned we do, but not exactly why. I guess I will have to wait for the next episode :)

[@AlcyonEldara]

Probably not, the proof is really technical.

[@daviddelaney2407]

You don't actually need infinities ... but you DO need something outside of {the natural numbers} plus {what can be proved by Peano-based arithmetic}. Infinite ordinals is one of the least complicated-to-see ways to prove it that is still outside of that. Someone else mentioned analysis, and there's certainly other ways - but this video in under 12 minutes gave you the basics to see how the infinite-ordinals proof worked. (If it's not obvious: each time you change to the next integer up after subtracting 1, having the base get bigger means the exponents you end up with get =smaller=. Incrementally, but consistently. Replacing the base number at a given step with omega shows you this gets-(slowly)-smaller progression, and at the same time lets you have a strictly-decreasing sequence of infinite ordinals ... which has to, in a finite number of steps, end up at zero.) ((The proof for the last bit is left to the reader, but a key step in it is that any infinite ordinal is always only a finite number of steps in some series counted UP from: the biggest limit ordinal less than or equal to it. If I have that right.))((In other words, if you do make an endless series of ordinals, and its limit is an ordinal that's NOT a finite number of steps etc., then that ordinal you landed on after "finishing" the endless series? Is itself a limit ordinal ... and thus 0 steps up from the biggest limit ordinal not greater than it.))((So if you're counting down, you keep hitting limit ordinals, and you can't keep counting down infinitely without doing so ... and each time you do adds only a finite number of counting-down steps. And you can't count infinitely down the limit ordinals, they've got a smallest infinite one in them -- aleph-null -- and the same sort of reasoning applies to them.)) --Dave, brain hurts, next comment prz

[@Tensoren]

Hii Kelsey, great job! I have one question. Would you say that the Axiom of Choice is indispensable in the proof you 've presented?

[@math__man]

Why does a decreasing sequence of ordinals always go to 0 in a finite amount of steps? What about: Omega Omega-1 Omega-2 . . . 0 It eventually goes to zero but it takes omega steps which isn’t finite. Thanks to anyone who answers.

[@TheBasikShow]

Omega minus one does not exist! Omega is what’s known as a “limit ordinal”, which just means an ordinal that isn’t preceded by any ordinal. To see why these need to exist, think of the reason ordinals exist in the first place: to order things. If you get “ωth” place in a race, that means that an infinite sequence of people finished the race, and then you did. More specifically, you were the fastest out of all the people who were beaten by infinitely many others. Given this, what would (ω-1)st place mean? It means you came in just before ωth place… but, by definition of ωth place, this means you came in after finitely many other people. So, ω-1 would need to be a finite number. But that would imply that ω is a finite number, and so we have a contradiction. In short, all the terms in an ordinal sequence must be non-negative in order to make sense, which is why decreasing ordinal sequences must eventually end.

[@MOSMASTERING]

Stunning videos. Love from the UK. PBS is awesome!

[@joeq6683]

PBS, it is definitely possible to write a program to calculate Goodstein sequences. The only problem is that numbers will get too big. Some of them might even exceed 2^63, the largest 64 bit number. You can get around this though by saving bigger numbers as another variable type like a string. But then there are more problems. Then you have to calculate the next numbers in the sequence in parts.You could also try scaling numbers down to be very small, and then upscale them for display. That is the general process of how I would go about doing this.

[@samuelbull5997]

So interesting, the idea that it explains the finite instead of merely informing it!

[@AliHSyed]

Hey awesome video! I never miss an opportunity to appreciate how beautiful and elusive maths can be.

[@DarkestValar]

First of all, awesome video, i love this series and the presenter handles the subjects very well. In this comment section, there seems to be alot of stress about the need for infinite ordinals. Because apparently the presenter did not specify why. The actual proof of kirby paris theorem is quite technical and uses math jargon that is inaccessible to a laymen. Still, i will try to explain why they are needed in plain english: suppose X is allowed to be any number but is restricted to some huge, finite number (for example, grahams number) We can use the same proof strategy using X for replacement instead of omega. However, the 12th goodsteins sequence beats graham's number (it actually beats the 2^2^2^2^2th step of the sequence to get grahams number, of which it is the 64th step). So while X can work for sequences 1-11, there will be some point in the process to prove g(12) where you will be required to do a "replace the base n with X" but you can't, because it would violate the restriction we imposed at the beginning. Of course, you could try lifting the restriction on X to some higher finite number, but it would only buy you some time before some goodstein sequence higher up eventually depletes the usefulness of X

[@Chalisque]

A curious observation (and I've tended to become quite a cantankerous finitist since they days I saw this stuff during my Ph.D.), is that: for a given specific goodstein sequence, if we know in advance that it will halt in under G steps, then we can replace all the omegas in the traditional proof with this big number G, and we will see a descending sequence of integers, which must hit zero. The role omega serves is, in a sense, as a placeholder for a sufficiently large integer which we can't work out until the proof is done, for a specific sequence. Furthermore, what this 'big number G' is, is not bounded for all starting points. (We have a 'for all x there exists y such that Delta0 stuff' type statement, i.e. Pi1, which is not Sigma1 or Sigma0... something like that.) This reminds me of Friedman's TREE thing.

[@flymypg]

I just listened to and read Francis Su's speech "Mathematics for Human Flourishing" (mathyawp.wordpress.com/2017/01/08/mathematics-for-human-flourishing/). It brought me nearly to tears. It connected things I hadn't thought about before, and it helps explain why I love this channel.

[@pbsinfiniteseries]

Yes!! I watched him deliver that speech. It was amazing. A room full of thousands of mathematicians, all wiping their eyes.

[@flymypg]

The precepts Su expounds upon apply to many other areas. Math may be the most abstract, making the aspects of flourishing more apparent by contrast. As a software engineer I may have a more direct connection between my personal "flourishing" and that which happens in the world. Thanks to Su, I'm able to look at my own work and thought in new ways.

[@h4nginggarden]

BobC sus

[@LeducNi]

Hello, great video. But at 6:10 the conversion of 1279 in omega notation is wrong. The exponent of the second term should but 3 (or omega?) but certainly not 4 since you transformed all 4 into omegas.

[@57op]

Great video, and very informative/instructive. Please keep up the good work.

[@DundG]

@PBS infinite series: is it possible to also provide a link to the music played in the background? The Sounds really fit to study. Thanks ^^

[@patrickmccartney2418]

It seems I missed a step. You presented the rules for the sequence that were straight forward, but at 5:39 the sequence for 1279 does not seem to follow the rules. We suddenly have multipliers of 3 and decreasing exponents. Where did the 3s come from? Why do we now have decreasing exponents that were not in the rules?

[@cyberneticqualanaut7207]

Which branch of mathematics do these topics fit under? Which classes would cover these topics?

[@madvorakCZ]

Given that it is not provable in Peano arithmetic and using the Gödel's *completeness* theorem, can we conclude that there must exist a model of Peano arithmetic in which the sequence does *not* converge?

[@tomkerruish2982]

I believe so, in that non-standard models of arithmetic are (AFAICT) not well-ordered. For example, we could start with a hydra body with a non-standard number of heads coming directly off it. We can then chop its heads to our (finite) heart's content and never finish.

[@edwardlau892]

So, when we write the number in hereditary base-n notation and do the following steps in Goodstein Sequence, are we essentially describing it as a mathematical Hydra with a regeneration capacity of n+1 (i.e. The Hydra would regenerate n+1 heads in our next step)?

[@HarDiMonPetit]

Awesome! I wonder if this proof provides a wonderfully typical example of the Gödel theorem because the Goodstein sequences and the question of their convergence to 0 can be formulated in Peano's axiomes whereas the proof cannot. Could you tell ?

[@justinward3679]

BIJECTION!! I mean, OBJECTION!!

[@y__h]

Frege and Geach would like to have words with you.

[@frankschneider6156]

SURJECTION granted

[@MrWazzup987]

Whoa their is no need for that thisis math court no needed to get kinky with you relations

[@asielsmith6007]

omg, did you just assume my jection!

[@Ricocossa1]

Not granted, your opinion is surjective

[@antoniolewis1016]

Is this inadequacy of Peano arithmetic related to Godel's theorem? In the sense that even though the hydra theorem is true, it is not provable with only Peano arithmetic?

[@leidem]

Kind of, yes. Gödel's (first) incompleteness theorem says that any axiom system which has at least the same strength as Peano arithmetic (so for instance Peano arithmetic itself) cannot be both consistent and complete. In particular this means that if you trust that Peano arithmetic isn't contradictory, then there are true statements about numbers which Peano arithmetic cannot prove. Here we get an actual concrete example of this, the hydra theorem!

[@christosvoskresye]

Wow! I did not know that any concrete examples had been found.

[@plasmaballin]

Yes. It is one of the few "natural" statements about numbers that has been shown to be unprovable from firsg order Peano arithmetic.

[@alexanderreusens7633]

I have a crazy idea: what if instead of changing all n's to n+1, you change it to the total value. So if you start with 13: 2^(2+1)+2^2+1 next step: 13^(13+1)+13^13+1-1, which is something like 4.24 *10^15 then, write that in base 13 and change all 13s with 4.24*10^15 and write that number back in base 4.24*10^15 Will it still terminate in a finite terms? I think it would, because ordinals are infinity and as long as you subtract 1 every time, the ordinal term should decrease.

[@justunderreality]

My guess is that you would run into a base 1 and get stuck

[@alexanderreusens7633]

noah schaefferkoetter Why do you think that?

[@justunderreality]

alexander reusens working backwards, if you end at 0 then the second to last iteration would have to be x represented in base n - 1 = 0 so that means that x is 1 in some base b. so while converting the second to last step to the last step means that you would have to write 1 in base 1. This brings up a bigger problem... increasing the base to a higher number gives you no problem... but going to a lower could make it meaningless. Moreover, always increasing the base will always make a new number... but having the base be dependent on the total gives a real potential for a cycle.

[@alexanderreusens7633]

noah schaefferkoetter I see, but how would lowering the base make it meaningless (not counting 1 as a base)? But yeah, making it depend on the total can make it more chaotic and unpredictable (At least, that's what I was hoping). That's why i called it a *crazy* idea :)

[@justunderreality]

alexander reusens suppose you eventually got 9 in base b. To take the next step you must write 9 (as a single digit) in base 9... specifically 9 * 10^0 -> 9 * 9^0... but 9 cannot be a coefficient in base 9, since the only meaningful values are 0 through 8. This will happen with any single valued transformation.

[@frankschneider6156]

+ PBS Infinite Series Hey, where has this channel been the past 5 years ? I mean you ROCK. You're possibly even better than PBS Space Time. Call me fanboy if you want to, but what you are doing here is absolutely outstanding.

[@bobbyharper8710]

These videos are great to play at sleep time.

[@Kaepsele337]

What am I missing in "Every decreasing sequence of ordinals reaches zero in a finite number of steps."? What about a_n = omega - n ?

[@goodwillhart]

You actually can't subtract a finite number from omega. Here's why. If we were able to take alpha = omega - 1 for example, that would mean there exists an ordinal alpha such that omega = alpha + 1 (this is basically the definition of subtraction). But this would mean that omega is the successor of an ordinal. But omega is *defined* to be an ordinal that is not a successor of anything! It is what is called a limit ordinal. It is the first ordinal bigger than all finite ordinals. If you step down from omega, you must choose a finite ordinal to step down to. You can't just step down by 1. Note that this is different to cardinal arithmetic, where (assuming the axiom of choice) it does often make sense to define subtraction of cardinals. I think aleph_0 - 1 = aleph_0, though I didn't explicitly check this, so don't quote me. The basic arithmetic operations are the same for finite ordinals and cardinals, but for infinite ordinals and cardinals, they are quite different.

[@Kaepsele337]

Ahh thanks, that makes sense. Now it seems reasonable that decreasing sequences go to zero in a finite number ob steps. Also "I think aleph_0 - 1 = aleph_0" - goodwillhart :P

[@drdca8263]

Guest6265+ Omega - n isn't an ordinal if n is a positive integer. Ordinals can be understood as each being the set of all smaller ordinals, with the empty set corresponding to 0. (So, omega is the set of natural numbers, omega plus 1 is the set of all natural numbers and also omega, and so on.) A sequence of descending ordinals is then a sequence of sets, where each one is an element of the previous one. Most commonly used versions of set theory use an axiom of foundation, saying that there is no infinite sequence of sets where each one is an element of the previous one. to go back to your question, omega-5 isn't an ordinal, unlike omega+5. It is a "surreal number", so you can do stuff with it if you want though. [ignore, unimportant stuff, badly written]But I should note that in terms of set theory based definitions, the omega of the surreal numbers is not the same set as the usual "directly from set theory" omega, though all the ordinals can be treated the same as the ordinals of the surreal numbers. This is pretty much the same sort of thing as how the 3 of the rational numbers "is really" the pair of 3 and 1 (from the integers), and isn't the same set as 3 from the integers. But it is safe to treat them as the same so long as we don't care about them as sets. And all this is assuming that we are using set theory as our foundation, as well as certain ways of defining these things within set theory, which we don't really have to, because there are other ways to define them which are equivalent, and different foundations we could use that would still have equivalent results. tl;dr of this ignore block: If you are using the surreal number definition of ordinals, treating them like sets won't have the same convenient things as if you are using "normal" ordinals as sets. otherwise they act the same way.[/ignore]

[@goodwillhart]

Ha ha, quoting me I see! :-)

[@drdca8263]

goodwillhart sorry, didn't see you had replied when I decided to reply edit: oh that was response to the other person, whoops

[@davidwilkie9551]

Intuitively, I can vaguely understand why the observable universe is a limited bubble of bubbles, because the superposition of logarithmic bases in Superspin projects inflation and then condensation, all continuously contained in the one infinite timing and spacing connection. (This is why I'm not a Mathematician, but like to see how it's done)

[@jadduajones]

Gotta love those constant hand gestures!

[@safwatalshazly]

I made a python program that can generate the first terms of the Goodstein sequence starting from any number and another one to represent any number in Hereditary notation or the ordinary base n notation. this took me a while but it is done at last. I searched the web for any online sites for such thing but I did not find so far, so I made what you said :))

[@MichaelEvans-yq7xj]

Learning so much here. Amazing channel. Just went over to Wikipedia, en.wikipedia.org/wiki/Goodstein's_theorem So it's Kirby and Paris that invented the Hydra game! It's fun watching all this stuff come together. Awesome channel! Thanks!

[@enlightedjedi]

I also like the collatz conjecture!

[@MCtheMD]

Very nice! I'll see if we can get this to Jeff (Paris) - I think being in a popular maths video means you've made it :P Are we going to see more unprovability? Goedel incompleteness? Oooo Ramsey theory next? Combinatorics, phase transitions, unprovability in first and second order arithmetics... I'm very curious to see where you go next :P

[@MCtheMD]

PS - I'm sure Jeff knows he's made it, being a professor at a respected logic dept. in the next city to where I'm doing my PhD ;)

[@avitimushi1541]

Why do you make me feel that Mathematics is easily understandable? While in actuality I know it is not that simple. It needs a little more work as evidenced in my day to day activities. Thanks for being great explainer.

[@MKD1101]

can infinity also be defined like that through finite series?

[@keffbarn]

Wow, this was really interesting. Makes you really wonder what role infinity has in the bigger picture...

[@AJourneyIntoPhysics]

can you make a video on sequences and series and maybe their physical use?

[@charlieb8735]

This feels like a topology thing since it seems like you can map the sequence of cutting one head off of every tallest branch to one head on a non-regenerating hydra with one branch.

[@soumyadipsarkar7242]

Hey,,Kelsey,,,,can't we use this idea to prove collatz conjecture??

[@messyhair42]

Will you do an episode on p-adics?

[@CorvusLiberatus]

Hmmm. I wonder if a similar idea could be used to prove (or disprove) the Collatz conjecture.

[@goodwillhart]

It's definitely not a new idea. mathforum.org/kb/thread.jspa?forumID=13&threadID=1768742&messageID=6286052#6286052

[@danielsoltesz8523]

Good question. This idea doesn't work. Let me elaborate on this. Let me make the concept of omega clear here. It is a symbol with certain properties. Te essential property here is that it is bigger than any number! To make it easier, to think this way, let me rename it. I will give the symbol "omega" a new name, I'll call it "dog". So dog is larger than every integer, and both at hydra and at godstein sequences we have the following setup. You do two things to you object (which is either the hydra or the number at the goodstein sequence) you increase and decrease your object ( at the hydra: you chop a head and then it regrowssome new ones, at the sequence: You increase the numbers in the hereditary representation, but then you subtract one). The trick is that for small instances you observe that the increasing mechanism is stronger because your numbers increase. But there is a hidden structure behind both problems that actually reveals to us that the decreasing mechanism is the stronger. And here comes the dog in the picture, because when we introduce the dog, it eliminates the increasing effect. Since the dog is larger than any finite number, thus it is immediate that there is no increasing effect anymore, but the decreasing effect is still there. Let me give two examples where we lose some structure. Suppose that you have a very complicated hydra which has 100 heads on the top of each other and this is the largest head tower. Suppose that you chop of the topmost head. No matter how complicated the hydra is beneath it, you will never again have 100 heads on the top of each other. So in this sense you might have much more heads bot you lost some "structure". Again suppose that after some steps you arrived at a number: 10.000.000 and you have to do its base 10 hereditary representation. It is easy, it is 1*10^6. Now you change the base and subtract one, so you are at the number 11^6-1. but now you have to do base 11 hereditary representation, and observe that our new number is smaller than 11^6, so it won't have 11^6 in it. And it is not hard to see that from now on after we do our writing in base whatever hereditary notation, we will never have a term that looks like "base^6" we will only have smaller terms. Many smaller ones, but never actually a big one like this. So we lost some structure. The thing is that noone ever came up with "a structure" that we "lose" at the collatz iteration. In general, there problems when you the same thing over and over to an integer and you ask whether it will always do something, are HARD. these ones in the video are the few rare instances where scientists could come up with very clever ways to defeat these problems, but these are one of the "hardest bosses" out there. I also spent like half a year fighting such a boss with a teammate of mine, and we could only slightly damage that boss. :-)

[@mukkor]

I wrote a Matlab function that evaluates the first 'iter' iterations of the Goodstein sequence for the number 'num'. It should evaluate correctly as long as none of the sequence elements are greater than 2^64 - 1 (Approx. 1.8447 * 10^16). Paste this into a file called good_eval.m to use it in Matlab. When I ran goodstein = good_eval(15, 8) it reproduced the sequence at 4:33. function goodstein = good_eval(num, iter) goodstein(1) = uint64(num); base = 2; for k = 1:iter hbn = dec_to_hbn(goodstein(k), base); hbn = inc_hbn(hbn); [g, base] = hbn_to_dec(hbn); goodstein(k + 1) = g - 1; end function hbn = dec_to_hbn(num, base) residual = num; k = 0; hbn.base = base; hbn.coeff = []; hbn.exp = {}; while residual ~= 0 k = k + 1; if mod(residual, base^k) / base ^ (k - 1) ~= 0 hbn.coeff = [hbn.coeff, mod(residual, base^k) / base ^ (k - 1)]; n = length(hbn.coeff); hbn.exp{n} = k - 1; residual = residual - mod(residual, base^k); if hbn.exp{n} > base hbn.exp{n} = dec_to_hbn(hbn.exp{n}, base); end end end function [num, base] = hbn_to_dec(hbn) num = 0; for k = 1:length(hbn.exp) if isstruct(hbn.exp{k}) hbn.exp{k} = hbn_to_dec(hbn.exp{k}); end num = num + hbn.coeff(k) * hbn.base ^ (hbn.exp{k}); end base = hbn.base; function hbn = inc_hbn(hbn) for k = 1:length(hbn.exp) if isstruct(hbn.exp{k}) hbn.exp{k} = inc_hbn(hbn.exp{k}); elseif hbn.exp{k} == hbn.base hbn.exp{k} = hbn.base + 1; end end hbn.base = hbn.base + 1;

[@reesverleur9546]

I don't know if I'm missing something here but since Goodstein sequences and the mathematical hydra from the last episode are related would it be possible to map one onto the other? what I mean to say is how would you find the Goodstein sequence that described a particular hydra and by extension the hydra described by a particular Goodstein sequence? Is this even possible? or am I missing something?

[@humanidrome]

05:00 - It should be said that in the fourth step of G(3) there simply is no 4 to be exchanged by a 5 and therefore only 1 is substracted.

[@martixy2]

Is the Kirby-Paris theorem an example of Godel's incompleteness theorem?

[@niroe82]

Little out of subject; What is the very first soundtrack on this video please ? Just after the first PBS thing to be precise

[@Zoxesyr]

so the next video will be about 2nd order arithmetic and after that you will finally discuss Goedel's Incompleteness Theorem? you should look at numberphile and singing banana to see how they make esoteric math videos

[@Bayugv]

Im confused. After a little research, I've found no reference to the Kirby-Paris theorem, but a lot to the Paris-Harrington one. Has anything to do the first with the latter?

[@Khazam1992]

It's been a long time since I've seen a cool new method to prove intresting things. :) Thank you for the episode PS: it is pretty easy to write a program which calculate the terms of Hereditary.

[@Thee_Sinner]

I have no understanding of what I'm watching...but I enjoyed it

[@kcwidman]

DO A VIDEO WITH MATT PARKER. HE'LL WRITE YOU A PYTHON SCRIPT FOR YOUR QUESTION.

[@adarshkumar4336]

No. He'll write you an Excel code!

[@castillaanderson8530]

@@romanhredil3799 dude this is amazing code, can you write some for the beklemishev's worm?

[@Kizaoners1]

Wow MATH you are one of a kind... PLS tellme where I'm wrong!!!! Lim [n^n+1 ] + [n^n] = Lim [ n^n] * n^1 + n^n = (we divide it with n^n) = 1*0 + 1 = 1...

[@Kizaoners1]

I get that its the INF an

[@sciencehour9476]

Wow, your channel is growing so fast!!!

[@mc4444]

After looking at these huge Goodstein sequences I've noticed that some numbers have many consecutive zeros towards the end. They start to appear after 8 iterations and then repeat every 10 iterations. Is there an explanation why this happens? It doesn't seem intuitive to me. Also the ratio of the end of those numbers seems to tend to 1, though we know that the sequences aren't infinite.

[@MrDannyDetail]

I've got 5 minutes in (after a lengthy pause around 3:20) and thought I'd understood this, until I saw how many steps it was said to take for a sequence beginning with 4 to terminate. At the time of the pause I had worked out this sequence for a fair few steps, then by extending out the patterns I realised that as the multipliers for each base exponent just cycle down through a sequence you would eventually reach a point where all multipliers of base exponents (including for b^0 or 1) would be 0 and so the number would be zero. The only problem is that I must be doing something subtly different, whilst believing I'm doing the same as the video, since I reached a point at base 321 where I had only 1(321), at which point you increase the base and subtract 1 to get 321(1) at which point you then, in effect, just end up subtracting 1 for each step until it reaches 0 at base 643. Clearly I'm misunderstanding something? Though I am glad I didn't stick doggedly to working out every step of the sequence longhand until it terminated as that would presumably take many lifetimes.

[@ThreeWayDude]

Can we say that if a sequence decreases in value for infinity after each step then it must also decrease for all finite numbers as well and hence the sequence always ends for a finite value?

[@feynstein1004]

Also, I'd like to make a request: The Minkowski metric for spacetime diagrams is neither very representative nor intuitive. I think we need a better metric but I'm not a mathematician. I was wondering if someone could come up with one.

[@iftruth]

Why can't we just divide members of the sequence like "member / prev-member" and to find out growth ratios and see that this growth has negative acceleration? Can we build a proof technique based on that?

[@MeIsGurlNow]

Programmer Here : It's not hard at all to write a program for this , BUT it is really process consuming so needs a very powerful computer . for small numbers thou it would be quit easy even in a high level programming language like excel-macro-VBA or even in excel itself .

[@quarkyquasar893]

When she asked to choose a number, I choose 2, and already found out the result shown on 4:48 Lucky example for me. :P

[@donaldhobson8873]

Not at all hard. Python 3 def stein(n,d): s=0 i=0 while n>0: r=n%d n=n//d s+=r*(d+1)**stein(i,d) i+=1 return s def gs(n,d): return stein(n,d)-1 t=int("value?") print(t) for i in range(2,10): t=gs(t,i) print(t)

[@ccosm4587]

Wow, that's a lot simpler than what I was cooking up.

[@RayLux]

yesssssss

[@thomaseboland8701]

Slight (required) change to input statement: t=int(input("value?")) Suggested change to print statement to print times through the loop and result in Sceintific Notation (4 sig digits): print(i-1,'%.4E' %t) You can easily change the for i in range(2,10) loop to a while ( t> 0): loop and it will run and run and run until it hits back to zero. You will to set i = 2 to start and then increment i with i += 1 statement after print statement.

[@spudd86]

Does python3 automatically switch to arbitrary width integers? Because with how fast things grow you're going to have integer overflow VERY fast otherwise.

[@joshurlay]

I thought math was its own language but that seems much more difficult.

[@PitchWheel]

In Hereditary notation, so, we need n+1 digits? i.e. "3" is not actually part of the numbers in base 3, that contain only digits 0, 1 and 2... but it seems ligit to use it? As soon as you use it for exponentials at 2:45

[@AlcyonEldara]

She used this notation to keep it visual.

[@valken666]

My suggestion is to delve deeper into the meaning of why some formula is used like that. Like the hereditary base notation, why did you do all those things? Where can that formula be used? Give some examples. No need to explain to me, it's just that I think you can improve on that. Good video.

[@xeuszzz]

It is always the smallest term in the hereditary base representation that wanes with the decrement by one. If the smallest term is constant n smaller than the base x, it will wane to zero in n steps as base grows to x+n. If the smallest term is the base x (raised to 1), it will wane to zero in x+1 steps as the base grows to 2x+1. Hence x+n wanes to zero in x+2n+1 steps. Furthermore n*x wanes to zero in (2^n-1)*(x+1) steps. In x+1 steps x² (x>2) goes to x*b where b is current base 2x+1. From here, using formula for n*x, it wanes to zero in (2^x-1)*(2x+2) steps. It takes altogether (2^(x+1)-1)*(x+1) steps. For example, starting with base 3 it takes (2^4-1)*4=60 steps for smallest term 3² to wane to zero: 3², 3*4+3, ..., 3*7, 2*8+7, ..., 2*15, 16+15, ..., 31, 31, 30, ..., 0. Now the smallest term is n*63³, which is followed by (n-1)*64³+63*64²+63*64+63. We can always wane the smallest term to zero and then wane the next term and the next and the next. I think one should be able prove by induction that at some point you reach x^k for some base x and k

[@Lexivor]

The proof isn't general if you don't use omega. For any finite x, no matter how big, there will always be Goodstein sequences that contain terms that exceed x. By using omega, you can prove that all Goodstein sequences terminate, not just some of them.

[@drdca8263]

This is a very good video. Thanks!